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I have a question about an arc consistency network example. its nothing with code.

i have a network R with Variables X = {x1, x2, x3, x4, x5, x6, x7, x8} and Domains D = {1,2,3,4}. This are my constraints:

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I have the solution fromy my professor. But i dont understand what happend with C24 / why nothing happend with C24.

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I think D2 is 2,3,4 and D4 is default 1,2,3,4. If x2 is not equal x4 (C24), x4 should be 1. I hope someone could help me.

Torben G
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1 Answers1

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A constraint c(xi,xj) is arc-consistent iff for every value a∈Di there exists a value b∈Dj such that c(a,b) is true (and vice versa with i and j swapped).

If your constraint is xi < xj, then the situation Di={1,2,3}, Dj={1,2,3,4} is not arc-consistent because there is no xi∈{1,2,3} that makes xi < 1 true. To get arc-consistency, you must eliminate 1 from Dj.

However, with the constraint xi ≠ xj and Di={1,2,3}, Dj={1,2,3,4} you already have arc-consistency, because for every xj∈{1,2,3,4} there exists an xi∈{1,2,3} that makes xi ≠ xj true (and for every xi∈{1,2,3} there exists an xj∈{1,2,3,4} that makes xi ≠ xj true).

jschimpf
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  • first of all thanks for your comment and your help :) but could you explain it a little bit more because i dont really understand what you mean... sorry – Torben G Mar 23 '17 at 18:16
  • ah okay. so it is arc-consistency because xi∈ has only 3 values in his domain ( {1,2,3} ) and xj∈ has 4 ( {1,2,3,4} ). So xi ≠ xj is true because you have 3 values vs 4 values? and not 4 on one side and for on the other side? – Torben G Mar 24 '17 at 05:13
  • No, it has nothing to do with the number of elements in the domains. – jschimpf Mar 24 '17 at 23:26