During all my previous tournaments the Algorithm track participants were the majority, while this will not be the case this year, so the atmosphere may turn out quite different. Anyway, I'm sure it will be fun.
[posted from New York - JFK]
Saturday, May 30, 2009
TopCoder Open 2009
TopCoder Open onsite finals is an unique opportunity where one can blog about daily events and still stay in the style of this blog (which I hope to be exclusively about programming competitions). I'll try to post news/thoughts/videos here, although of course the official http://www.topcoder.com/tco09/blog/ will be more interesting and diverse :)
Tuesday, April 21, 2009
ICPC 2009 World Finals
Here're the problem statements: http://cm2prod.baylor.edu/resources/pdf/2009Problems.pdf (mirrored at http://77.41.63.3/icpc2009/statements/problems.pdf)
Here're the standings: http://zibada.ru/pcms/finals/
Here's the SnarkNews coverage: http://acm.math.spbu.ru/~snark/finals/
I'll try to post some analysis here.
Problem A.
First, we loop over 8! possible orderings for the landing of the planes. Then, we do binary search on the answer. Having fixed the answer, we simply land each plane in the chosen order at the earliest possible time.
Here're the standings: http://zibada.ru/pcms/finals/
Here's the SnarkNews coverage: http://acm.math.spbu.ru/~snark/finals/
I'll try to post some analysis here.
Problem A.
First, we loop over 8! possible orderings for the landing of the planes. Then, we do binary search on the answer. Having fixed the answer, we simply land each plane in the chosen order at the earliest possible time.
Problem B.
This problem requires careful checking of all boundary cases, and doesn't seem to require any particular insight. One just verifies the output of the original scheme on all given inputs, as well as the output of all "wrong" schemes that can be obtained from the given one by introducing one mistake (since there's not more than 19 gates, there's not more than 57 such schemes), and then checks if the output can be uniquely determined. But there surely is some trickiness to implementing that.
Problem C.
This is a very well-known problem if the ant moves along the surface of the cube; the octahedron doesn't change much. We try all possible ways to lay out the octahedron's sides on the plane (first place the first side, then place any of its neighbors next to it, and so on). The number of possible layouts should be on the order of hundreds or thousands. And in at least one layout, the shortest path will be just a straight segment - so we need to check those segments in all layouts, throw away those that go outside the layout, and then find the shortest one of the remaining. This solution should work as long as no shortest path passes through the same side twice; I don't have a proof for that, but intuitively it seems true. The implementation will require a LOT of accuracy, though.
Problem D.
One thing that comes to mind is to draw all possible layouts on paper, and figure out the formulas for the answer in each of them. But that's very error-prone, so I hope there's a more concise solution.
Problem E.
Take any vertex in the graph. Let's say it has a "nice left" if any path from the start to that vertex has the same cost. Let's say is has a "nice right" if any path from the end to that vertex has the same cost.
This can be determined by DFSen.
If there's a vertex without both nice left and nice right, then there's no solution: at least one of the roads to the right of it will have to be modified, and at least one of the roads to the left of it will have to be modified - so there will be a path with two modified roads.
Otherwise, the vertices split into 3 sets: A - vertices with nice left but no nice right, B - vertices with nice left and nice right, and C - vertices with nice right but no nice left. If both A and C are empty, the constraints are already satisfied and there's no need to change anything. Othewise, A will always have the start vertex, C will always have the end vertex.
Let's assume for now that B is empty. The above argument about no path with two modified roads tells us that the only way to achieve what's required is to modify the roads that lead from A to C. More specifically: take all pairs of vertices a from A and c from C, where there exists and edge from a to c. The cost of that edge plus the cost of the path from start to a and from c to end is a lower bound for the answer. Take the highest of those lower bounds, say H, and change the cost of each edge between a from A and c from C to H-cost(start,a)-cost(c,end). This will quite obviously be the minimal possible answer, and every path will have the same cost of H since it will pass along exactly one edge from A to C.
Now, how to deal with B? It turns out that we can simply put all vertices from B to C, and the above solution will still work - since it always leaves at least one route from start to end unchanged, it can't make a non-optimal answer; and it will always produce a correct answer since the only thing it requires is for all vertices from A have a "nice left" and all vertices from C have a "nice right".
