The New Sudoku Players' Forum

[denis_berthier]

It's interesting that while nazaz's puzzle is not uniquely solvable by removing the three unique digits and filling a house, the three dobrichev puzzles are (and all of them reduce to the same ER with the relabel trick)
[...]My 9.3 reduces to a 7.1

If it was just a matter of "relabelling", the SER wouldn't change and the lack of bivalue pairs would be preserved. See my remarks here: http://forum.enjoysudoku.com/endor-fins-1-2-ser-8-3-8-9-t38805-13.html


Here are all the bivalue pairs for the 4 "relabelled" puzzles.

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97.85.6..56.7..9.8..8.69.7589.5.6..7.56.7.89...79.856.6123857497.5.9..86.896.7.5. ED=7.6/7.6/2.6
f-5644  (bivalue 0 419 437 434)
f-5646  (bivalue 0 231 232 412)
f-5648  (bivalue 0 232 262 322)
f-5650  (bivalue 0 437 447 374)
f-5652  (bivalue 0 382 391 473)
f-5654  (bivalue 0 382 387 283)
f-5656  (bivalue 0 382 482 182)
f-5658  (bivalue 0 482 491 474)
f-5660  (bivalue 0 391 491 191)
f-5662  (bivalue 0 491 495 294)

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85.79.6..67.8..9.5..9.5618778.5.926.9.56..378..6.785.95.7.8.496.9..6785..689.57.. ED=7.6/7.6/2.6
f-5597  (bivalue 0 113 123 411)
f-5599  (bivalue 0 113 116 211)
f-5601  (bivalue 0 419 449 394)
f-5603  (bivalue 0 343 345 243)
f-5605  (bivalue 0 345 364 453)
f-5607  (bivalue 0 149 168 461)
f-5609  (bivalue 0 149 449 149)
f-5611  (bivalue 0 449 468 464)
f-5613  (bivalue 0 168 198 381)
f-5615  (bivalue 0 168 468 168)
f-5617  (bivalue 0 484 495 484)
f-5619  (bivalue 0 491 495 294)

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76.85.9..95.7..6.8..8.96.5768.9..57...5.678.9.795.8.6.816..579.523.79.8649768...5 ED=7.6/7.6/2.6
f-5656  (bivalue 0 331 332 413)
f-5658  (bivalue 0 332 352 323)
f-5660  (bivalue 0 332 432 132)
f-5662  (bivalue 0 432 452 324)
f-5664  (bivalue 0 443 452 444)
f-5666  (bivalue 0 352 452 152)
f-5668  (bivalue 0 479 487 494)
f-5670  (bivalue 0 184 196 481)
f-5672  (bivalue 0 184 187 281)
f-5674  (bivalue 0 184 484 184)
f-5676  (bivalue 0 484 487 284)
f-5678  (bivalue 0 187 487 187)

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73.4...62.12.7634..4623...746.7.2..3.7.6.342.253.4.67.68732...4.2..6473.394..72.6 ED=7.1/7.1/2.6
f-5599  (bivalue 0 513 583 335)
f-5601  (bivalue 0 545 555 455)
f-5603  (bivalue 0 151 181 311)
f-5605  (bivalue 0 555 559 255)
f-5607  (bivalue 0 976 984 489)
f-5609  (bivalue 0 181 183 471)
f-5611  (bivalue 0 181 581 181)
f-5613  (bivalue 0 581 583 475)
f-5615  (bivalue 0 183 583 183)
f-5617  (bivalue 0 884 889 288)
f-5619  (bivalue 0 984 989 289)
f-5621  (bivalue 0 889 898 498)

[999_Springs]

Hi 999_Springs
Thanks for this interesting reference.
I've checked the full list of 101 puzzles there. Only 3 have no bivalue pairs: 6, 7 and 76

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#76: 76.85.9..95.7.46.8..8.96.5768.9..57...5.678.9.795.8.618.6..579.53..79.86.9768...5

I had a look at dobrichev's puzzle collections: https://sites.google.com/site/dobrichev/sudoku-puzzle-collections. I didn't find any with 48 givens, but the link may be of interest per se.

thanks for the link, i didn't know dobrichev had his own website with puzzle collections on it - i'll look into that. you do mean 5, 6 and 76 right? the ones i pointed out were 5 and 6

there is a summary of overall skfr ratings for all 2432930 puzzles on the next page here

i also checked some of these ones if their ed>=7.8 which i believe is a necessary condition, and dobrichev's example puzzle for 9.4 also has no bivalue cell or bilocation:

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970680530680500097005079608890056070056007809407890056560710980709068005008905760   9.4   9.4   7.9

it's also the first I can remember with SER 9.0 that can't be solved with whips only or braids only: it is NOT in T&E(1).

it may be a very long time ago now but i definitely remember there was a puzzle someone posted that is not in T&E(1) but is SE 3.6 and solves with naked triplets. i remember checking that if you disable triplets, quads and ATE in SE it jumps to 9.5 (dynamic plus). does anyone here remember or have a reference to such a puzzle?

