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Solving different types of Sudokus with the FICO Xpress Optimizer

Description

We formulate and solve Sudoku puzzles as binary optimization problems.

Further explanation of this example: Xpress R Reference Manual

 sudoku_r.zip [download all files]

Source Files
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 sudoku.R [download]

sudoku.R

#####################################
# This file is part of the          #
# Xpress-R interface examples       #
#                                   #
#   (c) 2021 Fair Isaac Corporation #
#####################################
#' ---
#' title: "Solving Sudoku Puzzles"
#' ---
#'
## ----setup, include=FALSE-----------------------------------------------------
knitr::opts_chunk$set(echo = TRUE) knitr::opts_chunk$set(results = "hold")

#'
#' We formulate and solve Sudoku puzzles as binary optimization problems. This
#' example is inspired by the [Sudoku
#' example](http://roi.r-forge.r-project.org/use_case_sudoku.html) of the ROI
#' package in R, but also by the Sudoku enthusiasm of the author.
#'
#' First, we will create a binary formulation inside a function,
#' load_sudoku_model, where we first load the general model. We fix the initial
#' assignment by modifying the lower bounds of variables for which we are given the
#' solution.
#'
#' Secondly, we will extend the classical Sudoku by further constraints
#' that forbid the same element along diagonals.
#'
#' Thirdly, we will extend the model further to incorporate a magic square and
#' knight's move constraints to solve [this
#' puzzle]((https://www.youtube.com/watch?v=hAyZ9K2EBF0)) with the FICO Xpress
#' Optimizer.
#'
#' For all this, we use the sudokuAlt package.
#'
## ----Load packages------------------------------------------------------------
# load the xpress library
suppressMessages(library(xpress))

# load the Sudoku package
library(sudokuAlt)

# for data transformations
library(dplyr)
set.seed(123)
game <- makeGame()
plot(game)

#'
#' # Mathematical Formulation
#'
#' A classical Sudoku puzzle is formulated in a 9x9 grid of cells,
#' where the cells are subdivided into 3x3 subcells, the so-called "squares".
#' The goal is to fill all cells with the numbers 1 through 9, such
#' that each number appears only once in
#'
#' - each cell
#' - each row
#' - each square
#'
#' Let $R = \{1,\dots, 9\}$,
#' $C = \{1,\dots, 9\}$, and $S = \{S_{11}, S_{12}, \dots, S_{33}\}$,
#' where each $S_{ij}$ denotes a 9-element subset of $R\times C$,
#' e.g., $S_{13} = \{1,2,3\} \times \{7,8,9\}$ is the top right square of the Sudoku.
#'
#' In general
#' $$#' S_{ij} := \{3i - 2, 3i - 1, 3i\} \times \{3j - 2, 3j - 1, 3j\} \quad \forall i,j \in \{1,2,3\} #'$$
#' Let $L = \{1,\dots, 9\}$ denote the allowed symbols.
#' Now we are ready to formulate the binary program that encodes
#' Sudoku using 9 binary variables per cell,
#' 1 for each symbol. If $x_{rcl} = 1$, the symbol $l$ should
#' be placed in cell $(r,c)$ of the Sudoku board.
#'
#' Let $\emptyset \neq V_0 \subseteq R\times C \times L$
#' denote the given set of nonconflicting assignments.
#'
#' #' \begin{align} #' &\min & 0\\ #' & & \sum\limits_{l = 1}^9 x_{rcl} & = 1 & \forall r\in R, c \in C\\ #' & & \sum\limits_{r = 1}^9 x_{rcl} & = 1 & \forall c \in C, l \in L\\ #' & & \sum\limits_{c = 1}^9 x_{rcl} & = 1 & \forall r \in R, l \in L\\ #' & & \sum\limits_{(r,c) \in S_{ij}} x_{rcl} & = 1 & \forall i,j\in \{1,2,3\}, l \in L\\ #' & & x_{rcl} & = 1 & \forall (r,c,l) \in V_0\\ #' & & x_{rcl} & \in \{0,1\} & \forall r \in R, c \in C, l \in L\\ #' \end{align} #'
#'
#' In this case, we have no objective function. There are five sets of constraints. The first constraint group ensures that each cell contains exactly 1 symbol. The next three model the restriction that each symbol should appear exactly once
#' in each column, row, and square, respectively.
#' The last set of constraints fixes the initial assignment. Note that with an additional assignment, there are further variables that could be fixed to 0.
#' We do not add those restrictions to 0 to the model itself, because these additional fixings can be inferred immediately from the
#' initial assignments during presolving by the Optimizer.
#'
#'
#' # The Load Sudoku Function
#'
#' We declare a function load_sudoku_problem that takes as input an
#' XPRSprob object (which can be NULL, in which case we create a new one),
#' and a Sudoku game.
#'
## -----------------------------------------------------------------------------
#' loads a sudoku problem into the FICO Xpress Optimizer
#'
#' @param prob an existing XPRSprob or NULL, in which case we create a new problem
#' @param sudoku_game  a Sudoku Game created via sudokuAlt::makeGame()
#'
#' @return list object into which the Sudoku Game formulation was loaded.
#'        This list has two named attributes:
#'          -  \code{prob} is the same \code{prob} if \code{prob} was non-NULL,
#'            or a newly created XPRSprob object
#'          - \code{index.df} The index data frame that contains the initial assignment
#'            and variable indices for each row/column/label triplet r,c,l.
#'
load_sudoku_problem <- function(prob=NULL, sudoku_game) {

