Sudoku variant examples 2 #777
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Main thoughts:
- Obviously, I'm not checking the constraints in depth. But if there is a unique solution, adding an assert will convince me that it's correct.
- Plus, if solved in <5 seconds, can these examples tested by test_examples?
- CI is failing still
- Also I think you meant to request me for review, not assign me. The assigned person is the one doing the coding.
| (Count(board, n) <= 2) & | ||
| (Count(board, q) <= 1) | ||
| ( | ||
| (Count(board, b) + Count(board, r) + Count(board, n) + Count(board, q) <= 2+2+1+1 + 8-Count(board, p)) |
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fine for me to include in this PR, but normally this could be split up in separate PRs (although it's a little more effort)
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i'd update the comment to reflect the new/fixed constraint
| import numpy as np | ||
| import time | ||
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| # This CPMpy example solves a sudoku by KNT, which can be found on https://logic-masters.de/Raetselportal/Raetsel/zeigen.php?id=0009KE |
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if there's a copy-paste description, you could add it. These comments can become nice parts of the documentation if you add them as doc-strings at the top of the module/file (see e.g. chess-recognition)
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| def killer_cage(idxs, values, regions, total): | ||
| # the sum of the cells in the cage must equal the total |
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I can recommend getting in the habit to make these comments also into doc-strings
| # the sum of the cells in the cage must equal the total | ||
| # all cells in the cage must be in the same region | ||
| constraints = [] | ||
| constraints.append(cp.sum([values[r, c] for r,c in idxs]) == total) |
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just for fun, I've been trying more numpy-type features lately. Could this become cp.sum(values[idxs]) == total?
| print("The regions are:") | ||
| print(cell_regions.value()) | ||
| print("The region cardinals are:") | ||
| print(region_cardinals.value()) No newline at end of file |
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for all of these, a good assert is to add the (unique?) solution at the bottom, and check that value matches it. that way, constraints can be simply changed/improved and you immediately know whether the solution is still correct.
| if (r,c) == (2,8): | ||
| # The path starts on emitter. It doesn't have any previous cells so the first 3 vectors in the sequence must be fully -1. As for the the last 3, they are taken from the neighbour. Vice versa also applies. | ||
| constraints.append(cp.all([cp.all(sequence[r,c,:3].flatten() == -1)])) | ||
| constraints.append(cp.any([cp.all([path[nr,nc] == 2, sequence[r,c,3,0] == cells[nr,nc], sequence[r,c,4,0] == nr, sequence[r,c,5,0] == nc, cp.all(sequence[r,c,3,1:] == sequence[nr,nc,3,:19]), cp.all(sequence[r,c,4,1:] == sequence[nr,nc,4,:19]), cp.all(sequence[r,c,5,1:] == sequence[nr,nc,5,:19]), sequence[nr,nc,0,0] == cells[r,c], sequence[nr,nc,1,0] == r, sequence[nr,nc,2,0] == c, cp.all(sequence[nr,nc,0,1:] == sequence[r,c,0,:19]), cp.all(sequence[nr,nc,1,1:] == sequence[r,c,1,:19]), cp.all(sequence[nr,nc,2,1:] == sequence[r,c,2,:19])]) for nr, nc in neighbours])) |
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if these things represent some kind of path, could this path be kept in a separate data structure to make this somewhat more insightfull?
Other way to make this somewhat more readable is to use an auto-formatter like ruff on just this files. It will introduce a lot of newlines, but it will be much easier to read and edit.
| import cpmpy as cp | ||
| import numpy as np | ||
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| from cpmpy.tools.explain import optimal_mus, mus |
PR for the sudoku variants I made some time ago. I tried some of these with LLMs as well and they quite difficult so perhaps interesting as projects for the course as well? I also noticed a small mistake in the chess recognition example.