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Backward Checking Search BT, BM, BJ, CBJ

Backward Checking Search BT, BM, BJ, CBJ. An example problem. 1. 2. 3. 4. 5. Colour each of the 5 nodes, such that if they are adjacent, they take different colours. Representation. 1. 2. 3. 4. 5. variables v1, v2, v3, v4, and v5 domains d1, d2, d3, d4, and d5

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Backward Checking Search BT, BM, BJ, CBJ

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  1. Backward Checking Search BT, BM, BJ, CBJ

  2. An example problem 1 2 3 4 5 Colour each of the 5 nodes, such that if they are adjacent, they take different colours

  3. Representation 1 2 3 4 5 • 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)

  4. Chronological Backtracking (BT) 1 2 3 4 5 • (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)

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

  6. A Trace of BT (assume domain ordered {R,B,G}) 1 2 3 4 1 1 2 2 5 5 3 3 5 4 4 • 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 16 checks and 12 nodes

  7. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 v1 v2 v3 v4 v5 4 5

  8. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 v1 v2 v3 v4 v5 4 5

  9. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 v1 v2 v3 v4 v5 4 5

  10. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 v1 v2 v3 v4 v5 4 5

  11. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 v1 v2 v3 v4 v5 4 5

  12. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 v1 v2 v3 v4 v5 4 5

  13. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 5 4 v1 v2 v3 v4 v5

  14. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 5 4 v1 v2 v3 v4 v5

  15. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 5 4 v1 v2 v3 v4 v5

  16. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 5 4 v1 v2 v3 v4 v5

  17. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 5 4 v1 v2 v3 v4 v5

  18. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 5 4 v1 v2 v3 v4 v5

  19. A Tree Trace of BT (assume domain ordered {R,B,G}) 1 2 3 5 4 v1 v2 v3 v4 v5

  20. 1 v1 2 v2 v3 3 v4 4 v5 5 Another view past variables check back current variable future variables

  21. 1 2 3 4 5 Can you solve this? Try bt1 and/or bt2 with constraints C2 - load(“bt2”) - bt(V,D,C2) this is csp2

  22. 1 2 3 9 4 7 8 5 6 Thrashing? (csp3)

  23. 1 2 3 9 4 7 8 5 6 Thrashing? (csp3) 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

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

  25. 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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