# create a new prob object if none was specified
if (is.null(prob)) {prob <- createprob()}

problemdata <- list()

# create some sets to match the notation above
R <- 1:9
C <- 1:9
L <- 1:9

# create a convenient data frame that contains all combinations
# of R,C,and L to simplify indexing
index.df <- expand.grid(R,C,L)
names(index.df) <- c("R", "C", "L")

# infer the subcell for each row/column combination.
index.df$Si <- (index.df$R - 1) %/% 3 + 1
index.df$Sj <- (index.df$C - 1) %/% 3 + 1

# we introduce var index to label each of the 9 * 9 * 9 = 729 binary variables
n <- nrow(index.df)
index.df$VarIndex <- 1:n # We use the designGame function to query a data frame representation of the Sudoku. game.df <- sudokuAlt::designGame(sudoku_game) # head of typical game.df. Unassigned symbols show as LNA. # Row Col Square Symbol # 1 R3 C3 S11 L9 # 2 R1 C6 S12 L6 # 3 R2 C4 S12 L4 # 4 R1 C7 S13 L5 # 5 R1 C8 S13 L7 # 6 R3 C8 S13 L2 # We convert the game.df and merge the symbols game.df <- game.df[game.df$Symbol != "LNA",]
game.df$R <- as.integer(gsub("R", "", game.df$Row))
game.df$C <- as.integer(gsub("C", "", game.df$Col))
game.df$Si <- as.integer(gsub("S(.).", "\\1", game.df$Square))
game.df$Sj <- as.integer(gsub("S.(.)", "\\1", game.df$Square))
index.df <- merge(index.df, game.df, all.x = T)
index.df <- index.df[order(index.df$VarIndex),] #the number of constraints: 9 * 9 * 4 (324) nconss <- 9 * 9 * 4 problemdata$objcoef <- rep(0, n)

# We formulate the constraints as 4 sparse column matrices
# Each constraint group arranges the variables differently,
# namely across a major and a minor index.
to_sparse_matrix <- function(major, minor) {
considx <- 9 * (major - 1) + minor
sparse_matrix <- Matrix::Matrix(0, 9*9, n, sparse = T)
idxmatrix <- matrix(c(considx,1:n), ncol = 2)
sparse_matrix[idxmatrix] <- 1
sparse_matrix
}

# 1 symbol in each cell: Constraint index determined by row and column
cell.cons.mat <- to_sparse_matrix(index.df$R, index.df$C)

# each symbol only once per column: Constraint index determined by column and Label
col.cons.mat <- to_sparse_matrix(index.df$C, index.df$L)

# each symbol only once per row: Constraint index determined by row and Label
row.cons.mat <- to_sparse_matrix(index.df$R, index.df$L)
#
# each symbol only once in each square:
# Constraint index determined by label and index of square
# use the square indices Si and Sj to compute the minor indices ranging from 1:9
square.cons.mat <- to_sparse_matrix(index.df$L, 3*(index.df$Si - 1) + index.df$Sj) problemdata$A <- rbind(
cell.cons.mat,
col.cons.mat,
row.cons.mat,
square.cons.mat
)

# right hand sides. we only treat the cardinality constraint here.
# Initial assignment is realized later via bounds on variables.
problemdata$rhs <- rep(1, nconss) # declare all constraints as equations problemdata$rowtype <- rep("E", nconss)

