# Solving the Sunday Telegraph's Griddler using Python

For many years now, The Sunday
Telegraph has published a grid-based puzzle each week. Originally
this was called a **Nonogram**, although it is now called a
**Griddler**.

Basically, you are given an empty grid (typically 30 × 35 squares) together with the pattern of filled squares for each row and column of the grid. For example, for the first row this might be "5.3.10" meaning that there are three blocks – of lengths 5, 3 and 10 respectively – each separated from each other by one or more blank cells, and from the edge of the grid by zero or more blank cells. The objective of the puzzle is to reconstruct the original picture (the filled cells on the grid) by using the information about the block lengths.

I was recently inspired by a presentation by Michael Foord showing an elegant solution by Peter Norvig to the problem of finding "did you mean this?" alternatives to a search term in a search engine using surprisingly few lines of Python.

I saw a parallel between generating the alternative spellings and generating the possible fits of blocks in a row or column of a grid. I decided to write a Python program using this technique (syntax colouring courtesy of Vim).

## The Code

```
# Nonogram/Griddler solver
# Matthew Strawbridge
# November 2009
# --
# Creative Commons Attribution 2.0 UK: England & Wales
# http://creativecommons.org/licenses/by/2.0/uk/
import sys
# Define some character constants
BLANK = " "
FILL = "#"
UNKNOWN = "?"
# Example data taken from The Sunday Telegraph, 2009-11-01
ROWS = [[2,2], [18], [1,1], [1,3,1,1,3,1],\
[1,3,2,2,3,1], [1,2,5,3,1], [1,2,5,1,1,1], [1,3,1,1,1,1,1,1],\
[1,3,1,1,3,1], [1,2,1,1,3,1], [2,2,1,1,1,4,2], [2,3,1,1,1,14],\
[2,2,1,8,1], [5,5,1,1], [4,4,1,2,1,1,1], [6,1,2,1,1,1],\
[4,1,1,1,1,1], [2,1,1,1,1,1], [2,1,2,1,1,1], [2,1,2,1,1,1],\
[2,1,1,1,1,1], [2,1,1,1,1,1], [2,1,1,3,1], [2,1,1,3,1],\
[2,2,2], [2,2,2], [2,2,2], [4,3,3],\
[10,8], [14,8], [14,6], [4],\
[2], [2], [2]]
COLUMNS = [[12], [2,4], [1,2,2], [1,9,2,2],\
[1,9,2,3], [1,2,2,1,2,3], [1,2,3], [1,9,2,4],\
[1,3,16], [1,3,16], [1,3,2,4], [1,9,2,3],\
[1,2,3], [1,9,2,3], [1,3,2,2,2], [1,7,2,2],\
[2,4], [12], [16], [2,4],\
[2,4], [2,10,4], [2,2,2,4], [2,7],\
[2,10,7], [2,2,4], [2,10,4], [2,4],\
[2,4], [16]]
ROW_COUNT = len(ROWS)
COLUMN_COUNT = len(COLUMNS)
def min_width(blocks):
"""The minimum width into which the supplied blocks will fit
e.g. min_width([1,2,3]) = 8
"""
assert(len(blocks) > 0)
return sum(blocks) + len(blocks) - 1
def fit(blocks, size):
"""Return all possible ways of fitting the supplied blocks into the
supplied size.
"""
assert(len(blocks) > 0)
assert(size >= min_width(blocks))
if len(blocks) == 1:
return [BLANK * i + FILL * blocks[0] + BLANK * (size - blocks[0] - i) \
for i in range(size - blocks[0] + 1)]
else:
return [BLANK * (i - blocks[0]) + FILL * blocks[0] + BLANK + f2 \
for i in range(blocks[0], size - min_width(blocks[1:])) \
for f2 in fit(blocks[1:], size - i - 1)]
def solve_row(candidates):
"""Return a string formed by merging the candidate solutions.
At each position the character will be FILL if all candidates have
FILL in that position, BLANK if they all have BLANK or UNKNOWN
if there is a mixture.
"""
return [UNKNOWN if len(x) > 1 \
else x.pop() \
for x in [set(y) for y in transpose(candidates)]]
def matches(candidate, pattern):
"""Returns True if the candidate matches the pattern, False otherwise."""
assert(len(candidate) == len(pattern))
for i in range(len(pattern)):
if pattern[i] <> UNKNOWN and pattern[i] <> candidate[i]:
return False
return True
def print_grid(g):
"""Print out the grid (an array of arrays of characters)."""
for row in g:
print ''.join(row)
def transpose(g):
"""Returns the transposed version of the supplied matrix
(i.e. rows become columns and columns become rows).
"""
assert(g)
return map(lambda *row: list(row), *g)
def is_solved(candidate):
"""Returns True if the supplied candidate is a complete solution
(contains no UNKNOWN entries), False otherwise.
"""
return reduce(lambda x, y: x and not UNKNOWN in y, candidate, True)
def update_solution(grid, row, pattern):
"""Update the specified row of the grid from the pattern, overwriting
any non-UNKNOWN values from the pattern.
"""
for p in range(len(pattern)):
if pattern[p] <> UNKNOWN:
grid[row][p] = pattern[p]
def main():
# Initially fill the solution with a grid of UNKNOWNs
solution = [[UNKNOWN for x in range(column_count)] \
for y in range(row_count)]
# Generate all possible fits for the rows and columns
fits_r = [fit(row, COLUMN_COUNT) for row in ROWS]
fits_c = [fit(col, ROW_COUNT) for col in COLUMNS]
fits = [fits_r, fits_c]
row_col = 0 # current index into fits; 0 = row, 1 = column
while not is_solved(solution):
assert(row_col == 0 or row_col == 1)
# Filter out any impossible fits given the solution so far
for ri in range(len(fits[row_col])):
fits[row_col][ri] = filter(lambda x: matches(x, solution[ri]),\
fits[row_col][ri])
this_solution = solve_row(fits[row_col][ri])
update_solution(solution, ri, this_solution)
row_col = abs(row_col - 1) # toggle 0, 1
solution = transpose(solution)
if row_col == 0:
transpose(solution)
print_grid(solution)
```

## Output

## ## ################## # # # ### # # ### # # ### ## ## ### # # ## ##### ### # # ## ##### # # # # ### # # # # # # # ### # # ### # # ## # # ### # ## ## # # # #### ## ## ### # # # ############## ## ## # ######## # ##### ##### # # #### #### # ## # # # ###### # ## # # # #### # # # # # ## # # # # # ## # ## # # # ## # ## # # # ## # # # # # ## # # # # # ## # # ### # ## # # ### # ## ## ## ## ## ## ## ## ## #### ### ### ########## ######## ############## ######## ############## ###### #### ## ## ##

## Conclusion

The algorithm, which generates every possible fit, is slower than some of the alternative ways of tackling this type of puzzle. In particular, it probably doesn't scale very well. However, it does work in a few minutes for the standard size of puzzle, and the code is reasonably concise.