Monday, June 13, 2016

Code Jam and Distributed Code Jam 2016 solutions

Code Jam

Since the official analysis page still says "coming soon", I thought I'd post my solutions to both the Code Jam and Distributed Code Jam here.

Teaching Assistant

We can start by observing that when we take a new task, it might as well match the mood of the assistant: doing so will give us at least 5 points, not doing so will give us at most 5. Also, there is no point in getting to the end of the course with unsubmitted assignments: if we did, we should not have taken that assignment, instead submitting a previous assignment (a parity argument proves that there was a previous assignment). So what we're looking for is a way to pair up days of the course, obeying proper nesting, scoring 10 points if the paired days have the same mood and 5 points otherwise.

That's all we need to start on the small case, which can be done with a standard DP. For every even-length interval of the course, we compute the maximum score. We iterate to pick the day to match up to the first day of the interval, then use previous results to find the best scores for the two subintervals this generates.

That's O(N³), which is too slow for the large case. It turns out that a greedy approach works: every day, if the mood matches the top of the stack, submit, otherwise take a new assignment. The exception is that once the stack height matches the number of remaining days, one must submit every day to ensure that the stack is emptied.

I haven't quite got my head around a proof, but since it worked on the test case I had for the small input I went ahead and submitted it.

Forest university

The very weak accuracy requirement, and the attention drawn to it, is a strong hint that the answer should be found by simulation rather than analytically. Thus, we need to find a way to uniformly sample a topological walk of the forest. This is not as simple as always uniformly picking one of the available courses. Consider one tree in the forest: it can be traversed in some number of ways (all of them starting with the root of the tree), and the rest of the forest can be traversed in some number of ways, and then these can be interleaved arbitrarily. If we consider all ways to interleave A items and B items, then A/(A+B) of them will start with an element of A. Thus, the probability that the root of a particular tree will be chosen is proportional to the size of that tree.

After picking the first element, the available courses again form a forest, and one can continue. After uniformly sampling a sequence of courses, simply check which of the cool words appear on the hat.

Rebel against the empire

I consider this problem to be quite a bit harder than D. It's easier to see what to do, but reams of tricky code.

For the small case, time is not an issue because the asteroids are static. Thus, one just needs to find the bottleneck distance to connect asteroids 0 and 1. I did this by adding edges from shortest to longest to a union-find structure until 0 and 1 were connected (ala Kruskal's algorithm), but it can also be done by priority-first search (Prim's algorithm) or by binary searching for the answer and then checking connectivity.

For the large case we'll take that last approach, binary searching over the answers and then checking whether it is possible. Naturally, we will need to know at which times it's possible to make jumps between each pair of asteroids. This is a bit of geometry/algebra, which gives a window of time for each pair (possibly empty, possibly infinite). Now consider a particular asteroid A. In any contiguous period of time during which at least one window is open, it's possible to remain on A, regardless of S, by simply jumping away and immediately back any time security are about to catch up. Also, if two of these intervals are separated in time by at most S, they can be treated as one interval, because one can sit tight on A during the gap between windows. On the other hand, any period longer than S with no open windows is of no use.

The key is to ask for the earliest time at which one can arrive at each interval. This can be done with a minor modification of Dijkstra's algorithm. The modification is that the outgoing edges from an interval are only those windows which have not already closed by the time of arrival at the interval.


I liked this problem, and I really wish the large had been worth fewer points, because I might then have taken it on and solved it.

The key is to realise that the good strings don't really matter. Apart from the trivial case of the bad string also being a good string, it is always possible to solve the problem by producing a pair of programs that can produce every string apart from the bad one.

For the small case, we can use 0?0?0?0? (N repetitions) for the first program, and 111 (N-1 repetitions) for the second (but be careful of the special case N=1!) Each of the 1's can be interleaved once into the first program to produce a 1 in the output, but because there are only N-1 1's, it is impossible to produce the bad string.

For the large case, the same idea works, but we need something a bit more cunning. The first program is basically the same: alternating digits and ?'s, with the digits forming the complement of the bad string. To allow all strings but the bad string to be output, we want the second program to be a string which does not contain the bad string as a subsequence, but which contains every subsequence of the bad string as a subsequence. This can be achieved by replacing every 1 in the bad string by 01 and every 0 by 10, concatenating them all together, then dropping the last character. Proving that this works is left as an exercise for the reader.

