Tweaking the Secret Election code

While waiting to see the results of this week’s Riddler, I started thinking about my solution to the problem. In my last post, I outlined two ways of solving the problem: one by hand on paper, and one using C++ and recursion with memoization.

If you look at the way I solved the puzzle by hand, you might recognize it as a form of dynamic programming (my coworker thinks I’m obsessed with finding ways to use dynamic programming; mea culpa). So I decided to write a version of the SimulateElection function using it instead of recursion. Here it is:

// initialize winner array with candidates
char winner[CANDIDATE_COUNT+1] = "";
for(int i = 0; i < CANDIDATE_COUNT; i++)
winner[i] = 'A' + i;

for(int cand = CANDIDATE_COUNT - 1; cand > 0; cand--)
// now simulate election for cand vs. winners in each 'row'
for(int i = 0; i < cand; i++)
winner[i] = RawVote(winner[i], winner[cand]);

Pretty short! The key here is to start our array of winners with the candidates in the order in which their secret elections happen, so “ABCDE”. Then in each round, starting from the last election, we hold a raw election between the current value we are evaluating, and the values in each of the preceding indices. So:

“ABCDE” – Starting values. Now we pit E vs. first four values, A, B, C, and D.

“ABEDE” – Values after round 1, because E defeated C, but lost to A, B, and D. Now pit D vs. values in first 3 spots, A, B, and E.

“DDDDE” – Values after round 2, because D defeated A, B, and E. Now pit 3rd value, D, vs. first two spots, D and D.

“DDDDE” – Values after round 3, Now pit 2nd value, D, vs. 1st value A.

“DDDDE” – We are done. Our first value, D, is the winner.

And this is how it goes when we allow candidate A to select candidate E as a proxy:

“ABCDE” – Starting values.

“AEEEE” – After round 1. Candidate E defeated everyone but A. Rounds 2 and 3 will pit E vs. E. The final round will pit A vs. E again, and A will win.

I think it’s easier to look at the recursive code and understand what is going on conceptually. Sometimes it’s hard to see how dynamic programming is “working” when you examine the code, but that might be my own bias.

I did some performance testing on the dynamic programming vs. the recursion. I also made a version of SimulateElection that used recursion but NOT memoization, meaning that all recursive branches were visited. The complexity for the recursive algorithm is n-factorial with n storage, but using memoization reduces it to n-squared with n-squared storage. Using dynamic programming is still n-squared time, but only n storage. Although dynamic programming and memoization are both n-squared time, there’s a little less overhead in the dynamic programming. Here are some time trial results for all three methods:

Elapsed ticks 100000 trials using recursion and memoization = 355 ms.
Elapsed ticks 100000 trials using recursion = 476 ms.
Elapsed ticks 100000 trials using dynamic programming = 283 ms.

All of them are pretty fast because the data is so small. I had to run 100,000 trials just to get decent numbers to compare.

Full source code here.

Author: Wiesman

Husband, father, video game developer, liberal, and perpetual Underdog.

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