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Examples for Discrete Constraint Programming

Examples for Discrete Constraint Programming. Belaid MOA UVIC, SHRINC Project. Examples. Map coloring Problem Cryptarithmetic N-queens problem Magic sequence Magic square Zebra puzzle Uzbekian puzzle A tiny transportation problem Knapsack problem graceful labeling problem.

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Examples for Discrete Constraint Programming

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  1. Examples for Discrete Constraint Programming Belaid MOA UVIC, SHRINC Project

  2. Examples • Map coloring Problem • Cryptarithmetic • N-queens problem • Magic sequence • Magic square • Zebra puzzle • Uzbekian puzzle • A tiny transportation problem • Knapsack problem • graceful labeling problem

  3. Map Coloring Problem (MCP) • Given a map and given a set of colors, the problem is how to color the map so that the regions sharing a boundary line don’t have the same color.

  4. Map Coloring Problem (MCP) • MCP can be viewed as a graph coloring problem: a b b c a d c d

  5. Modeling MCP • MCP can be Modeled as a CSP: • A set of variables representing the color of each region • The domain of the variables is the set of the colors used. • A set of constraints expressing the fact that countries that share the same boundary are not colored with the same color.

  6. Modeling (MCP) b d a c e f g • Example - Australia • Variables = {a,b,c,d,e,f,g} • Domain = {red, blue, green} • Constraints: • a!=b, a!=c • b!=c, b!=d • c!=e, c!=f • e!=d, e!=f

  7. Coding MCP • OPL studio - Ilog Solver: 54 solutions found enum Country={a,b,c,d,e,f,g}; enum Color = {red, green, blue}; var Color color[Country]; solve{ color[a]<>color[b];color[a]<>color[c]; color[b]<>color[c];color[b]<>color[d]; color[c]<>color[e];color[c]<>color[f]; color[e]<>color[d];color[e]<>color[f]; color[b]<>color[c];color[b]<>color[d]; };

  8. Coding MCP

  9. Coding MCP • ECLiPSe :- lib(fd). coloured(Countries) :- Countries=[A,B,C,D,E,F,G], Countries :: [red,green,blue], ne(A,B), ne(A,C), ne(B,C), ne(B,D), ne(C,E), ne(C,F), ne(E,D), ne(E,F), labeling(Countries). ne(X,Y) :- X##Y.

  10. Coding MCP

  11. Cryptarithmetic • The problem is to find the digits corresponding to the letters involved in the following puzzle: SEND + MORE MONEY

  12. Cryptarithmetic Modeling • A CSP model for Cryptarithmetic problem: • Variables={S,E,N,D,M,O,R,Y} • Domain={0,1,2,3,4,5,6,7,8,9} • Constraints • The variables are all different • S!=0, M!=0 • S*103+E*102+N*10+D {SEND} + M*103+O*102+R*10+E {MORE}= M*104+O*103+N*102+E*10+Y {MONEY}

  13. Coding Cryptarithmetic • OPL Studio - ILog Solver enum letter = {S,E,N,D,M,O,R,Y}; range Digit 0..9; var Digit value[letter]; solve{ alldifferent(value); value[S]<>0; value[M] <>0; value[S]*1000+value[E]*100+value[N]*10+value[D]+ value[M]*1000+value[O]*100+value[R]*10+value[E] = value[M]*10000+value[O]*1000+value[N]*100+value[E]*10+value[Y] };

  14. Coding Cryptarithmetic

  15. Coding Cryptarithmetic • ECLiPSe :- lib(fd). sendmore(Digits) :- Digits = [S,E,N,D,M,O,R,Y], Digits :: [0..9], alldifferent(Digits), S #\= 0, M #\= 0, 1000*S + 100*E + 10*N + D + 1000*M + 100*O + 10*R + E #= 10000*M + 1000*O + 100*N + 10*E + Y, labeling(Digits).

