Tough Sudoku for 18/March/2006

First posting! Posting First?
Happy St. Patrick's Day

I'm second, that's new!

Not so tough today. Goodbye!

Looks easy today and fills easily till 46...
Now what?

I'll think about that tomorrow...off to bed...

9:10

Cute kids! They look very happy! Have a great St. Patrick's Day!

UP to 26; bc5=1 elim dfgh5=1; UP to 38;
a46=2 elim a379=2;UP to 45; def8=4 elim df7e9=4; gi1=5 elim e1=5; e89=5 elim d78f78=5; ai7=45 elim i7=2; UP i9=2 to 46; bc1=2 elim e1=2; e1=4==e1=1--f3=1==f3=2--f8=2==f8=4 elim f2e8=4; UP to 51; de8=2 elim f7=2; UP to 54; d4=8==d4=4--d2=4==d2=7--d3=7==g3=7--g3=8 ==g8=8 elim d8=8; UP to end. Depth 4.

Too many mistakes.

Another way to solve:
1) Start 23. Unique Possibilities to 26.
2) Locked 1's forbid 1 at DFGH5. UP to 38.
3)Locked 2's forbid 2 at A379. UP to 45.
4) Locked 2's forbid 2 at E1. Locked 5's forbid 5 at E1. Locked 8's forbid 8 at H2.
Danger - ALS solution.
The Locked Set e1=14, h1=143, h2=43 forbids 4 at i1.
5) UP 81
Depth 3, sets 9.

Sorry I cannot count - depth 3, sets 8.

10:16 New personal best!

Steve, can you please explain step 4:
'The Locked Set e1=14, h1=143, h2=43 forbids 4 at i1'?

16:01

For those needing an explanation of my step 4 above:
e1,h1,h2 = (134). This is a 'bent' triple. Any cell which sees all three of these cannot contain 134. Thus, 134 cannot exist at g1,i1. Of these possible eliminations, one is available, i1=4. No more complex then a triple, same depth, not easy to fc.

thx Steve, it seems so logic now :)

Forthe purposes of the elimination, the fact that e1 does not contain 3 and h2 does not contain 1 are immaterial. If they did, the elimination would still exist. This is exactly the same logic one uses with a naked triple.

Sorry, I am mistaken in my last comment above. It is very important that 3's are weak across the sets:
(s1) = 34 at h2. s2 = 134 at h1, 14 at e1.
3's are weakly linked across s1,s2 forbids 4 at i1, h1. (As one of the two sets must be locked)

the same bent triple also eliminated 3 in h1, right?

in g1 I meant

As an fc like logic chain.
S1= (h2=34) S2=(e1=14, h1=134).
3 in S1 -- 3 in S2 implies either S1 is locked, or S2 is locked implies 4 in S1 or 4 in S2 forbids i1=4.
The only strong links considered are the cells themselves.

As an FC:
h2=4 == h2=3 -- h1=3 == (h1=14, e1=14) forbids i1=4.

Steve - what is a 'bent' triple?

As an FC above, it appears to be depth 4. Nevertheless, One can expsose the depth as follows:
h2=4 == h2=3 -- h1=3 == (h1=14) -- i1=4 or e1=1== e1=4 forbids i1=4

Steve just explained it, while you were refreshing :)

thanks Nic, but the comments are way over my head. its like a foreign language to me

Rob - it is a phrase I coined. It is seen in many cases of depth 3 fc's, and fc-like configurations. A Y wing is sort of a bent triple. This elimination is 'Y-wing' style. It is most easily recognized though as an Almost Locked Set elimination, for which GB has written a nice theorem - I wrote one also, but his is far superior. It is based on the simple idea of counting. Here, one has a choice of calling S1= h2, S2= e1,h1 and noting weak 3's across the sets - OR S1= h1,h2 and S2= e1 and no

Steve: congratulations on finding a neat step 4 which makes your solution much superior to mine. Bruno and I have found a way to write such a step in the matrix form:
e1=4 e1=1
h2=4 h2=3
h1=4 h1=1 h1=3
In this matrix every row is a strong set, and in every column except the first one every two entries are in conflict with each other. The conclusion is that the first column is also a strong set. This works for matrices of any size and covers lots of cases where the original FCs

cut-off : noting weak 1's across the sets. In either case, One set or the other is locked. Thus, one or the other must contain 4's. OK?

it is a foreign language to me too, but with some help from people like Steve , gb, DJ, Andre (thanks guys) we can get there

The last sentence was cut off: This works for matrices of any size and covers lots of cases where the original FCs don't exist. BTW, the FCs can also be represented in this form. Also the matrix should be
e1=4 e1=1 0
h2=4 0 h2=3
h1=4 h1=1 h1=3

Thanks Andrei! The matrix is a much neater way of detailing the native weak links which produce the strong 4's.

BTW - yesterday's puzzle could benefit from such matrices - although one might need to link the matrices.... How is that best done?

Much to easy. Even an old 'piker' like me solved this one easily

12:17 Great kids in the picture. Happy weekend, everyone!

Ah yes now I might understand! The idea is to make the matrix large enough to reveal wink links - I suppose I am confused on how to expose non-native weak links in the matrix, if they be needed. This topic warrants some study....

Bravo, Steve.

Andrei, Let me make sure I know how you form your matrices. You have n strong sets (cells in this case) with n colors. Each row is a strong set; each column is a color. The theorem is that if in every column but one, each pair of entries is in conflict, then the one column is a strong set? I'll have to check out your paper again.

I fear perhaps that the matrix idea is beyond my learning... I am Not sure that I really have a handle on it yet - It will give me something to muse about as I drive about later today!

BTW - I am still of the opinion that Sudoku is essentially a Three dimensional game, and that the struggles to represent proofs are the result of using essentially two dimensional representations. If I still knew how to represent transformations, I think I could illustrate this opinion more clearly.

13'21. G'Maen, all. First time in several days that I've been able to complete tough in one sitting. I suppose it's an easy tough? And have we Ashley Violet again? If so, who is the boy?