Simon Willison’s Weblog

sudoku-in-python-packaging (via) Absurdly clever hack by konsti: solve a Sudoku puzzle entirely using the Python package resolver!

First convert the puzzle into a requirements.in file representing the current state of the board:

git clone https://github.com/konstin/sudoku-in-python-packaging
cd sudoku-in-python-packaging
echo '5,3,_,_,7,_,_,_,_                                        
6,_,_,1,9,5,_,_,_
_,9,8,_,_,_,_,6,_
8,_,_,_,6,_,_,_,3
4,_,_,8,_,3,_,_,1
7,_,_,_,2,_,_,_,6
_,6,_,_,_,_,2,8,_
_,_,_,4,1,9,_,_,5
_,_,_,_,8,_,_,7,9' > sudoku.csv
python csv_to_requirements.py sudoku.csv requirements.in

That requirements.in file now contains lines like this for each of the filled-in cells:

sudoku_0_0 == 5
sudoku_1_0 == 3
sudoku_4_0 == 7

Then run uv pip compile to convert that into a fully fleshed out requirements.txt file that includes all of the resolved dependencies, based on the wheel files in the packages/ folder:

uv pip compile \
  --find-links packages/ \
  --no-annotate \
  --no-header \
  requirements.in > requirements.txt

The contents of requirements.txt is now the fully solved board:

sudoku-0-0==5
sudoku-0-1==6
sudoku-0-2==1
sudoku-0-3==8
...

The trick is the 729 wheel files in packages/ - each with a name like sudoku_3_4-8-py3-none-any.whl. I decompressed that wheel and it included a sudoku_3_4-8.dist-info/METADATA file which started like this:

Name: sudoku_3_4
Version: 8
Metadata-Version: 2.2
Requires-Dist: sudoku_3_0 != 8
Requires-Dist: sudoku_3_1 != 8
Requires-Dist: sudoku_3_2 != 8
Requires-Dist: sudoku_3_3 != 8
...

With a !=8 line for every other cell on the board that cannot contain the number 8 due to the rules of Sudoku (if 8 is in the 3, 4 spot). Visualized:

So the trick here is that the Python dependency resolver (now lightning fast thanks to uv) reads those dependencies and rules out every package version that represents a number in an invalid position. The resulting version numbers represent the cell numbers for the solution.