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

11 posts
• Page **1** of **1**

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

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

|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

- wuem
**Posts:**3**Joined:**23 August 2005

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.

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.

- Jeff
**Posts:**708**Joined:**01 August 2005

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.

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.

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

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.

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.

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

Sue

An example for better expression of double forcing chains is:

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.

- Jeff
**Posts:**708**Joined:**01 August 2005

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

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

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*]`

- Scott H
**Posts:**73**Joined:**28 July 2005

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

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

- stuartn
**Posts:**211**Joined:**18 June 2005

stuartn wrote: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*]`

- Scott H
**Posts:**73**Joined:**28 July 2005

11 posts
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