"""
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 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 * '-')
"""
See https://en.wikipedia.org/wiki/Change-making_problem
"""
import satx
k = 13
coins = [1, 5, 10, 20, 50, 100, 200]
opt = k
while True:
satx.engine(sum(coins).bit_length(), cnf_path='aux.cnf')
x = satx.vector(size=len(coins))
assert satx.dot(x, coins) == k
assert sum(x) < opt
if satx.satisfy('slime'):
opt = sum(x)
print(opt, x)
else:
break