yes i am interested add me
On 11/19/05, Infinity <[EMAIL PROTECTED]> wrote:
>
> Hi all,
> The group is already there. Please contribute.
> "Art Of Computer Programming Knuth" this is the the group.
>
>
--
Arise awoke stop not till the goal is reached
This is what I came up with: (this is technically C++ though)
void f(int i, int j, int k, bool ri, bool rj, bool rk)
// ri , rj, rk-> whether to recurse on i,j,k or not respectively
{
if (!ri && !rj && !rk)
return;
if ( (i == 0) && ri)
{
f(i
What I meant by "everyone knows the celebrity" is that if C is the
celebrity then the call "knows(P,C)" returns true for all P. But no one
knows who C is. I guess now it's clear.
I'm trying to convert this:
for(int i=0; i<5; ++i)
for(int j=0; j<5; ++j)
for(int k=0; k<5; ++k)
cout << i+1 << j+1 << k+1 << endl;
Into a recursive function. I came up with:
void permute(int n){
for(int i=0; i<5; ++i){
A[n]=i+1;
Sorry, but the original statement of the celebrity problem said
"everyone knows" the celebrity. Either the original statement had an
error, (which I now suspect is the case. Maybe he meant "only one
person knows"), or your post here has an error.
Which is correct?
After visiting Wolfram's site I
Hi all,
The group is already there. Please contribute.
"Art Of Computer Programming Knuth" this is the the group.
Hi Abhi,
About Sol2: The only question that can be asked to P is "do u know Q?",
for some Q. No one knows who the celebrity is, it's our job to find out
by asking these questions. How we eleminate a person for each question
is not clear to me, can you explain that. I think the assumption can be
