zip stringlengths 19 109 | filename stringlengths 4 185 | contents stringlengths 0 30.1M | type_annotations listlengths 0 1.97k | type_annotation_starts listlengths 0 1.97k | type_annotation_ends listlengths 0 1.97k |
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archives/1098994933_python.zip | matrix/tests/test_matrix_operation.py | """
Testing here assumes that numpy and linalg is ALWAYS correct!!!!
If running from PyCharm you can place the following line in "Additional Arguments" for the pytest run configuration
-vv -m mat_ops -p no:cacheprovider
"""
# standard libraries
import sys
import numpy as np
import pytest
import logging
# Custom/loca... | [] | [] | [] |
archives/1098994933_python.zip | networking_flow/ford_fulkerson.py | # Ford-Fulkerson Algorithm for Maximum Flow Problem
"""
Description:
(1) Start with initial flow as 0;
(2) Choose augmenting path from source to sink and add path to flow;
"""
def BFS(graph, s, t, parent):
# Return True if there is node that has not iterated.
visited = [False]*len(graph)
queue=... | [] | [] | [] |
archives/1098994933_python.zip | networking_flow/minimum_cut.py | # Minimum cut on Ford_Fulkerson algorithm.
test_graph = [
[0, 16, 13, 0, 0, 0],
[0, 0, 10, 12, 0, 0],
[0, 4, 0, 0, 14, 0],
[0, 0, 9, 0, 0, 20],
[0, 0, 0, 7, 0, 4],
[0, 0, 0, 0, 0, 0],
]
def BFS(graph, s, t, parent):
# Return True if there is node that has not iterated.
visited = [Fals... | [] | [] | [] |
archives/1098994933_python.zip | neural_network/back_propagation_neural_network.py | #!/usr/bin/python
# encoding=utf8
'''
A Framework of Back Propagation Neural Network（BP） model
Easy to use:
* add many layers as you want ！！！
* clearly see how the loss decreasing
Easy to expand:
* more activation functions
* more loss functions
* more optimization method
Author: Stephen Lee
Git... | [] | [] | [] |
archives/1098994933_python.zip | neural_network/convolution_neural_network.py | #-*- coding: utf-8 -*-
'''
- - - - - -- - - - - - - - - - - - - - - - - - - - - - -
Name - - CNN - Convolution Neural Network For Photo Recognizing
Goal - - Recognize Handing Writting Word Photo
Detail：Total 5 layers neural network
* Convolution layer
* Pooling layer
... | [] | [] | [] |
archives/1098994933_python.zip | neural_network/perceptron.py | '''
Perceptron
w = w + N * (d(k) - y) * x(k)
Using perceptron network for oil analysis,
with Measuring of 3 parameters that represent chemical characteristics we can classify the oil, in p1 or p2
p1 = -1
p2 = 1
'''
import random
class Perceptron:
def __init__(self, sample, exit, learn_r... | [] | [] | [] |
archives/1098994933_python.zip | other/anagrams.py | import collections, pprint, time, os
start_time = time.time()
print('creating word list...')
path = os.path.split(os.path.realpath(__file__))
with open(path[0] + '/words') as f:
word_list = sorted(list(set([word.strip().lower() for word in f])))
def signature(word):
return ''.join(sorted(word))
word_bysig = ... | [] | [] | [] |
archives/1098994933_python.zip | other/binary_exponentiation.py | """
* Binary Exponentiation for Powers
* This is a method to find a^b in a time complexity of O(log b)
* This is one of the most commonly used methods of finding powers.
* Also useful in cases where solution to (a^b)%c is required,
* where a,b,c can be numbers over the computers calculation limits.
* Done using iterati... | [] | [] | [] |
archives/1098994933_python.zip | other/binary_exponentiation_2.py | """
* Binary Exponentiation with Multiplication
* This is a method to find a*b in a time complexity of O(log b)
* This is one of the most commonly used methods of finding result of multiplication.
* Also useful in cases where solution to (a*b)%c is required,
* where a,b,c can be numbers over the computers calculation l... | [] | [] | [] |
archives/1098994933_python.zip | other/detecting_english_programmatically.py | import os
UPPERLETTERS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
LETTERS_AND_SPACE = UPPERLETTERS + UPPERLETTERS.lower() + ' \t\n'
def loadDictionary():
path = os.path.split(os.path.realpath(__file__))
englishWords = {}
with open(path[0] + '/dictionary.txt') as dictionaryFile:
for word in dictionaryFile.read... | [] | [] | [] |
archives/1098994933_python.zip | other/euclidean_gcd.py | # https://en.wikipedia.org/wiki/Euclidean_algorithm
def euclidean_gcd(a, b):
while b:
t = b
b = a % b
a = t
return a
def main():
print("GCD(3, 5) = " + str(euclidean_gcd(3, 5)))
print("GCD(5, 3) = " + str(euclidean_gcd(5, 3)))
print("GCD(1, 3) = " + str(euclidean_gcd(1, 3))... | [] | [] | [] |
archives/1098994933_python.zip | other/fischer_yates_shuffle.py | #!/usr/bin/python
# encoding=utf8
"""
The Fisher–Yates shuffle is an algorithm for generating a random permutation of a finite sequence.
For more details visit
wikipedia/Fischer-Yates-Shuffle.
"""
import random
def FYshuffle(LIST):
for i in range(len(LIST)):
a = random.randint(0, len(LIST)-1)
b = r... | [] | [] | [] |
archives/1098994933_python.zip | other/frequency_finder.py | # Frequency Finder
# frequency taken from http://en.wikipedia.org/wiki/Letter_frequency
englishLetterFreq = {'E': 12.70, 'T': 9.06, 'A': 8.17, 'O': 7.51, 'I': 6.97,
'N': 6.75, 'S': 6.33, 'H': 6.09, 'R': 5.99, 'D': 4.25,
'L': 4.03, 'C': 2.78, 'U': 2.76, 'M': 2.41, 'W': 2.36,
... | [] | [] | [] |
archives/1098994933_python.zip | other/game_of_life.py | '''Conway's Game Of Life, Author Anurag Kumar(mailto:anuragkumarak95@gmail.com)
Requirements:
- numpy
- random
- time
- matplotlib
Python:
- 3.5
Usage:
- $python3 game_o_life <canvas_size:int>
Game-Of-Life Rules:
1.
Any live cell with fewer than two live neighbours
dies, as if caused by under-popu... | [] | [] | [] |
archives/1098994933_python.zip | other/linear_congruential_generator.py | __author__ = "Tobias Carryer"
from time import time
class LinearCongruentialGenerator(object):
"""
A pseudorandom number generator.
"""
def __init__( self, multiplier, increment, modulo, seed=int(time()) ):
"""
These parameters are saved and used when nextNumber() is called.
... | [] | [] | [] |
archives/1098994933_python.zip | other/nested_brackets.py | '''
The nested brackets problem is a problem that determines if a sequence of
brackets are properly nested. A sequence of brackets s is considered properly nested
if any of the following conditions are true:
- s is empty
- s has the form (U) or [U] or {U} where U is a properly nested string
- s has the form VW whe... | [] | [] | [] |
archives/1098994933_python.zip | other/palindrome.py | # Program to find whether given string is palindrome or not
def is_palindrome(str):
start_i = 0
end_i = len(str) - 1
while start_i < end_i:
if str[start_i] == str[end_i]:
start_i += 1
end_i -= 1
else:
return False
return True
# Recursive method
def r... | [] | [] | [] |
archives/1098994933_python.zip | other/password_generator.py | """Password generator allows you to generate a random password of length N."""
from random import choice
from string import ascii_letters, digits, punctuation
def password_generator(length=8):
"""
>>> len(password_generator())
8
>>> len(password_generator(length=16))
16
>>> len(password_genera... | [] | [] | [] |
archives/1098994933_python.zip | other/primelib.py | # -*- coding: utf-8 -*-
"""
Created on Thu Oct 5 16:44:23 2017
@author: Christian Bender
This python library contains some useful functions to deal with
prime numbers and whole numbers.
Overview:
isPrime(number)
sieveEr(N)
getPrimeNumbers(N)
primeFactorization(number)
greatestPrimeFactor(number)
smallestPrimeFacto... | [] | [] | [] |
archives/1098994933_python.zip | other/sierpinski_triangle.py | #!/usr/bin/python
# encoding=utf8
'''Author Anurag Kumar | anuragkumarak95@gmail.com | git/anuragkumarak95
Simple example of Fractal generation using recursive function.
What is Sierpinski Triangle?
