Online Round 3 October 14, 2006 Match summaryIn the last online round before an onsite event, coding skills are not the only factor involved. Nerves can easily disrupt the thought process and ruin everything. Those who had been in this situation before certainly had an advantage. Only 28 coders solved more than one problem succesfully, and a reasonably fast submission on the easy problem was enough to advance to the onsite competition. The five contestants that solved all three problems deserve to be mentioned: krijgertje (the winner), kalinov, Petr, Eryx and tomek.The problems were a bit more difficult than a regular Division 1 round, as should be the case. The first problem caused problems for many coders. Most submitted it in the end, but the submissions were evenly distributed across all 75 minutes. Many submissions also went down in challenge phase or the system test. The medium problem had a small performance trap, which caused many to resubmit their solution. Instead, solutions that weren't tested enough generally went down in the challenge phase or system test. The last problem required a fair amount of coding, and the submissions started to appear fairly late in the contest. In the end, only seven coders solved it, and all of them were among the highest ranked. The ProblemsDiamondsUsed as: Division One  Level One:
In the first part of the solution, we determine the size of the diamond for each coordinate x, y containing a '#'. This can be done in several ways. Perhaps the easiest is to find the '.' closest to x, y (using the Manhattan distance). The radius of the diamond equals this shortest distance, if we take into account that all cells outside the given grid only contains '.'. In the second part, we need to determine the maximum radii sum. The examples showed that it's not always desirable to pick the largest diamond. However, the size of the input allows us to test all pairs of possible diamond centers (there are at most 50^{2} * 50^{2} such pairs). Given two diamonds, we need to determine if they intersect. This is done similarly to how you determine if two circles intersect, the difference being that we here use the Manhattan distance rather than the Euclidean distance. So, if the Manhattan distance between the center of the diamonds plus one is less than the sum of their radii, the diamonds intersect. If they do intersect, we have to decrease the size of the diamonds so that they are just less than the limit. See Petr's solution for a clean implementation of this algorithm. A possible corner case is when an empty diamond has to be used. The last example in the problem statement covered this case, however, so there shouldn't have been any nasty surprises. WordGuessingGameUsed as: Division One  Level Two:
The unusual limitation that there were at most 18 guesses should hint that an exponential solution is required (always look at the constraints to get a hint on how to solve the problem). The problem can be solved by an exhaustive search, using memoization. Recursively, we try all 26 letters at the first position. For each letter, we try all 26 letters at the second position, and so on. But since the maximum word length is 9 characters, this yields 26^{9} combinations, which obviously is way too much. This is where the memoization part comes in. At any time during this recursive process, each guessed word can be considered to be either "even" or "odd", meaning that an even or odd number of the letters left must be correct for that particular guessed word. When we try a letter at some position, all guessed words matching that letter should switch their evenodd status, since now one letter less must be correct (and hence the parity change). If, at some point during the recursion, we are at the same letter position and have the same set of even and odd words as at some previous point, then we don't have to go down that path again unless we found some solution last time (since we must return all solutions). Thus the state space in the search is at most 9 * 2^{18}, and each state is only visited once, except when we know that some path down this road contains a solution (and this happens at most 100 times, because the problem constraint specified that there were at most 100 solutions). Inside the recursive function, we test all 26 letters, and for each letter do the necessary parity switches. This parity switching stuff can be done without any loop, which is essential if the solution is to run in time. To do this, the set of even and odd words must be represented during the recursion as a bitmask. Before the recursion starts, we calculate the effect of trying some specific letter at some specific position. That is, we calculate a bitmask where each bit set marks a guessed word that should switch parity. During the recursion, switching parities can then be done using bitwise xor on the current mask with a precalculated value. This speedup trick is probably what caused the majority of solutions to fail, and was an intended trap from my side... See kalinov's solution for a short and elegant implementation of this algorithm. TopologicalEquivalenceUsed as: Division One  Level Three:
Each figure can be represented by a graph where each endpoint of a line segment corresponds to a vertex. The constraints guarantee that each such graph will have at most 8 vertices, which will be useful later on. So the first part of the problem is to create these graphs. This can be done by first mapping all distinct points to some global id, and then create a big adjacency matrix where each edge between two ids represents a line segment. Which point belongs to which figure can then be deduced using a depth first search. Now to the tricky part of the problem. When exactly are two figures equivalent, and how do we determine it from the graphs representing the figures? Many different graphs, with a different number of vertices, can all represent the same figure. The key lies in finding a canonical graph for each figure. That is, some graph that all other graphs representing a topological equivalent figure can be reduced or transformed into. It should be obvious that splitting a line segment into two parts doesn't affect the figures topological feature, even though it does introduce a new vertex in the graph. So we want to reverse this operation. That is, any vertex that has exactly two edges should be contracted. This means that the vertex should be removed from the graph, and the two edges going from the vertex should be replaced by a single edge connecting the two surrounding vertices. This procedure of finding vertices with exactly two neighbors should then continue until no more such vertices are found. Some care must be taken so we don't reduce the graph completely. We might end up with a graph containing a single vertex with a single edge connecting the vertex to itself; this graph should not be reduced any further. This graph reduction is the only operation needed to get a canonical graph. To determine the number of unique graphs, we can compare all pairs of graphs and see if they are isomorphic. Another method is to find a canonical representation of the set of isomorphic graphs, so that two isomorphic graphs map to the same canonical representation. The internal adjacency matrix of a graph is a 8x8 binary matrix. This matrix can be represented by an unsigned 64 bit integer. By testing all permutations of the vertices in the graph, we potentially get 8! = 40320 distinct adjacency matrices. Among them, we can simply pick the smallest unsigned integer as the representation of this graph. This value is then inserted in a set, and the answer to the problem is the size of this set. This algorithm was used by krijgertje, whose solution is also the cleanest of all successful solutions to this problem (if you don't mind the extensive use of C++ #define macros). 
