Halloween Themed Math Puzzles

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Happy Halloween From  The Muse Garden!

In the tradition of my Holiday Math Puzzles, I’m here with an appropriately themed puzzle for this time of year.

Candy Distribution

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It’s that time of year all right. You’re out and about, trick or treating with your friends or family and when you come home, you decide to dump all the candy out on the floor to sort through it. But then, as siblings often do, you begin to bicker about who has “more” than the other. In fact, there are some candies that you really like, and some you don’t like. You’d rather have a bunch of chocolate than a bunch of peppermints, for instance.  But wait a minute! Of course you can’t like the same thing! Your sibling actually likes peppermints!

Here’s a table of the different candies you have. Each candy has a “value” to it; that is, how much you “want” it. Try to split up the candies such that you and your sibling both have as equal value as possible at the end. And no fighting!

Candy Quantity Your Value (per piece) Sibling’s Value (per piece)
Candy Corn 150 25 50
Peppermints 50 5 50
Peanut Butter Cups 10 100 75
Hershey Bars 25 50 10
Kit Kat 20 75 30
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Chopsticks Game – A Combinatorial Challenge

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So I don’t know if anyone else is familiar with this game, but I just remember playing it with friends in middle school and it occurred to me the other day that it would be an interesting game to analyze combinatorially, and perhaps write a game playing algorithm for. This game can be found in more detail here: http://en.wikipedia.org/wiki/Chopsticks_(hand_game)

Players: 2+.

Rules of Play: Players each begin with two “piles” of points, and each pile has 1 point to begin with. We used fingers to represent this, one finger on each hand.

On each turn, a player can choose to do one of two things:

  1. Send points from one of the player’s pile to one of the opponents piles. So if Player 1 wanted to send 1 point to Player 2’s left pile, then Player 2 would have 2 points in the left pile and Player 1 would still have 1 point in his left pile. Player 1 does not lose points, they are simply “cloned” over to the opponents pile.
  2. If the player has an even number of points in one pile and zero points in the other pile, the player may elect to split his points evenly between the two piles. This consumes the player’s turn. Example: if Player 1 has (0     4) then he can use his turn to split his points, giving him (2     2).

If a player gets exactly 5 points in either pile, that pile loses all of its points and reverts to 0. However, if points applied goes over 5 (such as adding 2 points to a 4 point pile), then the remainder of points are added. (meaning that 4 + 2 = 1). The opponent simply gets points mod 5.

If a player gets to 0 points in both their piles, then they lose. The last person that has points remaining wins.
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Okay, so let’s break this down. Here’s an example game for those of you that are more visually oriented (follow the turns by reading left to right, moves are marked with red arrows):

Okay, so let’s point out a few things about this game.

  • On turn 4, player 2 adds 3 points to player 1’s 2 points, making 5. The rules state that any pile with exactly 5 points reverts to zero.
  • On turn 7, player 2 adds 3 points to 3 points. 3+3=6 as we all know, but 6 \equiv 1 \mod 5, so player 1 now has one point in his pile.
  • On turn 13, player 2 decides to split his points, turning his one pile of 4 into two piles of 2. This consumes his turn.
  • On turn 15, player 2 adds 2 points to player 1’s 3, thus reverting his pile to zero. Player 1 now has no more points to play with, so player 2 wins.

We can think about this game as a combinatorial problem. What are the optimal positions to play? How would one program a computer to play this game? I plan to create an interactive web game where players can try this for themselves.

A Genetic Algorithm for Computing Ramsey Numbers: Update

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All the 78 possible friends-strangers graphs with 6 nodes. For each graph the red/blue nodes shows a sample triplet of mutual friends/strangers.

In my last post on this topic, I discussed how I was working on a genetic algorithm to search mathematical graphs for elusive properties called Ramsey Numbers. (For a refresher on genetic algorithms,  visit here, and for a refresher on Ramsey Numbers, visit here). I’ve been doing some work on it since then (check out the code here), and I thought I would describe some improvements and further progress I’ve made in this area.

