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antony · 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

daveliney · Posted: Fri Jul 29, 2005 11:07 pm Post subject: Over-eager method F mistake?

Starting with:

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

karat · 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

arsoncupid · 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 middle-most 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.

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

arsoncupid · 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.

melvincutter

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.

InsaneProdigy · 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.

ednotover · 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.

Olliminatore · 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.

Ruud · Site Admin

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.
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Olliminatore · 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. Smile

_________________
Germany

Ruud · Site Admin

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 trilocation-implication-chains. 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.
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