Basics of Kenken Puzzles

If you don't know the rules of KenKen then you can refer to www.kenken.com for an excellent source of daily puzzles which also explains the rules. I'll summarise it as best I can, it's a puzzle where you need to fill in each cell such that each row and cell contains a unique sequence of numbers respectively. If you are more familiar with Sudoku, it's similar in terms of the fact that each row/column can only contain one 1, one 2, one 3.. one 9. In KenKen's case, you are allowed to use numbers 1,2...,n. Where n is the size of the puzzle. However, unlike Sudoku you must ensure each 'cage' (bolded subareas on the puzzle) adds, multiplies, subtracts or divides to the target number listed on the a top-left corner of the cage. Subtraction and Divisions can be done in any order, so 2 / 4 and 4 / 2 both match 2/ (division). However, the constraints of unique numbers must be adhered to at all times. For more information, just refer to www.kenken.com.

Today we will be solving and walking through a simple 6x6 KenKen puzzle (from www.kenken.com). There are several important concepts which will make your puzzle solving much simpler. KenKen can seem daunting at first, but it's a solvable puzzle without guessing! There is always a way to deduce/use logic to fill in the cells without trial and error. I hope to highlight some of these through this given KenKen puzzle:

So, how do we get started? We don't exactly have any free single 1x1 cage to immediately write down. Usually if there is a simple 1x1 cage you should write down those numbers first because they can give hints to surrounding cages in similar rows/columns.

Now note that today we are solving a 6x6 puzzle, hence we are only allowed to fill in numbers 1 to 6. This is important to note because it limits the number of possibilities for each cage. One of the first things I tend to look for first - is a cage in which I immediately know what numbers are contained in those cages but not necessarily the right order. For a 6x6 puzzle, you should look primarily for 11+ and 5- cages. Why is this? It's because there is only one possibility for each of the cages, namely 5 and 6, and 1 and 6 respectively. (Question: If it was a 8x8 puzzle, what numbers should you be looking for?) Don't worry too much about the ordering at the moment - the correct ordering will come as we fill the puzzle. Noting these down we get:

We can also note several one possibility cages, 15x = 3 * 5. There is no other way of multiplying two numbers less than 6 to equal 15. There's also another subtle one, 36x. How many possibilities does this have? Well it seems like there may be a few, 6*6*1, 6*3*2, 3*3*4. However, in our puzzle 36x exists in a row which means there can be no repeated terms in the factoring! Hence we can rule out 6*6*1 and 3*3*4 being valid possibilities for the 36x cage. This just leaves a single possibility, 6*3*2! Anymore cages that only have one possibility? 20x also looks tempting - let's see, 5*4*1 and 5*2*2. But wait! Again, 20x is on a row and can't have any repeated elements so it just leaves one possibility: 5*4*1! There may be more but let's just put these on the puzzle first:

You may note that we actually haven't even filled a single entry yet! This is quite normal, you'll slowly realise that many possibilities will start collapsing when you fill a few in!

Now we will employ a more tricky technique which is immensely useful in solving KenKen puzzles. In Mathematics, this is called the Pigeonhole Principle. Essentially, the Pigeonhole Principle states that if you have n items and wish to place them into m boxes with n > m, then at least 1 box will contain more than 1 item. How on earth are we going to apply this to the puzzle? We need to look in a vertical manner - note how the 3 rightmost columns (4th to 6th columns) already contain 4 possibilities of having a 5 in them? Wait! 4 possibilities in 3 columns that violates the rules of the puzzle! Hence, one of these must not belong in the 3 rightmost columns. Now look at the bottom-most 11+, the 5 in this cage can exist on the 3rd column! Therefore, if this puzzle has a solution the 5 must belong in the 3rd column. So let's fill this in...