# all upper bounds are 1
problemdata$ub <- rep(1, n) # fix the initial assignment by setting variable lower bounds initial.assignment <- which(sprintf("L%d", index.df$L) == index.df$Symbol) problemdata$lb <- rep(0, n)
problemdata$lb[initial.assignment] <- 1 # all columns are binary problemdata$columntypes <- rep("B", n)

# assign column names to identify the variables when writing the problem to a file
problemdata$colname <- sprintf("x_R%dC%dL%d", index.df$R, index.df$C, index.df$L)

problemdata$probname <- "Sudoku" # this call also returns prob from the function list(prob=xpress::xprs_loadproblemdata(prob, problemdata=problemdata), index.df=index.df ) } #' #' #' # Solving a Sudoku with Xpress #' #' We call load_sudoku_problem on the game that we created at the beginning #' of this tutorial. #' #' The problem is ready to be solved. We enable console output and #' solve the Sudoku using mipoptimize. #' ## ----solve-------------------------------------------------------------------- problist <- load_sudoku_problem(sudoku_game = game) prob <- problist$prob

print(prob)
setoutput(prob)
mipoptimize(prob)

#'
#' # Plot the Result
#'
#' We want to convert the binary solution into a 9-by-9 matrix containing the
#' entries 1--9.
#'
#' Therefore, we use the index data frame. Recall that each row $i$ of this data
#' frame corresponds to the $i$'th column of the Sudoku model.
#' We filter those index rows for which the corresponding binary variable
#' has a solution value of 1.
#'
#' We use the dplyr function arrange to sort the filtered entries by row and column.
#' The corresponding symbols are then used to fill a matrix representation
#' of the binary solution.
#'
#' Finally, we use the plot functionality of the sudokuAlt package
#' to display the solution.
#'
## ----plottheresult------------------------------------------------------------
# translate the MIP solution into a solution of the game:
solution <- getmipsol(prob)$x # nice trick to enable Sudoku style plotting from ROI example: to_sudoku_solution <- function(solution, index.df) { rowbyrowsol <- index.df %>% dplyr::filter(near(solution,1)==TRUE) %>% dplyr::arrange(R, C) %>% dplyr::select(L) matrix_solution <- matrix(rowbyrowsol$L, 9,9, byrow = T)
sudoku_solution <- structure(matrix_solution, class = c("sudoku", "matrix"))
sudoku_solution
}

sudoku_solution <- to_sudoku_solution(solution, problist$index.df) par(mfrow = c(1,2)) plot(game) plot(sudoku_solution) #' #' # Add Diagonals #' #' A Sudoku with diagonals constrains the symbols along the two main diagonals. #' The main diagonal$D^+$is characterized as all cells #'$(r,c)\in R \times C$where$r = c$. #' The antidiagonal$D^-$is characterized as$(r,c) \in R \times C$such that$r+c = 10$. #' #' #' ## Mathematical notation #' #' The$18additional constraints for the diagonal and anti diagonal, using the same variable notation as before, then read #' #' #' \begin{align} #' & & \sum\limits_{r=c} x_{rcl} & = 1 & \forall l \in L\tag{D+}\\ #' & & \sum\limits_{r+c=10} x_{rcl} & = 1 & \forall l \in L\tag{D-}\\ #' \end{align} #' #' #' #' ## Game Creation #' #' In order to use these diagonal constraints, #' we first create a new game with fewer initial assignments. #' The default has about 20 initial assignments. #' Together with diagonal constraints, this may result in an infeasible MIP. #' ## ----diagonals_create_game---------------------------------------------------- # create a game with slightly less initial assignments than before. sudoku_game <- sudokuAlt::makeGame(gaps = 68) # load it using our function. problist <- load_sudoku_problem(sudoku_game = sudoku_game) probdiag <- problistprob
index.df <- problist$index.df print(probdiag) setoutput(probdiag) #' #' ## Main Diagonal #' #' We add the constraints for the main diagonal via addrows. #' The main diagonal consists of all cells whose #' row and column index are equal. #' The sorting via arrange ensures the correct grouping of #' all variables that belong to 1 symbol #' required by addrows. #' #' Remember that the arguments start and colind to addrows must be zero-based. #' In addition, start has length 10 (nrows + 1), where the last entry contains the total #' number of elements. #' ## ----adding_main_diagonal----------------------------------------------------- add_main_diagonal <- function(prob, index.df) { diagonal.df <- index.df %>% dplyr::filter(R == C) %>% dplyr::arrange(L) prob <- addrows( prob = prob, rowtype = rep("E", 9), start = c((9 * (0:8)), 81), rhs = rep(1, 9), colind = (diagonal.df$VarIndex) - 1,
rowcoef = rep(1,81)
)

prob
}

#'
#' ## Anti Diagonal
#'
#' In contrast to the diagonal, only the filter condition in the below selection
#' is modified for the anti diagonal.
#'
## ----adding_antidiagonal------------------------------------------------------

add_anti_diagonal <- function(prob, index.df) {
antidiagonal.df <- index.df %>%
filter(R + C == 10) %>%
dplyr::arrange(L)

