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Space complexity of recursive function
Discuss Space complexity of recursive function in the C Programming forum on Dev Shed. Space complexity of recursive function C programming forum discussing all C derivatives, including C#, C++, Object-C, and even plain old vanilla C. These languages are low level languages, and used on projects such as device drivers, compilers, and even whole computer operating systems.
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January 23rd, 2013, 09:07 AM
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Space complexity of recursive function
Hi,
this function has to check if an array is not descending it has to be recursive and it's space complexity has to be O(log(n))
does it meet these requirements?
Code:
int is_sorted(int a[ ], int n)
{
int check=n-1;
if(n==1 || n==0) return 1;
if(a[check]<a[check-1])return 0;
else return(is_sorted(a,n-1));
}
if not. can anyone propose how to meet those demands?
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January 23rd, 2013, 09:30 AM
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Banned ;)
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Join Date: Nov 2001
Location: Woodland Hills, Los Angeles County, California, USA
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Well, it is recursive, but it appears to be O(N), not O(log(N))
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January 23rd, 2013, 10:33 AM
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Every element must be examined! Therefore the algorithm must be O(n) or worse. [edit]O(n) time complexity. The problem statement specifies SPACE complexity. This is possible.[/edit]
(Or show my logic error.)
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Last edited by b49P23TIvg : January 23rd, 2013 at 11:43 AM.
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January 23rd, 2013, 11:07 AM
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so this is a HW assignment and after asking they say there is a way to make it's space complexity O(log(n)). any ideas on how?
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January 23rd, 2013, 11:42 AM
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Contributing User
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Oh, space complexity, not time complexity.
Sure, and since the algorithm must be recursive it's important to keep the space complexity small.
An O(log(n)) space complexity algorithm divides intervals in half.
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January 23rd, 2013, 11:45 AM
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ok so how is it done?
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January 23rd, 2013, 12:40 PM
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Contributing User
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Code:
int is_sorted(int a[ ], int n) {
if (n < 2) return 1;
if (2 == n) return a[0] <= a[1];
return is_sorted(a,n/2) && is_sorted(a+n/2,n/2);
}
approximately like this. My usual warning applies: my programs don't work the first time and I didn't test this. Certainly could have off by one errors, and certainly could have major logic flaws.
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January 23rd, 2013, 12:57 PM
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first of all thanks,it works perfectly...can you please explain why?
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January 23rd, 2013, 02:58 PM
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Contributing User
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Quote: | Originally Posted by daniig first of all thanks,it works perfectly...can you please explain why? | Are you quite sure you tried it with ordered and unordered lists of length prime number? Like 17.
Sprinkle some useful diagnostic print statements and I'm sure you'll figure out why it works or not. To help with the diagnostics, rewrite as
Code:
int is_sorted(int a[], int n,int level) {
/* useful print statement here, include level */
if (n < 2) return 1;
if (2 == n) return a[0] <= a[1];
return is_sorted(a,n/2,level+1) && is_sorted(a+n/2,n/2,level+1);
}
and call is_sorted with a known value for level. 0, perhaps.
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January 23rd, 2013, 06:26 PM
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Banned ;)
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Join Date: Nov 2001
Location: Woodland Hills, Los Angeles County, California, USA
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Quote: | Originally Posted by daniig first of all thanks,it works perfectly...can you please explain why? |
That means you didn't test it all that well. Try and understand what the code is doing and put a few printf() calls to see if you can figure out what's going on.
HINT: Is {5, 20, 7, 15} sorted or not? How about {5, 20, 7, 34, 15}
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January 24th, 2013, 12:40 AM
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yeah I've actually found other cases it doesn't work that well...but the original "why" was cause i don't understand what the function is doing not why should i check it.
so 2 questions:
1. what is it doing?
2.how to change it so it'll work for all cases?
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January 24th, 2013, 08:35 AM
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Contributing User
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a+n/2 is the same as
&a[n/2]
This might help.
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