15 puzzle solver: Difference between revisions

Content added Content deleted
(Added 11l)
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<lang 11l>-V
<lang 11l>-V
Nr = [3, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3]
nr = [3, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3]
Nc = [3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2]
nc = [3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2]


T Solver
T Solver
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.n2[n + 1] = .n2[n] - a + (a << 16)
.n2[n + 1] = .n2[n] - a + (a << 16)
.n3[n + 1] = Char(‘d’)
.n3[n + 1] = Char(‘d’)
.n4[n + 1] = .n4[n] + Int(Nr[Int(a >> g)] > .n0[n] I/ 4)
.n4[n + 1] = .n4[n] + Int(:nr[Int(a >> g)] > .n0[n] I/ 4)


F fG()
F fG()
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.n2[n + 1] = .n2[n] - a + (a >> 16)
.n2[n + 1] = .n2[n] - a + (a >> 16)
.n3[n + 1] = Char(‘u’)
.n3[n + 1] = Char(‘u’)
.n4[n + 1] = .n4[n] + Int(Nr[Int(a >> g)] < .n0[n] I/ 4)
.n4[n + 1] = .n4[n] + Int(:nr[Int(a >> g)] < .n0[n] I/ 4)


F fE()
F fE()
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.n2[n + 1] = .n2[n] - a + (a << 4)
.n2[n + 1] = .n2[n] - a + (a << 4)
.n3[n + 1] = Char(‘r’)
.n3[n + 1] = Char(‘r’)
.n4[n + 1] = .n4[n] + Int(Nc[Int(a >> g)] > .n0[n] % 4)
.n4[n + 1] = .n4[n] + Int(:nc[Int(a >> g)] > .n0[n] % 4)


F fL()
F fL()
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.n2[n + 1] = .n2[n] - a + (a >> 4)
.n2[n + 1] = .n2[n] - a + (a >> 4)
.n3[n + 1] = Char(‘l’)
.n3[n + 1] = Char(‘l’)
.n4[n + 1] = .n4[n] + Int(Nc[Int(a >> g)] < .n0[n] % 4)
.n4[n + 1] = .n4[n] + Int(:nc[Int(a >> g)] < .n0[n] % 4)


F fY()
F fY()