Question

So I have a homework to make a flowchart of an algorithm printing first N terms of Fibonacci sequence. It’s not hard, sure, but the teacher told us this could be done in as little as six flowchart “balloons”. And here is the problem – I wanted to do it this way but I just don’t seem to be able to… Namely, I think the shortest way is to check if N>2 – if not, we have to check if it’s 1 or 2 and print 0 or 1 respectively. Only after that can we use the “regular” F(n)=F(n-1)+F(n-2) formula – otherwise, it would crash. Writing more formally:

1. input N
2. N>2?
   • No: check if it’s 1. If so, print 0 and stop.
   • No: check if it’s 2. If so, print 0 and 1 and stop.
   • Yes: proceed to 3
3. Let i=3. While i < N: fib(i)=fib(i-1)+fib(i-2) and print fib(i).
4. Stop.

The problem is, I suppose it would take something around 10 boxes to make this, if not more. What could be the shorter way, then? All the algorithms I find online tend to presume we would get only N more than 2 which may not be the case. Could you please help?

EDIT: OK, I tweaked it to 8 boxes and think it’s as little as one can go. Something like this:

1. Input N
2. Let older=0, younger=1 (respectively: the (n-2)th and (n-1)th term of the series), current=1, i=2.
3. Is N<=1?
   • Yes: output older, end.
   • No: proceed to 4
4. Output older, younger.
5. i < N?
   • Yes: current = older+younger, older = younger, younger = current, i+=1. Print current, go to 5.
   • No: end.

Could something be tweaked further here?

Answer 1

1. Input N
2. Let (x1, x2) = (0, 1)
3. Is N ≤ 0?
   • Yes: Stop
   • No: Proceed to step 4
4. Print x1
5. Let (x1, x2, N) = (x2, x1 + x2, N – 1), and proceed to step 3.

If “Stop” needs to be its own step, that would make step 6.