# Enigmatic Code

Programming Enigma Puzzles

## Enigma 303: Some dominoes

From New Scientist #1451, 11th April 1985 [link]

Most of a full set of dominoes has been arranged in a 7 × 7 block with a hole in the middle. Please mark in the boundaries between the dominoes. There is only one answer.

[enigma303]

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### One response to “Enigma 303: Some dominoes”

1. Jim Randell 17 August 2015 at 9:01 am

You can use the same program used to solve Enigma 179 for this one. All that needs to be changed are the data describing the grid.

This Python program runs in 41ms.

```from enigma import chunk, printf

# grid dimensions (columns, rows)
(N, M) = (7, 7)

# number of dominoes in the grid
D = (N * M - 1) // 2

# marker for unoccupied squares
X = -99

grid = [
0, 6, 0, 4, 4, 4, 3,
0, 2, 3, 2, 2, 2, 5,
0, 5, 4, 6, 6, 0, 5,
4, 1, 1, X, 4, 1, 1,
4, 5, 6, 5, 5, 0, 5,
6, 3, 3, 3, 2, 6, 5,
2, 1, 0, 6, 1, 6, 4,
]

# update grid <g> placing domino <n> at <i>, <j>
def update(g, i, j, n):
g = list(g)
g[i] = g[j] = n
return g

# g = grid
# n = label of next domino (1 to D)
# D = number of dominoes to place
# ds = dominoes already placed
def solve(g, n, D, ds):
# are we done?
if n > D:
# output the pairings
for r in chunk(g, N):
print(r)
print()
else:
# find the next unassigned square
for (i, d) in enumerate(g):
if d < 0: continue
(y, x) = divmod(i, N)
# find placements for the domino
js = list()
# horizontally
if x < N - 1 and not(g[i + 1] < 0): js.append(i + 1)
# vertically
if y < M - 1 and not(g[i + N] < 0): js.append(i + N)
# try possible placements
for j in js:
d = tuple(sorted((g[i], g[j])))
if d not in ds:
solve(update(g, i, j, -n), n + 1, D, ds.union([d]))
break

solve(grid, 1, D, set())
```

Solution: The arrangement of the dominoes is shown below: