Problem source: Topcoder, SRM 513, Div 1 - 500

Link: http://www.topcoder.com/stat?c=problem_statement&pm=11500 (Log in is required to see the problem)

In this problem you have a board having N * M cells. Each cell has a symbol in it's back. Each symbol is in exactly two cells in the board. So there will be (N * M) / 2 symbols. In each move you can turn two cells one by one to see the symbols behind them. If these two symbols differ, both the cell turned back to hide the symbols again. If both the symbols are same, they will remain same and never turn back again. What is the expected number of moves to make all the symbols visible, if you have the perfect memory?

Note that, having perfect memory, you'll never forget what was the symbol in a cell which you saw, but turned back. In this problem it's guaranteed that N * M will be even.

I couldn't solve this problem in real contest. I thought in each turn both of the cells will be turned simultaneously. But in the problem statement it was clearly mentioned that, two cells will be turned one by one. It seems that I am not the only one who got this wrong.

This problem can be solved by dynamic programming. Our state will be (total, seen, turn). Here,

total = number of symbols for which both cells are not seen yet i.e. at least one cell is unseen. So we are actually considering (2 * total) cells.

seen = number of symbols partially seen i.e. number of cells of which exactly one cell is seen.

turn = Is this the first turn or second turn of the move? We know each move consists of two turns.

Initially, total = (N * M) / 2, seen = 0, and turn = FIRST.

Boundary condition: If total = 0, we don't have any symbol left to consider, so expected number will be zero.

In state (total, seen, turn), we have (total * 2 - seen) cells we didn't see yet. Lets say unseen = total * 2 - seen.

Now if the turn = FIRST, here we will always open an unseen cell and two things can happen after turning a cell on.

1. With p1 probability the cell contains a symbol which is already seen in another cell.

2. With p2 probability the cell contains a new symbol which is not visible before.

Here,

p1 = seen / unseen

p2 = 1.0 - p1

If (1) happens, we will just turn on the other symbol we already saw, and both will remain open forever. So our new state will be (total - 1, seen - 1, FIRST) and one move will be completed.

If (2) happens, new state will be (total, seen + 1, SECOND)

exp_move(total, seen, FIRST) = p1 * (1 + exp_move(total - 1, seen - 1, FIRST)) +

p2 * (exp_move(total, seen + 1, SECOND)

Note that, in first option we completed a move, so added +1, but in second option move is not done yet.

If turn = SECOND, we'll open another unseen cell. Three things can happen

1. With probability p1, symbol in the SECOND turn matches with the symbol of the FIRST turn.

Here p1 = 1 / unseen

2. With probability p2, symbol in the SECOND move matches with a symbol which is previously seen, but not seen in the first turn of this move. So,

p2 = seen / unseen - p1 = (seen - 1) / unseen

3. With probability p3, symbol in the SECOND move is completely a new symbol.

p3 = 1 - p2 - p3

If (1) happens, this pair will remain visible, so our state will be (total - 1, seen - 1, FIRST) and one move will be completed.

If (2) happens, this pair will be turned back, but we know both cells of the symbol of last turn. So we'll open both symbol in next move, so two moves will be completed in this case and new state will be (total - 1, seen - 1, FIRST).

If (3) happens, just another seen added. So state will be (total, seen + 1, FIRST), and one move will be completed.

exp_move(total, seen, SECOND) = p1 * ((exp_move(total - 1, seen - 1, FIRST) + 1) +

p2 * ((exp_move(total - 1, seen - 1, FIRST) + 2) +

p3 * ((exp_move(total, seen + 1, FIRST) + 1)

My solution in C++:

#include <cstdio>

#include <cmath>

#include <cstdlib>

#include <cctype>

#include <algorithm>

#include <map>

#include <string>

#include <vector>

#include <iostream>

#include <sstream>

#include <queue>

#include <cstring>

#include <set>

#include <ctime>

#include <cfloat>

using namespace std;

typedef long long i64;

#define I64 "%lld"

#define rep(i,n) for (i=0; i < (n); ++i)

#define all(c) (c).begin(),(c).end()

bool done[1500][1500][2];

double memo[1500][1500][2];

enum{FIRST = 0, SECOND};

double exp_move(int total, int seen, int turn) {

double &ret = memo[total][seen][turn];

if (done[total][seen][turn]) return ret;

done[total][seen][turn] = true;

int unseen = total * 2 - seen;

if (total == 0) return ret = 0;

ret = 0.;

// open one unseen cell randomly

if (turn == FIRST) {

double p1 = 0, p2 = 0;

// Probability of it's pair is already seen is p1

if (seen > 0) {

p1 = (double) seen / (double) unseen;

ret += p1 * (exp_move(total - 1, seen - 1, FIRST) + 1);

}

// Probability of it's pair is not seen yet is p2

if (unseen > seen) {

p2 = 1.0 - p1;

ret += p2 * (exp_move(total, seen + 1, SECOND));

}

}

else {

double p1 = 0, p2 = 0, p3 = 0;

// Probability of it's pair is already open is p1

p1 = (double) 1 / (double) unseen;

ret += p1 * (exp_move(total - 1, seen - 1, FIRST) + 1);

// Probability of it's pair is seen but not open is p2

if (seen > 1) {

p2 = (double) (seen - 1) / (double) unseen;

ret += p2 * (exp_move(total - 1, seen - 1, FIRST) + 2);

}

// Probability of it's pair is unseen is p3

if (unseen > seen) {

p3 = 1.0 - p1 - p2;

ret += p3 * (exp_move(total, seen + 1, FIRST) + 1);

}

}

return ret;

}

class PerfectMemory {

public:

double getExpectation(int N, int M) {

return exp_move((N * M) / 2, 0, FIRST);

}

};