TC SRM 487 D1-500 KiwiJuice

Category: Dynamic Programming
URL: http://www.topcoder.com/stat?c=problem_statement&pm=11019

class KiwiJuice {
public:
   int theProfit(int, vector <int>, vector <int>);
};

int dp[51][1<<16];
vector<int> data;
vector<int> price;
int _C;

// f(current volume of the special bottle, bitmask of the used bottles)
int f(int i, int mask) {
   // caching mechanism
   if (dp[i][mask] != -1) return dp[i][mask];
   // base case
   if (mask+1 == (1 << (data.size()))-1) {
      return price[i];
   }
   int res = 0;
   for (int j = 1; j < data.size(); j++) {
      if (mask & (1 << j)) continue;   // already used
      // the next bottle to process
      int next = data[j];
      int d1 = 0;
      if (i + next >= _C) {
         // one of the bottles is full, the other is i+next-_C volume
         d1 = price[_C] + f(i+next-_C, mask | (1 << j));
      } else {
         // change to i + next (and used up the empty for bottle[j])
         d1 = price[0] + f(i+next, mask | (1 << j));
      }
      // don't add more to our current bottle
      // we swap with bottle j and "use" up our bottle
      int d2 = f(next, mask | (1 << j)) + price[i];
      
      // take the maximum of the decision branches
      res >?= max(d1,d2);
   }
   return dp[i][mask] = res;
}

int KiwiJuice::theProfit(int C, vector <int> bottles, vector <int> prices) {
   price = prices;
   int res = 0;
   _C = C;
   memset(dp,-1,sizeof(dp));
   data = bottles;
   return res = f(data[0],0);   // arbitrarily assign bottle 0 to be our special bottle
}

TC SRM 487 D1-250 CarrotJumping

Category: Brute Force, Mathematics
URL: http://www.topcoder.com/stat?c=problem_statement&pm=11022

class CarrotJumping {
public:
   int theJump(int);
};

long long MOD = 1000000007;

// compute 2^n (mod MOD)
long long pow(int v) {
   if (v == 0) return 1;
   if (v == 1) return 2;
   long long x = pow(v/2);
   if (v % 2 != 0) return (x * x * 2) % MOD;
   return (x * x) % MOD;
}

int CarrotJumping::theJump(int init) {
   int res = 100001;
   for (int a = 0; a <= 100000; a++) {
      for (int b = 0; b < 3; b++) {
         long long v = pow(a*3 + b*2);
         long long c = (v * init + (v - 1 + MOD) % MOD) % MOD;
         if (c == 0) {
            res <?= (a + b);
         }
      }
   }
   return res >= 100001 ? -1 : res;
}

SPOJ - Assignments (ASSIGN)

URL: http://www.spoj.pl/problems/ASSIGN/
Category: Dynamic Programming, Bit Manipulation

The problem is quite simple: to calculate the number of different assignments of n topics to n students provided that every student gets exactly 1 topic that he likes. Each student likes different topics which is given in the format of a matrix. The essence of the problem lies in optimisation, a brute force algorithm will clearly time out if that isn't already evident. As a note, the maximum number of assignments for a 20x20 (filled with 1's) matrix is a massive 2,432,902,008,176,640,000. Since we are enumerating we can easily eliminate greedy as a design approach. Naturally we turn to dynamic programming to somehow reduce the running time to an acceptable amount.

As with all dynamic programming problems, we first define the state. The most natural and obvious state is:

D(s,u) = amount of ways to assign up to student s with a bitmask of subjects u used.

The space complexity of this DP state is O(N * 2^N). However, a observation allows us to reduce the memory footprint by a factor of N, giving a complexity of O(2^N). How does this work? The unique property of the problem is that the number of students and subjects are the same, so we can deduce that when we are up to student V, there are also exactly V bits set in the bitmask. Hence, we can simply just use the bitmask to determine which student we are up to!

Now let's define the recurrence relation:

F(u) = sum of all i { F(u | (1 << i) } where student count_of_bits(u) likes subject i

The base case is,

F((1 << total_students)-1) = 1

Note we have done something implicit. That is, we do not define the base case for where we have "left over" subjects that weren't assign to a student - this is implicitly handled by our looping.

We then code a memoised solution using the recurrence above:

long long memo[1<<20];

int count_bits(int n) {
   int r = 0;
   while (n > 0) { r += (n&1); n >>= 1; }
   return r;
}

long long func(int topicMask) {
   int idx = count_bits(topicMask);
   if (idx >= size) {
      if (topicMask != (1 << size)-1) return 0;
      return 1;
   }
   if (memo[topicMask] != -1) return memo[topicMask];
   long long res = 0;
   for (int i = 0; i < size; i++) {
      if (adjmat[idx][i] == 0 || topicMask & (1 << i)) {
         continue;
      }
      res += func(topicMask | (1 << i));
   }
   return memo[topicMask] = res;
}

Unfortunately, this times out in SPOJ, i.e. it exceeds over 20 seconds for all test cases. This is where bottom-up dynamic programming shines, given that our algorithm is probably close to the time limit (you can check by running on your own machine and noting the speed - if it's quite fast but times out in SPOJ, you are within a constant time factor of passing). Now turning it into a bottom-up dynamic programming is a bit more tricky than usual. The problem lies with the order on which to evaluate.

If we set the same base case as before with,

F((1 << total_students)-1) = 1

We need to somehow traverse backwards whilst also ensuring that we have computed all DP states that are one bit away from the one we are calculating. Turns out due to binary number representation, traversing from (1 << total_students)-1 to 0 is sufficient. This is a rather easy proof and hence left to the exercise of the reader (just list them out for 4-bit numbers and you should see the connection).

The implementation is detailed below:

int adjmat[21][21];
int size;
long long memo[1<<20];

int count_bits(int n) {
   int r = 0;
   while (n > 0) { r += (n&1); n >>= 1; }
   return r;
}

int main() {
   memset(adjmat,0,sizeof(adjmat));
   ios_base::sync_with_stdio(false);
   int nT;
   cin >> nT;
   while (nT-- && cin >> size) {
      memset(adjmat,0,sizeof(adjmat));
      memset(memo,0,sizeof(memo));
      for (int i = 0; i < size; i++) {
         for (int j = 0; j < size; j++) {
            int val; cin >> val;
            adjmat[i][j] = val;
         }
      }
      memo[(1<<size)-1] = 1;
      for (int j = (1 << size)-1; j >= 0; j--) {
         int idx = count_bits(j);
         for (int k = 0; k < size; k++) {
            if (adjmat[idx][k] == 0 || (j & (1 << k))) continue;
            memo[j] += memo[j | (1 << k)];
         }
      }
      cout << memo[0] << "\n";
   }
   return 0;
}

We can shave off some extra time by using a constant time bit counting algorithm (you can google it):

#include <iostream>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>

using namespace std;

int adjmat[21][21];
int size;
long long memo[1<<20];

unsigned numbits(unsigned i)
{
    const unsigned MASK1  = 0x55555555;
    const unsigned MASK2  = 0x33333333;
    const unsigned MASK4  = 0x0f0f0f0f;
    const unsigned MASK8  = 0x00ff00ff;
    const unsigned MASK16 = 0x0000ffff;

    i = (i&MASK1 ) + (i>>1 &MASK1 );
    i = (i&MASK2 ) + (i>>2 &MASK2 );
    i = (i&MASK4 ) + (i>>4 &MASK4 );
    i = (i&MASK8 ) + (i>>8 &MASK8 );
    i = (i&MASK16) + (i>>16&MASK16);

    return i;
}

int main() {
   memset(adjmat,0,sizeof(adjmat));
   ios_base::sync_with_stdio(false);
   int nT;
   cin >> nT;
   while (nT-- && cin >> size) {
      memset(adjmat,0,sizeof(adjmat));
      memset(memo,0,sizeof(memo));
      for (int i = 0; i < size; i++) {
         for (int j = 0; j < size; j++) {
            int val; cin >> val;
            adjmat[i][j] = val;
         }
      }
      memo[(1<<size)-1] = 1;
      for (int j = (1 << size)-1; j >= 0; j--) {
         int idx = numbits(j);
         for (int k = 0; k < size; k++) {
            if (adjmat[idx][k] == 0 || (j & (1 << k))) continue;
            memo[j] += memo[j | (1 << k)];
         }
      }
      cout << memo[0] << "\n";
   }
   return 0;
}

TC TCO10 Qualification Round 2

Overview:

This was the second qualification round for TCO10 with around 1300 competitors. As 600 competitors advance through to the next round (provided that at least 600 score a positive score), it provided ample opportunity to qualify for Round 1. The problems were rather simple with two greedy problems and a simple DP problem as the 1000 pointer. Having said that, the system test claimed a decent portion of the 500 pointer solutions due to a corner case which wasn't covered in the examples.

Problem 1 (250 Points): JingleRingle
Category: Greedy
URL: http://www.topcoder.com/stat?c=problem_statement&pm=10896

This problem required us to calculate the maximum profit we can make by buying jingles and selling them to various buyers. We are assumed to have infinite wealth so we can essentially buy all the jingles but we also want to maximise our profits.

The first key observation is to note that we can only use each buyer or seller once. The second key observation is that we only buy jingles from sellers if we can find a suitable buyer in which we can make a net profit (after tax). Now the question that remains is which sellers do we buy from and whom do we sell it to. Since each seller only gives us 1 jingle and we can only sell back 1 jingle at a time, this constraint essentially means that we can match a seller to a buyer (i.e. we are buying at a lower price and re-selling it at a higher price for a net profit). A simple greedy strategy would be then to buy at the lowest price and selling it to the highest price for maximum profit.

Does this greedy strategy work? Let's informally prove it. As we need to match a seller and buyer, if we buy from N people then we also have to sell it to N people. To maximise our profit we need to maximise our revenue, so the optimal choice is to choose the N highest buyers. Then to minimise our expenses we need to choose the N lowest sellers of jingles. It turns out it doesn't matter what order we pair the buyers and sellers as the total profit is the same if we choose the N highest buyers and N lowest sellers. Now all we need is to determine the value of N. Our greedy algorithm keeps pairing the highest available buyer to the lowest available seller until we reach a situation where we no longer profit, this is the value of N. If we assume this is not the value of N, then we have to somehow match a lower buyer to a higher seller or vice versa. But our value of N is at least as good as this on all cases hence our deduced value of N is correct.

A straightforward implementation:

class JingleRingle {
public:
   int profit(vector <int>, vector <int>, int);
};

int calcTax(int p, int tax) {
   return (p * tax) / 100;
}

int JingleRingle::profit(vector <int> buyOffers, 
      vector <int> sellOffers, int tax) {
   int res = 0;
   int buyJ = buyOffers.size()-1;
   sort(buyOffers.begin(),buyOffers.end());
   sort(sellOffers.begin(),sellOffers.end());
   for (int i = 0; i < sellOffers.size(); i++) {
      if (buyJ >= 0 && buyOffers[buyJ] - calcTax(buyOffers[buyJ],tax) 
         - sellOffers[i] >= 0) {
         res += buyOffers[buyJ] - calcTax(buyOffers[buyJ],tax) - sellOffers[i];
      }
      buyJ--;
   }
   return res;
}

Problem 2 (500 Points): FuzzyLife
Category: Greedy, Brute Force
URL: http://www.topcoder.com/stat?c=problem_statement&pm=10911

This problem has us calculate the maximum number of live cells on an infinite 2D plane after one iteration of time. The 2D plane is consisted of cells which are considered either dead or alive at a given slice in time. However, there are unknown cells denoted as '?' and our job is to determine whether each of these cells should be considered alive or dead for the next evolution in time to have the maximum number of alive cells.

The rules are:
- Each live cell with less than 2 or more than 3 live neighbours are dead
- Each dead cell with exactly 3 live neighbours becomes alive
- All other cell remain the same

Each cell has exactly 8 neighbours (even the ones on the "border" of the input because of the infinite plane) - 2 horizontal, 2 vertical and 4 diagonal.

The main observation required for this problem is the constraints of the problem allow us to adopt a greedy strategy and consider the neighbouring cells of a '?' as its own problem subinstance. As no cell will have more than 1 neighbour of type '?' this means that every cell is at most affected by 1 unknown cell. All we need to do is try each '?' cell by placing a '1' and a '0' in its place and simulate how many alive cells we are left with on the next iteration. We then simply choose the maximum of these two choices.

One tricky case is that we also need to take into account the '?' cell, if we place a '1' in it's position does it become alive or dead in the next iteration - vice versa if we place a '0'. We need to add this to our alive count before we compare which one is better. The other tricky case was to add a padding of dead cells around the grid as some of these can turn into alive cells (which need to be included in the answer) - this case however, was covered in the example cases. After we have filled in each of the '?' cells we can then do a single iteration pass and work out the number of alive cells.

The implementation using the ideas above:

class FuzzyLife {
public:
   int survivingCells(vector <string>);
};

// movement vectors
int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};

int FuzzyLife::survivingCells(vector <string> grid) {
   int res = 0;
   // pad the grid with extra 0's around the perimeter
   string str(grid[0].size()+2,'0');
   for (int i = 0; i < grid.size(); i++) {
      grid[i] = '0' + grid[i] + '0';
   }
   grid.insert(grid.begin(),str);
   grid.push_back(str);
   for (int i = 0; i < grid.size(); i++) {
      for (int j = 0; j < grid[i].size(); j++) {
         if (grid[i][j] == '?') {
            // compute options ('?' = 1)
            grid[i][j] = '1';
            int c1 = 0;
            for (int m = 0; m < 8; m++) {
               int neighX = i + dx[m];
               int neighY = j + dy[m];
               if (neighX < 0 || neighY < 0 || neighX >= grid.size() 
               || neighY >= grid[0].size()) continue;
               int alive = 0;
               for (int n = 0; n < 8; n++) {
                  int mX = neighX + dx[n];
                  int mY = neighY + dy[n];
                  if (mX < 0 || mY < 0 || mX >= grid.size() 
                  || mY >= grid[0].size()) continue;
                  if (grid[mX][mY] == '1') alive++;
               }
               if (grid[neighX][neighY] == '1' && (alive == 2 || alive == 3)) c1++;
               else if (grid[neighX][neighY] == '0' && (alive == 3)) c1++;
            }
            // consider the square itself under '?' = 1
            int selfAlive = 0;
            for (int n = 0; n < 8; n++) {
               int selfX = i + dx[n];
               int selfY = j + dy[n];
               if (selfX < 0 || selfY < 0 || selfX >= grid.size() 
               || selfY >= grid[0].size()) continue;
               if (grid[selfX][selfY] == '1') selfAlive++; 
            }
            if (selfAlive == 2 || selfAlive == 3) c1++;
            // compute options ('?' = 0)
            grid[i][j] = '0';
            int c2 = 0;
            for (int m = 0; m < 8; m++) {
               int neighX = i + dx[m];
               int neighY = j + dy[m];
               if (neighX < 0 || neighY < 0 || neighX >= grid.size() 
               || neighY >= grid[0].size()) continue;
               int alive = 0;
               for (int n = 0; n < 8; n++) {
                  int mX = neighX + dx[n];
                  int mY = neighY + dy[n];
                  if (mX < 0 || mY < 0 || mX >= grid.size() 
                  || mY >= grid[0].size()) continue;
                  if (grid[mX][mY] == '1') alive++;
               }
               if (grid[neighX][neighY] == '1' && (alive == 2 || alive == 3)) c2++;
               else if (grid[neighX][neighY] == '0' && (alive == 3)) c2++;
            }
            // consider the square itself under '?' = 0
            selfAlive = 0;
            for (int n = 0; n < 8; n++) {
               int selfX = i + dx[n];
               int selfY = j + dy[n];
               if (selfX < 0 || selfY < 0 || selfX >= grid.size() 
               || selfY >= grid[0].size()) continue;
               if (grid[selfX][selfY] == '1') selfAlive++; 
            }
            if (selfAlive == 3) c2++;
            // choose the one with better live cells
            if (c1 >= c2) grid[i][j] = '1';
         }
      }
   }   
   // iterate one period and determine number of alive
   for (int i = 0; i < grid.size(); i++) {
      for (int j = 0; j < grid[i].size(); j++) {
         int alive = 0;
         for (int n = 0; n < 8; n++) {
            int mX = i + dx[n];
            int mY = j + dy[n];
            if (mX < 0 || mY < 0 || mX >= grid.size() 
            || mY >= grid[0].size()) continue;
            if (grid[mX][mY] == '1') alive++;
         }
         if (grid[i][j] == '1' && (alive == 2 || alive == 3)) res++;
         else if (grid[i][j] == '0' && alive == 3) res++;
      }
   }
   return res;
}

Problem 3 (1000 Points): HandlesSpelling
Category: Dynamic Programming, String Manipulation
URL: http://www.topcoder.com/stat?c=problem_statement&pm=10864

This problem required us to play a game and work out the maximum score based on a set of choices we are forced to make. The game is that we have a string of letters in which we can "stamp" a set of pre-determined strings onto it. We need to maximise the score based on A^2 - B where A is the longest contiguous sequence of letters covered by the stampings and B is the number of uncovered letters.

For example, if the string we are given is HELLO and we are given 3 badges namely: E, HE, L. We can stamp them as follows: {HE}{L}{L}O. We stamped HE with the second badge and the 2 L's with the third badge. The value of A is 4 because we have a contiguous sequence of stampings (HELL) and B is 1 because we didn't have a way to stamp the letter O. Hence the score given this set of operations is 4^2 - 1 = 15, which turns out to be the maximum score we can get.

One strategy is to determine the maximum value of A. Why? A is where our score gets maximised, no other measure increases our score except for the value of A so it's only natural that we want to get A as high as possible. We also note that the maximum contiguous sequence (i.e. maximum value A) is in the optimal solution. Why? Consider two configurations:

xxxxx....----n----....xxxxxxxx
xxxxx--m--.......--m--xxxxxxxx

Here the x's denote similar configurations, so we only need to focus on maximising the centre part of m's, n's and dots. We let n be the maximum contiguous length possible in our string, let that be denoted as length x. Let's assign m to be length x-1. Then our goal is to prove that it is always optimal to choose n over the 2 m's.

Then the score for our first configuration is:
F = (x * x) - (x-1) - (x-1) = x^2 - 2x + 2 (length x ^ 2 minus two sections of x-1 uncovered)

The score for our second configuration is:
G = (x-1) * (x-1) - x = x^2 - 3x + 1 (length (x-1) ^ 2 minus one section of x uncovered)

Hence for all positive values of length x, the score for the first is better than the second. In fact, this is a stricter argument than what is required, as if the sections of n and m were disjoint they'll both be covered, hence there must be some conflict between n and m which forces us to choose between one or the other. Our argument will hold for such a scenario.

Our first task is then to find the maximum contiguous sequence of our string - this can be done easily through basic dynamic programming. Then we are tasked with trying all configurations which have the same maximum contiguous sequence to determine the minimum number of uncovered letters. Again, this can be done via a similar dynamic programming algorithm where we divide our main string into a left and right sub-string and count the maximum number of covered cells we can get. Then we simply subtract this by the size of the left and right sub-string respectively to get the number of uncovered letters.

The implementation which uses the strategy above can be seen as:

class HandlesSpelling {
public:
   int spellIt(vector <string>, vector <string>);
};

#define UNI -12345678
#define INF 12345678

vector<string> _badges;
string _S;

// returns true if str is a prefix to S starting at idx
bool isPrefix(int idx, const string& str, string S) {
   if (idx + str.size() > S.size()) return false;
   for (int i = idx; i < idx + str.size(); i++) {
      if (S[i] != str[i-idx]) return false;
   }
   return true;
}

int prefix[1001][51];   // true if badge j is a prefix at index i
int A[1001];   // keeps track of the longest sequence ending with index i
int B[1001];   // left hand side
int C[1001];   // right hand side
int maxSize;

// maximise coverage using ptr as the memoisation table
int func(int idx, int* ptr) {
   int res = 0;
   if (idx >= maxSize) return 0;
   if (ptr[idx] != -1) return ptr[idx];
   for (int i = 0; i < _badges.size(); i++) {
      if (prefix[idx][i] && idx + _badges[i].size() <= maxSize) {
         res >?= func(idx + _badges[i].size(), ptr) + _badges[i].size();
      }
   }
   res >?= func(idx+1, ptr);
   return ptr[idx] = res;
}

int HandlesSpelling::spellIt(vector <string> parts, vector <string> badges) {
   string data;
   for (int i = 0; i < parts.size(); i++) data += parts[i];
   _badges = badges;
   _S = data;
   // pre-compute the prefixes for speed
   for (int i = 0; i < data.size(); i++) {
      for (int j = 0; j < badges.size(); j++) {
         if (isPrefix(i, badges[j], _S)) { prefix[i][j] = 1; }
      }
   }
   // determine the longest contiguous sequence in our string
   for (int i = _S.size()-1; i >= 0; i--) {
      for (int j = 0; j < _badges.size(); j++) {
         if (isPrefix(i, _badges[j], _S)) {
            A[i] >?= A[i+_badges[j].size()] + _badges[j].size();
         }
      }
   }
   int longest = 0;
   for (int i = 0; i < _S.size(); i++) {
      longest >?= A[i];
   }
   // choose the best score out of all combinations which use A[i] == longest
   int res = UNI;
   for (int i = 0; i < _S.size(); i++) {
      if (A[i] == longest) {
         // branch off to left and right
         string left = _S.substr(0, i);
         string right = _S.substr(i+A[i]);
         // reset memo table for each instance
         memset(B,-1,sizeof(B));
         memset(C,-1,sizeof(C));
         int lOffset = 0, rOffset = 0;
         maxSize = i;
         if (left.size() > 0) lOffset = left.size() - func(0,B);
         maxSize = _S.size();
         if (right.size() > 0) rOffset = right.size() - func(i+A[i],C);
         res >?= (A[i] * A[i]) - (lOffset) - (rOffset);
      }
   }
   return res;
}

USACO Training Gateway - "A Game"

Problem Statement:

"Consider the following two-player game played with a sequence of N positive integers (2 <= N <= 100) laid onto a game board. Player 1 starts the game. The players move alternately by selecting a number from either the left or the right end of the sequence. That number is then deleted from the board, and its value is added to the score of the player who selected it. A player wins if his sum is greater than his opponents.

Write a program that implements the optimal strategy. The optimal strategy yields maximum points when playing against the "best possible" opponent. Your program must further implement an optimal strategy for player 2."

Strategy:

This is a simple dynamic programming problem which involves the theme about games (two players) which is quite common. As no number in the sequence can be skipped, it is sufficient enough to compute one player's score and derive the other player's score by subtracting it from the total sum of the sequence on the board. This allows us to condense our DP state by a factor of 2 by keeping track of only one player's progress.

As with all DP problems we first define the DP state. Here an obvious one can be defined as:

D(n,m) = the best score for player 1 in which the board state is a contiguous subsequence between index n and m. Player 1's turn.
E(n,m) = the best score for player 1 in which the board state is a contiguous subsequence between index n and m. Player 2's turn.

As the maximum board size is only 100, an O(n^2) space complexity easily fits within the memory limits. Next we define our recurrence relation:

For the current player we can choose either the leftmost or the rightmost block, i.e. A[n] or A[m]. The next turn in the game is player 2's turn which is denoted by the E(x,y) relation. The obvious goal of player 2 is to minimise the gain of player 1 which will result in a higher score for him/herself.

Using the two recurrences we can combine them into one to get (using direct substitution from E(n,m) to D(n,m)):

Hence, we have constructed our recurrence for player 1. We keep track of the total sum of the sequence of the board during the input/parsing phase. Player 2's score is simply total sum - D(0, size-1). Lastly, we need to handle our base cases. If n = m, then we are left with one choice - choosing A[n]. If n > m, which means we have finished the game as the current board state is non-existent, then we return 0 as neither player can choose any more integers from the sequence. Hence explicitly the base cases are:

Implementation:

Using the idea above we generate the simple memoisation algorithm:

int dp[101][101];
vector<int> vec;

int func(int n, int m) {
   if (n > m) return 0;
   if (n == m) return dp[n][m] = vec[n];
   if (dp[n][m] != -1) return dp[n][m];
   int res = 0;
   res = max(min(func(n+2,m),func(n+1,m-1))+vec[n],
      min(func(n+1,m-1),func(n,m-2))+vec[m]);
   return dp[n][m] = res;
}

int main() {
   ifstream inFile("game1.in");
   ofstream outFile("game1.out");
   int N; inFile >> N;
   int sum = 0;
   for (int c = 0; c < N; c++) {
      int val; inFile >> val;
      sum += val;
      vec.push_back(val);
   }
   memset(dp,-1,sizeof(dp));
   outFile << func(0,N-1) << " " << sum - func(0,N-1) << "\n";
   return 0;
}

TC SRM 459 D2-1000 (ParkAmusement)

Category: Dynamic Programming, Graph Theory
URL: http://www.topcoder.com/stat?c=problem_statement&pm=10723

The problem describes having N landings in the amusement park. The landings can be terminating (i.e. the ride ends at this landing) which are either proper exits or crocodile ponds. The landings can also be non-terminating which means that you will slide down further or until you are stalled with no valid adjacent landings. All of the N landings have different heights and we can only slide down from a taller landing to a shorter one (by the laws of gravity). We start at a random landing and we wish to calculate the probability that we started at a specific landing given that we only went through exactly K different pipes to get safely home (i.e. at a proper exit and not at either a crocodile pond or a dead-end landing).