[coloin]

It's interesting that while nazaz's puzzle is not uniquely solvable by removing the three unique digits and filling a house, the three dobrichev puzzles are (and all of them reduce to the same ER with the relabel trick):

hmmmm... if you remove the 3 unique digits ... you get a subpuzzle with 24 solutions... filling up a complete row now gives "a" solution ... but a different one from the original puzzle ..

[coloin]

edit: coloin, if you are looking for a puzzle with no bivalue or bilocation, ed=6.6 will not cut it, because of the way se works the first step has to be in at least the middle of the nishio range 7.8+ unless it is a triplet or a swordfish. that's because forcing chains under se's construction rely on the existence of them to find eliminations. so 18s are the wrong place to look. turns out you were going in the wrong direction and 48s were the answer

Thanks for that explanation ....Well it seemed the right place to look .. but even the diamond 20 puzzles had a few bivalue/bilocal clues..... It seems that low clue puzzles need to be slightly unbalanced - so that a solution is possible.

But those non-minimal 48 clue puzzles .... with no bivalue .... they are bewildering for a paper solver - im glad i didnt have to try to solve them !

[mith]

It's interesting that while nazaz's puzzle is not uniquely solvable by removing the three unique digits and filling a house, the three dobrichev puzzles are (and all of them reduce to the same ER with the relabel trick):

hmmmm... if you remove the 3 unique digits ... you get a subpuzzle with 24 solutions... filling up a complete row now gives "a" solution ... but a different one from the original puzzle ..

Yes, see Endor Fins 1/2 for a discussion of this. The solution to the new puzzle can be (uniquely) mapped back to the solution to the original puzzle.

[coloin]

I see ...... dobrichev puzzles have 24 sol with the unique clues removed ... [4!=24] so essentially all the solutions might be the the same [ and indeed they are !!! ] mapped by the 3 [plus the other 1] clues.
It doesnt work for nazaz's puzzle because it has 144 sol with the unique clues removed ... and you therefore are stuck with the 6 ED solutions when you complete a house with clues ....

[denis_berthier]

I've checked the full list of 101 puzzles there. Only 3 have no bivalue pairs: 6, 7 and 76
you do mean 5, 6 and 76 right? the ones i pointed out were 5 and 6

Yes, 5, 6 and 76. I've corrected my post.

it's also the first I can remember with SER 9.0 that can't be solved with whips only or braids only: it is NOT in T&E(1).

it may be a very long time ago now but i definitely remember there was a puzzle someone posted that is not in T&E(1) but is SE 3.6 and solves with naked triplets. i remember checking that if you disable triplets, quads and ATE in SE it jumps to 9.5 (dynamic plus). does anyone here remember or have a reference to such a puzzle?

Yes; now I remember. And the reason is simple: some rare cases of (Naked, Hidden or Suoer-HIdden) Triplets are not covered by whips or braids. It may therefore not necessarily be in T&E(1).The same is true for rare cases of Quads.
I mentioned a case in [PBCS], section 8.8.2: a Swordfish not subsumed by whips or braids:

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..1..2..3....1..4.2..4..5....6..7..8.5.....2.9..3..4....8..1..5.9..6....1..9..7.. SER 3.8

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RESOLUTION STATE:
   45678     4678      1         5678      5789      2         689       6789      3         
   35678     3678      3579      5678      1         35689     2689      4         2679     
   2         3678      379       4         3789      3689      5         16789     1679     
   34        1234      6         125       2459      7         139       1359      8         
   3478      5         347       168       489       4689      1369      2         1679     
   9         1278      27        3         258       568       4         1567      167       
   3467      23467     8         27        2347      1         2369      369       5         
   3457      9         23457     2578      6         3458      1238      138       124       
   1         2346      2345      9         23458     3458      7         368       246

It has 2 swordfishes and then singles to the end:

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swordfish-in-columns: n4{c3 c6 c9}{r9 r5 r8} ==> r9c5 ≠ 4, r9c2 ≠ 4, r8c1 ≠ 4, r5c5 ≠ 4, r5c1 ≠ 4
swordfish-in-columns: n9{c3 c6 c9}{r3 r2 r5} ==> r5c7 ≠ 9, r5c5 ≠ 9, r3c8 ≠ 9, r3c5 ≠ 9, r2c7 ≠ 9
stte

In section 8.8.2, there's also an interesting example solved by quads+whips but not by whips alone.