  16. Coding Cryptarithmetic

  17. N-Queens Problem • The problem is to put N queens on a board of NxN such that no queen attacks any other queen. • A queen moves vertically, horizontally and diagonally

  18. Modeling N-Queens Problem • The queens problem can be modeling via the following CSP • Variables={Q1,Q2,Q3,Q4,...,QN}. • Domain={1,2,3,…,N} represents the column in which the variables can be. • Constraints • Queens not on the same row: already taken care off by the good modeling of the variables. • Queens not on the same column: Qi != Qj • Queens not on the same diagonal: |Qi-Qj| != |i-j|

  19. Coding N-Queens Problem • OPL Studio - ILog Solver int n << "number of queens:"; range Domain 1..n; var Domain queens[Domain]; solve { alldifferent(queens); forall(ordered i,j in Domain) { abs(queens[i]-queens[j])<> abs(i-j) ; }; };

  20. Coding N-Queens Problem

  21. Coding N-Queens Problem • ECLiPSe :-lib(fd). nqueens(N, Q):- length(Q,N), Q::1..N, alldifferent(Q), ( fromto(Q, [Q1|Cols], Cols, []) do ( foreach(Q2, Cols), param(Q1), count(Dist,1,_) do Q2 - Q1 #\= Dist, Q1 - Q2 #\= Dist ) ), search(Q). search([]). search(Q):- deleteff(Var,Q,R), indomain(Var), search(R).

  22. Coding N-Queens Problem

  23. Magic sequence problem • Given a finite integer n, the problem consists of finding a sequence S = (s0,s1,…,sn), such that si represents the number of occurrences of i in S. • Example: • (2, 0, 2, 0) • (1,2,1,0)

  24. Modeling MSP • MSP can be modeled by the following CSP • variables: s0,s1, …,sn-1 • Domain:{0,1,…,n} • constraints: • number of occurences of i in (s0,s1, …,sn-1) is si. • Redundant constraints: • the sum of s0,s1, …,sn-1 is n • the sum of i*Si, i in [0..n-1] is n

  25. Coding MSP • OPL - ILog Solver int n << "Number of Variables:"; range Range 0..n-1; range Domain 0..n; int value[i in Range ] = i; var Domain s[Range]; solve { distribute(s,value,s); //global constraint s[i]=sum(j in Range)(s[j]=i) sum(i in Range) s[i] = n; //redundant constraint sum(i in Range) s[i]*i = n; //redundant constraint };

  26. Coding MSP • ECLiPSe :- lib(fd). :- lib(fd_global). :- lib(fd_search). solve(N, Sequence) :- length(Sequence, N), Sequence :: 0..N-1, ( for(I,0,N-1), foreach(Xi, Sequence), foreach(I, Range), param(Sequence) do occurrences(I, Sequence, Xi) ), N #= sum(Sequence), % two redundant constraints N #= Sequence*Range, search(Sequence, 0, first_fail, indomain, complete, []).%search procedure

  27. Magic square problem • A magic square of size N is an NxN square filled with the numbers from 1 to N2 such that the sums of each row, each column and the two main diagonals are equal. • Example:

  28. Modeling Magic square pb • Magic square problem can be viewed as a CSP with the following properties: • Variables: the elements of the matrix representing the square • Domain: 1..N*N • Constraints: • magic sum = sum of the columns = sum of the rows = sum of the down diagonal = sum of the up diagonal • Remove symmetries • Redundant constraint: • magic sum = N(N2+1)/2

  29. Coding Magic square pb • OPL Studio: int N << "The length of the square:"; range Dimension 1..N; range Values 1..N*N; int msum = N*(N*N+1)/2; //magic sum var Values square[1..N,1..N]; solve { //The elements of the square are all different alldifferent(square); //the sum of the diagonal is magic sum msum = sum(i in Dimension)square[i,i]; //the sum of the up diagonal is magic sum

