dailysudoku.com

[Max Beran]

Terry

But you have changed the puzzle. We were given a "7" at r5c7 in the original layout and you have changed that.

Someone-somewhere - do you have a solution as all my attempts to bifurcate have led nowhere?

[Max Beran]

I tried harder and found a solution. It can be reached by two "invalid" moves: r9c3=6 (this allows 47 cells to be completed); then r9c8=1 lets you finish the job.

Not much fun doing them this way though.

[Terry]

You're right, I copied the original puzzle incorrectly. My old eyes have trouble with all those little numbers! I'll go back and try it again and let you know how I do.

A couple of points: As I said earlier, I just started doing these puzzles, so I don't know what you all mean by "invalid" moves or "bifurcating." Sometimes I get to a point where I have to approach the puzzle like a chess game and think through the logical consequences of alternate moves. This doesn't seem any less logical to me than other approaches.

Someone in this thread suggested writing a program to solve puzzles. I'm not a programmer, so I wouldn't have any idea how to do that, and even if I could, it doesn't sound like much fun to me. I'd rather think the puzzles through myself. But, to each his own.

BTW: I'm a person of the female persuasion. My name confuses people, but Terry can be a woman's name as well as a man's name.

I'll get back to you after I've had a go at the puzzle with the correct numbers.
Terry

[Terry]

How's this? Hope I copied it correctly this time. The incorrect one I did earlier was actually harder, but this one was certainly a challenge.

2 6 7 5 4 9 1 3 8
9 8 3 6 2 1 4 5 7
5 4 1 7 8 3 2 6 9
7 5 8 3 9 2 6 4 1
3 1 4 8 5 6 7 9 2
6 2 9 1 7 4 3 8 5
1 9 5 4 6 7 8 2 3
4 3 2 9 1 8 5 7 6
8 7 6 2 3 5 9 1 4

Terry

[someone_somewhere]

Terry,

That is nice. You give us the solution, and we should find probable the initial position with the minimum numbers. It should have around 17 such numbers.
That is a challange. Could be done ;-)

see u,

[Max Beran]

On further inspection it works out with just the one "invalid" move (by which I mean an arbitrary trial and error choice from among the candidates that has no logic behind it) not two as I suggested.

So the "6" at r9c3 remains as before the "invalid" move. But you then reach a point where you have 47 cells completed and a lot of cells with just the two candidates. There is an xy-chain starting at r1c1 and winding its way round to r5c9 that allows an elimination in r5c1. There may be other chains that have the same effect. This is a much more satisfactory way of proceeding to my purist mind which gives the result that Ms Terry already posted.

This is a good website for collecting together the principal logical techniques and giving them names:

http://www.simes.clara.co.uk/programs/sudokutechniques.htm

I wonder, someone-somewhere, if it is always the case that the minimum layouts like on the website you talked about need at least one trial and error move. This suggests another question; what is the minimum number of filled cells that produce a puzzle that can be solved by logic alone? That's another one for the mathematicians.

[Max Beran]

I've answered my own question about whether puzzles with few cells filled in will tend to need trial and error at some point. The following puzzle has only 17 starter cells but can be completed with simple logic.

xxx|xxx|x12
xxx|x35|xxx
xxx|6xx|x7x
---------------
7xx|xxx|3xx
xxx|4xx|8xx
1xx|xxx|xxx
--------------
xxx|12x|xxx
x8x|xxx|x4x
x5x|xxx|6xx

Note that there are no 9's at all given here.

[Terry]

Nice puzzle, Max. If I copied the darn thing right, this is what I get:

6 7 3 8 9 4 5 1 2
9 1 2 7 3 5 4 8 6
8 4 5 6 1 2 9 7 3
7 9 8 2 6 1 3 5 4
5 2 6 4 7 3 8 9 1
1 3 4 5 8 9 2 6 7
4 6 9 1 2 8 7 3 5
2 8 7 3 5 6 1 4 9
3 5 1 9 4 7 6 2 8

Terry

[someone_somewhere]

hi,
At least 85% out of the 19197 known 17 cells Sudokus, can be solved by LOGIC.

For the 19197 "17 cells", see the page Minimum Sudoku:
http://www.csse.uwa.edu.au/~gordon/sudokumin.php

see u,

[dlg]

[dlg]

[chobans]

r5c3 and r4c9 are fixed numbers. Your alternate solution changed those numbers. So the alternate solution is wrong.

[Guest]

[naz]

Hi dlg,

you nearly did it.Terry also made once same mistake.He corrected it when pointed out.I think second solution is almost certain.Try again and i am sure you will be able to get it.

I would also love to solve it now.

cheers,

naz