The New Sudoku Players' Forum

[yzfwsf]

Andrew Stuart's solver [sudokuwiki.org] detects the symmetry of givens. He refers to this technique as 'Gurth's Theorem'.
May I suggest that YZF_Sudoku should have this technique added?

Andrew's solver detects this one, but if the puzzle is morphed a bit more (e.g; by circular swap of bands 1,2,3 to 2,3,1), it detects no longer the "symmetry", actually the automorphism, (nor would I have detected it manually). It doesn't detect the "stick symmetry" either (whether morphed or not).
What would be useful in a solver, would be detecting automorphic puzzles, for any automorphism.

As Cenoman said, to support automorphic puzzles, it is necessary to enumerate various isomorphisms of the puzzle, which will take a long time, and I am not sure if it is enumerating the various possible combinations of rows and columns,6^8=1679616 types in total.

[denis_berthier]

Andrew Stuart's solver [sudokuwiki.org] detects the symmetry of givens. He refers to this technique as 'Gurth's Theorem'.
May I suggest that YZF_Sudoku should have this technique added?

Andrew's solver detects this one, but if the puzzle is morphed a bit more (e.g; by circular swap of bands 1,2,3 to 2,3,1), it detects no longer the "symmetry", actually the automorphism, (nor would I have detected it manually). It doesn't detect the "stick symmetry" either (whether morphed or not).
What would be useful in a solver, would be detecting automorphic puzzles, for any automorphism.

As Cenoman said, to support automorphic puzzles, it is necessary to enumerate various isomorphisms of the puzzle, which will take a long time, and I am not sure if it is enumerating the various possible combinations of rows and columns,6^8=1679616 types in total.

And I'm not sure it would be a meaningful technique for a manual solver. If the puzzle's hidden symmetry is too far from being visible, no one will be able to notice it (unless warned in advance, which makes it useless: in this case, why not propose directly the morph of the puzzle that makes the symmetry visible?).
"Too far from being visible" leaves some room for "reasonable" automorphisms.

I don't remember exactly when Mauricio published the original form of the W31 puzzle of this thread, but I reported it in [PBCS1, 2012]. So, it took at least 8 years, some happy random morph I did (with no idea of symmetry in mind) and Cenoman's bright insight to notice the symmetry!

[yzfwsf]

Code: Select all

Whip[31]: Supposing 8r1c3 will result in 7 to disappear in Column 1 => r1c3<>8
    8r1c3 - 8c7(r1=r2) - 8c1(r2=r4) - 8c6(r4=r3) - 4b2(p9=p1) - 7c4(r1=r2) - 5b2(p4=p2) - 5b3(p2=p6) - r4c9(5=9) - r4c5(9=2) - r4c6(2=4) - 9b5(p3=p6) - 3c6(r5=r8) - 2b8(p6=p9) - 6c6(r9=r2) -
    r3c5(6=9) - 9c8(r3=r1) - 9c7(r1=r9) - r4c7(9=5) - 5r5(c8=c1) - 5r8(c1=c8) - 7c8(r8=r3) - 7c9(r3=r9) - 7c3(r9=r8) - 2c3(r8=r2) - r2c1(2=1) - 1r8(c1=c5) - 6b8(p5=p2) - 6b9(p3=p4) - 6b3(p1=p9) -
    6c1(r3=r1) - 7c1(r1=.)
Whip[14]: Supposing 4r4c6 will result in 9 to disappear in Box 1 => r4c6<>4
    4r4c6 - 4c7(r4=r5) - 4c4(r5=r1) - 4c3(r1=r6) - 8c3(r6=r2) - 2r2(c3=c1) - 1b1(p4=p5) - r6c2(1=6) - 6r5(c3=c8) - 9r5(c8=c6) - r2c6(9=6) - r3c6(6=8) - r3c5(8=9) - 9c8(r3=r1) - 9b1(p2=.)
Whip[13]: Supposing 8r6c4 will result in 2 to disappear in Row 6 => r6c4<>8
    8r6c4 - 8c3(r6=r2) - 2r2(c3=c1) - 8c1(r2=r4) - 8c6(r4=r3) - 4c6(r3=r5) - 3c6(r5=r8) - 3c4(r7=r5) - 1b5(p4=p8) - 8c5(r6=r7) - 8r1(c5=c7) - 3c7(r1=r7) - 2r7(c7=c8) - 2r6(c8=.)
Whip[11]: Supposing 1r7c4 will result in 1 to disappear in Box 9 => r7c4<>1
    1r7c4 - 8r7(c4=c5) - 1c5(r7=r6) - 8b5(p8=p3) - 2b5(p3=p2) - 9b5(p2=p6) - r2c6(9=6) - r3c5(6=9) - 9c8(r3=r1) - 9r2(c9=c2) - 1c2(r2=r9) - 1b9(p9=.)
Whip[12]: Supposing 4r1c4 will result in 8 to disappear in Box 4 => r1c4<>4
    4r1c4 - 4c6(r3=r5) - 4c3(r5=r6) - 8c3(r6=r2) - 2r2(c3=c1) - 1b1(p4=p5) - r6c2(1=6) - r6c9(6=3) - r6c4(3=1) - r9c4(1=5) - r9c2(5=7) - 7r4(c2=c1) - 8b4(p1=.)
Hidden Single: 4 in b2 => r3c6=4
Whip[6]: Supposing 1r6c4 will result in 8 to disappear in Box 4 => r6c4<>1
    1r6c4 - 4c4(r6=r5) - 4c7(r5=r4) - 2b6(p1=p8) - r6c5(2=8) - 8r3(c5=c1) - 8b4(p1=.)
Whip[6]: Supposing 9r2c9 will result in 9 to disappear in Column 6 => r2c9<>9
    9r2c9 - 9c8(r3=r5) - r5c6(9=3) - 3c4(r6=r7) - 8r7(c4=c5) - 8b5(p8=p3) - 9c6(r4=.)
Whip[6]: Supposing 9r2c7 will result in 9 to disappear in Column 6 => r2c7<>9
    9r2c7 - 9c8(r3=r5) - r5c6(9=3) - 3c4(r6=r7) - 8r7(c4=c5) - 8b5(p8=p3) - 9c6(r4=.)
Whip[7]: Supposing 1r6c5 will result in all candidates in cell r5c4 being impossible => r6c5<>1
    1r6c5 - 1c4(r5=r9) - 1c9(r9=r7) - 1c2(r7=r2) - 9r2(c2=c6) - r5c6(9=3) - r6c4(3=4) - r5c4(4=.)
stte