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109 lines
3.3 KiB
Python
109 lines
3.3 KiB
Python
#ProjectEuler/Python/Problem14.py
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#Matthew Ellison
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# Created: 01-31-19
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#Modified: 07-24-21
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"""
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The following iterative sequence is defined for the set of positive integers:
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n → n/2 (n is even)
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n → 3n + 1 (n is odd)
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Which starting number, under one million, produces the longest chain?
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"""
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#Unless otherwise listed, all of my non-standard imports can be gotten from my pyClasses repository at https://bitbucket.org/Mattrixwv/pyClasses
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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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from Problems.Problem import Problem
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class Problem14(Problem):
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#Variables
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__topNum = 1000000 - 1 #The largest number that you will check against the chain
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#Functions
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#Constructor
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def __init__(self) -> None:
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super().__init__("Which starting number, under one million, produces the longest chain using the itterative sequence?")
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self.maxLength = 0
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self.maxNum = 0
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#Operational functions
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#Solve the problem
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def solve(self) -> None:
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#If the problem has already been solved do nothing and end the function
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if(self.solved):
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return
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#Start the timer
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self.timer.start()
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#Loop through all number <= topNum and check them against the series
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for currentNum in range(1, self.__topNum + 1):
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currentLength = self.checkSeries(currentNum)
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#If the current number has a longer series than the max then the current becomes the max
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if(currentLength > self.maxLength):
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self.maxLength = currentLength
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self.maxNum = currentNum
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#Stop the timer
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self.timer.stop()
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#Throw a flag to show the problem is solved
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self.solved = True
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#This function follows the rules of the sequence and returns its length
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def checkSeries(self, num: int) -> int:
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length = 1 #Start at 1 because you need to count the starting number
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#Follow the series, adding 1 for each step you take
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while(num > 1):
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if((num % 2) == 0):
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num = num / 2
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else:
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num = (3 * num) + 1
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length += 1
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return length
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#Reset the problem so it can be run again
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def reset(self) -> None:
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super().reset()
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self.maxLength = 0
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self.maxNum = 0
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#Gets
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#Returns the result of solving the problem
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def getResult(self) -> str:
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self.solvedCheck("result")
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return f"The number {self.maxNum} produced a chain of {self.maxLength} steps"
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#Returns the length of the requested chain
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def getLength(self) -> int:
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self.solvedCheck("length of the longest chain")
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return self.maxLength
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#Returns the starting number of the requested chain
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def getStartingNumber(self) -> int:
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self.solvedCheck("starting number of the longest chain")
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return self.maxNum
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"""Results:
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The number 837799 produced a chain of 525 steps
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It took an average of 26.591 seconds to run this problem through 100 iterations
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"""
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