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That's true (and clever) but, unfortunately, doesn't allow us to solve the puzzle. It would be interesting to see a puzzle that could not be solved were it not for this new technique.

- Sue De Coq
**Posts:**93**Joined:**01 April 2005

I don't understand this, but looking at the example I think it MAY help with a puzzle I'm stuck on at the moment.

Can someone tell me what is a locked set (and ALMOST locked set)? And if someone who understands what bennys is saying could please rephrase (almost-newbie here) so I can follow it better, I'd be very grateful.

Can someone tell me what is a locked set (and ALMOST locked set)? And if someone who understands what bennys is saying could please rephrase (almost-newbie here) so I can follow it better, I'd be very grateful.

- Shazbot
**Posts:**220**Joined:**24 September 2005

HTH to clarify things:

The fact that 3 and 8 are candidates at Z means that we cannot have X=3 and Y=8. However, this is exactly what we would have if 2 and 5 were removed as candidates from X and Y. Note that if the cell r2c1 were to contain a 2 or a 5, r1c1 would have to contain a 5 or a 2 (i.e. the other one), whereupon we would be left with a contradiction. Therefore, r2c1 cannot contain a 2 or a 5.

Note that the candidates for r3c3 are 1 and 8. (Restrictions in Column 3 remove the seemingly plausible 2 as a candidate).

My solver has to revert to a forced chain in order to solver the published puzzle.

The fact that 3 and 8 are candidates at Z means that we cannot have X=3 and Y=8. However, this is exactly what we would have if 2 and 5 were removed as candidates from X and Y. Note that if the cell r2c1 were to contain a 2 or a 5, r1c1 would have to contain a 5 or a 2 (i.e. the other one), whereupon we would be left with a contradiction. Therefore, r2c1 cannot contain a 2 or a 5.

Note that the candidates for r3c3 are 1 and 8. (Restrictions in Column 3 remove the seemingly plausible 2 as a candidate).

My solver has to revert to a forced chain in order to solver the published puzzle.

- Sue De Coq
**Posts:**93**Joined:**01 April 2005

I guess it's hard to follow without seeing the rest of the candidates pencilled in. I don't follow the reasoning behind r3c3 <> 2, and wouldn't remove 2 and 5 from X and Y unless they were eliminated via a different method. Maybe this is a bit beyond me (and maybe THAT'S why I'm still stuck on that puzzle!)

- Shazbot
**Posts:**220**Joined:**24 September 2005

Easy stuff first. Here's a copy of the puzzle that will fit on the screen for stuartn. The 'W' is my addition:

Here's solver log output that explains why 2 isn't a candidate for r3c3 and gives a forced chain that cracks the problem:

shazbot - 2 and 5 are removed from r2c1, not X and Y.

Now some chat. Here's a slightly more general way to look at the technique introduced by bennys. Note that since the values 2, 3, 5 and 8 fill the cells W, X, Y and Z, they form a disjoint subset (or quad). Normally, we only identify disjoint subsets that lie in a single sector and allow us to make immediate eliminations. However, even when no immediate eliminations are possible, disjoint subsets enforce useful restrictions. Consider the candidates 2 and 5 in Box 1. Were they to be placed anywhere other than in cells W, X and Y, we would no longer have four candidates with which to fill the disjoint subset {W,X,Y,Z}. On the other hand, the candidates 3 and 8 cannot immediately be eliminated from Box 1 because they are candidates for Z, which, crucially, lies outside Box 1.

The general rule here is 'Once a Disjoint Subset has been identified, the subset values should be removed as candidates from other cells in any sector that contains a subset cell, unless that subset value should occur in a subset cell outside of that sector'.

Does that make sense? It will give me something to code over the weekend!

- Code: Select all
`W 6 9 | 1 4 3 | 8 . 7`

. X 7 | 9 . . | . . 4

4 Y . | 7 . . | . 9 3

-------+-------+------

9 . . | 8 5 7 | 3 6 1

7 Z . | 2 1 6 | 9 4 5

6 1 5 | 3 9 4 | 2 7 8

-------+-------+------

. 7 . | 5 8 9 | 4 1 .

8 . . | 6 7 1 | 5 3 .

. . . | 4 3 2 | 7 8 .

Here's solver log output that explains why 2 isn't a candidate for r3c3 and gives a forced chain that cracks the problem:

- Code: Select all
`The values 1, 3, 6 and 8 occupy the cells r3c3, r5c3, r7c3 and r9c3 in some order.`

- The moves r3c3:=2 and r7c3:=2 have been eliminated.

Consider the chain r3c2-5-r3c6~8~r3c3-1-r9c3-1-r9c1-5-r9c2.

When the cell r3c2 contains the value 5, so does the cell r9c2 - a contradiction.

Therefore, the cell r3c2 cannot contain the value 5.

shazbot - 2 and 5 are removed from r2c1, not X and Y.

Now some chat. Here's a slightly more general way to look at the technique introduced by bennys. Note that since the values 2, 3, 5 and 8 fill the cells W, X, Y and Z, they form a disjoint subset (or quad). Normally, we only identify disjoint subsets that lie in a single sector and allow us to make immediate eliminations. However, even when no immediate eliminations are possible, disjoint subsets enforce useful restrictions. Consider the candidates 2 and 5 in Box 1. Were they to be placed anywhere other than in cells W, X and Y, we would no longer have four candidates with which to fill the disjoint subset {W,X,Y,Z}. On the other hand, the candidates 3 and 8 cannot immediately be eliminated from Box 1 because they are candidates for Z, which, crucially, lies outside Box 1.

The general rule here is 'Once a Disjoint Subset has been identified, the subset values should be removed as candidates from other cells in any sector that contains a subset cell, unless that subset value should occur in a subset cell outside of that sector'.

Does that make sense? It will give me something to code over the weekend!

- Sue De Coq
**Posts:**93**Joined:**01 April 2005

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