The New Sudoku Players' Forum

[r.e.s.]

The first two irregular toroidals that I posted were done entirely "by hand" -- which is a fun challenge in itself, involving finding a solution-path as part of the process. However, I just now installed Ruud's SumoCue program and used it "interactively" to make this one ...



(Here's a black & white version. )

It has exactly one solution according to SumoCue, but I haven't yet found a nice solution-path for it.

The following will paste directly into SumoCue ...

SumoCueV1
=0J0=0J1=0J1=0J8=7J2=0J2=0J3=0J3=0J3
=0J0=0J0=0J0=0J6=0J4=5J4=0J3=2J3=0J3
=0J6=0J6=0J6=0J6=0J4=0J4=9J3=0J3=0J3
=0J6=5J5=0J5=0J5=0J5=0J4=0J4=3J4=0J6
=0J6=0J7=2J7=0J7=0J5=0J5=0J4=0J4=8J6
=0J8=0J8=0J7=0J7=0J7=0J5=0J2=0J2=0J8
=0J1=0J8=3J8=0J7=6J7=0J5=0J2=0J0=0J1
=0J1=0J1=0J8=0J8=0J7=6J5=0J2=0J0=0J1
=1J0=0J1=0J1=0J8=0J2=0J2=0J2=0J0=0J0

A cute fact about toroidals: In a plane tiled with a toroidal sudoku, every 9x9 subgrid is the "same" toroidal puzzle.

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

[r.e.s.]

I still haven't found a nice solution-path for this one. Although with LoL quite a few candidates can be eliminated, I've managed to place only one digit -- 6[r3c9]. For anyone interested, the complete solution grid is <here>.

[Pyrrhon]

I used no LoL up to this stage:



Additional I know R4C6 = R5C5. Have you some LoL or another step found that can be applied here?

Pyrrhon

[r.e.s.]

Have you some LoL or another step found that can be applied here?

Yes, after several applications of LoL, and one forcing chain, the initial grid becomes ...



I won't detail the steps at this point (and I haven't sorted out which parts are not needed if we begin with your partial solution), but I applied LoL to each of the following "starting sets of units": c78, c567, c2345, c5678, r23; and 2[r3c1] is eliminated by a forcing chain on 2's. (I hadn't done your elimination of 5[r3c4].)

I would be very surprised if there weren't other ways to get the same results.

EDIT(terminology): The forcing chain has seven nodes, and should probably
not be called "simple", due to the grouped cells r6789c7 in one of the nodes:

2[r3c1]--2[r3c5]==2[r4c7]--2[r6789c7]==2[r9c5]--2[r9c9]==2[r1c1]--2[r3c1]
=> 2[r3c1] is False

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

[Pyrrhon]

There is a group cell interaction between R5 and R3C4. R3C4 is buddie of all cells with candidate 5 in R5 and itself it is not in R5, so it can't be 5.

You must not minimize your solution path. The question is with which step you have found that R1C1 <> 5 and R2C2 <> 3.

Pyrrhon

[r.e.s.]

The question is with which step you have found that R1C1 <> 5 and R2C2 <> 3.

LoL(r23):
(5[r1c789] OR 5[r5c1]) => 5[r1c1] is False

Supposing you meant to write r9c9<>3 instead of r2c2<>3 ...

LoL(c78):
(3[r12c9] OR 3[r9c5]) => 3[r9c9] is False

[Pyrrhon]

No I meant R1C3 <> 3 and forgot R9C9. Sorry. But I have it now.

Pyrrhon