[yzfwsf]

[quote="denis_berthier"]
It has 2 swordfishes and then singles to the end:

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swordfish-in-columns: n4{c3 c6 c9}{r9 r5 r8} ==> r9c5 ≠ 4, r9c2 ≠ 4, r8c1 ≠ 4, r5c5 ≠ 4, r5c1 ≠ 4
swordfish-in-columns: n9{c3 c6 c9}{r3 r2 r5} ==> r5c7 ≠ 9, r5c5 ≠ 9, r3c8 ≠ 9, r3c5 ≠ 9, r2c7 ≠ 9
stte

For fun,it have one step solution as follow:

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Grouped Continuous Nice Loop: 9r4c5 = r4c78 - r5c9 = r23c9 - r1c78 = r1c5 - r23c6 = (9-4)r5c6 = r89c6 - r7c5 = r7c12 - r89c3 = r5c3 - r4c12 = (4-9)r4c5 => r4c5<>2,r4c5<>5,r5c6<>6,r5c6<>8,r5c1<>4,r5c5<>4,r8c1<>4,r9c2<>4,r9c5<>4,r2c7<>9,r3c5<>9,r3c8<>9,r5c5<>9,r5c7<>9

[denis_berthier]

It has 2 swordfishes and then singles to the end:

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swordfish-in-columns: n4{c3 c6 c9}{r9 r5 r8} ==> r9c5 ≠ 4, r9c2 ≠ 4, r8c1 ≠ 4, r5c5 ≠ 4, r5c1 ≠ 4
swordfish-in-columns: n9{c3 c6 c9}{r3 r2 r5} ==> r5c7 ≠ 9, r5c5 ≠ 9, r3c8 ≠ 9, r3c5 ≠ 9, r2c7 ≠ 9
stte

For fun,it have one step solution as follow:

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Grouped Continuous Nice Loop: 9r4c5 = r4c78 - r5c9 = r23c9 - r1c78 = r1c5 - r23c6 = (9-4)r5c6 = r89c6 - r7c5 = r7c12 - r89c3 = r5c3 - r4c12 = (4-9)r4c5 => r4c5<>2,r4c5<>5,r5c6<>6,r5c6<>8,r5c1<>4,r5c5<>4,r8c1<>4,r9c2<>4,r9c5<>4,r2c7<>9,r3c5<>9,r3c8<>9,r5c5<>9,r5c7<>9

For fun also, as mentioned in PBCS, it has a very simple g-whip[2] 2-step solution:

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g-whip[2]: r7n4{c5 c123} – c3n4{r8 .} ==> r5c5 ≠ 4
g-whip[2]: r1n9{c5 c789} – c9n9{r2 .} ==> r5c5 ≠ 9
stte

[nazaz]

What do you think of my original, much weaker (therefore much harder to disprove) conjecture that, after Singles, there is no unsolved puzzle with no bivalue and no trivalue cell?

I expect it's true even when strengthened: after Singles, there is no unsolved puzzle with no bivalue rc-cell and no trivalue rc-cell. You can find puzzles where each unsolved "rc-cell" has >=4 candidates, having as few as 6 solutions (example: 12..3...4........5..421..3...23.....4..5..31..3...124..4...6.5.2...534..6.5.4...3). Perhaps that's the limit.

[denis_berthier]

What do you think of my original, much weaker (therefore much harder to disprove) conjecture that, after Singles, there is no unsolved puzzle with no bivalue and no trivalue cell?

I expect it's true even when strengthened: after Singles, there is no unsolved puzzle with no bivalue rc-cell and no trivalue rc-cell. You can find puzzles where each unsolved "rc-cell" has >=4 candidates, having as few as 6 solutions (example: 12..3...4........5..421..3...23.....4..5..31..3...124..4...6.5.2...534..6.5.4...3). Perhaps that's the limit.

