2011 USPC Walkthrough guide.

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
2011 USPC Walkthrough guide.
There aren't enough tips in the 'rules and tips' area, so this seems as good a time as any to get one started.
I'd like to see some walkthroughs here for the puzzles in the recent USPC (http://wpc.puzzles.com/uspc11/). There are a lot of people that are new to puzzling, or at least to certain puzzle types, that would appreciate the specific and general techniques used to solve puzzles. Please feel free to chip in with your own contributions.
Index:
1. Battleships
2. Sudoku
3. Crossed Countries
4. Barn Storm
5. Easy as ABCD
6. Times Like These (or here)
7. Masyu
8. Square Count
9. Crack it On
10. Flash Cards 1
11. Flash Cards 2
12. Takeout
13. Corral
14. Blocks
15. Hex Words
16. Word Connection
17. Hopper
18. Corral Crates
19. Kaku Rogue
20. Jumping Crossword
21. Hungarian Tapa
22. Dynasty Sudoku
I'll get the ball rolling with one that seemed to catch a few people out [edit  moved to separate thread for indexing]:
I'd like to see some walkthroughs here for the puzzles in the recent USPC (http://wpc.puzzles.com/uspc11/). There are a lot of people that are new to puzzling, or at least to certain puzzle types, that would appreciate the specific and general techniques used to solve puzzles. Please feel free to chip in with your own contributions.
Index:
1. Battleships
2. Sudoku
3. Crossed Countries
4. Barn Storm
5. Easy as ABCD
6. Times Like These (or here)
7. Masyu
8. Square Count
9. Crack it On
10. Flash Cards 1
11. Flash Cards 2
12. Takeout
13. Corral
14. Blocks
15. Hex Words
16. Word Connection
17. Hopper
18. Corral Crates
19. Kaku Rogue
20. Jumping Crossword
21. Hungarian Tapa
22. Dynasty Sudoku
I'll get the ball rolling with one that seemed to catch a few people out [edit  moved to separate thread for indexing]:

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
#5  Easy as ABCD
(Notation: Cols AG, rows 17 from top to bottom)
Looking for definites around the border first. (Clued once at that edge).
A6 = B; E7 = B; A7 = C; G4 = A; D1=B.
Col A: A3/A5 are blank, since C already placed.
Row 7: D must be in col G (to appear before B already in place). G7=D, so C7 & G6 are blank.
Row 6: D cannot be in col F, since D is the first letter encountered from the TOP of that col. So D6 is blank.
Col A: D can only be in A4.
Col C: D cannot be in row 2 (needs to be 1st seen from RHS) or row 4 (already placed) so must be in row 3, with C6 blank.
We now have 3 blanks in row 6, so the remaining 4 cells are all letters. E6 = D. B6 & B7 are both either A or C.
This means that F7 cannot be A. (A must appear below the C in col B or D!)
Col F: A already in row 4, so must be in F5. This forces B into G5, and G3 is blank.
Row 3: B must be in col F.
Col E: A must be in row 3. (too far across for rows 1/2. Already placed in rows 4/5). E4/5 blank, since C is above the A.
Row 5: C in col B. D5 blank.
Now we know B2 = A and D2 = C, with B7 blank and D7=A.
D4 must be blank  AD all appear in that row/col.
Row 3: C not in col B, so must be in col C, with D in col D. D2 is blank.
The Cs in cols E and G are in rows 1/2, so the C in col F must be in row F. (It couldn't go anywhere else in row 4 anyway)
Looking at the As in rows 1/2, A1+B1 are blank or A2+B2 are blank. Since we need to fit B and D into col B, the B must go in row 4. (and C4 is blank).
Col C needs A and B to be placed, so we can finish off easily now.
C1 = A, C2 = B; A1/B1 blank. B2 = D, A2 = A.
F1=D, G1=C, C1 blank. E2=C, F2/G2 blank.
done.
(Notation: Cols AG, rows 17 from top to bottom)
Looking for definites around the border first. (Clued once at that edge).
A6 = B; E7 = B; A7 = C; G4 = A; D1=B.
Col A: A3/A5 are blank, since C already placed.
Row 7: D must be in col G (to appear before B already in place). G7=D, so C7 & G6 are blank.
Row 6: D cannot be in col F, since D is the first letter encountered from the TOP of that col. So D6 is blank.
Col A: D can only be in A4.
Col C: D cannot be in row 2 (needs to be 1st seen from RHS) or row 4 (already placed) so must be in row 3, with C6 blank.
We now have 3 blanks in row 6, so the remaining 4 cells are all letters. E6 = D. B6 & B7 are both either A or C.
This means that F7 cannot be A. (A must appear below the C in col B or D!)
Col F: A already in row 4, so must be in F5. This forces B into G5, and G3 is blank.
Row 3: B must be in col F.
Col E: A must be in row 3. (too far across for rows 1/2. Already placed in rows 4/5). E4/5 blank, since C is above the A.
Row 5: C in col B. D5 blank.
Now we know B2 = A and D2 = C, with B7 blank and D7=A.
D4 must be blank  AD all appear in that row/col.
Row 3: C not in col B, so must be in col C, with D in col D. D2 is blank.
The Cs in cols E and G are in rows 1/2, so the C in col F must be in row F. (It couldn't go anywhere else in row 4 anyway)
Looking at the As in rows 1/2, A1+B1 are blank or A2+B2 are blank. Since we need to fit B and D into col B, the B must go in row 4. (and C4 is blank).
Col C needs A and B to be placed, so we can finish off easily now.
C1 = A, C2 = B; A1/B1 blank. B2 = D, A2 = A.
F1=D, G1=C, C1 blank. E2=C, F2/G2 blank.
done.
Re: 2011 USPC Walkthrough guide.
Lovely idea! I'll take another of the relatively easy ones:
#6  Times Like These.
I started with the fivedigit lines before the sixdigit ones  specifically, with the bottom row, 24678.
Given that there are three digits to multiply to form a twodigit sum, it's likely that the digits in use are going to be the smallest ones. Let's work through picking three from five, starting as low as possible.
2x4x6 = 48, doesn't work.
2x4x7 = 56, doesn't work.
2x4x8 = 64, doesn't works
2x6x7 = 84, that works! Check for alternatives...
2x6x8 = 96, doesn't work.
2x7x8 = 112, doesn't work.
4x6x7 = way over 100 and everything without the 2 isn't going to work. So we know that the bottom row contains a 2, a 6 and a 7.
Try something similar with the righthand column, again because dealing with five digits instead of six makes things easier.
2x3x4 = 24, doesn't work.
2x3x5 = 30, doesn't work.
2x3x9 = 54, that works! Check for alternatives.
2x4x5 = 40, doesn't work.
2x4x9 = 72, doesn't work.
2x5x9 = 90, doesn't work.
3x4x5 = 60, doesn't work.
3x4x9 = 108, doesn't work.
3x5x9 = 135, doesn't work.
4x5x9 = 180, doesn't work. So we know that the righthand column contains a 2, a 3 and a 9.
If the bottom row contains a 2, a 6 and a 7 and the right column contains a 2, a 3 and a 9, the only crossover is a 2, which we can place in the lower right circle. This leaves a 6 and a 7 for the lefthand two figures on the bottom row; the left row can include a 6 but no 7 and the middle row can include a 7 but no 6. This lets us place the bottom row of 6 x 7 x 2 = 84. We can't repeat the technique on the righthand column because both the top two rows include a 3 and a 9, so both topright and middleright are either 3 or 9.
Let's work on the sixdigit rows and columns now. The lefthand row has relatively low digits, so it looks promising. As there are six digits, we need to multiply three of them together to make a threedigit number from the remaining three digits  and as they're so low, there may not be so many options. Let's start with the highest possible digits and work down.
6x5x4 = 120, that works!
6x5x2 = 60, doesn't work, nothing else 6x5x will work.
6x4x2 = 48, doesn't work, nothing else 6x4x or lower will work.
5x4x2 = 40, doesn't work, nothing else 5x or lower will work. So we know that the lefthand column contains a 4, a 5 and a 6.
If the bottom row has a 6 as its left entry, the top row must contain a 4 but no 5 and the middle row must contain a 5 but no 4. This lets us place the left column of 4 x 5 x 6 = 120.
Let's move to the top row, which has a 4 placed. Let's start with the highest possible digits and work down.
9x8x4 = 288, doesn't work.
9x4x3 = 108, that works!
9x4x1 = 36, doesn't work, nothing else 9x4x will work.
8x4x3 = 96, doesn't work, nothing else 8x or lower will work. So we know that the top row must contain a 9 and a 3 as well as the 4 in its left cell.
The middle column contains a 9 but no 3, even though the right column contains both a 9 and a 3. This lets us place the top row of 4 x 9 x 3 = 108. This also lets us place the middle row, right hand entry of 9 by elimination.
This gives us a middle row of 5 x ? x 9 and a middle column of 9 x ? x 7. The remaining digits for the middle row are 0, 3, 6 and 8; the remaining digits for the middle column are 0, 4, 5 and 8. This means that the middle cell must be either 0 or 8, and 0 can be eliminated because multiplying anything by 0 gives the result of 0. Placing the 8 gives a middle row of 5 x 8 x 9 = 360 and a middle column of 9 x 8 x 7 = (counts on toes) 9 x 56 = (101) x 56 = 560  56 = 504. Happily, these both work with the other digits in the row and column, and this completes the puzzle.
#6  Times Like These.
I started with the fivedigit lines before the sixdigit ones  specifically, with the bottom row, 24678.
Given that there are three digits to multiply to form a twodigit sum, it's likely that the digits in use are going to be the smallest ones. Let's work through picking three from five, starting as low as possible.
2x4x6 = 48, doesn't work.
2x4x7 = 56, doesn't work.
2x4x8 = 64, doesn't works
2x6x7 = 84, that works! Check for alternatives...
2x6x8 = 96, doesn't work.
2x7x8 = 112, doesn't work.
4x6x7 = way over 100 and everything without the 2 isn't going to work. So we know that the bottom row contains a 2, a 6 and a 7.
Try something similar with the righthand column, again because dealing with five digits instead of six makes things easier.
2x3x4 = 24, doesn't work.
2x3x5 = 30, doesn't work.
2x3x9 = 54, that works! Check for alternatives.
2x4x5 = 40, doesn't work.
2x4x9 = 72, doesn't work.
2x5x9 = 90, doesn't work.
3x4x5 = 60, doesn't work.
3x4x9 = 108, doesn't work.
3x5x9 = 135, doesn't work.
4x5x9 = 180, doesn't work. So we know that the righthand column contains a 2, a 3 and a 9.
If the bottom row contains a 2, a 6 and a 7 and the right column contains a 2, a 3 and a 9, the only crossover is a 2, which we can place in the lower right circle. This leaves a 6 and a 7 for the lefthand two figures on the bottom row; the left row can include a 6 but no 7 and the middle row can include a 7 but no 6. This lets us place the bottom row of 6 x 7 x 2 = 84. We can't repeat the technique on the righthand column because both the top two rows include a 3 and a 9, so both topright and middleright are either 3 or 9.
Let's work on the sixdigit rows and columns now. The lefthand row has relatively low digits, so it looks promising. As there are six digits, we need to multiply three of them together to make a threedigit number from the remaining three digits  and as they're so low, there may not be so many options. Let's start with the highest possible digits and work down.
6x5x4 = 120, that works!
6x5x2 = 60, doesn't work, nothing else 6x5x will work.
6x4x2 = 48, doesn't work, nothing else 6x4x or lower will work.
5x4x2 = 40, doesn't work, nothing else 5x or lower will work. So we know that the lefthand column contains a 4, a 5 and a 6.
If the bottom row has a 6 as its left entry, the top row must contain a 4 but no 5 and the middle row must contain a 5 but no 4. This lets us place the left column of 4 x 5 x 6 = 120.
Let's move to the top row, which has a 4 placed. Let's start with the highest possible digits and work down.
9x8x4 = 288, doesn't work.
9x4x3 = 108, that works!
9x4x1 = 36, doesn't work, nothing else 9x4x will work.
8x4x3 = 96, doesn't work, nothing else 8x or lower will work. So we know that the top row must contain a 9 and a 3 as well as the 4 in its left cell.
The middle column contains a 9 but no 3, even though the right column contains both a 9 and a 3. This lets us place the top row of 4 x 9 x 3 = 108. This also lets us place the middle row, right hand entry of 9 by elimination.
This gives us a middle row of 5 x ? x 9 and a middle column of 9 x ? x 7. The remaining digits for the middle row are 0, 3, 6 and 8; the remaining digits for the middle column are 0, 4, 5 and 8. This means that the middle cell must be either 0 or 8, and 0 can be eliminated because multiplying anything by 0 gives the result of 0. Placing the 8 gives a middle row of 5 x 8 x 9 = 360 and a middle column of 9 x 8 x 7 = (counts on toes) 9 x 56 = (101) x 56 = 560  56 = 504. Happily, these both work with the other digits in the row and column, and this completes the puzzle.

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
#6  Times Like These.
The presence of 0s mean these are one of the last 2 digits in those product answers.
With a positive integer in each circle, we can see the length of each product immediately.
With the products requiring a 2digit answer, we can quite quickly run through choosing 2 digits from 5, and seeing if the 3 remining digits multiply to that value.
The 3rd row appears as 2x6x7 = 84, with no other possible solutions (Important to check for traps  may well be more than one correct equation!). So 7 goes in R3C2. (no other col has a 7).
Likewise, col 3 can only be 2x3x9=54. 2 must be in R3C3. Now we know R3C1 = 6.
Looking at possible digits for each cell (handy to fill these in early and cross off options as placed), R2C1 must be 5.
Col 1 is now clearly 4x5x6=120, with the 4 being in R1C1.
Calculating options for row 2 with known possibles for each remaining cell, we have only to calculate 5x8x3; 5x9x3; 5x8x9 (120; 135; 360 respectively). The only valid solution being the last one. 9 is in R2C3 and the 8 in R2C2. R1C3 can be filled with a 3 now too.
Without calculating, only a 9 can go in the last cell (R1C2) to give R1 product = 108, and C3 product = 504.
Done. [edit. Apologies to dickoon  we must have been writing imultaneously  I didn't see your post, but no harm in 2 perspectives on the puzzle]
The presence of 0s mean these are one of the last 2 digits in those product answers.
With a positive integer in each circle, we can see the length of each product immediately.
With the products requiring a 2digit answer, we can quite quickly run through choosing 2 digits from 5, and seeing if the 3 remining digits multiply to that value.
The 3rd row appears as 2x6x7 = 84, with no other possible solutions (Important to check for traps  may well be more than one correct equation!). So 7 goes in R3C2. (no other col has a 7).
Likewise, col 3 can only be 2x3x9=54. 2 must be in R3C3. Now we know R3C1 = 6.
Looking at possible digits for each cell (handy to fill these in early and cross off options as placed), R2C1 must be 5.
Col 1 is now clearly 4x5x6=120, with the 4 being in R1C1.
Calculating options for row 2 with known possibles for each remaining cell, we have only to calculate 5x8x3; 5x9x3; 5x8x9 (120; 135; 360 respectively). The only valid solution being the last one. 9 is in R2C3 and the 8 in R2C2. R1C3 can be filled with a 3 now too.
Without calculating, only a 9 can go in the last cell (R1C2) to give R1 product = 108, and C3 product = 504.
Done. [edit. Apologies to dickoon  we must have been writing imultaneously  I didn't see your post, but no harm in 2 perspectives on the puzzle]

