fliegendewurst.eu/static/assets/logelei_off_by_one_sums.py

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2022-11-30 16:48:51 +00:00
"""
Python program to solve a logic puzzle.
Full blog post: https://fliegendewurst.github.io/z3-logic-puzzle-solving.html
Puzzle description: https://www.zeit.de/zeit-magazin/2022/32/logelei
🄯 FliegendeWurst 2022
This work is licensed under a Creative Commons Attribution 4.0 International License.
http://creativecommons.org/licenses/by/4.0/
"""
from z3 import *
import math
s = Solver()
# n*n grid
n = 6
grid = [[FreshInt() for x in range(n)] for y in range(n)]
def print_solution(m, print_grid=True):
for k in range(num_sums):
print(" " * (3*num_sums + 2), end="")
for x in range(n):
val = m[sums_v[x][k]].as_long()
if val == -1:
print(" ", end=" ")
else:
print(f"{val: >2}", end=" ")
print()
print()
print(" " * (3*num_sums + 1), end="")
print("-" * (3*n))
y = 0
for row in grid:
for sum_h in sums_h[y]:
val = m[sum_h].as_long()
if val == -1:
print(" ", end=" ")
else:
print(f"{val: >2}", end=" ")
print("|", end=" ")
x = 0
for cell in row:
val = m[cell].as_long()
if val == 0 or not print_grid:
val = " "
print(f"{val: >2}", end=" ")
x += 1
print()
y += 1
def distinct_if_nonzero(x):
for i in range(len(x)):
for j in range(i+1, len(x)):
cell_i = x[i]
cell_j = x[j]
s.add(If(And(cell_i != 0, cell_j != 0), cell_i != cell_j, True))
# empty grid cell = 0
# other numbers have to be in [1,2,3,4,5]
# no duplicate digits in any row / column
for row in grid:
for cell in row:
s.add(cell >= 0)
s.add(cell <= 5)
distinct_if_nonzero(row)
for column in [[grid[y][x] for y in range(n)] for x in range(n)]:
distinct_if_nonzero(column)
# each row and column has at most math.ceil(n/2) segments
# if there are less, they are indicated by -1
num_sums = math.ceil(n / 2)
sums_h = [[FreshInt() for i in range(num_sums)] for y in range(n)]
sums_v = [[FreshInt() for i in range(num_sums)] for x in range(n)]
# for each pattern of used digits in a row, determine the sums
# iterate over each row of the grid
for y in range(n):
# iterate over each binary pattern
for p in range(2**n):
# match condition of this pattern
mc = True
# will contain a list of variables for each segment
segments = []
start_next_segment = True
# iterate over each grid cell
for x in range(n):
# check that binary digit in the pattern
if (p >> x) & 1 == 1:
mc = And(mc, grid[y][x] != 0)
if start_next_segment:
segments.append([grid[y][x]])
start_next_segment = False
else:
segments[-1].append(grid[y][x])
else:
mc = And(mc, grid[y][x] == 0)
start_next_segment = True
# specify sums value
for k in range(num_sums):
if k < len(segments):
off_by_one = Or(
sums_h[y][k] == sum(segments[k]) + 1,
sums_h[y][k] == sum(segments[k]) - 1
)
s.add(If(mc, off_by_one, True))
else:
s.add(If(mc, sums_h[y][k] == -1, True))
# for each pattern of used digits in a columns, determine the sums
for x in range(n):
for p in range(2**n):
# match condition
mc = True
segments = []
start_next_segment = True
for y in range(n):
if (p >> y) & 1 == 1:
mc = And(mc, grid[y][x] != 0)
if start_next_segment:
segments.append([grid[y][x]])
start_next_segment = False
else:
segments[-1].append(grid[y][x])
else:
mc = And(mc, grid[y][x] == 0)
start_next_segment = True
# specify sums value
#print(f"{p:b}"[::-1], segments)
for k in range(num_sums):
if k < len(segments):
off_by_one = Or(
sums_v[x][k] == sum(segments[k]) + 1,
sums_v[x][k] == sum(segments[k]) - 1
)
s.add(If(mc, off_by_one, True))
else:
s.add(If(mc, sums_v[x][k] == -1, True))
res = s.check()
print(res)
def add_spec(spec_h, spec_v):
for a, b in zip(spec_h, sums_h):
for x, y in zip(a, b):
s.add(x == y)
for a, b in zip(spec_v, sums_v):
for x, y in zip(a, b):
s.add(x == y)
spec_h = [
[2, 13, -1],
[5, 3, 6],
[8, 7, -1],
[4, 4, 1],
[5, 5, -1],
[9, 0, -1]
]
spec_v = [
[11, -1, -1],
[4, 5, -1],
[7, 7, -1],
[2, 7, 3],
[4, 4, 5],
[10, 2, -1]
]
s.push()
add_spec(spec_h, spec_v)
res = s.check()
print(res)
while res == sat:
m = s.model()
print_solution(m)
# find another solution, if possible
c = False
for row in grid:
for cell in row:
c = Or(c, cell != m[cell].as_long())
s.add(c)
res = s.check()
print("no other solutions")
s.pop()
# try to find another puzzle with a unique solution
while True:
s.push()
print("new push")
print(s.check())
m = s.model()
# condition c: keep found off-by-one sums configuration
# condition c2: to ban this sums configuration
c = True
c2 = False
for l in [*sums_h, *sums_v]:
for x in l:
c = And(c, x == m[x].as_long())
c2 = Or(c2, x != m[x].as_long())
s.add(c)
# try to find another solution
c = False
for row in grid:
for cell in row:
c = Or(c, cell != m[cell].as_long())
s.add(c)
res = s.check()
print(res)
if res == unsat:
print("found another puzzle with unique solution")
print_solution(m, print_grid=False)
input("continue? [press enter]")
s.pop()
s.add(c2)