After Singles for this puzzle, the resolution state is:

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standard rc-view:
Physical rows are rows, physical columns are columns. Data are digits.
   1         2         6789      6789      3         5         6789      789       4         
   789       6789      3         4         6789      789       1         2         5         
   789       5         4         2         1         789       6789      3         6789     
   789       1         2         3         6789      4         5         789       6789     
   4         6789      6789      5         2         789       3         1         6789     
   5         3         6789      6789      6789      1         2         4         6789     
   3         4         1         789       789       6         789       5         2         
   2         789       789       789       5         3         4         6         1         
   6         789       5         1         4         2         789       789       3         

The following representations may be used e.g. to more easily spot
rn-, cn- or bn- bivalue pairs (also named bilocal pairs), mono-typed-chains,
hidden subsets and fishes (which will appear as naked subsets in the proper space).

rn-view:
Physical rows are rows, physical columns are digits. Data are columns.
   1         2         5         9         6         347       3478      3478      3478     
   7         8         3         4         9         25        1256      1256      1256     
   5         4         8         3         2         79        1679      1679      1679     
   2         3         4         6         7         59        1589      1589      1589     
   8         5         7         1         4         239       2369      2369      2369     
   6         7         2         8         1         3459      3459      3459      3459     
   3         9         1         2         8         6         457       457       457       
   9         1         6         7         5         8         234       234       234       
   4         6         9         5         3         1         278       278       278       

cn-view:
Physical rows are columns, physical columns are digits. Data are rows.
   1         8         7         5         6         9         234       234       234       
   4         1         6         7         3         25        2589      2589      2589     
   7         4         2         3         9         156       1568      1568      1568     
   9         3         4         2         5         16        1678      1678      1678     
   3         5         1         9         8         246       2467      2467      2467     
   6         9         8         4         1         7         235       235       235       
   2         6         5         8         4         13        1379      1379      1379     
   5         2         3         6         7         8         149       149       149       
   8         7         9         1         2         3456      3456      3456      3456     

bn-view:
Physical rows are blocks, physical columns are digits. Data are positions in a block.
   1         2         6         9         8         35        3457      3457      3457     
   8         7         2         4         3         15        1569      1569      1569     
   4         5         8         3         6         179       1279      1279      1279     
   2         3         8         4         7         569       1569      1569      1569     
   9         5         1         3         4         278       2678      2678      2678     
   5         7         4         8         1         369       2369      2369      2369     
   3         4         1         2         9         7         568       568       568       
   7         9         6         8         5         3         124       124       124       
   6         3         9         4         2         5         178       178       178             


There are trivalue cells in in each of the rc, rn, cn and bn spaces. There are even bivalue cells in the rn, cn and bn spaces. It's not a counter-example to my original (no bi, no tri) conjecture.

I also believe that it is impossible to find a puzzle that has none of these in any of the 4 2D-spaces after Singles. But I have no idea how to prove this and it may be very hard to find a counter-example.
I put the "after Singles" condition, because anything before the initial Singles is quite irrelevant to the resolution of a puzzle.

[nazaz]

It's not a counter-example to my original (no bi, no tri) conjecture.

It's not supposed to be. It's a demonstration of what I suspect is the minimum solution count associated with the stronger claim, call it X4(rc), that there is no uniquely solvable puzzle in which each unsolved "rc-cell" has >=4 candidates. Proving that claim would also prove your conjecture, so it's an attractive target to go after.

However, random trials suggest the easiest way to prove your conjecture might be to focus on X4(bn) instead.

(Update, 21 March.) I've confirmed the much weaker result that there's no uniquely solvable puzzle in which each unsolved constraint has >=5 candidates. The bn constraints bite hard: each digit must appear 0, 1, 3-in-a-band (or stack) or 9 times in the puzzle, making it difficult to place the clues at all, let alone pin down a unique solution. If you also demand at least 17 clues, no two digits unclued and that no band (/stack) has two empty rows (/columns) then -- someone should check -- there are only 17886 puzzles up to isomorphism and none has a unique solution.

(Update, later.) Let's say in a 999991110 puzzle with one each of clues {6,7,8} that the 33 unclued and three 6,7,8-clued cells are the "nucleus". The nucleus falsifies the bivalue conjecture if no unclued cell sees more than one of the 6,7,8-clued cells AND there is only one way to fill it out with the remaining eight 6's, eight 7's and eight 8's. There are essentially 2182 ways of doing this, if I counted correctly.