This problem is best split into two parts. The first is to calculate the probability we reach home from a specific starting point. The second is to calculate the probability that out of all the viable starting points we chose the specific starting landing. We are already given the adjacency matrix in terms of a vector/array of strings which we proceed to convert into a more usable format. We first keep track of exits - i.e. whether or not they are proper exits (home) or crocodile ponds. This is done by specifically checking whether or not landings[i][i] is '0' or not (based on the problem definition). If it's not '0', then the current vertex is an exit and we proceed to determine which type. Otherwise, we add the edges into our adjacency list. This is shown below:

vector<vector<int> > adjlist;   // adjacency list
int table[51];   // the exit table for each vertex, 0 if it's not an exit, 1 if it's home exit, 2 if it's a pond exit

...
double ParkAmusement::getProbability(vector <string> landings, int startLanding, 
      int K) {
   adjlist = vector<vector<int> >(landings.size(), vector<int>());
   for (int i = 0; i < landings.size(); i++) {
      if (landings[i][i] != '0') {
         // exit
         if (landings[i][i] == 'P') table[i] = 2;   // pond
         else table[i] = 1; // exit
         continue;
      }
      for (int j = 0; j < landings[i].size(); j++) {
         if (landings[i][j] == '1') adjlist[i].push_back(j);
      }
   }
...

Once we have parsed the input into a better format, we need to find out the probability of reaching home from a fixed starting vertex/landing. We can accomplish this through a DFS-like recursive function by keeping track of how many steps we have left to reach home. This involves keeping track of which vertex/landing we are currently at and how many hops left we must use. The base case is when the number of steps left is zero, if we are at an exit and it's the proper one (i.e. an 'E' in the matrix) then we have reached home - otherwise we have failed in our mission.

We utilise basic probability calculations to compute our chances of succeeding. If we arrive at a landing that has M out-going edges, then the probability of choosing a particular edge is 1/M. Therefore, the probability we will survive is simply the sum of 1/M * f(edge i, K-1) for all out-going edges i. The base cases are obvious, probability of surviving is 1.0 if we reach an 'E' exit otherwise it's 0.0. The function below illustrates this concept:

double calcPr(int startNode, int K) {
   if (K == 0) {
      // check if it's the exit
      if (table[startNode] != 1) return 0.0;
      return 1.0;
   }
   double res = 0.0;
   // note: if there are no more valid out-going edges it defaults to a 
   //    0.0 probability
   int poss = adjlist[startNode].size();
   for (int i = 0; i < adjlist[startNode].size(); i++) {
      res += (1.0 / poss) * calcPr(adjlist[startNode][i], K-1);
   }
   return res;
}

However, we can optimise this by memoising it. Otherwise we may run into trouble with time limits for the larger cases. This is easily done by setting up a cache table and marking it with a sentinel value to determine whether the state has been computed or not. Since no probability can be negative, we can exploit this fact.

double dp[51][51];

double calcPr(int startNode, int K) {
   if (dp[startNode][K] >= 0.0) return dp[startNode][K];
   if (K == 0) {
      if (table[startNode] != 1) return dp[startNode][K] = 0;
      return dp[startNode][K] = 1.0;
   }
   double res = 0.0;
   int poss = adjlist[startNode].size();
   for (int i = 0; i < adjlist[startNode].size(); i++) {
      res += (1.0 / poss) * calcPr(adjlist[startNode][i], K-1);
   }
   return dp[startNode][K] = res;
}

Now, we have finished the first step of the problem. The second step involves calculating out of the many possible landings we have chosen and given that we did arrive home safely, what is the probability we chose startLanding? The easiest way to think of this is to think geometrically. If we let a circle C be the total sum of the probabilities that we arrived safely from all possible landings, we can scale this to 1.0 as we know for certainty that we arrived safely (given in the problem). Now we simply need to calculate how much probability was contributed by starting from startLanding. This is simply calcPr(startLanding, K) / totalSum. This can be visualised as:

double tot = 0.0;
for (int i = 0; i < 51; i++) for (int j = 0; j < 51; j++) dp[i][j] = -1.0;
for (int i = 0; i < landings.size(); i++) {
   tot += calcPr(i, K);
}
return calcPr(startLanding, K) / tot;