Now the bottom-most 11+ cage must have 6 in the unfilled cell. Now you notice that it also fills in the other 11+ cage in the 3rd row! Because the number 6 exists in the 4th column, it means the 3rd row 11+ cage must have number 6 in the 5th column. This then cascades to 36x! The number 6 for 36x (2*3*6) must go to the rightmost column otherwise it will violate the other 2 columns that we just filled in. For Sudoku solvers this should be in your familiar territory! That one observation allowed us to chain-complete a portion of the puzzle:

What else can we fill in? It's important to consider new point of views after you have filled in a set of entries because by filling in correct positions for cages - it reduces the possibilities for cages that are related to it (relation comes from the single 1 to 6 constraint on the rows/columns). If we observe the 3/ (division) cage on the rightmost column, previously it has the possibility of 3/1, 6/2. However now that we have filled in 6 on the rightmost column, the only possibility is 3/1. If you jot this down on the cage, you notice that we can now fill in 15x because the number 3 in 3*5=15 cannot belong to the rightmost column anymore so hence we have uniquely determined the correct position. Again, a lot of cascading solutions - by solving the 15x, we now know the position of the 5 in 20x and the position in 3x (validate it for yourself). If you fill in the details you'll get something like the following:

Now it looks much better! At this stage, we can fill in 24x because we know the product of its divisors cannot contain a 3 as it will violate the first row only having one 3 (it's taken by the 15x). 24x = 1*4*6. Note that it also can't be 2*2*6 because it exists in a single row and we can't have duplicate elements because it'll violate the rules. Note that the 4- cage on the leftmost column can be uniquely identified to be 6-2. How did we do this?

There are actually multiple ways to view this - the simplest is to note that the first row possibilities are completely filled except for one number. We have used 1,4,6 for 24x and 3,5 for 15x. That leaves only the possibility of 2 in the left and uppermost cell. Now we can't put negative numbers so the only possibility for the other cell in the cage is 6 - making 6-2=4.

The other way to view this will prove to be more useful when solving larger KenKen puzzles. Note that on both row 1 and row 2 we have used the number 5 already. 4- only has two possibilities for a 6x6 KenKen puzzle, 6-2 and 5-1. Hence if 5 cannot go in either of the cells in the cage then it cannot be 5-1, leaving 6-2. Let's fill these in:

So now we have an interesting pattern. the 12x cage is a L-shaped figure. This actually provides useful information that you may not first realise. By filling in the 6 on the 2nd row, we know for certainty that the remainder of the 2nd row is numbers 2,3. Hence we can automatically deduce the 3rd row/3rd column cell to be 2 as shown above. Again, there are multiple ways to view this. We can view 12x as a cage and note that we have 2,3 inside the cage already. This leaves us the unique number remaining to be 2 (12 / (2*3)).

For any L-shaped cage, if there are a repeated element (in this case the number 2), then they will always go on the ends of the L-shape. The reason for this is simple, if they were placed any other way they would conflict and violate the unique number rule. Filling in these numbers yields the following partially complete puzzle:

There is another tip that is useful. If you look at the 4th column in which we have filled in (1,5,3,6). We can actually uniquely and correctly identify the 4th number in the first row by noting that we reduced 24x to 1*4*6. But 2 of these numbers have already been used in the 4th column, hence the missing and correct number to fill in is '4'. This gives the partially solved puzzle:

Give it a try to solve the remainder of the puzzle! Happy Solving! =)

[Spoilers Ahead]
Note:
If you get stuck, I've included the solution below:

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;
}

Discrete Processes: Binomial Tree Model

After reading a bit from Baxter and Rennie's book about discrete processes, I decided to implement the basic binomial tree described in the book. It's a build onto the binomial branch model, in essence it's the same thing but done on a recursive scale. It is mostly used for evaluating the cost of an option via arbitrage. The main downfall to discrete process modelling computationally is that it becomes exceedingly difficult to scale it for problems of large instances. Unlike continuous processes which are evaluated through approximation, discrete processes models each tiny step and for an accurate representation would require a lot of memory and computational time.

