# Simplicity

Today I taught a pre-algebra lesson on ordering real numbers. In discussing this topic, we determined that in order to compare or order numbers, it is helpful to format them in a similar fashion. For instance, it is difficult to compare a fraction to a decimal without first representing the fraction as a decimal, or the decimal as a fraction.

I told the students that they could could compare decimals to decimals, fractions to fractions, or percentages to percentages, just so long as they used the same form for all of their numbers. I then recommended that they change everything to a decimal representation, because it was the easiest form to work with.

Much to my surprise, the students seemed almost offended that I was trying to make math easier for them. They thought that I was cheating them by not showing them how to change everything to fractions, then compare those fractions [1]. I was, quite frankly, shocked that they would feel this way. I loved it when teachers showed me short cuts, or tried to make things easier.

Then I got to thinking. Why would the students feel cheated? It is possible that their very understanding of what mathematics is, and how it is conducted, is fundamentally flawed. These students have been taught a bunch of symbol manipulations in school. Addition, multiplication, proportions, solving linear equations—all of these are fundamentally just symbol manipulations. When they got good at one kind of manipulation, they were taught another, more complicated manipulation. So it seems that they expect to be taught even more symbol manipulations, and the difficulty of those manipulations is a measure of how much they have learned.

In short, these students seem to think that mathematics is all about arcane manipulations of numbers and other tokens.

In reality, most mathematicians would probably prefer to work without symbols as much as possible. The fact that mathematics is full of ugly notation doesn’t imply that mathematicians like that ugly notation, only that there are times when that notation is the simplest and clearest way of conveying an idea. If a simpler way could be found, most mathematicians would probably rejoice!

A good example of this comes from a problem that I encountered a while back. The basic setting is as follows:

An arcade is planning a Pong tournament. The rules are fairly simple: two players will each take a controller, and they will play until one player wins a match. The tournament will be single elimination, meaning that the winning player of each match is promoted to the next round, and the loser is eliminated.

The arcade needs to close at 11 in the evening, so the tournament organizers need to ensure that they have enough time to conduct the competition. Unfortunately, they only have one Pong machine, so only one match can take place at a time. If 53 people are signed up to compete, and each match takes about 5 minutes, when do the organizers need to start the tournament?

Basically, the problem comes down to figuring out how many matches need to be played. The rest of the problem is just arithmetic. There are at least three ways of solving this problem that I can think of.

## Solution I

An obvious or naive approach might be to actually write out the elimination table. From there it is a simple matter of counting the matches. Unfortunately, this solution is a bit time consuming, and might cause problems if people drop out of the tournament, or enter late. This approach will get you the answer, but it is clearly not ideal.

## Solution II

We could try applying some arithmetic to the problem. Generally, we want to eliminate half of the players in every round, until we get to a championship round with just two players. Working backwards from the final round, each previous round should have double the number of players and matches. This means that we would have a round of two players (the finals), a round of four players (the semi-finals), a round of eight players (the quarter-finals), and so on. As long as the number of players is a power of 2, it is easy to count the number of matches: add up powers of 2 until we reach half the number of players (the first round played will have half as many matches as players, as each match consists of two players). For instance, if there were 64 competitors, then we would need to play $$2^0+2^1+2^2+2^3+2^4+2^5 = 1+2+4+8+16+32 = 63$$ matches.

If the number of competitors is not a power of 2, then we have a little problem, but it can be fixed. In the problem presented above, there are 53 competitors. That means that we can play rounds with 1, 2, 4, 8, and 16 matches. However, in the round that should have 32 matches, we don’t have enough players to fill it out. Instead, we want to play enough matches to get the pool of players down to 32, so that we can play 16 matches in the next round. A total of 53 players minus 32 players in the next round gives us a 21 players that need to be eliminated. We need to play 21 matches in order to eliminate those players. So the total number of matches will be $$(2^0+2^1+2^2+2^3+2^4)+21 = (1+2+4+8+16)+21 = 52$$.

This solution is pretty complicated, but it does have some advantages over the first method. First, once you know how it is done, it is pretty quick. Simply add up powers of 2 until just before you exceed the total number of players, then add extra matches to eliminate just enough players to get down to a power of 2. Second, the solution is generalizable, in the sense that if more players join, we don’t have to do a whole lot of work. We can just plug the new number into a formula and go.

## Solution III

The last solution takes a slightly different approach. Both of the above solutions look at the problem from the point of view of how many players there will be in each round, and how many rounds need to be played. Instead, why don’t we consider the losers?

Each time a match is played, there is a winner and a loser. As each match has exactly one loser, and as each loser will win exactly one match then be eliminated, there is a correspondence between losers and matches. In fact, there are exactly as many losers as there are matches. If 53 players enter, then there is 1 winner and 52 losers. Hence 52 matches need to be played.

In general, the number of matches played will be one less than the number of players, so this solution is also very generalizable, and quite easy to use. What more could we ask for?

I suspect that many non-mathematicians (and even some mathematicians) might jump on solution II as the “correct” solution. It has everything that we are taught to think of as math: lots of notation, a complex argument, and tons of numbers. On the other hand, I would argue that solution II is a “bad” solution for exactly those reasons. Solution III is much easier to follow, and gets the same result in a fraction of the time.

The moral of the story is this: mathematicians try to make clear, logical arguments. One can use mathematical notation to obscure a topic, or make it hard to understand, and there are times when mathematical ideas are so complicated and abstract that arcane notation is the only possible way to get a handle on things, but the heart of mathematics is simplicity. Simple arguments are preferred over complex arguments. Simple proofs are better than complex proofs. Any time a problem can be made simpler, that problem is made easier to understand, and easier to work with in the future.

In mathematics, one of the most important mantras is “keep is simple, stupid.”

## Notes:

[1]
NB: Both of these topics have been presented to these students before, though not together. Changing numbers to fractions was part of last week’s material, while comparing fractions using a common denominator was covered earlier in the year (as well as in lower grades).
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