Combining all of the above steps, we yield the final code:

class ParkAmusement {
public:
   double getProbability(vector <string>, int, int);
};

vector<vector<int> > adjlist;
int table[51];
double dp[51][51];

double calcPr(int startNode, int K) {
   if (dp[startNode][K] >= 0.0) return dp[startNode][K];
   if (K == 0) {
      if (table[startNode] != 1) return dp[startNode][K] = 0;
      return dp[startNode][K] = 1.0;
   }
   double res = 0.0;
   int poss = adjlist[startNode].size();
   for (int i = 0; i < adjlist[startNode].size(); i++) {
      res += (1.0 / poss) * calcPr(adjlist[startNode][i], K-1);
   }
   return dp[startNode][K] = res;
}

double ParkAmusement::getProbability(vector <string> landings, int startLanding, 
      int K) {
   adjlist = vector<vector<int> >(landings.size(), vector<int>());
   for (int i = 0; i < landings.size(); i++) {
      if (landings[i][i] != '0') {
         // exit
         if (landings[i][i] == 'P') table[i] = 2;   // pond
         else table[i] = 1; // exit
         continue;
      }
      for (int j = 0; j < landings[i].size(); j++) {
         if (landings[i][j] == '1') adjlist[i].push_back(j);
      }
   }
   double tot = 0.0;
   for (int i = 0; i < 51; i++) for (int j = 0; j < 51; j++) dp[i][j] = -1.0;
   for (int i = 0; i < landings.size(); i++) {
      tot += calcPr(i, K);
   }
   return calcPr(startLanding, K) / tot;
}

Applications of Floyd-Warshall's Algorithm

We will expand on the last post on Floyd-Warshall's algorithm by detailing two simple applications. The first is using the algorithm to compute the transitive closure of a graph, the second is determining whether or not the graph has a negative cycle.

Transitive closure is simply a reachability problem (in terms of graph theory) between all pairs of vertices. So if we compute the transitive closure of a graph we can determine whether or not there is a path from vertex x to vertex y in one or more hops. Unlike the shortest path problem we aren't concerned on how long it takes to get there, only whether or not if we can eventually get there. Obviously you can simply not modify our original algorithm and assume if the distance between vertex x to vertex y is at least "infinity" then there is no way to get from vertex x to vertex y, otherwise there is a way to get there. This perfectly solves the reachability problem.

There is cleaner approach of computing the transitive closure using a slight modification of Warshall's algorithm. Instead of letting no edge between x to y be denoted by infinity, we can let it be 0 (or false). Any weighted edge is simply reduced to 1 (or true). Now instead of adding distances we use the binary-AND operation. If graph[x][k] is true and graph[k][y] is also true, then there is a way to connect x to y. Any other combination will result in failing to connect vertex x to vertex y using an intermediate node k. So graph[x][k] & graph[k][y] correctly fits our desired behaviour. As we just need 1 such intermediate node k to connect vertex x to vertex y, we combine it with the binary-OR operation as we don't care which intermediate vertex it requires to get from x to y so long as there is a path that allows us to get there.

The modified algorithm is the following:

// warshall's algorithm for transitive closure
for (int k = 1; k <= N; k++) {
   for (int i = 1; i <= N; i++) {
      for (int j = 1; j <= N; j++) {
         // only need one connection for it to be reachable, hence the 
         //    OR operation.
         // need both adjmat[i][k] and adjmat[k][j] to be connected for i to go 
         //    to j using intermediate node k.
         adjmat[i][j] |= (adjmat[i][k] & adjmat[k][j]);
      }
   }
}
// post-condition if adjmat[i][j] is 1 then there is a path from vertex i to j

We can make an obvious small optimisation by only running the inner loop if adjmat[i][k] is true, otherwise all the inner computations would fail anyway.

// warshall's algorithm for transitive closure - attempt 2
for (int k = 1; k <= N; k++) {
   for (int i = 1; i <= N; i++) {
      // optimise here by reducing the number of inner loops
      if (adjmat[i][k]) {
         for (int j = 1; j <= N; j++) {
            adjmat[i][j] |= (adjmat[i][k] & adjmat[k][j]);
         }
      }
   }
}
// post-condition if adjmat[i][j] is 1 then there is a path from vertex i to j

The second application is using it to detect negative cycles in a graph. How does Floyd-Warshall relate to this? Since the algorithm calculates the shortest path between all pairs of vertices we can use the "shortest path" for vertex i to itself to determine if there is a cycle. If we can reach from vertex i to itself with a cost less than 0 then we can keep looping through to successively generate shorter paths - hence the term negative cycle. Therefore it is sufficient to just check whether or not adjmat[i][i] < 0 for all vertices. This is used in many other algorithms such as the cycle-cancelling algorithm for minimum cost flows.

// warshall's algorithm for detecting negative cycles
bool cycle = false;
for (int k = 1; k <= N; k++) {
   for (int i = 1; i <= N; i++) {
      for (int j = 1; j <= N; j++) {
         adjmat[i][j] = min(adjmat[i][j], adjmat[i][k] + adjmat[k][j]);
      }
      if (adjmat[i][i] < 0) cycle = true;
   }
}

Floyd-Warshall All-Pairs Shortest Path Algorithm

There are many notable algorithms to calculate the shortest path between vertices in a graph. Most are based on single source to a set of destination vertices. Examples of such famous algorithms include Dijkstra's, Bellman-Ford and even breadth first search for weightless graphs. However, sometimes we wish to calculate the shortest paths between all pairs of vertices. An easy way to calculate this is to loop through each possible pair of vertices and run a Dijkstra through it. However, there exists a more efficient way to calculate this in n^3 time complexity.

