## TC SRM 478

Been a while since I last posted… I’ve fallen behind as was to be expected.

Q1) Given carrots are placed at locations 0 mod 1billion 7, and from any given location there are 2 options to go to, 4x+3 or 8x+7 and at most 100thousand movements are allowed, what is the minimum number of movements to get from starting position x, to a carrot.

A1) Obviously this problem is about transitions modulo 1 billion and 7.  The trick is needing to rule out trying all 2^100000 options.   1 billion and 7 is too large to compute all transitions, so a simple memotization won’t work.

The first step is to realise that order of application doesn’t matter. A(B(x)) == B(A(x)). That gets rid of almost all of the solution space.  Furthermore A(A(A(x)))=B(B(x)) and since we are trying to minimise moves the ideal solution will always choose 2 8x rather than 3 4x.  From there it is a simple matter of repeated application of B and trying 0 1 or 2 applications of A. (Technically you need to verify that another application of B isn’t equal of 2 appliations of A are equal given the statements made so far, but A(A(x))!=B(x) for all x so that is not a problem)

Another solution is to notice that all transition states available are of the form 2^(n)*(x+1)-1, and that every possible n is reachable except n=1.  Furthermore since the order of application doesn’t matter, and a greedy approach of using 8x as often as possible therefore is the ideal scenario. Therefore the answer is n/3 if n is a multiple of 3, or n/3+1 otherwise, for the first n which results in a 0 mod 1 billion and 7 answer. (if 2 mod 3 then apply 4x, if 1 mod 3 then apply 4x twice from one less 8x.)

Q2) Given up to 15 bottles each with capacity C and a list of initial contents levels (all integer), and a price list for how much a bottle with any specific content level is worth determine maximum profit given the ability to pour 1 bottle into another until it is empty or the other is full. Maximum C is 50 and maximum price is 1million.

A2) Each pouring operation either results in a bottle which is empty or one which is full (potentially both).  Empty and full bottles are of no interest for future pouring operations, so there are at most 14 pouring operations that will be performed. Giving a search space of 14! – which is too large.

A dynamic programming approach might seem like a good idea, all we have to do is determine the maximum for each subset and combine them.  But a closer look shows that it doesn’t make sense because we don’t know the contents levels in bottles as they may be neither full, empty nor their initial value.

So a different approach is required.  If a subset of the bottles all pour, one into another in some order.  They all end up either full or empty except for possibly 1.  Given we know the total contents and the capacity of each one, div gives you the number of full bottles, and mod gives you the capacity of the last possibly non-empty one, the rest are empty.  Presto we can trivially calculate the maximum for each subset where they all pour.  Therefore we can determine the maximum for a specific subset, it is the maximum for all partitions of pouring.  Number of partitions of 15 elements is still ~1billion so we can’t brute force those.

The maximum of n elements is therefore found by considering each possible non-empty subset as all pour, and the maximum of the rest.  This has a running time of less than 2^29, but it seems likely to time-out in the worst case.

Next approach – For each subset one of the elements will have content j, the rest are already sold.  You can generate a subset 1 larger by combining the unsold one with an unused one, or selling the unsold one and keeping the new one as unsold.  This has a state space of 2^15*50, but only an average of about 15 options to consider for each spot.  This gives running time <2^25 which is clearly going to run in time.

Q3) Given a list of the count of each type of apple in up to 50 boxes, determine the probability of selecting each specific apple type at random if you first randomly select a non-empty subset of the boxes to further select from randomly. (Up to 50 types of apple and each count is <200, no box is empty)

A3) The dual random selection is kind of annoying as the number of subsets is up to 2^50.  So obviously we can’t consider each scenario and average.

Dynamic programming might be a possibility. Maybe determine the probabilities using the first n boxes?  Except how do you expand from n boxes to n+1.  This appears to be impossible.  To modify the probabilities you need to know how many apples there were before.  Hmm, that <200 limitation is pretty small, gives a maximum of 500k apples in any subset.  Each transition only depends on the one before, so we don’t need to keep everything in memory.  So probabilities using the first n boxes with a total of k apples.  That is 2meg ram for just one probabilitiy, so we’ll have to do each probability separately, which I think is fine…

Well maybe a bit slow.

A faster way to do it is to break down the problem differently so you can accumulate the probabilities for each item simultaneously without requiring any additional ram.  One breakdown is to consider the answer as being the sum of count of apples of specific type in a box divided by total apples in subset, times number of ways to create that total with a subset that contains the specific box divided by total number of subsets, for each possible box and total number of apples in a subset.  Efficient calculation of number of ways to create total with a subset that contains the box means that running time is proportional to 500k times 50 worst case, as each apple type can be accumulated at the same time, because the part which requires the giant array is constant over each apple type.  Unlike the previous calculation where the giant array is specific to the apple type, which is 50 times slower. (Although apparently an efficient implementation can pass system tests… without this further optimization)