Topic 3: MiniZinc (Version of 10th November 2015)

Transcription

1 Topic 3: (Version of 10th November 2015) Jean-Noël Monette ASTRA Research Group on Combinatorial Optimisation Uppsala University Sweden Course 1DL449: for Combinatorial Optimisation

2 Outline Course 1DL for Combinatorial Optimisation

3 Overview is a declarative language to model combinatorial problems using constraints. supports the separation of the model the data. A couple (model, data) is called an instance. A instance is solved by: 1 Compiling the instance into a FlatZinc model. 2 Solving the FlatZinc model using one of many existing backend solvers. Course 1DL for Combinatorial Optimisation

4 Model A model is comprised of the following items: Parameter declarations Variable declarations Predicate declarations Function declarations s Objective Output comm Course 1DL for Combinatorial Optimisation

5 Model A model is comprised of the following items: Parameter declarations Variable declarations Predicate declarations Function declarations s Objective Output comm Note: uses the word parameter for any constant identifier. We follow their terminology in this topic. Course 1DL for Combinatorial Optimisation

6 Types for Parameters is strongly typed offers the following types for parameters: int: integer bool: Boolean float: floating-point number (not covered further) string: string of characters set of τ: set of elements of type τ, with τ being int, bool, float, or string array[ρ] of τ: possibly multidimensional array of elements of type τ (anything except array). The range ρ is an integer interval l..u in each dimension. The declaration int: N declares a parameter of type integer with name N. One can also write par int: N to stress that it is a parameter. Course 1DL for Combinatorial Optimisation

7 Types for Variables Variables are declared by prefixing var to the type, e.g., var int: N declares a variable N that must be assigned an integer value. The only variable types are: int bool float (not covered further) set of int One can also have arrays of elements of any variable type. Course 1DL for Combinatorial Optimisation

8 Subtyping Parameters (par) can be used whenever a var is expected. This extends to arrays: A function expecting an array[int] of var int can be passed an array[int] of int. The types bool int are two distinct types. One can coerce from bool to int using the function bool2int. This coercion is automatic but, for clarity, should be made explicit. Course 1DL for Combinatorial Optimisation

9 Option Variables supports option variables. An option variable is a variable that can also take the special value. A variable is declared optional with the keyword opt. For example, var opt 1..4: x; declares a variable that can take values in {1, 2, 3, 4} { }. We will not cover the use of option variables in this course. However, one can meet them: in the documentation: var int is a subtype of var opt int, in error messages: this is probably a sign that your model is too complicated. Course 1DL for Combinatorial Optimisation

10 Declaring Parameters Variables 1 int: N = 4; 2 int: M; 3 M = 10; 4 set of int: prime = {2,3,5,7,11,13}; 5 var int: X; 6 var 0..59: minute = X + N; 7 var set of prime: taken; All parameters must be given a fixed value, but declaration assignment can be separated. Variables can be constrained at declaration. The domain of a variable can be restricted by replacing the type declaration by a set of values of that type: X can take any integer value. minute must take an integer value between taken must be a subset of {2,3,5,7,11,13}. Course 1DL for Combinatorial Optimisation

11 Literals Boolean: true false Integers: In decimal, hexadecimal, or octal format Sets: Between braces: {1,3,5}, or as intervals Arrays: Between square brackets: [6,3,1,7] 2D arrays: A vertical bar is used before, between, after each row: [ 11,12,13,14 21,22,23,24 31,32,33,34 ] No special literals for arrays of higher dimensions. Note: The indices of arrays start at 1 by default. Course 1DL for Combinatorial Optimisation

12 Array Set Comprehensions Arrays sets can be built with comprehensions using the notations [σ γ] {σ γ} where σ is an expression evaluated for each element generated by the generator γ. The generator introduces one or more identifiers with values drawn from one or more sets of integers, optionally followed by a test. 1 [x * 2 x in 1..8] = [2,4,6,8,10,12,14,16] 2 [x + 2*y x in 1..3, y in 1..4] = [3,5,7,9,4,6,8,10,5,7,9,11] 3 [x * y x, y in 1..3 where x < y] = [2,3,6] 4 {x + 2*y x in 1..3, y in 1..4} = {3,4,5,6,7,8,9,10,11} Course 1DL for Combinatorial Optimisation

