The New Sudoku Players' Forum

[Donkarko]

Screenshot_20190802-193646__01.jpg (152.14 KiB) Viewed 968 times

[SpAce]

Hi Donkarko,

It seems that this is not a trivial puzzle that could be solved with just a single slightly advanced technique. Several of them are needed. The default solving path of Hodoku uses eight non-basic moves with six techniques (X-Wing, Swordfish, Turbot Fish, Empty Rectangle, W-Wing, XYZ-Wing). The SudokuWiki online solver (default settings) uses eleven non-basic moves with also six but slightly different techniques (Simple Coloring, XY-Chain, Swordfish, X-Cycle, Y-Wing, X-Wing). I recommend you check out those solvers and try learn from the moves they use. Both sites have descriptions of those techniques as well.

This is not going to help you, but just for fun, here's my manually found single-step solution:

Code: Select all

.-----------------------.------------------------.---------------.
| cd26     8      5     |  9       7       b46   |  1   24    3  |
|   7      36     4     |  2368    368      1    |  5   28    9  |
| ef23     1      9     |  2348    348      5    |  7   248   6  |
:-----------------------+------------------------+---------------:
|   34568  3456  i3(6)8 |  348-6   1        9    |  48  7     2  |
|   3468   346    2     |  7       348-6  ab4[6] |  9   5     1  |
|   1      9      7     | c48      5        2    |  6   3    d48 |
:-----------------------+------------------------+---------------:
|   9      2    gh68    |  5     gi4(6)     3    | f48  1     7  |
|   4568   456   h168   |  146     2        7    |  3   9    e48 |
|  g34     7     h13    |  14      9        8    |  2   6     5  |
'-----------------------'------------------------'---------------'

(6)r5c6 = (6,4)r15c6 - (6|4)r1c1,r6c4 = (24)r1c1,r6c9 - (2|4)r3c1,r8c9 = (34)r3c1,r7c7 - (3)r9c1|(46)r7c53 = (318-6)r987c3 = (6)r4c3&(6)r7c5 => -6 r4c4,r5c5; stte

In other words, the chain proves that either r5c6 is 6 or both r4c3 and r7c5 are 6s. Either way, there can be no 6 in r4c4 nor r5c5. Once you eliminate them, the rest solves with singles. However, that's obviously quite a complicated way to solve this, and I won't even try to explain it further. The sensible approach is to use multiple simpler steps.

[SCLT]

SpAce, that is an astonishing single-step solution - even more so given it doesn't require memory! Here's the shortest solution I could find consisting of simpler steps:

Turbot Fish: 6r2c2 = r1c1 - r1c6 = r5c6 => -6 r5c2
Franken Swordfish: 4 ; r689 ; c49,b7 => -4 r34c4
W-Wing: (6=4)r5c6 - r1c6 = r3c5 - (4=6)r7c5 => -6r5c5
XYZ-Wing: 4;38 @ r5c5; r5c2,r6c4 => -4 r5c6

[SpAce]

SpAce, that is an astonishing single-step solution - even more so given it doesn't require memory!

Thanks, SCLT! It's kind of fun to play with those kinds of chains. However, it's largely luck if a split chain can be easily synchronized to avoid using memory. I guess pretty much anything can be written without memory, but it's likely to get ugly and redundant if the parallel chains are very unbalanced. Here I was happily surprised that they ended up joining quite nicely without any major kludges. The key was figuring out this part:

(3)r9c1|(46)r7c53 = (318-6)r987c3

Originally I saw the left chain ending as (38)r97c3 - (3|8=6)r4c3, but that would have been uglier and harder to sync with the other. The HT saved the day.

Here's the shortest solution I could find consisting of simpler steps:

Turbot Fish: 6r2c2 = r1c1 - r1c6 = r5c6 => -6 r5c2
Franken Swordfish: 4 ; r689 ; c49,b7 => -4 r34c4
W-Wing: (6=4)r5c6 - r1c6 = r3c5 - (4=6)r7c5 => -6r5c5
XYZ-Wing: 4;38 @ r5c5; r5c2,r6c4 => -4 r5c6

Very nice strategic moves, all working towards enabling the checkmate with the XYZ-Wing! The Turbot might even be moved to the third place, to highlight that the Franken is needed to enable the W-Wing, and then the W-Wing and the Turbot are needed to open up the XYZ-Wing. (Then again, the Turbot is the easiest, so it's logical to start with it.)

Good of you to spot the finless Franken, too. I've been looking for some samples of those. Of course it's clearly a much more advanced move than the others, so it's a bit out of place. I know it's very unlikely that I would have spotted it as such, though I'd love to (that's why I'm collecting those samples). I would have probably seen its eliminations through a grouped X-Loop instead:

(4)r6c4 = r6c9 - r8c9 = r7c7 - r7c5 = (4)r89c4 - loop => -4 r34c4 (aka Mutant Swordfish: (4)r6b89\r7c49)

However, that's not necessarily a beginner move either, so it might be a good idea to find some alternate path with all moves about the same level. Do you have any idea about that? I guess one option is to replace the Franken with two moves (after your Turbot -- now it really needs to be first):

(Turbot)
* W-Wing: (4=3)r5c2 - r4c3 = r9c3 - (3=4)r9c1 => -4 r45c1,r8c8
* Swordfish (4)r689\c149 => -4 r34c4
(W-Wing)
(XYZ-Wing)

That would be five relatively simple moves. (At least in theory. I've never been good at spotting even normal Swordfishes, so I'd still probably see the grouped X-Chain first.) Then again, that seems to be Hodoku's default solution exactly, if the useless steps are removed. I'd like the Franken, if only I could spot them! (This one seems pretty easy, though, at least once seen.)