The model suggests having a portfolio of stocks and bonds. This simple combination allows us to synthesize the derivative by replicating desired values. The key point is the probability of up and down moves from a given stock at a given time does not influence our ability to generate the desired values. For example, we can say for certainty that if the stock moved up then down and then up again we would reach our desired value f(n). We use an algebraic relation from the next stock up and down prices to determine our claim value. We evaluate the claim value from the back of the tree where the payoff/desired value is guaranteed and move our way back to our initial node. This claim value at the initial node is what we should price the option at.

Now if the market was to move up or down, we need to adjust our proportion of stocks and bonds to compensate. It's essentially like a board game, when the stock moves we make an appropriate adjustment to ensure that the next move (either up or down) is within our described tree regardless of the outcome. At the end of the time period, we are guaranteed to get the claim value for the given sequence of stock movements from t = 0 to t = n, where n is the number of time ticks (i.e. depth of the binomial tree). For more information refer to Baxter and Rennie's Financial Calculus book.

#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <algorithm>
#include <cmath>

using namespace std;

// represent the stock as a value in time
class Stock {
public:
   double price;
   double prUp;
   double prDown;
   Stock() { }
   Stock(double p, double up, double down) : price(p), prUp(up), prDown(down) { }
};

vector<double> fTable;   // represents the claim tree
vector<Stock> graph;   // represents the recombinant tree
int sz;

#define EPS 1e-09

// claims are calculated as:
//      q = (stock_now - stock_down) / (stock_up - stock_down)
//      f = q * f_up + (1 - q) * f_down
// base conditions can be computed based on valuation of the option at t=0 versus the
// payoff given a sequence of situations (when the stock goes up/down for t=0 to t=n).
double computeClaims(int node) {
   int baseNode = (sz-1) >> 1;
   if (fTable[node] >= 0) {
      // memoisation
      return fTable[node];
   }
   if (node >= baseNode) {
      // base case
      return fTable[node] = max(0.0, graph[node].price - graph[1].price);
   }
   // compute
   double q = (graph[node].price - graph[node*2+1].price) / 
      (graph[node*2].price - graph[node*2+1].price);
   fTable[node] = q * computeClaims(node*2) + (1 - q) * computeClaims(node*2+1);
   return fTable[node];
}

// define an recombinant tree and reconstruct the option value
int main() 
{
   int N;   // the depth of the tree
   cin >> N;
   graph = vector<Stock>((1 << N) + 1);
   
   // read in the recombinant tree (stated in order of time)
   // file format:
   // stockprice prUp
   // stockprice2 prUp2
   // etc.

   int totalCnt = (1 << N);
   for (int i = 0; i < totalCnt-1; i++) {
      double stockPrice, prUp, prDown;
      cin >> stockPrice >> prUp;
      prDown = 1 - prUp;
      graph[i+1] = Stock(stockPrice, prUp, prDown);
   }

   // reconstruct the option price by backtracking
   fTable = vector<double>((1 << N) + 1, -1.0);
   sz = (1 << N) + 1;
   double optionPrice = computeClaims(1);
   cout << "Option price is " << optionPrice << "\n";
   
   // run simulations (optional)
   // file format:
   //   sequence of {up, down}'s equal to N-1.
   // note: stockHolding is calculated by:
   //   s = (f_up - f_down) / (s_up - s_down)
   string status;
   string lastStatus = "-";
   int cnt = 0;
   int curTime = 0;
   double stockHolding = 0.0;
   double bondHolding = 0.0;
   int curBranch = 1;
   while (cin >> status && cnt++ <= N-1) {
      // output the status
      cout << "Time: " << curTime << " Last Jump: " << lastStatus << 
         " Stock Price: " << graph[curBranch].price << 
         " Option Value: " << fTable[curBranch] << " Stock Holding: " 
         << stockHolding << " Bond Holding: " << bondHolding << "\n";
      curTime++;
      lastStatus = status;
      stockHolding = (fTable[curBranch*2] - fTable[curBranch*2+1]) / 
         (graph[curBranch*2].price - graph[curBranch*2+1].price);
      bondHolding = fTable[curBranch] - (stockHolding * graph[curBranch].price);
      if (status == "up") {
         curBranch = (curBranch * 2);
      } else {
         curBranch = (curBranch * 2) + 1;
      }
   }
   if (cnt > 0) {
      cout << "Time: " << curTime << " Last Jump: " << lastStatus << 
         " Stock Price: " << graph[curBranch].price << 
         " Option Value: " << fTable[curBranch] << " Stock Holding: " 
         << stockHolding << " Bond Holding: " << bondHolding << "\n";
   }
   return 0;
}

