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Published byRosa Brindle Modified over 2 years ago

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Backward Checking Search BT, BM, BJ, CBJ

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An example problem Colour each of the 5 nodes, such that if they are adjacent, they take different colours

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Representation variables v1, v2, v3, v4, and v5 domains d1, d2, d3, d4, and d5 domains are the three colours {R,B,G} constraints v1 v2 v1 v3 v2 v3 v2 v5 v3 v5 v4 v5 Assign a value to each variable, from its domain, such that the constraints are satisfied CSP = (V,D,C)

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Chronological Backtracking (BT) (0) label(variable:vi) (1) let consistent := false (2) assign vi a value x from its domain di (3) consistent := true (4) while consistent, check against past variables vh if inconsistent(vh,vi) then consistent := false (5) if not consistent remove x from di if di is not empty go to (2) (6) return consistent (0) unlabel(variable:vi) (1) restore domain di (set to original value) (2) let h be i-1 (3) remove the value of vh from domain dh (3) if dh is empty then return false else return true (0) BT(integer:i) (1) if i > N then return(true) (2) if i = 1 and d_1 is empty then return(false) (3) if label(v[i]) then bt(i+1) else unlabel(v[i]) bt(i-1)

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Three Different Views of Search a trace a tree past, current, future

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V[1] := R V[2] := R check(v[1],v[2]) fails V[2] := B check(v[1],v[2]) good V[3] := R check(v[1],v[3]) fails V[3] := B check(v[1],v[3]) good check(v[2],v[3]) fails V[3] := G check(v[1],v[3]) good check(v[2],v[3]) good V[4] := R V[5] := R check(v[2],v[5]) good check(v[3],v[5]) good check(v[4],v[5]) fails V[5] := B check(v[2],v[5]) fails V[5] := G check(v[2],v[5]) good check(v[3],v[5]) fails backtrack! V[4] := B V[5] := R check(v[2],v[5]) good check(v[3],v[5]) good check(v[4],v[5]) good solution found A Trace of BT (assume domain ordered {R,B,G}) checks and 12 nodes

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v5

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v5

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v5

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v5

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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A Tree Trace of BT (assume domain ordered {R,B,G}) v1 v2 v3 v4 v

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Another view v1 v2 v3 v4 v5 current variable past variables future variables check back

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Can you solve this? Try bt1 and/or bt2 with constraints C2 - load(“bt2”) - bt(V,D,C2) this is csp2

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Thrashing? (csp3)

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V1 = R v2 = B v3 = G v4 = R v5 = B v6 = R v7 = B v8 = R v9 = conflict The cause of the conflict with v9 is v4, but what will bt do? Find out how it goes with bt3 and csp4

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Where’s the code? See clairExamples/bt*.cl and clairExample/csp*.cl

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questions Why measure checks and nodes? What is a node anyway? Who cares about checks? Why instantiate variables in lex order? Would a different order make a difference? How many nodes could there be? How many checks could there be? Are all csp’s binary? How can we represent constraints? Is it reasonable to separate V from D? Why not have variables with domains What is a constraint graph? How can (V,D,C) be a graph?

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