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zoltag Expert
Joined: 01 Apr 2006 Posts: 126

Posted: Mon Jul 09, 2007 4:05 am Post subject: Sashimi Swordfish 


Code:  ....
 5 129 7  19 8 6  124 149 3 
 6 3 189  4 2 19  7 89 5 
 189 1289 4  3 7 5  128 1689 68 
:++:
 2 7 1689  15689 3 189  158 168 4 
 18 5 3  2 16 4  9 7 68 
 4 189 168  15689 169 7  158 3 2 
:++:
 189 1689 5  1689 1469 2  3 48 7 
 7 168 2  168 146 3  48 5 9 
 3 4 89  7 5 89  6 2 1 
'''' 
Here SQ tells me there is a vertical Sashimi Swordfish.
I cannot see how r2c3 is impacted by this, yet that is the cell with an elimination. I would really like some help on this! 

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JeanChristophe Addict
Joined: 23 Apr 2007 Posts: 92 Location: Belgium

Posted: Mon Jul 09, 2007 8:21 am Post subject: 


Not sure this is the same swordfish found by SQ, but it makes eliminations on R2C3.
R9, C5, N4 and R6, C3, N8 forms a Generalized Swordfish on 9 > not elsewhere in R6, C3, N8
i.e. there are 3 strong links on 9 in R9C36, R67C5 & R4C3=R6C2
Each of them is weak linked to another, forming a closed loop: R6C25, R49C3, R7C5R9C6. Since they form a closed loop, one of the end must be 9: either R9C6 & R6C5 & R4C3 = 9 or R9C3 & R6C2 & R7C5 = 9. Whichever the case they must a 9 in R6C25 > R6C4<>9, R49C3 > R2C3<>9, R7C5R9C6 > R7C4<>9
Then:
XYWing on 8 with pivot R2C6, pincers R9C6 & R2C3 > R9C3 <> 8 = 9
The rest are singles and a naked pair. 

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rep'nA Hooked
Joined: 19 Jan 2007 Posts: 49 Location: Union City, California

Posted: Tue Jul 10, 2007 8:47 pm Post subject: Re: Sashimi Swordfish 


Code: 
**
 5 129& 7  19 8 6  124 149 3 
 6* 3* 189  4 2 19  7 89 5 
 189& 1289& 4  3 7 5  128 1689 68 
++
 2 7 1689  15689 3 189  158 168 4 
 18 5 3  2 16 4  9 7 68 
 4 189* 168  15689 169* 7  158 3 2 
++
 189* 1689* 5  1689 1469* 2  3 48 7 
 7 168 2  168 146 3  48 5 9 
 3 4 89  7 5 89  6 2 1 
**

If the 9's in columns 1,2 and 5 were in the starred cells, then we would have a swordfish and would be be allowed to eliminate 9 from r2c3 (among other places). The fact that there are no 9's in r2c12 does not change this fact. However, there are 9's in r1c2, r3c1 and r3c2, which obstructs the swordfish. But since these are in the same box as r2c3, we may still eliminate 9 from r2c3. _________________ "Obviousness is always the enemy to correctness."Bertrand Russell 

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zoltag Expert
Joined: 01 Apr 2006 Posts: 126

Posted: Tue Jul 10, 2007 9:20 pm Post subject: Re: Sashimi Swordfish 


rep'nA wrote:  Code: 
**
 5 129& 7  19 8 6  124 149 3 
 6* 3* 189  4 2 19  7 89 5 
 189& 1289& 4  3 7 5  128 1689 68 
++
 2 7 1689  15689 3 189  158 168 4 
 18 5 3  2 16 4  9 7 68 
 4 189* 168  15689 169* 7  158 3 2 
++
 189* 1689* 5  1689 1469* 2  3 48 7 
 7 168 2  168 146 3  48 5 9 
 3 4 89  7 5 89  6 2 1 
**

If the 9's in columns 1,2 and 5 were in the starred cells, then we would have a swordfish and would be be allowed to eliminate 9 from r2c3 (among other places). The fact that there are no 9's in r2c12 does not change this fact. However, there are 9's in r1c2, r3c1 and r3c2, which obstructs the swordfish. But since these are in the same box as r2c3, we may still eliminate 9 from r2c3. 
What you post may be true. Does it matter whether r2c12 were "givens"? 

