Require integer optimization variables to take unique values

Question

I used lp solve to solve a linear programming equation and the solution gives a vector

> lp("max", obj, con, ineqs, rhs, all.int=TRUE,)$solution
[1]  5  0 13 11  4  0  1 11  0

which is fine, but I want each entry in this vector to be an integer between 1-9 and each integer to only be used once. e.g like what the vector looks like underneath.

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

Is there any way this can be done? Thank you in advance!

EDIT

This is the code I have used for the lp function

 obj<-c(1,1,1,1,1,1,1,1,1)
con<-matrix(c(1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,0,0,0,0,1,1,0,1,1,0,0,0,0,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,1,1),nrow=5,byrow=TRUE)
ineqs<-c("=", "=", "=", "=", "=")
rhs<-c(45,20,17,27,15)

Basically what this does is it solves this optimisation problem for the 3x3 grid:

x1 x2 x3
x4 x5 x6
x7 x8 x9

Where the constraints are x1+x2+x4+x5=20, x2+x3+x5+x6=17, x4+x5+x7+x8=27, x5+x6+x8+x9=15, each x must be an integer between 1 and 9 and each x must be unique.

Is your expected result in any way related to the example output you show as output of the lp model? If so, how? — talat
– talat, Commented Nov 5, 2014 at 16:14
Hello, I edited the questions so that you can understand what the lp function does. the constraints where "x must be an integer between 1 and 9 and each x must be unique", is where I think i'm going wrong. — calculator
– calculator, Commented Nov 5, 2014 at 16:58

josliber · Accepted Answer · 2014-11-05 19:10:54Z

The problem with your formulation is that all you're doing is constraining the sum of the values to be 45; there are many sets of 9 integers that sum to 45 but don't take values 1 through 9.

Instead of formulating this with 9 integer-valued variables, you'll probably find it easier to formulate this with 81 binary-valued variables x_ij, where each variable indicates if x_i (in the original formulation) takes value j.

# Helper functions to index variables
unit <- function(idx) as.numeric((1:9) == idx)
ivals <- function(i) { ret <- rep(0, 81) ; ret[(9*i-8):(9*i)] <- 1:9 ; ret }
ivars <- function(i) rep(unit(i), each=9)
jvars <- function(j) rep(unit(j), 9)

# Setup and solve optimization model
obj <- rep(0, 81)
const <- rbind(do.call(rbind, lapply(1:9, jvars)),  # Each value once
               do.call(rbind, lapply(1:9, ivars)),  # Each var once
               ivals(1) + ivals(2) + ivals(4) + ivals(5),
               ivals(2) + ivals(3) + ivals(5) + ivals(6),
               ivals(4) + ivals(5) + ivals(7) + ivals(8),
               ivals(5) + ivals(6) + ivals(8) + ivals(9))
ineqs <- rep("=", 22)
rhs <- c(rep(1, 18), 20, 17, 27, 15)
library(lpSolve)
res <- lp("max", obj, const, ineqs, rhs, all.bin=TRUE)
apply(matrix(res$solution, nrow=9), 2, which.max)
# [1] 3 7 5 6 4 1 9 8 2

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