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Updated getPrimes and getNumPrimes to improve efficiency

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Matthew Ellison
2019-01-30 21:33:32 -05:00
parent 681a709e19
commit ad7a69ebc4
1 changed files with 23 additions and 7 deletions
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30
Algorithms.py
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@@ -1,7 +1,7 @@
#Python/myClasses/Algorithms.py #Python/myClasses/Algorithms.py
#Matthew Ellison #Matthew Ellison
# Created: 1-27-19 # Created: 1-27-19
#Modified: 1-29-19 #Modified: 1-30-19
#This is a file that contains a few algorithms that I have used several times #This is a file that contains a few algorithms that I have used several times
import math import math
@@ -20,11 +20,19 @@ def getPrimes(goalNumber: int):
primes.append(2) primes.append(2)
#We can now start at 3 and skip all even numbers, because they cannot be prime #We can now start at 3 and skip all even numbers, because they cannot be prime
for possiblePrime in range(3, goalNumber + 1, 2): #Need goalNumber + 1 to account for goalNumber being prime for possiblePrime in range(3, goalNumber + 1, 2): #Need goalNumber + 1 to account for goalNumber being prime
#See if the current possible prime is divisible by any prime less than it. If not it is a prime itself #Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
for primeNum in primes: primesCnt = 0
if((possiblePrime % primeNum) == 0): #We can safely assume that there will at lease be 1 element in the primes list because of 2 being added before the loop
topPossibleFactor = math.ceil(math.sqrt(possiblePrime))
while(primes[primesCnt] <= topPossibleFactor):
if((possiblePrime % primes[primesCnt]) == 0):
foundFactor = True foundFactor = True
break break
else:
primesCnt += 1
#Check if the index has gone out of range
if(primesCnt >= len(primes)):
break
#If you didn't find a factor then the current number must be prime #If you didn't find a factor then the current number must be prime
if(not foundFactor): if(not foundFactor):
@@ -50,12 +58,20 @@ def getNumPrimes(numberOfPrimes: int):
#Loop through every odd number starting at 3 until you reach the correct number of entries looking for a prime number #Loop through every odd number starting at 3 until you reach the correct number of entries looking for a prime number
possiblePrime = 3 #Holds the next possible prime number possiblePrime = 3 #Holds the next possible prime number
while((len(primes) < numberOfPrimes) and (possiblePrime > 0)): while((len(primes) < numberOfPrimes) and (possiblePrime > 0)):
#Loop through all primes we have already found, checking for a factor #Loop through all primes we have already found, up to sqrt(possiblePrime), checking for a factor
for primeNum in primes: primesCnt = 0
#We can safely assume that there will at lease be 1 element in the primes list because of 2 being added before the loop
topPossibleFactor = math.ceil(math.sqrt(possiblePrime))
while(primes[primesCnt] <= topPossibleFactor):
#If you find a factor the number is not a prime so raise the flag and break the loop #If you find a factor the number is not a prime so raise the flag and break the loop
if((possiblePrime % primeNum) == 0): if((possiblePrime % primes[primesCnt]) == 0):
foundFactor = True foundFactor = True
break break
else:
primesCnt += 1
#Check if the index has gone out of bounds and break the loop if it has
if(primesCnt >= len(primes)):
break
#If you don't find a factor then this number is prime so add it to the list #If you don't find a factor then this number is prime so add it to the list
if(not foundFactor): if(not foundFactor):