>
> Given the conditions for the subgraphs, it's always possible to create
> a graph that limits the size of all the acceptable subgraphs to less
> than
> three nodes.
>
In the problem, there are no constraints on the cardinality of the
subgraphs. It could have a subgraph with all nodes (e.g. if
(reformatting last email)
>
> these as 1 or 4 or something in between?
>
> Also does a single node count as a tree?
>
These are good questions. In my problem, a binary tree is different if
the set of nodes are different. For example:
a We have 9 different binary trees: {a}, {
>
> these as 1 or 4 or something in between?
>
> Also does a single node count as a tree?
>
These are good questions. In my problem, a binary tree is different if
the set of nodes are different. For example:
a We have 9 different binary trees: {a}, {b}, {c},
{a,b}, {b,c}, {b,
Yes, you're right. It depends on the topology of the graph. Do you
have any references for the upper or lower bound? Or even an
asymptotic of form O(2^k)?
Bruno
On Tue, May 13, 2008 at 12:28 PM, Geoffrey Summerhayes
<[EMAIL PROTECTED]> wrote:
>
>
>
>
> On May 12, 8:20 pm, brunotavila <[EMAIL PR
Look to the problem as merge of two sorted array, thus, use the same technique the merge sort uses (linear time).Regards,BrunoOn 8/4/06, deadlock
<[EMAIL PROTECTED]> wrote:How to basically merge the two BST??
--~--~-~--~~~---~--~~
You received this message because
To sort an array A you could use Radix Sort, which complexity is O(d*n), where d = log2(max(A)). So,sort the first array -> O(d*n)sort the second array -> O(e*n)compare the two arrays -> O(n)the solution : (d+e+1)*O(n) = O(n)
Use it only if (d+e+1) is smaller than the hash constant.Bruno Avila2006/
Try looking for
Component Labelling Algorithms, which is a two-pass classic algorithm
that put an label for each connected group of pixels (component) and,
thus, you can also find the total of components (blobs) and their areas.Bruno2006/1/26, H. <[EMAIL PROTECTED]>:
Today in a data structures clas
