Tao of Coding

Broad Update

Yes, this shall be a blog updated once every few months to say ‘I should write more’. Whatever.

ACM and AI
Starting with the current purpose of this blog, the ACM Local competition was last week; and I successfully succeeded at succeeding. First place by one problem. I went to the last two ACM contests with one of my teammates, and the other one I have a fair amount of confidence in. We shall see what happens. But, as a humorous aside to that: I very nearly implemented a Sudoku solver in under 30 minutes. Had I stopped asking myself ‘is it worth it to put in this much effort into a problem you probably won’t solve before the contest is over’; I probably could have solved it. Probably not, but maybe. I solved it using a horribly hacked together Constraint Satisfaction based Solver. For that interesting 30-50 minutes of my life, I would like to thank Robin Murphy (AI Prof), and Peter Norvig (Algorithms God/AI Book Author); both of which suggested Sudoku solving as an application of CSP. I’ll probably clean it up later and post it… as it stands currently it makes me want to cry, but it theoretically should work. (Theoretically because the test data was, in fact, invalid).

Education- Dr. Tao?
On a different note, I think I’m going to go for a PhD. This conclusion was got to by an interesting failure in my ability to communicate. As part of a follow up email, I had written the line ‘assume I go for a PhD’, which led to the thought ‘If I graduate and go into industry with a Bach, I’m going to end up drifting from job to job with mounds of uncertainty and stress’ (drifting coming from the popular consensus that a programmer should expect to change jobs every 2-3 years). If I get a PhD, there’s a lot less uncertainty in terms of job (at least I hope), I have the opportunity to interact with really incredibly frustratingly annoyingly intelligent folk, and if I go the academia route, I can torture a class of Yankee’s with ‘Howdy Class’. How can you not like that?

As a side note, I suspect the reason I was never comfortable with the thought of Grad School before is because I had never really come across something that peaked my interests. There’s a popular quote by Dijkstra: “Computer Science is no more about Computers then Astronomy is about Telescopes”. I have this odd sense that it is often misinterpreted from it’s original context. Computer Science is about the mathematical basis behind computing: proving algorithms correct, classifying problems as NP-hard, finding the most efficient way to find a correct solution, all about the theory. To me, that is what Computer Science Grad School is/was about. And I hate that definition. Mostly because it feels like something from the Ivory Tower- occasionally interesting results with only limited real world applicability. I want to generate solutions to real problems. Is making a formally proven OS kernel useful? Maybe to some, but spending 5 years to develop 10,000 lines of proven code just feels like a waste. I’ll probably regret those statements later, but, such is life.

Further side note: one thing I’ve heard fairly commonly is that in general, you don’t recoup the cost of a PhD. A master’s is worth it, but a PhD won’t earn you more over a lifetime. My response: Who cares? If you’re making a career or education choice based soley on ‘will it pay off in the end’, then you’re doing something wrong. Education (and career) should be driven off personal interest, and as an auxiliary, will it earn enough money such that you won’t live in a cardboard box. Going into advanced education because it will buy your retirement home a heated pool is just plain wrong.

Shiny New Toy- Arduino
Yet another different tact, I have a shiny new toy. An Arduino. With a product name which I can neither remember, pronounce, nor spell. I can at least remember/spell Arduino… pronunciation is still off. I blame the Italians. I got it in part because I wanted to get involved in the electronics/embedded side of computing, and mostly because it provided a convenient way to solve a problem my brother was having.

My brother, through what I assume were family connections, was hired to fix an outdoor LED sign for a bank. The company that manufactured the sign had gone out of business, but he was able to get some schematics for how the individual boards in the sign were wired. While he could take the boards down, ohm out connections, and so on, he didn’t have a convenient way of actually testing that either the repaired sign worked, or finding where issues might lie in malfunctioning boards.

I psuedo-come-in at this point. He asks me if I can write a program that would be able to test the boards. The boards themselves are driven off 4 inputs- 2 serial lines, a clock, and a latch. In the sign they’re driven off of what is probably a parallel port on a 386 running windows 98. I’m not sure if you’ve noticed, future reader, but most computers these days don’t come with parallel ports. So I go through the thought process of ‘how could I go about this’; serial ports don’t really work, as at most you might be able to control voltage on two lines; we could go through the trouble of finding and installing a parallel port expansion on a testing computer, and try to write a program that would drive the port/board. And then I thought “wasn’t there something Kyle mentioned not long ago; an Arduino?”.

It fits the bill perfectly. So the short ending to this is: the first day I got it, I broke it out, did the standard “Make an LED flash” thing to see if everything was working; and the second day I built a controller for the sign boards. A bit later I added some simple configuration bits, a button to switch between testing modes, and hooked up the 9V battery so the computer became completely unnecessary. My brother has since stolen it for testing, until he can get his own.

