The New Sudoku Players' Forum

[wuem]

but not by trial&error
+---+---+---+
|142|536|789|
|758|...|..6|
|963|...|25.|
+---+---+---+
|279|685|1..|
|384|192|675|
|516|473|928|
+---+---+---+
|42.|36.|...|
|637|...|..2|
|89.|..4|.6.|
+---+---+---+
I don't hope already

[Karyobin]

Angus's stops hinting and Wayne's gets really angry about this one.

[stuartn]

R2C5 can't be a 4. The chain bombs out at R3C9 - and it all falls apart after that.

stuartn

[Jeff]

This grid can be solved logically in 2 steps:

1) xyz-wing: r2c6-r8c6-r8c4:19-189-89 => r7c6<>9 => r7c8=9 => r8c8=14

2) xyz-chain: xyz-wing=r8c8-r2c8-r2c7:14-143-43, but when r8c8=4, r4c9=4, therefore elimination footprint is extended to column 9 => r3c9<>4

The rest is simple.

[Sue De Coq]

Consider the chains r8c7-4-r8c8 and r8c7-4-r2c7-3-r2c8-3-r4c8~4~r8c8.
When the r8c7 doesn't contain the value 4, one chain states that the r8c8 contains the value 4 while the other says it doesn't - a contradiction.
Therefore, the r8c7 must contain the value 4.

[stuartn]

Sue - any chance of re-posting using the same operands as Jeff? It'll make it clearer for all. Ta everso

stuartn

[Sue De Coq]

Assume r8c7<>4.

A. r8c7-r8c8:458-149 so r8c7<>4 => r8c8=4
B. r8c7-r2c7-r2c8-r4c8-r8c8:458-34-134-34-149 so r8c7<>4 => r2c7=4 => r2c7<>3 => r2c8=3 => r4c8<>3 => r4c8=4 => r8c8<>4.

A and B are contradictory, so the initial assumption must be false.

I hope that's clearer.

[Jeff]

Sue

An example for better expression of double forcing chains is:

Code: Select all

r8c8=5 => r8c1<>5
r8c8=4 => r8c3=7 => r9c1=5 => r8c1<>5
Therefore, r8c1<>5.

[Scott H]

Using the notation and ideas here, Sue's chain can be written

Code: Select all

[A,87:4*] <4> [88:4*] <-4> [48:43] <-3> [28:3*] <3> [27:34] <4> [B,87:4*]

From the end labels of this chain you can quickly read off the relation it enforces, "A=4 or B=4". Since A=B=r8c7, this chain proves r8c7=4.

This can also be expressed logically as a double implication chain starting from anywhere in the middle, or as a contradiction starting from either end of the chain. All such choices are arbitrary, so I really like this way of clearly showing a forcing chain and its effect indepent of any implication chain derived from it.

This puzzle seems to have many forcing chains. Another is

Code: Select all

[A,28:3*] <-3> [27:34] <-4> [39:4*] <4> [49:43] <3> [48:3*] <-3> [B,28:3*]

forcing "A<>3 or B<>3". Since A=B=r2c8 this proves r2c8<>3.

[stuartn]

Another way -

Assume 3,5 <>4

=> 3,9 = 4 => 4,9 = 3 and 2,7 = 3 and 2,8 <> 3
=> 2,8 = 1 and 4,8 = 3 - which is inconsistent - 4,9 and 4,8 can't both be 3.

How would you write that using the terminology above? (and joining the nodes we get what looks suspiciously like a tadpole - I therefore have the honour of christening this shape... blah blah..)

stuartn

http://www.brightonandhove.org/Sudolinks.htm

[Scott H]

Another way -

Assume 3,5 <>4

=> 3,9 = 4 => 4,9 = 3 and 2,7 = 3 and 2,8 <> 3
=> 2,8 = 1 and 4,8 = 3 - which is inconsistent - 4,9 and 4,8 can't both be 3.

How would you write that using the terminology above?

I'd first note 3,5 is peripheral to the reasoning, so the deductions might as well start at 3,9. When you exclude or include a candidate, two or more chains meet there or a chain starts and ends there. Here, 3,9=4 causes your contradiction, so look for chains from 3,9. A contradiction like "both nodes in a row can't be 3" mean there's a 3 or -3 edge in a forcing chain between those nodes. Note that your internal inference 2,8=1 is unneeded, since 2,8<>3 => 4,8=3 already. Throwing out the 2,8=1 red herring, your forcing chain is

Code: Select all

[A,3,9:4*] <-4> [27:43] <-3> [28:3*] <3> [48:3*] <-3> [49:34] <-4> [A:4*]

The chain enforces A<>4 or A<>4, i.e. A=r3c9<>4.