Friday, October 24, 2008

Solutions for previous problems

I can't resist to complete my previous questions with answers.

First, about the integral of f' squared being not less than the integral of f squared.

First, I'll write the proof:

0<=int((f'(x)-cot(x)*f(x))^2)=int(f'(x)^2)-2*int(f'(x)*f(x)*cot(x))+int(cot(x)^2*f(x)^2)=A-2B+C, where cot(x) is cos(x)/sin(x).
Let's simplify each of A,B and C by itself. The most interesting is B:
B=int(f'(x)*f(x)*cot(x)), and integration by parts yields B=f(3)*f(3)*cot(3)-f(0)*f(0)*cot(0)-int(f(x)*f'(x)*cot(x)+f(x)*f(x)*(1-cot(x)^2))=-B-int(f(x)*f(x)*(1-cot(x)^2)), thus -2B=int(f(x)*f(x)*(1-cot(x)^2)), and substituting that to the original inequality we get 0<=A-2B+C=int(f'(x)^2)+int(f(x)^2*(1-cot(x)^2))+int(f(x)^2*cot(x)^2)=int(f'(x)^2)-int(f(x)^2), q.e.d.

We've used that 0=f(0)*f(0)*cot(0) (well, we don't have cot(0) really, but we should interpret this as a limit from right), but that's easy because f(0)=0 and it has a derivative, thus is at most linear around zero, thus f(0)*f(0) is at most quadratic, and cot(x) is 1/linear, thus together they still stay at most linear.

This proof won't work if we substitute 3 with something larger than pi (because cot won't be defined at pi, and that integration by parts wouldn't be possible), and it helps to see that if we replace 3 with pi exactly, then equality would be achieved when f(x)=K*sin(x) (one can see from the above proof that when 3 is 3, equality is possible only if f(x) is always zero), and moreover, when 3 is replaced with something more than pi (say, k), then sin(x*pi/k) would provide a counterexample.

But what's bad (and interesting :)) about this proof is that starts from the end. It takes a seemingly weird combination of f'(x) and f(x), and that combination turns out to be exactly what we need. But how do we invent that? I don't know. As Jedal points in comments to the original problem statement, this has to do with f'(x)^2-f(x)^2 having some physical meaning when f(x) is a position of a body attached to a spring as a function of time, and sin(x) being the real trajectory of that body, thus delivering the extremum of that function. But I can't say I understand much of that.

Second, about the problem being solvable faster than O(n^2).

When we look closer at the sum being calculated, we see that everything except gcd is a polynomial on w and h, and sum of poly(w,h) over w from 0 to n is a polynomial over h and n, and repeating that for h, we'll get a polynomial over n and m that can be evaluated in O(1). But how to deal with gcd? We're looking at a sum of (n-w+1)*(m-h+1)*gcd(w,h). Let's iterate over w, then this is reduced to sum of (m-h+1)*gcd(w,h). Since we're already doing O(n) iterations and want to be under O(n^2) total, we need to do that sum faster than O(n). And here's how: let's group its terms where gcd(w,h)=k. Since k is a divisor of w, there can be only so much of those (I can't invent an exact formula now), so if we count all the terms with gcd(w,h)=k in O(1), then we're done. We need to count the amount of values for h such that gcd(w,h)=k, and the sum of those values. When we replace "gcd(w,h)=k" by "k divides gcd(w,h)", the answer is simple - we just need to examine values of form k*a, where a is integer. But then, we can obtain the values we need via inclusion-exclusion principle, and while this isn't O(1), it's still fast enough: the number of iterations for each w will be comparable to the amount of pairs (a,b) such that a divides b and b divides w, which is still much less than w (for each p^k in the prime decomposition of w we get (k+1)*(k+2)/2 choices for a and b, which is much less than p^k for large p's).

Phew. That was long, difficult to understand, impractical, and whatever (and quite probably wrong somewhere).

And last but not least, here is the screencast from SRM 422. It might be funny when I try to Google the solution for the hard problem, and when I fix my solution in last minutes by adding many random tries until a second runs out :)

Streaming: I believe you need at least Flash 9 to view that.
Download: or or (when my server goes down) (this is in Russian, but the only thing you need to do is to solve the CAPTCHA just above the green button, and then press the green button; there might also be a checkbox on Firefox and IE asking to install a toolbar - uncheck it). The file is about 50M.

If you have problems viewing, please tell - I'm still trying to figure out the right setup for this.

Wednesday, October 22, 2008

XXI St. Petersburg State University Championship - problem C

I've participated in a contest this weekend in St. Petersburg, and I'd like to explain (at least) one problem from that contest - I think it may be interesting even for people not very much into programming competitions.

The problem is: how many parallelograms are there with vertices having integer coordinates, x from 0 to n, y from 0 to m (both inclusive)? n and m are given as input, and both don't exceed 2000 (so an O(n^2) algorithm is almost surely OK, while O(n^3) or more is almost surely too much).