So, I feel good about that. One thing I’ll probably post soon is a list of projects to go along with it. Y’Know, to go along with the other list of projects I’m not making progress on. I might also post more details about the sign, as it was a fun first project/jaunt into Computer Engineering.

Deutschland
I’m also planning to go to Bonn next Summer as part of a study abroad program. I’m trying to learn some German for that; and it’s coming along slowly. And that’s it. That’s all. See y’all later.

ACM 2008 SCUSA Wormholes- 2nd Edition

Last time we left our time travelling hero’s, they were using a repeated applications of the Bellman-Ford algorithm (at time complexity O(|V| * |E|), where |E| is on par with |V|, and number of requests is probably also on the order of |V|, giving a total of O(|V|^3)). Our valiant programmer desperately seeks to apply an the Floyd-Warshall algorithm to the problem, just so that the time spent reading the analysis of algorithms book isn’t wasted.

The secret to successfully applying the Floyd-Warshall algorithm to this problem is the same secret to getting reasonable performance out of the previous solution. You can’t look at each individual year, you instead must look at time spans between meaningful events. That brings the number of vertices down from many thousand to at most 301 (starting year + 200 distinct wormhole years + 100 mission years). It has the downside of making the process of finding weights a fair bit more complex:

public int[,] BuildInitialWeights(List<int> distinctVerticies)
{
    int[,] weights = new int[distinctVerticies.Count, distinctVerticies.Count];
    for(int i=0; i<distinctVerticies.Count; i++)
        for(int j=0; j<distinctVerticies.Count; j++)
        {
            if (i == j) weights[i,j] = 0;
            else weights[i,j] = Int32.MaxValue;
        }
 
    //assumes vertices are in order, from first year to last
    for (int i = 1; i < distinctVerticies.Count; i++)
    {
        int start = distinctVerticies[i - 1];
        int end = distinctVerticies[i];
        weights[i-1, i] = end - start;
    }
    foreach (int[] years in wormholes)
    {
        int i = distinctVerticies.IndexOf(years[0]);
        int j = distinctVerticies.IndexOf(years[1]);
        int weight;
        if (years[1] > years[0]) weight = (years[1] - years[0]) / 2;
        else weight = (years[1] - years[0]) / 4;
        weights[i, j] = weight;
    }
    return weights;
}

Here we’re building a 2D matrix of weights. Informally we’re mapping each year in the distinct vertices list to its index, then using that index as coordinates in the weight matrix. Weights are constructed by first setting all weights to their default values (0 to get to current node, max value for other nodes). Then default weights for moving forward with time are assigned. Finally, wormholes are accounted for. Wormhole accounting is done last as it will be, by definition, the shortest path between two points. This has the advantage that only one path is ever considered between two points, where as the previous solution allowed for multiple paths.

public List<int> BuildDistinctVerticies(List<int> addedYears)
{
    List<int> years = new List<int>(addedYears);
    foreach (EndPoint wormhole in wormholes)
    {
        years.Add(wormhole.Start);
        years.Add(wormhole.End);
    }
    years.Sort();
    for (int i = 1; i < years.Count; i++)
        if (years[i - 1] == years[i])
            years.RemoveAt(i--);
    return years;
}

This is the simple method for building all distinct vertices. It starts by taking all mission and starting years, adds each end point from a wormhole to the list, then sorts it and removes duplicates. Again we get into ‘it would be nice to have a Sorted Set’ (Should appear in .NET 4; or, alternatively, has always been in Java… accursed Java with its nicer Collections libraries…).

public int[,] AllPairs(List<int> missionYears, out List<int> distinctVerticies)
{
    distinctVerticies = BuildDistinctVerticies(missionYears);
    int[,] currentWeights = BuildInitialWeights(distinctVerticies);
    int[,] nextWeights = new int[distinctVerticies.Count, distinctVerticies.Count];
    for (int i = 0; i < distinctVerticies.Count; i++)
    {
        Extend(currentWeights, nextWeights, i);
        int[,] temp = currentWeights;
        currentWeights = nextWeights;
        nextWeights = temp;
    }
    return currentWeights;
}
 
public void Extend(int[,] previosLevel, int[,] thisLevel, int k)
{
    for(int i=0; i<previosLevel.GetLength(0); i++)
        for(int j=0; j<previosLevel.GetLength(1); j++)
        {
            int iToJ = previosLevel[i,j];
            int iToK = previosLevel[i,k];
            int kToJ = previosLevel[k,j];
            if (iToK != Int32.MaxValue && kToJ != Int32.MaxValue)
                thisLevel[i,j] = Math.Min(iToJ, iToK + kToJ);
            else thisLevel[i,j] = iToJ;
        }
}

I’m about 90-100% certain I misspelled previous. Ignoring that, here we can see Floyd-Warshall algorithm implemented. I don’t fully understand how it works. I’ve read that the ‘extension’ element is simulating whether or not the shortest path from i to j will go through k; but at this hour of the morning I still can’t quite wrap my head around how it works. Something to learn for the new school year.