With an example file input based on the book examples (pipe it in):

4
100 0.75
120 0.75
80 0.25
140 0.75
100 0.25
100 0.75
60 0.25
160 0.75
120 0.25
120 0.75
80 0.25
120 0.75
80 0.25
80 0.75
40 0.25
up
up
down

TC SRM 454 D1-500/D2-1000 (NumbersAndMatches)

Category: Dynamic Programming
URL: http://www.topcoder.com/stat?c=problem_statement&pm=10709

We are given a number which is composed of a sequence of digits. Each digit is represented by a combination of matches represented in a certain position to represent the digit. We are given the option of using up to K different match moves and are asked to calculate the number of different integers we can construct without discarding or adding any new matches to our given number.

First, we decompose the problem into several parts. The first involves somehow representing each digit in our program. A simple way to do this is to use a 2D integer array in which index (i,j) is equal to 1 if the digit i has a match in position j. An alternative way is to use a bitmask with the bit representation of the j-th bit set to 1 if there is a match in position j.

Next, we need to somehow compute the transitional costs between two different digits. For example, we need to quantify how many moves we would use up if we convert say a 6 to a 3. Since we are only allowed a limited number of moves its obvious we need a way to keep track of how many we have used up. One possible way is to determine how many places the two numbers differ. Then the number of moves we need to make is equivalent to exactly half of this.

There is another issue we need to deal with, that is the situation where the number of matches used to construct two digits differ. For example, the digit 6 uses 6 matches whereas digit 3 only uses 5 matches. We need to keep track of how much surplus (or deficit) matches we get from transitioning from one digit to another. This is important as we cannot add or remove matches from the final number - i.e. the number of matches used in constructing a unique integer must be the same as the original integer. Now we need to keep track of two things when we consider transforming one digit to another: the number of moves it takes and the difference in matches we have used so far.

We can define the cost function as follows:

To calculate the surplus/deficit it's quite simple, just calculate the number of matches used for each digit and subtract one from the other depending on which direction the transformation is taken. The above cost function ensures that we don't double count the number of moves required from moving a match from a surplus digit to a deficit digit. A way to visualise this is we "precharge" the surplus with an additional move but we are also granted a free move since this surplus is already counted once so the actual number of moves decreases if we need to import more matches from another transformation.

Lastly, we need to somehow compute all the ways we can construct unique integers given that we keep the number of matches used the same (i.e. at the end of our decision algorithm the number of surplus matches is exactly 0) as well as using less than or equal to K moves (which we subtract along the way). This suggests a recursive algorithm from potentially changing the first digit to something else, depending on the transformation we are left with either more/less/equal number of matches and the number of available moves to be less or equal to what we started with. We must also handle the case where we "borrow" matches from the latter part of the integer, so we must also handle cases where our match balance is negative. This can be done by computing an offset to keep the total number positive (to use for our caching purposes).

To efficiently compute the number of ways we need to memoise or apply bottom up DP to our algorithm. Note that the number of unique states is rather small so computational time isn't an issue. An implementation of the above ideas is shown below (using memoisation):

class NumbersAndMatches {
public:
   long long differentNumbers(long long, int);
};

int mat[10][7] = {
  {1,1,1,0,1,1,1}, // 0
  {0,0,1,0,0,1,1}, // 1
  {1,0,1,1,1,0,1}, // 2
  {1,0,1,1,0,1,1}, // 3
  {0,1,1,1,0,1,0}, // 4
  {1,1,0,1,0,1,1}, // 5
  {1,1,0,1,1,1,1}, // 6
  {1,0,1,0,0,1,0}, // 7
  {1,1,1,1,1,1,1}, // 8
  {1,1,1,1,0,1,1}  // 9
};