13 Iteration Syntactic Sugar sum(i, j in 1..5)(i*j) is syntactic sugar for: sum([i*j i, j in 1..5]) This works for any function or predicate that takes an array as unique argument. Course 1DL for Combinatorial Optimisation

14 Iteration Syntactic Sugar sum(i, j in 1..5)(i*j) is syntactic sugar for: sum([i*j i, j in 1..5]) This works for any function or predicate that takes an array as unique argument. In particular: forall(i in 1..9)(x[i+1] = x[i] + y[i]) is syntactic sugar for: forall([x[i+1] = x[i] + y[i] i in 1..9]) forall(array[int] of var bool: b) is a predicate that is true if all the variables in b are true. Course 1DL for Combinatorial Optimisation

15 Array Manipulation Changing array dimensions ranges (the numbers of elements must match): array1d(5..10,[ 3,2 5,4 6,1 ]) Indices start at 5. array2d(1..2,1..3,[2,7,3,7,4,9]) Tip: Try to keep your ranges starting at 1. It is simpler to read a model that follows the usual convention. There has been some subtle bugs when arrays are not indexed from 1. Concatenation: [1,2] ++ [3,4] Course 1DL for Combinatorial Optimisation

16 s A constraint is the constraint keyword followed by a Boolean expression that must be true in any solution. constraint x < y; constraint sum(q) = 0 /\ alldifferent(q); constraint if x < y then x = y else x > y endif; Course 1DL for Combinatorial Optimisation

17 s A constraint is the constraint keyword followed by a Boolean expression that must be true in any solution. constraint x < y; constraint sum(q) = 0 /\ alldifferent(q); constraint if x < y then x = y else x > y endif; What does the following mean? constraint x = x + 1; Course 1DL for Combinatorial Optimisation

18 s A constraint is the constraint keyword followed by a Boolean expression that must be true in any solution. constraint x < y; constraint sum(q) = 0 /\ alldifferent(q); constraint if x < y then x = y else x > y endif; What does the following mean? constraint x = x + 1; Minizinc is a declarative language: this line is unsatisfiable. Course 1DL for Combinatorial Optimisation

19 Objective The solve item gives the objective of the problem: solve satisfy; The problem is a satisfaction problem. solve minimize x; The objective is to minimise the value of variable x. solve maximize abs(x)* y; The objective is to maximise the value of expression abs(x)* y. Course 1DL for Combinatorial Optimisation

20 Objective The solve item gives the objective of the problem: solve satisfy; The problem is a satisfaction problem. solve minimize x; The objective is to minimise the value of variable x. solve maximize abs(x)* y; The objective is to maximise the value of expression abs(x)* y. does not have direct support for multi-objective optimisation: either multiple objectives must be aggregated as a weithted sum, or the multi-objective nature must be hled outside of. Course 1DL for Combinatorial Optimisation

21 Output The output item describes what must be printed upon finding a solution. The keyword output is followed by an array of strings. output [show(x)]; output ["solution:"] ++ [if x[i] > 0 then show(2 * x[i]) ++ ", " else "_, " endif i in 1..10]; The function show returns a string representing the value of the given expression. The operator ++ concatenates two strings or two arrays. "x = \(x), " is equivalent to "x = "++show(x)++", ". Course 1DL for Combinatorial Optimisation

22 Tests Conditional: if θ then φ 1 else φ 2 endif Generator: [x x in ρ where θ ] The expressions φ 1 φ 2 must have the same type. Note that the Boolean expression θ can depend on the values of variables. With great power comes great responsibility. (see Section 4) Course 1DL for Combinatorial Optimisation

