Why does this function works? - r

Question

with this function you can calculate the fibonacci sequence with a recursive function, but i am not sure why this works, i marked at which position i struggled, can someone explain me this code?
fib <- function(n){
if (n == 0) return(0)
if (n == 1) return(1)
seq <- integer(n) # at this point i didnt understand much at all
seq[1:2] <- 1
calc <- function(n) {
if (seq[n] != 0) return(seq[n])
seq[n] <<- calc(n-1) + calc(n-2)
seq[n]
}
calc(n)
}

Answer 1

In the course of the recursive function evaluation, fib() ends up getting called many times with the same n. One way to make this computation faster is to use memoization, which saves a record of values for which the function has previously been called, and the return values. Using #AnoushArivanR's fib function:
system.time(fib(30))
## user system elapsed
## 3.987 0.000 3.987
library(memoise)
fib <- memoise(fib)
system.time(fib(30))
## user system elapsed
## 0.004 0.000 0.004
In fact, now that I look at your code above, I believe it's doing exactly this — but it's definitely hard to understand! (Your question might have been better received if you explained that this was what you're trying to do ...)

Answer 2

Note
This implementation of fib function has been cited from Mastering Software Development in R by Dr. Roger D. Peng who taught me so much and I am forever grateful to him.
As mentioned above this code is poorly written and has been unnecessarily complicated. Here is a more simpler version. We first check that the n value is not less than 0, then as the first two elements of the sequence are necessary for the series calculation to start (each element being the sum of two previous elements in the series) we set them as 0 and 1 for n == 1 and n == 2 respectively. Then we use recursion which is a technique that a function calls itself from its body creating a series of repetitive computations until the maximum number of n is reached. So for example for n == 3 the function calls it self by calling fib(1) and fib(2) both of which have already been set and so on. Then for fib(4) the function calls both fib(3) and fib(2). fib(2) is already set and fib(3) will be calculated by summing fib(2) and fib(1) and ...
I hope this explanation has helped you get your mind around the idea.
fib <- function(n){
stopifnot(n > 0)
if(n == 1) {
return(0)
} else if(n == 2) {
return(1)
} else {
fib(n - 1) + fib(n - 2)
}
}
fib(7)
8
But as some of the calculations are computed more than once for example both fib(6) and fib(5) calculate fib(4) the execution of the function gets slower. For the sake of optimization your code has saved every fib(n) output into and empty vector called seq so before any computation takes place it checks whether the value for seq[n] has already been computed or not. If so, it will be used and if not it will be computed again. This technique is called memoization and whenver a new seq[n] is calculated it will be the nth element of the seq vector and for this purpose we make use of <<- called complex assignment operator as we are modifying an object in the parent environment of the function.