{ "cells": [ { "cell_type": "markdown", "id": "dbe7ced4", "metadata": { "tags": [ "CSP", "easy", "AllDifferent", "Intension", "complex", "AllDifferentMatrix" ] }, "source": [ "# Problem *Sudoku*" ] }, { "cell_type": "markdown", "id": "28437889", "metadata": {}, "source": [ "This well-known problem is stated as follows: fill in a grid using digits ranging from 1 to 9 such that:\n", "- all digits occur on each row\n", "- all digits occur on each column\n", "- all digits occur in each $3 \\times 3$ block (starting at a position multiple of 3) " ] }, { "attachments": { "sudoku.png": { "image/png": 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} }, "cell_type": "markdown", "id": "aac61629", "metadata": {}, "source": [ "Here is an illustration: \n", "" ] }, { "cell_type": "markdown", "id": "ddb8e078", "metadata": {}, "source": [ "To build a CSP (Constraint Satisfaction Problem) model, we need first to import the library PyCSP$^3$:" ] }, { "cell_type": "code", "execution_count": 1, "id": "e9ad9c6e", "metadata": {}, "outputs": [], "source": [ "from pycsp3 import *" ] }, { "cell_type": "markdown", "id": "9db7f45e", "metadata": {}, "source": [ "In a first step, we shall build a Sudoku grid from no data/clues (i.e., from an empty grid)." ] }, { "cell_type": "markdown", "id": "c4b9334c", "metadata": {}, "source": [ "We start our CSP model with a two-dimensional array $x$ of $9 \\times 9$ variables, each variable having $\\{1,2,\\dots,9\\}$ as domain. " ] }, { "cell_type": "code", "execution_count": 2, "id": "a6a7d3de", "metadata": {}, "outputs": [], "source": [ "# x[i][j] is the value at row i and col j\n", "x = VarArray(size=[9, 9], dom=range(1, 10))" ] }, { "cell_type": "markdown", "id": "7d27a4c7", "metadata": {}, "source": [ "We can display the structure of the array, as well as the domain of the first variable (remember that all variables have the same domain)." ] }, { "cell_type": "code", "execution_count": 3, "id": "50dbf0a3", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[\n", " [x[0][0], x[0][1], x[0][2], x[0][3], x[0][4], x[0][5], x[0][6], x[0][7], x[0][8]]\n", " [x[1][0], x[1][1], x[1][2], x[1][3], x[1][4], x[1][5], x[1][6], x[1][7], x[1][8]]\n", " [x[2][0], x[2][1], x[2][2], x[2][3], x[2][4], x[2][5], x[2][6], x[2][7], x[2][8]]\n", " [x[3][0], x[3][1], x[3][2], x[3][3], x[3][4], x[3][5], x[3][6], x[3][7], x[3][8]]\n", " [x[4][0], x[4][1], x[4][2], x[4][3], x[4][4], x[4][5], x[4][6], x[4][7], x[4][8]]\n", " [x[5][0], x[5][1], x[5][2], x[5][3], x[5][4], x[5][5], x[5][6], x[5][7], x[5][8]]\n", " [x[6][0], x[6][1], x[6][2], x[6][3], x[6][4], x[6][5], x[6][6], x[6][7], x[6][8]]\n", " [x[7][0], x[7][1], x[7][2], x[7][3], x[7][4], x[7][5], x[7][6], x[7][7], x[7][8]]\n", " [x[8][0], x[8][1], x[8][2], x[8][3], x[8][4], x[8][5], x[8][6], x[8][7], x[8][8]]\n", "]\n", "Domain of any variable: 1..9\n" ] } ], "source": [ "print(x)\n", "print(\"Domain of any variable: \", x[0][0].dom)" ] }, { "cell_type": "markdown", "id": "9c67a728", "metadata": {}, "source": [ "Interestingly, with a single constraint *AllDifferentMatrix*, we can impose that all digits are different on each row and column of $x$." ] }, { "cell_type": "code", "execution_count": 4, "id": "3369bd46", "metadata": {}, "outputs": [], "source": [ "satisfy(\n", " # imposing distinct values on each row and each column\n", " AllDifferent(x, matrix=True)\n", ");" ] }, { "cell_type": "markdown", "id": "ded2198b", "metadata": {}, "source": [ "By calling the function *solve()*, we can check that the problem is satisfiable (SAT). We can also display the found solution. Here, we call the function *values()* that collects the values assigned to a specified list of variables." ] }, { "cell_type": "code", "execution_count": 5, "id": "c268f917", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[1, 2, 3, 4, 5, 6, 7, 8, 9]\n", "[2, 