Unification Algorithm Idea: eliminate disagreements Consider the set - PowerPoint PPT Presentation

Unification Algorithm

Idea: eliminate ‘disagreements’ Consider the set of expressions { P ( a ) , P ( x ) } – The expressions ‘disagree’ on a and x – The disagreement set is { a, x } – Since x is a variable, we can eliminate this disagreement by using the substitution θ = { x/a } – Then P ( a ) θ = P ( x ) θ = P ( a ) Unification algorithm 1

Idea: eliminate ‘disagreements’ Consider the set of expressions { P ( a ) , P ( x ) } – The expressions ‘disagree’ on a and x – The disagreement set is { a, x } – Since x is a variable, we can eliminate this disagreement by using the substitution θ = { x/a } – Then P ( a ) θ = P ( x ) θ = P ( a ) Definition The disagreement set for a set W of expressions is obtained by find- ing the first position (starting from the left) at which not all expressions in W have the same symbol. We then extract from each expression in W the sub- expression that begins with the symbol occupying that position. The set of these sub-expressions is the disagreement set for W . Exercise Find the disagreement set for W = { P ( x, f ( y, z )) , P ( x, a ) , P ( x, g ( h ( k ( x )))) } Unification algorithm 1

Unification algorithm Step 1 Set k = 0 , W 0 = W and σ 0 = ε (empty substitution) Step 2 If W k is a singleton, STOP: σ k is an MGU for W . Otherwise, find the disagreement set D k for W k . Step 3 If there are x k , t k ∈ D k such that x k is a variable that does not occur in t k , then go to step 4; otherwise STOP: the set W is not unifiable. Step 4 Set σ k +1 = σ k { x k /t k } and W k +1 = W k { x k /t k } ( = W σ k +1 ) Step 5 Set k = k + 1 and go to step 2 Unification algorithm 2

Unification algorithm Step 1 Set k = 0 , W 0 = W and σ 0 = ε (empty substitution) Step 2 If W k is a singleton, STOP: σ k is an MGU for W . Otherwise, find the disagreement set D k for W k . Step 3 If there are x k , t k ∈ D k such that x k is a variable that does not occur in t k , then go to step 4; otherwise STOP: the set W is not unifiable. Step 4 Set σ k +1 = σ k { x k /t k } and W k +1 = W k { x k /t k } ( = W σ k +1 ) Step 5 Set k = k + 1 and go to step 2 Examples Find MGUs for the following sets: – W 1 = { P ( a, y ) , P ( x, f ( b )) } – W 2 = { P ( a, x, f ( g ( y ))) , P ( z, f ( z ) , f ( u )) } – W 3 = { Q ( f ( a ) , g ( x )) , Q ( y, y ) } – W 4 = { Q ( a, x, f ( x )) , Q ( a, y, y ) } – W 5 = { Q ( x, y, z ) , Q ( u, h ( v, v ) , u ) } Unification algorithm 2

Unification algorithm: termination The unification algorithm always comes to a stop after finitely many steps – For otherwise there would be an infinite sequence W σ 0 , W σ 1 , W σ 2 , . . . – Each step eliminates one variable from W – Which is a contradiction since W contains finitely many variables Unification algorithm 3

Unification algorithm: termination The unification algorithm always comes to a stop after finitely many steps – For otherwise there would be an infinite sequence W σ 0 , W σ 1 , W σ 2 , . . . – Each step eliminates one variable from W – Which is a contradiction since W contains finitely many variables Unification algorithm: correctness Theorem If W is a finite unifiable set of expressions, then the unification algorithm always comes to a stop and returns an MGU for W . Proof Suppose θ is a unifier for W . By induction on k one can show that there is a substitution λ k such that θ = σ k λ k . Unification algorithm 3