Does that make sense? :)
Problem F.
First, let's learn how to calculate the length of one fence required to enclose the given set of points. To find that fence, we find the convex hull of that set of points, and then "extend" it to the outside by the given margin. The result may have a complex structure of straight and circular segments, but its total length is easily computed: the total length of straight segments is equal to the perimeter of the convex hull, and the total length of circular segments is equal to one circle, 2*pi*margin (because you have to "rotate" by 2*pi to complete the cycle - when you're moving in straight line, your direction doesn't change).
The rest of the problem is standard: we do a DP that calculates the minimum possible cost for each subset of points, by choosing a subset of that subset to be contained by one fence, and taking the previously-computed DP value for the rest of the points.
Problem G.
I think this problem requires one just to implement the game's rules carefully. The number of possible states in the game is reasonably small: the position is characterized by the number of cards still lying in the row, the cards held by both players, if any, and the layout of the "top line" of the house of cards - which will always include not more than 8 cards. This gives us a rough estimate of C(26,10)*26 states, which is roughly 100 million - but in reality much less of the states will be reachable, since the top line only contains 8 cards in one case, and so on.
Problem H.
If we introduce variables x1,x2,... for the decisions on the bills, then satisfying all ministers can be formulated via statements of form "if x5 is false, then x7 is true" (since if at least one decision contradicts minister's choices, then all his other choices must be respected). That gives us an instance of 2-CNF satisfiability problem, and its solution is well-known. Checking which variables have the same value in any solution is also quite routinely added to that well-known solution (for example, we fix that variable and run the satisfiability checking again).
Problem I.
Another implementation problem here. One should understand the problem statement, and then simulate all resizes going from the outermost window inside.
Problem J.
We can try solving this with a binary search + DP. We binary search over the answer, then do a DP on subtrees. For each subtree, consider all possible roundings that don't contradict the chosen answer (that is, the paths completely inside the subtree don't change by more than that answer), and from each of those roundings, we're interested in the lowest and highest differences for paths from the root of that subtree to its vertices (these two values are the only ones that affect paths that pass through the root of that subtree). So the DP can be: for a subtree T and a lower bound L on the difference on the paths from its root, what is the lowest possible upper bound U on that difference? This should give us only at most of 100*30*100 states, and I think the processing of all states can be done in at most 100*30*100 time (since there're only 100 edges, and "updating" along each edge will take one scan along a 30*100 array that has the U's for each L), thus even a binary search outside still leaves the running time reasonable.
I know this is very unclear, but I think the only way to figure out the details is to actually start coding the solution, and I'm too lazy for that :)
A shot at clarifying: for a subtree, there're 3 parameters:
1) the maixmum modulus of delta for paths inside that subtree
2) the maximum delta for paths from the root of that subtree into the subtree
3) the minimum delta for paths from the root of that subtree into the subtree
Without the binary search, we'd be forced to add the 1st parameter to the DP, and that would make it too slow. That's why we use binary search outside.
Problem K.
Consider the chain of transformation in the optimal answer. Consider the change(s) that affect the leftmost character of the string. Between those changes, the transformation is split into independent parts. That gives us the possibility of the following DP: consider the set of all suffixes of length L of all given strings (S,T, and all transformation strings). There're at most 202 of those for each L. Now, let's calculate the best way to transform each of those strings to each other, given that we know such answers for smaller values of L. First, we account for the transformations that don't involve the whole L characters - we can just take the value from the L-1 step. And then, there are transformations with the whole L characters - they are given by rules of length exactly L. And then, we need to combine those two kinds with a shortest-paths algorithm, like the Floyd-Warshall. That gives us 202^3 running time for each L, so about 160 million very simple operation for each testcase - that should be enough.
So that makes about 170 minutes for all solutions with some interruptions :) It's nice that they don't have problems that are theoretically impossible to solve during the contest.
I think the problems are nice and continue the usual trend of World Finals problems. Thanks for all judges for composing this contest, and for KTH and ACM for organizing it!