1, 4, 3, 6, 5, 8, 9, 7]\n", "[3, 4, 1, 2, 7, 8, 9, 5, 6]\n", "[4, 3, 2, 1, 8, 9, 6, 7, 5]\n", "[5, 6, 7, 8, 9, 1, 2, 3, 4]\n", "[6, 5, 8, 9, 1, 7, 3, 4, 2]\n", "[7, 8, 9, 5, 2, 3, 4, 6, 1]\n", "[8, 9, 6, 7, 4, 2, 5, 1, 3]\n", "[9, 7, 5, 6, 3, 4, 1, 2, 8]\n" ] } ], "source": [ "if solve() is SAT:\n", " for i in range(9):\n", " print(values(x[i]))" ] }, { "cell_type": "markdown", "id": "eebe3754", "metadata": {}, "source": [ "Note that we get a structured string representation of multi-dimensional lists when printing them directly." ] }, { "cell_type": "code", "execution_count": 6, "id": "b760c35e", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[\n", " [1, 2, 3, 4, 5, 6, 7, 8, 9]\n", " [2, 1, 4, 3, 6, 5, 8, 9, 7]\n", " [3, 4, 1, 2, 7, 8, 9, 5, 6]\n", " [4, 3, 2, 1, 8, 9, 6, 7, 5]\n", " [5, 6, 7, 8, 9, 1, 2, 3, 4]\n", " [6, 5, 8, 9, 1, 7, 3, 4, 2]\n", " [7, 8, 9, 5, 2, 3, 4, 6, 1]\n", " [8, 9, 6, 7, 4, 2, 5, 1, 3]\n", " [9, 7, 5, 6, 3, 4, 1, 2, 8]\n", "]\n" ] } ], "source": [ "if solve() is SAT:\n", " print(values(x))" ] }, { "cell_type": "markdown", "id": "a74a9947", "metadata": {}, "source": [ "On can observe that some digits occur several times in the same block. This is why we add a list of constraints *AllDifferent*. " ] }, { "cell_type": "code", "execution_count": 7, "id": "a3376f3f", "metadata": {}, "outputs": [], "source": [ "satisfy(\n", " # imposing distinct values on each block\n", " AllDifferent(x[i:i + 3, j:j + 3]) for i in [0, 3, 6] for j in [0, 3, 6]\n", ");" ] }, { "cell_type": "markdown", "id": "aa00992c", "metadata": {}, "source": [ "Note how the notation $x[i:i + 3, j:j + 3]$ extracts a list of variables corresponding to a block of size $3 \\times 3$ in $x$. This is similar to notations used in package NumPy. We can check this:" ] }, { "cell_type": "code", "execution_count": 8, "id": "b1e87fbb", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[\n", " [x[0][0], x[0][1], x[0][2]]\n", " [x[1][0], x[1][1], x[1][2]]\n", " [x[2][0], x[2][1], x[2][2]]\n", "]\n" ] } ], "source": [ "print(x[0:3,0:3])" ] }, { "cell_type": "markdown", "id": "7f656a4d", "metadata": {}, "source": [ "We can control the scope (i.e., involved variables) of all posted constraints by displaying their internal representation. " ] }, { "cell_type": "code", "execution_count": 9, "id": "6a58887d", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "allDifferent(matrix:x[][])\n", "allDifferent(list:[x[0][0], x[0][1], x[0][2], x[1][0], x[1][1], x[1][2], x[2][0], x[2][1], x[2][2]])\n", "allDifferent(list:[x[0][3], x[0][4], x[0][5], x[1][3], x[1][4], x[1][5], x[2][3], x[2][4], x[2][5]])\n", "allDifferent(list:[x[0][6], x[0][7], x[0][8], x[1][6], x[1][7], x[1][8], x[2][6], x[2][7], x[2][8]])\n", "allDifferent(list:[x[3][0], x[3][1], x[3][2], x[4][0], x[4][1], x[4][2], x[5][0], x[5][1], x[5][2]])\n", "allDifferent(list:[x[3][3], x[3][4], x[3][5], x[4][3], x[4][4], x[4][5], x[5][3], x[5][4], x[5][5]])\n", "allDifferent(list:[x[3][6], x[3][7], x[3][8], x[4][6], x[4][7], x[4][8], x[5][6], x[5][7], x[5][8]])\n", "allDifferent(list:[x[6][0], x[6][1], x[6][2], x[7][0], x[7][1], x[7][2], x[8][0], x[8][1], x[8][2]])\n", "allDifferent(list:[x[6][3], x[6][4], x[6][5], x[7][3], x[7][4], x[7][5], x[8][3], x[8][4], x[8][5]])\n", "allDifferent(list:[x[6][6], x[6][7], x[6][8], x[7][6], x[7][7], x[7][8], x[8][6], x[8][7], x[8][8]])\n" ] } ], "source": [ "print(posted())" ] }, { "cell_type": "markdown", "id": "a616e12c", "metadata": {}, "source": [ "Now, we can generate a valid Sudoku grid." ] }, { "cell_type": "code", "execution_count": 10, "id": "078be930", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[\n", " [1, 2, 3, 4, 5, 6, 7, 8, 9]\n", " [4, 5, 6, 7, 8, 9, 1, 2, 3]\n", " [7, 8, 9, 1, 2, 3, 4, 5, 6]\n", " [2, 1, 4, 3, 6, 5, 8, 9, 7]\n", " [3, 6, 5, 8, 9, 7, 2, 1, 4]\n", " [8, 9, 7, 2, 1, 4, 3, 6, 