  30. Coding Magic square pb msum = sum(i in Dimension)square[i,N-i+1]; //the sum of the rows and columns are all magic sum forall(i in Dimension){ msum = sum(j in Dimension)square[i,j]; msum = sum(k in Dimension)square[k,i]; }; //remove symmetric behavior of the square square[N,1] < square[1,N]; square[1,1] < square[1,N]; square[1,1] < square[N,N]; square[1,1] < square[N,1]; };

  31. Magic square problem

  32. Coding Magic square pb • ECLiPSe: :- lib(fd). magic(N) :- Max is N*N, Magicsum is N*(Max+1)//2, dim(Square, [N,N]), Square[1..N,1..N] :: 1..Max, Rows is Square[1..N,1..N], flatten(Rows, Vars), alldifferent(Vars),

  33. Coding Magic square pb %constraints on rows ( for(I,1,N), foreach(U,UpDiag), foreach(D,DownDiag), param(N,Square,Sum) do Magicsum #= sum(Square[I,1..N]), Magicsum #= sum(Square[1..N,I]), U is Square[I,I], D is Square[I,N+1-I] ), %constraints on diagonals Magicsum #= sum(UpDiag), Magicsum #= sum(DownDiag),

  34. Coding Magic square pb %remove symmetry Square[1,1] #< Square[1,N], Square[1,1] #< Square[N,N], Square[1,1] #< Square[N,1], Square[1,N] #< Square[N,1], %search labeling(Vars), print(Square). Unfortunately, ECLiPSe ran for a long time without providing any answer. ECLiPSe had also the same behavior for a similar code from ECLiPSe website.

  35. A Tiny Transportation problem x 11 x x 12 1n x 2 21 a 2 b x 2 2 22 x 2n • How much should be shipped from several sources to several destinations Demand Qty Supply cpty Source Qty Shipped Destination a 1 b 1 1 1 : : : : n a m b m n

  36. A Tiny Transportation problem 2 2 3 • 3 plants with known capacities, 4 clients with known demands and transport costs per unit between them. 200 500 1 1 400 Find qty shipped? 300 300 3 400 4 100

  37. Modeling TTP • TTP can be modeled by a CSP with optimization as follows: • Variables: A1,A2,A3,B1,B2,B3,C1, C2, C3,D1,D2,D3 • Domain: 0.0..Inf • constraints: • Demand constraints: A1+A2+A3=200; B1+B2+B3=400; C1+C2+C3=300; D1+D2+D3=100; • Capacity constraints: A1+B1+C1+D1500; A2+B2+C2+D2300; A3+B3+C3+D3400; • Minimization of the objective function: 10*A1 + 7*A2 + 11*A3 + 8*B1 + 5*B2 + 10*B3 + 5*C1 + 5*C2 + 8*C3 + 9*D1 + 3*D2 + 7*D3

  38. Coding TTP • OPL Studio - Cplex Solver var float+ productA[Range];var float+ productB[Range]; var float+ productC[Range]; var float+ productD[Range]; minimize 10*productA[1] + 7*productA[2] + 11*productA[3] + 8*productB[1] + 5*productB[2] + 10*productB[3] + 5*productC[1] + 5*productC[2] + 8*productC[3] + 9*productD[1] + 3*productD[2] + 7*productD[3] subject to { productA[1] + productA[2] + productA[3] = 200; productB[1] + productB[2] + productB[3] = 400; productC[1] + productC[2] + productC[3] = 300; productD[1] + productD[2] + productD[3] = 100; productA[1] + productB[1] + productC[1] + productD[1]<=500 productA[2] + productB[2] + productC[2]+productC[2] <= 300; productA[3] + productB[3] + productC[3] + productD[3] <= 400; };

