shuffle card deck issues in language agnostic [closed]

Answer 1

Shuffle card deck evenly in C++

#include <algorithm>

class Deck {
  // each card is 8-bit: 4-bit for suit, 4-bit for value
  // suits and values are extracted using bit-magic
  char cards[52];
  public:
  // ...
  void shuffle() {
    std::random_shuffle(cards, cards + 52);
  }
  // ...
};

Complexity: Linear in N. Exactly 51 swaps are performed. See http://www.sgi.com/tech/stl/random_shuffle.html

Testing:

  // ...
  int main() {
    typedef std::map<std::pair<size_t, Deck::value_type>, size_t> Map;
    Map freqs;    
    Deck d;
    const size_t ntests = 100000;

    // compute frequencies of events: card at position
    for (size_t i = 0; i < ntests; ++i) {
      d.shuffle();
      size_t pos = 0;
      for(Deck::const_iterator j = d.begin(); j != d.end(); ++j, ++pos) 
        ++freqs[std::make_pair(pos, *j)]; 
    }

    // if Deck.shuffle() is correct then all frequencies must be similar
    for (Map::const_iterator j = freqs.begin(); j != freqs.end(); ++j)
      std::cout << "pos=" << j->first.first << " card=" << j->first.second 
                << " freq=" << j->second << std::endl;    
  }

As usual, one test is not sufficient.

Answer 2

Tnilsson, I've also uset the Fisher-Yates shuffle :). I'm very interested dough, about the testing part :)

Answer 3

Product of everything except current in C

void product_except_current(int input[], int index[], int out[], 
                            int len) {
  int prod = 1, nzeros = 0, izero = -1;

  for (int i = 0; i < len; ++i) 
    if ((out[i] = input[index[i]]) != 0)
      // compute product of non-zero elements 
      prod *= out[i]; // ignore possible overflow problem
    else {
      if (++nzeros == 2) 
         // if number of zeros greater than 1 then out[i] = 0 for all i
         break; 
      izero = i; // save index of zero-valued element
    }

  //  
  for (int i = 0; i < len; ++i)  
    out[i] = nzeros ? 0 : prod / out[i];                               

  if (nzeros == 1)
    out[izero] = prod; // the only non-zero-valued element
}

Answer 4

Tnilsson, I agree that YXJuLnphcnQ solution is arguably faster, but the idee is the same. I forgot to add, that the language is optional in the first problem, as well as int the second.

You're right, that calculationg zeroes, and the product int the same loop is better. Maybe that was the thing.

Answer 5

Vaibhav, unfortunately we have to assume, that there could be a 0 in the input table.

Answer 6

For the first one, first calculate the product of entire contents of input, and then for every element of index, divide the calculated product by input[index[i]], to fill in your result array.

Of course I have to assume that the input has no zeros.