I am writing a method that returns the nth element in the Fibonacci sequence, but am running into an unexpected end error.
def fib_seq(n)
a = [0]
n.times do |i|
if i==0
a[i] = 0
else if i==1
a[i] = 1
else
a[i] = a[i-1] + a[i-2]
end
end
return a[n]
end
puts fib_seq(4)
Any tip on what I can be screwing up on?
Assuming you are trying to return the nth (and not the (n-1)th, i.e. fib(1) = 0 NOT fib(0) = 0).
I fixed it by changing:
else if i==1
to
elsif i==1
(AND)
return a[n]
to
return a[n - 1]
So your final code should look like:
def fibSeq(n)
a = [0]
n.times do |i|
if i==0
a[i] = 0
elsif i==1
a[i] = 1
else
a[i] = a[i-1] + a[i-2]
end
end
return a[n-1]
end
puts fibSeq(4)
Per your comment below, the following code will work:
def fibSeq(n)
a = [0]
(n+1).times do |i|
if i==0
a[i] = 0
elsif i==1
a[i] = 1
else
a[i] = a[i-1] + a[i-2]
end
end
return a[n]
end
puts fibSeq(4)
If you want to output the fibs as a list then use:
return a[0..n]
Instead of
return a[n]
elsif is what a c-style programmer (let's just say that means C, C++, Java, C#, etc...) would know as else if. I believe that in ruby else if actually does two things it, (1) acts as the else from the above if statement, then it (2) acts as an if.
Very optimised code
Version 1
def fib(n, cache)
return 1 if n <= 2
return cache[n] if cache[n]
cache[n] = fib(n - 1, cache) + fib(n - 2, cache)
end
cache = {}
puts fib(5,cache)
Version 2
def fib(n)
return 1 if n <= 2
fib_index = 3
a, b = 1, 1
while fib_index <= n
c = a + b
a = b
b = c
fib_index += 1
end
c
end
puts fib(100)
If you go recursive this is the seudo-code:
fib(n) {
if n = 0 -> return 0
else if n = 1 -> return 1
else -> return fib(n-1) + fib (n-2)
}
lowercase_with_underscoresfor method and variables.