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 Nick70
 Joined: 08 Jun 2005  Posts: 160  :   Items 

Posted: Tue Jul 26, 2005 6:59 am Post subject: Re: XYWing 


mathrec wrote:  The XYWing is a particularly easy logical loop to search for by hand. I appreciate that you've given it a name. 
I didn't invent the name, it was originally proposed here.
mathrec wrote:  The XYWing is a special case of the exposed loop, depending only on simple elimination. 
I would say the opposite is true: XYWing is more general.
mathrec wrote:  The challenge2 puzzle is not so kind. There are no exposed loops in that puzzle, since there are no wxxyyz pivots at all. 
The challenge2 can be solved using several double forcing chains. It is solvable without backtracking, but definitely very hard for a human. 

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 antony
 Joined: 22 Jul 2005  Posts: 13  :   Items 

Posted: Fri Jul 29, 2005 4:33 am Post subject: generic method D 


I would suggest a new, more generic phrasing for Method D :
Look for N digits that are candidates in only N cells of a group ("unique subset"). Then the N digits must be in the N cells, and you can remove other candidates from those N cells. And similarly:
Look for N digits that are the only candidates in N cells of a group ("disjoint subset"). Then the N digits must be in the N cells, and you can remove those N digits from other cells in the group.
Example:
123  123  234  456  45  56  7  8  9
The two aspects of the rule help removing the 4 in the third cell:
1,2,3 appear in the first three cells only (unique subset) > 4 can be removed in those cells
Only 4,5,6 appear in the last three cells (disjoint subset) > 4 can be removed in the other cells 

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 daveliney
 Joined: 29 Jul 2005  Posts: 1  :   Items 

Posted: Fri Jul 29, 2005 11:07 pm Post subject: Overeager method F mistake? 


Starting with:
Code:  9  1 7  5 2 8
7  5 2  4
5 2  9  7 6
++
4  5  8 1 7
5 8  1  2 9 6
9 1  2 8  3 5 4
++
5  1  6 3
6  5  7
3 1  2 6  8 5 
9__1_7528+_7_5_2_4_+_52_9_76_+__4_5_817+58___1296+_91_28354+__5_1_63_+_6___5_7_+31_2_6_85
the solver starts using Method F and four steps in says:
"Found a double match for 8 in columns 3 and 5 and columns B and H
.. Removing number 8 from [B1] using Solve Method F"
However setting B1 to 8 does not cause there to be no possible values for all of B3, B5, H3 and H5. For example:
Code:  9  1 7  5 2 8
8 7 3  5 6 2  4
5 2  9  7 6
++
4  5  8 1 7
5 8  1  2 9 6
9 1  2 8  3 5 4
++
5  1  6 3
6 8  4 5  7
3 1  2 6  8 5 
As it turns out those values don't work but not for reasons that Method F was looking for, as I see it.
The problem seems to be that in the cycle of four cells there are five unique entries available (34689). Is this a problem with Method F, or is my logic off? 

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 mathrec
 Joined: 15 Jul 2005  Posts: 10  :  Location: Carlsbad, CA  Items 

Posted: Thu Aug 11, 2005 12:41 am Post subject: 


Quote:  However setting B1 to 8 does not cause there to be no possible values for all of B3, B5, H3 and H5. For example: 
Code:  9  1 7  5 2 8
8 7 3  5 6 2  4
5 2  9  7 6
++
4  5  8 1 7
5 8  1  2 9 6
9 1  2 8  3 5 4
++
5  1  6 3
6 8  4 5  7
3 1  2 6  8 5 
It does, however, cause there to be no place for an 8 in column 5. That's what XWings (AKA Solve Method F) does. It identifies situations where a choice like B1=8 would cause one of columns 3 or 5 to have no place for an 8. 

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 karat
 Joined: 14 Aug 2005  Posts: 1  :   Items 

Posted: Sun Aug 14, 2005 12:28 pm Post subject: 


Karat
I think that the solution is toyour chalenge (http://www.sudokusolver.co.uk/challenge.html)
is simply that you missed a hidden chain on row E7, E8,E9 (49,24,(1)29)
so you have a 1 at E5 and then you can solve the puzzle. Sorry I can't code it (I didnt't wrote a line of code in 15 years).
Reuben 

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 arsoncupid
 Joined: 22 Nov 2005  Posts: 10  :   Items 