# note that only the data frame that provides column indices changes between the two calls to addrows().
prob <- addrows(
prob = prob,
rowtype = rep("E", 9),
start = c((9L * (0:8)), 81),
rhs = rep(1, 9),
colind = (antidiagonal.df$VarIndex) - 1, rowcoef = rep(1,81) ) prob } #' #' #' ## Solve with Diagonal Constraints #' #' We invoke the solution process and display the solution as before. Note that this time, the solution process is no longer trivial, #' as the Branch-and-Bound search actually has to explore a small search tree for finding a solution. #' #' As for the simple Sudoku without diagonal constraints,we use the convenience #' function to to_sudoku_solution to convert the binary representation #' obtained from the Optimizer into a 9-by-9 matrix representation #' for plotting. #' ## ----solve_with_diagonals----------------------------------------------------- probdiag <- add_main_diagonal(probdiag, index.df) probdiag <- add_anti_diagonal(probdiag, index.df) mipoptimize(probdiag) summary(probdiag) solution <- getmipsol(probdiag)$x

sudoku_solution <- to_sudoku_solution(solution, index.df)
par(mfrow = c(1,2))
plot(sudoku_game)
plot(sudoku_solution)

#'
#'
#' # Solving a Sudoku with Only 4 Digits
#'
#' Finally, we turn our attention towards the following masterpiece
#' of Sudoku puzzles. In his Youtube channel
#' [Cracking the Cryptic](https://www.youtube.com/watch?v=hAyZ9K2EBF0),
#' Simon Anthony solves a Sudoku with only 4 given digits. We recommend
#' watching this video.
#'
#' The rules that apply to this Sudoku are as follows.
#'
#' * Classical rules (each digit once per row, column, and square)
#' * Diagonals, see Section above
#' * The central square must also form a *magic square*, meaning that
#'   the digits in each row and column of the central square must sum up
#'   to the same number.
#' * Cells that are a knight's move apart (in chess) must not contain the same digit.
#'
#' We would like to solve this puzzle using Xpress. We have already established
#' the building blocks for the classical rules (via load_sudoku_problem)
#' and for the diagonals.
#'
#' How do we model a magic square? The contribution of a single digit in cell
#' $r,c$ can be expressed as $\sum_{l=1}^{9} l \cdot x_{rcl}$ because
#' we know that exactly one of the 9 binary variables is set to 1.
#'
#' The digit sum in the top row of the central square (row 4) can then be expressed
#' as $\sum\limits_{c=4}^{6}\sum\limits_{l=1}^{9} l \cdot x_{4cl}$.
#' Since the total sum of all 3 rows (columns) in the central square must be equal to 45,
#' the digits in each row (column) must sum up to 15.
#'
#' We arrive at the following additional magic square constraints:
#'
#' #' \begin{align} #' & & \sum\limits_{c=4}^{6}\sum\limits_{l=1}^{9} l \cdot x_{rcl} & = 15 & \forall r \in \{4,5,6\}\\ #' & & \sum\limits_{r=4}^{6}\sum\limits_{l=1}^{9} l \cdot x_{rcl} & = 15 & \forall c \in \{4,5,6\}\\ #' \end{align} #'
#' We declare a function that receives a prob and an index data frame
#' and adds the 6 magic square constraints.
#'
## ----Adding a Magic Square----------------------------------------------------
add_magic_square <- function(prob, index.df) {
# consider only the central square S_22
central.square.df <- index.df %>% filter(Si == 2) %>% filter(Sj == 2)

# sort by row
central.square.df.by.row <- central.square.df %>% arrange(R)
# add the constraints for each row
prob <- addrows(prob, rowtype = c("E", "E", "E"), rhs = c(15,15,15), start = c(0,27,54,81),
colind = central.square.df.by.row$VarIndex - 1, rowcoef = central.square.df.by.row$L)

