Constraint solvers are a great attraction of many Prolog systems, and in fact they are often an important reason for buying a fast commercial Prolog system. For

Constraint solvers are a great attraction of many Prolog systems, and in fact they are often an important reason for buying a fast commercial Prolog system. For example, here is a solution of the first puzzle shown in the article, using Scryer Prolog and its CLP(ℤ) solver, i.e., constraint logic programming over integers: :- use_module(library(clpz)). solution(N) :- N in 0..10_000, N #= 13*_, N #= 19*_, N #= 37*_. Note how seamlessly constraint programming blends into Prolog: We simply state these constraints as relations that must hold for logic variables, just as always in Prolog! In fact, plain Prolog can be regarded as a particular instance of constraint logic programming (CLP), namely as constraint logic programming over Herbrand terms, abbreviated as CLP(H). Sample query and answer: ?- solution(N), indomain(N). N = 0 ; N = 9139. We see from this answer that there are in fact two solutions to the puzzle: 0 and 9139. Prolog reports all solutions on backtracking. A nice feature of the Scryer Prolog toplevel is that we can simply press "a" to enumerate all solutions. For example, if we increase the upper bound to 100_000 as mentioned in the course of the article, we get: ?- N in 0..100_000, N #= 13*_, N #= 19*_, N #= 37*_, indomain(N). N = 0 ; N = 9139 ; N = 18278 ; N = 27417 ; N = 36556 ; N = 45695 ; N = 54834 ; N = 63973 ; N = 73112 ; N = 82251 ; N = 91390. So, 11 solutions (not 10). Hence, the sample code shown in the article should probably change: model.NewIntVar(1, ...) to army = model.NewIntVar(0, ...) in order to correctly model the equation shown before the code. The knapsack problem shown in the article can also be easily expressed and solved: ?- Vs = [Bread,Meat,Beer], Vs ins 0..sup, Bread + 3*Meat + 7*Beer #=< 19, Popularity #= 3*Bread + 10*Meat + 26*Beer, labeling([max(Popularity)], Vs). yielding solutions in decreasing order of popularity: Vs = [2,1,2], Bread = 2, Meat = 1, Beer = 2, Popularity = 68 ; Vs = [5,0,2], Bread = 5, Meat = 0, Beer = 2, Popularity = 67 ; Vs = [0,4,1], Bread = 0, Meat = 4, Beer = 1, Popularity = 66 ; Vs = [1,1,2], Bread = 1, Meat = 1, Beer = 2, Popularity = 65 ; ... . This answer shows that the optimal solution with popularity 68 is unique in this case. The program output in the article shows "3712 popularity", yet the text also mentions 68: The number 3712 stems from using 1_000 as the capacity. It is notable how compactly and conveniently such tasks can be expressed and solved with Prolog.