[英]Array of Z3 Counter variables in Python
I would like to create a Z3 array ( A ) whose elements are Z3 Int variables and each of those elements has a fixed range.我想创建一个 Z3 数组( A ),其元素是 Z3 Int 变量,并且每个元素都有一个固定的范围。 Also, I wanted to initialize the values of these elements with 0. Now let there is a simple function x1 + x2 + x3 = 6, where 0 <= x1, x2, x3 <= 4, where x1, x2, x3 are Z3 Int variables.另外,我想用 0初始化这些元素的值。现在让有一个简单的 function x1 + x2 + x3 = 6,其中 0 <= x1,x2,x3 <= 4,其中 x1,x2,x3 是 Z3整数变量。 Next, I want to increase the values at x1, x2, and x3 index positions (with the values of x1, x2, and x3 which are decided by Z3) of A by 1.接下来,我想将 A 的 x1、x2 和 x3 索引位置的值(x1、x2 和 x3 的值由 Z3 决定)增加 1。
Eg:例如:
My question is how to initialize a Z3 variable with its lowest possible value and use it as a counter variable?我的问题是如何用其最低值初始化 Z3 变量并将其用作计数器变量?
Note: The max.注意:最大。 and min.和分钟。 possible values of x1, x2, and x3 are always the lowest and the highest index of A , respectively. x1、x2 和 x3 的可能值始终分别是A的最低和最高索引。
General Considerations:一般注意事项:
Given a value assignment (x1, x2, x3) = (v1, v2, v3)
, any permutation of (v1, v2, v3)
is also a solution of the problem.给定一个赋值(x1, x2, x3) = (v1, v2, v3)
, (v1, v2, v3)
v1, v2, v3) 的任何排列也是问题的解决方案。 In some circumstances, it might be convenient to break symmetries in the search space by requiring, for example, ∀ i, j.((i < j) => (x_i <= x_j))
.在某些情况下,通过要求∀ i, j.((i < j) => (x_i <= x_j))
来打破搜索空间中的对称性可能会很方便。
It is much easier to think of each location A_i
as the number of all x_j
assigned with value i
.将每个位置A_i
视为分配有值i
的所有x_j
的数量要容易得多。
Python + z3: Python + z3:
from z3 import *
def model_to_str(m):
fmt = "\t A:= {}; x := {};"
A_tuple = tuple([m.eval(el) for el in A_list])
x_tuple = tuple([m.eval(el) for el in x_list])
return fmt.format(A_tuple, x_tuple)
###########
# PROBLEM #
###########
s = Solver()
# declarations
A_length = 5
A_list = [Int('A_{0}'.format(idx)) for idx in range(0, A_length)]
x_list = [Int('x_{0}'.format(idx)) for idx in range(0, 3)]
# A's constraints
for i, A_i in enumerate(A_list):
s.add(A_i == Sum([If(x_j == i, 1, 0) for x_j in x_list]))
s.add(A_i <= 2)
# x's constraints
s.add(6 == Sum(x_list))
for x_i in x_list:
s.add(0 <= x_i)
s.add(x_i < A_length)
# Symmetry-breaking constraints:
# (remove symmetry solutions)
#s.add(x_list[0] <= x_list[1])
#s.add(x_list[0] <= x_list[2])
#s.add(x_list[1] <= x_list[2])
# Get one solution:
if s.check() == sat:
print("Random Solution:")
print(model_to_str(s.model()))
# Get all solutions:
s.push()
print("\nEnumerate All Solutions:")
while s.check() == sat:
m = s.model()
print(model_to_str(m))
s.add(Or([x_j != m.eval(x_j) for x_j in x_list]))
s.pop()
# Enumerate all possible assignments
print("\nTry All Possible Assignments:")
for x_0_val in range(0, A_length):
for x_1_val in range(0, A_length):
for x_2_val in range(0, A_length):
values = (x_0_val, x_1_val, x_2_val)
s.push()
s.add([x_j == x_j_val for x_j, x_j_val in zip(x_list, values)])
if s.check() == sat:
print(model_to_str(s.model()))
else:
print("\t\tx := {} is unsat!".format(values))
s.pop()
Output: Output:
Random Solution:
A:= (1, 0, 1, 0, 1); x := (4, 0, 2);
Enumerate All Solutions:
A:= (1, 0, 1, 0, 1); x := (2, 0, 4);
A:= (1, 0, 1, 0, 1); x := (2, 4, 0);
A:= (0, 1, 1, 1, 0); x := (2, 1, 3);
A:= (0, 1, 1, 1, 0); x := (2, 3, 1);
A:= (1, 0, 0, 2, 0); x := (3, 3, 0);
A:= (1, 0, 1, 0, 1); x := (4, 0, 2);
A:= (0, 2, 0, 0, 1); x := (4, 1, 1);
A:= (1, 0, 1, 0, 1); x := (4, 2, 0);
A:= (0, 1, 1, 1, 0); x := (3, 2, 1);
A:= (1, 0, 1, 0, 1); x := (0, 2, 4);
A:= (0, 1, 1, 1, 0); x := (1, 2, 3);
A:= (1, 0, 1, 0, 1); x := (0, 4, 2);
A:= (0, 1, 1, 1, 0); x := (1, 3, 2);
A:= (0, 2, 0, 0, 1); x := (1, 4, 1);
A:= (1, 0, 0, 2, 0); x := (0, 3, 3);
A:= (0, 2, 0, 0, 1); x := (1, 1, 4);
A:= (1, 0, 0, 2, 0); x := (3, 0, 3);
A:= (0, 1, 1, 1, 0); x := (3, 1, 2);
Try All Possible Assignments:
x := (0, 0, 0) is unsat!
