Huang Lie Jun edited subsection_A_B_Since_A__.tex  about 8 years ago

Commit id: 4dcd892d4a20b40dded75571765064d7e9d64d7b

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\(D_B = \{0, 1, 2, 3, 4\}\)  \(D_C = \{0, 1, 2, 3, 4\}\) \subsection{\( . (B, A)\)}  Since \(B = A - 1\) is the constraint, Domain \(D_B\) will be reduced to \{0, 1, 2, 3\} to be make arc \((B, A)\) arc-consistent.  State:  \(D_A = \{1, 2, 3, 4\}\)  \(D_B = \{0, 1, 2, 3\}\)  \(D_C = \{0, 1, 2, 3, 4\}\)  \subsection{\( . (B, C)\)}  Since \(B = 2C\) is the constraint, Domain \(D_B\) will be reduced to \{0, 2\} to be make arc \((B, C)\) arc-consistent.  State:  \(D_A = \{1, 2, 3, 4\}\)  \(D_B = \{0, 2\}\)  \(D_C = \{0, 1, 2, 3, 4\}\)  \subsection{\( . (C, B)\)}  Since \(C = \frac{B}{2}\) is the constraint, Domain \(D_C\) will be reduced to \{0, 1\} to be make arc \((C, B)\) arc-consistent.  State:  \(D_A = \{1, 2, 3, 4\}\)  \(D_B = \{0, 2\}\)  \(D_C = \{0, 1\}\)