# Solve sudokus with miniKanren

I am reminded today of Peter Norvig's blog post on solving every sudoku puzzle using constraint propagation and search. In this essay we will revisit the problem using the python implementation of miniKanren, kanren, a domain-specific language for logic programming.

First a short refresher about sudokus. A sudoku puzzle is a square grid of 81 spaces which we subdivide in 9 squares each containing 9 spaces. For every valid sudoku grid:

- The rows are a permutation of the digits from 1 to 9;
- The columns are a permutation of the digits from 1 to 9;
- The squares are a permutation of the digits from 1 to 9;

The following represents a (partially filled) grid with one row, one column, and one square completed:

6 1 4 | | 9 2 8 | 7 6 5 | 4 3 1 5 3 6 | | ------+-------+------ 4 | | 5 | | 6 | | ------+-------+------ 7 | | 8 | | 9 | |

Peter's solution is *imperative*: we implement a series of procedures that will give us the missing values in a Sudoku puzzle. The *nature* of the Sudoku, the rules, are hidden in this code. Here we will instead implement the rules of the Sudoku, i.e. the *relations* that the units must respect and let miniKanren search for the grids that respect these rules (given some hints) for us. Sounds too good to be true?

## Solving sudoku rows

Let us work on a simpler problem first: a sudoku row. We assume that we have a vector with 9 spaces already filled with a few hints. To represent a valid sudoku row the vector must be the permutation of the digits from 1 to 9. Let us consider the following puzzle as an example:

```
row = (0, 0, 0, 1, 4, 0, 0, 9, 8)
```

Where the "0" values are placeholders. We can tell `kanren`

these are placeholders using *logic variables*, whose value it will try to compute later:

from kanren import var row_expr = tuple(r if r > 0 else var() for r in row) row_expr

~_{6} |
~_{7} |
~_{8} |
1 | 4 | ~_{9} |
~_{10} |
9 | 8 |

We now need to express the relation that the elements of the vector must follow. In `kanren`

we express these relations in the form of *goals*. `kanren`

comes with a handy `permuteo`

function which, when applied to two arguments, produces a goal that states that these two arguments must be a permutation of one another.

from kanren import permuteo DIGITS = tuple(range(1, 10)) goal = permuteo(row_expr, DIGITS) goal

<function permuteo.<locals>.permuteo_goal at 0x7f412b2fc280>

We are now ready to ask `kanren`

to find the missing values for us using the `run`

function, which takes an expression—a data structure with logic variables whose value we want to compute—one or several goals and returns results. Here we ask `kanren`

for 3 solutions:

from kanren import run results = run(3, row_expr, goal) results

2 | 3 | 5 | 1 | 4 | 6 | 7 | 9 | 8 |

2 | 3 | 5 | 1 | 4 | 7 | 6 | 9 | 8 |

2 | 3 | 6 | 1 | 4 | 5 | 7 | 9 | 8 |

As you can see, once we have described the relations, `kanren`

does all the heavy lifting for us and searches for solutions that satifsy the relations. Let us now move back to our initial problem: solving sudoku grids.

## Solving sudoku grids

We store the sudoku grid as a 81-element list. Since constraints are defined on rows, columns and squares we need to be able to extract those units from a grid:

def get_rows(grid): """Return a grid's rows.""" columns_idx = [[9 * i + j for j in range(9)] for i in range(9)] return tuple(tuple(grid[i] for i in column) for column in columns_idx) def get_columns(grid): """Return a grid's columns.""" rows_idx = [[i + 9 * j for j in range(9)] for i in range(9)] return tuple(tuple(grid[i] for i in row) for row in rows_idx) def get_squares(grid): """Return a grid's squares.""" squares_idx = [ [9 * i + j for i in a for j in b] for a in ((0, 1, 2), (3, 4, 5), (6, 7, 8)) for b in ((0, 1, 2), (3, 4, 5), (6, 7, 8)) ] return tuple(tuple(grid[i] for i in square) for square in squares_idx)

