Periodic Last Digits of Fibonacci Numbers

Here’s something I was playing around with the other day.

Consider the Fibonacci sequence: 0, 1, 1, 2,3,5,8,13,21, \dots. F(n) = F(n-1)+F(n-2). I wanted to see if there was a pattern regarding the values of the ones digit in the fibonacci sequence. This is what I found:

The ones digits of the Fibonacci numbers are periodic, with period 60. This means that if you continue the sequence and look only at the ones digits of the numbers, they begin to repeat the same pattern after the 60th number. In addition, every 15th number in the sequence has a ones digit of zero. I wonder why this is. I can’t seem to find any other patterns at the moment. It probably has something to do with mod.

Here is the list of the 60 digit period (ones digits only):

1
1
2
3
5
8
3
1
4
5
9
4
3
7
0
7
7
4
1
5
6
1
7
8
5
3
8
1
9
0
9
9
8
7
5
2
7
9
6
5
1
6
7
3
0
3
3
6
9
5
4
9
3
2
5
7
2
9
1
0
---this is where the cycle starts over
1
1
2
3
5
8
...

Can you find any other interesting properties?

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A Genetic Algorithm for Computing Ramsey Numbers: Update

Friends_strangers_graph

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

800px-Jigsaw_puzzle_01_by_Scouten

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.

yeqx2

Graph A

Graph B

Graph B

The Class NP

cookies

“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!