  39. Coding TTP • OPL Studio - Cplex Solver: Solution found Optimal Solution with Objective Value: 6200.0000 productA[1] = 100.0000 productA[2] = 0.0000 productA[3] = 100.0000 productB[1] = 100.0000 productB[2] = 300.0000 productB[3] = 0.0000 productC[1] = 300.0000 productC[2] = 0.0000 productC[3] = 0.0000 productD[1] = 0.0000 productD[2] = 100.0000 productD[3] = 0.0000

  40. Coding TTP • ECLiPSe :- lib(eplex_cplex). main1(Cost, Vars) :- Vars = [A1, A2, A3, B1, B2, B3, C1, C2, C3, D1, D2, D3], Vars :: 0.0..inf, A1 + A2 + A3 $= 200, B1 + B2 + B3 $= 400, C1 + C2 + C3 $= 300, D1 + D2 + D3 $= 100, A1 + B1 + C1 + D1 $=< 500, A2 + B2 + C2 + D2 $=< 300, A3 + B3 + C3 + D3 $=< 400, optimize(min(10*A1 + 7*A2 + 11*A3 + 8*B1 + 5*B2 + 10*B3 + 5*C1 + 5*C2 + 8*C3 +9*D1 + 3*D2 + 7*D3), Cost). It didn’t run! calling an undefined procedure cplex_prob_init(…)

  41. Zebra puzzle • Zebra is a well known puzzle in which five men with different nationalities live in the first five house of a street. They practice five distinct professions, and each of them has a favorite animal and a favorite drink, all of them different. The five houses are painted in different colors. The puzzle is to find who owns Zebra. • Zebra puzzle has different statements. We’ll handle two. • This kind of Puzzle is usually treated as an instance of the class of tabular constraint satisfaction problems in which we express the problem using tables. • In this presentation we show how to solve it using CSP languages: OPL studio and ECLiPSe.

  42. Modeling Zebra puzzle • In crucial step in Modeling Zebra puzzle is finding the decision variables. In our case, we consider: • variables: persons, colors, pets, drinks, tobaccos. • Domain: 1..5 (representing the five houses) • Constraints: Expressed directly from the statement of the puzzle.

  43. Coding Zebra puzzle • OPL Studio - ILog Solver enum people {Englishman, Spaniard, Ukrainian, Norwegian, Japanese}; enum drinks {coffee, tea, milk, orange, water}; enum pets {dog, fox, zebra, horse, snails}; enum tabacco {winston, kools, chesterfields, luckyStrike, parliaments }; enum colors {ivory, yellow, red , green, blue}; range houses 1..5;//{first, second, third, fourth, fifth}; var houses hseClr[colors]; var houses hsePple[people]; var houses hsePts[pets]; var houses hseDrks[drinks]; var houses hseTaba[tabacco];

  44. Coding Zebra puzzle solve{ hsePple[Englishman] = hseClr[red]; hsePple[Spaniard]=hsePts[dog]; hseDrks[coffee] = hseClr[green]; hsePple[Ukrainian] = hseDrks[tea]; hseClr[green] = hseClr[ivory]+1; hseTaba[winston] = hsePts[snails]; hseTaba[kools]=hseClr[yellow]; hseDrks[milk]=3; hsePple[Norwegian] = 1; hseTaba[chesterfields]=hsePts[fox]+1 \/ hseTaba[chesterfields]=hsePts[fox]-1; hseTaba[kools] = hsePts[horse]+1 \/ hseTaba[kools] = hsePts[horse]-1 ; hseTaba[ luckyStrike] = hseDrks[orange]; hsePple[ Japanese] = hseTaba[parliaments]; hsePple[Norwegian] = hseClr[blue]+1 \/ hsePple[Norwegian] = hseClr[blue]-1; //slient constraints-Global constraint alldifferent alldifferent(hsePple);//forall(ordered i,j in people) hsePple[i] <> hsePple[j]; alldifferent(hsePts);//forall(ordered i,j in pets) hsePts[i] <> hsePts[j]; alldifferent(hseClr);//forall(ordered i,j in colors) hseClr[i] <> hseClr[j]; alldifferent(hseDrks);//forall(ordered i,j in drinks) hseDrks[i] <> hseDrks[j]; alldifferent(hseTaba);//forall(ordered i,j in tabacco) hseTaba[i] <> hseTaba[j]; };

