def sort(lst1, lst2):
"""
First, the variables of the collection π
π°ππΈπ°π±π»π΄π 2 correspond to a permutation of the variables of π
π°ππΈπ°π±π»π΄π 1. Second, the variables of π
π°ππΈπ°π±π»π΄π 2 are sorted in increasing order.
"""
_, ys = satx.permutations(lst1, len(lst1))
satx.apply_single(lst2, lambda i, t: t == ys[i], indexed=True)
satx.apply_dual(lst2, lambda a, b: a <= b)
import satx
satx.engine(bits=80, cnf_path='4d_perfect_euler_bricks.cnf', simplify=True, signed=True)
a = satx.integer()
b = satx.integer()
c = satx.integer()
d = satx.integer()
p = satx.integer()
q = satx.integer()
r = satx.integer()
s = satx.integer()
t = satx.integer()
u = satx.integer()
v = satx.integer()
satx.apply_single([a, b, c, d, p, q, r, s, t, u, v], lambda x: x > 0)
assert a ** 2 + b ** 2 == p ** 2
assert a ** 2 + c ** 2 == q ** 2
assert b ** 2 + c ** 2 == r ** 2
assert a ** 2 + d ** 2 == s ** 2
assert b ** 2 + d ** 2 == t ** 2
assert c ** 2 + d ** 2 == u ** 2
assert a ** 2 + b ** 2 + c ** 2 + d ** 2 == v ** 2
if satx.satisfy(solver='./slime', log=True):
print(a, b, c, d, p, q, r, s, t, u, v)
else:
print('Infeasible...')
import satx
satx.engine(bits=80, cnf_path='3d_perfect_euler_bricks.cnf', simplify=True, signed=True)
a = satx.integer()
b = satx.integer()
c = satx.integer()
p = satx.integer()
q = satx.integer()
r = satx.integer()
s = satx.integer()
satx.apply_single([a, b, c, p, q, r, s], lambda x: x > 0)
assert a ** 2 + b ** 2 == p ** 2
assert a ** 2 + c ** 2 == q ** 2
assert b ** 2 + c ** 2 == r ** 2
assert a ** 2 + b ** 2 + c ** 2 == s ** 2
if satx.satisfy(solver='./slime', log=True):
print(a, b, c, p, q, r, s)
else:
print('Infeasible...')
"""
Dad wants one-cent, two-cent, three-cent, five-cent, and ten-cent stamps.
He said to get four each of two sorts and three each of the others, but I've
forgotten which. He gave me exactly enough to buy them; just these dimes."
How many stamps of each type does Dad want? A dime is worth ten cents.
-- J.A.H. Hunter
"""
import satx
satx.engine(10, cnf_path='aux.cnf')
# x is the number of dimes
x = satx.integer()
# s[i] is the number of stamps of value 1, 2, 3, 5 and 10 according to i
s = satx.vector(size=5)
satx.apply_single(s, lambda t: t.is_in([3, 4]))
# 26 is a safe upper bound
assert x <= 26
assert satx.dot(s, [1, 2, 3, 5, 10]) == x * 10
while satx.satisfy('slime'):
print(s, x)
"""
See http://en.wikibooks.org/wiki/Puzzles/Arithmetical_puzzles/Digits_of_the_Square
There is one four-digit whole number x, such that the last four digits of x^2
are in fact the original number x. What is it?
"""
import satx
satx.engine(30, cnf_path='aux.cnf')
# x is the number we look for
x = satx.integer()
# d[i] is the ith digit of x
d = satx.vector(size=4)
satx.apply_single(d, lambda t: t.is_in(range(10)))
assert 1000 <= x < 10000
assert satx.dot(d, [1000, 100, 10, 1]) == x
assert (x * x) % 10000 == x
if satx.satisfy('slime'):
print(x, d)
else:
print('Infeasible...')
There are 5 non trivial numbers for base 10, and the highest such number is formed of 5 digits.
Below, the model is given for base 10.