This is kind of classical problem if we replace parallelograms with triangles. But it turns out that parallelograms allow for some more interesting solutions. There's an approach that is the same for both parallelograms and triangles. What's the problem with counting those parallelograms directly? Well, there may be too much - on the order of n^6 (the parallelogram is basically defined by 3 of its vertices) - so we need to somehow count many at once. How can we do that? The most obvious way would be to count parallelograms that differ only by a shift together. But even that would only get us n^4 variants. To do better, we need to somehow group different parallelograms together. The most straightforward way to do this is to count all parralelograms that have the same bounding box together. Why? Because all those parallelograms have the same amount of possible shifts (if the bounding box is [x1,x2]*[y1,y2], then we have (n-x2+x1+1)*(m-y2+y1+1) possible shifts), so the answer for the problem can be calculated as: sum (w from 0 to n) of sum (h from 0 to m) of num(0,0,w,h)*(n-w+1)*(m-h+1), where num(x1,x2,y1,y2) is the number of different parallelograms with the given bounding box. If we can get that number in O(1) , then we're done - we have a O(n^2) solution!

For a parallelogram ABCD to have the given rectangle (w times h) as a bounding box, the parallelogram must have a vertex on each of the sides of this rectangle. There're two choices:
1) one of the vertices of the parallelogram (say, A) coincides with a vertex of the rectangle. Then it's easy to see that the opposite vertex C of the parallelogram coincides with the opposite vertex of the rectangle, and the rest of the parallelogram is defined by the position of the third vertex B of the parallelogram. The number of possibilities for B is just the number of points inside the rectangle, minus the 2 points already taken, and minus the points that lie on the segment AC. And we need to divide the resulting number by 2 to account for B and D being interchangeable, and multiply it by 2 to account for two possible choices for which vertices of the rectangle A and C go to. Thus, we get: ((w+1)*(h+1)-gcd(w,h)-1)/2*2=(wh+w+h-gcd(w,h)). (The reason why gcd appears here is left to the reader)
2) Each side of the rectangle has one vertex of parralleogram strictly inside it. Then, the parallelogram is defined by two of its vertices, A and B, as C and D will be located in symmetric positions, so there're (w-1)*(h-1) choices here.

So, we'd have a O(n^2) solution if we knew the values for gcd(w,h). But, we can calculate all such gcd's in O(n^2) using Dynamic Programming and the fact that gcd(w,h)=gcd(w-h,h) for w>h. So we're done!

There's another O(n^2) solution that groups parallelograms in a different manner: we can group them by the location of their center, which always has half-integer coordinates. Given the location of the center, the parallelogram is defined by two of its vertices, and those two vertices can be placed arbitrarily inside some rectangle, minus the cases where they lie on the same line with the center.

But explaining those solutions is not the main point of this post. I'm writing this because, when I was asked recently at a TC Spotlight Session what kind of mathematical background is needed to succeed in programming competitions, my answer was "One thing that is absolutely crucial is to be able to think 'mathematically'", and I had difficulty explaining what that means. I believe this problem, and this solution, is an excellent example, and explanation, of that phrase. It's crucial to be able to do several quite easy reduction steps until your problem splits into several quite easy problems, and to carefully do the reductions without omitting any important cases (like when the gcd appeared in the above solution). To be able to 'explore' some given direction deeply, and check if it yields a solution or not (instead of just trying to apply some known ideas one by one).

Does this make sense?

And also, can you solve the above problem faster than O(n^2)?

Thursday, October 16, 2008

A short math problem instead of a long philosophical text this time! Here's a problem that was given by one of our math lecturers in the University, and I was quite impressed with its solution at that point.

Suppose f(x) is a good function (say, has n-th derivative for every n) on [0,3], f(0)=f(3)=0. Prove or disprove: int(der(f)^2)>=int(f^2), where int() means integral from 0 to 3, and der(f) is the derivative of f. Can you solve it? [right, right, I'm testing if anyone actually reads this :)]

And as a bonus, here's a random blurry picture from my recent trip to London for Google Code Jam (I've tried to find a good picture that has me on it, but it appears there isn't any). London is exciting!

Wednesday, October 15, 2008

IOI 2002: Frog

This time I'll try to remember my encounter with problem Frog from IOI 2002 ( This looks interesting to me because I remember long discussion with teammates after the contest about an efficient algorithm, we spent maybe several hours and could come up with something OK but difficult to write. Of course none of us thought he got it during the contest (but I'm not sure if some of us didn't actually get a full score because we overestimated our algorithm complexities). I was wondering if this problem will look difficult to me now, 6 years and hundreds of competitions later.