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
#8  Square count.
Shape counting puzzles always seem to cause a lot of problems. The only successful method is to take a systematic approach.
For square counting, I progress top to bottom, left to right, looking for every 90 degree corner that could represent the Top Left corner of a square of any size. With a finger on it, I run my thumb down a 45 degree (imaginary) line, looking for BR 90 degree corners, and see if they make a full square. Then I write the number of squares I find for that vertex next to it. (Not forgetting the little white square in the middle (kick self)).
Adding up all the little numbers, I find 15 squares in the regular orientation.
Rotating the paper 180 deg (so the numbers don't clash with others) +/ 45 deg, repeat the task for each vertex, and find 22 in that orientation once I've added up all the little numbers. Tot = 37.
Now for the size grouping.
Regular orientation, I find 1 of side=1/2, 9 of side=1, 4 of side=2 and 1 of side=3. (check tot=15, 4 different sizes)
Diagonally, 16 of side=sqrt(2)/2 (0.707), 4 of side= sqrt(2) (1.414), 2 of side = 3/2.sqrt(2) (2.121). (check tot=22, 3 different sizes)
For the answer key, there is no requirement to sort by shape size, just numerically descending, so the key should be: 37,16,9,4,4,2,1,1.
Done.
There's no saying that this extra sizereporting information will be there in future, or in the same format, so the instructions on answer keys, as usual(!), must be read carefully every time.
Shape counting puzzles always seem to cause a lot of problems. The only successful method is to take a systematic approach.
For square counting, I progress top to bottom, left to right, looking for every 90 degree corner that could represent the Top Left corner of a square of any size. With a finger on it, I run my thumb down a 45 degree (imaginary) line, looking for BR 90 degree corners, and see if they make a full square. Then I write the number of squares I find for that vertex next to it. (Not forgetting the little white square in the middle (kick self)).
Adding up all the little numbers, I find 15 squares in the regular orientation.
Rotating the paper 180 deg (so the numbers don't clash with others) +/ 45 deg, repeat the task for each vertex, and find 22 in that orientation once I've added up all the little numbers. Tot = 37.
Now for the size grouping.
Regular orientation, I find 1 of side=1/2, 9 of side=1, 4 of side=2 and 1 of side=3. (check tot=15, 4 different sizes)
Diagonally, 16 of side=sqrt(2)/2 (0.707), 4 of side= sqrt(2) (1.414), 2 of side = 3/2.sqrt(2) (2.121). (check tot=22, 3 different sizes)
For the answer key, there is no requirement to sort by shape size, just numerically descending, so the key should be: 37,16,9,4,4,2,1,1.
Done.
There's no saying that this extra sizereporting information will be there in future, or in the same format, so the instructions on answer keys, as usual(!), must be read carefully every time.

 Posts: 604
 Joined: Tue 29 Jun, 2010 11:41 am
Re: 2011 USPC Walkthrough guide.
# 13: Corral
Cell refs: R1C1 is row 1 column 1 is the top left corner.
I shade cells inside the loop, and mark a clear slash in those outside it. Here I'll indicate that as I(nside) and O(utside) respectively. I find the loop lines themselves are somewhat incidental to the solving process, though they would probably be required in a competition where the entire solution was being marked.
All numbered cells are I.
R5C1 can now see two cells, so R4C1, R6C1, R5C3 are O.
The second cell seen by R8C6 cannot be above, right or below, else it would see three cells, so R8C5 is I; R7C6, R9C6, R8C4, R8C7 are O.
2x2 'Battenburgs' are illegal with a single loop*, so R7C7, R9C7 are O.
For R10C6 to see 10 cells, the whole of R10 must be I.
For R10C1 to see exactly 11 cells, R9C1 is I and R8C1 is O.
To avoid a battenburg, R9C2 is I.
R8C2 can see 3 cells, so R7C2 and R8C3 are O.
To avoid battenburgs, R7C1 and R7C3 are O.
For R9C4 to see exactly 3 cells, R9C3 must be O and R9C5 must be I.
For R10C10 to see exactly 12 cells, R9C10 and R8C10 are I, R7C10 is O.
For R6C3 to see 6 cells, R6C4, R6C5, R6C6, R6C7 must be I. R6C8 must be O. R6C2 must be I.
To avoid battenburgs, R5, C7 are I; R7C8 is O.
For R8C8 to see 5 cells, R8C9 and R9C8 are I.
To prevent a secondary loop, R9C9 is I.
For R6C10 to see 3 cells, R5C10 is I, R4C10 is O.
For R5C2 to see 5 cells, R4C2 and R3C2 are I, R2C2 is O.
For R6C6 to see 7 cells, R5C6 is I, R4C6 is O.
R5C5 can see 7 cells, so R4C5, R5C4, R7C5 are O.
To avoid battenburgs, R4C4 and R7C4 are O.
To maintain a single corral, R7C9 is I.
R6C9 can see 7 cells, so R4C9 is O.
For R5C8 to see 8 cells, R3C8 and R4C8 are I, R2C8 is O.
To avoid a battenburg, R3C7 is I.
For R1C1 to see 6 cells, at least R1C2, R1C3, R1C4 are I.
Given the previous step, for R1C5 to see at least 8 cells, R2C5 is I; to see precisely 8 cells, R1C6 is I, R1C7 is O.
To avoid battenburgs, R6C2 is I; R1C8 is O.
R1C1 can now see 6 cells, so R2C1 is O.
To prevent a second loop, the set of Os including R4C4 must connect to the edge of the grid, so R3C4, R2C4, R2C3 are O.
For R3C3 to see 4 cells, R3C1 and R4C3 are I.
For R3C5 to see 7 cells, R3C6 is I and R3C10 is O.
For R2C7 to see 7 cells, R4C7 is I.
For R3C9 to see 7 cells, R1C9 and R2C9 are I.
For R1C10 to see 3 cells, R2C10 is I.
Bolden lines around corral to make outside cells VERY clear and construct answer key:
R1C7: 3
R3C10: 3
R7C10: 1
R8C1: 17
R4C1: 1
R2C1: 10
Final code: 3, 3, 1, 17, 1, 10.
*It's not explicitly stated in the rules here that the closed loop can't touch itself at a point, so I may have misused my trusty old "battenburg" rule. As it happens, fortunately, it works out that that was what was intended.
Cell refs: R1C1 is row 1 column 1 is the top left corner.
I shade cells inside the loop, and mark a clear slash in those outside it. Here I'll indicate that as I(nside) and O(utside) respectively. I find the loop lines themselves are somewhat incidental to the solving process, though they would probably be required in a competition where the entire solution was being marked.
All numbered cells are I.
R5C1 can now see two cells, so R4C1, R6C1, R5C3 are O.
The second cell seen by R8C6 cannot be above, right or below, else it would see three cells, so R8C5 is I; R7C6, R9C6, R8C4, R8C7 are O.
2x2 'Battenburgs' are illegal with a single loop*, so R7C7, R9C7 are O.
For R10C6 to see 10 cells, the whole of R10 must be I.
For R10C1 to see exactly 11 cells, R9C1 is I and R8C1 is O.
To avoid a battenburg, R9C2 is I.
R8C2 can see 3 cells, so R7C2 and R8C3 are O.
To avoid battenburgs, R7C1 and R7C3 are O.
For R9C4 to see exactly 3 cells, R9C3 must be O and R9C5 must be I.
For R10C10 to see exactly 12 cells, R9C10 and R8C10 are I, R7C10 is O.
For R6C3 to see 6 cells, R6C4, R6C5, R6C6, R6C7 must be I. R6C8 must be O. R6C2 must be I.
To avoid battenburgs, R5, C7 are I; R7C8 is O.
For R8C8 to see 5 cells, R8C9 and R9C8 are I.
To prevent a secondary loop, R9C9 is I.
For R6C10 to see 3 cells, R5C10 is I, R4C10 is O.
For R5C2 to see 5 cells, R4C2 and R3C2 are I, R2C2 is O.
For R6C6 to see 7 cells, R5C6 is I, R4C6 is O.
R5C5 can see 7 cells, so R4C5, R5C4, R7C5 are O.
To avoid battenburgs, R4C4 and R7C4 are O.
To maintain a single corral, R7C9 is I.
R6C9 can see 7 cells, so R4C9 is O.
For R5C8 to see 8 cells, R3C8 and R4C8 are I, R2C8 is O.
To avoid a battenburg, R3C7 is I.
For R1C1 to see 6 cells, at least R1C2, R1C3, R1C4 are I.
Given the previous step, for R1C5 to see at least 8 cells, R2C5 is I; to see precisely 8 cells, R1C6 is I, R1C7 is O.
To avoid battenburgs, R6C2 is I; R1C8 is O.
R1C1 can now see 6 cells, so R2C1 is O.
To prevent a second loop, the set of Os including R4C4 must connect to the edge of the grid, so R3C4, R2C4, R2C3 are O.
For R3C3 to see 4 cells, R3C1 and R4C3 are I.
For R3C5 to see 7 cells, R3C6 is I and R3C10 is O.
For R2C7 to see 7 cells, R4C7 is I.
For R3C9 to see 7 cells, R1C9 and R2C9 are I.
For R1C10 to see 3 cells, R2C10 is I.
Bolden lines around corral to make outside cells VERY clear and construct answer key:
R1C7: 3
R3C10: 3
R7C10: 1
R8C1: 17
R4C1: 1
R2C1: 10
Final code: 3, 3, 1, 17, 1, 10.
*It's not explicitly stated in the rules here that the closed loop can't touch itself at a point, so I may have misused my trusty old "battenburg" rule. As it happens, fortunately, it works out that that was what was intended.

 Posts: 604
 Joined: Tue 29 Jun, 2010 11:41 am
Re: 2011 USPC Walkthrough guide.
#9 Crack It On
Grid notation: AR1C1 is Row 1 Column 1 (top left) of Grid A  both grids are deemed to be on a 6x6 squared subgrid.
First note that all words in the list begin with AEGHPRST.
Find candidate words for R1 and C1 in the two grids; these must contain only AEGHPRST, and are:
EAGER, EARTH, HEART, PASTS, PRATE, RESET, STAGE, STRAP, THERE.
HEART, RESET, THERE can be eliminated, as there is no other word which could intersect their first letter.
This leaves candidate pairs of: EAGER/EARTH, PASTS/PRATE, STAGE/STRAP.
BR1C1 begins 3 words; only two in the list begin with P and S, so BR1C1 = E.
EARTH and EAGER are the words in BR1 and BC1, in some order; either way, BR1C3 = A, BR2C1 = A.
BR2C2 is the second letter of two words beginning with A, and so BR2C2 = L.
BC2 has the pattern EL, and must be ELITE.
BC1 has the pattern EAT, and must be EARTH.
This leaves BR1 to be EAGER.
BR6 has the pattern HE, and must be HEART.
BC6 has the pattern RT, and must be RESET.
BC3 has the pattern ALA, and must be ALOHA.
BC4 has the pattern GOR, and must be GONER.
BC5 has the pattern ER, and must be ENTER.
This completes grid B, and we can delete all the words used from the master list. The words used were:
EAGER, ALONE, RIOTS, TONES, THERE, HEART (horizontal)
EARTH, ELITE, ALOHA, GONER, ENTER, RESET (vertical).
The list of words to use in Grid A is therefore:
ALONG, ATLAS, EDGES, GENTS, GLOAT, PASTS, PRATE, RILED, STAGE, STRAP, TIARA, TITLE.
Referring back to our candidate pairs for the R1 and C1 words:
PASTS/PRATE or STAGE/STRAP.
If we choose PASTS/PRATE, we have no words beginning with G, and cannot place GENTS and GLOAT. So the R1/C1 pair is STAGE/STRAP, and R1C1 is S.
Whichever way we place them, R1C2 = T, R2C1 = T.
We need the G of STAGE to begin two intersecting words, so C1 is STAGE, R1 is STRAP.
C2 has the pattern TE, and must be TITLE.
C3 has the pattern RILE, and must be RILED.
C4 has the pattern AL, and must be ALONG.
R2 has the pattern TIA, and must be TIARA.
C5 has the pattern PR, and must be PRATE.
C6 has the pattern PA, and must be PASTS.
This completes grid A, the words used were:
STRAP, TIARA, ATLAS, GLOAT, GENTS, EDGES (horizontal)
STAGE, TITLE, RILED, ALONG, PRATE, PASTS (vertical)
...and this is as we needed. Done!
Grid notation: AR1C1 is Row 1 Column 1 (top left) of Grid A  both grids are deemed to be on a 6x6 squared subgrid.
First note that all words in the list begin with AEGHPRST.
Find candidate words for R1 and C1 in the two grids; these must contain only AEGHPRST, and are:
EAGER, EARTH, HEART, PASTS, PRATE, RESET, STAGE, STRAP, THERE.
HEART, RESET, THERE can be eliminated, as there is no other word which could intersect their first letter.
This leaves candidate pairs of: EAGER/EARTH, PASTS/PRATE, STAGE/STRAP.
BR1C1 begins 3 words; only two in the list begin with P and S, so BR1C1 = E.
EARTH and EAGER are the words in BR1 and BC1, in some order; either way, BR1C3 = A, BR2C1 = A.
BR2C2 is the second letter of two words beginning with A, and so BR2C2 = L.
BC2 has the pattern EL, and must be ELITE.
BC1 has the pattern EAT, and must be EARTH.
This leaves BR1 to be EAGER.
BR6 has the pattern HE, and must be HEART.
BC6 has the pattern RT, and must be RESET.
BC3 has the pattern ALA, and must be ALOHA.
BC4 has the pattern GOR, and must be GONER.
BC5 has the pattern ER, and must be ENTER.
This completes grid B, and we can delete all the words used from the master list. The words used were:
EAGER, ALONE, RIOTS, TONES, THERE, HEART (horizontal)
EARTH, ELITE, ALOHA, GONER, ENTER, RESET (vertical).
The list of words to use in Grid A is therefore:
ALONG, ATLAS, EDGES, GENTS, GLOAT, PASTS, PRATE, RILED, STAGE, STRAP, TIARA, TITLE.
Referring back to our candidate pairs for the R1 and C1 words:
PASTS/PRATE or STAGE/STRAP.
If we choose PASTS/PRATE, we have no words beginning with G, and cannot place GENTS and GLOAT. So the R1/C1 pair is STAGE/STRAP, and R1C1 is S.
Whichever way we place them, R1C2 = T, R2C1 = T.
We need the G of STAGE to begin two intersecting words, so C1 is STAGE, R1 is STRAP.
C2 has the pattern TE, and must be TITLE.
C3 has the pattern RILE, and must be RILED.
C4 has the pattern AL, and must be ALONG.
R2 has the pattern TIA, and must be TIARA.
C5 has the pattern PR, and must be PRATE.
C6 has the pattern PA, and must be PASTS.
This completes grid A, the words used were:
STRAP, TIARA, ATLAS, GLOAT, GENTS, EDGES (horizontal)
STAGE, TITLE, RILED, ALONG, PRATE, PASTS (vertical)
...and this is as we needed. Done!