// defines the total number of matches used for digit i
int tot[10] = {6, 3, 5, 5, 4, 5, 6, 3, 7, 6};   

long long dp[19][155][155];
pair<int,int> costTab[10][10];
string num;

#define SURPLUS_MOD 72

long long func(int idx, int moves, int surplus) {
   // offset due to the possibility of negative surplus
   int surplusMod = surplus + SURPLUS_MOD;
   if (idx >= num.size()) {
      // requires the surplus amount of matches to be 0 otherwise its not valid
      if (moves >= 0 && surplus == 0) return 1;
      return 0;   
   }
   if (dp[idx][moves][surplusMod] != -1) return dp[idx][moves][surplusMod];
   long long res = 0;
   // try each digit
   int curDig = num[idx] - '0';
   for (int dig = 0; dig <= 9; dig++) {
      int c = costTab[curDig][dig].first;
      int d = costTab[curDig][dig].second;
      if (moves - c >= 0) {
         res += func(idx+1, moves - c, surplus + d);
      }
   }
   return dp[idx][moves][surplusMod] = res;
}

long long NumbersAndMatches::differentNumbers(long long N, int K) {
   long long res = 0;
   memset(dp,-1,sizeof(dp));
   // compute the cost function for every digit -> digit transition
   for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
         int c = 0;
         for (int k = 0; k < 7; k++) {
            if (mat[i][k] != mat[j][k]) c++;
         }
         int s = tot[i] - tot[j];
         costTab[i][j] = make_pair((c+s)/2, s);
      }
   }
   // decompose the number into a string
   long long vN = N;
   while (vN > 0) {
      num += ((vN % 10) + '0');
      vN /= 10;
   }
   reverse(num.begin(),num.end());
   // calculate the number of ways
   return res = func(0, K, 0);
}

Matrix Exponentiation

Matrix Exponentiation is a useful technique that can be applied to a wide variety of problems. Most commonly it is used for efficiently solving linear recurrence problems and as such can be used in any problem that can be represented as a linear recurrence. The main problem lies with the fact that it is inefficient when we naively evaluate them. Let's consider an easy example that you should know about: Fibonacci numbers.

The sequence is defined by:

It's easy enough to increasingly evaluate it 1 by 1 each time until we hit our targetted n-th Fibonacci number. However, let's say n was 1 billion - we need at least 1 billion computations of the sequence to derive the answer. If we represent the sequence in terms of a matrix we discover a clever optimisation which can be applied.

The correctness can be verified by a simple matrix multiplication. What's more important to note is that the next two sequence of Fibonacci numbers can be represented like the one above.

If we expand:

This is equivalent to:

We can continue expanding and eventually we see a remarkable observation. We can always reduce it to the power of our matrix multiplied by our initial conditions. This yields the recurrence below:

 

Since exponentiation can be done in logarithmic time we have essentially reduced our Fibonacci computation speed from linear to logarithmic. A huge difference in speed for large numbers! As an exercise, write a logarithmic Fibonacci term generator - as the numbers do get quite large, feel free to modulo it with an appropriate number (like 10^8).

We now expand our knowledge to a slightly more complicated linear recurrence. This is based on the problem "Number Sequences": http://acm.tju.edu.cn/toj/showp2169.html

To solve this recurrence we simply just do the same thing as we did with Fibonacci with a slight modification. Fibonacci is really a special case of the above recurrence where x = 1 and y = 1 and a0 = a1 = 1. Note the prime observation below:

If we matrix multiply - we yield the correct recurrence for both Fn and Fn-1. Therefore we simply need to change our original Fibonacci matrix of [ 1 1, 1 0 ] to [ x y, 1 0] and the initial conditions from being always 1 and 1 (F1 and F0 respectively) to [a1 a0]. We then simply use matrix exponentiation to calculate the correct term, as always we apply modulo arithmetic to keep the number representable with integers.