We can now express the relations that define a valid sudoku grid in the form of a `kanren`

goal. In addition to `permuteo`

we will use the `lall`

which produces a goal that states that all the relations passed as arguments must be satisfied at the same time. The solution is fairly simple, and very elegant:

from kanren import lall def sudoku(grid): """Define the Sudoku grid as a set on constraints on the rows, columns and squares. Parameter --------- grid A list that contains hints and logic variables ("unknown values"). """ rows = get_rows(grid) columns = get_columns(grid) squares = get_squares(grid) return lall( lall(*(permuteo(r, DIGITS) for r in rows)), lall(*(permuteo(c, DIGITS) for c in columns)), lall(*(permuteo(s, DIGITS) for s in squares)) )

We can now theoretically solve any sudoku grid using the following function (we only ask for one solution):

def solve_sudoku(grid): # Flatten the grid grid_flat = [int(elem) for row in grid for elem in row.split(" ")] assert len(grid_flat) == 81 assert max(grid_flat) == 9 grid_expr = tuple(val if val > 0 else var() for val in grid_flat) result = run(1, grid_expr, sudoku(grid_expr)) return result[0]

Let us make sure that our code can solve a trivially easy problems where only a few values are missing:

ridiculously_easy_puzzle = ( "5 0 4 6 7 8 0 1 2", "6 0 2 1 0 5 3 4 8", "1 9 8 3 4 2 5 6 7", "8 5 9 7 0 1 4 2 3", "4 0 6 8 5 3 7 9 0", "7 1 3 9 2 4 8 5 6", "9 6 1 5 3 7 2 8 4", "2 8 7 4 1 9 6 3 5", "3 0 5 2 0 6 0 7 9", ) result = solve_sudoku(ridiculously_easy_puzzle) for i in range(9): row = [str(r) for r in result[9*i: 9*i+9]] print(" ".join(row))

5 3 4 6 7 8 9 1 2 6 7 2 1 9 5 3 4 8 1 9 8 3 4 2 5 6 7 8 5 9 7 6 1 4 2 3 4 2 6 8 5 3 7 9 1 7 1 3 9 2 4 8 5 6 9 6 1 5 3 7 2 8 4 2 8 7 4 1 9 6 3 5 3 4 5 2 8 6 1 7 9

This actually works! And since miniKanren does the search for us our solution is much more expressive than Peter's. It is obvious reading the code what the *rules* of the sudoku are. This is the beauty of relational programming and DSLs like miniKanren.

## Limitation

While `kanren`

could find the solution for a ridiculously easy grid, the computation hangs on the following (supposedly easy) problem:

easy_puzzle = ( "0 0 3 0 2 0 6 0 0", "9 0 0 3 0 5 0 0 1", "0 0 1 8 0 6 4 0 0", "0 0 8 1 0 2 9 0 0", "7 0 0 0 0 0 0 0 8", "0 0 6 7 0 8 2 0 0", "0 0 2 6 0 9 5 0 0", "8 0 0 2 0 3 0 0 9", "0 0 5 0 1 0 3 0 0", ) result = solve_sudoku(easy_puzzle)

…while Peter Norvig's code solves most grids in less than 1 second. To understand why Peter's solution is orders of magnitude faster than ours, we need to understand how their respective search strategies differ. This is currently out my depth, so I started implementing the examples from The Reasoned Schemer to better understand the way miniKanren works. Expect more blog posts (and more lisp!) on the topic; maybe we can eventually find a way to make `kanren`

's search performance closer to Peter's code's in this example.

## Conclusion

This is a toy example, and miniKanren can be used for much more than solving sudoku puzzles. In Aemcmc we use miniKanren extensively for several purposes, such as:

- Describe relations between probabilistic models. The relation between centered and non-centered representations in the loc-scale families, conjugacy relations, etc. which can be used to rewrite probabilistic models written in Aesara so they are easier to sample;
- Match samplers to models.

The possibilities are endless, and to my knowledge haven't been explored much in python! This why I am particularly excited about where Aesara, Aeppl and Aemcmc are going—and why you should too. The promise is to make probabilistic programming in python a lot more *efficient* and *easy to use* that it currently is.