The distinct verticies are passed as an out parameter to the caller, so that they have the map from index to year available to them.

List<int> years = new List<int>(missions);
years.Add(startingYear);
int[,] distances = timeline.AllPairs(years, out years);
 
foreach(int destination in missions)
{
    int i = years.IndexOf(startingYear);
    int j = years.IndexOf(destination);
    int toDistance = distances[i, j];
    int fromDistance = distances[j, i];
    if(toDistance==Int32.MaxValue || fromDistance==Int32.MaxValue)
        Console.WriteLine("IMPOSSIBLE");
    else
        Console.WriteLine(toDistance+fromDistance);
}

Finally, the relevant changes to the problem solving method. The down side to using this approach is that there isn’t a direct mapping from year to index; so a linear search is performed to get that mapping. It’s not really an issue, considering what else is being done, but it is a bit unfortunate.

Analysis
Psuedo-analysis might be more accurate. Comparing this to the previous solution, I like the fact that the algorithm seems more straight forward, and that in the end we end up with a massive table of possible paths, should our time travelling agents ever need to do strange other things. It also has a slight performance advantage over the straight version from last time. (It’s about on par with not continuing the Bellman Ford algorithm if no relaxations occur, though that might be in part due to poor implementation of that algorithm).

That said, in its current form, it adds about 30 lines of code, and the indirect mapping from index to representative year feels ungood. In the end, the question for ACM problems is ‘which could you code fastest with the lowest probability of error’, and looking at the two, I think the Bellman Ford would be a bit easier, specifically because it uses a series of edges, rather then a weight table. I could be wrong though, and the performance level does seem to be in favor of Floyd Warshall.

Side Rant
I seem to be rather poor at a very specific bit of contestifying. That is seeing points where you can discard unimportant information. In this problem, it was my attempt to treat every year as a vertex. In a problem about a disk defragmenter, I was tempted to treat the disk as a series of sectors rather then a series of filled and unfilled ranges (would have massively increased computation time to simulate, and probably over complicated the code; didn’t get to find out due to contest ending). I find that this seems like a common theme to me: I constantly attempt the simplest solution that could possibly work, and these problems are more often then not tuned to make that ‘the wrong way’. I suspect this comes from my many years of employment, where you’re not necessarily seeking the best solution, but rather just something that works well enough to get by. That is, of course, part of why I started this crazy blog: to get better at recognizing when I’m going down an incredibly wrong path.

Full Source Code for this solution.

ACM 2008 SCUSA Wormholes

South Central USA has to be one of the odder acronyms in common use… a single character away from SCUBA no less.

The last time I went to ACM, this problem was delegated to one of my team mates, and for one reason or another we never got it solved. I was reading through an Analysis of Algorithms book and came across a section on All Pairs Shortest Path and thought ‘This would apply perfectly and elegantly to this situation’. (I was wrong, but we’ll get to that soon). (Also, judge data can be had at this location, and I have been cheating miserably).

Synopsis
You work at Time Travel Inc. in the payroll department. Sick of keeping up with all the math related to time travel, you contracted yourself (while in college) to develop an application to determine how much a person has aged during their time travel, and whether or not their last journey was even physically possible. More specifically, time travel is acheived by portals which appear at specific moments in time, and exit at other specific moments, either aging or youthifying the human in the process, according to a specific formula; and ignoring the concept that you could die, or be negative years old.

Analysis
This is a graph traversal problem with negative weight directed edges. The goal is to find the minimum weight path that takes you from a start, to a destination, and back to a start; or equivelantly, two graph traversals.

As I said, I had read a bit on All Pairs Shortests Paths and was interested to see how it would handle this. Specifically, I was thinking ‘if you treat each year as a vertex, then each vertex has a weight one edge to the next year; and if a wormhole is present, an edge to that destination year. At this point in time, I thought the input was constricted to a few hundred years near the 2000’s (where else would you time travel to?). The actual range of years is from 1-10,000. 10,000 verticies, 2 10,000 x 10,000 matricies for storing path weights… no. Not gonna work. (As we’ll see in a bit, this is actually a bit of a lie).