5]\n", " [5, 3, 1, 6, 4, 2, 9, 7, 8]\n", " [6, 4, 2, 9, 7, 8, 5, 3, 1]\n", " [9, 7, 8, 5, 3, 1, 6, 4, 2]\n", "]\n" ] } ], "source": [ "if solve() is SAT:\n", " print(values(x))" ] }, { "cell_type": "markdown", "id": "de18c3ae", "metadata": {}, "source": [ "In a second step, we decide to take into account the clues as those in the illustration above. When 0 is present in a cell, it means that no clue is given for this cell." ] }, { "cell_type": "code", "execution_count": 11, "id": "f1f06cda", "metadata": {}, "outputs": [], "source": [ " clues = [\n", " [0, 2, 0, 5, 0, 1, 0, 9, 0],\n", " [8, 0, 0, 2, 0, 3, 0, 0, 6],\n", " [0, 3, 0, 0, 6, 0, 0, 7, 0],\n", " [0, 0, 1, 0, 0, 0, 6, 0, 0],\n", " [5, 4, 0, 0, 0, 0, 0, 1, 9],\n", " [0, 0, 2, 0, 0, 0, 7, 0, 0],\n", " [0, 9, 0, 0, 3, 0, 0, 8, 0],\n", " [2, 0, 0, 8, 0, 4, 0, 0, 7],\n", " [0, 1, 0, 9, 0, 7, 0, 6, 0]\n", " ]" ] }, { "cell_type": "markdown", "id": "27faca96", "metadata": {}, "source": [ "We can post an unary constraint *Intension* for any given clue. " ] }, { "cell_type": "code", "execution_count": 12, "id": "76874335", "metadata": {}, "outputs": [], "source": [ "satisfy (\n", " # imposing clues\n", " x[i][j] == clues[i][j] for i in range(9) for j in range(9) if clues[i][j] > 0\n", ");" ] }, { "cell_type": "markdown", "id": "b63a399f", "metadata": {}, "source": [ "We can control that such instantiations are well defined by displaying, for example, the internal representation of the last posted constraint." ] }, { "cell_type": "code", "execution_count": 13, "id": "c6027d04", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "intension(function:eq(x[8][7],6))\n" ] } ], "source": [ "print(posted(-1,-1))" ] }, { "cell_type": "markdown", "id": "c53cfc8b", "metadata": {}, "source": [ "We can now find a solution for that grid." ] }, { "cell_type": "code", "execution_count": 14, "id": "5a0cbf07", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "[\n", " [4, 2, 6, 5, 7, 1, 3, 9, 8]\n", " [8, 5, 7, 2, 9, 3, 1, 4, 6]\n", " [1, 3, 9, 4, 6, 8, 2, 7, 5]\n", " [9, 7, 1, 3, 8, 5, 6, 2, 4]\n", " [5, 4, 3, 7, 2, 6, 8, 1, 9]\n", " [6, 8, 2, 1, 4, 9, 7, 5, 3]\n", " [7, 9, 4, 6, 3, 2, 5, 8, 1]\n", " [2, 6, 5, 8, 1, 4, 9, 3, 7]\n", " [3, 1, 8, 9, 5, 7, 4, 6, 2]\n", "]\n" ] } ], "source": [ "if solve() is SAT:\n", " print(values(x))" ] }, { "cell_type": "markdown", "id": "6306d9ea", "metadata": {}, "source": [ "Most of the time, there is only one solution for a Sudoku puzzle. We can check this here:" ] }, { "cell_type": "code", "execution_count": 15, "id": "96debbdd", "metadata": {}, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Number of solutions: 1\n" ] } ], "source": [ "if solve(sols=ALL) is SAT:\n", " print(\"Number of solutions: \", n_solutions())" ] }, { "cell_type": "markdown", "id": "05b7ff70", "metadata": {}, "source": [ "Finally, we give below the model in one piece. Here the data is expected to be given by the user (in a command line). " ] }, { "cell_type": "raw", "id": "4700d783", "metadata": { "raw_mimetype": "text/x-python" }, "source": [ "from pycsp3 import *\n", "\n", "clues = data # if not 0, clues[i][j] is a value imposed at row i and col j\n", "\n", "# x[i][j] is the value at row i and col j\n", "x = VarArray(size=[9, 9], dom=range(1, 10))\n", "\n", "satisfy(\n", " # imposing distinct values on each row and each column\n", " AllDifferent(x, matrix=True),\n", "\n", " # imposing distinct values on each block tag(blocks)\n", " [AllDifferent(x[i:i + 3, j:j + 3]) for i in [0, 3, 6] for j in [0, 3, 6]],\n", "\n", " # imposing clues tag(clues)\n", " [x[i][j] == clues[i][j] for i in range(9) for j in range(9)\n", " if clues and clues[i][j] > 0]\n", ")" ] } ], "metadata": { "kernelspec": { "display_name": "Python 3 (ipykernel)", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.7.12" } }, "nbformat": 4, "nbformat_minor": 5 }