>>The Sierpinski triangle (also with the original orthography Sierpinski), also called the Sierpinski gasket or the Si... | [] | [] | [] |
archives/1098994933_python.zip | other/tower_of_hanoi.py | def moveTower(height, fromPole, toPole, withPole):
'''
>>> moveTower(3, 'A', 'B', 'C')
moving disk from A to B
moving disk from A to C
moving disk from B to C
moving disk from A to B
moving disk from C to A
moving disk from C to B
moving disk from A to B
'''
if height >= 1:
... | [] | [] | [] |
archives/1098994933_python.zip | other/two_sum.py | """
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0,... | [] | [] | [] |
archives/1098994933_python.zip | other/word_patterns.py | import pprint, time
def getWordPattern(word):
word = word.upper()
nextNum = 0
letterNums = {}
wordPattern = []
for letter in word:
if letter not in letterNums:
letterNums[letter] = str(nextNum)
nextNum += 1
wordPattern.append(letterNums[letter])
return '... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_01/sol1.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
>... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol2.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol3.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""
This solution is based on the pattern that the successive numbers in the
series... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol4.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol5.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
"""A straightforward pythonic solution using list comprehension"""
def solution(n):
"""Returns the sum of... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_01/sol6.py | """
Problem Statement:
If we list all the natural numbers below 10 that are multiples of 3 or 5,
we get 3,5,6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below N.
"""
def solution(n):
"""Returns the sum of all the multiples of 3 or 5 below n.
>>> solution(3)
0
>... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_02/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_02/sol1.py | """
Problem:
Each new term in the Fibonacci sequence is generated by adding the previous two
terms. By starting with 1 and 2, the first 10 terms will be:
1,2,3,5,8,13,21,34,55,89,..
By considering the terms in the Fibonacci sequence whose values do not exceed
n, find the sum of the even-valued terms. e.g. for n=1... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_02/sol2.py | """
Problem:
Each new term in the Fibonacci sequence is generated by adding the previous two
terms. By starting with 1 and 2, the first 10 terms will be:
1,2,3,5,8,13,21,34,55,89,..
By considering the terms in the Fibonacci sequence whose values do not exceed
n, find the sum of the even-valued terms. e.g.... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_02/sol3.py | """
Problem:
Each new term in the Fibonacci sequence is generated by adding the previous
two terms. By starting with 1 and 2, the first 10 terms will be:
1,2,3,5,8,13,21,34,55,89,..
By considering the terms in the Fibonacci sequence whose values do not exceed
n, find the sum of the even-valued terms. e.g. for n=1... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_02/sol4.py | """
Problem:
Each new term in the Fibonacci sequence is generated by adding the previous two
terms. By starting with 1 and 2, the first 10 terms will be:
1,2,3,5,8,13,21,34,55,89,..
By considering the terms in the Fibonacci sequence whose values do not exceed
n, find the sum of the even-valued terms. e.g. for n=1... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_03/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_03/sol1.py | """
Problem:
The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor
of a given number N?
e.g. for 10, largest prime factor = 5. For 17, largest prime factor = 17.
"""
import math
def isprime(no):
if no == 2:
return True
elif no % 2 == 0:
return False
sq = int(math.... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_03/sol2.py | """
Problem:
The prime factors of 13195 are 5,7,13 and 29. What is the largest prime factor
of a given number N?
e.g. for 10, largest prime factor = 5. For 17, largest prime factor = 17.
"""
def solution(n):
"""Returns the largest prime factor of a given number n.
>>> solution(13195)
29
>>> solution... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_04/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_04/sol1.py | """
Problem:
A palindromic number reads the same both ways. The largest palindrome made from
the product of two 2-digit numbers is 9009 = 91 x 99.
Find the largest palindrome made from the product of two 3-digit numbers which
is less than N.
"""
def solution(n):
"""Returns the largest palindrome made from the prod... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_04/sol2.py | """
Problem:
A palindromic number reads the same both ways. The largest palindrome made from
the product of two 2-digit numbers is 9009 = 91 x 99.
Find the largest palindrome made from the product of two 3-digit numbers which
is less than N.
"""
def solution(n):
"""Returns the largest palindrome made from the prod... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_05/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_05/sol1.py | """
Problem:
2520 is the smallest number that can be divided by each of the numbers from 1
to 10 without any remainder.
What is the smallest positive number that is evenly divisible(divisible with no
remainder) by all of the numbers from 1 to N?
"""
def solution(n):
"""Returns the smallest positive number that is ... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_05/sol2.py | """
Problem:
2520 is the smallest number that can be divided by each of the numbers from 1
to 10 without any remainder.
