Hi,
1. multiply them as you would do on a paper with O(n^3) multplications
2. use Strassen (a recursive approach) with O(n^{2.81}) multiplications
Regards,
Daniel
mahmoud kiki wrote:
> I need two algorithms to multiplicat two matrix in C++
> and thank you so much
>
> -
Shishir , have a look at my example ... there is no repetition of
points in it. Your formula wont work for it.
-Dhyanesh
On 4/6/06, shishir <[EMAIL PROTECTED]> wrote:
>
> Well the formula works only for the corner end points as starting and
> ending node and that too on the diagonal ends.
> Its
I need two algorithms to multiplicat two matrix in C++ and thank you so much
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Also: no lookup tables or 2D arrays. I'm after a solution using lists
and only lists.
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Hi,
0 < x < y < z < N-1 is the requirement.
Lei
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I'm looking for an algorithm which achieves the following:
Given a directed, unweighted graph G and nodes g[i] and g[j] in G, find
the shortest path from g[i] to g[j]. Use a single breadth-first
traversal of G, marking visited nodes as you go. This is essentially a
tree traversal problem.
The t
Well the formula works only for the corner end points as starting and
ending node and that too on the diagonal ends.
Its not a general formula for any set of starting/ending nodes.
@Dhyanesh
I think the problem clearly states that the nodes can only be traversed
once i.e. no repetition.
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[EMAIL PROTECTED] wrote:
> Is the solution always x = N-4, y = N-3, z = N-2 ?
>
> Suppose we are looking for x and y to minimize the sum.
> Sum = a[i]*x + a[j]*y, where 0 <= i <= x < j <= y <= N.
> It is always bigger than a[i]*x + a[j]*x, because x < y and all numbers
> are positive.
> We have t
This works only if you assume that you do not go backwards i.e. your
max path length is N+M-2. If you can go backwards, then I think that
there are more paths than just these.
e.g. in 3x3 grid you can have path length more than 4
(1,1) -> (1,2) -> (2,2) -> (2,1) -> (3,1) -> (3,2) -> (3,3)
I hav
Check out the problem posted at this link. This was some contest problem posted some time back. Was just trying to solve this problem, but could not come up with an algo better than a brute force. Can you suggest something better?
http://synapse.da-iict.org/events/algorythmus2.htmlThanks,-Vijendra
Can't understand man!!! can u explain and also does your formula take into account the position of starting and ending points or is it just about the corner points that u r talking abt
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There's a small error in my formula. The formula holds for a NxM grid
(N horizontal and M vertical lines) where N= n+1 and M = m+1, which
essentially boils down to
C(N+M-2, N-1) = C(N+M-2,M-1).
This should work fine.
-Shishir
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You received t
missed the m*n in the last fnxn call in recursion area...PLZ CORRECT
No, it's definitely not going the right way..
i wonder if this can be done using that formula...
here's this algo i just thought.
Suppose u've got m horizontal lines and n vertical and u give each of them a
No, it's definitely not going the right way..
i wonder if this can be done using that formula...
here's this algo i just thought.
Suppose u've got m horizontal lines and n vertical and u give each of them an index value. i.e. a 3x2 grid would be like...
___!_(1)_! (2)__
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