23 Operators Boolean: not, /\, \/, <->, ->, <-, xor, forall, exists, xorall, iffall, clause, bool2int Integer: +, -, *, div, mod, abs, pow, min, max, sum, product, =, <, <=, =>, >,!=,.. Sets: union, intersect, diff, symdiff, card, in, subset, superset, set2array, array_union, array_intersect Strings: ++, concat, join Arrays: length, index_set, index_set_1of2, index_set_2of2,..., index_set_6of6, array1d, array2d,..., array6d Course 1DL for Combinatorial Optimisation

24 remarks The order of model items does not matter. One can include other files: include "globals.mzn"; The following functions are useful for debugging models: Assert: assert(θ,"error message", φ). If the constant expression θ is false, aborts with the error message, otherwise returns φ. Trace: trace("message", φ). Prints the message returns φ. Course 1DL for Combinatorial Optimisation

25 Outline Course 1DL for Combinatorial Optimisation

26 offers a large collection of predefined functions predicates to increase the level at which models are described (see Topic 4: s). It is also possible for the modellers to define their own functions predicates. Course 1DL for Combinatorial Optimisation

27 Function Predicate Declarations 1 function int: double(int: x) ; 2 function var int: double(var int: x) ; 3 4 predicate positive(var int: x) ; 5 function var bool: negative(var int: x); are functions returning a var bool. Function ( hence predicate) names can be overloaded. Course 1DL for Combinatorial Optimisation

28 Function Definitions Akin to functional programming, the body of a function is an expression of the same type as the returned value. 1 function int: double(int: x) = 2 * x; 2 function var int: double(var int: x) = 2 * x; 3 4 predicate positive(var int: x) = x > 0; 5 function var bool: negative(var int: x)= x < 0; Possible to use if then else endif, forall, exists. Course 1DL for Combinatorial Optimisation

29 Let Expressions One can introduce local identifiers with a let expression constrain them. 1 function int: double(int: x) = 2 let { int: y = 2 * x; } in y; 3 4 function var int: double(var int: x) = 5 let { var int: y = 2 * x; } in y; 6 7 function var int: double(var int: x) = 8 let { var int: y; 9 constraint y = 2 * x; 10 } in y ; The second third functions are equivalent each invocation introduces a new (existentially quantified) variable into the model. Course 1DL for Combinatorial Optimisation

30 s in Let Expressions What is the difference between the following two definitions? 1 predicate positiveprod(var int: x, var int: y)= 2 let { var int: z; constraint z = x * y; 3 } in z > 0; 4 5 predicate positiveprod(var int: x, var int: y)= 6 let { var int: z; 7 } in z = x * y /\ z > 0; Course 1DL for Combinatorial Optimisation

31 s in Let Expressions What is the difference between the following two definitions? 1 predicate positiveprod(var int: x, var int: y)= 2 let { var int: z; constraint z = x * y; 3 } in z > 0; 4 5 predicate positiveprod(var int: x, var int: y)= 6 let { var int: z; 7 } in z = x * y /\ z > 0; Their behaviour is different in a negative context, e.g., not positiveprod(a,b): The first one then ensures a * b = z /\ z <= 0 The second one then ensures a * b!= z \/ z <= 0 The second one leaves a b unconstrained. Course 1DL for Combinatorial Optimisation

32 Using There are many advantages to using functions predicates in a model: Software engineering good practice: Reusability Readability Modularity The model might be solved more efficiently: Better common subexpression elimination Definitions can be solver-specific Course 1DL for Combinatorial Optimisation

33 Outline Course 1DL for Combinatorial Optimisation

34 Some examples by Guido Tack Course 1DL for Combinatorial Optimisation

35 Zinc Essence Essence OPL see IBM website AIMMS GAMS AMPL Mosel-Xpress see FICO website SMT-LIB Course 1DL for Combinatorial Optimisation

36 Outline Course 1DL for Combinatorial Optimisation

37 Solving a Model Two phases: 1 Compile (or flatten) the model into a solver-specific FlatZinc model 2 Interpret the FlatZinc model with a constraint solver, also called backend FlatZinc is a low-level subset of : No quantifiers or array comprehensions No function calls Only built-in constraint predicates of the targeted solver Only variables of types supported by the targeted solver s only on (arrays of) variables parameters Course 1DL for Combinatorial Optimisation

