Skip to main content

Problem 57 Project Euler Solution with python

Square root convergents

It is possible to show that the square root of two can be expressed as an infinite continued fraction.
√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
By expanding this for the first four iterations, we get:
1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.
In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?


Before we start with the solution, have a look at the following links:

Mathematics Stack: Convergent of square root of 2
Convergent Wolfram

You can also have a look at Wikipedia Continued fraction if you want.

According to the links, if p/q is is the first convergent then the next convergent can be written as (p+2q)/(p+q).

So simply we will start with 1/1 where p = 1 and q = 1 and iterate 1000 times and find how many times the length of p is greater than q.

Program

You can download the source code from Github Gist pep57.py

Output

Summary

This problem is again simple. I am satisfied with the execution time and haven't tried to optimize the code. I also liked the clean and clear solution I have written. One can still optimize the code. 

 Please excuse and correct me if my grammar is wrong or in an ambiguous way. 

 Comment in the comment box below if you have any doubt or didn't understand anything. I will be glad to help you.

 Please do comment in the comment box below if you have found any typo or have a different solution or a better program or have a suggestion. I will be very happy to view each of them.

 You can also contact me.

 Thank you. Have a nice day😃!
Labels:

Popular posts from this blog

Project Euler Problem 67 Solution with Python

Maximum path sum II By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23. 3 7 4 2 4 6 8 5 9 3 That is, 3 + 7 + 4 + 9 = 23. Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.
Read more

Problem 60 Project Euler Solution with python

Prime pair sets The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property. Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime. This problem is j u st a brute force problem. If you have come here because you don't know the limit upto which you will h ave to gener ate the prime numbers t hen go ahe ad and t r y with 10,000 . When I first start ed solving the problem I chose 1 million(beca use most of the problem s on project E uler have this limit ), but it took very long for the computer to fin d the solution. After searching on the internet then I found many people choosing 10, 000 so I have changed my in put f rom 1 million to 10000 and the output was f ast. He...
Read more

Project Euler Problem 66 Solution with python

Diophantine equation ¶ Consider quadratic Diophantine equations of the form: $$ x^{2} – Dy^{2} = 1 $$ For example, when $D = 13$, the minimal solution in $ x $ is $ 649^{2} – 13 \times 180^{2} = 1 $ It can be assumed that there are no solutions in positive integers when $ D $ is square. By finding minimal solutions in $ x $ for $ D = {2, 3, 5, 6, 7} $, we obtain the following: $$ 3^{2} – 2×2^{2} = 1 $$ $$ 2^{2} – 3×1^{2} = 1 $$ $$ 9^{2} – 5×4^{2} = 1 $$ $$ 5^{2} – 6×2^{2} = 1 $$ $$ 8^{2} – 7×3^{2} = 1 $$ Hence, by considering minimal solutions in $ x $ for $ D ≤ 7 $, the largest $ x $ is obtained when $ D = 5 $. Find the value of $ D ≤ 1000 $ in minimal solutions of $ x $ for which the largest value of $ x $ is obtained.
Read more