What is the smallest positive number that is evenly divisible(divisible with no
remainder) by all of the numbers from 1 to N?
"""
""" Euclidean GCD Algorithm """
def gcd(x, y):
return x if y ==... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_06/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_06/sol1.py | # -*- coding: utf-8 -*-
"""
Problem:
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natur... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_06/sol2.py | # -*- coding: utf-8 -*-
"""
Problem:
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natur... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_06/sol3.py | # -*- coding: utf-8 -*-
"""
Problem:
The sum of the squares of the first ten natural numbers is,
1^2 + 2^2 + ... + 10^2 = 385
The square of the sum of the first ten natural numbers is,
(1 + 2 + ... + 10)^2 = 552 = 3025
Hence the difference between the sum of the squares of the first ten natur... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_07/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_07/sol1.py | # -*- coding: utf-8 -*-
"""
By listing the first six prime numbers:
2, 3, 5, 7, 11, and 13
We can see that the 6th prime is 13. What is the Nth prime number?
"""
from math import sqrt
def isprime(n):
if n == 2:
return True
elif n % 2 == 0:
return False
else:
sq = int(sqrt(n))... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_07/sol2.py | # -*- coding: utf-8 -*-
"""
By listing the first six prime numbers:
2, 3, 5, 7, 11, and 13
We can see that the 6th prime is 13. What is the Nth prime number?
"""
def isprime(number):
for i in range(2, int(number ** 0.5) + 1):
if number % i == 0:
return False
return True
def solution(... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_07/sol3.py | # -*- coding: utf-8 -*-
"""
By listing the first six prime numbers:
2, 3, 5, 7, 11, and 13
We can see that the 6th prime is 13. What is the Nth prime number?
"""
import math
import itertools
def primeCheck(number):
if number % 2 == 0 and number > 2:
return False
return all(number % i for i in ra... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_08/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_08/sol1.py | # -*- coding: utf-8 -*-
"""
The four adjacent digits in the 1000-digit number that have the greatest
product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
1254069874715852386305071569329... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_08/sol2.py | # -*- coding: utf-8 -*-
"""
The four adjacent digits in the 1000-digit number that have the greatest
product are 9 × 9 × 8 × 9 = 5832.
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
1254069874715852386305071569329... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_09/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_09/sol1.py | """
Problem Statement:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
def solution():
"""
Returns the product of ... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_09/sol2.py | """
Problem Statement:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
def solution(n):
"""
Return the product of a,... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_09/sol3.py | """
Problem Statement:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
"""
def solution():
"""
Returns the product of... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_10/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_10/sol1.py | """
Problem Statement:
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
"""
from math import sqrt
def is_prime(n):
for i in range(2, int(sqrt(n)) + 1):
if n % i == 0:
return False
return True
def sum_of_primes(n):
if n > 2:
... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_10/sol2.py | """
Problem Statement:
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
"""
import math
from itertools import takewhile
def primeCheck(number):
if number % 2 == 0 and number > 2:
return False
return all(number % i for i in range(3, int(math.sqrt(... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_10/sol3.py | """
https://projecteuler.net/problem=10
Problem Statement:
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million using Sieve_of_Eratosthenes:
The sieve of Eratosthenes is one of the most efficient ways to find all primes
smaller than n when n is smaller than 10 millio... | [
"int"
] | [
374
] | [
377
] |
archives/1098994933_python.zip | project_euler/problem_11/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_11/sol1.py | """
What is the greatest product of four adjacent numbers (horizontally,
vertically, or diagonally) in this 20x20 array?
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 ... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_11/sol2.py | """
What is the greatest product of four adjacent numbers (horizontally,
vertically, or diagonally) in this 20x20 array?
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 ... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_12/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_12/sol1.py | """
Highly divisible triangular numbers
Problem 12
The sequence of triangle numbers is generated by adding the natural numbers. So
the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten
terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_12/sol2.py | """
Highly divisible triangular numbers
Problem 12
The sequence of triangle numbers is generated by adding the natural numbers. So
the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten
terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_13/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_13/sol1.py | """
Problem Statement:
Work out the first ten digits of the sum of the following one-hundred 50-digit
numbers.
"""
def solution(array):
"""Returns the first ten digits of the sum of the array elements.
