tomek wins Room 3

Discuss this match
Thursday, May 4, 2006
Introduction by lbackstrom

Semifinal round 3 proved to be the most exciting so far with tight races and a slew of challenges. The match started out slowly with many competitors taking over half an hour to submit the easy problem. The medium problem didn't prove much easier and it was about an hour before anyone had two submissions. John Dethridge tried to skip the medium after he submitted the easy, but he quickly gave up, and went back, getting the fastest time on the medium. Going into the challenge phase, John Dethridge had the lead, followed by ploh, tomek, and halyvin, each less than a challenge away.

The challenges phase yielded no fewer than 16 challenges, 6 of which were successful. With 2 of the successful challenges, tomek moved up into first place, while ploh and John Dethridge both dropped 25 points. The system testing phase proved uneventful by comparison, and the top 6 remained unchanged: tomek and JD move on to the finals, while ploh, reid, liympanda, and JongMan will have to fight it out in the wildcard round.


by brett1479

As done in Apostol's wonderful calculus text, we define the nth Taylor operator (it takes a function and produces the corresponding finite Taylor series). This problem explores how derivatives and integrals interact with the Taylor operator (and requires no knowledge of calculus). To solve the task at hand, we recursively descend into the given nested formula keeping track of the number of D's and I's we have accumulated, and cancelling appropriately (no sense in having both D's and I's). Upon finding a T, we compute the maximum number of D's (or I's) that can be pushed deeper without violating the numeric constraints. When we hit the deepest nesting level of the expression, we are left with an 'f'. As we climb out of our recursive calls, we construct the return value by surrounding the current formula with the correct number of D's, I's, and T's that were accumulated at that level (of the recursion).


by radeye

This problem has two parts: calculating the number of configurations, and handling the symmetry. Consider the number of consecutive straight sections as we go around the circle as six numbers a, b, c, d, e, and f. In order for the loop to be closed, a+b must equal d+e, and b+c must equal e+f. These two equations also imply a third, that c+d=f+a. Using these equations, we can set any four of the variables (say, a, b, c, and d) and compute the other two. Thus one possible solution is to iterate over all possible values for a, b, c, and d, compute the other two, and if they are nonnegative, consider it a found configuration. This runs too slowly.

We can eliminate one of the loops by giving only a, b, and c values, and then computing the range of values d can take on that would result in a valid configuration. These equations can be calcuated by considering the equations above, keeping d as a variable, and solving them for e and f nonnegative, and the sum of all six numbers less than straight. The lower bound on d is the greater of 0 and a-c, and the upper bounds on d is the lesser of a+b and n-a-2*(b+c). Sum up the count of all possible configurations, and call this n1.

This lets us enumerate all configurations ignoring symmetry. To handle symmetry, we calculate in addition to the above number the count of positions that have 6-symmetry (all six 60 degree rotations are identical); 3-symmetry (all three 120 degree rotations are identical), and 2-symmetry (both 180 degree rotations are identical). We call these values n6, n3, and n2, respectively. Each of these can be calculated very quickly because the number of free variables is reduced to 1, 2, and 3, respectively.

To calculate the final answer, consider that every position whose greatest symmetry is 1-symmetry has been counted six times in n1 and no times in any of n2, n3, or n6. Every position whose greatest symmetry is 2-symmetry has been counted three times in n1 and three times in n2. Every position whose greatest symmetry is 3-symmetry has been counted twice in n1 and twice in n3. Every position that exhibits 6-symmetry has been counted once in each of n1, n2, n3, and n6. If sn is the number of positions that have a greatest symmetry of n, then we have the following equations:
n1 = 6 * s1 + 3 * s2 + 2 * s3 + s6
n2 = 3 * s2 + s6
n3 = 2 * s3 + s6
n6 = s6
The answer we desire is s1+s2+s3+s6, and we can easily solve these equations to find that that answer is simply (n1+2*n2+3*n3+2*n6)/6.


by radeye

Operating on compressed representations of data is frequently key to high-performance algorithms. Sparse matrices and Binary Decision Diagrams are just two examples of common data structures that are just compressed representations of more natural formats. Indeed, memoization itself simply compresses the expression graph. It is no surprise, then, that memoization is key to this problem.

The first step to the solution of this problem is to ignore the compression, and just consider calculating a string match by walking the haystack character by character. At any given point, the state is the maximum number of characters in the needle that have been successfully matched. For every new character in haystack we see, we calculate the next state based on the current state and the new character. When we reach a state that indicates we have matched all characters, we have found our match.

Because of needles like "aabab" and input like "aabaabab", we cannot simply start over at the beginning of needle every time we find a mismatched character. Instead, when we find a mismatch, we want to find out the longest prefix of needle that is a suffix of the part of haystack we have examined so far. There are many ways to compute this state machine. Perhaps the easiest is to simply use a loop like the following:
      for (int i=0; i<needle.length(); i++) {
         String p = needle.substring(0, i) ;
         char c = needle.charAt(i) ;
         for (int j=0; j+i<=needle.length(); j++)
            if (needle.substring(j, j+i).equals(p))
               nextState[j+i][c] = (i+1) ;
This works great so long as we have no compression. As it turns out, it is easy to add compression too; just consider a character that represents a compressed string as another character, and calculate what the next state is for a given input state. With the potential of compressed strings representing huge haystacks, this may seem to run too slow, but memoization makes it easily run in time.

The only remaining complication is remembering the offset where we found the needle. To do this we simply calculate how many total uncompressed characters we have consumed, and when we find a match, return that value less the length of the needle. In order to keep the code clean, we can use an exception to return the final match position. Note that we can memoize both length and search easily.
   long length(char a) { // memoize this for speed
      if (!inDictionary(a))
         return 1 ;
      long r = 0 ;
      for (char c : expansion(a))
         r += length(c) ;
      return r ;
   int search(String haystack, long pos, int state) throws Found {
      for (char c : haystack) {
         if (state == needle.length())
            throw new Found(pos-needle.length()) ;
         if (not_calculated(nextState[state][c]))
            nextState[state][c] = search(expansion(c), pos, state) ;
         state = nextState[state][c] ;
         pos += length(c) ;
      if (state == needle.length())
         throw new Found(pos-needle.length()) ;
The rest is straightforward; if we get an exception indicating the string was found, we extract the found position from that exception and return it; otherwise, we return -1.