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Time complexity of algorithm
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Thank you for a quick reply. I have a long program with me. I wonder it will take many days to calculate its complexity.
any tutorial or website to help me...i want to get the time complexity of this algo using summation and big o...i really find it difficult to understan...can you help me please solve step by step
add=0;
for (a=1 to n) do
for b=1 to a*a do
for c=1 to b do
add=add+c;
anyone please help
add=0;
for (a=1 to n) do
for b=1 to a*a do
for c=1 to b do
add=add+c;
anyone please help
Last edited by vbcielle; Mar 11th, 2007 at 3:58 am.
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Join Date: Jun 2005
Posts: 2,047
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Yes, there are tutorials and websites to help you. Try reading the rest of this thread...
All my posts may be redistributed under the GNU Free Documentation License.
I think in this case, addition is based on for loops.
And there are 3 loops in your given algo.
An exmaple of three loops is also present on this thread or of a similar kind of thread.
I suggest you read them first and you will be able to find a solution by yourself.
And there are 3 loops in your given algo.
An exmaple of three loops is also present on this thread or of a similar kind of thread.
I suggest you read them first and you will be able to find a solution by yourself.
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Originally Posted by Rashakil Fol 
Yes, there are tutorials and websites to help you. Try reading the rest of this thread...
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Originally Posted by fahimarif
I think in this case, addition is based on for loops.
And there are 3 loops in your given algo.
An exmaple of three loops is also present on this thread or of a similar kind of thread.
I suggest you read them first and you will be able to find a solution by yourself.
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Originally Posted by vbcielle
any tutorial or website to help me...i want to get the time complexity of this algo using summation and big o...i really find it difficult to understan...can you help me please solve step by step
add=0;
for (a=1 to n) do
for b=1 to a*a do
for c=1 to b do
add=add+c;
anyone please help
Looks like n^5 to me. The outermost loop executes n times, the next-inner one on the order of a^2, which in the worst case is n^2, and the inner-most to that same n^2 worst-case.
Multiply these out to find the minimum number of executions and they will be on the order of n^5.
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Originally Posted by Gotcha
Verifying if anyone could give me a hand solving this excersices:
Instructions:
Find the Big Oh for the times which is excecuted x : = x + 1. You should justify your answer:
I : = N
WHILE I >= 1 DO
BEGIN
FOR J : = 1 TO N DO
X : = X + 1
I = [I/2]
END
Ya , I think the solution to the given program is correct. The outer loop divids the sampling increment by 2. Hence it is log N and the inner loop is straight forward. Its N.
so solution is O(NlogN)
so solution is O(NlogN)
Can someone help on this one please?
If I have this:
Would the time complexity of this be 2n^2 therefore O(n^2)?
Am I counting it correctly? An if statement condition within the second FOR loop is run n times and the second FOR loop is run n times, therefore 2n and then the first FOR loop runs n times, therefore (2n * n) operations?
If I have this:
public static boolean find(int [] a, int k) {
for (i = 0; i < a.length; i++) {
for (j = 0; j < a.length; j++) {
if ((a[i] + a[j]) == k) {
return true;
}
}
}
return false;
}Would the time complexity of this be 2n^2 therefore O(n^2)?
Am I counting it correctly? An if statement condition within the second FOR loop is run n times and the second FOR loop is run n times, therefore 2n and then the first FOR loop runs n times, therefore (2n * n) operations?
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