Wednesday, January 19, 2022

JongMan wins the Wildcard round!

Discuss this match
Thursday, June 28, 2007
Introduction by Olexiy

For more minute-by-minute analysis, check out Petr's latest blog.

JongMan and tomek advance to the finals. They'll join Jan_Kuipers, Eryx, Vitaliy, tomekkulczynski, bmerry and darnley in the finals tomorrow, starting at 1:30 PDT. Congratulations and good luck to the finalists!


by brett1479

Here we must compute the coefficients of the required power series. The problem statement gives the formula for each coefficient in terms of previously computed coefficients. The real work here is the rational arithmetic, which consists of fraction manipulations, and GCD for reducing. Fortunately, the constraints allow everything to be done with 64-bit types, which removes many potential headaches.


by brett1479

Suppose we perform a sequence of folds. At each stage, the entire state is described by a partition of the paper into a grid of rectangles containing the respective thicknesses. If we only make folds along vertical axes, the paper is partitioned into a collection of vertical strips of varying thicknesses. Analogously, if we make only folds along horizontal axes, the paper will be partitioned into horizontal strips. If we make both kinds of folds, we can look at each kind in isolation. Each strip will have an associated thickness. The rectangle formed by the intersection of a vertical and horizontal strip will have thickness equal to the product of the thicknesses of the strips. This can be proved by induction on the number of folds made.

Knowing that the vertical and horizontal folds can be treated in isolation, we now solve the one dimensional problem (ie., assume we only had folds along vertical axes). At each step in the process we will maintain a range object. The range object will store a sequence of integers representing the edges of the paper and the strip boundaries. In addition, it will store thicknesses of the associated strips. Each time a (valid) fold is made, the coordinate of the fold axis becomes an element of the sequence of integers in the range object. Looping over the old range elements, we construct a new range object accommodating the changes associated with the fold (reflecting coordinates across the folding axis as needed). This can be done in time linear in the size of the old range object. In addition, we also update the associated thicknesses. If we process all of the folds in this manner, we will have a final range object. The highest thickness is what we desire. The return value is the product of the highest thicknesses for the vertical and horizontal ranges.


by brett1479

We will solve this problem using memoization/dynamic programming. The key question to ask when each integer is added to the list is whether it should also be added to the cache in order to achieve the optimal cost. The locality constraint allows us to consider only the previous 10 (actually, this is overkill) decisions we have made when deciding on the current element. This is because the number of remaining queries is at least halved at each add so, for a particular element, after 10 adds no more queries will occur. With knowledge of the previous 10 decisions, we can quickly compute the cost trade-off of adding the current integer to the cache:

  1. Cost of adding the current element to the cache: Previously added integers that have not been added to the cache will incur an extra penalty of 1 on queries occurring after the current element is in the cache. Previously added integers which were added to the cache will not incur this cost. All elements added after the current will incur an added penalty of 1 for queries. To speed up these computations, the number of each kind of query occurring after an add can be precomputed prior to the memoized recursion.
  2. Cost of not adding the current element to the cache: Each query for the current element will have to go through the actual list, and thus we must add the cost of searching through the actual list for each future query. Note that since each element we do add to the cache accounts for all future associated penalties, we don't have to worry about any cache penalties in this case.
Summing up what has been done, we recurse over the list of added elements using a 10 bit mask of the previous 10 decisions, and the current position in the list as our memoization key. To speed up our computations at each stage of the recursion, we use the following precomputed structures:
  • queries[n][prevInd] : stores the number of queries for the kth added integer following the addition of the nth added integer. Here k = n - prevInd.
  • afterqueries[n] : stores the number of queries for integers added after the nth integer

This allows us to quickly compute all penalties associated with a cache add.