The New Sudoku Players' Forum

[RickM]

this is from sudoku.com, which says all their puzzles are solvable without guessing or allowing branches... but I'm stuck.

{67}{679}{469}{57}{2}{1}{8}{3457}{347}
{1278}{178}{248}{57}{6}{3}{1457}{9}{47}
{3}{17}{5}{8}{4}{9}{6}{17}{2}
{268}{4}{2689}{3}{589}{7}{259}{258}{1}
{178}{13789}{89}{2}{589}{6}{34579}{34578}{34789}
{5}{3789}{289}{1}{89}{4}{2379}{6}{3789}
{9}{68}{7}{46}{3}{2}{14}{148}{5}
{68}{2}{3}{46}{1}{5}{479}{478}{4789}
(4}{5}{1}{9}{7}{8}{23}{23}{6}

I've just registered - I hope this is in a usable format.

[SteveF]

Have a close look at box 4, you should be able to see a naked quad.

This should allow you to enter a candidate in column 2.

[RickM]

Box 4 being:

{268}{4}{2689}
{178}{13789}{89}
{5}{3789}{289} ?

I can't find it..

[Dusty Chalk]

Have a close look at box 4, you should be able to see a naked quad.

This should allow you to enter a candidate in column 2.

He left out a step of indirection -- the naked quad doesn't yield the candidate, it removes candidates from box four. The changes to the two boxes in column 2 of box four yield a naked trio in column 2 which -- once you've eliminated those from elsewhere in column 2 -- yields the candidate.

[RickM]

Okay... I found the 1-3-7 triple in box 4... guess I was blind...
but I'm still stumped...

[Dusty Chalk]

Okay... I found the 1-3-7 triple in box 4... guess I was blind...
but I'm still stumped...

But that causes another 1-3-7 triple in column 2...

[RickM]

I used the 1-3-7 triple to put a 9 in c2r1...

but I never did find the quad...

[RickM]

Thanks for being patient... but please explain in detail... I eliminated the 8s and 9s from c1r5, c2r5, c2r6. That still leaves me 7s in 4 cells in r2? How does that result in a second triple?

It did leave a naked 9 in c2r1, so I can go ahead with the solution, but I am interested in the logic you guys are applying.

[RickM]

Sorry... meant 4 7s in c2 above...

[SteveF]

The initial quad I found is in cells r4c1, r4c3, r5c3 and r6c3. The only candidates for these four cells are the 4 values of 2, 6, 8 and 9.

As Dusty Chalk correctly points out, this allows you to remove 2, 6, 8, and 9 as candidates from the other cells in box 4.

My next step was to put a 9 in r1c2, it is the only place a 9 can go in column 2, which is the step I think you have found.

I think the triple that others have referred to is in r3c2, r5c2, r6c2? However once you have placed a 9 in r1c2 you don't actually need this, a number of 'only one possibility for a cell' situations lead on from this.

[RickM]

Thanks. I have to say it's amazing how long one can stare at one of these puzzles and not see something. And how once something like that quad is pointed out, how it then jumps out at you.

Was my technique valid, or did I just get lucky? I figured since R5C1, R5C2, and R6C2 were the only cells that contained 1, 3, & 7, that I could eliminate the 8s and 9s from those cells.

[SteveF]

Perfectly valid. Some time back there was a thread in this forum discussing the two alternative ways of looking at the situation.