  45. Coding Zebra puzzle % The Englishman lives in a red house. % The Spaniard owns a dog. % The Japanese is a painter. % The Italian drinks tea. % The Norwegian lives in the first house on the left. % The owner of the green house drinks coffee. % The green house is on the right of the white one. % The sculptor breeds snails. % The diplomat lives in the yellow house. % Milk is drunk in the middle house. % The Norwegian's house is next to the blue one. % The violinist drinks fruit juice. % The fox is in a house next to that of the doctor. % The horse is in a house next to that of the diplomat. % % Who owns a Zebra, and who drinks water? %

  46. Coding Zebra puzzle :- lib(fd). zebra :- % we use 5 lists of 5 variables each Nat = [English, Spaniard, Japanese, Italian, Norwegian], Color = [Red, Green, White, Yellow, Blue], Profession = [Painter, Sculptor, Diplomat, Violinist, Doctor], Pet = [Dog, Snails, Fox, Horse, Zebra], Drink = [Tea, Coffee, Milk, Juice, Water], % domains: all the variables range over house numbers 1 to 5 Nat :: 1..5, Color :: 1..5, Profession :: 1..5, Pet :: 1..5, Drink :: 1..5,

  47. Coding Zebra puzzle % the values in each list are exclusive alldifferent(Nat), alldifferent(Color), alldifferent(Profession), alldifferent(Pet), alldifferent(Drink), % and here follow the actual constraints English = Red, Spaniard = Dog, Japanese = Painter, Italian = Tea, Norwegian = 1, Green = Coffee, Green #= White + 1, Sculptor = Snails, Diplomat = Yellow, Milk = 3, Dist1 #= Norwegian - Blue, Dist1 :: [-1, 1], Violinist = Juice, Dist2 #= Fox - Doctor, Dist2 :: [-1, 1], Dist3 #= Horse - Diplomat, Dist3 :: [-1, 1],

  48. Coding Zebra puzzle % put all the variables in a single list flatten([Nat, Color, Profession, Pet, Drink], List), % search: label all variables with values labeling(List), % print the answers: we need to do some decoding NatNames = [English-english, Spaniard-spaniard, Japanese-japanese, Italian-italian, Norwegian-norwegian], memberchk(Zebra-ZebraNat, NatNames), memberchk(Water-WaterNat, NatNames), printf("The %w owns the zebra%n", [ZebraNat]), printf("The %w drinks water%n", [WaterNat]). Answer from ECLiPSe after 0.02s The japanese owns the zebra The norwegian drinks water

  49. Uzbekian Puzzle • An uzbekian sales man met five traders who live in five different cities. The five traders are: • {Abdulhamid, Kurban,Ruza, Sharaf, Usman} The five cities are : • {Bukhara, Fergana, Kokand, Samarkand, Tashkent} Find the order in which he visited the cities given the following information: • He met Ruza before Sharaf after visiting Samarkand, • He reached Fergana after visiting Samarkand followed by other two cities, • The third trader he met was Tashkent, • Immediately after his visit to Bukhara, he met Abdulhamid • He reached Kokand after visiting the city of Kurban followed by other two cities;

  50. Modeling Uzbekian Puzzle • The uzbekian puzzle can formulated within the CSP framework as follows: • Variables: order in which he visited each city and met each trader • Domain:1..5 • constraints: • He met Ruza before Sharaf after visiting Samarkand, • He reached Fergana after visiting Samarkand followed by other two cities, • The third trader he met was Tashkent, • Immediately after his visit to Bukhara, he met Abdulhamid • He reached Kokand after visiting the city of Kurban followed by other two cities;

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