Pattern Matching - Prolog vs. Haskell

Question

This is not a homework question, rather an exam study guide question. What is the difference between pattern matching in Prolog Vs Haskell?

I've done some research and reading up on the theories behind them doesn't really give me a solid understanding between the two. I read that in Prolog, pattern matching is different because it has the ability to unify variables and thus be able to deduce through resolution and spit out the possible answer

eg ?- [a,b] = [a,X]    X = b 

Now I'm not sure how to display pattern matching in Haskell. I know that the same query above shown in Prolog will not work in Haskell because Haskell cannot unify like Prolog. I remember somewhere that to get the same answer in Haskell, you have to explicitly tell it through guards.

I know that I am pretty close to understanding it, but I need someone to break it down Barney style for me so I can FULLY understand it and explain it to a 12 year old. This has been bugging me for quite some time and I can't seem to find a solid explanation.

By the way the example shown above was just to display to you guys what I've learned so far and that I'm actually trying to find an answer. My main question does not relate to the examples above but rather a complete understanding on the difference between the two.

Answer 1

Prolog pattern matching is based on unification, specifically the Martelli-Montanari Algorithm (minus the occurs check, by default). This algorithm matches values of the same position, binding variables on one side to a value at corresponding position on the other side. This kind of pattern matching could work both ways, therefore in Prolog you could use arguments as both input and output. A simple example, the length/2 predicate. We could use this to (comment explains the query):

?- length([],0).      % is the length of empty list zero? ?- length([a,b,c],X). % what's the length of list consisting of a,b and c? ?- length(L,5).       % give me all lists that have length of 5 

Haskell pattern matching is a one way matching, to bind variables to different parts of given value. Once bound, it carries out the corresponding action (the right hand side). For example, in a function call, pattern matching may decide which function to call. e.g.:

sum []     = 0 sum (x:xs) = x + sum xs 

the first sum binds empty list, while the second binds a list of at least 1 element. Based on this, given sum <a list>, the result could be either 0 or x + sum xs depending on whether sum <a list> matches sum [] or sum (x:xs).

Answer 2

The difference between pattern matching as in Haskell and Prolog's unification stems from the fundamentally different rôle of variables in both languages.

In Haskell, variables hold values. Concrete values. Such a value might not have been computed yet, and it might even be ⊥, but otherwise it is one concrete value. In Haskell you cannot take a variable and only state some properties of its value first.

So pattern matching always means that a concrete value is matched against a pattern which contains some variables. The outcome of such a matching is either failure or a binding of the variables to concrete values. In Haskell this is further restricted to avoid the need of general comparison, which would imply that class Eq is defined for the terms being matched.

In Prolog, however, variables may refer to a set of possible solutions. Variables may occur anywhere - also somewhere between other values. Unification now ensures that the stated equalities still hold and the result is represented optimally, i.e. the most general unifier is computed.

 | ?- length(L,5).                       L = [_,_,_,_,_] | ?- length(L,5), maplist(=(E),L). L = [E,E,E,E,E] 

So Prolog does not answer here with concrete values like L = [1,1,1,1,1] or L = [[],[],[],[],[]] but gives the most general unifier as an answer which contains all those concrete values.