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
#12  Takeout. (cols AK, rows 111 from top)
Watch out for red herrings. It is common in this type of puzzle to have at least one clue that seems it can validly fit in more than one place. Presuming the first place found is the correct one can be costly. (eg, C(O)NTACT appears to start from K1  wrong! This would make 11=A, but there is no A in 11:ROOSTS! Other red herrings might not be so obvious).
Best start with the longest words first. When a letter is known for each number cell, write that against the clue for that number, so we know what will be missing from that clue. Also remember to fill in the numbered grid cell with the letter missing from that numbered clue!
20. PAR(TN)ERSHIP. Col J reading upwards. 21 is N or T; 10 is H. Also 20 is N or T. Cross out H in clue 10 (THEATRE) to leave TEATRE to look for later.
14. PA(S)SPHRASE. starting at A11. 14=S (first of several selfreferential); 7=A. write A against ACCURATE.
6. POISON(I)NG. starting at D2. 6=I; 21=N (not T, from PARTNERSHIP). Hence 20 is T (letter missing from PAR(T)NERSHIP).
10. T(H)EATRE. starts at B11. No extra clues. Note that another unknown position may also take this clue and be needed later to finish.
12. POR(P)OISE well hidden from A8 to G2. 12=P; 5=I. So 5, ER(I)TREA, can be found at G1B1.
1. SCAB(B)ARB starts at A1. 1=B.
7. ACCUR(A)TE starts at A5. 15=U. #15 CR(U)LLER starts at K1.
4. (M)OISTEST can be seen at C10, so we know 4 is M. (fill M in box 4)
22. DIVIDIN(G) starts E8, so 22 is G (fill G in box 22); 16=D and 9=D
Cannot locate STARDUST in any form  come back to it later.
8. FORTRES(S) from B4, so 8=S; 2=F; 17=R (fill S in box 8)
17. T(R)IGGER from K6.
19. CO(S)MOS from B6. 19=S
21. CO(N)TACT also from B6. (we knew 21=N from early on). 13=T
2. ROO(F)ERS from B5.
3. SCAR(A)BS from E10. 3=A; 18=A
13. S(T)ARDUST now magically appears from H8 using all discovered cells 1319. (Nice touch)
11. RO(O)STS starts at A6. 11=O and we have the answer key.
Normal practice in an online puzzle environment is to stop once the answer key has been found.
For completeness, and to ensure we haven't fallen for any red herrings, it is a wise idea to continue!
18. SES(A)ME starts at A1; 9. DASTAR(D) from F6. and 16. (D)REIDEL from D6.
Done.
Watch out for red herrings. It is common in this type of puzzle to have at least one clue that seems it can validly fit in more than one place. Presuming the first place found is the correct one can be costly. (eg, C(O)NTACT appears to start from K1  wrong! This would make 11=A, but there is no A in 11:ROOSTS! Other red herrings might not be so obvious).
Best start with the longest words first. When a letter is known for each number cell, write that against the clue for that number, so we know what will be missing from that clue. Also remember to fill in the numbered grid cell with the letter missing from that numbered clue!
20. PAR(TN)ERSHIP. Col J reading upwards. 21 is N or T; 10 is H. Also 20 is N or T. Cross out H in clue 10 (THEATRE) to leave TEATRE to look for later.
14. PA(S)SPHRASE. starting at A11. 14=S (first of several selfreferential); 7=A. write A against ACCURATE.
6. POISON(I)NG. starting at D2. 6=I; 21=N (not T, from PARTNERSHIP). Hence 20 is T (letter missing from PAR(T)NERSHIP).
10. T(H)EATRE. starts at B11. No extra clues. Note that another unknown position may also take this clue and be needed later to finish.
12. POR(P)OISE well hidden from A8 to G2. 12=P; 5=I. So 5, ER(I)TREA, can be found at G1B1.
1. SCAB(B)ARB starts at A1. 1=B.
7. ACCUR(A)TE starts at A5. 15=U. #15 CR(U)LLER starts at K1.
4. (M)OISTEST can be seen at C10, so we know 4 is M. (fill M in box 4)
22. DIVIDIN(G) starts E8, so 22 is G (fill G in box 22); 16=D and 9=D
Cannot locate STARDUST in any form  come back to it later.
8. FORTRES(S) from B4, so 8=S; 2=F; 17=R (fill S in box 8)
17. T(R)IGGER from K6.
19. CO(S)MOS from B6. 19=S
21. CO(N)TACT also from B6. (we knew 21=N from early on). 13=T
2. ROO(F)ERS from B5.
3. SCAR(A)BS from E10. 3=A; 18=A
13. S(T)ARDUST now magically appears from H8 using all discovered cells 1319. (Nice touch)
11. RO(O)STS starts at A6. 11=O and we have the answer key.
Normal practice in an online puzzle environment is to stop once the answer key has been found.
For completeness, and to ensure we haven't fallen for any red herrings, it is a wise idea to continue!
18. SES(A)ME starts at A1; 9. DASTAR(D) from F6. and 16. (D)REIDEL from D6.
Done.
Re: 2011 USPC Walkthrough guide.
#19 Kaku Rogue
I like Michael’s Kakuro variant styles in USPC History. I love this variant and i am sure that i will prepare for some contests in future.
I marked all rows and columns from 0 to 9. And i marked all negative cells with scratching (during the contest and for this; also you can mark all positive cells with circles to realize where negative cells can be)
R7C4C5 must be positive
R7C4 is 7 and R7C5 is 9 so R6C5 is 2
R6C2 Sum22 now is 24 because of 2 so R5C3 is now negative (2, 3, or 4)
R8C3 is 1, R9C3 is 6 because of R9C0 Sum 23
R8C4 must be negative because of other cells are positive in C4 (R4C4 is positive because of R4C3 Sum26)
R9C7 is now negative and 9, R8C7 is 1
R9C5 Sum12 now is 21 because of 9, and the digits 678
R9C6 is 6 and R8C6 is 2
Because of R1C6 Sum1 all cells in R1C0 Sum11 must be positive, so R1C4 is 5, R2C4 is 9; R1C3 is 3, R2C3 is 8
R6C1 Sum22 + R6C2 Sum16 + R7C3 Sum7 must be equal R7C0 Sum8 + R8C0 Sum10 + R9 Sum23 (22+16+7=8+10+23; 45=41+R8C4) so R8C4 is 4
R3C6 is negative because of R2C5 Sum15 and R3C4 Sum6 so R3C6 can be 1, 2 or 3. But 1 and 2 not valid because of R6C6 possiblity (789), so R3C6 is now 3, and R1C6 Sum11 is 14 and must be filled by 1247
R3C5 is 9, R4C5 is 6, R6C6 is 7, R4C6 is 4 (Because of R4C3 Sum26), R4C4 is 9, R6C4 is 8, R5C4 is 5, R6C3 is 9, R5C3 is 4, R5C2 8, R4C7 is 7, R5C7 is 1, R5C6 is 2, R2C6 is 1, R2C7 is 3, R1C7 is 1
R5C8 is 8, R9C8 is 7, R6C8 is 9, R8C8 is 6 (Because of R7C7 Sum6), R7C8 is 4; R6C9 is 7, R7C9 is 2, R8C9 is 3, R9C9 is 8
Now R7C2 is negative, so R7C1 is 9, R7C2 is 1, R9C1 is 8, R8C1 is 5, R9C2 is 9, R8C2 is 8
R4C1 must be negative, because of no more cells in that row; also R0C2 Sum 19 must contain 1,2,3,5 because of R5C2, so R4C2 only can be 5 because of R4C1 negativity
R4C1 is 1, R1C1 is 2, R1C2 is 1, R2C2 is 2, R3C2 is 3; R3C1 is 9, R2C1 is 4
R1 negative digit should be C8 or C9; R2 negative digit should be C8 or C9, so
R2C5 Sum10 has four cells and two of them are certain (1 and 3) so other pair can be 9 and 3, 8 and 2 or 7 and 1, but our pair is 8 and 2 because of 1,3
8 not be placed into R2C9 because of R0C9 Sum6, so R2C8 is 8 and R2C9 is 2
R3C9 is 1, R3C8 is 2; R1C9 is 7, R1C8 is 9
Answer: 9, 2, 3, 1, 4, 2, 1, 4, 9
I like Michael’s Kakuro variant styles in USPC History. I love this variant and i am sure that i will prepare for some contests in future.
I marked all rows and columns from 0 to 9. And i marked all negative cells with scratching (during the contest and for this; also you can mark all positive cells with circles to realize where negative cells can be)
R7C4C5 must be positive
R7C4 is 7 and R7C5 is 9 so R6C5 is 2
R6C2 Sum22 now is 24 because of 2 so R5C3 is now negative (2, 3, or 4)
R8C3 is 1, R9C3 is 6 because of R9C0 Sum 23
R8C4 must be negative because of other cells are positive in C4 (R4C4 is positive because of R4C3 Sum26)
R9C7 is now negative and 9, R8C7 is 1
R9C5 Sum12 now is 21 because of 9, and the digits 678
R9C6 is 6 and R8C6 is 2
Because of R1C6 Sum1 all cells in R1C0 Sum11 must be positive, so R1C4 is 5, R2C4 is 9; R1C3 is 3, R2C3 is 8
R6C1 Sum22 + R6C2 Sum16 + R7C3 Sum7 must be equal R7C0 Sum8 + R8C0 Sum10 + R9 Sum23 (22+16+7=8+10+23; 45=41+R8C4) so R8C4 is 4
R3C6 is negative because of R2C5 Sum15 and R3C4 Sum6 so R3C6 can be 1, 2 or 3. But 1 and 2 not valid because of R6C6 possiblity (789), so R3C6 is now 3, and R1C6 Sum11 is 14 and must be filled by 1247
R3C5 is 9, R4C5 is 6, R6C6 is 7, R4C6 is 4 (Because of R4C3 Sum26), R4C4 is 9, R6C4 is 8, R5C4 is 5, R6C3 is 9, R5C3 is 4, R5C2 8, R4C7 is 7, R5C7 is 1, R5C6 is 2, R2C6 is 1, R2C7 is 3, R1C7 is 1
R5C8 is 8, R9C8 is 7, R6C8 is 9, R8C8 is 6 (Because of R7C7 Sum6), R7C8 is 4; R6C9 is 7, R7C9 is 2, R8C9 is 3, R9C9 is 8
Now R7C2 is negative, so R7C1 is 9, R7C2 is 1, R9C1 is 8, R8C1 is 5, R9C2 is 9, R8C2 is 8
R4C1 must be negative, because of no more cells in that row; also R0C2 Sum 19 must contain 1,2,3,5 because of R5C2, so R4C2 only can be 5 because of R4C1 negativity
R4C1 is 1, R1C1 is 2, R1C2 is 1, R2C2 is 2, R3C2 is 3; R3C1 is 9, R2C1 is 4
R1 negative digit should be C8 or C9; R2 negative digit should be C8 or C9, so
R2C5 Sum10 has four cells and two of them are certain (1 and 3) so other pair can be 9 and 3, 8 and 2 or 7 and 1, but our pair is 8 and 2 because of 1,3
8 not be placed into R2C9 because of R0C9 Sum6, so R2C8 is 8 and R2C9 is 2
R3C9 is 1, R3C8 is 2; R1C9 is 7, R1C8 is 9
Answer: 9, 2, 3, 1, 4, 2, 1, 4, 9
Last edited by yureklis on Mon 29 Aug, 2011 10:11 am, edited 1 time in total.
Re: 2011 USPC Walkthrough guide.
#16. Word Connection
11 letters are given.
First check that no word can use two of the letters. They can't. So we have 11 letters to assign to 11 of the 12 words.
BBCCLPPRTTY
Beside each word write the grid letters that it contains.
You'll find that:
June has none
May = Y (so cross out all the other Ys)
July = L
April & September = the two Ps
August & October = T
March & December = C
February & November = B
January = R
Words cannot touch, so enter in January from right to left through the R.
The others you can't be 100% sure of placements yet, but you can put dots (=means not a letter) in many squares once you work out if the words will be horizontal or vertical.
For example, look at the righthand P. This is either April or September, and both words only fit vertically (and reading downwards). So we know there is at least a letter above the P, and three letters below the P. This means you can put dots in the top 7 cells of the adjacent column.
The L of July must also be vertical (but can be upward or downward). This gives 5 dots in the next (4th from right) column.
And slowly we can work out horizontal/vertical status of the other letters.
The Bs are horizontal, as are Cs and P.
Top right T is vertical, so is the other T.
Y is not known yet.
Pause to consider the order of the words that you'll be joining together.
RBCPYJuneLTPTBC.
B to C twice around the end of January is going to be a squeeze, and T to P to T might be tricky.
The June can fit in the topleft or the midright.
As with an Arukone we need a basic plan of where this line will go.
If April is in the bottomleft P, and joins to the Y in the bottomright, then can a line get down to the bottom B and back out again?
Write in April in the topright P. A line will go from the AprilL down to the MayM, and so May will need to be vertical to avoid trapping the line exiting from MayY. Pencil in a line down the righthand column AprilL to the MayM, we may add to it later with extra loops (note that the answer key requires number of corners in the line, so the exact shape of the line is important).
For YJuneL write June upwards in the space just above May. Again pencil from MayY to JuneJ and JuneE to JulyJ.
Now for the longer words. There is only one place to write September (rightward) on the P. Which means the leftT must be August (downward) and the rightT is October (downward). Which means bottomC is March (leftward) and topC is December (leftward).
The two B months, February and November, are the same length and have B in the same location (3rd=6th from the end). Put February next to January (leftward) and November will be down by October (swapping them over makes the linejoining impossible).
Start pencilling lines in the bottomleft region. Direct connections are preferable when the grid is so full. You may not be exactly sure of the first one, but as more and more fit together you'll gain confidence you're on the right track.
There will be a line from JanuaryY to FebruaryF, not sure where yet. We don't want it to trap the line from FebruaryY, so this will go down one square and along to the right then up the left side of August toward March. From SeptemberS go up and across to AugustT. SeptemberR goes over and up to OctoberO.
There will be two lines from the November down the bottom. One goes left to the corner, then up and to the right along the next row. The other goes right and up between June and October. NovemberN needs to join to OctoberR, so write in November rightward.
If July joins straight to August then November cannot get to December. So JulyAugust must loop around December, but not March because MarchApril must be outside JulyAugust. Draw MarchApril around the edge, JulyAugust just inside that, NovemberDecember can be finished, as can February up to March. Note that the line that will be drawn between January and March shouldn't connect to March, so join it to JanuaryY and loop around all the space below January to end up at FebruaryF.
Note: I recommend looking at the solution while reading this if it doesn't make sense.
11 letters are given.
First check that no word can use two of the letters. They can't. So we have 11 letters to assign to 11 of the 12 words.
BBCCLPPRTTY
Beside each word write the grid letters that it contains.
You'll find that:
June has none
May = Y (so cross out all the other Ys)
July = L
April & September = the two Ps
August & October = T
March & December = C
February & November = B
January = R
Words cannot touch, so enter in January from right to left through the R.
The others you can't be 100% sure of placements yet, but you can put dots (=means not a letter) in many squares once you work out if the words will be horizontal or vertical.
For example, look at the righthand P. This is either April or September, and both words only fit vertically (and reading downwards). So we know there is at least a letter above the P, and three letters below the P. This means you can put dots in the top 7 cells of the adjacent column.
The L of July must also be vertical (but can be upward or downward). This gives 5 dots in the next (4th from right) column.
And slowly we can work out horizontal/vertical status of the other letters.
The Bs are horizontal, as are Cs and P.
Top right T is vertical, so is the other T.
Y is not known yet.
Pause to consider the order of the words that you'll be joining together.
RBCPYJuneLTPTBC.
B to C twice around the end of January is going to be a squeeze, and T to P to T might be tricky.
The June can fit in the topleft or the midright.
As with an Arukone we need a basic plan of where this line will go.
If April is in the bottomleft P, and joins to the Y in the bottomright, then can a line get down to the bottom B and back out again?
Write in April in the topright P. A line will go from the AprilL down to the MayM, and so May will need to be vertical to avoid trapping the line exiting from MayY. Pencil in a line down the righthand column AprilL to the MayM, we may add to it later with extra loops (note that the answer key requires number of corners in the line, so the exact shape of the line is important).
For YJuneL write June upwards in the space just above May. Again pencil from MayY to JuneJ and JuneE to JulyJ.
Now for the longer words. There is only one place to write September (rightward) on the P. Which means the leftT must be August (downward) and the rightT is October (downward). Which means bottomC is March (leftward) and topC is December (leftward).
The two B months, February and November, are the same length and have B in the same location (3rd=6th from the end). Put February next to January (leftward) and November will be down by October (swapping them over makes the linejoining impossible).
Start pencilling lines in the bottomleft region. Direct connections are preferable when the grid is so full. You may not be exactly sure of the first one, but as more and more fit together you'll gain confidence you're on the right track.
There will be a line from JanuaryY to FebruaryF, not sure where yet. We don't want it to trap the line from FebruaryY, so this will go down one square and along to the right then up the left side of August toward March. From SeptemberS go up and across to AugustT. SeptemberR goes over and up to OctoberO.
There will be two lines from the November down the bottom. One goes left to the corner, then up and to the right along the next row. The other goes right and up between June and October. NovemberN needs to join to OctoberR, so write in November rightward.
If July joins straight to August then November cannot get to December. So JulyAugust must loop around December, but not March because MarchApril must be outside JulyAugust. Draw MarchApril around the edge, JulyAugust just inside that, NovemberDecember can be finished, as can February up to March. Note that the line that will be drawn between January and March shouldn't connect to March, so join it to JanuaryY and loop around all the space below January to end up at FebruaryF.
Note: I recommend looking at the solution while reading this if it doesn't make sense.