The basis of this algorithm lies with dynamic programming. As with all dynamic programming algorithms we need to first define the DP state. The key factor in this algorithm is to consider an intermediate node k of which to connect two vertices i and j. We check if the path from i to k and then from k to j improves our current shortest path length from vertex i to j. If it does, we update - otherwise we don't. Using this definition our DP state becomes:

D(i,j,k) = shortest path length from vertex i to j considering a set of intermediate nodes from vertex 1 to k

We don't need to explicitly label our vertices as this is done implicitly through our algorithm. Now we need to define our recurrence relation. For a given instance of D(i,j,k) we can calculate the shortest path by considering all intermediate vertices from 1 to k. The proposed distance is easily calculated as D(i,k,k-1) + D(k,j,k-1) by definition of an intermediate node. Formally,

F(i,j,k) = min { F(i,j,k-1), F(i,k,k-1) + F(k,j,k-1) }

We have to determine whether or not the new intermediate node k actually helps achieve a shortest path by comparing it to the previous computation without considering intermediate node k, i.e. F(i,j,k-1).

As with all recurrence relations, a base case is required. If we don't consider any intermediate nodes at all, then the shortest path between two vertices is simply the direct edge between vertex i and j. In other words, the weight of the (i,j) entry in the adjacency matrix. So now we have fully defined our recurrence relation with the proper base case:

F(i,j,k) = min { F(i,j,k-1), F(i,k,k-1) + F(k,j,k-1) }
F(i,j,0) = w(i,j)

To calculate this in a bottom-up manner is actually quite simple. First we note the order dependency of the recurrence relation. It can be seen that a particular instance D(i,j,k) can depend on D(i,0,k-1) to D(N,j,k-1) where N is the number of vertices. So it's obvious we need to compute the previous intermediate k values first (i.e. 1, 2, .., k-1 before calculating k). The order of i and j does not matter in our case because we would have computed all k-1 (i,j) pairs before. Using this information we yield an extremely elegant solution:

// warshall's algorithm
for (int k = 1; k <= N; k++) {
   for (int i = 1; i <= N; i++) {
      for (int j = 1; j <= N; j++) {
         adjmat[i][j] = min(adjmat[i][j], adjmat[i][k] + adjmat[k][j]);
      }
   }
}

Let's apply this algorithm to a simple graph theory problem.

TopCoder SRM 301 D1-450 (EscapingJail)
URL: http://www.topcoder.com/tc?module=ProblemDetail&rd=9822&pm=6222

The problem states there are a set of prisoners chained up and we need to determine the maximum distance that a pair of prisoners can be from each other. It's obvious that the maximum length that two prisoners can be apart is defined by the shortest chain separating a connected path between the two prisoners. Sound familiar? This is the exact definition of a shortest path problem! We simply need to find the shortest distance between every pair of prisoner and calculate the maximum of all such pairs.

How do we handle cases in which the prisoners are not chained together (i.e. there is no edge between them)? We can set a sentinel value to denote that the prisoners aren't chained, if we set this to 0 it would be wrong because then it would interfere with our minimum distance calculation. A simple option is to set this to a high/infinity value and if we know the distance from prisoner i to prisoner j exceeds or equals this value then we know they are unbounded. For this problem, we need to return a value of -1 for unbounded distance.

The only remaining part of the problem is translating the given string array into the adjacency matrix. This is a simple exercise of converting a character into the suitable value - if it's a space then we set the distance to be infinity since they aren't chained together. All that leaves is to use the Warshall algorithm and compute the maximum of the pairs (and remembering to return -1 if the distance between at least two prisoners is unbounded).

The implementation is as follows:

class EscapingJail {
public:
   int getMaxDistance(vector <string>);
};

#define INF (1 << 27)

int translate(char ch) {
   if (isdigit(ch)) return ch - '0';
   if (islower(ch)) return ch - 'a' + 10;
   if (isupper(ch)) return ch - 'A' + 36;
   return INF;
}

int adjmat[51][51];

int EscapingJail::getMaxDistance(vector <string> chain) {
   int res = 0;
   for (int i = 0; i < chain.size(); i++) {
      for (int j = 0; j < chain[i].size(); j++) {
         adjmat[i][j] = translate(chain[i][j]);
      }
   }
   for (int k = 0; k < chain.size(); k++) {
      for (int i = 0; i < chain.size(); i++) {
         for (int j = 0; j < chain.size(); j++) {
            adjmat[i][j] = min(adjmat[i][j], adjmat[i][k] + adjmat[k][j]);
         }
      }
   }
   for (int i = 0; i < chain.size(); i++) {
      for (int j = 0; j < chain.size(); j++) {
         if (i == j) continue;
         if (adjmat[i][j] >= INF) return -1;
         res = max(res, adjmat[i][j]);
      }
   }
   return res;
}