Pretty Print JSON From The Command Line

JSON can come in all shapes and sizes, and sometimes you want to see it in a structured format that’s easier on the eyes. This is called pretty printing. But how do you accomplish that, especially if you have a really large JSON file? While there are some converter tools online to show json files in a pretty format, they can get very slow and even freeze if your file is too large. Let’s do it locally.

Requirements: Have Python installed, and a terminal environment.

cat file.json | python -m json.tool > prettyfile.json

And that’s all there is to it! Python comes built in with a JSON encoding/decoding library, and you can use it to your advantage to get nice formatted output. Alternatively, if you are receiving JSON from an API or HTTP request, you can pipe your results from a curl call directly into this tool as well.

Hope this helps!

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Halloween Themed Math Puzzles

free-halloween-powerpoint-background-8

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

Halloween-Candy1

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

4 Is The Magic Number – Riddle

So fishes56 alerted me to this riddle in a comment here, and if you didn’t see it I wanted to share it here in a separate post cause its a little fun thing to think about.

12 is 6, 6 is 3, 3 is 5, 5 is 4, 4 is 4.

4 is the magic number, why?

68 is 10, 10 is 3, 3 is 5, 5 is 4, 4 is 4.

4 is the magic number, why?

26 is 9, 9 is 4, 4 is 4. 4 is the magic number, why?

Give it a try. 🙂

The Collatz Conjecture and Hailstone Sequences: Deceptively Simple, Devilishly Difficult

Here is a very simple math problem:

If a number is even, divide the number by two. If a number is odd, multiply the number by three, and then add one. Do this over and over with your results. Will you always get down to the value 1 no matter what number you choose?

Go ahead and test this out for yourself. Plug a few numbers in. Try it out. I’ll wait.

Back? Good. Here’s my example: I start with the number 12.

12 is even, so I divide it by 2.

12/2=6.

6/2=3.

3 is odd, so we multiply it by 3 then add 1.

(3*3)+1=10.

10/2=5.

(5*3)+1=16.

16/2=8.

8/2=4.

4/2=2.

2/2=1.

And we have arrived at one, just like we thought we would. But is this ALWAYS the case, for ANY number I could possibly dream of?

This may sound easy, but in fact it is an unsolved mathematics problem, with some of the best minds in the field doing research on it. It’s easy enough to check low values, but numbers are infinite, and can get very, very, very large. How do we KNOW that it works for every single number that exists?

That is what is so difficult about this problem. While every single number we have checked (which is a LOT) ends up at 1 sooner or later (some numbers can bounce around for a very long time, hence why they are called “hailstone sequences”), we still have no method to prove that it works for every number. Mathematicians have been looking for some method to predict this, but despite getting into some pretty heavy mathematics in an attempt to attack this problem, we still do not know for sure.

This problem is known as the Collatz Conjecture and is very interesting to mathematicians young and old because it is so easy to explain and play with, yet so tough to exhaustively prove. What do you think? Will this problem ever be solved? And what would be the implications if it was?

Here is some simple Python code that can display the cycles for any number you type in:

#!/usr/bin/python
num = int(input("Enter a number: "))
while num!=1:
if num%2 == 0:
num = num/2
else:
num = (num*3)+1
    print num

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?

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!