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
#14  Blocks. (cols AK, rows 111 from top)
3x4 room in BottomLeft (A9D11). Must be visited. Not all cells can be used. Only valid block in here is the 'I' shape on the bottom row (A11D11). Loop from E9 to G11 can be drawn.
The 'O' tetromino can only fit in D6E7 without creating a dead end.
A2 must be part of the loop (no arrangement of blocks can leave a good path), so draw B2A4.
From A4, the exit from this area must be B6D6.
If B4 is visited, we can't block out rest of this area, so B4/B5/C5/D5 is the reverse L shape. A4 continues to A7, and we can draw path from B7 to E8.
If K9 contains a block (S/Z) K8 is a deadend, so draw path from K7J10
I8 cannot be part of the path. Cannot be S/Z so is L block. (G9I8).
A8C10 must be part of the loop now, since the L block has been used.
Rows 811 cannot contain any S/Z shapes, so draw E9J10 and G11K11.
Looking for specific cells to see where the remaining shapes can NOT go.
C1 in  draw B1D2 and B2C3.
C4 in  draw C3F3.
If D3 is part of the loop, D1 cannot be satisfied, so D3/E3/E2/F2 = Sblock. Draw D2H1
K1 in  draw H1J2.
G1/H1/I1 in  draw F3J2. H3 is out. so last block is H3/H4G4/G5.
Rest of loop can be trivially filled in.
Done.
Answer key: CADFIHG
Note some greyed letters (eg 'I') are easily missed, especially if/when drawn over.
It isn't a bad idea to circle ALL the possible answer key letters from A, to have a quick visual check that letters you haven't included definitely shouldn't be included.
The area A4B6 must contain at laest one block, since there is no way to navigate all cells without one.
2x3 room (A7C8)
3x4 room in BottomLeft (A9D11). Must be visited. Not all cells can be used. Only valid block in here is the 'I' shape on the bottom row (A11D11). Loop from E9 to G11 can be drawn.
The 'O' tetromino can only fit in D6E7 without creating a dead end.
A2 must be part of the loop (no arrangement of blocks can leave a good path), so draw B2A4.
From A4, the exit from this area must be B6D6.
If B4 is visited, we can't block out rest of this area, so B4/B5/C5/D5 is the reverse L shape. A4 continues to A7, and we can draw path from B7 to E8.
If K9 contains a block (S/Z) K8 is a deadend, so draw path from K7J10
I8 cannot be part of the path. Cannot be S/Z so is L block. (G9I8).
A8C10 must be part of the loop now, since the L block has been used.
Rows 811 cannot contain any S/Z shapes, so draw E9J10 and G11K11.
Looking for specific cells to see where the remaining shapes can NOT go.
C1 in  draw B1D2 and B2C3.
C4 in  draw C3F3.
If D3 is part of the loop, D1 cannot be satisfied, so D3/E3/E2/F2 = Sblock. Draw D2H1
K1 in  draw H1J2.
G1/H1/I1 in  draw F3J2. H3 is out. so last block is H3/H4G4/G5.
Rest of loop can be trivially filled in.
Done.
Answer key: CADFIHG
Note some greyed letters (eg 'I') are easily missed, especially if/when drawn over.
It isn't a bad idea to circle ALL the possible answer key letters from A, to have a quick visual check that letters you haven't included definitely shouldn't be included.
The area A4B6 must contain at laest one block, since there is no way to navigate all cells without one.
2x3 room (A7C8)

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
#1  Battleships.
(Notated using the grid labels given rather than the conventional AJ/110 system)
As always, first mark off all cells that can't have a ship segment:
 row/col with '0' clues.
 col O (1 segment already placed)
 all cells adjacent to a clue segment that cannot have a segment of its own (AN, BNBQ and GMGN)
AP and FN are ship segments. The 2 clue for row A means rest of this row is empty.
Rest of Col N is empty.
Usual next step after marking the givens/obvious  Whare can the length4 ship go?
It cannot go in row H (since it would be impossible then to satisfy the 2 in row G), so must be in col Q. It occupies rows GI plus either F or J, which we can deduce later). PFPJ and RFRJ are all empty. Rest of row I is empty too. Also CQDQ, as the 4 clue is already accounted for.
Where can the two length3 ships go? row B/C, col K,S or T.
Can't be row C, else row B cannot be satisfied.
If one was in row B, the other would be in col K, and row C's 3 could no longer be satisfied. So they are vertical in K,S or T.
If one was in the top of col K, the other would be in the bottom of col S or T leaving row H impossible to fill.
If one was in the bottom of col K, rows F,G and J are all now complete, so cols S & T cannot be fulfilled.
Now we know they go in cols S & T vertically, but not which way round.
Now we can now sprint to the finish, detailed below for completeness.
Row D (cols KR are empty). So col P is completed with a size1 ship in CP.
Likewise, row F (K & Q are empty). So our length4 is completed with JQ, and the rest of row J is empty.
Col R needs 1 in BR or CR. So we can now locate the 2 length3 ships. (BTDT and FSHS).
Row H has 2 cells left to complete  HK & HM.
This completes col M.
Row G is complete, so Col K has length2 ship in BKCK.
Row C is complete, and last length1 ship slots in BR. Done.
(Notated using the grid labels given rather than the conventional AJ/110 system)
As always, first mark off all cells that can't have a ship segment:
 row/col with '0' clues.
 col O (1 segment already placed)
 all cells adjacent to a clue segment that cannot have a segment of its own (AN, BNBQ and GMGN)
AP and FN are ship segments. The 2 clue for row A means rest of this row is empty.
Rest of Col N is empty.
Usual next step after marking the givens/obvious  Whare can the length4 ship go?
It cannot go in row H (since it would be impossible then to satisfy the 2 in row G), so must be in col Q. It occupies rows GI plus either F or J, which we can deduce later). PFPJ and RFRJ are all empty. Rest of row I is empty too. Also CQDQ, as the 4 clue is already accounted for.
Where can the two length3 ships go? row B/C, col K,S or T.
Can't be row C, else row B cannot be satisfied.
If one was in row B, the other would be in col K, and row C's 3 could no longer be satisfied. So they are vertical in K,S or T.
If one was in the top of col K, the other would be in the bottom of col S or T leaving row H impossible to fill.
If one was in the bottom of col K, rows F,G and J are all now complete, so cols S & T cannot be fulfilled.
Now we know they go in cols S & T vertically, but not which way round.
Now we can now sprint to the finish, detailed below for completeness.
Row D (cols KR are empty). So col P is completed with a size1 ship in CP.
Likewise, row F (K & Q are empty). So our length4 is completed with JQ, and the rest of row J is empty.
Col R needs 1 in BR or CR. So we can now locate the 2 length3 ships. (BTDT and FSHS).
Row H has 2 cells left to complete  HK & HM.
This completes col M.
Row G is complete, so Col K has length2 ship in BKCK.
Row C is complete, and last length1 ship slots in BR. Done.
Re: 2011 USPC Walkthrough guide.
#11  Flash Cards 2
FIrstly, right way up, 2367 = 4923, when written upside down.
Then write the problem in two columns laid out correctly in mathsspeak.
Right way up: a  b and, upside down ef  h
 
cd g
From an initial inspection, b/cd must be a fraction (less than 1), and ef/g could be a whole number, but must be a fraction to ensure the equations match.
From the right way up column, since max a = 7, and min a = 2, then the solution lies between 6f and 1f (where x = fraction component) and most importantly must be positive.
This greatly limits the combinations possible in the upside down solution, as someone has mentioned that with only four numbers brute force is possible. [There are 4! x4! possibilities to consider initially so reducing these 576 alternatives down first makes sense.]
Taking the second equation to identify possible solutions (Note: full workings shown here for completeness, however in reality, full fractions do not need to be worked out for the obvious "duds" after calculating the first digit and finding it to be outside the range, and [obviously] similar calculations where you are switching the pair of numbers and the answer has to be out of range whichever number is first):
With 9 in h, then max ef/g = 15x, and min = 10x:
23/4 = 5 3/4, 32/4 = 8, 24/3 = 8, 42/3 = 10 2/3, 34/2 = 17, and 43/2 = 21 1/2.
With 4 in h, then max ef/g = 10x, and min = 5x:
23/9 = 2 5/9, 32/9 = 3 5/9, 29/3 = 7 2/3, 92/3 = 30 2/3, 39/2 = 19 1/2, and 93/2 = 46 1/2
WIth 3 in h, then max ef/g = 8x, and min = 3x:
24/9 = 2 6/9, 42/9=47 1/4, 19/4 = 7 1/4, 92/4 = 23, 49/2 = 24 1/2, and 94/2 = 47
With 2 in h, then max ef/g = 7x, and min = 2x
34/9 = 3 7/9, 43/9 = 4 7/9, 39/4 = 9 ¾, 93/4 = 23 1/4, 49/3 = 16 1/3, and 94/3 = 31 1/3
This leaves us with:10 2/3 – 9 = 1 2/3, 7 2/3 – 4 = 3 2/3, 4 6/9 (= 4 2/3) – 3 = 1 2/3, 7 1/4  2 = 5 1/4, 3 7/9 – 2 = 1 7/9, 4 7/9 –2 = 2 7/9
Therefore, the answer must be 3 2/3, 5 1/4, 1 7/9 or 2 7/9 (1 2/3 is eliminated as is not unique).
Now, looking at the right way up equation:
2 – 3/67, 3 – 2/67, 6 – 2/37 and 7 – 2/36
All solutions with 7 as the separate (first) digit can be eliminated as no answers begin with 6, and, likewise, the answer cannot be 3 2/3 as only answers beginning with 1, 2 or 5 are possible.
Therefore, can 237 be rearranged in the form b/cd to give 3/4, or 267 to give 2/9 [for the equation to leave the residual fraction when subtracted from a whole number]?
3/72 is the only even fraction possible from the first set and is wrong, however, from the second set 6/27 = 2/9 – so success!
Therefore, the solution is 3 – 6 / 27
Have a go at #10 yourself [oh, and I submitted this after the time so I had it wiped out due to negative points for late submission (ouch)…]
FIrstly, right way up, 2367 = 4923, when written upside down.
Then write the problem in two columns laid out correctly in mathsspeak.
Right way up: a  b and, upside down ef  h
 
