Algorithm to send to par-CodePudding

I have a large list of data that I am plotting, however my lists of data can change rather dynamically. It can range from 1 to 20 plots. Now 20 plots is pretty ugly to plot but rarely will I get any more than 10.

So essentially, here's the reasoning behind my algorithm:

For n = 1:
(1,1)
n=2
(1,2)
n=3
(2,2)
n=4
(2,2)
n=5
(2,3)
n=6
(2,3)
n=7
(3,3)
n=8
(3,3)
n=9
(3,3)
..
..

So we add an extra row and once that matches n by the number of plots, then we add 1 to the column, and when the number of plots matches n we add 1 to the row, and continue this repeatedly. So for example, when n=7 we wait until n=9 to fill maximum plots and when n=10 we should get (3,4).

I am having difficulties figuring out the algorithm for this, and would kindly appreciate some help figuring it out!

CodePudding user response：

The following function implements the algorithm as you describe it for a given number of plots n

plot_size <- function(n) {
  
  for(i in seq(n)) {
    if(i == 1) start <- c(1, 1)
    else if(start[1] == start[2] & i > prod(start)) start[2] <- start[2]   1
    else if(start[2] > start[1] & i > prod(start)) start[1] <- start[1]   1
  }
  return(start)
}

We can check it produces the correct output by printing the values from 1 to 20.

lapply(1:20, plot_size)
#> [[1]]
#> [1] 1 1
#> 
#> [[2]]
#> [1] 1 2
#> 
#> [[3]]
#> [1] 2 2
#> 
#> [[4]]
#> [1] 2 2
#> 
#> [[5]]
#> [1] 2 3
#> 
#> [[6]]
#> [1] 2 3
#> 
#> [[7]]
#> [1] 3 3
#> 
#> [[8]]
#> [1] 3 3
#> 
#> [[9]]
#> [1] 3 3
#> 
#> [[10]]
#> [1] 3 4
#> 
#> [[11]]
#> [1] 3 4
#> 
#> [[12]]
#> [1] 3 4
#> 
#> [[13]]
#> [1] 4 4
#> 
#> [[14]]
#> [1] 4 4
#> 
#> [[15]]
#> [1] 4 4
#> 
#> [[16]]
#> [1] 4 4
#> 
#> [[17]]
#> [1] 4 5
#> 
#> [[18]]
#> [1] 4 5
#> 
#> [[19]]
#> [1] 4 5
#> 
#> [[20]]
#> [1] 4 5

^{Created on 2022-06-07 by the reprex package (v2.0.1)}