Sunday, March 22, 2009
TCO R1, R2 and R3 screencasts
This is another interesting-only-for-TopCoder-players post.
I've liked the hard problem from R3 a lot. I think it is an excellent example of how dynamic programming problems can be very non-standard (and the medium problem from the same round provides an example of how they can be very standard :)) and beautiful. The regular appearance of such problems makes me sure that we'll never run out of ideas in algorithm competitions. Although I'm biased for sure.
Saturday, March 14, 2009
A couple recent FAILs
I've recently lost my 1st TopCoder rating to ACRush and have already fell quite a bit behind. This is mostly due to his superb performance (great job Lou!), but it also highlighted several of the weaknesses of the choices I've made for my competing style: namely, using C# and expecting the time limit to have some margin, and not having any pre-written code to use in the competition.
All 3 last matches made me think again about those questions. In TCO09 Round 1 the hard problem turned out to have a rather strict time limit, and thus my asymptotically correct solution timed out slightly (I believe it ran within 3 seconds) and I had to resubmit, placing only 10th. SRM 436's hard was about fast multiplication of polynomials, and as I don't understand Fast Fourier Transformation good enough to code it and don't have it prewritten, I've resorted to the Karatsuba algorithm and again, my solution was very little over the time limit, so I've spent a lot of time until finally getting it under the time limit (and again, it turned out that this solution was expected to be correct by the problem authors), scored very low and placed only 15th. Today, me failing the medium problem was totally my fault (I couldn't invent the correct solution) - but even if it didn't fail, I would place below Lou because he had a pre-written implementation of the Dinic maximum flow algorithm for the hard problem, and it turned out that one could run that 2500 times under the time limit. This was not possible with the stupid maximum flow algorithm I've used, and thus I had to spent a bit more time implementing a slightly more complex approach.
I still believe that all this happening in a row is just a bad coincidence, and in the long term my strategy has more pros than cons. Next couple of months will show :)
Monday, March 9, 2009
Self-referential sentence solution
OK, here's the sentence:
This sentence has seven hundred and seventeen letters, one hundred and fifty-six words, forty-six commas, one dot, six dashes, one hundred and forty-nine spaces, eighty-two quotes, two words "This", five words "and", two words "commas", one word "dash", two words "dashes", two words "dot", one word "eight", two words "eighteen", two words "eighty", one word "eleven", one word "fifteen", two words "fifty", two words "five", three words "forty", four words "four", one word "fourteen", two words "has", four words "hundred", two words "letters", two words "nine", one word "nineteen", one word "ninety", eighteen words "one", two words "quotes", two words "sentence", two words "seven", two words "seventeen", one word "seventy", four words "six", one word "sixteen", one word "sixty", two words "spaces", one word "ten", two words "thirteen", two words "thirty", three words "three", one word "twelve", two words "twenty", twenty-one words "two", thirteen words "word" and thirty-one words "words".
Now, how to build that one? Amazingly, the following very simple approach works. Start with some string. Build a sentence that describes it. Now build a sentence that describes the previous sentence. And so on, until at one point you find that the sentences stop changing. It's true that this can never happen - but it turns out that trying out several initial strings is enough to find one that leads to a solution. It's true that the solution will be excessively long (e.g., all [one word "smth"] clauses look really unnecessary, don't they?), but who cares?
Here's the code:
http://77.41.63.3/blog/self-referential/SelfReferentialSentence.html
http://77.41.63.3/blog/self-referential/NumberSpelling.html
This sentence has seven hundred and seventeen letters, one hundred and fifty-six words, forty-six commas, one dot, six dashes, one hundred and forty-nine spaces, eighty-two quotes, two words "This", five words "and", two words "commas", one word "dash", two words "dashes", two words "dot", one word "eight", two words "eighteen", two words "eighty", one word "eleven", one word "fifteen", two words "fifty", two words "five", three words "forty", four words "four", one word "fourteen", two words "has", four words "hundred", two words "letters", two words "nine", one word "nineteen", one word "ninety", eighteen words "one", two words "quotes", two words "sentence", two words "seven", two words "seventeen", one word "seventy", four words "six", one word "sixteen", one word "sixty", two words "spaces", one word "ten", two words "thirteen", two words "thirty", three words "three", one word "twelve", two words "twenty", twenty-one words "two", thirteen words "word" and thirty-one words "words".
