The New Sudoku Players' Forum

[r.e.s.]

I haven't seen any like this, so I decided to make a "greater-than" jigsaw sudoku ...



(Here's a large-format version.)

The greater-than ('>') signs must be obeyed, as well as the usual rules for sudoku with irregular boxes. There is exactly one solution.

[2007-04-05: Updated link addresses.]

[Smythe Dakota]

Cool. Your puzzle even has diagonal symmetry.

Next task: Create one where the greater-than signs are ALSO preserved (or perhaps reversed) under diagonal symmetry.

Bill Smythe

[Smythe Dakota]

Yikes -- your puzzle is a killer. After 3 hours I've been able to place only nine 1's, seven 2's, and one 9. Maybe I'm overlooking something.

Bill Smythe

[Smythe Dakota]

.... Next task: Create one where the greater-than signs are ALSO preserved (or perhaps reversed) under diagonal symmetry. ....

I just realized that this is likely to result in either dual (mirror-image) solutions, no solutions, or (in the "reversed" case) anti-symmetry in the solution. By anti-symmetry, I mean that if there is an X in one cell, then there would be a 10-X in the diametrically opposite cell. In particular, there would be a 5 in the middle cell (r5c5).

Bill Smythe

[Ruud]

Yikes -- your puzzle is a killer. After 3 hours I've been able to place only nine 1's, seven 2's, and one 9. Maybe I'm overlooking something.

Bill Smythe

There is a second 9 that can be placed with no special tricks.

The next step is a little Nishio on the 3's. When you start at the right-top, you will soon be able to place all 3's. And the last two 2's. From there, it gets easier and easier, 4 - 5 - 6 - 7.

Really nice puzzle.

Ruud.

[motris]

A great puzzle r.e.s - I did it in about 40 minutes which seems on par for me for "regular" greater than/less than variants which are typically amongst the hardest variants. Somehow the steps I took for placements here were different than in just the regular ones, but it was a fun puzzle to tackle with a slightly tweaked solving style.

The route I took for those who may want to get over their sticking points is summarized here: Using the standard starting point of marking all the maxima/minima cells, you can place all of the 1's in this puzzle and I placed one quite obvious 2 before deciding to do something a little different. I thought the irregular geometry might make the puzzle a lot more prone to candidate placement tests (so basically Nishio-like reduction or, in some cases, "coloring"). As it seemed the high end was harder to place, this is where I then decided to methodically start, tackling the puzzle from the 9's down. For example, in the center piece, 9's could initially go in R4C6, R5C5, and R7C4. However, if you look at R5C5, you'll see that a 9 here makes it impossible to put a 9 in the piece just below it and to the right. If you then check R7C4 you'll see that this leads to a contradiction if you propagate 9's about 3 pieces. So there is a 9 in R4C6. Continue like this to solve all the 9's. Then, put in all the cells that can hold the 8's. Coloring and/or Nishios again seemed to be a functional route to place the 8's. The 7's had a little twist in that the two paths shared a forced spot telling you a place there had to be a 7, which then worked backwards to settle the 7's. Then I got a little sloppy and made some errors placing the 6's twice which slowed me for awhile but eventually when I was very careful they fell the same way and from there I just started putting 5's and 2's and 3's and 4's where the pieces forced them to be and finished.

Thomas Snyder

[r.e.s.]

I'm happy that the puzzle is being enjoyed! For anyone getting stuck, here are a few general solving-tips ...

(1) Since this kind of puzzle requires an ordinary mortal to pencil-mark all the candidates, do this right away: For each cell, count how many of its neighbors it's greater-than (call this count G) and how many it's less-than (call this count L) -- then the candidates for that cell can be reduced to the range [G+1,...,9-L].

(2) Apply the inequalities to further reduce the candidate-ranges in each unit (row/column/box): If two neighboring cells have ranges like [min1,...,max1] > [min2,...,max2], then the inequality requires min1 > min2 and max1 > max2 -- so look for violations to correct, which will eliminate the offending candidates (thus shrinking a range each time). Certain easily-recognised "inequality flow" patterns will become apparent.

(3) In each unit, look at the mins that equal 1 (that is, currently the least candidate in the unit) ... If there is only one of those, that cell is solved as a 1, eliminating any other 1's in the unit -- each elimination possibly producing new inequality violations that would eliminate still more candidates. Likewise for the maxs that equal 9 (that is, currently the greatest candidate in the unit) ... If there is only one of those, that cell is solved as a 9. And so on. Every time a candidate is eliminated, be sure to check its neighbors for inequality violations whose correction will eliminate additional candidates.

(4) Use the least- and greatest-candidates in a unit to make eliminations in other units (e.g. line-line and line-box intersections). Eventually, of course, candidates other than mins/maxs may be eliminated. Apply the Law of Leftovers and other sudoku strategies as usual.

[Smythe Dakota]

.... The next step is a little Nishio on the 3's. ....

Yes, the 3's were the sticking point. Every row, column, and square (what do you call those things in a jigsaw version?) had at least two cells with candidate 3's. Finally, I assumed one of them, which precluded a couple others, which forced another, etc. After about 12 steps I reached a contradiction, so was finally able to eliminate the original possibility. From there the 3's fell into line. As it turned out, the 4's, etc, were easier.

Whew!

Bill Smythe

[Smythe Dakota]

R.E.S.: For your next puzzle, how about complex integers? Every cell should contain a+bi, where a=1,2,3 and b=1,2,3. Each wall between cells would contain two comparative operators, one for the real part and one for the imaginary part -- and each could take on three values, greater than, less than, or equal to. (Of course you could never have equal-to in both the real and imaginary parts simultaneously.)

If this comparison operator makes the puzzle too easy or explicit (or too hard to invent because too many possibilities are eliminated automatically), you could get rid of equal-to in favor of just greater-or-equal and less-or-equal.

Bill Smythe

[r.e.s.]

.... The next step is a little Nishio on the 3's. ....

Yes, the 3's were the sticking point. Every row, column, and square (what do you call those things in a jigsaw version?) had at least two cells with candidate 3's. Finally, I assumed one of them, which precluded a couple others, which forced another, etc. After about 12 steps I reached a contradiction, so was finally able to eliminate the original possibility. From there the 3's fell into line. As it turned out, the 4's, etc, were easier.

Whew!

There are clearly very many solution-paths for this puzzle, and the one I took, though different from the one motris used, seems quite comparable in using only very basic moves. Of course I could have made a mistake, but anyway the first part of the solution path I used, lettered in solving-order A-Za-k, is <here> (this solves all the 1's first, then seven 2's, all the 3's, five 9's, ...).

For your next puzzle, how about complex integers?

I think I'll pass on that -- but you go right ahead!