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Assignment 2 - Q3, Q5

Posted by Anonymous User 
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Anonymous User
Assignment 2 - Q3, Q5
May 14, 2008 07:52PM
There's allot of posts going out on assignment 2, and here's mine...

Can't help thinking there is something amiss with Q3, the recursive function does seem to achieve anything.

running funcRec(5,'A'winking smiley simply prints B - I get it, it converts (casts) the char v into a int, add 1 and cast this into a char before displaying it.

the recursive call manages to decrement u each round. Anyway, lemme know if there supposed to be more...

In terms of Question 5, optimise fibonacci recursive by taking the thinking behind the itterative approach. The problem here is that fibonacci requires tracking to two values. 0 1 1 2 3 5 8 13 21 ... you need to have 0+1 = 1, and 1+1=2, we all know the sequence. But to keep track of two variables when only 1 is returned by a function is a little tricky? hence the original recursive calculates previous and the one before the previous.

I've tried a few approaches and have not solved this one yet, maybe this is something stupid like

int fibonacci( int, int * )

and the function can now "return" two variables... will see what I can come up with - not looking to be spoonfed the answer, just looking for a hint...
avatar Re: Assignment 2 - Q3, Q5
May 16, 2008 05:17PM
Q3 : Yes, As far as I can see the function is a bit stupid.

Q5 : I hate how this question was asked. I'm really not happy with the sample code. Hint from my solution... *cough* array *cough*.
Anonymous User
Re: Assignment 2 - Q3, Q5
May 16, 2008 05:37PM
This is what I managed to come up with... but it's skewed, each result is a fibonacci number but off by n*2...

I gave up and submitted like as is. Agree, array will work...

// recursive function
int fibonacci( int n, int &oldprev )
int prev;

if( n == 0 )
oldprev = 0;
return 0;
if( n == 1 )
oldprev = 1;
return 1;

prev = fibonacci( n-1, oldprev );
oldprev = oldprev+prev;

return oldprev+prev;

// driver function
int fibonacci( int n )
int oldprev;
return fibonacci( n, oldprev );
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