38 Unroll all quantifiers array comprehensions Inline all function predicate calls Evaluate all expressions not depending on the value of a variable Replace unsupported variable types Decompose complex expressions, introducing new variables Course 1DL for Combinatorial Optimisation

39 Unroll all quantifiers array comprehensions Inline all function predicate calls Evaluate all expressions not depending on the value of a variable Replace unsupported variable types Decompose complex expressions, introducing new variables produces solver-specific models: Not all solvers have the same set of built-in constraint predicates Different solvers may use different function constraint predicate definitions. Course 1DL for Combinatorial Optimisation

40 : Examples 1 constraint forall(i in 1..4)( 2 x[i] = i 3 ); Course 1DL for Combinatorial Optimisation

41 : Examples 1 constraint forall(i in 1..4)( 2 x[i] = i 3 ); 1 constraint forall([x[i] = i i in 1..4]); Course 1DL for Combinatorial Optimisation

42 : Examples 1 constraint forall(i in 1..4)( 2 x[i] = i 3 ); 1 constraint forall([x[i] = i i in 1..4]); 1 constraint forall([x[1]=1,x[2]=2,x[3]=3,x[4]=4]); Course 1DL for Combinatorial Optimisation

43 : Examples 1 constraint forall(i in 1..4)( 2 x[i] = i 3 ); 1 constraint forall([x[i] = i i in 1..4]); 1 constraint forall([x[1]=1,x[2]=2,x[3]=3,x[4]=4]); 1 constraint x[1]=1 /\ x[2]=2 /\ x[3]=3 /\ x[4]=4; Course 1DL for Combinatorial Optimisation

44 : Examples 1 constraint forall(i in 1..4)( 2 x[i] = i 3 ); 1 constraint forall([x[i] = i i in 1..4]); 1 constraint forall([x[1]=1,x[2]=2,x[3]=3,x[4]=4]); 1 constraint x[1]=1 /\ x[2]=2 /\ x[3]=3 /\ x[4]=4; 1 constraint x[1]=1; 2 constraint x[2]=2; 3 constraint x[3]=3; 4 constraint x[4]=4; Course 1DL for Combinatorial Optimisation

45 : Examples (2) 1 int: x = 4; 2 var 0..10: y; 3 constraint if x < 8 then y < x 4 else y = x + 1 endif; Course 1DL for Combinatorial Optimisation

46 : Examples (2) 1 int: x = 4; 2 var 0..10: y; 3 constraint if x < 8 then y < x 4 else y = x + 1 endif; 1 var 0..10: y; 2 constraint if 4 < 8 then y < 4 3 else y = endif; Course 1DL for Combinatorial Optimisation

47 : Examples (2) 1 int: x = 4; 2 var 0..10: y; 3 constraint if x < 8 then y < x 4 else y = x + 1 endif; 1 var 0..10: y; 2 constraint if 4 < 8 then y < 4 3 else y = endif; 1 var 0..10: y; 2 constraint y < 4; Course 1DL for Combinatorial Optimisation

48 : Examples (2) 1 int: x = 4; 2 var 0..10: y; 3 constraint if x < 8 then y < x 4 else y = x + 1 endif; 1 var 0..10: y; 2 constraint if 4 < 8 then y < 4 3 else y = endif; 1 var 0..10: y; 2 constraint y < 4; 1 var 0..3: y; Course 1DL for Combinatorial Optimisation

49 : Examples (3) 1 var 0..10: x; 2 var 0..10: y; 3 constraint if x < 8 then y < x 4 else y = x + 1 endif; Course 1DL for Combinatorial Optimisation

50 : Examples (3) 1 var 0..10: x; 2 var 0..10: y; 3 constraint if x < 8 then y < x 4 else y = x + 1 endif; 3 var bool: b1; 4 var bool: b2; 5 var bool: b3; 6 constraint b1 = (x < 8); 7 constraint b2 = (y < x); 8 constraint b3 = (y = x + 1); 9 constraint if b1 then b2 else b3 endif; Course 1DL for Combinatorial Optimisation