# sort by column
central.square.df.by.col <- central.square.df %>% arrange(C)
# add the constraints for each column
prob <- addrows(prob, rowtype = c("E", "E", "E"), rhs = c(15,15,15), start = c(0,27,54,81),
colind = central.square.df.by.col$VarIndex - 1, rowcoef = central.square.df.by.col$L)

# return prob with the added constraints
prob
}

#'
#'
#' Consider the cell $r,c$ where $r \in R$ and $c \in C$. A knight in
#' chess may move by either 2 horizontal and 1 vertical
#' or 1 horizontal and 2 vertical steps. Denote by $K(r,c)\subset R\times C$ the neighbors
#' of a cell $r,c$ with respect to the knight's move.
#'
#' We have $K(r,c)$ consisting of up to eight neighboring cells of the form
#'
#' $$#' K(r,c) = (\{(r \pm 2, c \pm 1)\} \cup \{(r \pm 1, c \pm 2)\}) \cap R\times C #'$$
#'
#' With this definition, we formulate the knight's move constraints compactly as follows:
#'
#' #' \begin{align} #' & & x_{rcl} + x_{r'c'l} & \leq 1 & \forall r,c,l \in R\times C\times L, (r',c') \in K(r,c)\\ #' \end{align} #'
## ----Adding Knight\'s Move Constraints----------------------------------------
get_knights_move_neighbors <- function(r,c) {
# create all possible moves, possibly outside the feasible range

# note that this traversal will encounter each pair of conflicting variables x,y
# twice. The first time when all neighboring cells of x are traversed,
# and a second time when those around y are determined.
# The resulting duplicate rows are detected by Xpress in presolving
neighbors <- matrix(c(
r - 2, c - 1,
r + 2, c - 1,
r - 2, c + 1,
r + 2, c + 1,
r - 1, c - 2,
r + 1, c - 2,
r - 1, c + 2,
r + 1, c + 2
), byrow = T, ncol=2)

# filter elements outside the feasible range and return
neighbors[(neighbors[,1]>=1) & (neighbors[,1]<=9)  & (neighbors[,2]<=9)   & (neighbors[,2]>=1),]
}

# new we can create the knight's move constraints
add_knights_move_constraints <- function(prob, index.df) {
# one additional constraint for each r,c,l

# we establish the R,C,L order in index.df.sorted.
# An element r,c,l has the index 81 * (r - 1) + 9 * (c - 1) + l
index.df.sorted <- index.df %>% arrange(R,C,L)

for (row in 1:9) {
for (col in 1:9) {
neighbors <- get_knights_move_neighbors(row, col)

# add constraints that forbid the same symbol in the neighbors and this cell
# one for each symbol
for (symbol in 1:9) {
cellvaridx <- index.df.sorted$VarIndex[81 * (row - 1) + 9 * (col - 1) + symbol] neighborvarindex <- index.df.sorted$VarIndex[81 * (neighbors[,1] - 1) + 9 * (neighbors[,2] - 1) + symbol]

# add one row per neighbor
for (n in neighborvarindex) {
prob <- addrows(prob, rowtype = "L", rhs = 1, start = c(0,2),
colind = c(cellvaridx, n) - 1,
rowcoef = c(1,1))
}
}
}
}

prob
}

#'
#' We are ready to create the game from Anthony's video. First, we have to create the game by hand.
#'
## -----------------------------------------------------------------------------
# Make a fresh game and override the given symbols by the four digits from the video.
game4digits <- makeGame()
for (i in 1:9) for (j in 1:9) game4digits[i,j] <- NA
game4digits[4,1] <- 3; game4digits[4,2] <- 8; game4digits[4,3] <- 4; game4digits[9,9] <- 2
plot(game4digits)

#'
#' Now we load all the constraints into a fresh XPRSprob object.
#'
## ----Load all constraints-----------------------------------------------------
problist <- load_sudoku_problem(prob = createprob(), sudoku_game = game4digits)
prob4digits <- problist$prob # add diagonals prob4digits <- add_main_diagonal(prob4digits, problist$index.df) %>%
add_anti_diagonal(problist$index.df) %>% # add magic square add_magic_square(problist$index.df) %>%
# add knight's move
add_knights_move_constraints(problist$index.df) setoutput(prob4digits) summary(mipoptimize(prob4digits)) #' #' Here is the solution together with the original game. #' ## ----Plot the solution alongside the game------------------------------------- solution <- getmipsol(prob4digits)$x

solution4digits <- to_sudoku_solution(solution, problist\$index.df)
par(mfrow = c(1,2))
plot(game4digits)
plot(solution4digits)

#'
#'

#'