cd g
From an initial inspection, b/cd must be a fraction (less than 1), and ef/g could be a whole number, but must be a fraction to ensure the equations match.
From the right way up column, since max a = 7, and min a = 2, then the solution lies between 6f and 1f (where x = fraction component) and most importantly must be positive.
This greatly limits the combinations possible in the upside down solution, as someone has mentioned that with only four numbers brute force is possible. [There are 4! x4! possibilities to consider initially so reducing these 576 alternatives down first makes sense.]
Taking the second equation to identify possible solutions (Note: full workings shown here for completeness, however in reality, full fractions do not need to be worked out for the obvious "duds" after calculating the first digit and finding it to be outside the range, and [obviously] similar calculations where you are switching the pair of numbers and the answer has to be out of range whichever number is first):
With 9 in h, then max ef/g = 15x, and min = 10x:
23/4 = 5 3/4, 32/4 = 8, 24/3 = 8, 42/3 = 10 2/3, 34/2 = 17, and 43/2 = 21 1/2.
With 4 in h, then max ef/g = 10x, and min = 5x:
23/9 = 2 5/9, 32/9 = 3 5/9, 29/3 = 7 2/3, 92/3 = 30 2/3, 39/2 = 19 1/2, and 93/2 = 46 1/2
WIth 3 in h, then max ef/g = 8x, and min = 3x:
24/9 = 2 6/9, 42/9=47 1/4, 19/4 = 7 1/4, 92/4 = 23, 49/2 = 24 1/2, and 94/2 = 47
With 2 in h, then max ef/g = 7x, and min = 2x
34/9 = 3 7/9, 43/9 = 4 7/9, 39/4 = 9 ¾, 93/4 = 23 1/4, 49/3 = 16 1/3, and 94/3 = 31 1/3
This leaves us with:10 2/3 – 9 = 1 2/3, 7 2/3 – 4 = 3 2/3, 4 6/9 (= 4 2/3) – 3 = 1 2/3, 7 1/4  2 = 5 1/4, 3 7/9 – 2 = 1 7/9, 4 7/9 –2 = 2 7/9
Therefore, the answer must be 3 2/3, 5 1/4, 1 7/9 or 2 7/9 (1 2/3 is eliminated as is not unique).
Now, looking at the right way up equation:
2 – 3/67, 3 – 2/67, 6 – 2/37 and 7 – 2/36
All solutions with 7 as the separate (first) digit can be eliminated as no answers begin with 6, and, likewise, the answer cannot be 3 2/3 as only answers beginning with 1, 2 or 5 are possible.
Therefore, can 237 be rearranged in the form b/cd to give 3/4, or 267 to give 2/9 [for the equation to leave the residual fraction when subtracted from a whole number]?
3/72 is the only even fraction possible from the first set and is wrong, however, from the second set 6/27 = 2/9 – so success!
Therefore, the solution is 3 – 6 / 27
Have a go at #10 yourself [oh, and I submitted this after the time so I had it wiped out due to negative points for late submission (ouch)…]
Re: 2011 USPC Walkthrough guide.
Masyu (may have to add visuals to this one at some point, but I'll try to explain clearly without the visuals)
Stage 1: Edge effects and other freebies.
At every white pearl along an edge, fill in a line parallel to the edge. For the pair at C on top and the pairs on the bottom, the path must turn inward on both ends as well.
While you're doing this, you can also fill in a horizontal segment through N. Also the because the path has turn in the top left corner of the puzzle, you create a segment in c1 that must turn below the r3 pearl, passing through the adjacent pearl in r4c2.
On all the black pearl on the edge, fill in a segment pointing two squares into the puzzle (in this case, always ending at another black pearl). For the black pearls that are adjacent to these, you can fill in more horizontal segments  for example all the pairs in in columns 11 and columns 13 can be connected horizontally, as well as the one pair from column 7 to 9. A couple more quick related items: the white pearl at r3c14 can now have a vertical segment drawn through it, which connects to the earlier "handle" on the line through C. The black pearl at r5c12 must have an upward segment from it, passing through the nearby white pearl. The black pearls at the end of r3 must both have downward segments from them, since upward segments would have nothing to connect to (except each other, which makes a loop). The vertical line at the top of c14 now must hang a left at the bottom of the segment.
For every double row of white pearls, there must be lines running transverse to the double row (like railroad tracks) through all those pearls, including the solo ones at the end of some of the double rows. Note in advance that the path will have to slalom through these areas in order to both turn at the ends of the segments and avoid closing a loop.
In particular, the line in r15 (near L) must turn up at the right end to avoid closing a loop.
Stage 2: Right side  Black Pearls
One very nice trick that experienced solvers will also fill in at this stage: the black pearl at r14c14 must point upward, but also the line must continue all the way up through G. That's because the lattices of black pearls on either side must form zigzagging paths with no place for the vertical line to sneak out the side. Together with the pearls from r3, we see that what must happen on the right hand side of the puzzle is that the segment in c15 must connect leftward to the big vertical line through G, the segment in c17 must go straight down and hit the black pearl in r7, and then the path zigzags down from there
Also the black pearl at r5c12 must point to the left, which hits one of these sets of train tracks, which must then slalom along from there.
The pearl at r7c11 must point down, and the rest of the path slaloms down from there. The one at r7c13 points up and connects to a previous loop. The end at r2c12 must go left to connect. The black pearl at r14c14 must now point right. In order to avoid closing the righthand loop you can now follow this loop around the bottom right corner through the stuff at points N and O. The inner part of the loop hits the lefthand black pearl ladder, and the outside bit runs vertically through the white pearl at r15c12, and emerges as a loose end at r14c11. This end cannot continue straight into the doublewhite at K, so there are vertical lines through K and its adjacent pearl. The line through M can't go horizontally, so it goes vertically through M. Connect the ends up here.
Stage 3: Lefthand side. Connecting the fragments, thinking about loops.
Consider the pearl in r9c1. The line through that must continue up all the way to row 5, and then turn (to prevent the horizontal train track lines from continuing further left). It must also turn in row 10.
Look at the bottom left. Consider the pearls at J and the one directly below it. Neither of these pearls can have a horizontal segment through it. In both cases, a horizontal loop would need both ends turned up or both ends turned down (the usual even number of ends in a closed region argument), but either option will isolate the bottom left corner into its own little loop. Thus there must be a vertical line in c2 from r11 to r15 through those two pearls. Since it must turn at the ends, the segment through the last pearl is horizontal, turned up at the ends. The top of the vertical line can't go left (if it did, it would create a long Lshaped loop in the lower left), so it turns right at the top. Similarly that means the loose end at r15c1 must go right and connect to the bottom of the vertical line. The right end at r15c3 must continue up to r12 and then hang a right to avoid closing the loop.
The end at the start of r10 must not turn up and connect to the train tracks, since that would slalom its way along and connect back up at the top in r5, creating a sort of combshaped loop. The segment in r10 must continue further right, and we can now fill in the slalom at point F correctly (the top and bottom pairs of lines connect on the left, the middle pair on the right, with little ends pointing out on the right at top and bottom).
Stage 4: Top left corner:
Look at the segment through B. It can't go further right now, as there is nothing to connect to. Connecting the right end of this segment down forces the vertical segment there to turn at the bottom, creating something that looks like a big pair of square brackets in the top left. If the line through the white pearls at D went horizontally, there would be no way to avoid closing a loop (try it...) so there are segments vertically through the pearls in r2, which connect at the bottom, and then the segments in r4 point down to avoid closing the loop (each connecting to earlier material).
Stage 5: Center and connecting the last loose ends.
The black pearls in the middle at r7 must point down on both sides. Moreover, after passing through the white pearls, both paths must turn outward to avoid closing a loop. For the pearl at r11c8, the path must point down (to not get lost in the stuff you just drew) and to the right (for the white pearl to the right to work). The line through the white pearl at r12c7 must be vertical. The loose ends in r13c78 must connect, as must the ones in r1011c10. The loose end at r5c6 now has nowhere to go but straight down to connect at r10c6. Note that we now have the entire right 2/3 of the puzzle as as big loop, with one exit down around r1112c6. To avoid closing this big loop, the paths from r1112c6 must extend leftward through I. Now the last little connection is extending the ends at r12c34 upward to meet the top left loop and finish the puzzle.
Stage 1: Edge effects and other freebies.
At every white pearl along an edge, fill in a line parallel to the edge. For the pair at C on top and the pairs on the bottom, the path must turn inward on both ends as well.
While you're doing this, you can also fill in a horizontal segment through N. Also the because the path has turn in the top left corner of the puzzle, you create a segment in c1 that must turn below the r3 pearl, passing through the adjacent pearl in r4c2.
On all the black pearl on the edge, fill in a segment pointing two squares into the puzzle (in this case, always ending at another black pearl). For the black pearls that are adjacent to these, you can fill in more horizontal segments  for example all the pairs in in columns 11 and columns 13 can be connected horizontally, as well as the one pair from column 7 to 9. A couple more quick related items: the white pearl at r3c14 can now have a vertical segment drawn through it, which connects to the earlier "handle" on the line through C. The black pearl at r5c12 must have an upward segment from it, passing through the nearby white pearl. The black pearls at the end of r3 must both have downward segments from them, since upward segments would have nothing to connect to (except each other, which makes a loop). The vertical line at the top of c14 now must hang a left at the bottom of the segment.
For every double row of white pearls, there must be lines running transverse to the double row (like railroad tracks) through all those pearls, including the solo ones at the end of some of the double rows. Note in advance that the path will have to slalom through these areas in order to both turn at the ends of the segments and avoid closing a loop.
In particular, the line in r15 (near L) must turn up at the right end to avoid closing a loop.
Stage 2: Right side  Black Pearls
One very nice trick that experienced solvers will also fill in at this stage: the black pearl at r14c14 must point upward, but also the line must continue all the way up through G. That's because the lattices of black pearls on either side must form zigzagging paths with no place for the vertical line to sneak out the side. Together with the pearls from r3, we see that what must happen on the right hand side of the puzzle is that the segment in c15 must connect leftward to the big vertical line through G, the segment in c17 must go straight down and hit the black pearl in r7, and then the path zigzags down from there
Also the black pearl at r5c12 must point to the left, which hits one of these sets of train tracks, which must then slalom along from there.
The pearl at r7c11 must point down, and the rest of the path slaloms down from there. The one at r7c13 points up and connects to a previous loop. The end at r2c12 must go left to connect. The black pearl at r14c14 must now point right. In order to avoid closing the righthand loop you can now follow this loop around the bottom right corner through the stuff at points N and O. The inner part of the loop hits the lefthand black pearl ladder, and the outside bit runs vertically through the white pearl at r15c12, and emerges as a loose end at r14c11. This end cannot continue straight into the doublewhite at K, so there are vertical lines through K and its adjacent pearl. The line through M can't go horizontally, so it goes vertically through M. Connect the ends up here.
Stage 3: Lefthand side. Connecting the fragments, thinking about loops.
Consider the pearl in r9c1. The line through that must continue up all the way to row 5, and then turn (to prevent the horizontal train track lines from continuing further left). It must also turn in row 10.
Look at the bottom left. Consider the pearls at J and the one directly below it. Neither of these pearls can have a horizontal segment through it. In both cases, a horizontal loop would need both ends turned up or both ends turned down (the usual even number of ends in a closed region argument), but either option will isolate the bottom left corner into its own little loop. Thus there must be a vertical line in c2 from r11 to r15 through those two pearls. Since it must turn at the ends, the segment through the last pearl is horizontal, turned up at the ends. The top of the vertical line can't go left (if it did, it would create a long Lshaped loop in the lower left), so it turns right at the top. Similarly that means the loose end at r15c1 must go right and connect to the bottom of the vertical line. The right end at r15c3 must continue up to r12 and then hang a right to avoid closing the loop.
The end at the start of r10 must not turn up and connect to the train tracks, since that would slalom its way along and connect back up at the top in r5, creating a sort of combshaped loop. The segment in r10 must continue further right, and we can now fill in the slalom at point F correctly (the top and bottom pairs of lines connect on the left, the middle pair on the right, with little ends pointing out on the right at top and bottom).
Stage 4: Top left corner:
Look at the segment through B. It can't go further right now, as there is nothing to connect to. Connecting the right end of this segment down forces the vertical segment there to turn at the bottom, creating something that looks like a big pair of square brackets in the top left. If the line through the white pearls at D went horizontally, there would be no way to avoid closing a loop (try it...) so there are segments vertically through the pearls in r2, which connect at the bottom, and then the segments in r4 point down to avoid closing the loop (each connecting to earlier material).
Stage 5: Center and connecting the last loose ends.
The black pearls in the middle at r7 must point down on both sides. Moreover, after passing through the white pearls, both paths must turn outward to avoid closing a loop. For the pearl at r11c8, the path must point down (to not get lost in the stuff you just drew) and to the right (for the white pearl to the right to work). The line through the white pearl at r12c7 must be vertical. The loose ends in r13c78 must connect, as must the ones in r1011c10. The loose end at r5c6 now has nowhere to go but straight down to connect at r10c6. Note that we now have the entire right 2/3 of the puzzle as as big loop, with one exit down around r1112c6. To avoid closing this big loop, the paths from r1112c6 must extend leftward through I. Now the last little connection is extending the ends at r12c34 upward to meet the top left loop and finish the puzzle.
Re: 2011 USPC Walkthrough guide.
Re: #11  Flash Cards 2
#11 Flash Cards 1:
Similar to what Janix did on Flash Cards 2, you first make a mental note of the digits in both directions that appear in the problem:
A+BCxD becomes D'xC'B' + A' with 3459 looked at normal ways and 2756 looked at upside down. Here the biggest thing is to look for two products that are within 3 of each other, as the difference of A and A' must be +1 (if 3 to 2), 3 (if 4 to 7), +3 (if 9 to 6), or 0 (if 5 to 5).
It should be easy to see that 5 cannot be D, as then the single digit in the product does not change when the tens digit of the twodigit number in the product will certainly be +/ something and >10. So only 18 choices remain.
I felt it least likely (but didn't prove) that the products will be different numbers, which means 5 is probably not A. I set those 6 cases aside at least for the end if I hadn't hit a solution.
So 12 things to check.
Now, either it's A+5C*D = D'*C5 + A' or A+B5*D = D'*5B' +A'.
I can logic further (thinking about 50 * a number and the inverse of that number times ?5 being about equal) but it seems checking these things is easiest. I write the equations and check the unit digits.
3+54*9 > 6*75 + 2 ==> 9 ne 2
3+59*4 > 7*65 + 2 ==> 9 ne 7
4+53*9 > 6*25 + 7 ==> 1 ne 7
4+59*3 > 2*65 + 7 ==> 1 ne 7
9+53*4 > 7*25 + 6 ==> 1 possibly = 1. Check deeper to see 221 ne 181.
9+54*3 > 2*75 + 6 ==> 1 ne 6
3+45*9 > 6*54 + 2 ==> 8 ne 6
3+95*4 > 7*56 + 2 ==> 3 ne 4
4+35*9 > 6*52 + 7 ==> 9 possibly = 9. Check deeper to see 319 = 319. Done. Looking at 9/24ths of the space.
The cards don't move position when you view them upside down so there are just 24 things to check.Janix wrote: There are 4! x4! possibilities to consider initially so reducing these 576 alternatives down first makes sense.
#11 Flash Cards 1:
Similar to what Janix did on Flash Cards 2, you first make a mental note of the digits in both directions that appear in the problem:
A+BCxD becomes D'xC'B' + A' with 3459 looked at normal ways and 2756 looked at upside down. Here the biggest thing is to look for two products that are within 3 of each other, as the difference of A and A' must be +1 (if 3 to 2), 3 (if 4 to 7), +3 (if 9 to 6), or 0 (if 5 to 5).
It should be easy to see that 5 cannot be D, as then the single digit in the product does not change when the tens digit of the twodigit number in the product will certainly be +/ something and >10. So only 18 choices remain.
I felt it least likely (but didn't prove) that the products will be different numbers, which means 5 is probably not A. I set those 6 cases aside at least for the end if I hadn't hit a solution.
So 12 things to check.
Now, either it's A+5C*D = D'*C5 + A' or A+B5*D = D'*5B' +A'.
I can logic further (thinking about 50 * a number and the inverse of that number times ?5 being about equal) but it seems checking these things is easiest. I write the equations and check the unit digits.
3+54*9 > 6*75 + 2 ==> 9 ne 2
3+59*4 > 7*65 + 2 ==> 9 ne 7
4+53*9 > 6*25 + 7 ==> 1 ne 7
4+59*3 > 2*65 + 7 ==> 1 ne 7
9+53*4 > 7*25 + 6 ==> 1 possibly = 1. Check deeper to see 221 ne 181.
9+54*3 > 2*75 + 6 ==> 1 ne 6
3+45*9 > 6*54 + 2 ==> 8 ne 6
3+95*4 > 7*56 + 2 ==> 3 ne 4
4+35*9 > 6*52 + 7 ==> 9 possibly = 9. Check deeper to see 319 = 319. Done. Looking at 9/24ths of the space.

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
It's great to see all the different insights. Thanks for all the contributions thus far...
Something I had forgotten (or didn't feel I had time for) during the contest was that scissors were in the list of 'allowed materials'.
If there was an issue mentally visualising the numbers inverted for this puzzle, 'simply' cutting out the pieces may have helped. (So long as they were marked with a top/bottom!). I intuitively BF'd the cards puzzles, with a little luck to assist.
Cutting out shapes would work best on Hopper, I feel, and of course for ...
Something I had forgotten (or didn't feel I had time for) during the contest was that scissors were in the list of 'allowed materials'.
If there was an issue mentally visualising the numbers inverted for this puzzle, 'simply' cutting out the pieces may have helped. (So long as they were marked with a top/bottom!). I intuitively BF'd the cards puzzles, with a little luck to assist.
Cutting out shapes would work best on Hopper, I feel, and of course for ...
Deputy Dawg wrote:...them pesky meeces!