x := (0, 0, 1) is unsat!
x := (0, 0, 2) is unsat!
x := (0, 0, 3) is unsat!
x := (0, 0, 4) is unsat!
x := (0, 1, 0) is unsat!
x := (0, 1, 1) is unsat!
x := (0, 1, 2) is unsat!
x := (0, 1, 3) is unsat!
x := (0, 1, 4) is unsat!
x := (0, 2, 0) is unsat!
x := (0, 2, 1) is unsat!
x := (0, 2, 2) is unsat!
x := (0, 2, 3) is unsat!
A:= (1, 0, 1, 0, 1); x := (0, 2, 4);
x := (0, 3, 0) is unsat!
x := (0, 3, 1) is unsat!
x := (0, 3, 2) is unsat!
A:= (1, 0, 0, 2, 0); x := (0, 3, 3);
x := (0, 3, 4) is unsat!
x := (0, 4, 0) is unsat!
x := (0, 4, 1) is unsat!
A:= (1, 0, 1, 0, 1); x := (0, 4, 2);
x := (0, 4, 3) is unsat!
x := (0, 4, 4) is unsat!
x := (1, 0, 0) is unsat!
x := (1, 0, 1) is unsat!
x := (1, 0, 2) is unsat!
x := (1, 0, 3) is unsat!
x := (1, 0, 4) is unsat!
x := (1, 1, 0) is unsat!
x := (1, 1, 1) is unsat!
x := (1, 1, 2) is unsat!
x := (1, 1, 3) is unsat!
A:= (0, 2, 0, 0, 1); x := (1, 1, 4);
x := (1, 2, 0) is unsat!
x := (1, 2, 1) is unsat!
x := (1, 2, 2) is unsat!
A:= (0, 1, 1, 1, 0); x := (1, 2, 3);
x := (1, 2, 4) is unsat!
x := (1, 3, 0) is unsat!
x := (1, 3, 1) is unsat!
A:= (0, 1, 1, 1, 0); x := (1, 3, 2);
x := (1, 3, 3) is unsat!
x := (1, 3, 4) is unsat!
x := (1, 4, 0) is unsat!
A:= (0, 2, 0, 0, 1); x := (1, 4, 1);
x := (1, 4, 2) is unsat!
x := (1, 4, 3) is unsat!
x := (1, 4, 4) is unsat!
x := (2, 0, 0) is unsat!
x := (2, 0, 1) is unsat!
x := (2, 0, 2) is unsat!
x := (2, 0, 3) is unsat!
A:= (1, 0, 1, 0, 1); x := (2, 0, 4);
x := (2, 1, 0) is unsat!
x := (2, 1, 1) is unsat!
x := (2, 1, 2) is unsat!
A:= (0, 1, 1, 1, 0); x := (2, 1, 3);
x := (2, 1, 4) is unsat!
x := (2, 2, 0) is unsat!
x := (2, 2, 1) is unsat!
x := (2, 2, 2) is unsat!
x := (2, 2, 3) is unsat!
x := (2, 2, 4) is unsat!
x := (2, 3, 0) is unsat!
A:= (0, 1, 1, 1, 0); x := (2, 3, 1);
x := (2, 3, 2) is unsat!
x := (2, 3, 3) is unsat!
x := (2, 3, 4) is unsat!
A:= (1, 0, 1, 0, 1); x := (2, 4, 0);
x := (2, 4, 1) is unsat!
x := (2, 4, 2) is unsat!
x := (2, 4, 3) is unsat!
x := (2, 4, 4) is unsat!
x := (3, 0, 0) is unsat!
x := (3, 0, 1) is unsat!
x := (3, 0, 2) is unsat!
A:= (1, 0, 0, 2, 0); x := (3, 0, 3);
x := (3, 0, 4) is unsat!
x := (3, 1, 0) is unsat!
x := (3, 1, 1) is unsat!
