[algogeeks] Re: permuttaion

[Caio Dias]

OK,

On Thu, Feb 21, 2008 at 8:44 AM, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:

  hi,

  can some one help me in writing an algorithm for finding permuttaion


I will assume that you want all permutation :

let's call:
-  S your set of number
-  pos your current position

** WARNING: Untested code follow:  =)
PERM(pos, S)
   if pos == len(S):
   print S
   for i = pos to len(S):
   swap(S[pos],S[i])
   PERM(pos+1,S)
   swap(S[pos],S[i])

To find all permutation, just call PERM(0,S).

Hope this help,
Caio Valentim

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[algogeeks] Re: factorial

[Caio Dias]

Hi,

On Feb 13, 2008 1:11 AM, Dave [EMAIL PROTECTED] wrote:

 Sum log 1 + log 2 + log 3 + ... + log n, where log represents the
 base-10 logarithm function. The ceiling of the sum gives the number of
 digits.

 Dave

Yeh, that work fine. Only to explain a bit more to Hariharan:

**let N your number

I)- First, make cases by hand and convince yourself  that
the number of digits in a number is equals to exponent of the minor
power of 10(ten) which is
bigger then N.

II)- Remember of the logarithm property learned at high school:

- log(x.y) = log(x) + log(y)

III)- N! = N.(N-1).(N-2)1 = log10(N!)= log10(N.(N-1).(N-2)1)
= log10(N) + log10(N-1) + log(N-2)+...+1

Best regards,
Caio  Valentim

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[algogeeks] Re: Graph problem

[Caio Dias]

On 11/30/07, John [EMAIL PROTECTED] wrote:

 Given a DAG and two vertices s and t , give a linear time  number of
 paths between s and t in G.
 How does topological sort help in finding the same ?

Hi Joh,
you can do a dynamic programming algorithm Your state is S[x] = number
of path going from s to x, s is the initial vertex. It's easy to see
that equation holds :
S[x] = sum { S[v], for  all v neighbor of x }

Think about base cases and make a memorized function to get the answer.
Hope this help,
Caio

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