Linear integer programming in R?

Question

I have a vector of 100 integers, x1, where each element of the vector is in [1, 7]. I want to find, using R, another vector of 100 elements, x2, also containing integers ranging from 1 to 7, that minimizes the following function:

abs(sum(x2 - x1 > 1) - 50)

In other words, I want as close to 50% of the x2 values to be larger than the x1 values by more than 1. There are no other constraints, except that x2 must be between 1 and 7.

Here is the data for x1:

> dput(x1)
c(3L, 4L, 2L, 5L, 1L, 3L, 2L, 5L, 2L, 2L, 1L, 2L, 2L, 2L, 3L, 
5L, 3L, 3L, 3L, 1L, 5L, 2L, 2L, 7L, 2L, 4L, 2L, 2L, 3L, 3L, 1L, 
5L, 2L, 4L, 3L, 2L, 4L, 3L, 3L, 3L, 3L, 2L, 2L, 3L, 3L, 3L, 3L, 
2L, 3L, 4L, 3L, 2L, 4L, 3L, 2L, 3L, 3L, 4L, 5L, 3L, 2L, 3L, 3L, 
3L, 4L, 2L, 4L, 4L, 4L, 2L, 4L, 2L, 4L, 3L, 2L, 3L, 3L, 2L, 3L, 
3L, 3L, 2L, 2L, 4L, 3L, 2L, 6L, 5L, 5L, 3L, 3L, 2L, 3L, 2L, 3L, 
3L, 4L, 3L, 2L, 2L)

You mean to minimize the absolute value of the function? Also, you must reorder x1 or can choose any number from 1 to 7? — Ric, Jul 21 '23 at 06:23
Yes, sorry. You are correct. I've edited the question. :) x1 and x2 cannot be shuffled (they belong to the same person). — Edward, Jul 21 '23 at 06:24
Homework questions with no effort at creating code is often met with resistance and requests to show effort. — IRTFM, Jul 21 '23 at 06:31
This is not homework. It is work work. And I am not familiar with integer programming in R, so I do not know how to even begin doing this. — Edward, Jul 21 '23 at 06:32
In your simple case if there are no other constraints, `x2 <- ifelse(1:100 %in% which(x1 <= 5)[1:50], x1 +2, x1)` solves the question — Ric, Jul 21 '23 at 06:34
Hmmm - I guess that's true. I think I need to reformulate the problem better. >.< Thank you @Ric ! — Edward, Jul 21 '23 at 06:38

score 3 · Accepted Answer · answered Jul 21 '23 at 06:48

Simulate samples x2 and find the function's minimum, which is zero.

x1 <- c(3L, 4L, 2L, 5L, 1L, 3L, 2L, 5L, 2L, 2L, 1L, 2L, 2L, 2L, 3L, 
        5L, 3L, 3L, 3L, 1L, 5L, 2L, 2L, 7L, 2L, 4L, 2L, 2L, 3L, 3L, 1L, 
        5L, 2L, 4L, 3L, 2L, 4L, 3L, 3L, 3L, 3L, 2L, 2L, 3L, 3L, 3L, 3L, 
        2L, 3L, 4L, 3L, 2L, 4L, 3L, 2L, 3L, 3L, 4L, 5L, 3L, 2L, 3L, 3L, 
        3L, 4L, 2L, 4L, 4L, 4L, 2L, 4L, 2L, 4L, 3L, 2L, 3L, 3L, 2L, 3L, 
        3L, 3L, 2L, 2L, 4L, 3L, 2L, 6L, 5L, 5L, 3L, 3L, 2L, 3L, 2L, 3L, 
        3L, 4L, 3L, 2L, 2L)

fobj <- function(x, y, const = 50L) abs(sum( (y - x) > 1L) - const)

R <- 1000L

mat <- replicate(R, sample(7, length(x1), TRUE))
vec <- apply(mat, 2, fobj, x = x1)
i_min <- which(vec == min(vec))

