Take a look at xkcd's election predictor. It scrapes a bunch of election outcome probabilities per state from intrade and uses them to provide a prediction for the overall election. It does so by using a Monte Carlo simulation, given the probabilities, it runs a mock election assuming each state outcome is independent a whole bunch of times with those probabilities and see what happens. In his code, he runs the simulation 100,000 times. On my laptop, this takes about 4 seconds and yields only an approximation with error of about .001. Aww, poor physicist, if only Randall was a computer scientist, maybe he'd see how to compute it exactly and much more quickly.

So, how can we compute this efficiently? The key to my program running quickly is that total number of electoral votes (538) is much smaller than the total number of possible election outcomes (2^50).

Imagine we have a partially computed an election in which k states have already voted, and that p_k,i is the probability that the democrats have won exactly i electoral votes after the first k elections have run. Furthermore, let v_k be the number of electoral votes for state k and e_k be the probability state that state k elects a democrat.

How can we compute p_k,i? Well, this simple little recurrence will do it.
p_k,i = e_k * p_{k-1,i - vk} + (1 - e_k) * p_{k-1, i}. Basically, with probability e_k the state votes democrat and has i - v_k votes left, otherwise, with probability 1 - e_k it votes republican (assuming two party system), and the democrat still has i votes left. For this problem, we only need about 539 * 51 table entries (we could even do something clever and push the needed space down to 539 entries, but I'll leave that as an exercise for the reader).

After I implemented this, the computation runs much more quickly than one second and provides an exact answer.

Looking at the code a bit more, I found a (slightly refactored) snippet of code like this. I assume the code is supposed to output the outcomes in order. Take a look at the comment, is it simpler than sorting? Is it even correct? I challenge both accounts.

def render_outcome(pdem, prep, ptie):
    dstring="Obama:  <span style=\"color: #0000FF\">"+str(round(pdem,1))+"</span>"
    rstring="McCain: <span style=\"color: #FF0000\">"+str(round(prep, 1))+"</span>"
    tstring="Tie: <span style=\"color: #888888\">"+str(round(ptie, 1))+"</span>"

    if pdem>prep and prep>ptie:
        return dstring+"\n"+rstring+"\n"+tstring
    if prep>pdem and pdem>ptie:
        return rstring+"\n"+dstring+"\n"+tstring
    #just in case ... (easier to do this than a sort)
    if pdem>ptie and ptie>prep:
        return dstring+"\n"+tstring+"\n"+rstring
    if prep>ptie and ptie>pdem:
        return rstring+"\n"+tstring+"\n"+dstring
    if ptie>pdem and pdem>prep:
        return tstring+"\n"+dstring+"\n"+rstring
    if ptie>prep and prep>pdem:
        return tstring+"\n"+rstring+"\n"+dstring
    return tstring+"\n"+rstring+"\n"+dstring

Here is my alternative implementation.

def render_outcome2(pdem, prep, ptie):
    dstring="obama:  <span style=\"color: #0000ff\">"+str(round(pdem,1))+"</span>"
    rstring="mccain: <span style=\"color: #ff0000\">"+str(round(prep, 1))+"</span>"
    tstring="tie: <span style=\"color: #888888\">"+str(round(ptie, 1))+"</span>"
    l = [(pdem, dstring), (prep, rstring), (ptie, tstring)]
    return '\n'.join(pair[1] for pair in sorted(l, reverse=True))

Mine definitely seems simpler. It relies on the natural sorting order of python tuples to get the messages sorted in the right order.

Is his implementation correct? Well.. notice all of those < operators (not <=). What happens with ties?

>>> print states.render_outcome(.4, .3, .3)
Tie: 0.3
McCain: 0.3
Obama:  0.4

Uh oh..

In all fairness, quoting the page, Randall says "I made this tool to help me understand the race, especially on election night." I am sure he just wanted to get things done, and not have some nerd nitpick at all of the code. The Monte Carlo simulation is a bit easier to code than the dynamic program I posted and it gets things done. His code basically works. I don't think he actually sucks at programming, I just wanted to put some blood on the pages for reddit. Furthermore, I was thinking about using this problem as an interview question, but after trying it on a few of my coworkers (who all have at least a BS in computer science from a nice university), I think it's a bit too hard.

From a reddit post I made a month ago.

There are n! paths in a graph of size n. I can find the optimal TSP solution in O(n^2 2^n) time. Therefore, it is possible to do better than examine every path to find a TSP solution.

http://www.algorithmist.com/index.php/Traveling_Salesperson_Problem

And today's xkcd comic.



Suspicious?

I just wish I had a modicum of artistic ability, I swear I've got about five ideas for good xkcd style comics.

I write this because I am slacking with the typsetting of my not sufficiently challenging Fundamental Algorithms class at NYU.

Web comic #1, Alien Loves Predator. I think anyone who has spent any significant amount of time in New York City can relate to this webcomic about the damn metrocard getting rejected by the turnstile.

Coming in at #2, is indexed, which typically consist of witty and amusing diagrams drawn on index cards. My personal favorite is this beauty, a complete graph on seven vertexes, where each vertex is a deadly sin, and edges are combinations of sins. For example, the pairing of gluttony and envy is high metabolism.

Last but not least, the one true web comic, xkcd. I have so many xkcd episodes that I really like, so instead, I'll link a favorite of someone else. Inanity of comments as a function of proximity to cats. Oh so true.

And I am sad! I tried to organize a bunch of Googlers to go to Cambridge for the xkcd meetup next week, but everyone fell through. There will be no one to travel with, and no one to split a hotel with. Oh well.

And if you have any desire to experience pop culture and vulgarity, you have to see Superbad. The muscles in my face responsible for smiling were literally worn out after the show was over.