Backward Checking SearchBT, BM, BJ, CBJ

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 tracea tree

past, current, future

• 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})

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16 checks and 12 nodes

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A Tree Trace of BT (assume domain ordered {R,B,G})

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v1

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v5

A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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A Tree Trace of BT (assume domain ordered {R,B,G})

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Another view

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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

Thrashing? (csp3)

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Thrashing? (csp3)

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V1 = Rv2 = Bv3 = Gv4 = Rv5 = Bv6 = Rv7 = Bv8 = Rv9 = conflict

The cause of the conflict with v9is v4, but what will bt do?

Find out how it goes with bt3 and csp4

Where’s the code?

See clairExamples/bt*.cl and clairExample/csp*.cl

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?