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208 lines
6.4 KiB
Python
208 lines
6.4 KiB
Python
#Python/pyClasses/NumberAlgorithms.py
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#Matthew Ellison
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# Created: 07-21-21
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#Modified: 07-21-21
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#This file contains my library of number functions
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"""
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Copyright (C) 2021 Matthew Ellison
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This program is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as published by
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the Free Software Foundation, either version 3 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>.
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"""
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import itertools
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import math
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#Generate an infinite sequence of prime numbers using the Sieve of Eratosthenes
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#Based on code by David Eppstein found at https://code.activestate.com/recipes/117119/
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def primeGenerator():
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#Return 2 the first time, this lets us skip all even numbers later
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yield 2
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#Dictionary to hold the primes we have already found
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dict = {}
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#Start checking for primes with the number 3 and skip all even numbers
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for possiblePrime in itertools.count(3, 2):
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#If possiblePrime is not in the dictionary it is a new prime number
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#Return it and mark its next multiple
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if possiblePrime not in dict:
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yield possiblePrime
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dict[possiblePrime * possiblePrime] = [possiblePrime]
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#If possiblePrime is in the dictionary it is a composite number
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else:
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#Move each witness to its next multiple
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for num in dict[possiblePrime]:
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dict.setdefault(num + num + possiblePrime, []).append(num)
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#We no longer need this, free the memory
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del dict[possiblePrime]
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#Generate an inifinite sequence of fibonacci numbers
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def fibGenerator():
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#Set this so the first returned number is 1 and the second is also 1
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a, b = 0, 1
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while(True):
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yield b
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a, b = b, a + b
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#This function returns a list with all the prime numbers <= goalNumber
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def getPrimes(goalNumber: int) -> list:
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gen = primeGenerator() #The prime number generator
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primes = [] #The list of prime numbers
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#If the number is <= 1 return a blank list
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#?Should this throw an exception if goalNumber < 0
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if(goalNumber <= 1):
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return primes
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#There is at least 1 prime number
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primes.append(next(gen))
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#Loop until you find all the prime numbers requested
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while(primes[len(primes) - 1] < goalNumber):
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primes.append(next(gen))
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if(primes[len(primes)- 1] > goalNumber):
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primes.pop()
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return primes
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#This function gets a certain number of primes
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def getNumPrimes(numberOfPrimes: int) -> list:
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gen = primeGenerator() #The prime number generator
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primes = [] #The list of prime numbers
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#If the number is < 1 return a blank list
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#?Should this throw an exception if numberOfPrimes < 0
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if(numberOfPrimes < 1):
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return primes
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#Count how many primes you are adding to the list and stop when you reach the requested length
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for _ in range(1, numberOfPrimes + 1):
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primes.append(next(gen))
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#Return the list with all the prime numbers
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return primes
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#This function determines whether a number is prime
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def isPrime(possiblePrime: int) -> bool:
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if(possiblePrime <= 3):
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return possiblePrime > 1
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elif(((possiblePrime % 2) == 0) or ((possiblePrime % 3) == 0)):
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return False
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for cnt in range(5, int(math.sqrt(possiblePrime)) + 1, 6):
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if(((possiblePrime % cnt) == 0) or ((possiblePrime % (cnt + 2)) == 0)):
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return False
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return True
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#This is a function that returns all the factors of goalNumber
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def getFactors(goalNumber: int) -> list:
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prime_factors_list = []
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while goalNumber % 2 == 0:
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prime_factors_list.append(2)
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goalNumber /= 2
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for i in range(3, int(math.sqrt(goalNumber))+1, 2):
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if goalNumber % i == 0:
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prime_factors_list.append(i)
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goalNumber /= i
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if goalNumber > 2:
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prime_factors_list.append(int(goalNumber))
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prime_factors_list.sort()
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return prime_factors_list
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#This function returns all the divisors of goalNumber
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def getDivisors(goalNumber: int) -> list:
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divisors = []
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#Start by checking that the number is positive
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if(goalNumber <= 0):
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return divisors
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#If the number is 1 return just itself
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elif(goalNumber == 1):
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divisors.append(1)
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return divisors
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#Start at 3 and loop through all numbers < (goalNumber / 2 ) looking for a number that divides it evenly
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topPossibleDivisor = math.ceil(math.sqrt(goalNumber))
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possibleDivisor = 1
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while(possibleDivisor <= topPossibleDivisor):
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#If you find one add it and the number it creates to the list
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if((goalNumber % possibleDivisor) == 0):
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divisors.append(possibleDivisor)
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#Account for the possibility sqrt(goalNumber) being a divisor
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if(possibleDivisor != topPossibleDivisor):
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divisors.append(goalNumber / possibleDivisor)
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#Take care of a few occations where a number was added twice
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if(divisors[len(divisors) - 1] == (possibleDivisor + 1)):
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possibleDivisor += 1
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possibleDivisor += 1
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#Sort the list before returning for neatness
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divisors.sort()
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#Return the list
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return divisors
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#This function returns the numth Fibonacci number
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def getFib(goalSubscript: int) -> int:
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gen = fibGenerator() #The fibonacci number generator
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#Loop through the numbers up to the subscript we want
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for _ in range(1, goalSubscript):
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next(gen)
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#The next number is F(goalSub), return it
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return next(gen)
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#This function returns a list of all Fibonacci numbers <= num
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def getAllFib(goalNumber: int) -> list:
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gen = fibGenerator() #The Fibonacci number generator
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fibNums = [] #A list to save the Fibonacci numbers
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#If the number is <= 0 return an empty list
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if(goalNumber <= 0):
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return fibNums
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#There is at least one number in the list now
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fibNums.append(next(gen))
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#Loop to generate the rest of the Fibonacci numbers
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while(fibNums[len(fibNums) - 1] <= goalNumber):
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fibNums.append(next(gen))
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#At this point the most recent number is > goalNumber, so remove it
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fibNums.pop()
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return fibNums
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#This function returns the GCD of the two numbers sent to it
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def gcd(num1: int, num2: int) -> int:
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while((num1 != 0) and (num2 != 0)):
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if(num1 > num2):
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num1 %= num2
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else:
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num2 %= num1
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return num1 | num2
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#Returns the factorial of the number passed in
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def factorial(num: int) -> int:
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fact = 1
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for cnt in range(1, num + 1):
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fact *= cnt
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return fact
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#Converts a number to its binary equivalent
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def toBin(num: int) -> str:
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#Convert the number to a binary string
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return "{0:b}".format(num)
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