Generating "Non-Random" Numbers in R? - r

Question

I know how to generate 100 random numbers in R (without replacement):
random_numbers = sample.int(100, 100, replace = FALSE)
I was now curious about learning how to generate 100 "non random" numbers (without replacement). The first comes to mind is to generate a random number, and the next number will be the old number + 1 with a probability of 0.5 or an actual random number with probability 0.5. Thus, these numbers are not "fully random".
This was my attempt to write this code for numbers in a range of 0 to 100 (suppose I want to repeat this procedure 100 times):
library(dplyr)
all_games <- vector("list", 100)
for (i in 1:100){
index_i = i
guess_sets <- 1:100
prob_i = runif(n=1, min=1e-12, max=.9999999999)
guess_i = ifelse(prob_i> 0.5, sample.int(1, 100, replace = FALSE), guess_i + 1)
guess_sets_i <- setdiff(guess_sets_i, guess_i)
all_games_i = as.list(index_i, guess_i, all_games_i)
all_games[[i]] <- all_games_i
}
all_games <- do.call("rbind", all_games)
I tried to make a list that stores all guesses such that the range for the next guess automatically excludes numbers that have already been guessed, but I get this error:
Error in sample.int(1, 100, replace = FALSE) :
cannot take a sample larger than the population when 'replace = FALSE'
Ideally, I am trying to get the following results (format doesn't matter):
index_1 : 5,6,51,4,3,88,87,9 ...
index_2 77,78,79,2,65,3,1,99,100,4...
etc.
Can someone please show me how to do this? Are there easier ways in R to generate "non-random numbers"?
Thank you!
Note: I think an extra line of logic needs to be added - Suppose I guess the number 100, after guessing the number 100 I must guess a new random number since 100+1 is not included in the original range. Also, if I guess the number 5, 17 then 4 - and after guessing 4, the loop tells me to guess 4+1, this is impossible because 5 has already been guessed. In such a case, I would also have to guess a new random number?

Answer 1

It would be tricky to make your algorithm very efficient in R... it doesn't lend itself nicely to vectorization. Here's how I'd write it directly as a for loop:
semirandom = function(n) {
safe_sample = function(x, ...) {
if(length(x) == 1) return(x)
sample(x, ...)
}
result = numeric(n)
result[1] = sample.int(n, size = 1)
for(i in 2:length(result)) {
if(runif(1) < .5 &&
result[i - 1] < n &&
!((result[i - 1] + 1) %in% result)) {
result[i] = result[i - 1] + 1
} else {
result[i] = safe_sample(x = setdiff(1:n, result), size = 1)
}
}
result
}
# generate 10 semirandom numbers 5 times
replicate(semirandom(10), n = 5)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 6 4 4 2 6
# [2,] 3 5 5 3 7
# [3,] 4 3 6 4 5
# [4,] 5 1 2 5 2
# [5,] 7 9 3 6 3
# [6,] 9 10 10 1 1
# [7,] 10 2 8 9 4
# [8,] 2 8 1 8 10
# [9,] 1 7 9 10 9
# [10,] 8 6 7 7 8

Answer 2

You get the error cannot take a sample larger than the population when 'replace = FALSE' because you attempt to extract 100 values from a vector of length one without replacement.
The following draws numbers between 1 and 100, draws each number not more than once, has a 50 percent chance of drawing the previous number + 1 and a 50 percent chance of drawing another random number, if the previous number + 1 has not been drawn yet, and a 100 percent chance to draw another random number, if the previous number + 1 has been drawn.
i <- sample.int(100, 1)
j <- i
for(x in 1:99) {
if((i + 1L) %in% j) {
i <- sample((1:100)[-j], 1L)
} else {
if(runif(1L) > 0.5 || i == 100L) {
i <- sample((1:100)[-j], 1L)
} else {
i <- i + 1L
}
}
j <- c(j, i)
}