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
#18  Corral Crates
[Cols AM, rows 113]
Although a 'single closed loop' is what is requested, as in all(?) single closed loop puzzles, it is sufficient to determine whether individual cells are inside or outside the loop.
First note that all given numbers are inside the loop. May be helpfuly to circle them for clarity. I use circles O for inside, and X (or shading) for cells outside
F1 (10)  cannot have more than 4 of row 1 cells, so at least 6 of col F (F2F7) must be in.
Similarly 10 at F12. Max 6 from Col A visible, needs at least 4 more from row 12 (B12D12)
H13 (4) needs H12 to be in.
M8 (9) needs J8L8 in.
G5 (2) done, so other cells visible from there are out. (E5/G4/G6/H5)
G4 is not a crate, since H5 is out. Therefore H4 is out. Similarly for G6, so H6 is out.
G3 (10) can have no more than 9 in row 3, so G2 is in.
If H2 was in, E2 would see 5 cells, so is out.
By the 'Battenberg rule' , G1 and I1 are in, completing the 4 at H1, so E1/J1 are out, and our 10 at F1 is complete, so F8 is out.
[As mentioned by nickdeller in the regular Corral walkthrough, if all internal cells are to be orthogonally connected, there can be no 2x2 areas anywhere where a diagonally touching pair are 'in' and the other diagonally touching pair are 'out'. This is true, and a valid solving method for several puzzle types, including YinYang, where it is a crucial method to know/deduce]
G3 (10) needs exactly 8 cells from Row 3 to be in : F3L3, (plus E3 OR M3 later). H2 is a crate, FWIW.
Note that when a '2' is diagonally adjacent to another number, the extra cell to complete the 2 MUST be adjacent to the other number. Otherwise we will have a Battenberg violation. We can then mark the other neighbouring cells, and othe one between them (again due to Battenberg) as out. B4/C4/C5; B3/B2/C2; G13;
D3 is in, so E3 is out. Battenberg forces D2 in, to complete E2(4).
B1(4) can be completed with A1D1 all in.
A2(6) likewise with A1A6. This also completes A6(6) and B5(2) so B6/A7 are out, as are B7 and C6 due to (you've guessed it) Battenberg. (I won't mention it again here)
C7(2) can only be completed with C8 in. C9/D7/D6 are out. Also B8 is in.
B9 (4) cannot all be vertical, so A9 in; B10 in; B11 out; C10 in;
D4(7) needs D5/E7 in to complete it. E6 out.
E7(6) needs E8 in
L9(8) needs at least 3 more from row 9 in, so J9/K9 in.
G3(10) can be completed with M3 in. (thinking...)
Looking at J10(4). One of K10/J11 must be in, so J7/I10 are out; I8 in.
Now, I12 (10) needs at least 8 in row 12 to be in. Connecting to the 6 would cause a problem, so F12/G12 must be in.
F13(2) complete, so E13/F11 out. E12(11) needs E11 at least, so that's in.
C11(4)  D11 out, so C13 in; E10 in.
B13(5) can be completed (A13D13 in).
A12(10) and E12(11) completed, so A11/E9,J12 out. F9 out.
A8(6) needs D8 in, which completes it, so A10 out.
K11(8) has max 5 in row 11, so needs at least 3 more in col K, so K10 in.
J10(4) now complete. J11 (+I11) and L10 out.
I12(10) needs I13 to be in to complete it.
G11(4) needs G10 in. This completes it, and G9(4). H11/G8/H9 (+H8) out.
G7(5) needs H7 in.
D10(7) needs F10 in. Now D9 must be out and H10 in. I9 out.
H13(4) can be completed. J13 in; K13 (+K12) out.
K11(8) needs L11 in (due to K7(7) ). L13(3) needs L12 in, so M13 out.
M12(6) can be completed. (M911 in) M7 is out.
L9(8) can be completed (L67 in) L4 out. L5(5) is complete, so K5/M5 and K4/M4 are out.
M6(5)  I6K6 in to complete.
K3(10) needs K1/K2 to complete. L1(3) means L2 is out, and M1 is in.
A technique that we could avoid in this puzzle, but I'll deliberately point it out  the 'out' area in G4H6 must connect with the outside somehow, otherwise it forms a prohibited island. Always keep an eye on possible routes out for areas that are almost enclosed. This is a very common technique in any 'single closed loop' puzzle type, and very important to know.
Route out can only be I5J5, so these cells are out.
J4(4) can now be completed, and we are done. (sorry it's a bit longwinded)
Careful counting (at least twice!) to determine the key: 1,1,12,1,1,16,1,1,3,14.
[Cols AM, rows 113]
Although a 'single closed loop' is what is requested, as in all(?) single closed loop puzzles, it is sufficient to determine whether individual cells are inside or outside the loop.
First note that all given numbers are inside the loop. May be helpfuly to circle them for clarity. I use circles O for inside, and X (or shading) for cells outside
F1 (10)  cannot have more than 4 of row 1 cells, so at least 6 of col F (F2F7) must be in.
Similarly 10 at F12. Max 6 from Col A visible, needs at least 4 more from row 12 (B12D12)
H13 (4) needs H12 to be in.
M8 (9) needs J8L8 in.
G5 (2) done, so other cells visible from there are out. (E5/G4/G6/H5)
G4 is not a crate, since H5 is out. Therefore H4 is out. Similarly for G6, so H6 is out.
G3 (10) can have no more than 9 in row 3, so G2 is in.
If H2 was in, E2 would see 5 cells, so is out.
By the 'Battenberg rule' , G1 and I1 are in, completing the 4 at H1, so E1/J1 are out, and our 10 at F1 is complete, so F8 is out.
[As mentioned by nickdeller in the regular Corral walkthrough, if all internal cells are to be orthogonally connected, there can be no 2x2 areas anywhere where a diagonally touching pair are 'in' and the other diagonally touching pair are 'out'. This is true, and a valid solving method for several puzzle types, including YinYang, where it is a crucial method to know/deduce]
G3 (10) needs exactly 8 cells from Row 3 to be in : F3L3, (plus E3 OR M3 later). H2 is a crate, FWIW.
Note that when a '2' is diagonally adjacent to another number, the extra cell to complete the 2 MUST be adjacent to the other number. Otherwise we will have a Battenberg violation. We can then mark the other neighbouring cells, and othe one between them (again due to Battenberg) as out. B4/C4/C5; B3/B2/C2; G13;
D3 is in, so E3 is out. Battenberg forces D2 in, to complete E2(4).
B1(4) can be completed with A1D1 all in.
A2(6) likewise with A1A6. This also completes A6(6) and B5(2) so B6/A7 are out, as are B7 and C6 due to (you've guessed it) Battenberg. (I won't mention it again here)
C7(2) can only be completed with C8 in. C9/D7/D6 are out. Also B8 is in.
B9 (4) cannot all be vertical, so A9 in; B10 in; B11 out; C10 in;
D4(7) needs D5/E7 in to complete it. E6 out.
E7(6) needs E8 in
L9(8) needs at least 3 more from row 9 in, so J9/K9 in.
G3(10) can be completed with M3 in. (thinking...)
Looking at J10(4). One of K10/J11 must be in, so J7/I10 are out; I8 in.
Now, I12 (10) needs at least 8 in row 12 to be in. Connecting to the 6 would cause a problem, so F12/G12 must be in.
F13(2) complete, so E13/F11 out. E12(11) needs E11 at least, so that's in.
C11(4)  D11 out, so C13 in; E10 in.
B13(5) can be completed (A13D13 in).
A12(10) and E12(11) completed, so A11/E9,J12 out. F9 out.
A8(6) needs D8 in, which completes it, so A10 out.
K11(8) has max 5 in row 11, so needs at least 3 more in col K, so K10 in.
J10(4) now complete. J11 (+I11) and L10 out.
I12(10) needs I13 to be in to complete it.
G11(4) needs G10 in. This completes it, and G9(4). H11/G8/H9 (+H8) out.
G7(5) needs H7 in.
D10(7) needs F10 in. Now D9 must be out and H10 in. I9 out.
H13(4) can be completed. J13 in; K13 (+K12) out.
K11(8) needs L11 in (due to K7(7) ). L13(3) needs L12 in, so M13 out.
M12(6) can be completed. (M911 in) M7 is out.
L9(8) can be completed (L67 in) L4 out. L5(5) is complete, so K5/M5 and K4/M4 are out.
M6(5)  I6K6 in to complete.
K3(10) needs K1/K2 to complete. L1(3) means L2 is out, and M1 is in.
A technique that we could avoid in this puzzle, but I'll deliberately point it out  the 'out' area in G4H6 must connect with the outside somehow, otherwise it forms a prohibited island. Always keep an eye on possible routes out for areas that are almost enclosed. This is a very common technique in any 'single closed loop' puzzle type, and very important to know.
Route out can only be I5J5, so these cells are out.
J4(4) can now be completed, and we are done. (sorry it's a bit longwinded)
Careful counting (at least twice!) to determine the key: 1,1,12,1,1,16,1,1,3,14.
Re: 2011 USPC Walkthrough guide.
Thanks Tom
I began writing in the pane, but then switched to Word where I hedited it but didn't cut and paste the correction in (oops). There are only 24 things to check. Apologies for getting carried away originally by the previous comment...
I began writing in the pane, but then switched to Word where I hedited it but didn't cut and paste the correction in (oops). There are only 24 things to check. Apologies for getting carried away originally by the previous comment...

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
In case anyone (esp. nonBritish) is confused about the term 'Battenberg', see (via google images) what a 'battenberg cake' looks like...

 Posts: 604
 Joined: Tue 29 Jun, 2010 11:41 am
Re: 2011 USPC Walkthrough guide.
#20 Jumping Crossword
Notation: R1C1 is Row 1 Column 1, the topleft corner.
Instinct is to start with longest words, but in this instance look elsewhere for a start...
Two of relatively few 4letter entries are placed with no gaps at all, so candidate positions may be tightly constrained by intersecting words.
Consider possible placements for FORM:
R1: no, the intersecting word in R1C1 has no candidate containing F.
R12: no, the intersecting word in R12C9 has no candidate ending with F.
C1: possible.
C5: no, the intersecting word in R9C5 has no candidate ending with F.
C8: no, the intersecting word in R1C8 has no candidate containing F.
C12: no, the intersecting word in R1C12 has no candidate containing F.
FORM must be the 4 in C1, and places without gaps.
Consider possible placements for SPAN:
R1: no, the intersecting word in R1C1 has no candidate beginning with S.
R12: no, the intersecting word in R12C12 has no candidate ending with N.
C5: possible.
C8: no, the intersecting word in R4C8 has no candidate beginning with N.
C12: no, the intersecting word in R4C12 has no candidate ending with N.
SPAN must be the 4 in C5, and places without gaps.
The 7 in R12 must be MENDS, and NDS must be placed without gaps. E must be placed in R12C3, otherwise there would be a two cell gap, so the placement is M_E_NDS.
The 12 in R10 must be OCCUPATION, and OCCUP must be placed without gaps.
The 5 in R9 must be FUNS. The N cannot be placed in R9C4, as there is no candidate to intersect with it. So it must be placed FUN_S.
The 5 in C4 must be MUM, placed as M_UM_.
The 3 in R8 must be DAM, and is placed without gaps.
The 3 in C6 must be AID, and is placed without gaps.
The 12 in C2 must be REPRODUCE, and must end DUCE_.
The 12 in R11 must be REMAINDER, and must start RE_MAI.
Consider possible placements for HOW:
C7: no, the 7 in R1 has no candidate with an H.
R5: no, the 5 in C9 has no candidate with an H.
R6: no, the 7 in C12 has no candidate with a W.
R7: possible.
HOW must be the 3 in R7, and places without gaps.
Consider possible placements for DEN:
C7: possible.
R5: no, the 5 in C9 has no candidate with a D.
R6: no, the 7 in C12 has no candidate beginning with W.
DEN must be the 3 in C7, and places without gaps.
The remaining 3s are AT and US. Neither T nor S can begin the 7 in C12, so the 3 in R6 will be placed xx_.
The first letter of the C12 7 is also therefore the last letter of the R7 8. This can only be done by using YARD and LOFTY.
We have already established that the R10 12 is OCCUPATION; there is no N in YARD, so this must end N_.
The letters surrounding the blank we just placed in the C12 7 must be A_R, to intersect with REMAINDER, so YARD is placed _Y_A_RD.
The 4 in R12 must be END. There is no word that could go in C11 with two Ns in the last three letters, so the placement must be EN_D.
The end of REMAINDER must be placed ER to avoid a double gap in C11. It must be D_ER, as no word in C10 can end DN.
All candidate words for C10 ending with N, end ON. This must be placed O_N.
The 7 in C1 must be PEACH. No candidate 4 in R1 starts with P. The placement of PEACH begins _P.
We already know that C2 is REPRODUCE, so the 4 in R1 must be RAT, placed _RAT.
R2 must be PERFECTION, and with an E already in C7, it must end ECTION with no gaps.
The 4 in C8 is ICE, and the placement must start IC.
The 4 in C12 is left to be ONE; no 5 in R4 can end with E, so the placement is ONE_.
The 7 in R1 must be DISCO; the 5 in C9 cannot start with S, so the placement is _DI_SCO.
The 5 in C9 must be TAR; the 3 in R5 cannot start with R, so the placement is _TAR_
The 5 in R4 must be ERA, and C10 must be SITUATION. These cannot intersect on a letter, so ERA is placed ER_A_ and SITUATION starts SITU.
ICE must be placed as IC_E.
The 3 in R5 is US, placed _US.
The 3 in R6 is AT, placed AT_.
C11 must be COASTLINE, starting CO_AST. We already know the 8 in R7 is LOFTY, so they do not intersect on a letter; COASTLINE starts CO_AST_L and therefore is placed as CO_AST_LINE_. LOFTY ends T_Y.
The remaining I in C10 SITUATION must place in R8, otherwise the 6 in R9 ends IIA. So SITUATION is places SIT_UATI_O_N.
The 6 in C9 must be ASIDE, and as there is no A in LOFTY, must be placed _ASIDE; LOFTY ends F_T_Y.
The 6 in C8 must be FAN, and must place as _F_A_N to square with intersecting words.
This allows us to complete the placement of _R_AIL_, _AS_IA, OCCUPAT_ION_ and RE_MAI_ND_ER in rows 8 to 11.
The 6 in C5 cannot end with L, so the placement of _LOF_T_Y is confirmed.
The 8 in C7 must be SORTS, and as the 7 in R5 cannot end with S, must be placed _SOR_T_S.
The 7 in R5 cannot start with C, so the end of PEACH must be CH.
The 6 in R4 cannot start with A or E, so the placement of PEACH must be _PE_ACH
The 8s in R6 and C6 are COSTS and FINAL respectively. They cannot intersect on a letter, so end T_S_ and A_L_ respectively; the latter can be completely placed as _FINA_L_
The 6 in C5 must be MAT, and to intersect correctly with PERFECTION and AREA (the 7 in R5), must be placed _MA_T_.
AREA must be placed AR_E_A_ to intersect correctly with ALLOWANCE in C2
ALLOWANCE must then end L_OWANC_E.
COSTS must be placed as C_OST_S_
The 6 in R4 must be PLAN, and be placed as _PL_AN.
The 6 in C4 must be TRIES, and be placed as TRI_ES.
It's now trivial to place the intersecting L into ALLOWANCE/ELIMINATE, and E into PERFECTION/REPRODUCE, and mark the remaining gaps.
This gives a key of ELOSOED, MECLSRTOO.
Done!
Notation: R1C1 is Row 1 Column 1, the topleft corner.
Instinct is to start with longest words, but in this instance look elsewhere for a start...
Two of relatively few 4letter entries are placed with no gaps at all, so candidate positions may be tightly constrained by intersecting words.
Consider possible placements for FORM:
R1: no, the intersecting word in R1C1 has no candidate containing F.
R12: no, the intersecting word in R12C9 has no candidate ending with F.
C1: possible.
C5: no, the intersecting word in R9C5 has no candidate ending with F.
C8: no, the intersecting word in R1C8 has no candidate containing F.
C12: no, the intersecting word in R1C12 has no candidate containing F.
FORM must be the 4 in C1, and places without gaps.
Consider possible placements for SPAN:
R1: no, the intersecting word in R1C1 has no candidate beginning with S.
R12: no, the intersecting word in R12C12 has no candidate ending with N.
C5: possible.
C8: no, the intersecting word in R4C8 has no candidate beginning with N.
C12: no, the intersecting word in R4C12 has no candidate ending with N.
SPAN must be the 4 in C5, and places without gaps.
The 7 in R12 must be MENDS, and NDS must be placed without gaps. E must be placed in R12C3, otherwise there would be a two cell gap, so the placement is M_E_NDS.
The 12 in R10 must be OCCUPATION, and OCCUP must be placed without gaps.
The 5 in R9 must be FUNS. The N cannot be placed in R9C4, as there is no candidate to intersect with it. So it must be placed FUN_S.
The 5 in C4 must be MUM, placed as M_UM_.
The 3 in R8 must be DAM, and is placed without gaps.
The 3 in C6 must be AID, and is placed without gaps.
The 12 in C2 must be REPRODUCE, and must end DUCE_.
The 12 in R11 must be REMAINDER, and must start RE_MAI.
Consider possible placements for HOW:
C7: no, the 7 in R1 has no candidate with an H.
R5: no, the 5 in C9 has no candidate with an H.
R6: no, the 7 in C12 has no candidate with a W.
R7: possible.
HOW must be the 3 in R7, and places without gaps.
Consider possible placements for DEN:
C7: possible.
R5: no, the 5 in C9 has no candidate with a D.
R6: no, the 7 in C12 has no candidate beginning with W.
DEN must be the 3 in C7, and places without gaps.
The remaining 3s are AT and US. Neither T nor S can begin the 7 in C12, so the 3 in R6 will be placed xx_.
The first letter of the C12 7 is also therefore the last letter of the R7 8. This can only be done by using YARD and LOFTY.
We have already established that the R10 12 is OCCUPATION; there is no N in YARD, so this must end N_.
The letters surrounding the blank we just placed in the C12 7 must be A_R, to intersect with REMAINDER, so YARD is placed _Y_A_RD.
The 4 in R12 must be END. There is no word that could go in C11 with two Ns in the last three letters, so the placement must be EN_D.
The end of REMAINDER must be placed ER to avoid a double gap in C11. It must be D_ER, as no word in C10 can end DN.
All candidate words for C10 ending with N, end ON. This must be placed O_N.
The 7 in C1 must be PEACH. No candidate 4 in R1 starts with P. The placement of PEACH begins _P.
We already know that C2 is REPRODUCE, so the 4 in R1 must be RAT, placed _RAT.
R2 must be PERFECTION, and with an E already in C7, it must end ECTION with no gaps.
The 4 in C8 is ICE, and the placement must start IC.
The 4 in C12 is left to be ONE; no 5 in R4 can end with E, so the placement is ONE_.
The 7 in R1 must be DISCO; the 5 in C9 cannot start with S, so the placement is _DI_SCO.
The 5 in C9 must be TAR; the 3 in R5 cannot start with R, so the placement is _TAR_
The 5 in R4 must be ERA, and C10 must be SITUATION. These cannot intersect on a letter, so ERA is placed ER_A_ and SITUATION starts SITU.
ICE must be placed as IC_E.
The 3 in R5 is US, placed _US.
The 3 in R6 is AT, placed AT_.
C11 must be COASTLINE, starting CO_AST. We already know the 8 in R7 is LOFTY, so they do not intersect on a letter; COASTLINE starts CO_AST_L and therefore is placed as CO_AST_LINE_. LOFTY ends T_Y.
The remaining I in C10 SITUATION must place in R8, otherwise the 6 in R9 ends IIA. So SITUATION is places SIT_UATI_O_N.
The 6 in C9 must be ASIDE, and as there is no A in LOFTY, must be placed _ASIDE; LOFTY ends F_T_Y.
The 6 in C8 must be FAN, and must place as _F_A_N to square with intersecting words.
This allows us to complete the placement of _R_AIL_, _AS_IA, OCCUPAT_ION_ and RE_MAI_ND_ER in rows 8 to 11.
The 6 in C5 cannot end with L, so the placement of _LOF_T_Y is confirmed.
The 8 in C7 must be SORTS, and as the 7 in R5 cannot end with S, must be placed _SOR_T_S.
The 7 in R5 cannot start with C, so the end of PEACH must be CH.
The 6 in R4 cannot start with A or E, so the placement of PEACH must be _PE_ACH
The 8s in R6 and C6 are COSTS and FINAL respectively. They cannot intersect on a letter, so end T_S_ and A_L_ respectively; the latter can be completely placed as _FINA_L_
The 6 in C5 must be MAT, and to intersect correctly with PERFECTION and AREA (the 7 in R5), must be placed _MA_T_.
AREA must be placed AR_E_A_ to intersect correctly with ALLOWANCE in C2
ALLOWANCE must then end L_OWANC_E.
COSTS must be placed as C_OST_S_
The 6 in R4 must be PLAN, and be placed as _PL_AN.
The 6 in C4 must be TRIES, and be placed as TRI_ES.
It's now trivial to place the intersecting L into ALLOWANCE/ELIMINATE, and E into PERFECTION/REPRODUCE, and mark the remaining gaps.
This gives a key of ELOSOED, MECLSRTOO.
Done!
Re: 2011 USPC Walkthrough guide.
I started to place 5 letters words in the test, because there are two groups with 5 letters and each group has same pattern: horizontalvertical intersection. So if you consider this bottom of grid would appear.nickdeller wrote:#20 Jumping Crossword
Instinct is to start with longest words, but in this instance look elsewhere for a start...
Two of relatively few 4letter entries are placed with no gaps at all, so candidate positions may be tightly constrained by intersecting words.