Posted: Tue Nov 22, 2005 5:16 am Post subject: illogic puzzle 14 


I've solved illogic puzzle #14 by hand. It did involve one guess, so this won't provide a new logical method by itself. But, puzzle 14 was listed as having 4 guesses required. With only one guess required, solving the puzzle could be simplified to solving for this one square.
The absolute middle square is either a 2 or a 3 (just from finding the givens and simplifying with mundane methods). By guessing a 2 in that square, and working into the upper and lower middle boxes (3x3). then out left from those, an inconsistency is found: an 8 in the second row in both the top and bottom boxes.
So, that middlemost square must be a 3. From there simple logic completes the puzzle.
Your logic program correctly solves the puzzle with only the 3 added to the starting arrangement.
...
When solving it by hand, I opted to guess the absolute middle square because the entire middle column of boxes are almost exclusively populated with 2's and 3's  a lot would resolve from guessing this square.
I guessed a 2 after looking a couple levels deep into both the 2 and 3 guess ... after the 2 showed it was more quickly fruitful. Usually, for me, the branch with more resolution is the more likely to fail. 

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 arsoncupid
 Joined: 22 Nov 2005  Posts: 10  :   Items 

Posted: Tue Nov 22, 2005 6:41 pm Post subject: Re: illogic puzzle 14 


arsoncupid wrote:  By guessing a 2 in that square, and working into the upper and lower middle boxes (3x3). then out left from those, an inconsistency is found: an 8 in the second row in both the top and bottom boxes. 
I meant the second column, not row. arg 

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 arsoncupid
 Joined: 22 Nov 2005  Posts: 10  :   Items 

Posted: Tue Nov 22, 2005 8:45 pm Post subject: Method E 


I've been looking at your site some more ... your Method E is the method I describe above, I think, that excludes the 2. I've found this is called Nishio here, and I believe it is a generalized algorithm applicable in many different problems. 

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 melvincutter
 Joined: 23 Dec 2005  Posts: 1  :   Items 

Posted: Fri Dec 23, 2005 11:49 pm Post subject: This is correct but you are missing 1 thing... 


I found this site yesterday and figured out the formula for problem 1 last night. This response seems to be the closest to accuracy but one point needs to also be added.
He/She said:
"In general:
 Find a locked pair {xy}.
 Find two cells forming a square with the locked pair, one containing {zx} and one {zy}.
 Set the cell of the pair next to {zy} to {x} and the other one to {y}."
However, it must also be noted that {zx} and {zy} are the only possibilities that {z} can be; i.e. {z}cannot go anywhere else within the 3X3 box. 

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 InsaneProdigy
 Joined: 29 Dec 2005  Posts: 1  :   Items 

Posted: Thu Dec 29, 2005 6:31 am Post subject: Remote Pairs 


Someone may have already brought this up but the method described in "Solve Method E Explanation  by Mark Summerville" can be simplified some to solve that puzzle.
Rule: No cell that lies at the intersection of two cells that form an exclusive pair on a single number may contain that number.
Considering only the number 7 we find:
1A & 1H are exclusive on 7 (i.e. only one or the other can be a 7)
1H & 4H are exclusive on 7 (i.e. only one or the other can be a 7)
4H & 6G are exclusive on 7 (i.e. only one or the other can be a 7)
therefore 1A & 6G are exclusive on 7 (i.e. only one or the other can be a 7)
therefore 6A (the cell at the intersection of row A and column 6) may not contain the value 7 and must in fact be a 6. Your solver finishes this puzzle once that cell is resolved.
Roughly, the pseudocode would be:
1. For each cell:
A. For each possible value:
i. Check the row/column/box for the current cell to see if there is an exclusive pair (i.e. only 2 valid positions for the current value)
ii. Increment the step counter.
iii. If the step counter is even then this is an exclusive pair with the original cell. The current value can be excluded from the intersections of the base cell and the current cell.
iv. Using this new cell repeat steps i.  iv. until no new exclusive pair can be found. 

