Explanation. Standard caveat: don’t look here if you are trying to do these yourself.
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
Getting a little trickier already. I initially attempted this one by a recursive brute-force search of each step in the sequence;
## recursively define the Fibonacci sequence
fibonacci <- function(x) {
y <- ifelse(x > 1, fibonacci(x-1)+fibonacci(x-2), x)
return(y)
}
## values in fibonacci < 4e6
z <- 4e6L
## check i=20,30,40
fibonacci(20)
## [1] 6765
fibonacci(30) # takes over 8 seconds
## [1] 832040
# fibonacci(40) # takes too long
But of course these solutions are supposed to be calculable in about a minute so I’ve taken a wrong turn.
It quickly became obvious that I was needlessly re-calculating each step to add another, which is silly, as this explicitly needs all of them each time. I decided to store the sequence as a data.frame
to keep the iteration number alongside it. Note that I use the more correct definition of the sequence which starts with 1, 1, 2. That’s not going to be an issue here, as we’re summing the even values anyway.
## this is silly, why recompute every time?
fibonacci.seq <- data.frame(n=integer(), f=integer())
fibonacci.seq[1,] <- c(1,1)
fibonacci.seq[2,] <- c(1,1)
w <- 2
f <- fibonacci.seq$f[w]
while(f < z) {
w <- w + 1
fibonacci.seq[w,] <- data.frame(n=w, f=fibonacci.seq$f[w-1]+fibonacci.seq$f[w-2])
f <- fibonacci.seq$f[w]
}
head(fibonacci.seq)
## n f
## 1 1 1
## 2 1 1
## 3 3 2
## 4 4 3
## 5 5 5
## 6 6 8
tail(fibonacci.seq)
## n f
## 29 29 514229
## 30 30 832040
## 31 31 1346269
## 32 32 2178309
## 33 33 3524578
## 34 34 5702887
## sum of even values
sum(fibonacci.seq[fibonacci.seq$f %% 2 == 0, "f"])
## [1] 4613732
### CORRECT
devtools::session_info()
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## setting value
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## ctype en_AU.UTF-8
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## date 2019-08-13
##
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