# any of these columns is a solution to the problem
# mat[, i_min]

# show a few solutions
mat[, head(i_min)]
#>        [,1] [,2] [,3] [,4] [,5] [,6]
#>   [1,]    6    7    5    4    4    4
#>   [2,]    1    5    6    3    6    2
#>   [3,]    7    1    2    4    3    2
#>   [4,]    3    7    5    5    2    7
#>   [5,]    4    4    7    6    4    7
#>   [6,]    2    1    2    5    1    1
#>   [7,]    5    5    6    6    2    2
#>   [8,]    7    7    2    6    7    6
#>   [9,]    6    6    7    6    4    2
#>  [10,]    7    1    4    6    4    1
#>  [11,]    2    3    5    6    7    2
#>  [12,]    6    3    4    4    7    6
#>  [13,]    6    1    6    5    2    3
#>  [14,]    4    7    7    1    4    3
#>  [15,]    5    6    7    2    1    2
#>  [16,]    5    1    6    7    4    1
#>  [17,]    3    4    2    5    4    3
#>  [18,]    7    1    5    7    7    3
#>  [19,]    3    1    4    5    7    7
#>  [20,]    1    5    7    4    4    5
#>  [21,]    4    4    2    1    6    5
#>  [22,]    3    5    2    5    2    7
#>  [23,]    2    5    6    6    1    7
#>  [24,]    3    1    5    3    1    6
#>  [25,]    7    1    7    2    2    2
#>  [26,]    1    3    4    7    3    6
#>  [27,]    4    5    1    6    7    1
#>  [28,]    6    4    4    5    7    1
#>  [29,]    1    5    2    1    1    7
#>  [30,]    5    3    7    3    7    3
#>  [31,]    6    7    1    4    4    2
#>  [32,]    2    6    2    3    5    4
#>  [33,]    5    6    5    7    5    6
#>  [34,]    6    7    7    2    2    6
#>  [35,]    1    4    7    7    6    1
#>  [36,]    5    4    6    3    1    7
#>  [37,]    7    4    1    1    4    6
#>  [38,]    7    1    7    7    4    5
#>  [39,]    4    1    4    7    5    7
#>  [40,]    7    5    2    4    7    3
#>  [41,]    2    3    7    3    4    6
#>  [42,]    3    6    4    4    7    7
#>  [43,]    2    2    1    7    4    6
#>  [44,]    7    7    3    2    5    7
#>  [45,]    3    1    5    3    5    4
#>  [46,]    7    5    5    3    4    6
#>  [47,]    4    4    2    5    5    7
#>  [48,]    4    3    4    6    2    4
#>  [49,]    4    4    6    7    6    5
#>  [50,]    5    6    2    1    4    2
#>  [51,]    6    5    1    4    3    1
#>  [52,]    1    5    4    3    3    3
#>  [53,]    6    6    7    3    7    2
#>  [54,]    5    6    7    4    7    5
#>  [55,]    2    6    6    7    4    5
#>  [56,]    2    3    4    7    2    5
#>  [57,]    4    3    7    3    5    5
#>  [58,]    7    2    3    5    7    5
#>  [59,]    1    4    1    2    5    6
#>  [60,]    1    2    6    3    3    6
#>  [61,]    3    2    4    2    2    5
#>  [62,]    6    5    7    6    5    7
#>  [63,]    5    3    1    1    7    7
#>  [64,]    2    1    1    4    6    5
#>  [65,]    6    7    2    2    7    5
#>  [66,]    2    3    1    5    4    6
#>  [67,]    5    2    3    1    4    3
#>  [68,]    2    6    1    5    3    2
#>  [69,]    2    4    4    3    6    1
#>  [70,]    6    6    2    3    6    5
#>  [71,]    3    4    5    3    5    5
#>  [72,]    7    4    4    6    7    6
#>  [73,]    6    6    5    3    6    1
#>  [74,]    3    6    4    5    5    3
#>  [75,]    4    3    3    3    7    4
#>  [76,]    7    6    6    2    4    7
#>  [77,]    5    7    2    5    1    5
#>  [78,]    4    4    6    4    2    3
#>  [79,]    7    7    1    6    6    5
#>  [80,]    4    1    6    1    4    5
#>  [81,]    1    6    5    3    5    4
#>  [82,]    7    5    3    6    3    2
#>  [83,]    5    5    1    5    5    5
#>  [84,]    3    4    3    7    2    6
#>  [85,]    1    4    6    3    6    3
#>  [86,]    7    1    7    5    2    6
#>  [87,]    2    4    6    6    2    7
#>  [88,]    5    7    6    5    5    5
#>  [89,]    1    2    7    7    3    2
#>  [90,]    6    7    4    2    2    1
#>  [91,]    1    5    2    3    5    6
#>  [92,]    5    3    7    7    2    5
#>  [93,]    3    5    7    5    2    7
#>  [94,]    6    1    5    5    3    6
#>  [95,]    6    3    6    7    3    6
#>  [96,]    4    7    7    7    2    4
#>  [97,]    2    7    6    3    7    7
#>  [98,]    5    2    4    1    6    2
#>  [99,]    5    2    1    7    6    3
#> [100,]    3    7    3    6    5    7

tbl <- table(vec)

# around 3% of the samples is a solution
proportions(table(vec))
#> vec
#>     0     1     2     3     4     5     6     7     8     9    10    11    12 
#> 0.036 0.075 0.054 0.059 0.090 0.081 0.079 0.079 0.088 0.076 0.073 0.059 0.041 
#>    13    14    15    16    17    18    19    20    21 
#> 0.035 0.030 0.016 0.008 0.008 0.001 0.007 0.004 0.001
# plot the results, the bar we want is the first
barplot(tbl)

^{Created on 2023-07-21 with reprex v2.0.2}

Could be! Let me check carefully first – Edward Jul 21 '23 at 07:10 — Edward, Jul 21 '23 at 07:10
I think that's good enough. Thanks @Rui. :) – Edward Jul 21 '23 at 08:32 — Edward, Jul 21 '23 at 08:32

score 2 · Answer 2 · answered Jul 21 '23 at 21:08

I don't think you need to solve the problem from a optimization perspective. Once you have observed the properties of x1, you can easily "construct" a x2 that meets your objective.

For example, you can focus on the elements that are less than 6 in x1 such that you have margin to make x2[k] - x1[k] > 1. The code below is one possible implementation

x2 <- x1
idx <- sample(which(x1 < 6), 50)
x2[idx] <- pmin(x2[idx] + sample(2:7, length(idx), TRUE), 7)

and you can check that

> abs(sum(x2 - x1 > 1) - 50)
[1] 0

Linear integer programming in R?

2 Answers2