CODEWITHSUNDEEP
rcollatz
MrrMan
MrrMan

Reputation: 158

Return Collatz Conjecture for matrix with n<-1:30 with R loop

I am new to R. There is plenty on here and the rest of the web around writing functions to return the Collatz conjecture for a given integer.

As such, i have managed to create a function that will: 

numseq <- function(num)
{
  if(num == 1)
    return(NULL)

  if(num %% 2 == 0)
    return(num / 2)

  return(3 * num + 1)
}

seqfetch <- function(input)
{
  result <- input
  while(input != 1)
  {
    input <- numseq(input)
    result <- c(result, input)
  }
  return(result)
}

seqfetch(15)

However i have a vector nums<-1:30 and am trying to loop through each element of it and run the function against it, outputting the values for each iteration and each value of n.

I am aiming for:

n = 2: [1] 2 1

n = 3: [1] 3 10 5 16 8 4 2 1

...

...

...

n = 30: [1] 30 15 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1

And so on...

I dont know a lot about R and have tried wrapping the whole thing in a loop. I have also tried the same thing with a matrix of one column. I get an error saying that only the first element will be used.

Can anyone help with the loop please? Or am i going about it the wrong way?

Thanks

Upvotes: 0

Views: 105

Answers (2)

Rui Barradas
Rui Barradas

Reputation: 76683

Here is another way, rewriting numseq as a recursive function.

numseq <- function(num) {
  if(num == 1)
    num
  else if(num %% 2 == 0)
    c(num, numseq(num / 2))
  else
    c(num, numseq(3 * num + 1))
}

Now all you have to do is

lapply(1:30, numseq)

Or, in the form of a function,

seqfetch2 <- function(input){
  lapply(input, numseq)
}

seqfetch2(1:30)

Upvotes: 1

MrrMan
MrrMan

Reputation: 158

I managed to acheive the desired output using sapply(). By wrapping function 2 in sapply() as below:

sapply(1:30, function(input)
{
  result <- input
  while(input != 1)
  {
    input <- numseq(input)
    result <- c(result, input)   
  }
  return(result)
}
)

I acheived the desired output of:

[[1]]
[1] 1

[[2]]
[1] 2 1

[[3]]
[1]  3 10  5 16  8  4  2  1

[[4]]
[1] 4 2 1

[[5]]
[1]  5 16  8  4  2  1

[[6]]
[1]  6  3 10  5 16  8  4  2  1

[[7]]
 [1]  7 22 11 34 17 52 26 13 40 20 10  5 16  8  4  2  1

[[8]]
[1] 8 4 2 1

[[9]]
 [1]  9 28 14  7 22 11 34 17 52 26 13 40 20 10  5 16  8  4  2  1

And so on through to 30.

Upvotes: 0

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