51 : Examples (3) 1 var 0..10: x; 2 var 0..10: y; 3 constraint if x < 8 then y < x 4 else y = x + 1 endif; 3 var bool: b1; 4 var bool: b2; 5 var bool: b3; 6 constraint b1 = (x < 8); 7 constraint b2 = (y < x); 8 constraint b3 = (y = x + 1); 9 constraint if b1 then b2 else b3 endif; 9 constraint b1 -> b2; 10 constraint (not b1) -> b3; Course 1DL for Combinatorial Optimisation

52 : Examples (3) 1 var 0..10: x; 2 var 0..10: y; 3 var bool: b1; 4 var bool: b2; 5 var bool: b3; 6 constraint b1 = (x < 8); 7 constraint b2 = (y < x); 8 constraint b3 = (y = x + 1); 9 constraint b1 -> b2; 10 constraint b1 \/ b3; Course 1DL for Combinatorial Optimisation

53 : Examples (3) 1 var 0..10: x; 2 var 0..10: y; 3 var bool: b1; 4 var bool: b2; 5 var bool: b3; 6 constraint b1 = (x < 8); 7 constraint b2 = (y < x); 8 constraint b3 = (y = x + 1); 9 constraint b1 -> b2; 10 constraint b1 \/ b3; 6 constraint int_lt_reif(x,8,b1); 7 constraint int_lt_reif(y,x,b2); 8 constraint int_lin_eq_reif([1,-1],[y,x],1,b3); 9 constraint bool_imply(b1,b2); 10 constraint array_bool_or([b1,b3],true); Course 1DL for Combinatorial Optimisation

54 Outline Course 1DL for Combinatorial Optimisation

55 MZN model FZN model MZN data mzn2fzn backend solution Course 1DL for Combinatorial Optimisation

56 MZN model MZN data MZN solution mzn2fzn FZN model output specification solns2out backend FZN solution Course 1DL for Combinatorial Optimisation

57 MZN model MZN data MZN solution PP definitions FZN model mzn2fzn output specification solns2out backend FZN solution PP: predefined predicates Course 1DL for Combinatorial Optimisation

58 MZN model MZN data MZN solution PP definitions for backend X mzn2fzn FZN model for backend X output specification solns2out backend X FZN solution PP: predefined predicates Course 1DL for Combinatorial Optimisation

59 MZN model MZN data MZN solution PP PPdefinitions for for backend GecodeX mzn2fzn FZN FZNmodel for forbackend GecodeX output specification solns2out backend fzn-gecode X FZN solution Course 1DL for Combinatorial Optimisation

60 MZN model PP PPdefinitions for for backend GecodeX FZN FZNmodel for forbackend GecodeX MZN data mzn2fzn output specification MZN solution mzn-gecode solns2out backend fzn-gecode X FZN solution Course 1DL for Combinatorial Optimisation

61 MZN model MZN data MZN solution PP PPdefinitions PPdefinitions for for backend GecodeX for fzn2smt mzn2fzn mzn2smt FZN FZN model FZNmodel for forbackend GecodeX fzn2smt output specification solns2out backend fzn-gecode X fzn2smt FZN solution Course 1DL for Combinatorial Optimisation

62 : Model Once, Solve Everywhere MZN model MZN data MZN solution PP PPdefinitions PPdefinitions forpp for backend definitions GecodeX for fzn2smt for... mzn2fzn mzn-... FZN FZN model FZNmodel forfzn forbackend model GecodeX fzn2smt for... output specification solns2out backend fzn-gecode X fzn2smt fzn-... FZN solution Course 1DL for Combinatorial Optimisation

