Artificial IntelligenceIntelligence

Ian Gentipg@cs.st-and.ac.uk

Constraint Programming 2

Artificial IntelligenceIntelligence

Part I : Arc ConsistencyPart II: MACPart III: Special kinds of constraintsPart IV: Formulation

Constraint Programming 2

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Constraint Satisfaction Problems

CSP = Constraint Satisfaction ProblemsA CSP consists of:

a set of variables, X for each variable xi in X, a domain Di

Di is a finite set of possible values

a set of constraints restricting tuples of values if only pairs of values, it’s a binary CSP

A solution is an assignment of a value in Di to each variable xi such that every constraint satisfied

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Formal Definition of Constraints

A constraint Cijk… involving variables xi, xj, xk …

is any subset of combinations of values from Di, Dj, Dk …

I.e. Cijk... Di x Dj x Dk …

indicating the allowed set of values

Most constraint programming languages/toolkits allow a number of ways to write constraints: e.g. if D1 = D2 = {1,2,3} …

{ (1,2), (1,3), (2,1), (2,3), (3,1), (3,2) } x1 x2

CtNeq(x1,x2)

I’ll use whatever notation seems right at the time

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Arc Consistency

Last time, we saw forward checking extremely simple form of reasoning

if variable x has value v& constraint involves variables x-y& constraint disallows x=v, y = u,then remove value u from domain of y

This time, look at a more complex form of reasoning arc consistency

In graph theory, arc = edge In constraints, arc will be a (directed) constraint

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Arc Consistency

Arc consistency is based on a very simple concept if we can look at just one constraint and see that x=v is

impossible … obviously we can remove the value x=v from consideration

How do we know a value is impossible? If the constraint provides no support for the valuee.g. if Dx = {1,4,5} and Dy = {1, 2, 3}

then the constraint x > y provides no support for x=1 we can remove x=1 from Dx

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Arc Consistency Propagation

When we remove a value from Dx, we may get new removals because of it

E.g. Dx = {1,4,5}, Dy = {1, 2, 3}, Dz= {2, 3, 4, 5} x > y, z > x As before we can remove 1 from Dx, so Dx = {4,5}

But now there is no support for Dz = 2,3,4

So we can remove those values, Dz = {5}, so z=5

Before AC applied to y-x, we could not change Dz

This can cause a chain reaction

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Establishing Arc Consistency

If we consider every constraint and full propagation, there are only two possibilities: the final problem will be arc-consistent

every value in every domain is supported by some other value in every constraint

we have established arc consistency at some point some variable will have an empty domain

no value can be given to that variable the problem is insoluble due to arc consistencywe have seen a domain wipe out

A number of algorithms can be used to establish AC

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AC3

AC3 is one of many algorithms (1 - 7 and lots more!)Only consider binary constraints with 2 variablesFor this algorithm, we revise directed arcs

I.e. when considering constraint x-y we remove values from ywhen considering y-x we remove values from x

of course both constraints are really the same one

AC3 can be done in polynomial time, em3

(e = number of constraints in problem, m = domain size) emphasises that AC does not solve CSP’s

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AC3

Set Q = List of all arcs (directed constraints) While (Q not empty)

remove the first arc y-x from Q revise the arc y-x

I.e. if all values in Dx supported by element of Dy do nothing

else remove any values from Dx not supported by y

for all constraints involving x (including x-y)– add the arc x-y to Q if not already a member

If Dx empty, fail (domain wipe out)

If no Domain empty, succeed (established AC)

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Exercises

Describe AC3 as a search algorithm in terms of the generic search algorithm so what’s the search problem?

Consider the constraints x < y, y < z, z < x obviously inconsistent, and AC on its own can prove this Show what AC3 does when Dx= Dy= Dz={1,2,3,4,5,6,7,8,9}

Describe a constraint satisfaction problem which … is not arc-consistent but we can establish AC but the problem has no solution

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Maintaining AC (MAC)

Like any other propagation, we can use AC in search I.e. search proceeds as follows:

establish AC at the root when AC3 terminates, choose a new variable/value re-establish AC given the new variable choice (I.e. maintain

AC) repeat; backtrack if AC gives domain wipe out

The hard part of implementation is undoing effects of AC

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Special kinds of Consistency

Some kinds of constraint lend themselves to special kinds of arc-consistency

Consider the all-different constraint the named variables must all take different values not a binary constraint can be expressed as n(n-1)/2 not-equals constraints

We can apply (e.g.) AC3 as usualBut there is a much better option

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All Different

Suppose Dx = {2,3} = Dy, Dz = {1,2,3}

All the constraints xy, yz, zx are all arc consistent e.g. x=2 supports the value z = 3

the single ternary constraint AllDifferent(x,y,z) is not! We must set z = 1

Generalised Arc Consistency applies to n-ary constraints (but complicated and expensive)

A special purpose algorithm exists for All-Different to establish GAC in efficient time

Special purpose propagation algorithms are vital

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Formulation of CSP’s

All-different is an example of the importance of formulation all-different(x,y,z) much better than xy, yz, zx even though logically equivalent

In general, it’s hard to find the best formulationRemember DONALD + GERALD = ROBERTThe formulation I gave had just 2 constraints

all-different and a complicated arithmetic constraint

All-different fine, but neither FC nor MAC can do much with the arithmetic constraint

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Cryptarithmetic Revisited

FC cannot propagate until only one variable left in constraint

AC cannot propagate until only two variables leftWhen coded in ILOG Solver, search backtracks

8018 timesHow can we formulate the problem better?

Hint: we’d like to consider the sum in each column separately

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DONALD + GERALD = ROBERT

One solution is to add more variables to the problemVariables C1, C2, C3, C4, C5

Ci represents carry from previous column i DCi = {0,1}

Now we can express more constraints D + D = 10*C1 + T C1 + L + L = 10*C2 + R C2 + A + A = 10*C3 + E C3 + N + R = 10*C4 + B C4 + O + E = 10*C5 + O C5 + D + G = R

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This shouldn’t work ?!?

We’ve made the problem bigger, so how can it help?Before, there were 93 107 possibilities

now there are 25 = 32 times as many!

The constraints now involve fewer variables constraint propagation can happen sooner variables can be set sooner (reduced to one value) domain wipe out & backtracking occurs earlier

In ILOG Solver, this encoding needs only 212 down from 8,018 if that doesn’t impress you, call it minutes (or hours)