And the thing is - when I look at it now, it is painfully obvious. The authors seem to expect a N2 solution, with N equal to 5000, and give us 64MB of memory (restrictions come from But hey, there's the obvious N3 solution: for every pair of marks (a,b) we check if the path that starts with a and has b as the first hit continues correctly until the outside (and that going to the left of a gets us outside immediately). To avoid counting the same path twice, we can order all marks by x-coordinate and then by y-coordinate, and only look for pairs with a less than b. It's not clear if this algo implemented with hashtables will actually hit the time bound and not get full score; but there's no need to check that, as this solution is very easily transformed to an N2 DP: we can compute the same boolean D[a,b] as above (does the path continue correctly beyond b until the end of the field) for each pair (a,b) with a less than b, but without requiring anything about what precedes a on that path (and only take that into account when constructing the answer). To compute that, we find the next mark c on the path a-b-..., and if that mark is present and the D[b,c] is true or c is outside the field, then D[a,b] is true. The only issue is to find c in constant time. This can be done with hashtables, but again, there's no need: if we traverse the values for b (with the value for a fixed) in our order (first by x-coordinate, then by y-coordinate), then we should traverse the values for c in the same order, thus only requiring O(N) operations for each a, thus getting O(N2) in total. In terms of memory this uses only 25M bools, which is compressable to just about 3M, but there's no need as we have 64M. And the code should be so clean and simple that I won't even write it here (does anyone want me to?). In other terms, I would spent at most 20-30 minutes on such problem today.

But one might say, wasn't this expected? Don't the contestants' knowledge and problem difficulty constantly raise and try to pursue one another? True. But, it amazed me how there's no (I mean, totally no) clever algorithm involved, just a simple DP with a simple optimization that reduces O(N) to amortized O(1). I can swear we used to do much more complicated DPs in high school without any trouble. Is the amortization idea so new that it was very difficult 6 years ago? I doubt that it's the only reason. Looking back, I can think of at least 2 more reasons:
  1) I expect that we were too nervous to find that at the actual IOI and after that. But the point is, we didn't feel that way! At least I remember myself being fairly confident and determined then.
  2) We were reluctant to inject the ideas that we knew into algorithms that we knew, changing them significantly. This is more of a mathematical drawback, so both the university education and solving problems in a tougher timeframe at the ICPC trainings should have eliminated that difficulty.

  Any more ideas?

Friday, October 10, 2008

Screencast and challenging

For those of you who don't follow the TopCoder forums - I've been screencasting my SRM experiences, in a hopeless assumption that people might find it interesting.

Recollection of events for the first part of the challenge phase:
00:40 into the video - open the code, read it
01:53 - spot a bug: the coder doesn't distinguish "player X will win" from "only player X or nobody can win". Start thinking about the challenge case.
02:40 - how should the challenge case work? The first player should have two options; in one of them, the situation "only player 1 or nobody can win" should arise, so that the code mistakes by thinking that person 1 has a forced win and chooses that move; but in reality, there should be another move that allows some other person to win. The problem statement lets us build practically any graph for the game, so we can choose arbitrary number of stones removed for those two moves; let's say that those moves take 1 and 2 stones, and let's say the initial amount of stones is 5 (hopefully that will be enough).
02:50 - one of the cases should make other person win; let's make it so that the second player wins instantly by taking 3 stones from 3.
02:56 - and we should have two variants (either player 1 wins or nobody wins) in the second case; so that gives us two possible moves again; not to overlap with the existing positions, let's make them 2 and 3 from 4.
02:59 - one of them should lead to "nobody wins" situation, so we leave position with 2 stones with no jumps. Position with 1 stone should have the first person win, so assuming we have only 2 players, that position should have a winning move: 1.
03:15 - so is this gonna work? No, it isn't; player 1 should still be able to win after both initial moves, otherwise even without a forced win he should choose only one of them according to the problem statement. So we should modify the "3" position to allow the first player to win. But how?
03:24 - we'll need a couple more positions to achieve this; let's insert them.
03:39 - we'll need to adjust existing jumps; but...
04:08 - we can't achieve the current goal with only two players, as we should have a position when player 1 AND some other player can win; if some other player is player 2, then he will certainly win if given the choice; thus, player 2 should choose whether player 1 or player 3 can win. But, first of all, now that we have 3 players, we should alter the "player 1 or nobody wins" position. For player 1 to win, we now need two extra moves after player 2's move. We achieve that by having position 2 have one move to position 1, and position 3 having no moves; Player 2 now has two moves, 3 and 4, from position 6 - the first one leads to a draw, and the second leads to two more moves performed and player 1 winning.
04:52 - and how do we make player 2 choose whether player 1 or player 3 wins in the other case? He should choose whether there're 1 or 2 moves left; luckily, we already have such positions, so we just create appropriate moves from position 5: move 3 that leads to position 2 and player 1 winning, and move 4 that leads to position 1 and player 3 winning.
05:10 - re-verifying the case shows that everything should work; challenge phase is short and others may challenge this as well, so let's not re-verify once again and click "OK".
05:15 - enter two more numbers required and challenge!
05:17 - successful! And the expected and actual results are what I expected them to be, so the case should be OK. Cool, let's look if the others have made the same mistake.

SRM 420:
