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motris
motris
07 March 2009 @ 12:20 am
3/7/09 - Here Comes The SunSun  
So here we are at Saturday; time for the hardest puzzle of the week, the puzzle to outdo the famously difficult "Saturday Themeless" NYT KenKen puzzle.

Difficulty: HardHard 8x8 KenKen



Unfortunately, my puzzle is themed unlike the NYT's puzzle. I will no longer ever write a themeless (or at least untitled) logic puzzle when I don't have to. It may not be easy for all logic puzzle constructors to do this - and I don't mean to set the bar unfairly high for my peers - but I post so few puzzles here that I might as well demonstrate the possibilities when I can.

For this Saturday puzzle, there is most definitely a theme involving the region shapes - which also critically affects solving. While the puzzle is consistent with my new rules regarding operators, it is also consistent with "KenKen" rules if you must use them, specifically regarding multi-cell regions. I'd just rather have any logical write-up posted here say "15 in 3-cells isn't division because 15 is way too big" than "division can't be in 3-cell cages" if you must choose one of the rule sets.

Fun Fact: While researching this series, but after I made my prediction of a NYT 900 number, I found out the London Times actually DOES have such a premium-rate number for its KenKen! Since this puzzle is very hard, and you might get stuck, I'll offer the free one-time-only option of just clicking here to get all the operations or here to see the answer. In the future, this level of detail will cost $1 per click.
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[info]veep on March 7th, 2009 10:18 am (UTC)
Fun!
Very nice and definitely my favorite (though I liked the interactions between the Union Jack corners a lot). I even solved it in pen... on the third copy I printed.

I enjoyed not just solving it but getting the "Aha"s where I saw what you were doing and how. The best Nikoli puzzles do this well, and I find it very satisfying.

motris[info]motris on March 7th, 2009 02:42 pm (UTC)
Re: Fun!
Thanks a lot for these comments. With this puzzle, I definitely wanted to have a fairly narrow solving path, where you might not see where to go, but could eventually deduce things if you looked hard enough. If there is one thing I'm still reluctant to do though in these constructions, its apply multiplication to get very large numbers, so it maybe still has a very kakuro-like feel because of the large number of addition-based cages.
The Dan[info]thedan on March 7th, 2009 08:29 pm (UTC)
Re: Fun!
Very satisfying. The step on which I was stalled the longest was comparing the bottom square of the lower-right 15-cell to the top three squares in the 27-cell. I was moving pretty smoothly up until then, and once I got that the rest fell out pretty quickly. Well done.
devjoe[info]devjoe on March 7th, 2009 11:07 pm (UTC)
Well you've already posted the solution, but here is my solving path (part 1):

The obvious starting place is the 6-cell 10. This number is too big to be division or subtraction, and too small for multiplication, so it must be a sum of three 1s, two 2s, and a 3. The 3 must go at the corner of the two lines of three cells, and the 2 at the other bend, with 1s in the two isolated cells, and 1,2 in the other two cells.

This means that the 3-cell 15 directly above this cannot be multiplication, since it cannot contain a 1. It must be a sum, but the numbers are pretty wide open.

The horizontal 12 in the 7th row now contains only numbers 3 and larger, so it can only be the sum of 3+4+5. This means the vertical 12 next to this has at least 6 in its upper cell and at least 4 in the lower one; this must be a sum 7+5 or 8+4.

The 8 can only be a product 1x2x4 or a sum 1+3+4 or 1+2+5. In any of these cases, since we have already placed 1s in its last two columns, the 1 goes into the its first cell. The middle of this region can only be 4 or 5, so the last cell is 2 or 3.

The 13 in the bottom row can only be the sum 6+7 or 5+8. If the vertical 12 is 7+5, then only 6+7 is possible. In that case, the first two numbers in the bottom row are 4 and 8 in some order; otherwise, they are the other pair of 6+7 and 5+8. This leaves a sum of 14 or 15 for the other four numbers in the 27, one of which (because it is in the 7th row) is 6 or more. This means the three cells of the 27 in the 6th row add to 9 or less, and 1 is already used, so they are exactly 2,3,4, and the rest of the 27 contains the 6, 4, and 8. This forces the 6+7 in the 13 and 7+5 in the vertical 12, and the bottom of the lower right 15 is an 8. The middle number of this 15 must be 5 or more, and since we cannot put a 1 in here, it must be 5, with a 2 above it.

The 80 is too large to be anything besides multiplication. There must be a 5, four factors of 2 (either 2,8; 4,4; or 2,2,4), and two or three 1s. But there is already a 1 in row 2, so there can only be two 1s, and it must contain 1,1,2,2,4,5 in some order, with one of the 1s in the lowest cell. Since this forces three 4s in the first two columns between the 80 and 27, the horizontal 12 must have its 4 at the right. This also forces three 2s between these two regions, so nothing else has a 2 in these columns.

In the second column, 1, 6, and 7 are not possible in the bottom three cells. 6 and 7 additionally cannot go into the 80, so they must go into the other cells; at least one of these must go into the 15, which must therefore be a sum. But they cannot both go, since then the last number would be a 2, and we are already full of 2s here. So the 4 must have a 6 or 7 in column 2, making it a difference 6-2 or 7-3. But that row already has a 2, so it can only be 7-3.

Since the 80 has to have a 2 in each of the first two rows, the 14, the 8, and the 2-cell 10 cannot have a 2, making the 14 be 8+6 in some order, the 8 1+4+3 in that order, and the 10 a sum. Since there is also a 4 in the column with the 10, and a 3 in the second row, the sum can only be 3+7 in that order, and this forces the 13 to be 7+6 in this order.

The 1+4+3 entry forces the 80 to have a 4 in the top row and a 5 in the second row. The 5 and 7 in the top row must go into the last two cells, and an existing 5 in the 15 in column 7 forces them into the order 7,5.

The 1 has a 6, 7, or 8 in its lower cell, and can only be a difference. The existing 5 in its column and the 7 in row 5 prevent that lower cell from being a 6 or 8, so it is a 7.

Now in the first three columns we have, in some arrangement, 1,1,2,2,4,5 in the 80; 3,7 in the 4; 2,3,4,6,4,8 in the 27; 3,5 in the 12; 7 in the 13, and 6 and two other digits in the 15. All of the 2s, 3s, and 4s are used, so the only way to complete the 15 is with a 1 and an 8; the 1 must go in the bottom cell. This forces the last 1 in the 80 to the upper left corner, and the bottom row of the 27 is in the order 8,4. The remaining unplaced digits are all 5 or larger, so the 2 must be a difference 7-5 or 8-6, but there is already a 6 in its column, so it is 7-5. The two leftmost cells of the 57 contain the remaining digits, 6 and 8.
devjoe[info]devjoe on March 7th, 2009 11:08 pm (UTC)
continued...

Now we can finally attempt to solve for the digits in the 57. Its first column contains 6 and 8. Its second contains 1, 2, 5, and 8. Its third contains 2, 3, 4, and either 6 or 8. Its last column contains two digits each 5 or larger. The minimum for this is 56, so its third column has a 6 (with the 8 going into the top row) and its last column contains 5 and 7, forcing an 8 immediately above in the leftmost cell of the 33.

The 33 now contains a 5, a 7, an 8, a cell in the second row that is either 6 or 8, and two more cells in the third row which have at a minimum 4 and 3. The minimum values for all these cells are required to reach the sum 33, so they can be placed, and finally the placements for the 15 in the second column are known. So the unknown number in the 1 is an 8, and eliminations let us place the remaining numbers.

12438675
28571436
76152843
51823764
37614528
43286157
65347281
84765312
cyrebjr[info]cyrebjr on March 8th, 2009 04:00 am (UTC)
Per click?
If you really want to charge per click, you should add an intermediate page. Not so that it takes two clicks instead of one; rather, so that solvers can't view the solution by hovering over the link.
lunchboy[info]lunchboy on March 8th, 2009 06:30 pm (UTC)
Lots of fun, and a lot to dig into. Nice easy break-in with the 10 in the lower right and then it got very tough indeed. Key moments were narrowing down the various alternative layouts of the 80 cell and spotting the intersection of min/max possibilities of the top three and bottom three squares of the 27 cell.
[info]hoskeebo on May 27th, 2009 02:33 pm (UTC)
Here comes the Sun
Ouch! That hurt! Got it (the Kenken part) after several hours. Thanks for a morning's worth of fun.
But I must admit, I don't get the theme. Will give it some more thought.
Your posts have added very much to my understanding and appreciation of the art of the puzzle. You rock!
Brian Brian (a.k.a. Professor Homunculus at http://mathmojo.com/chronicles