[algogeeks] Re: reverse all the bits in a unsigned integer

[[EMAIL PROTECTED]]

On Mar 21, 7:40 am, hijkl [EMAIL PROTECTED] wrote:
 . How to reverse all the bits in a unsigned integer? an signed
 integer (you have to keep the sign of the integer)?

For an unsigned int, the following algorithm can be used (I believe
you can find in in the Hacker's Delight book by H.Warren)  The idea
is to to swap bits with increasing distance each time:

uint bit_reverse (uint x)
{
  x = ((x  0x)  1) | ((x  0x)  1);
  x = ((x  0x)  2) | ((x  0x)  2);
  x = ((x  0x0f0f0f0f)  4) | ((x  0xf0f0f0f0)  4);
  x = ((x  0x00ff00ff)  8) | ((x  0xff00ff00)  8);
  x = ((x  0x)  16) | ((x  0x)  16);
  return x;
}

The function doesn't have if's and should work pretty fast.  For 64-
bit numbers the solution is similar (masks are of double width and yet
another operation, shifting by 32 bits before return.

Regards,
Igor


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[algogeeks] A hanoi based problem

[Mahdi]

hello!
It's a problem with all specifications of hanoi( 3 towers and and
ordered rings) except that rings are not all in the same tower. They
are distributed in 3 towers ( in each tower they are ordered ) and we
want to bring them all in tower C ( assuming towers A,B,C). We want
the solution with minimum time.
I thank if you can answer it! (perhaps with a pseudocode)


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[algogeeks] Re: count the 1 bits in integer

[vim]

how abt my code

#includestdio.h
int count1bitsn(int n)
{
   int count=0;
   while(n)
   {
   count+=n1;
   n=n1;
}
}
int main()
{
  int n;
  printf(Enter the value of n\n);
  scanf(%d,n);
  printf(Answer=%d\n,count1bitsofn(n));
}






On Mar 22, 12:39 am, Karthik Singaram L [EMAIL PROTECTED]
wrote:
 The standard linkhttp://graphics.stanford.edu/~seander/bithacks.html


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[algogeeks] Re: count the 1 bits in integer

[vim]

how abt my code

#includestdio.h
int count1bitsn(int n)
{
   int count=0;
   while(n)
   {
   count+=n1;
   n=n1;
}
return count;
}

int main()
{
  int n;
  printf(Enter the value of n\n);
  scanf(%d,n);
  printf(Answer=%d\n,count1bitsofn(n));

}

On Mar 22, 12:39 am, Karthik Singaram L [EMAIL PROTECTED]
wrote:
 The standard linkhttp://graphics.stanford.edu/~seander/bithacks.html


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[algogeeks] Re: 2D arrays

[Luciano Pinheiro]

char a [n][n];
int i, j, m;

m = n/2;
if ( n % 2 == 0) m--;

for(i = 0; i = m; i++) {
   for(j = 0; j = i; j++) {
  a[i][j] = '*';
  a[n - i - 1][j] = '*';
   }
   for(j = i + 1; j  n; j++) {
   a[i][j] = '-';
   a[n - i - 1][j] = '-';
   }
}

2007/3/18, BiGYaN [EMAIL PROTECTED]:


 Please don't post homework questions here. This problem is too simple
 for a group calling themselves Algorithm Geeks.

 I hope most of you'll agree to this and put a stop to this practice.


 


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[algogeeks] Re: traverse the tree layer by layer.

[tuesday]

u can use a queen in bfs traverse

void visit(Node* r)
{

}
void BF_Traverse(Node* r)
{
   queenNode* q;
   q.push(r);
   while(!q.empty())
   {
  Node* t = *q.front();
  q.pop();
  visit(t);
  if(t-left !=0)
 q.push(t-left);
  if(t-right!=0)
 q.push(t-right);
   }
}


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