2022 WPF Puzzle GP

[pinkagape]

Thanks James, that’s very helpful. I’ve not found a reliable technique with them yet - very slow going for me.

[detuned]

Round 7 preliminary results:

3. Freddie Hand 821.5 (perhaps this will change due to partial bonus?)
7 Neil Zussman 648
26 James McGowan 460

Adam Bissett, Matthew White and Tom Collyer are all just outside the top 50 (separated by a mere 7 points!), and Jeremy Kong also made the top 100.

[pinkagape]

Incredible competition for that fourth UK space! Well played gentlemen.

Did anybody else find that 90 point Araf much easier than the 90 points suggested? I didn't attempt it in the time - too many broken Arafs in my past to consider it - but after the time, I spotted the trick very quickly and then it was a very quick solve - I'd be surprised if it took me more than 5 minutes. Or is this the way, and I should be spotting tricks like this on all puzzles?!

[mwhite]

Wow, that was very close - well done everyone, especially Adam.

I find Araf very frustrating - whenever I try it during live competition it always breaks spectacularly but I can get on quite well if I just do them without the time pressure. Haven't tried that particular example yet though.

[pinkagape]

Ok, any ideas for Arrows Cipher from the next round? How do you even begin the example?

[Mblount]

So Freddie's post on the "Arrows" topic mentions the "octagon rule" for puzzles without all given clues, which I've been having a think about (particularly how "regularish" a shape must be to work).

If I understand correctly, the idea is that if you label corners as being "odd" or "even", then all four directions through an "odd" vertex must also go through an "even" vertex, and vice versa, so that an arrow pointing (or not) in a certain direction contributes equally to the odd and even totals. This seems to work for octagons of "edge length" 1 (counting diagonally adjacent cells as having edge length 1), and would probably scale up to other edge lengths.

I don't know if there are any other configurations that work. A hexagonal one (eg a square R35,C35 plus vertices either at R4,C26 or C4,R26) looked promising at first but I'm not sure how helpful it is - the totals don't
match so at best, it could possibly be used to figure out differences between vertices or sets of vertices.

[Mblount]

I've had a bit more of a look at the "hexagon" configuration and it might be of some use. Below I've labelled 6 vertices A and B depending on parity. The only directions that don't cancel (i.e. which don't go through 1 shaded and 1 unshaded vertex) are horizontal through lines 1,2,4 and 5.

OOAOO
OBOBO
OOOOO
OAOAO
OOBOO


The result is that the number of horizontal arrows in rows 1 and 4 relates those in rows 2 and 5 according to the difference between the totals of the A and B vertices:

Row 1 + 2*row 4 - 2×row2 + row 5 = A total - B total

For the cryptic arrows puzzle, this means that the totals can differ by at most 8. Perhaps more importantly, it means that if you shift the vertices to the left and to the right, then the difference between A and B should remain unchanged.