The New Sudoku Players' Forum

[bennys]

Code: Select all

+-------------+-------------+-------------+
|*139 6   8   | 14  5   2   |^49  7   13  |
|*13  124 127 | 6   17  9   |^48  5   138 |
|*159*14  157 | 147 8   3   | 2   149 6   |
+-------------+-------------+-------------+
| 4   12  6   | 157 127 15  | 3   8   9   |
| 8   5   9   | 3   4   6   | 1   2   7   |
| 7   3   12  | 9   12  8   | 5   6   4   |
+-------------+-------------+-------------+
| 26  7   3   | 15  9   145 | 68  14  28  |
| 156 8   145 | 2   3   7   | 469 149 15  |
| 125 9   145 | 8   6   14  | 7   3   125 |
+-------------+-------------+-------------+

A={R1C1,R2C1,R3C1,R3C2}
B={R1C7,R2C7}
x=9
z=4
now 9 is NOT restricted common however 5 can appear only on R3C1 and so we get
that the Almost locked sets xz rule works
and r3c8<>4.

[Carcul]

This deduction can also be made by the following multiple implication nice loop (Jeff, correct me if I am wrong):

[r3c8]-4-[r3c2](-1-[r1c1])-1-[r2c1]-3-[r1c1]-9-[r1c7]-4-[r3c8]
=> r3c8<>4.

Carcul

[Jeff]

Nice observation as usual, Bennys. I noticed that cells r3c1 and r2c7 are not needed in the nice loops. You may like to check your ALS to see whether these 2 cells are needed too.

This deduction can also be made by the following multiple implication nice loop (Jeff, correct me if I am wrong):

Your multiple nice loop is spot on. Alternatively, this deduction can be expressed as a triple chain.

Trivalue tripod at r1c1
[r3c8]-4-[r3c2]-1-[r2c1]-3-[r1c1]-9-[r1c7]-4-[r3c8]
[r3c8]-4-[r3c2]-1-[r1c1]-9-[r1c7]-4-[r3c8]
All imply r3c8<>4.

[ronk]

now 9 is NOT restricted common however 5 can appear only on R3C1 and so we get
that the Almost locked sets xz rule works
and r3c8<>4.

I don't see how the 5 only in r3c1 helps.

Either r1c1<>9 or r1c7<>9
If r1c7<>9, then r1c7=4, r3c8<>4
If r1c1<>9, then A = {r1c1,r2c1,r3c1,r3c2} = {13459}

... which is still only an almost locked set ... meaning we don't yet know which value *is not* part of the locked value set. IOW I don't see why the 5 *must be* part of the final locked set A. Would you please clarify?

I was able to take the grid to ...

Code: Select all

 
 *--------------------------------------------------*
 | 139  6    8    | 14   5    2    | 49   7    13   |
 | 123 *124 *127  | 6   ^17   9    | 48   5    138  |
 | 159 *14   157  | 147  8    3    | 2    149  6    |
 |----------------+----------------+----------------|
 | 4    12   6    | 157  127  15   | 3    8    9    |
 | 8    5    9    | 3    4    6    | 1    2    7    |
 | 7    3    12   | 9    12   8    | 5    6    4    |
 |----------------+----------------+----------------|
 | 26   7    3    | 15   9    145  | 68   14   28   |
 | 156  8    145  | 2    3    7    | 469  149  15   |
 | 125  9    1245 | 8    6    14   | 7    3    125  |
 *--------------------------------------------------*
sets A = {r2c2,r2c3,r3c2}
     B = {r2c5}
     x = 7
     z = 1
for r2c1<>1

bennys, how did you do the eliminations r2c1<>2 and r9c3<>2?

Carcul or Jeff, how would your write the nice loop chain(s) for the above elimination of r2c1<>1?

[Carcul]

Hi Ronk.

Carcul or Jeff, how would your write the nice loop chain(s) for the above elimination of r2c1<>1?

The following simple nice loop will prove r2c1<>1:

[r2c1]=3=[r2c9]=8=[r2c7]=4=[r2c2]-4-[r3c2]-1-[r2c1].

It is also possible to write a triple implication chain with a trivalue tripod at r3c8 that also proves that r2c1<>1: whichever value populates cell r3c8, r2c1 cannot be 1.

Regards, Carcul

[ronk]

The following simple nice loop will prove r2c1<>1:

[r2c1]=3=[r2c9]=8=[r2c7]=4=[r2c2]-4-[r3c2]-1-[r2c1].