New features:

  • colorings dumped to a file at the end of each run
  • ability to load in data sets from file, further refining of the data than starting from scratch each time

The next problem I ran up against while working through this was that even if I am able to load in previously analyzed data, I still only have one fitness function that checks a static set of edges. As I see it, there are two ways to solve this:

  • Make the current fitness function dynamic; that is, it tests a different set of edges every time. However, this is counterproductive to the purpose of the program “eliminating” certain sets of edges in each “round”. However, this would be easier to maintain than the other option, which is
  • Make a “FitnessHandler” method that takes in a value for which method to run, and uses that to determine what set of edges to test. However, this would lead to a lot of extra code and overhead. I’m thinking having a static variable at the beginning of each run with what “fitness method” to start on, so that it doesn’t have to start on round one each time.

I haven’t fully decided which of these I will go with. I feel like the second one fills my purpose of methodically “weeding out” the improbable graphs, but its going to be a lot of extra work. Oh well, nothing worthwhile ever came easy…

Leave a note here or on my github if you have suggestions!

Introduction to P vs. NP

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Its a Millennium Problem (reward $1,000,000) and the subject of a movie. But what is this mysterious problem, and what makes it so hard?

Creation vs. Verification

Pop quiz: Choose which one of these tasks takes longer.

  • A) someone hands you a jigsaw puzzle and asks you if it is finished or not
  • B) someone hands you a jigsaw puzzle and asks you to put it together

Answer: BWhile you can determine if a puzzle is finished or not in a split second (it is easy to tell if pieces are missing), it is not so easy to put all those pieces together yourself. We have problems like this in computer science and mathematics as well. Think about this: what if there was a method that could solve a jigsaw puzzle as fast as you could check if its right? Wouldn’t that be amazing? That’s what this problem seeks to find out: if such a method exists.

The Class P

There is a class of problems that can be solved in “polynomial time”, and we call this P. Polynomial time means that as the complexity of the problem increases, the time it takes to solve increases at a rate no greater than a polynomial would. Informally, we can say that for P problems, we can solve the problem “quickly”. (Polynomial time).

Here’s an example: Say we have a list of numbers in front of us, and we want to pick out the number that is the greatest. At the very worst, we could start at the beginning of the list and compare every number until we got to the end of the list. Yes, the list may be very very long. But the time it takes to check every number increases in a linear fashion. More numbers does not make it exponentially more difficult. If we compare the growth rates of linear time O(n) (graph A) and polynomial time O(n^2) (graph B) we can clearly see that solving this problem is quicker than polynomial time, thus it is in P.

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Graph A

Graph B

Graph B

The Class NP

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“Mom, she got more than me!”

The class of NP problems, put simply, can be verified in polynomial time. NP stands for “nondeterministic polynomial time” and harkens back to a computational construct known as a Turing Machine. Consider this example: a woman is dividing up cookies of different sizes into two groups for her two young children. Naturally, they are very picky about things being fair, so she needs to make sure that each child has exactly the same amount of cookies (by weight). While we can show that this problem can get quite hard to sit down and solve, if someone showed us two piles of cookies we could quickly add up the weights of the two piles and tell if they were the same or not. 

If you have only 2 cookies to separate out, sure, its easy. Put one cookie in each pile and that’s the best you can do. But what if you had 5 cookies? 10? Keep in mind that each cookie weighs a slightly different amount and we want the two groups of cookies to be equal in weight as much as possible. Adding more cookies to work with increases our options for separating them exponentially. This is not good. If she has 5 cookies, the number of different ways to separate them is 2^5 = 32, but if she doubles the amount of cookies to 10, the possibilities skyrocket because 2^{10} = 1024.

Let’s consider, for fun, the number of possibilities if she had 100 cookies! 2^{100}=1,267,650,600,228,229,401,496,703,205,376. This…is pretty large. But wait! Computers can do that in a heartbeat, right? They’re way faster than we are! This is all fine and good, until you realize that the number of seconds in the age of the entire universe is (only) 450,000,000,000,000,000. So even if our computer could go through thousands or millions of computations per second, it would still take longer than the entire age of the universe. This is what makes these sorts of problems so hard.

P = NP?

So now that you know a little about what P and NP are, you may be wondering what all this business about “P=NP” is. What does it mean, after all? If someone were to prove that P equals NP, this would mean that every problem in the class NP (the hard ones) can be solved in a polynomial time, just like the P problems. This would have huge implications for all sorts of applications, not just in the mathematical world. For instance, a cornerstone of many security systems rests on the difficulty of factoring very large numbers. If P=NP, it would mean that there exist very easy methods for factoring these numbers, and as such, financial systems everywhere would be vulnerable.

