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incorrect solution to problem 640. Used a heuristic approach, the card with the lowest probability out of the rolled options is turned face down, if there is no such card the card with the highest probability is turned face up. The expected value can be then calculated from the transition matrix of the markov chain. Very surprising this doesn't work! I guess it has to do with the fact some of the options are not independent, and this is where my intuition failed me. Works for n up to n=4
Problem had difficulty of 35%, solved using dynamic programming. Let t(n) be number of combinations of blocks such that last block ends in cell n. Then t(n) is the sum of t(k) for k<=n-l-1 for each possible length of the ending block l. Obviously t(3)=1. Obviously number of combinations is the sum of t(x) for x<=n +1 since we can also have no red blocks.
Problem had difficulty 35%. Modified code from 134 and iterated until the value was larger than 10^6
Surprisingly easy dynamic programming problem for 50% difficulty, solve time 30 minutes, answer in 2 ms.
Problem had difficulty of 70%, solved using logarithmic matrix exponentiation, before I knew about Cayley-Hamilton theorem. Run time: long, since the matrices become dense around millionth power, could be done much faster by calculating x^(10^18) modulo the characteristic equation.
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