63 Solvers programming CP: Gecode, SICStus Prolog, Choco, Google or-tools, JaCoP, Mistral,... Lazy-clause generation LCG: Opturion CPX, Chuffed, G12/CPX,... Mixed Integer programming MIP: G12/MIP, CPLEX, Gurobi,... Local search metaheuristics LS: OscaR/CBLS, EasyLocal++, YACS,... SAT: G12/SAT,... SAT modulo theories SMT: fzn2smt with Yices 2,... Hybrids: izplus, Minisat(ID),... Portfolios of solvers: sunny-cp,... Course 1DL for Combinatorial Optimisation

64 Solvers programming CP: Gecode, SICStus Prolog, Choco, Google or-tools, JaCoP, Mistral,... Lazy-clause generation LCG: Opturion CPX, Chuffed, G12/CPX,... Mixed Integer programming MIP: G12/MIP, CPLEX, Gurobi,... Local search metaheuristics LS: OscaR/CBLS, EasyLocal++, YACS,... SAT: G12/SAT,... SAT modulo theories SMT: fzn2smt with Yices 2,... Hybrids: izplus, Minisat(ID),... Portfolios of solvers: sunny-cp,... Solvers provided with the virtual machine are in red. Course 1DL for Combinatorial Optimisation

65 Example: Pigeonhole Problem Example (Pigeonhole) Place n pigeons into n 1 holes so that all pigeons are placed no two pigeons are placed in the same hole. This problem is trivially unfeasible. We will use this problem to show: how different solvers have different definitions for the same predicate; that it is often important for efficiency to use predefined predicates. Course 1DL for Combinatorial Optimisation

66 Pigeonhole Models Using ALLDIFFERENT 1 int: n; 2 array[1..n] of var 1..(n-1): hole; 3 constraint alldifferent(hole); 4 solve satisfy; Using binary disequalities 1 int: n; 2 array[1..n] of var 1..(n-1): hole; 3 constraint forall(p1 in 1..n, p2 in i+1..n) 4 (hole[p1]!= hole[p2]); 5 solve satisfy; Course 1DL for Combinatorial Optimisation

67 Predicate Definitions mzn-gecode predicate all_different_int (array[int] of var int: x); predicate int_ne(var int: x, var int: y); mzn2smt predicate all_different_int (array[int] of var int: x) = forall(i,j in index_set(x) where i < j) ( x[i]!= x[j] ); predicate int_ne(var int: x, var int: y); Course 1DL for Combinatorial Optimisation

68 mzn-g12mip predicate all_different_int (array[int] of var int: x) = let { array[int,int] of var 0..1: x_eq_d = eq_encode(x) } in forall(d in index_set_2of2(x_eq_d)) ( sum(i in index_set_1of2(x_eq_d)) ( x_eq_d[i,d] ) <= 1 ); predicate int_ne(var int: x, var int: y) = let { var 0..1: p } in x - y + 1 <= ub(x - y + 1) * (1 - p) /\ y - x + 1 <= ub(y - x + 1) * p; Course 1DL for Combinatorial Optimisation

69 mzn-g12mip (end) function array[int,int] of var int: eq_encode (array[int] of var int: x) = [...] predicate equality_encoding(var int: x, array[int] of var 0..1: x_eq_d) = x in index_set(x_eq_d) /\ sum(d in index_set(x_eq_d))(x_eq_d[d]) = 1 /\ sum(d in index_set(x_eq_d))(d*x_eq_d[d])=x; Course 1DL for Combinatorial Optimisation

70 Experimental Comparison Time in seconds to prove unsatisfiability: n = 10 ALLDIFFERENT disequalities mzn-gecode < 1 1 mzn-g12mip < mzn2smt Course 1DL for Combinatorial Optimisation

71 Experimental Comparison Time in seconds to prove unsatisfiability: n = 10 ALLDIFFERENT disequalities mzn-gecode < 1 1 mzn-g12mip < mzn2smt mzn-gecode with disequalities: n = 11: 12 sec. n = 12: 153 sec. mzn-gecode with ALLDIFFERENT: n = 11: < 1 sec. n = 100: < 1 sec. n = : 2 sec. mzn-g12mip with ALLDIFFERENT: n = 11: < 1 sec. n = 100: 3 sec. n = 300: 35 sec. Course 1DL for Combinatorial Optimisation