A:= (0, 1, 1, 1, 0); x := (3, 1, 2);
x := (3, 1, 3) is unsat!
x := (3, 1, 4) is unsat!
x := (3, 2, 0) is unsat!
A:= (0, 1, 1, 1, 0); x := (3, 2, 1);
x := (3, 2, 2) is unsat!
x := (3, 2, 3) is unsat!
x := (3, 2, 4) is unsat!
A:= (1, 0, 0, 2, 0); x := (3, 3, 0);
x := (3, 3, 1) is unsat!
x := (3, 3, 2) is unsat!
x := (3, 3, 3) is unsat!
x := (3, 3, 4) is unsat!
x := (3, 4, 0) is unsat!
x := (3, 4, 1) is unsat!
x := (3, 4, 2) is unsat!
x := (3, 4, 3) is unsat!
x := (3, 4, 4) is unsat!
x := (4, 0, 0) is unsat!
x := (4, 0, 1) is unsat!
A:= (1, 0, 1, 0, 1); x := (4, 0, 2);
x := (4, 0, 3) is unsat!
x := (4, 0, 4) is unsat!
x := (4, 1, 0) is unsat!
A:= (0, 2, 0, 0, 1); x := (4, 1, 1);
x := (4, 1, 2) is unsat!
x := (4, 1, 3) is unsat!
x := (4, 1, 4) is unsat!
A:= (1, 0, 1, 0, 1); x := (4, 2, 0);
x := (4, 2, 1) is unsat!
x := (4, 2, 2) is unsat!
x := (4, 2, 3) is unsat!
x := (4, 2, 4) is unsat!
x := (4, 3, 0) is unsat!
x := (4, 3, 1) is unsat!
x := (4, 3, 2) is unsat!
x := (4, 3, 3) is unsat!
x := (4, 3, 4) is unsat!
x := (4, 4, 0) is unsat!
x := (4, 4, 1) is unsat!
x := (4, 4, 2) is unsat!
x := (4, 4, 3) is unsat!
x := (4, 4, 4) is unsat!
MiniZinc + z3: MiniZinc + z3:
I want to share also this approach because it took me literally only 5 minutes to solve this problem in MiniZinc , much less than with z3py
.我也想分享这种方法,因为我在MiniZinc中解决这个问题实际上只花了5 分钟,比z3py
少得多。
[ Note: for this task, I am using the fzn2z3.py
script from the fzn2omt
project .] [注意:对于这个任务,我使用的是来自fzn2omt
项目的fzn2z3.py
脚本。]
File counter.mzn
: (general problem structure)文件counter.mzn
:(一般问题结构)
include "globals.mzn";
% PARAMETERS
set of int: DOMAIN_A;
set of int: INDEX_SET_A;
set of int: DOMAIN_X = INDEX_SET_A;
set of int: INDEX_SET_X;
% VARIABLES
array [INDEX_SET_A] of var DOMAIN_A: A;
array [INDEX_SET_X] of var DOMAIN_X: x;
% CONSTRAINTS
constraint sum(x) == 6;
constraint forall(i in INDEX_SET_A) (
A[i] = sum(j in INDEX_SET_X)(x[j] == i)
);
% SYMMETRY BREAKING CONTRAINT
constraint increasing(x);
solve satisfy;
( Note: I have enabled the symmetry breaking constraint on this problem just to reduce the number of solutions printed below.) (注意:我在这个问题上启用了对称破坏约束,只是为了减少下面打印的解决方案的数量。)
File counter.dzn
: (instance-specific data)文件counter.dzn
:(特定于实例的数据)
DOMAIN_A = 0..2;
INDEX_SET_A = 0..4;
INDEX_SET_X = 1..3;
Find one solution with z3
:使用z3
找到一种解决方案:
~$ mzn2fzn counter.mzn counter.dzn
~$ fzn2z3.py counter.fzn
x = array1d(1..3, [1, 1, 4]);
A = array1d(0..4, [0, 2, 0, 0, 1]);
----------
Find all possible solutions (only with optimathsat
):找到所有可能的解决方案(仅使用optimathsat
):
~$ mzn2fzn counter.mzn counter.dzn
~$ fzn2optimathsat.py --all-solutions counter.fzn
% allsat model
x = array1d(1..3, [1, 1, 4]);
A = array1d(0..4, [0, 2, 0, 0, 1]);
----------
% allsat model
x = array1d(1..3, [0, 2, 4]);
A = array1d(0..4, [1, 0, 1, 0, 1]);
----------
% allsat model
x = array1d(1..3, [0, 3, 3]);
A = array1d(0..4, [1, 0, 0, 2, 0]);
----------
% allsat model
x = array1d(1..3, [1, 2, 3]);
A = array1d(0..4, [0, 1, 1, 1, 0]);
----------
==========
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