Thanks, but would you do that for the almost locked sets A = {r2c2,r2c3,r3c2} and B = {r2c5}? Or is there no equivalent?

[Jeff]

but would you do that for the almost locked sets A = {r2c2,r2c3,r3c2} and B = {r2c5}?

Code: Select all

 
 *--------------------------------------------------*
 | 139  6    8    | 14   5    2    | 49   7    13   |
 | 123 *124 *127  | 6   ^17   9    | 48   5    138  |
 | 159 *14   157  | 147  8    3    | 2    149  6    |
 |----------------+----------------+----------------|
 | 4    12   6    | 157  127  15   | 3    8    9    |
 | 8    5    9    | 3    4    6    | 1    2    7    |
 | 7    3    12   | 9    12   8    | 5    6    4    |
 |----------------+----------------+----------------|
 | 26   7    3    | 15   9    145  | 68   14   28   |
 | 156  8    145  | 2    3    7    | 469  149  15   |
 | 125  9    1245 | 8    6    14   | 7    3    125  |
 *--------------------------------------------------*

Of course Ronk, it's just an wxyz-wing. It can be expressed as a weak grouped nice loop or a poly-implication chain.

Weak grouped nice loop:
[r2c1]-1-[r2c5]-7-[r2c2|r2c3|r3c2]-1-[r2c1] => r2c1<>1

Quadruple implication chain with trivalue tripods at r2c3 and r2c2:
[r2c1]-1-[r2c5]-7-[r2c3]-1-[r2c1]
[r2c1]-1-[r2c5]-7-[r2c3]-2-[r2c2]-1-[r2c1]
[r2c1]-1-[r2c5]-7-[r2c3]-2-[r2c2]-4-[r3c2]-1-[r2c1]
All imply r2c1<>1

[bennys]

Code: Select all

 *--------------------------------------------------*
 | 139  6    8    | 14   5    2    | 49   7    13   |
 | 123 *124 *127  | 6   *17   9    |*48   5    138  |
 | 159  14   157  | 147  8    3    | 2    149  6    |
 |----------------+----------------+----------------|
 | 4    12   6    | 157  127  15   | 3    8    9    |
 | 8    5    9    | 3    4    6    | 1    2    7    |
 | 7    3    12   | 9    12   8    | 5    6    4    |
 |----------------+----------------+----------------|
 |^26   7    3    | 15   9    145  |^68   14   28   |
 | 156  8    145  | 2    3    7    | 469  149  15   |
 | 125  9    1245 | 8    6    14   | 7    3    125  |
 *--------------------------------------------------*
     A = {r2c2,r2c3,r2c5,r2c7}
     B = {r7c1,r7c7}
     x = 8
     z = 2



[bennys]

In the ALS XZ rule we need that the x is restricted common.
Now when we have that any other common must appear in the union of A and B.
However if x is common but not restricted it does not work but if lets say in B the only place that is consistent with appearance also in A is the only place that other candidate lets say y can appear in B then we get.
If x not appear in A then z appear in A
If x not appear in B then z appear in B
if x appear in BOTH A and B then y will not appear in B and then we get that z appear in B.

[Myth Jellies]

Bennys,

Do you think this new rule is useful? As Jeff pointed out, you could have used the following which required no extras to the xz-rule to eliminate the same candidate 4 in r3c8.

Code: Select all

+-------------+-------------+-------------+
|*139 6   8   | 14  5   2   |^49  7   13  |
|*13  124 127 | 6   17  9   | 48  5   138 |
| 159*14  157 | 147 8   3   | 2   149 6   |
+-------------+-------------+-------------+
| 4   12  6   | 157 127 15  | 3   8   9   |
| 8   5   9   | 3   4   6   | 1   2   7   |
| 7   3   12  | 9   12  8   | 5   6   4   |
+-------------+-------------+-------------+
| 26  7   3   | 15  9   145 | 68  14  28  |
| 156 8   145 | 2   3   7   | 469 149 15  |
| 125 9   145 | 8   6   14  | 7   3   125 |
+-------------+-------------+-------------+

A={R1C1,R2C1,R3C2}
B={R1C7}
x=9
z=4


Do you have another example where the simpler xz-rule would not apply?

[bennys]

Actually I start to think that you can all ways remove the that cell from B and use the regular rule.