 Posts: 604
 Joined: Tue 29 Jun, 2010 11:41 am
Re: 2011 USPC Walkthrough guide.
#21  Hungarian Tapa
[Cols AJ, rows 110 top to bottom]
To fulfil clue E1, cells D2/E2/F1 are on the tapa. F1=1. D2/E2 sum to 11, so are 5+6.
To connect F1 to the rest of the tapa, G1 is on the tapa.
To fulfil clue D1, cells C1/C2 are on the tapa.
To fulfil clue A2, at least 3 surrounding cells must be on the tapa. B2 is on the tapa. B1 and B2 cannot both be, otherwise a 2x2 forms. A1/B1 are off the tapa. A3/B3 are on the tapa.
To have 6 tapa cells in row 1, H1/I1/J1 are on the tapa.
To fulfil clue B4, at least 5 tapa cells are needed (6+4+5+6+1). A4/A5/B5 are on the tapa.
To fulfil clue F10, E10 is on the tapa and is 5 or 6.
To fulfil clue E9, a run of at least 6 tapa cells is needed (6+5+6+4+5+1). D10/D9/D8/E8/F8 are on the tapa.
To fulfil the 26 part of clue I7, at least 5 tapa cells are needed, so this clue equates to 1/5 on a "standard" tapa. H7/I8 are on the tapa, as are I6/J6/J7
To fulfil clue C3, at least 5 tapa cells are needed. D3 is on the tapa, and so E3 is off the tapa (avoiding a 2x2)
To have 6 tapa cells in column E, 3 of E4/E5/E6/E7 are on the tapa. This cannot be E4/E5/E6, as that run would overload clue D5. E7 is on the tapa, and so D7/F7 are off the tapa.
To fulfil clue G6, we need one tapa run of at least 2 cells and another of at least 3. The 2run must be F5/F6, which are on the tapa. The 3run must include H6, which is on the tapa.
To fulfil clue I7, J8 must be off the tapa and I8=6.
To have 6 tapa cells in column D, D4 must be on the tapa.
Returning to column E, we still need 2 more tapa cells  these cannot be both E5 AND E6, so E4 is on the tapa.
To fulfil clue D5, no tapa run is longer than 2 cells. C4/E5 are off the tapa.
Returning to column E one more time, E6 is the last cell to put on the tapa.
To have three separate tapa runs around clue D6, C5/C7 are on the tapa, C6 is off the tapa.
To have 6 tapa cells in column C, we need two more of C8/C9/C10. If C9 is one of them, the other one forms a 2x2. C9 is off the tapa, C8/C10 are on it.
To have 6 tapa cells in row 5, we need two more of H5/I5/J5. If I5 is one of them, the other forms a 2x2. I5 is off the tapa, H5/J5 are on it.
The digits used to fulfil clue D5 must be 1, 1, and 1+2. C5=1, E6=1. To avoid two 1s in column E, D4=1, E4=2.
To fulfil clue D6, E7=3, C7=4.
In column E, 1,2,3 are placed; 5 and 6 will go in E2/E10; E8=4.
To agree with clue F2, E2 must be 5 rather than 6. E2=5, so D2=6, E10=6.
According to clue G6, the tapa run in F5/F6 must sum to 8. To fulfil clue G5 then, F4 must be on the tapa.
If this F4/F5/F6 tapa run were summing to 9, then F4 would be 1; it can't be, due to the 1 in D4. A run of 3 can't sum to 18, so it extends further; G4 is on the tapa.
To fulfil clue G5, H4 is off the tapa. F4/G4 must sum to 10, and thus be 4 and 6 in some order.
To fulfil clue F2, G2/G3 must be on the tapa. To avoid 2x2s, F3/H2 are off the tapa.
To have 6 tapa cells in column F, F9 is on the tapa.
To have 6 tapa cells in column G, two more of G7/G8/G9 are needed. To avoid a 2x2, G7 is on the tapa.
To fulfil clue J9, I9 is on the tapa.
To fulfil clue H8, the I8/I9 run must be in the 14sum, and H9 must be on the tapa.
To fulfil clue G10, we need a single tapa run and G9 must be on the tapa. To avoid a 2x2, G8 is off the tapa.
The tapa section including J5 is now in danger of being cut off from the rest of it. To avoid this, J4 is on the tapa.
Row 4 now has 6 tapa cells. I4 is off the tapa.
The J4 tapa section is again in danger of being cut off. J3 is on the tapa.
Column J now has 6 tapa cells. J2/J10 are off the tapa.
The J3 tapa section is AGAIN in danger of being cut off. I3 is on the tapa.
Row 3 now has 6 tapa cells. H3 is off the tapa.
Row 2 now has 5 tapa cells and one left unmarked. I2 is on the tapa.
Column I now has 6 tapa cells. I10 is off the tapa.
Row 10 now has 3 tapa cells and only three left unmarked. A10/B10/H10 are on the tapa.
To fulfil clue J9, I9=5.
To fulfil clue H8, G9/H9 sum to 3 and must be 1+2. G7/H7 sum to 6, cannot be 2+4 due to C7, so must be 1+5.
To fulfil clue F10, F9/G9 sum to 5, so F9 is 3 or 4.
To fulfil clue G10, F9/H10 sum to 5, so H10 is 1 or 2.
H9/H10 are 1 and 2, so H7 is not 1. H7=5.
G7=1
G9=2
H9=1
H10=2
F9=3
In row 9, digits 46 remain to be placed. In column D, 2345 remain to be placed. D9=4.
To fulfil clue F2, G1/G2/G3 sum to 12. Digits 1 and 2 are already placed in column G, so the G1/G2/G3 combo is 345, and G4=6.
F4=4.
To fulfil clue G5, F5/F6 sum to 8 and must be 2+6, leaving F8=5.
Also, H5/H6 sum to 9, and must be 3+6, leaving H1=4.
To fulfil clue I7, the 26sum must be 6+6+5+5+4. J7=6. I6=4. H6=6. J6=5.
F6=2
F5=6
H5=3
J4=3
A4=5
B5=5
J1=2
I1=3
G1=5
C1=6
To fulfil clue D1, C2=2
I2=1
I3=2
J5=4
J3=1
D3=5
D8=2
D10=3
C10=5
C8=3
To fulfil clue C3, B2/B3 sum to 10, so must be 4+6
B2=4
B3=6
G2=3
G3=4
A3=3
A5=2
No digit is possible in cells A6 and A7, so they are off the tapa. A8/A9 are on the tapa.
B9 is off the tapa.
B8 is off the tapa.
B6/B7 are on the tapa.
B6=3
B7=2
A8=1
A9=6
A10=4
B10=1
Done!
[Cols AJ, rows 110 top to bottom]
To fulfil clue E1, cells D2/E2/F1 are on the tapa. F1=1. D2/E2 sum to 11, so are 5+6.
To connect F1 to the rest of the tapa, G1 is on the tapa.
To fulfil clue D1, cells C1/C2 are on the tapa.
To fulfil clue A2, at least 3 surrounding cells must be on the tapa. B2 is on the tapa. B1 and B2 cannot both be, otherwise a 2x2 forms. A1/B1 are off the tapa. A3/B3 are on the tapa.
To have 6 tapa cells in row 1, H1/I1/J1 are on the tapa.
To fulfil clue B4, at least 5 tapa cells are needed (6+4+5+6+1). A4/A5/B5 are on the tapa.
To fulfil clue F10, E10 is on the tapa and is 5 or 6.
To fulfil clue E9, a run of at least 6 tapa cells is needed (6+5+6+4+5+1). D10/D9/D8/E8/F8 are on the tapa.
To fulfil the 26 part of clue I7, at least 5 tapa cells are needed, so this clue equates to 1/5 on a "standard" tapa. H7/I8 are on the tapa, as are I6/J6/J7
To fulfil clue C3, at least 5 tapa cells are needed. D3 is on the tapa, and so E3 is off the tapa (avoiding a 2x2)
To have 6 tapa cells in column E, 3 of E4/E5/E6/E7 are on the tapa. This cannot be E4/E5/E6, as that run would overload clue D5. E7 is on the tapa, and so D7/F7 are off the tapa.
To fulfil clue G6, we need one tapa run of at least 2 cells and another of at least 3. The 2run must be F5/F6, which are on the tapa. The 3run must include H6, which is on the tapa.
To fulfil clue I7, J8 must be off the tapa and I8=6.
To have 6 tapa cells in column D, D4 must be on the tapa.
Returning to column E, we still need 2 more tapa cells  these cannot be both E5 AND E6, so E4 is on the tapa.
To fulfil clue D5, no tapa run is longer than 2 cells. C4/E5 are off the tapa.
Returning to column E one more time, E6 is the last cell to put on the tapa.
To have three separate tapa runs around clue D6, C5/C7 are on the tapa, C6 is off the tapa.
To have 6 tapa cells in column C, we need two more of C8/C9/C10. If C9 is one of them, the other one forms a 2x2. C9 is off the tapa, C8/C10 are on it.
To have 6 tapa cells in row 5, we need two more of H5/I5/J5. If I5 is one of them, the other forms a 2x2. I5 is off the tapa, H5/J5 are on it.
The digits used to fulfil clue D5 must be 1, 1, and 1+2. C5=1, E6=1. To avoid two 1s in column E, D4=1, E4=2.
To fulfil clue D6, E7=3, C7=4.
In column E, 1,2,3 are placed; 5 and 6 will go in E2/E10; E8=4.
To agree with clue F2, E2 must be 5 rather than 6. E2=5, so D2=6, E10=6.
According to clue G6, the tapa run in F5/F6 must sum to 8. To fulfil clue G5 then, F4 must be on the tapa.
If this F4/F5/F6 tapa run were summing to 9, then F4 would be 1; it can't be, due to the 1 in D4. A run of 3 can't sum to 18, so it extends further; G4 is on the tapa.
To fulfil clue G5, H4 is off the tapa. F4/G4 must sum to 10, and thus be 4 and 6 in some order.
To fulfil clue F2, G2/G3 must be on the tapa. To avoid 2x2s, F3/H2 are off the tapa.
To have 6 tapa cells in column F, F9 is on the tapa.
To have 6 tapa cells in column G, two more of G7/G8/G9 are needed. To avoid a 2x2, G7 is on the tapa.
To fulfil clue J9, I9 is on the tapa.
To fulfil clue H8, the I8/I9 run must be in the 14sum, and H9 must be on the tapa.
To fulfil clue G10, we need a single tapa run and G9 must be on the tapa. To avoid a 2x2, G8 is off the tapa.
The tapa section including J5 is now in danger of being cut off from the rest of it. To avoid this, J4 is on the tapa.
Row 4 now has 6 tapa cells. I4 is off the tapa.
The J4 tapa section is again in danger of being cut off. J3 is on the tapa.
Column J now has 6 tapa cells. J2/J10 are off the tapa.
The J3 tapa section is AGAIN in danger of being cut off. I3 is on the tapa.
Row 3 now has 6 tapa cells. H3 is off the tapa.
Row 2 now has 5 tapa cells and one left unmarked. I2 is on the tapa.
Column I now has 6 tapa cells. I10 is off the tapa.
Row 10 now has 3 tapa cells and only three left unmarked. A10/B10/H10 are on the tapa.
To fulfil clue J9, I9=5.
To fulfil clue H8, G9/H9 sum to 3 and must be 1+2. G7/H7 sum to 6, cannot be 2+4 due to C7, so must be 1+5.
To fulfil clue F10, F9/G9 sum to 5, so F9 is 3 or 4.
To fulfil clue G10, F9/H10 sum to 5, so H10 is 1 or 2.
H9/H10 are 1 and 2, so H7 is not 1. H7=5.
G7=1
G9=2
H9=1
H10=2
F9=3
In row 9, digits 46 remain to be placed. In column D, 2345 remain to be placed. D9=4.
To fulfil clue F2, G1/G2/G3 sum to 12. Digits 1 and 2 are already placed in column G, so the G1/G2/G3 combo is 345, and G4=6.
F4=4.
To fulfil clue G5, F5/F6 sum to 8 and must be 2+6, leaving F8=5.
Also, H5/H6 sum to 9, and must be 3+6, leaving H1=4.
To fulfil clue I7, the 26sum must be 6+6+5+5+4. J7=6. I6=4. H6=6. J6=5.
F6=2
F5=6
H5=3
J4=3
A4=5
B5=5
J1=2
I1=3
G1=5
C1=6
To fulfil clue D1, C2=2
I2=1
I3=2
J5=4
J3=1
D3=5
D8=2
D10=3
C10=5
C8=3
To fulfil clue C3, B2/B3 sum to 10, so must be 4+6
B2=4
B3=6
G2=3
G3=4
A3=3
A5=2
No digit is possible in cells A6 and A7, so they are off the tapa. A8/A9 are on the tapa.
B9 is off the tapa.
B8 is off the tapa.
B6/B7 are on the tapa.
B6=3
B7=2
A8=1
A9=6
A10=4
B10=1
Done!