Still curious to further my learning, I read up on the Bellman-Ford algorithm, which is a single source shortest paths algorithm. It works on the assumption that the absolute longest path will go through every single vertex in V, and therefore you can determine the minimum cost to get anywhere by going through each edge, relaxing the edge’s destination weight if appropriate, and then repeating that |V| times. Complexity of |V|*|E|. That actually sort of worked, in the sense that it produced an answer for most problems in under a minute. But, under a minute isn’t quite ideal.

To get reasonable response times, you can’t simulate every single meaningless year. You have to focus on only the meaningful ones, and declare edges between them. If you’re in 1950 and want to get to 2050, yes, you can look at it as 100 1 year edges, but it would be less pain to look at it as 1 100 year edge… or, if a wormhole happens to open up in 2012, a 62 year edge followed by a 38 year edge. The downside to ignoring meaningless years is this adds a fair bit of complex non intuitive code, just for the purpose of filling in the ‘time moves forward at a constant rate’ bits.

With that said, you could apply the same reasoning to all pairs shortest path, so long as you only kept track of the vertexes you actually cared about.

Source Code
As this one is a tad bit long, and far from being the greatest, I’m only going to post bits and peices of it.

public static void ProcessProblem(int id, TextReader input)
{
	Console.WriteLine("DATA SET #" + id);
	Timeline timeline = new Timeline();
	int wormHoleCount = ReadInt(input);
	for (int i = 0; i < wormHoleCount; i++)
	timeline.AddWormhole(ReadInts(input));
	int startingYear = ReadInt(input);
	int missionCount = ReadInt(input);
	for (int m = 0; m < missionCount; m++)
	{
		int destination = ReadInt(input);
		int distance = timeline.ShortestPath(startingYear, destination);
		int returnDistance = timeline.ShortestPath(destination, startingYear);
		if(distance==Int32.MaxValue || returnDistance==Int32.MaxValue)
			Console.WriteLine("IMPOSSIBLE");
		else
			Console.WriteLine(distance+returnDistance);
	}
}

Above is the primary problem solving bit. We create a timeline, register wormholes, then go through each mission and determine if the destination can be reached from the starting point, and if the starting point can be reached from the destination. Note that we’re using the max value of an int as a ‘no path’ return code.

class Timeline
{
	private List<int[]> wormholes=new List<int[]>();
 
	public void AddWormhole(int[] wormhole)
	{
		wormholes.Add(wormhole);
	}
 
	public int ShortestPath(int sourceYear, int destYear)
	{
		List<Edge> edges = new List<Edge>();
		List<int> vertexes=new List<int>();
		vertexes.Add(sourceYear); vertexes.Add(destYear);
		foreach (int[] wormhole in wormholes)
		{
			vertexes.Add(wormhole[0]);
			vertexes.Add(wormhole[1]);
			edges.Add(new Edge(wormhole[0], wormhole[1], true));
		}
		vertexes.Sort();
		for (int i = 1; i < vertexes.Count; i++)
			if (vertexes[i] == vertexes[i - 1])
				vertexes.RemoveAt(i--);
		for (int i = 1; i < vertexes.Count; i++)
			edges.Add(new Edge(vertexes[i - 1], vertexes[i], false));
 
		Dictionary<int, int> distances = new Dictionary<int, int>();
		distances[sourceYear] = 0;
 
		for (int i = 0; i < vertexes.Count; i++ )
		foreach (Edge edge in edges)
		{
			if (!distances.ContainsKey(edge.start)) continue;
 
			int distance = distances[edge.start] + edge.weight;
			if (!distances.ContainsKey(edge.end) || distances[edge.end] > distance)
				distances[edge.end] = distance;
		}
		if (distances.ContainsKey(destYear))
			return distances[destYear];
		return Int32.MaxValue;
	}
}

And here we have the meat of the problem. The graph traversal algorithm/timeline class. The first portion of the ShortestPath method is soley responsible for building a list of edges (note, third parameter to edge constructor is ‘isWormhole’). Once that is constructed, the Bellman Ford algorithm is repeatedly applied to get the distance from the start point to every other point. Bellman Ford performance can actually be improved slightly by not continuing to iterate if no relaxations were needed on a given iteration. On my machine, execution time went from about 3s to .5s. (Also, as a fun aside, note the sad lack of a .NET Set class).

That’s all for now. I think I’ll reimplement this with an all pairs shortest path just to see if it would meet time constraints on execution; and make a post about that later. Also, I’m planning to try and deploy/install Wordpress, with some form of auto syntax highlighting magic included, to make this blog thing a bit easier. We’ll see how that goes. And, for a third thought, I finally got started on LC3D, after mere months. I’m working on the GUI now, getting camera to work right, everything rendering, etc. Then I can start in on the fun that is integrating EVERYTHING. Yay.

Full Source Code for First Solution.