The implementation below demonstrates this idea:

class Matrix {
public:
   long long a;
   long long b;
   long long c;
   long long d;
   Matrix() { }
   Matrix(long long _a, long long _b, long long _c, long long _d) : 
      a(_a), b(_b), c(_c), d(_d) { }
};

Matrix multiply(const Matrix& lhs, const Matrix& rhs) {
   long long a = lhs.a * rhs.a + lhs.b * rhs.c;
   long long b = lhs.a * rhs.b + lhs.b * rhs.d;
   long long c = lhs.c * rhs.a + lhs.d * rhs.c;
   long long d = lhs.c * rhs.b + lhs.d * rhs.d;
   return Matrix(a % 100, b % 100, c % 100, d % 100);
}

ostream& operator<<(ostream& out, const Matrix& M) {
   out << "a: " << M.a << " b: " << M.b << " c: " << M.c << " d: " << M.d;
   return out;
}

Matrix power(Matrix M, int n) {
   if (n == 1) return M;
   Matrix tmp = power(M, n/2);
   if (n % 2 != 0) {
      Matrix n = multiply(tmp,tmp);
      Matrix m = multiply(n, M);
      return m;
   }
   Matrix v = multiply(tmp,tmp);
   return v;
}

void output(int n) {
   if (n < 10) cout << "0";
   cout << n << "\n";
}

int main() {
   long long x, y, a0, a1, n;
   while (cin >> x >> y) {
      if (x == 0 && y == 0) break;
      cin >> a0 >> a1 >> n;
      if (n == 0) { output(a0%100); continue; }
      else if (n == 1) { output(a1%100); continue; }
      Matrix m = power(Matrix(x,y,1,0), n-1);
      output((m.a * a1 + m.b * a0) % 100);
   }
   return 0;
}

Now for a more advanced example, we use matrix exponentiation to determine the number of cycles with a length smaller than k in a given directed graph.

Problem: TourCounting
Source: TopCoder SRM 306 D1-1050
URL: http://www.topcoder.com/stat?c=problem_statement&pm=6386

An elegant way to calculate the number of paths from a source vertex u to a destination vertex v that takes exactly k steps is to use matrix multiplication. First let's define A to be the adjacency matrix with each entry representing the number of edges from u to v. The (u,v) entry of A^t is precisely the answer we are looking for. The way this works is similar to the way that Floyd-Warshall algorithm works. The matrix multiplication considers the connectivity between an intermediate node k and attempts to link u to k and k to v. Due to the multiplicative nature of matrix multiplication as opposed to an additive (for shortest paths) it determines the number of paths based on the multiplication principle from discrete mathematics.

To illustrate this principle, consider a snapshot of the matrix A^t and it's adjacency matrix which is represented as A^1:

If we were to calculate the number of paths that start from vertex 0 to itself (i.e. a cycle) that takes exactly t+1 steps. Note that we can access vertex 0 from either vertex 1 or vertex 2. Since at state t, we can reach vertex 1 from vertex 0 with 2 paths and we can reach vertex 2 from vertex 0 with 3 paths. It stands that we can now reach vertex 0 to itself from 2+3=5 paths. Note that this is a direct consequence of the matrix multiplication process - validate this yourself for the first row and column. If the adjacency matrix was changed such that there are two edges from vertex 1 to vertex 0 like the following:

Then we can reach vertex 0 in a total of 2*2 + 1*3 = 7 ways since we have an option of choosing one of the two edges from vertex 1 to vertex 0. Again, this is consequence of the matrix multiplication algorithm. The argument holds for every other cell in the matrix. So what's the advantage of this method? The prime factor is efficiency - we can calculate matrix powers in logarithmic number of matrix multiplications. We accomplish this through a similar algorithm to fast powering.