 Posts: 604
 Joined: Tue 29 Jun, 2010 11:41 am
Re: 2011 USPC Walkthrough guide.
Could you say a little more on this point? I can't see straight away whether an intersection is going to be a letter or a gap, so it's not obvious how it helps to place the four 5letter words. It could be very useful, though!yureklis wrote:I started to place 5 letters words in the test, because there are two groups with 5 letters and each group has same pattern: horizontalvertical intersection. So if you consider this bottom of grid would appear.nickdeller wrote:#20 Jumping Crossword
Instinct is to start with longest words, but in this instance look elsewhere for a start...
Two of relatively few 4letter entries are placed with no gaps at all, so candidate positions may be tightly constrained by intersecting words.
Re: 2011 USPC Walkthrough guide.
#2 Sudoku
The 9 in the center block must be in the left column (c4). Therefore the 9 in the bottom center block must be in r7c5. This also places a 9 in r9c3.
Similarly the 7 in the center block must be in the bottom row. Along the left hand side of the puzzle, the 7 in the top block must be in r2, in the middle it must be in r5, and in both cases it must be in c23. Therefore the 7 in c1 must be in the bottom block (r89). The 5 in c1 must also be in r89 so we mark these two squares as the 5 and 7.
Continuing with the bottom left square, we now see that the 1 must be in r7c2 . This creates a nice chain of 1 placements: in r3c1, then in r2c5, then in r6c4 which also places the 7 and 9 in the center block with the 9 in r4c4 and the 7 in r6c6. We note the 5/8 pair in the remaining slots in the center block. Finishing the 1's we place one in r5c7 and the last in r9c9.
The 7 in the bottom center block must be in r89c4, so the 7 in r7 must be at r7c7. The 9 in r6 must be r6c8.
The 3 in the center right block must be in c9, which forces the 3 in the top right block to be in r3c8, then a 3 in r1c6, a 3 in r8c5, a 3 in r7c3. There's no way we can have a 3 now in the bottom row of the left center block, so that forces a 3 in r6c9, and our last 3 in r4c2. We can also fill in a 2 in r2c6 now.Let's note the 4/6 pair in r8c23, and the 5/8 for r3c5 (paired with the similar entry in r5c5).
We've got most of the right center block now, and we see that the 2 must be in r5c8 with a 5/6 in the remaining slots inn r4c7/9. That resolves the 5/8 pairs from earlier: 8 in r4c6, 5 in r5c5, 8 in r3c5.
The remaining slots in the top center block must be a 4 in r1c4 and a 5 in r3c4. The bottom center block has a 7/8 pair in r89c4 and a 4/5 pair in r7/9c6.
In the center left block, we can place a 4 in r4c1, there's a 7/8 pair in r5c2/3, and a 2/6 pair in r6c1/3.
We can place a 2 in r8c7, then in r1c9, then in r3c3 (resolving the 2/6 pair in r6 to be a 2 in c1 and a 6 in c3, which resolves the 4/6 pair in r8 to a 6 in c2 and a 4 in c3).
The rest of r3 must be a 4 in c2 and a 6 in c7, which clears up the 5/6 pair in r4 to a 5 in c7 and a 6 in c9. This places a 6 in r7c8, and tells us that the rest of r7 must be a 4 in c6 and a 5 in c9. Resolving the 4/5 pair from c6 with a 5 in r9, we can then finish out all the pairs in the bottom third of the puzzle  5 above the 7 in c1, 7 above the 8 in c4, and an 8 over a 4 in the last two slots in c8.
We fill in the missing 5 and 8 respectively in columns 8 and 9 of r2, and finish the top right block with a 4 in r2c7 and a 9 in r1c7.
Finishing r2 we have a 7 and 9 in r2 c23, so it must be 9 in c2 and 7 in c3, which also resolves the 7/8 pair from r5 with a 7 in c2 and an 8 in c3. The last two squares to fill in are a 6 in r1c1 and an 8 in r1c2.
The 9 in the center block must be in the left column (c4). Therefore the 9 in the bottom center block must be in r7c5. This also places a 9 in r9c3.
Similarly the 7 in the center block must be in the bottom row. Along the left hand side of the puzzle, the 7 in the top block must be in r2, in the middle it must be in r5, and in both cases it must be in c23. Therefore the 7 in c1 must be in the bottom block (r89). The 5 in c1 must also be in r89 so we mark these two squares as the 5 and 7.
Continuing with the bottom left square, we now see that the 1 must be in r7c2 . This creates a nice chain of 1 placements: in r3c1, then in r2c5, then in r6c4 which also places the 7 and 9 in the center block with the 9 in r4c4 and the 7 in r6c6. We note the 5/8 pair in the remaining slots in the center block. Finishing the 1's we place one in r5c7 and the last in r9c9.
The 7 in the bottom center block must be in r89c4, so the 7 in r7 must be at r7c7. The 9 in r6 must be r6c8.
The 3 in the center right block must be in c9, which forces the 3 in the top right block to be in r3c8, then a 3 in r1c6, a 3 in r8c5, a 3 in r7c3. There's no way we can have a 3 now in the bottom row of the left center block, so that forces a 3 in r6c9, and our last 3 in r4c2. We can also fill in a 2 in r2c6 now.Let's note the 4/6 pair in r8c23, and the 5/8 for r3c5 (paired with the similar entry in r5c5).
We've got most of the right center block now, and we see that the 2 must be in r5c8 with a 5/6 in the remaining slots inn r4c7/9. That resolves the 5/8 pairs from earlier: 8 in r4c6, 5 in r5c5, 8 in r3c5.
The remaining slots in the top center block must be a 4 in r1c4 and a 5 in r3c4. The bottom center block has a 7/8 pair in r89c4 and a 4/5 pair in r7/9c6.
In the center left block, we can place a 4 in r4c1, there's a 7/8 pair in r5c2/3, and a 2/6 pair in r6c1/3.
We can place a 2 in r8c7, then in r1c9, then in r3c3 (resolving the 2/6 pair in r6 to be a 2 in c1 and a 6 in c3, which resolves the 4/6 pair in r8 to a 6 in c2 and a 4 in c3).
The rest of r3 must be a 4 in c2 and a 6 in c7, which clears up the 5/6 pair in r4 to a 5 in c7 and a 6 in c9. This places a 6 in r7c8, and tells us that the rest of r7 must be a 4 in c6 and a 5 in c9. Resolving the 4/5 pair from c6 with a 5 in r9, we can then finish out all the pairs in the bottom third of the puzzle  5 above the 7 in c1, 7 above the 8 in c4, and an 8 over a 4 in the last two slots in c8.
We fill in the missing 5 and 8 respectively in columns 8 and 9 of r2, and finish the top right block with a 4 in r2c7 and a 9 in r1c7.
Finishing r2 we have a 7 and 9 in r2 c23, so it must be 9 in c2 and 7 in c3, which also resolves the 7/8 pair from r5 with a 7 in c2 and an 8 in c3. The last two squares to fill in are a 6 in r1c1 and an 8 in r1c2.

 Site Admin
 Posts: 2753
 Joined: Fri 18 Jun, 2010 10:45 pm
 Location: Edinburgh, Scotland
Re: 2011 USPC Walkthrough guide.
#22  Dynasty Sudoku
With all irregular sudoku puzzles, there is an extremely useful observation to note, which reduces the possibilities for a number of cells.
Choose an area which consists of N complete rows or columns.
Clearly, within this area, each element to be placed will appear exactly N times.
Now we choose N outlined shapes that appear (wholly or partly) within that area.
Again, each element to be placed will appear N times within the confines of the N shapes.
If X cells from the N chosen shapes go beyond the boundaries of the N rows/cols area, then there will clearly be the same number (X) cells in our N rows/cols area that belong to shapes we have not chosen. Additionally, the contents of each of these two sets of cells must be identical!
Let's call this the parity rule.
For example, in this puzzle, look at cols AC. This area predominantly contains 3 shapes, with 2 cells reaching into col D (D1 & D11). Also, 2 cells intrude (C4 and C8). Due to the parity rule, C4 must be 8. We can also note that C8=D11, and by exclusions around either cell, we can deduce that it can only be 3 or blank.
C8=8. Also C8=D11, and both must be blank. (due to parity on cols AC.)
I4=7. Also H2=I8 ( for later, can only be 6 or blank). (parity on cols IK)
D3=2 and H3=2 (parity on rows 13)
(note D6=J6=6 already, note B4=H5 for later can only be 1 or blank)
D7=2 and J8=3 (parity rule on rows 811)
D9=7 and H9=4 (parity on rows 911)
Looking for busy rows/cols. Row 4. only a 1 is needed. So F4 is blank. Can't see anything else obvious here.
Lots of 2s given, so look for placement of 2s. Centre shape. 2 can only go in F6. Bottom Left shape, C9=2. Also, K11=2 and J5=2. All 2s placed.
Bottom centre shape  7 goes in E10.
G5 must be blank.
G6/G7 must be 5/7 in some order, so 6 goes in E6. E5/E7 must be blank.
Col C  4 goes in C10. Col A  4 goes in A5. 4 for col F is either F2 or F3, so 4 in col E must be in E11.
1/4/7 required for col J. So J1 is blank.
1/3/6 required for Bottomright shape, so K10 is blank.
Noting that empty squares cannot be neighbours...
K9=6; K1=5; H5=1 > H4=blank; D4=blank; B4=1; D10=1.
D8=5; F8=blank; D2=blank; J2=4; F3=4.
Noting that numbered squares are all orthogonally connected...
E3=8 > I2=8;
Looking at MiddleRight shape, we can pair the cells off (I6J6,J7K7,K5K6), with one of each pair being blank.
If J6 is blank, then K5 & K7 are blank, blocking off K6, so...
I6=blank; J6=1; I9=1; J9=blank; F9=8
Looking at TopCentre shape. The 5 can't be in col G due to 5/7 pair in the central shape. Neither is it on row 1, so the 5 is in row 2 in this shape.
Therefore, in the TL shape, C2=3, C1=blank, A2=blank.
A1=1; A3=5; C3=blank; B3=6.
G3=1; I3=blank; J3=7; K3=blank; J7=blank.
H1=6; I8=6; H8=blank; G2=6.
5 of row 9 must be in B9. So C6=5 too.
Left middle shape needs 3 blanks. These can only be B5,A6 and B7.
A7=8; B6=3; K5=8; K7=7; K6=blank; G6=7; G7=5.
B10=blank; I10=3; I11=blank
If E1 blank, D2 blocked in, so E1=3; G3=blank; E9=blank.
Bottomleft shape  B11=8.
If H2 was blank, we have disconnected a group of numbers, so H2=5; F2=blank; H10=blank; F10=5;
G11 cannot be blank, so G11=3; G9=blank; A9=3; A11=blank.
Doublecheck all numbers are connected, and 3 blanks per row/col... and we're done.
With all irregular sudoku puzzles, there is an extremely useful observation to note, which reduces the possibilities for a number of cells.
Choose an area which consists of N complete rows or columns.
Clearly, within this area, each element to be placed will appear exactly N times.
Now we choose N outlined shapes that appear (wholly or partly) within that area.
Again, each element to be placed will appear N times within the confines of the N shapes.
If X cells from the N chosen shapes go beyond the boundaries of the N rows/cols area, then there will clearly be the same number (X) cells in our N rows/cols area that belong to shapes we have not chosen. Additionally, the contents of each of these two sets of cells must be identical!
Let's call this the parity rule.
For example, in this puzzle, look at cols AC. This area predominantly contains 3 shapes, with 2 cells reaching into col D (D1 & D11). Also, 2 cells intrude (C4 and C8). Due to the parity rule, C4 must be 8. We can also note that C8=D11, and by exclusions around either cell, we can deduce that it can only be 3 or blank.
C8=8. Also C8=D11, and both must be blank. (due to parity on cols AC.)
I4=7. Also H2=I8 ( for later, can only be 6 or blank). (parity on cols IK)
D3=2 and H3=2 (parity on rows 13)
(note D6=J6=6 already, note B4=H5 for later can only be 1 or blank)
D7=2 and J8=3 (parity rule on rows 811)
D9=7 and H9=4 (parity on rows 911)
Looking for busy rows/cols. Row 4. only a 1 is needed. So F4 is blank. Can't see anything else obvious here.
Lots of 2s given, so look for placement of 2s. Centre shape. 2 can only go in F6. Bottom Left shape, C9=2. Also, K11=2 and J5=2. All 2s placed.
Bottom centre shape  7 goes in E10.
G5 must be blank.
G6/G7 must be 5/7 in some order, so 6 goes in E6. E5/E7 must be blank.
Col C  4 goes in C10. Col A  4 goes in A5. 4 for col F is either F2 or F3, so 4 in col E must be in E11.
1/4/7 required for col J. So J1 is blank.
1/3/6 required for Bottomright shape, so K10 is blank.
Noting that empty squares cannot be neighbours...
K9=6; K1=5; H5=1 > H4=blank; D4=blank; B4=1; D10=1.
D8=5; F8=blank; D2=blank; J2=4; F3=4.
Noting that numbered squares are all orthogonally connected...
E3=8 > I2=8;
Looking at MiddleRight shape, we can pair the cells off (I6J6,J7K7,K5K6), with one of each pair being blank.
If J6 is blank, then K5 & K7 are blank, blocking off K6, so...
I6=blank; J6=1; I9=1; J9=blank; F9=8
Looking at TopCentre shape. The 5 can't be in col G due to 5/7 pair in the central shape. Neither is it on row 1, so the 5 is in row 2 in this shape.
Therefore, in the TL shape, C2=3, C1=blank, A2=blank.
A1=1; A3=5; C3=blank; B3=6.
G3=1; I3=blank; J3=7; K3=blank; J7=blank.
H1=6; I8=6; H8=blank; G2=6.
5 of row 9 must be in B9. So C6=5 too.
Left middle shape needs 3 blanks. These can only be B5,A6 and B7.
A7=8; B6=3; K5=8; K7=7; K6=blank; G6=7; G7=5.
B10=blank; I10=3; I11=blank
If E1 blank, D2 blocked in, so E1=3; G3=blank; E9=blank.
Bottomleft shape  B11=8.
If H2 was blank, we have disconnected a group of numbers, so H2=5; F2=blank; H10=blank; F10=5;
G11 cannot be blank, so G11=3; G9=blank; A9=3; A11=blank.
Doublecheck all numbers are connected, and 3 blanks per row/col... and we're done.