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 ednotover
 Joined: 12 Jan 2006  Posts: 1  :   Items 

Posted: Thu Jan 12, 2006 11:28 pm Post subject: 


I see method E somewhat differently. I see it as a way to evaluate consistency of "fallout" trees when there's cyclical group overlap.
As Mark defined them, cyclical groups would just be sets of cells with related number pairs. However, it helps also to consider groups of three cells where two of the cells are pairs with a single shared number and the third is a triple of the numbers from the first two cells. This second type of grouping is determined completely if the number appearing in all three cells is chosen in any of those cells, or if any number is chosen in the cell with the triple.
(A1, A6, A9)
(H2, H3)
(H6, H7, H8)
(A1, H1)
(C4, D4, H4)
(A6, C4, C5)
(G3, H1)
By bringing those together where there's cell overlap, we have:
(A1, A6, A9) + (A1, H1) + (A6, C4, C5) + (G3, H1) + (C4, D4, H4)
(H2, H3)
(H6, H7, H8)
We need to evaluate only the first grouping, since the other two involve only a single group (there's no opportunity there for inconsistencies).
We simply check the possible values for the first cell in each evaluated grouping, and see what the fallout is. Build out the trees until all determinable cells in the grouping have been computed, or until an inconsistency is discovered.
Code: 
A1: 1

H1: 7
++
G3: 1 H4: 1
 
G5: 6 D4: 3
 
C5: 3 C4: 7
 
A6: 6 A6: 6
 
A9: 7 A9: 7

That's consistent; the only duplicated cell computations are A6 and A9, and the computed values agree.
When setting A1 to 7:
Code: 
A1: 7
++
H1: 1 A6: 6
++ 
G3: 7 H4: 7 C5: 3
 
C4: 3 C4: 7

It's inconsistent; the different branches of the A1: 7 fallout gave two different values for C4.
So A1 cannot be 7. 

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 Olliminatore
 Joined: 10 Feb 2006  Posts: 3  :   Items 

Posted: Fri Feb 10, 2006 6:48 am Post subject: Grid Illogical 8, ordinary solved 


Hello and greetings, I'm very new here.
I found a method to solve the Grid 8. (From the Illogical "Can't Solve By Logic".) Although this Method is (very) easy. It is also not included in the "Simple Sudoku" program by Angus.
I don't know how this method/technique is named, but I think it has already one.
Explanation: If a "nacked pair" appear in two boxes you can exclude bot numbers in the intersection of row and line.
Code:  8 3  2 9  7 6
6  1 8  5
 6 
++
5  4  8
7 9  3  6 2
6  9  1
++
 7 
1  6 5  8
5  9 8  4 3

In this grid the "nacked pair" 37 appear in R2C4; R6C8, so it can exclude the 3 on R2C8.
And then the riddle is complete solvable by ordinary method A/B.
How is the name of it? (or perhaps a new?)
[edit] img+


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 Ruud Site Admin
 Joined: 17 Sep 2005  Posts: 708  :  Location: Netherlands  Items 

Posted: Fri Feb 10, 2006 4:14 pm Post subject: 


There is no name for this technique.
Maybe because it does not work.
If it works in your example, then it could be called a lucky guess.
A naked pair must be connected in some way. The cells you show will not connect these 2 cells. If they would, they you would have remote pairs, a known technique.
If I am missing something, please explain carefully how your technique works. Currently I do not see it working.
Greetings, and welcome to the forum,
Ruud. _________________ Meet me at sudocue.net 

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 Olliminatore
 Joined: 10 Feb 2006  Posts: 3  :   Items 

Posted: Fri Feb 10, 2006 6:08 pm Post subject: 


Hello Ruud,
Yes you are right this method is'nt so easy as i thought and it worked not exactly as i explained. It seems (very) rare and lucky but not guessy.
In the 37 pair can not occur the 7 concurrently, so it must a 3 in the fields and I can exclude the 3 on R2C8.
Ok let depict (declare): It is like a "nacked box pair" which i can exclude a number in the row/col intersection, if I know that be the other number in this pair can't occur concurrently.
Thanks for (future) name suggestion. _________________ Germany 

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 Ruud Site Admin
 Joined: 17 Sep 2005  Posts: 708  :  Location: Netherlands  Items 

Posted: Fri Feb 10, 2006 6:42 pm Post subject: 


I took another look at your method, and I do see how it works, now. My previous answer was too quick.
The center box has 3 candidates for digit 7. No matter which of these candidates is placed, it will always result in the elimination of 7 in one of those {3,7} cells, which in turn eliminates the 3, forcing BOTH cells in your pair to 7.
it may be difficult to describe this pattern in terms other than trilocationimplicationchains. The elimination of the 3 would then be a verity or veracity (names coined by MadOverlord).
I do not think this will occur often enough to deserve its own name, but I may be mistaken. A lot of new techniques have been named at the players forum.
Ruud. _________________ Meet me at sudocue.net 

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