Distributed Code Jam


The code appears to be summing up every product of an element in A with an element in B. However, closer examination shows that those where the sum of the indices is a multiple of M (being the number of nodes) are omitted. However, once we pick an index modulo M for each, we're either adding all or none. The sum of all products of pairs is the same as the product of the sums. So, for each remainder modulo M, we can add up all elements of A, and all elements of B. Then, for each pair of remainders, we either multiply these sums together and accumulate into the grand total, or not. There are some details regarding the modulo 1000000007, but they're not difficult.

For the large case, we obviously need to distribute this, which can be done by computing each sum on a different node, then sending all the sums back to the master.


Here, a valid lisp program is just a standard bracket sequence (although it is required to be non-empty). It's well-known that a sequence is valid if and only if:
  • the nesting level (number of left brackets minus number of right brackets) of every prefix is non-negative; and
  • the nesting level of the whole sequence is zero.
If the nesting level ever goes negative in a left-to-right scan, then the point just before it went negative is the longest sequence that can be completed. This makes the small case trivial to implement.

To distribute this, we can do the usual thing of assigning each node an interval to work on. We can use a parallel prefix sum to find the nesting level at every point (there is lots of information on the internet about how to do this). Then, each node can find it's first negative nesting level, and send this back to a master to find the globally first one. We also need to know the total nesting level by sending the sum for each interval to the master, but that's already done as part of the parallel prefix sum.


I liked this one better than the Code Jam asteroids problem, but it was also rather fiddly. The small case is a straightforward serial DP: going from bottom to top, compute the maximum sum possible when ending one's turn on each location.

At first I thought that this could be parallelised in the same manner as Rocks or Mutex from last year's problems. However, those had only bottom-up and left-to-right propagation of DP state. In this problem, we have bottom-up, left-to-right and right-to-left. On the other hand, the left-to-right and right-to-left propagation is slow: at one most unit for every unit upwards. We can use this!

Each node will be responsible for a range of columns. However, let's say it starts with knowledge of the prior state for B extra columns on either side. Then after one row of DP iteration, it will correctly know the state for B-1 extra columns, and after B iterations it will still have the correct information for the columns its responsible for. At this stage it needs to exchange information with its neighbours: it will receive information about the B edge columns on either side, allowing it to continue on its way for another B rows.

There are a few issues involved in picking B. If it's too small (less than 60), then nodes will need to send more than 1000 messages. If it's too large, then nodes will do an excessive number of extra GetPosition queries. Also, B must not be larger than the number of columns a node is handling; in narrow cases we actually need to reduce the number of nodes being used.

Gas stations

The serial version of this problem is a known problem, so I'll just recap a standard solution quickly. At each km, you need to decide how much petrol to add. If there is a town within T km that is cheaper, you should add just enough petrol to reach it (the first one if there are multiple), since any extra you could have waited until you got there. Otherwise, you should fill up completely. A sweep algorithm can give the next cheaper town for every town in linear time.

What about the large case? The constraint is interesting: if we had enough memory, and if GetGasPrice was faster, there is enough time to do it serially! In fact, we can do it serially, just passing the baton from one node to the next as the car travels between the intervals assigned to each node.

What about computing the next cheaper station after each station? That would be difficult, but we don't actually need it. We only need to know whether there is a cheaper one within T km. We can start by checking the current node, using a node-local version of the next-cheapest array we had before. If that tells us that there is nothing within the current node, and T km away is beyond the end of the current node, then we will just find the cheapest within T km and check if that is cheaper than the current value. The range will completely overlap some nodes; for those nodes, we can use a precomputed array of the cheapest for each node. There will then be some prefix of another node. We'll just call GetGasPrice to fetch values on this node. Each iteration this prefix will grow by one, so we only need one call to GetGasPrice per iteration (plus an initial prefix). This means we're calling GetGasPrice at most three times per station (on average), which turns out to the fast enough. It's possible to reduce it to 2 by computing the minimum of the initial prefix on the node that already has the data, then sending it to the node that needs it.

There is one catch, which caused me to make a late resubmission. I was calling GetGasPrice to grow the prefix in the serial part of the code! You need to make all the calls up front in the parallel part of the code and cache the results.

1 comment:

Anand said...

Hi Bruce,

Could you do an analysis of IOI 2016 sometime?

Thanks :)