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Ron Moore Addict
Joined: 13 Aug 2006 Posts: 72 Location: New Mexico

Posted: Wed Jul 11, 2007 5:47 pm Post subject: Franken Fish 


zoltag,
For the grid that you show, I like JeanChristophe's "closed loop" idea in his post. I would use "AIC loop" or maybe "AIC ring," but of course the name isn't particularly important as long as you understand the idea. I think AIC loops are much easier to spot than a sashimi swordfish. For some fish, though, there may not be a AIC loop handy, such as for an ordinary, unfinnned 3 x 3 x 3 swordfish.
Like JeanChristophe, I'm going to deviate slightly from the specific question that you posed and consider a different approach. I recently started to get into exotic fish, by which I mean Frankenfish and mutant fish, and I notice that in your grid there is a Franken Swordfish for digit 9.
I remember that it took me a while to grasp the "sashimi" concept. So you may be muttering to yourself asking why I'm introducing even more new and possibly confusing terminology. Again, don't worry about the specific terms so much at this point. Initially, I'll try to avoid all the usual fishy jargon and explain things in terms of the basic defining constraints of Sudoku. If you bear with me, I think you'll find this interesting, as you can sometimes use this approach when no AIC loop is handy.
The following grid shows only digit 9 candidates in your full grid.
Code: 
····
 · X9 ·  9 · ·  · 9 · 
 · · 9  · · 9  · 9 · 
 X9 X9 ·  · · ·  · 9 · 
·++·
 · · 9  9 · 9  · · · 
 · · ·  · · ·  9 · · 
 · X9 ·  9 X9 ·  · · · 
·++·
 X9 X9 ·  9 X9 ·  · · · 
 · · ·  · · ·  · · 9 
 · · 9  · · 9  · · · 
···· 
Let's focus on the digit 9 candidates in columns 1, 2, and 5 of the Sudoku grid. These have been marked with an "X" in the grid above. Suppose we arrange these candidates in a matrix, as shown below. Initially, don't worry about the columns of this matrix, but only the rows. The candidates of column 1 of Sudoku grid have been listed in row 1 of the matrix. There are only two, r3c1 and r7c1. Since column 1 of the Sudoku grid must contain exactly one "9", exactly one of the candidate premises in row 1 of our matrix must be true. Similarly, in row 2 of our matrix, all candidates in column 2 of the Sudoku grid have been listed. There is a bit of a twist, in that r13c2 is a grouped candidate. r13c2 = 9 means that exactly one of r1c2, r3c2 is 9. Again, all possible candidates in column 2 have been listed in row 2 of our matrix, so exactly one of the candidate premises in row 2 of our matrix must be true. Finally, in row 3 of our matrix, all candidates in column 5 of the Sudoku grid appear, so exactly one of these candidate premises must be true. Since each row of our matrix must contain exactly one true premise, our matrix contains exactly three true premises.
Code: 
weakly linked weakly linked weakly linked
(box 1) (row 6) (row 7)
  
  
V V V
····
 r3c1 = 9   r7c1 = 9  < exactly 1 premise must be true (col 1)
·++·
 r13c2 = 9  r6c2 = 9  r7c2 = 9  < exactly 1 premise must be true (col 2)
·++·
  r6c5 = 9  r7c5 = 9  < exactly 1 premise must be true (col 5)
···· 
Now consider the columns of our matrix. There is some method to the madness in how the candidates have been arranged in columns. Notice that in column 1 of our matrix, all of the candidates (r3c1, r13c2) there lie in box 1 of the Sudoku grid. Thus, they are weakly linked  at most one of the candidates can be true. Similarly, all candidates which appear in the second column of our matrix lie in row 6 of the Sudoku grid, so at most one of those candidates can be true. Finally, for the third column of our matrix, all candidates which appear there lie in row 7 of the Sudoku grid, so at most one of those candidates can be true.
So each of the three columns of our matrix can contain at most one true premise. But we've established earlier that our matrix has exactly three true premises. Thus each column of our matrix must contain (exactly) one true premise.
From the first column of our matrix, we deduce that either r3c1 = 9 or r13c2 = 9. These are both in box 1, so any other candidate in box 1 can be eliminated. This gives the elimination of (9)r2c3, which was your original subject. But there's more  from column 2 of our matrix, we deduce that either r6c2 = 9 or r6c5 = 9, so that any other candidate in row 6 can be eliminated. Thus, (9)r6c4 is eliminated. From the last column of our matrix, we deduce that one of r7c125 is "9", so any other candidate in row 7 can be eliminated. So, (9)r7c4 is eliminated. In total, we've found the same eliminations resulting from JeanCristophe's AIC loop.
If you think about this a bit, you'll see that what makes such a matrix possible, in this case, is that all digit 9 candidates in columns 1, 2, and 5 of the Sudoku grid are contained within box 1, row 6, and row 7. In fishy jargon, it's often said that box 1, row 6, and row 7 "cover" all the digit 9 candidates in columns 1, 2, and 5, or that box 1, row 6, and row 7 form a "cover set" for columns 1, 2, and 5. Whenever we have a set of N houses, all of whose candidates are covered by another set of N houses, then we have the potential for a (perhaps exotic) fish. The first set of houses (the one which is covered) is often called the "base set." If no two houses of the base set share a common candidate, then we do in fact have a fish structure. In this case that's clearly true since all houses in the base set are columns. If you consider the reasoning behind the eliminations derived from the above matrix, you'll see that any candidate which is in the cover set, but not in the base set, can be eliminated. Thus, if you can identify suitable base and cover sets for some fish, you don't have to construct explicitly a matrix like the one above  you can work directly from that principle.
I'll spare you (and myself) the details here, but for general fish we can sometimes come up with even more eliminations. Some day you may want to take a look at this post.
A final note: In the Sudopedia, you will find the terms "defining set" and "secondary set," respectively, instead of "base set" and "cover set." 

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zoltag Expert
Joined: 01 Apr 2006 Posts: 126

Posted: Thu Jul 12, 2007 9:34 pm Post subject: Great help, thanks. 


Thank you for the helpful replies!
JeanChristophe, while I can follow what you wrote it is not a method I use.
It certainly finds the correct eliminations!
rep'nA, your method is the answer that I couldn't find but I do still wonder whether it would matter if r2c12 were "givens" rather than "solves"?
Ron Moore, what a great and lengthy post! 

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Ron Moore Addict
Joined: 13 Aug 2006 Posts: 72 Location: New Mexico

Posted: Sat Jul 14, 2007 12:07 am Post subject: Re: Great help, thanks. 


zoltag wrote: 
rep'nA, your method is the answer that I couldn't find but I do still wonder whether it would matter if r2c12 were "givens" rather than "solves"? 
zoltag,
Let me have a try at your question. The answer is no.
Once again, at the outset let me say that I think that sashimi swordfish, and especially this one, are tough to see, and I initially had some difficulty with the "sashimi" concept, so I can appreciate your difficulty with this.
First of all, be sure you understand the reasoning behind an elimination from a finned (but not sashimi) swordfish. Suppose, as rep'nA wrote, that digit 9 were a valid candidate at r2c1 and r2c2. These cells, along with others which repn'A marked in his grid, are marked with "*" in the diagram below. Observe that these cells would form an ordinary swordfish in columns 1, 2, and 5. Then we note the presence of digit 9 candidates in r1c2, r3c12 (marked with "#" in the diagram) invalidates this pattern, but all is not lost since we can consider these cells as fins to the swordfish. This is perhaps a bit different from the run of the mill finned swordfish in that fin cells exist in two columns (columns 1 and 2). That's quite acceptable, though, as long as all fin cells lie within a common box. Now, in this hypothetical configuration, if all cells in the fins  namely, r1c2, r3c12  were false, then we would have our ordinary swordfish in columns 1, 2, and 5, and (9)r2c3 would be eliminated by that swordfish. On the other hand, if any of the fin cells were true, then, since they are all in box 1, r2c3 sees them all so again (9)r2c3 would be eliminated.
Code: 
····
 · #9 ·  9 · ·  · 9 · 
 * * 9  · · 9  · 9 · 
 #9 #9 ·  · · ·  · 9 · 
·++·
 · · 9  9 · 9  · · · 
 · · ·  · · ·  9 · · 
 · *9 ·  9 *9 ·  · · · 
·++·
 *9 *9 ·  9 *9 ·  · · · 
 · · ·  · · ·  · · 9 
 · · 9  · · 9  · · · 
···· 
Now, although you know that r1c2 and r2c2 are not "9", suppose that you slyly added 9 as a candidate in those cells (in addition to the given or solved value already there), and submitted that complete candidate grid to this forum, asking someone to help you find some elimination to advance the solution. Then someone comes back with a reply that there's a finned swordfish in columns 1, 2, and 5 eliminating (9)r2c3. Would you then be entitled to say, "Your reasoning is faulty because I knew all along that neither r2c1 nor r2c2 is '9', so there's not 'really' a finned swordfish there"? No, of course not. The logic of the finned swordfish elimination in no way assumes that one of r2c1 or r2c2 must be 9. It only uses the fact that either some cell of the (unfinned) swordfish seen by r2c3 is 9, or one of the fin cells is 9.
So if it helps, temporarily pencil in the hypothetical 9 candidates (either physically or mentally). There's no harm in that, since we're only temporarily saying that 9 is a possibility, not a certainty, for those cells. Then note the finned swordfish, make the elimination from it, and then quickly remove those hypothetical 9's since you know, whether r2c12 were either givens or solved cells, that those 9's are false. 

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zoltag Expert
Joined: 01 Apr 2006 Posts: 126

Posted: Sat Jul 14, 2007 7:20 am Post subject: Re: Great help, thanks. 


Ron Moore wrote:  zoltag wrote: 
rep'nA, your method is the answer that I couldn't find but I do still wonder whether it would matter if r2c12 were "givens" rather than "solves"? 
zoltag,
Let me have a try at your question. The answer is no.

Ron, your answer is again lengthy and useful.
Thanks!
I am pretty sure that your response is showing what Ruud is using, although Ruud's explanations do quite suck. 

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