sample N个数中随机选M个

 

void RandomSelectMofN(int *a, int N, int M)...{
  // pls refer to Knuth's TAOCP
  int n;
  srand(unsigned(time(NULL))); 
  for(n=N;n>0;n--)...{
    if( rand()%n < M )...{  // assume bigrand.
      cout<<N-n<<endl;
      m--;
    }
  }
}

 

let's prove this algorithm:

for convenience ,change to for(int i=0;i<n;i++){}

now, we have randomly selected t elements in i elements,  then we want to selected the next one(named x).

there are C(n-i, m-t) kinds of choice to randomly select the last m-t elements.

and, there are C(n-i-1, m-t-1) kinds of choice to select the last m-t elements which contains the ith element(x).

so C(n-i, m-t) / C(n-i-1, m-t-1) = m-t / n-i, which is the probability of selecting ith element in our code.

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former wrong proof:

each element should be selected with a probability of m/n.

 now, we have randomly selected t elements in i elements,  then we want to selected the next one(named x).

Because there are N-i left, and we have to selected M-t more,

so the probability to selected x should be (M-t)/(N-i).