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Posted: Fri 12 Jul, 2019 8:42 am
This section of the forum isn't too active, but hopefully I'll get some replies! So, some questions about Arrows puzzles:
1. How do you solve them?
2. How do you solve them quickly?
3. Any useful notation?
I've attached an image of my solved copies of the two Arrows from this year's GP. The first I solved very painstakingly using an exhaustive notation that seems to be the only way I can visualise the logical steps. The second I made a couple of deductions at the start and then just guessed at a binary choice and the solution came out.
Obviously the second was much quicker, but the first felt much more reliable - I'm not sure I'd actually notice if I made a wrong guess because it's so easy to end up with too many arrows pointing at something without realising.
How does everyone else solve these?
Posted: Fri 12 Jul, 2019 9:18 pm
I am also terrible at arrows puzzles, but I do have a couple tricks up my sleeve which can help me solve them.
Firstly, the 'octagon rule' could come in handy for puzzles without all numbers given. This approximately says that for any regular-ish octagon (I am not sure of the exact statement because I worked this out from a Thomas Snyder blog), the sum of the numbers at vertices 1,3,5,7 must be the same as at 2,4,6,8. I don't think you can actually do anything with octagons though in puzzles with all given clues.
For 5 by 5 puzzles, it can be helpful to look at 3 by 3 square centred at the centre of the puzzle, since only 8 arrows can reach these clues. Looking at the 4-1-3-5 square, we observe that the sum is 13. Due to the 1 clue, at least 3 arrows cannot be directly horizontal or vertical (meaning they can point to at most 1 of the four clues). Since all arrows horizontal/vertical leads to a sum of 16, this means we can actually determine the direction of all 8 arrows. This means the 3 in R3C5 now becomes a 1, and this allows us to place vertical arrows in C5 to satisfy the 3 in R5C5. Now, we can make a similar deduction with the 5 in R2C3, and the puzzle quickly resolves after that.
As for the 6 by 6 puzzle, I am not sure of the best way to solve it (since I haven't found any standard configurations) The way I did it logically was first to consider the two 1's in the grid. There are two arrows (one at the top right, one bottom left) that have to point to at least one of the 1's, so this forces the two arrows in the bottom right to point diagonally to the left. This causes the 2's in R2C3 and R3C4 to become 1's, which might motivate thinking about the 5 in R4C1 next; indeed, if the third arrow down on the left column and the fourth arrow across in the bottom row both point to the 5, that makes the 6 in R5C4 unable to be completed. Therefore, both arrows in column 1 must point towards the 5, which makes the 3 in R5C3 resolvable (since R3C1 and R3C3 are both 0 now, and the leftmost arrow of the bottom row is pointing away from the 3), etc. The pseudo-V-shape formed by the 5 in R4C1, the 2 in R3C6 and the 6 in R5C4 may be a configuration to bear in mind.
I do not know of a better notation than the one you are using. It does seem pretty tedious though, so suggestions from others would be greatly appreciated!
Posted: Sat 13 Jul, 2019 5:37 am
More generally, I have found that with a lot of arrows puzzles, there is a fairly narrow solving path at the start, with each deduction leading on to the next. Typically, after finishing the opening of the puzzle, you will only need simple arrows logic to finish off.
To see the initial logic, it can be helpful to look for the following things:
-Notable configurations: eg the central 3 by 3 square in the 5 by 5 puzzle. Seeing these often involves a bit of creativity. After the initial break-in, the logic tends to become looking for the interaction between small and large numbers (obviously large numbers with 1’s or 0’s along the same row, column or diagonal, but also the interaction of multiple large numbers with small numbers, as we saw in the 6 by 6). It is also important, when you have placed a few arrows, to consider which cells those arrows can no longer point to. This is in fact an advantage of Sam’s notation.
Actually, knowing the very restricted kind of logic could mean that you could get away with no notation at all (or perhaps re-writing all clues somewhere in the middle of the solve). However, it would be important to know when cells have been reduced to 1 or 0: this could be done by circling and crossing out clues respectively, for example.
Posted: Mon 15 Jul, 2019 3:06 am
I have a fast notation.
I use a small ruler, and quickly draw a long line wherever I know an arrow points. Each 'row' can have 0, 1, or 2 lines along it. Initially it's not important to know which direction the line comes from.
e.g. let's say there is a 6 and a 1 that can see each other (same 'row'). Then I know the other three directions through the 6 must all have at least one line, so I draw those 3 lines in.
If I know which end the arrow comes from then I draw the line into that box (I don't bother putting arrowheads on the arrows!). If not, I put a mark outside the grid to say that later on this mark will need allocating. If a row has 2 lines then obviously both ends can be filled.
I then circle digits that have been finished, so no new lines can go through a circle. As the number of circles increase it becomes easier.
Look for larger numbers on the diagonals, where fewer arrows can get to.
Tips for #14:
1) Bottom-of-left and right-of-top must both point at the two cells that contain "1"s. So there is a diagonal line between them. And no other arrow can point at the 1s, so the two bottom-right arrows must point up-left.
or 2) If there was a horizontal line along the bottom, then the bottom-leftmost-6 could only have 5 lines maximum (2 horizontal, 1 in each other direction), so that's not possible.
After drawing in those, and the 6-1 rule mentioned above, I have made the red lines below. Then I see the main-diagonal-4 (R4C4) only has 4 possible lines left now, so a whole bunch of 2s also get circled and the end is nigh.
Tips for #15:
1) 5 on diagonal gets 3 lines.
2) a 5 that can see two 1s gets two lines.
3) a horizontal line along the top would stuff up the rightmost top arrows.
4) a vertical line in column 4 would break the top-right-5.
5) but most importantly, the 4 on the diagonal in R2C2 can only have 4 lines, so draw them all in.
Happy to answer further questions.
Posted: Mon 15 Jul, 2019 3:23 am
Originally I was only going to post about notation, but then decided to start solving the puzzles as an example.
I hadn't read Freddie's interesting posts at that point, so have repeated some of his observations in my own words.
I haven't thought about that 4-1-3-5 square logic before, a great start for 5x5 puzzles when it works.
Posted: Mon 15 Jul, 2019 10:23 am
Awesome notation James, thanks for sharing! Really helps to quickly visualise the logical steps.
Just for fun, I made a little arrows puzzle. Nothing too serious, but it does illustrate a configuration that could come up in some puzzles.
. 6 . . .
. . 2 2 .
4 . . . .
. 2 . . 4
. . 6 . .
Posted: Mon 15 Jul, 2019 3:21 pm
Thanks Freddie and James,
In a sense, I suppose that means the dismaying answer (on the whole) is that I'm not missing any amazing tricks - the techniques are roughly the ones I'm already using, I'm just not very good at them. Except that 5x5 trick which is nice to have in the back pocket!
James, that notation is far less cumbersome than the one I'm using. I expect it will take a bit of getting used to, but I'm grateful to you for it!
Freddie, that puzzle you created is very neat. I'm impressed that something with so few clues, all in the range 2-6, is possible, actually!