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One version of the riddle (from Wikipedia) begins: There are five houses. The Englishman lives in the red house. The Spaniard owns the dog.

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. The idea is that there are five houses, painted different colors, with occupants (one in each house) of different nationalities each of whom smokes a different brand, drinks a different beverage, and keeps a different pet. Given the “clues,” the puzzle asks which resident owns the zebra. Given that Mathematica is a rule-based language, it seems that there should be a way of solving the puzzle without resorting to brute-force (i.e. while loops or searching). Can Mathematica be used to elegantly deduce the solution? How? Rico Picone. asked May 3, 2014 at 6:29. Rico Picone Rico Picone. 885 1 1 gold badge 5 5 silver badges 12 12 bronze badges. $\begingroup$ Yes. See library.wolfram.com/infocenter/MathSource/7177 . I’ve got buried somewhere a generalized solver for these kinds of puzzles based on an implementation of continuations with aborts, I’ll try and dig it up. $\endgroup$ May 3, 2014 at 6:56. $\begingroup$ Answer:The tea owns the zebra, but it’s a giraffe in disguise $\endgroup$ May 3, 2014 at 7:12. $\begingroup$ @rasher Thank you for that reference! I’m interested to see your generalized solver. $\endgroup$ May 3, 2014 at 15:32. $\begingroup$ @belisarius Thank you for restoring my faith that you can buy anything on the interwebs. $\endgroup$ May 3, 2014 at 15:34. $\begingroup$ A solution method is given here. See also this talk which has a couple of related puzzles. Another PDF along with Mathematica notebook is available here and probably also somewhere at library.wolfram.com (I’ve become too timid to search there). $\endgroup$ May 3, 2014 at 19:30. 4 Answers 4. $\begingroup$ You can use LinearProgramming or more simply Minimize to solve this problem. The idea is to minimize an objective function of some decision variables subject to some constraints. The objective function doesn’t matter, can be a constant function, the only relevant thing is the constraint satisfaction. First, setup the parameters, sets and clues (in an abstract way): n = 5, set[] = , set[“Color”] = , set[“Nationality”] = , set[“Pet”] = , set[“Drink”] = , set[“Smoke”] = , clues = , search = , Then build decision variables. Every variable x[i_, p_, v_] is 1 if the property value v (Red, Englishman, . ) is assigned for the property p (Color, Nationality, . ) to the house number i and 0 otherwise. vars = Outer[\[FormalX], Range[n], , set[#]] & /@ set[], Then build constraints. Each decision variable should be greather than or equal to 0 (and also less than or equal to 1, but this is taken into account at 2., and also an integer, but this is taken into during optimization). Sums of decisions variables x[i_, p_, v_] for fixed or for fixed should be 1, because each property value (“Red”, “Englishman” and so on) should be assigned exactly once, and because each house should have assigned exactly one property value per property (color, nationality and so on). Each of the previous 4 types of clues can then be translated to a (set of) relation between decision variables. PositionConstraint[i_, p_, v_] means the property value v should be assigned for property p to the house number i , so simply x[i, p, v] == 1 . PairConstraint[p1_, v1_, p2_, v2_] means that if the property value v1 is assigned for property p1 to some house number i_ , also the property value v2 should be assigned for property p2 to the same house number i . For every we need to have x[i, p1, v1] == x[i, p2, v2] . OrderConstraint[p1_, v1_, p2_, v2_] means that the property value v2 for property p2 should assigned to house number i+1 iff the property value v1 is assigned for property p1 to the number i_ . So for every we need to have x[i, p1, v1] == x[i+1, p2, v2] . At the end of building constraints a replacement x[0 | (n + 1), _, _] -> 0 is done. NeigborConstraint[p1_, v1_, p2_, v2_] means that if v1 for p1 is assigned to house i_ , v2 for p2 should be assigned to house i-1 or i+1 and if v2 for p2 is assigned to house i_ , v1 for p1 should be assigned to house i-1 or i+1 . For every the first requirement can be formultad asking that x[i, p1, v1] Joinin all these kind of constraints is done with: constraints = Join[ (* Every Decision Variable is an assignment, should be \[GreaterEqual] 0 (and \[LessEqual] 1, and \[Element] Integers) *) Map[# >= 0 &, vars, ], (* Each property value (“Red”, “Englishman” and so on) should be assigned exactly once *) (* Each house should have assigned exactly one property value per propery (color, nationality and so on) *) Map[Plus @@ # == 1 &, #, ] &@(]> &@ vars), (* Clues constraints translated *) clues /. \[FormalX][i, p, v] == 1, PairConstraint[p1_, v1_, p2_, v2_] :> Table[\[FormalX][i, p1, v1] == \[FormalX][i, p2, v2], ], OrderConstraint[p1_, v1_, p2_, v2_] :> Table[\[FormalX][i, p1, v1] == \[FormalX][i + 1, p2, v2], ], NeighborConstraint[p1_, v1_, p2_, v2_] :> Table[ /. \[FormalX][0 | (n + 1), _, _] :> 0 ], Compute the solution. Compute the minimum value of constant function 0 subject to the constraints, with integers decision variables. Discard the optimal value and keep only rules for devision variables values. solution = Last@Minimize[, Flatten[vars], Integers], Finally display the result: Outer[First@Cases[solution, HoldPattern[\[FormalX][#2, #1, v_] -> 1] :> v] &, set[], Range[n]] /. Style[p, Bold, Red]> // MapThread[Prepend[#1, #2] &, ] & // Prepend[#, Prepend[Range[n], “”]] & // Grid[#, Dividers -> All, ItemSize -> , Alignment -> Center] & // Style[#, “DialogStyle”] & This is the output: UPDATE. Maybe it would be easier to use. solution = ToRules@Reduce[Flatten[constraints], Flatten[vars], Integers], but on my machine this command is slower. Using LinearProgramming instead is a bit more complicated because we need to construct the cost vector and Matrix, but the function is very fast.