This can also be applied to many logistical and optimization problems (see my posts on intractable problems here and here). For many of the NP problems, finding an “easy” method to solve one can be applied to any of the problems in that class. We don’t need to show an easy method to solve every problem in the world. We only need one. That shouldn’t be too hard, right? Unfortunately, researchers have been working on this problem for decades without success. There have been many attempted “proofs” but they all ultimately have fallen short. The problem is so difficult that the Clay Mathematics Institute is offering $1,000,000 for a correct proof of this problem one way or another.

Although an official answer to this question has not been decided one way or another, many professionals believe that P \neq NP. This means that problems in NP will always be inherently “harder” than problems in P, and at the worst case require an exhaustive (brute force) search to find a solution. No matter how good our computers get, these problems will still be very difficult for us to solve, especially at large cases.

I hope you enjoyed this little introduction to P and NP and if you have any comments, questions or corrections, please leave them in the comments below!

Intractable Problems — Part Two: Data Storage

This post continues my series on intractable problems. In this installment, I will talk about problems relating to Data Storage. As a refresher, remember that an intractable problem is one that is very computationally complex and very difficult to solve using a computer without some sort of novel thinking. I will discuss two famous problems related to Data Storage below, as well as provide a few examples and references.

Part Two — Data Storage

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Knapsack — given a set of items with weights w and values v, and a knapsack with capacity C, maximize the value of the items in the knapsack without going over capacity.

To start with, here is a small example that I refer to in this previous post. If you’re trying to place ornaments on a tree, you want to get the max amount of coverage possible. However, the tree can only hold so much weight before it falls over. What is the best way to pick ornaments and decorations such that you can cover as much of the tree as possible without it falling over?

I actually saw a really interesting video describing an example of this problem in this video (watch the first few minutes). The professor here sets up a situation of Indiana Jones trying to grab treasures from a temple before it collapses. He wants to get the most valuable treasures he can, but he can only carry so much.

Class: There are several different types of knapsack problems, but the most common one (the one discussed above) is one-dimensional knapsack. The decision problem (can we get to a value V without exceeding weight W?) is NP-Complete. However, the optimization problem (what is the most value we can get for the least weight?) is NP-Hard.

References: 

Wikipedia Page – General discussion of the Knapsack problem, different types, complexity, and a high level view of several algorithms for solving

Coursera Course on Discrete Optimization – The source for the above video and a great discussion of not only Knapsack but quite a few of these problems

Knapsack Problem at Rosetta Code – a good example data set and a variety of implementations in different languages

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TetrisPieces2

Bin Packing — given a set of items with weights w and a set of n bins with capacity c each, place the items into the bins such that the minimum amount of bins are used.

This problem is very similar to the knapsack problem found above, but this time we don’t care how much the items are worth. We just want to pack them in the smallest area possible. Solving this problem is invaluable for things like shipping and logistics. Obviously, companies want to be able to ship more with less space.

A more commonplace example could be thinking of this problem as Tetris in real life. Consider that you’re moving to a new place,  and you have you and your friends car in which to move things. How can you place all the items in your cars such that you take up space the most efficiently?

Some progress has been made on reasonably large data sets by using what is called the “first fit decreasing algorithm”. This means that you pick up an item, and place it into the first bin that it will fit in. If it can’t fit in any of the current bins, make a new bin for it. Decreasing means that before you start placing items, you sort them all from biggest to smallest. You probably do this in your everyday life. If you want to pack a box, you start with the big items, right? No need to put lots of small items on the bottom. By getting the big items out of the way first, you can be more flexible with the remaining space because you will then have smaller items.

Class: This problem is NP-Hard.

References:

Wikipedia Page – a high level description of the problem

First Fit Decreasing Paper (pdf) – this is a technical paper describing computational bounds for using the first fit decreasing algorithm. Not for beginners.

3D Bin Packing Simulation – looks like a resource for companies to use to pack boxes and such

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I hope this provided a little taste of why these problems are so important. If you know any other good resources please let me know.