Now, how to build that one? Amazingly, the following very simple approach works. Start with some string. Build a sentence that describes it. Now build a sentence that describes the previous sentence. And so on, until at one point you find that the sentences stop changing. It's true that this can never happen - but it turns out that trying out several initial strings is enough to find one that leads to a solution. It's true that the solution will be excessively long (e.g., all [one word "smth"] clauses look really unnecessary, don't they?), but who cares?
Here's the code:
http://77.41.63.3/blog/self-referential/SelfReferentialSentence.html
http://77.41.63.3/blog/self-referential/NumberSpelling.html
Friday, February 27, 2009
Self-referential sentence
Here's one funny problem back from 1998 summer training camp for Russian IOI team hopefuls. Construct a sentence of form:
This sentence has one hundred and ninety letters, twenty one word, twelve commas, one dot, seventy five spaces, fifteen quotes, one word "This", seven words "word" and two words "twenty".
Which is completely true and has nothing missing. At the camp, there were only so many participants, so the results were checked by hand with help of a computer program, and thus some leeway was allowed in interpreting the problem statement. However, the problem doesn't suffer if we fix what's needed precisely: the sentence should be like the above but obviously with different numbers (but still spelled in "canonical" English way) and all words listed with correct amounts in case-sensitive way, with no word from the sentence missing.
I couldn't solve it back in 1998, and the solution looked like a very innovative idea to me at that time. It doesn't anymore, but I hope some of you will enjoy thinking about it. Any suggestions on the approach? Any solutions? :)
// I know that this is a well-known idea, and Googling the title of this post reveals a lot of examples and solutions :)
// To give you some context: here's me at the similar camp a year later in 1999:
This sentence has one hundred and ninety letters, twenty one word, twelve commas, one dot, seventy five spaces, fifteen quotes, one word "This", seven words "word" and two words "twenty".
Which is completely true and has nothing missing. At the camp, there were only so many participants, so the results were checked by hand with help of a computer program, and thus some leeway was allowed in interpreting the problem statement. However, the problem doesn't suffer if we fix what's needed precisely: the sentence should be like the above but obviously with different numbers (but still spelled in "canonical" English way) and all words listed with correct amounts in case-sensitive way, with no word from the sentence missing.
I couldn't solve it back in 1998, and the solution looked like a very innovative idea to me at that time. It doesn't anymore, but I hope some of you will enjoy thinking about it. Any suggestions on the approach? Any solutions? :)
// I know that this is a well-known idea, and Googling the title of this post reveals a lot of examples and solutions :)
// To give you some context: here's me at the similar camp a year later in 1999:
Saturday, February 21, 2009
Screencasts since fall
This will probably only be interesting for TopCoder players, and they've probably already seen this on the forum, but anyway. Here're the screencasts for the SRMs that happened since my next-to-last post.
With problems not being very difficult, all came down to the challenge phase, and I was lucky to emerge in the 1st place in the end :) There was nothing particularly interesting about my challenges - I've had a pre-crafted case for the medium where I felt many can hit integer overflow, but only found one solution to challenge with it; and I've found a timeout case for a backtracking solution in the hard.
Nothing exceptional except stupid switch to C++ to use upper_bound that costed me quite a lot of time and points instead of just using binary search :)
This SRM highlights one of the techniques that I believe has improved my TopCoder results quite a bit - doing a stress-test against a stupid solution on random testcases. I've managed to resubmit the hard problem correctly because of it.
SRM 433 - no screencast - but just a comment.
Wasn't it stupid to do a last-second challenge on a possible hash collision when I was in first place with less than 25 points of margin? Yes it was.
With problems that seemed to suit me and a late resubmission by ACRush, there was no need in much challenge activity. And since I've chosen the wrong problem to challenge (who would've thought of so many ways to fail the easy? :)), the screencast should be rather boring in the challenge phase.
Very nice problems, but too slow thinking at 5AM. And very thorough example cases.
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