>>> import os
>>> sum = 0
>>> array = []
>>> with open(os.path.dirname(__file__) + "/num.txt","... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_14/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_14/sol1.py | # -*- coding: utf-8 -*-
"""
Problem Statement:
The following iterative sequence is defined for the set of positive integers:
n → n/2 (n is even)
n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
It can be seen ... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_14/sol2.py | # -*- coding: utf-8 -*-
"""
Collatz conjecture: start with any positive integer n. Next term obtained from
the previous term as follows:
If the previous term is even, the next term is one half the previous term.
If the previous term is odd, the next term is 3 times the previous term plus 1.
The conjecture states the s... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_15/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_15/sol1.py | """
Starting in the top left corner of a 2×2 grid, and only being able to move to
the right and down, there are exactly 6 routes to the bottom right corner.
How many such routes are there through a 20×20 grid?
"""
from math import factorial
def lattice_paths(n):
"""
Returns the number of paths possible in a n... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_16/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_16/sol1.py | """
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
"""
def solution(power):
"""Returns the sum of the digits of the number 2^power.
>>> solution(1000)
1366
>>> solution(50)
76
>>> solution(20)
31
>>> solution(15)
... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_16/sol2.py | """
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
What is the sum of the digits of the number 2^1000?
"""
def solution(power):
"""Returns the sum of the digits of the number 2^power.
>>> solution(1000)
1366
>>> solution(50)
76
>>> solution(20)
31
>>> solution(... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_17/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_17/sol1.py | """
Number letter counts
Problem 17
If the numbers 1 to 5 are written out in words: one, two, three, four, five,
then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
If all the numbers from 1 to 1000 (one thousand) inclusive were written out in
words, how many letters would be used?
NOTE: Do not count space... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_19/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_19/sol1.py | """
Counting Sundays
Problem 19
You are given the following information, but you may prefer to do some research
for yourself.
1 Jan 1900 was a Monday.
Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twe... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_20/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_20/sol1.py | """
n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
"""
def factorial(n):
fact = 1
for i in range(1, n + 1):
fact *= i
return... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_20/sol2.py | """
n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
"""
from math import factorial
def solution(n):
"""Returns the sum of the digits in the n... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_21/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_21/sol1.py | # -.- coding: latin-1 -.-
from math import sqrt
"""
Amicable Numbers
Problem 21
Let d(n) be defined as the sum of proper divisors of n (numbers less than n
which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and
each of a and b are called amicable numbers.
For exampl... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_22/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_22/sol1.py | # -*- coding: latin-1 -*-
"""
Name scores
Problem 22
Using names.txt (right click and 'Save Link/Target As...'), a 46K text file
containing over five-thousand first names, begin by sorting it into
alphabetical order. Then working out the alphabetical value for each name,
multiply this value by its alphabetical positio... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_22/sol2.py | # -*- coding: latin-1 -*-
"""
Name scores
Problem 22
Using names.txt (right click and 'Save Link/Target As...'), a 46K text file
containing over five-thousand first names, begin by sorting it into
alphabetical order. Then working out the alphabetical value for each name,
multiply this value by its alphabetical positio... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_234/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_234/sol1.py | """
https://projecteuler.net/problem=234
For an integer n ≥ 4, we define the lower prime square root of n, denoted by
lps(n), as the largest prime ≤ √n and the upper prime square root of n, ups(n),
as the smallest prime ≥ √n.
So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37. Let us
call an integer... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_24/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_24/sol1.py | """
A permutation is an ordered arrangement of objects. For example, 3124 is one
possible permutation of the digits 1, 2, 3 and 4. If all of the permutations
are listed numerically or alphabetically, we call it lexicographic order. The
lexicographic permutations of 0, 1 and 2 are:
012 021 102 120 201 210... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_25/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_25/sol1.py | # -*- coding: utf-8 -*-
"""
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_25/sol2.py | # -*- coding: utf-8 -*-
"""
The Fibonacci sequence is defined by the recurrence relation:
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
Hence the first 12 terms will be:
F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144
The... | [] | [] | [] |
archives/1098994933_python.zip | project_euler/problem_28/__init__.py | [] | [] | [] | |
archives/1098994933_python.zip | project_euler/problem_28/sol1.py | """
Starting with the number 1 and moving to the right in a clockwise direction a 5
by 5 spiral is formed as follows:
21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13
It can be verified that the sum of the numbers on the diagonals is 101.
What is the sum of the numbers o... | [] | [] | [] |
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