dailysudoku.com

[David Bryant]

Here's one for people who enjoy a challenge. I found this one on the Sad Man Software site. http://www.simes.clara.co.uk/programs/sudokutechniques.htm

[Katie]

Cracked it - and sent you the answer on a post message. Didn't want to post it here as that would spoil the fun for others.
Katie

[someone_somewhere]

Ye, I was just wondering what kind of swordfish it was in that 3-dt example of the site you mentioned. Could not find it, so I was thnking that there is a mistake in that position. Never crossed my mind to look for a "stinky fish".
BTY, I got to the following point for the condidate table:

267 26 8 37 9 23 1 4 5
5 3 1 6 4 28 2789 789 289
4 9 27 1578 2578 128 28 36 36
12679 12468 46 3789 2378 5 2789 13789 2389
279 5 3 789 1 289 6 789 4
1279 128 27 4 2378 6 5 13789 2389
1236 1246 456 13589 358 1389 89 689 7
136 16 56 13589 358 7 4 2 689
8 7 9 2 6 4 3 5 1

Here there are a lot of numbers where the fish is hidding. And when it involves more than one numbers, no wonder that I am color blind and could not spot it.
At least a better hint is wellcomed for a bind and old man, like me.
Thank you in advance,

[David Bryant]

OK, I will give you a better hint, since you were so kind as to give me some 100 year old single malt whisky. First, though, I should point out a very small error in your table: I think that "2389" in r4c9 ought to say "23689".

The numbers you should look for are "2" (in rows) and "9" (in columns). You can already see one of the pairs of "9"s, in column 6. Unfortunately, you still have too many "2"s on the board to spot the pairs in rows. So you need to get rid of some "2"s.

Try looking carefully at the middle left 3x3 box (the intersection of rows 4, 5, & 6 with columns 1, 2, &3). I think you will find that there are only two cells in this box where a "2" can actually be placed without forcing a contradiction somewhere. Or, to look at it another way, there are two cells in this box such that if one of them does not contain a "2", the other one must contain a "2". This pattern is, I think, a sort of "swordfish" that connects two cells in the same 3x3 box ... a very nice embellishment of the concept.

After you find that, you may be able to isolate a "hidden pair" in the same 3x3 box.

I hope you're having fun! dcb

[chobans]

Actually r4c9 is "2389". There is a 6 at r5c7 so 6 can't be one of the option at r4c9.

I... actually took different route of "swordfish". I was able to find THIRTEEN numbers first without problem. It's the next two that caused the problem. I tried to find "traditional" swordfish as described on that site but couldn't really locate one. Then I noticed one thing.

The number "2" in row 1, 5 and 7. In row 5 and 7, 2 appears only twice. But in row 1, it appears three times. But the columns in row 1 that number 2 can appear corresponds to the three columns for row 5 and 7. (column 1, 2 and 6). So I did some more research and found out that this is also a form of "swordfish".

Then after removing the number 2 from other cells in column 1, 2 and 6 (6 cells removed), I was able to get 8 at r2c6 and 1 at r3c6.

[David Bryant]

[chobans]

[geoff h]

I ended up with the same 34 numbers as "chobans"and "someone somewhere" before I got stuck. I eventually solved the puzzle, but only through a bit of trial and error involving the Nr 2s. I could easily see that placing the 2s was the key to the puzzle. However, as I said, it involved a little trial and error, which I don't like using.

However, David, it seems that using your swordfish solution, there is still a bit of trial and error involved. Is this correct? I only ever like placing a number in a cell when I know that logic dictates that that is where it must go. However, your swordfish method seems to suggest some trial and error is needed.

Is this correct or am I missing something?

Cheers.

[David Bryant]

There is still a piece missing, Geoff.

I really ought to count more carefully. In my original post I said there were twelve easy moves, then three very hard ones. As Chobans pointed out, that should have been 13 fairly simple moves, followed by three tough ones.

Chobans showed us an elegant way to make two of those moves. He didn't explain the third one. The missing piece is in the middle left box, specifically in row 4, column 1. We assert that that cell cannot validly contain the number "6", and proceed as follows. Here's the table as it stands at this point:

[geoff h]

Thanks for your reply David. As I said, I also found the 13 numbers fairly quickly and eventually solved the puzzle through a bit of trial and error in placing the 2s.

I suppose I'm being a bit pedantic, but I only like to put a number in a cell when that number is the only logical possibility and there is no guess work ( or assumptions ) at all. I understand your train of thought involving the chain of inference and thanks for sending the information.

I suppose I still don't like to make a start to the chain by guessing ( or assuming ) a certain number. However, as I said, maybe I'm being a bit too pedantic.

Thanks for your help.

Cheers.

[chobans]

[David Bryant]

[jiger]

David Bryant wrote:

>The only way to solve a Sudoku puzzle is by noticing the possible contradictions..

In r 1/c 4 there should obviously be a 3 or a 7 (after having introduced the first 13 digits)
If you try with a 7, you will end up with no rome for an 8 in the upper right 3x3 block. Thus it must be a 3.
The rest is more or less a piece of cake. Thus you only need to arrive at one difficult move after the first 13.

Jiger - beginner