I recently worked on a picross solver that solves puzzles from a screenshot of the puzzle.

The solver uses traditional image processing techniques (without machine learning) to recognise puzzle hints from a screenshot, and constraint programming to find a solution.

It is written in Python and relies on the scikit-image library for image processing and the Z3 Theorem Prover for constraint solving.

Though still a work in progress, the source code can be found on GitHub:

yi-jiayu/picrosser
Z3-powered picross solver
https://github.com/yi-jiayu/picrosser

Background

Picross, or nonograms, are a type of puzzle similar to sudoku which look like this:

Screenshot of a picross puzzle from Hungry Cat Picross on Android

Screenshot of a picross puzzle from Hungry Cat Picross on Android

The objective is to colour in all the cells on the grid according to the hints for each row and column (the numbers at the top and left side). Each hint indicates the number of cells which should be filled with a certain colour in the row or column it applies to. Circled hints mean that the cells of that colour should be contiguous.

Here’s what the solution to the puzzle above looks like:

The completed puzzle

The completed puzzle

Picross as a constraint satisfaction problem

Similar to a sudoku puzzle, a picross puzzle can be modelled as a constraint satisfaction problem, with the hints being the constraints on each row and column.

Boolean algebra

The following picross puzzle only has one colour. Hints which refer to consecutive cells are represented by negative numbers.

1 -2 2
2
1
-2

Assuming that each cell is a boolean variable that is true if the cell is filled, we can come up with a boolean expression which indicates whether a row or column is coloured correctly.

Let’s focus on a single row, with each cell given a label:

1 a b c

The hint is 1, so only one of cells A, B and C can be filled, and the other two will be empty. This can be written in boolean algebra as (a ∧ ¬b ∧ ¬c) ∨ (¬a ∧ b ∧ ¬c) ∨ (¬a ∧ ¬b ∧ c), which means:

a AND (NOT b) AND (NOT c)  -  a is coloured, but b and c are not
OR
(NOT a) AND b AND (NOT c)  -  b is coloured, but a and c are not
OR
(NOT a) AND (NOT b) AND c  -  c is coloured, but a and b are not

These are the only 3 possibilies for these 3 cells given the hint 1.

We can do the same for another row:

-2 a b c

The hint -2 means that there are two coloured cells in this row, but they also have to be connected. There are only two possibilities this time:

a AND b AND (NOT c)  -  a and b are coloured while c is not
OR
(NOT a) AND b AND c  -  b and c are coloured while a is not

How about the following row?

2 a b c

There’s only one way to have two cells filled without them connecting—a and c have to be filled but not b. However, there will be more possibilities with a longer row. Rather than enumerate all the possible combinations in boolean algebra, later we’ll take a shortcut and use other types of constraints available in Z3.

Finally, the case for a hint of 0 is trivial.

By applying these constraints to each row and column, we can solve for the correct colouring of the puzzle:

1 -2 2
2 x x
1 x
-2 x x

For a puzzle with multiple colours, each cell will have a boolean variable for each colour, and a constraint has to be added that each cell can only have one colour.

Using Z3

We can almost directly translate our problem for Z3 in Python with the z3-solver package:

from z3 import And, Bool, Not, Or, Solver, sat

a = Bool('a')
b = Bool('b')
c = Bool('c')

constraints = Or(
    And(a, b, Not(c)),
    And(Not(a), b, c),
)

s = Solver()
s.add(constraints)

if s.check() == sat:
    m = s.model()
    print('a:', m.evaluate(a))
    print('b:', m.evaluate(b))
    print('c:', m.evaluate(c))
else:
    print('unsat')

Bool creates a boolean variable, while And, Or and Not directly map to boolean algebra operations.

Running this, we get:

a: False
b: True
c: True

The solver only returns the first solution it finds. There are ways to find all possible solutions, however generally there should only be one solution to a given picross puzzle so that’s enough for us.

My actual solver implementation takes a map of puzzle attributes containing the number of rows, columns and colours in the puzzle and a nested array of hints for each row and colour, another one for the columns, and the actual colours to use to generate an image of the solution.

Here’s the puzzle from the start of the post represented in TOML:

nrows = 15
ncols = 10
ncolours = 4
rows = [
  [0, 0, -7, -3],
  [0, 0, -4, -6],
  [0, 0, -3, 7],
  [0, 0, -6, 4],
  [0, 0, -4, -6],
  [0, 1, 4, -5],
  [0, -3, -4, -3],
  [0, -4, -3, 3],
  [0, -5, -4, 1],
  [0, -6, -4, 0],
  [0, -7, -3, 0],
  [-2, -8, 0, 0],
  [1, -9, 0, 0],
  [0, -10, 0, 0],
  [-10, 0, 0, 0]
]
columns = [
  [3, 1, 5, 6],
  [2, -2, 6, 5],
  [1, -3, -8, -3],
  [1, -4, 7, 3],
  [1, -5, 4, 5],
  [1, -6, 2, 6],
  [1, -7, -4, -3],
  [1, -8, -4, -2],
  [1, -9, -3, -2],
  [1, -8, 3, -3]
]
colours = [[211, 175, 83], [136, 93, 0], [155, 151, 219], [255, 255, 255]]

It outputs a textual representation of the puzzle solution (squint and you can sort of see it):

[4, 4, 4, 3, 3, 3, 3, 3, 3, 3]
[4, 4, 4, 4, 4, 4, 3, 3, 3, 3]
[4, 4, 4, 4, 4, 4, 3, 3, 3, 4]
[4, 4, 3, 3, 3, 3, 3, 3, 4, 4]
[3, 3, 3, 3, 4, 4, 4, 4, 4, 4]
[3, 3, 3, 4, 4, 4, 4, 4, 2, 3]
[3, 3, 3, 3, 4, 4, 4, 2, 2, 2]
[4, 4, 3, 3, 3, 4, 2, 2, 2, 2]
[4, 3, 3, 3, 3, 2, 2, 2, 2, 2]
[3, 3, 3, 3, 2, 2, 2, 2, 2, 2]
[3, 3, 3, 2, 2, 2, 2, 2, 2, 2]
[1, 1, 2, 2, 2, 2, 2, 2, 2, 2]
[1, 2, 2, 2, 2, 2, 2, 2, 2, 2]
[2, 2, 2, 2, 2, 2, 2, 2, 2, 2]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

As well as an image:

The generated puzzle solution

Note that this is not the only way to formulate picross as a constraint satisfaction problem. For example, instea d of using boolean variables and separate grids for each colour, we could have a single grid and integer variables to represent colours.


In another post, I’ll describe how the solver recognises the puzzle hints from a screenshot.