Profiling of inefficient recursive Fibonacci series function

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I am asked to profile an intentionally inefficient code. Following code will overtime with a huge input:
function f = fibo(n)
if n <= 2
f = 1;
else
f = fibo(n-2) + fibo(n-1);
end
end
I used a persistente variable, to create an array that reflects the recursion calls:
function [f, trace]=fibo_trace(n,v)%with persistent variable 'trc'
persistent trc;
if isempty(trc)
trc=v;
end
if n<=2
trc=[trc n];
f=1;
else
trc=[trc n];
f=fibo_trace(n-2)+fibo_trace(n-1);
end
trace=trc;
end
That works flawlessly when I run the function with followin code:
[f trace] = fibo_trace(6,[])
And this is what the code returns:
f = 8
trace = 6 4 2 3 1 2 5 3 1 2 4 2 3 1 2,
which is OK, since the trace vector shows how inefficient the original code is. Now my problem is, it is an assigment for a MOOC platform and the solution with the persistent variable won't work because the the automated grader will run the function a number of times with random inputs and it will not clear the function. Without the persistent variable I must split the recursive calls in two code lines and I am struggling now to compute the nth fibonacci number:
Can anyone give a clue? The creation of the array works but I honestly don't know how to deal with the two recursive call lines to compute the fibonacci.

采纳的回答

Stephen23
Stephen23 2020-12-23
编辑:Stephen23 2020-12-23
Perhaps something like this:
[val,vec] = fibo(6)
val = 8
vec = 1×15
8 5 3 2 1 1 1 2 1 1 3 2 1 1 1
function [f,t] = fibo(n)
if n <= 2
f = 1;
t = f;
else
[f2,t2] = fibo(n-2);
[f1,t1] = fibo(n-1);
f = f2+f1;
t = [f,t1,t2];
end
end
It might be including the 1's repeatedly, so that gives you something to debug :)
  5 个评论
Carlos Rueda
Carlos Rueda 2021-8-13
It might be a bit confusing and it is a while ago since I did this. But here some intuition:
The first recursive call assigns a new value to trc taking the current value of trc as an asgument.
The second recursive call assigns again a new value to trace taking as argument the trc value from the first recursive call. Maybe it would be more illustrative to rewrite the code like this:
else
trc=[trc n];
a=n-1;
[a trc1]=fibo_trace(a-1,trc);
[n trc2]=fibo_trace(n-1,trc1);
f=n+a;
end
You try it and tell me.
Lusty Lloyd
Lusty Lloyd 2023-3-13
function [f, trace] = fibo_trace(n, v)
if isempty(v)
v = [];
end
v = [v n];
if n <= 2
f = 1;
else
[f1, v] = fibo_trace(n-2, v);
[f2, v] = fibo_trace(n-1, v);
f = f1 + f2;
end
trace = v;
end

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更多回答(2 个)

Rajith
Rajith 2023-12-17
function [f, v] = fibo_trace(n,v)
v(end+1) = n;
if n == 1 || n == 2
f = 1;
else
[f1, v] = fibo_trace(n-2,v);
[f2, v] = fibo_trace(n-1,v);
f = f1+f2;
end
end

Divyanshu
Divyanshu 2024-7-28
function [f, v] = fibo_trace(n,v)
v(end+1) = n;
if n == 1 || n == 2
f = 1;
else
[f1, v] = fibo_trace(n-2,v);
[f2, v] = fibo_trace(n-1,v);
f = f1+f2;
end
end

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