So now we know how to calculate the number of paths from a source vertex u to a destination vertex v that takes exactly k steps. The actual problem requires us to compute the number of cycles that are present with less than k steps. So for a given instance of the matrix A^t we want to sum the diagonal to yield the number of paths with exactly t steps. To calculate it for less than k steps we would normally have to compute each power of t and sum all the diagonals up, however this is not fast enough as the number of steps can be as much as 1 million. We can reduce this using a similar method to our matrix powering algorithm. First define a function f:

f(n) = represents the number of ways we can get from all vertices to other vertices using less than n steps (matrix form)

The main optimisation is reducing it from linear (summing all matrices less than n) to logarithmic. The huge hint in logarithmic is dividing the input space by a (usually) constant factor. Note that if we divide the space into two f(n/2) parts we are close to our answer but not quite there. The problem is that upper n/2 part is not symmetrical to the lower n/2 part. But wait! Given the simple mathematical property of powers a^(b+n) = a^b * a^n we can simply just multiply one of the f(n/2) parts by A^(n/2) to yield the correct upper half! With this our recurrence becomes:

Like our fast powering algorithm we need to distinguish between odd and even states. The easy way to make an odd number even is to subtract it by 1. We then just calculate one instance of the odd power and make all successively calls even. This maintains the correctness of the recurrence whilst also maintaining the time complexity.

Then it's simply a matter of implementation which is the easy part. Just note we need to use sensible modulo arithmetic to ensure that we don't overflow any of our computations.

class TourCounting {
public:
   int countTours(vector <string>, int, int);
};

class Matrix {
public:
   vector<vector<long long> > data;
   Matrix() { }
   Matrix(int n, int m) {
      data = vector<vector<long long> >(n, vector<long long>(m, 0));
   }
};

long long MOD;

Matrix operator*(const Matrix& lhs, const Matrix& rhs) {
   Matrix res(lhs.data.size(), rhs.data[0].size());
   for (int i = 0; i < lhs.data.size(); i++) {
      for (int j = 0; j < lhs.data[i].size(); j++) {
         for (int k = 0; k < rhs.data[0].size(); k++) {
            res.data[i][k] = (res.data[i][k] + lhs.data[i][j] * rhs.data[j][k])%MOD;
         }
      }
   }
   return res;
}

Matrix operator+(const Matrix& lhs, const Matrix& rhs) {
   Matrix res(lhs.data.size(),lhs.data[0].size());
   for (int i = 0; i < lhs.data.size(); i++) {
      for (int j = 0; j < lhs.data[0].size(); j++) {
         res.data[i][j] = (lhs.data[i][j] + rhs.data[i][j])%MOD;
      }
   }
   return res;
}

Matrix power(const Matrix& lhs, int P) {
   if (P == 1) return lhs;
   Matrix tmp = power(lhs, P/2);
   if (P % 2 != 0) {
      Matrix n = tmp * tmp;
      Matrix m = n * lhs;
      return m;
   }
   Matrix v = tmp * tmp;
   return v;
}

long long compute(const Matrix& m) {
   long long res = 0;
   for (int i = 0; i < m.data.size(); i++) res = (res + m.data[i][i]) % MOD;
   return res % MOD;
}

Matrix dp[1000001];
int visited[1000001];

Matrix f(const Matrix& ref, int n) {
   if (visited[n]) return dp[n];
   visited[n] = 1;
   if (n == 1) return dp[n] = ref;
   if (n % 2 != 0) {
      // odd
      Matrix v = f(ref, n-1) + power(ref, n);
      return dp[n] = v;
   }
   // even 
   Matrix vE = f(ref, n/2) * power(ref, n/2);
   Matrix vEE = vE + f(ref, n/2);
   return dp[n] = vEE;
}

int TourCounting::countTours(vector <string> g, int k, int m) {
   Matrix mat((int)g.size(),(int)g.size());
   for (int i = 0; i < g.size(); i++)
      for (int j = 0; j < g[i].size(); j++)
         mat.data[i][j] = g[i][j] == 'Y' ? 1 : 0;
   MOD = m;
   long long res = 0;
   Matrix rr = f(mat, k-1);
   res = compute(rr);
   return res;
}