"""
from math import ceil
import satx
# for base 10
n_digits = 5
satx.engine((10**n_digits).bit_length(), cnf_path='aux.cnf')
# n is a (non-trivial) Dudeney number
n = satx.integer()
# s is the perfect cubic root of n
s = satx.integer()
# d[i] is the ith digit of the Dudeney number
d = satx.vector(size=n_digits)
satx.apply_single(d, lambda t: t < 10)
assert 2 <= n < 10**n_digits
assert s < ceil((10**n_digits)**(1 / 3)) + 1
assert n == s * s * s
assert sum(d) == s
assert satx.dot(d, [10**(n_digits - i - 1) for i in range(n_digits)]) == n
while satx.satisfy('slime'):
print(n, s, d)
"""
See https://en.wikipedia.org/wiki/Verbal_arithmetic
A model for a general form of this problem is in CryptoPuzzle.py
"""
import satx
satx.engine(16, cnf_path='aux.cnf')
# letters[i] is the digit of the ith letter involved in the equation
s, e, n, d, m, o, r, y = letters = satx.vector(size=8)
satx.apply_single(letters, lambda t: t < 10)
# letters are given different values
satx.all_different(letters)
# words cannot start with 0
assert s > 0
assert m > 0
# respecting the mathematical equation
assert satx.dot([s, e, n, d], [1000, 100, 10, 1]) + satx.dot([m, o, r, e], [1000, 100, 10, 1]) == satx.dot([m, o, n, e, y], [10000, 1000, 100, 10, 1])
if satx.satisfy('slime'):
print(letters)
else:
print('Infeasible...')
"""
import itertools
import satx
n, m = 5, 8
opt = 1
while True:
print('OPTIMAL? : {}'.format(opt))
satx.engine(opt.bit_length(), cnf_path='aux.cnf')
# x[i][j] is the color at row i and column j
x = satx.matrix(dimensions=(n, m))
# at least one corners of different color for any rectangle inside the board
for i1, i2 in itertools.combinations(range(n), 2):
for j1, j2 in itertools.combinations(range(m), 2):
assert satx.one_of([x[i1][j1], x[i1][j2], x[i2][j1], x[i2][j2]]) != \
satx.one_of([x[i1][j1], x[i1][j2], x[i2][j1], x[i2][j2]])
satx.apply_single(satx.flatten(x), lambda t: t < opt)
if satx.satisfy('slime'):
print(x)
break
else:
opt += 1
See "Constraint Programming in Oz. A Tutorial" by C. Schulte and G. Smolka, 2001
A kid goes into a grocery store and buys four items.
The cashier charges $7.11, the kid pays and is about to leave when the cashier calls the kid back, and says
``Hold on, I multiplied the four items instead of adding them;
I'll try again;
Hah, with adding them the price still comes to $7.11''.
What were the prices of the four items?
"""
import satx
# 711 * 100 * 100 * 100 -> 30 bits
satx.engine(30, cnf_path='aux.cnf')
# x[i] is the price (multiplied by 100) of the ith item
x = satx.vector(size=4)
satx.apply_single(x, lambda t: t < 711)
# adding the prices of items corresponds to 711 cents
assert sum(x) == 711
# multiplying the prices of items corresponds to 711 cents (times 1000000)
assert x[0] * x[1] * x[2] * x[3] == 711 * 100 * 100 * 100
if satx.satisfy('slime'):
print(x)
else:
print('Infeasible...')
import satx
satx.engine(32, cnf_path='aux.cnf')
mapping = {1: 'r', 2: 'ry', 3: 'g', 4: 'y'}
R, RY, G, Y = 1, 2, 3, 4
table = [(R, R, G, G), (RY, R, Y, R), (G, G, R, R), (Y, R, RY, R)]
# v[i] is the color for the ith vehicle traffic light
v = satx.vector(size=4)
# p[i] is the color for the ith pedestrian traffic light
p = satx.vector(size=4)
satx.apply_single(v, lambda t: t.is_in([R, RY, G, Y]))
satx.apply_single(p, lambda t: t.is_in([R, G]))
for i in range(4):
assert satx.dot([v[i], p[i], v[(i + 1) % 4], p[(i + 1) % 4]],
[1, 10, 100, 1000]) == satx.one_of(
[satx.dot(t, [1, 10, 100, 1000]) for t in table])
while satx.satisfy('kissat'):
vv = [mapping[t.value] for t in v]
pp = [mapping[t.value] for t in p]
for a, b in zip(vv, pp):
print(a, b, end=', ')
print()
print(80 * '-')