查找具有指定 collatz 序列长度的编号-解网

问：

我需要制作一个程序来查找具有指定 collatz 序列长度的数字。但是，总是存在程序太慢的问题。例如，我目前能得到的最佳分数是 collatz 序列长度为 1200 的数字（我需要能够获得 collatz 序列长度为 1800 的数字）。

我尝试了很多不同的方法，但到目前为止最好的方法是尝试重新创建 collatz 数字树。这是来自 wiki 的示例。正如我之前所说，我需要能够得到一个 collatz 序列长度为 1800 的数字，但我不能得到超过 1200 的数字。

这是我目前的解决方案（我知道这很复杂，但到目前为止我尝试过的其他方法只能使 collatz 序列长度达到 500）：

A = int(input())

limit = 1000000000000000000

def runCollaz(ciag):

    steps = 0

    while ciag != 1:
        
        if (ciag % 2 == 0):
            ciag /= 2
        else:
            ciag *= 3
            ciag += 1
        steps+=1
    return steps

def makeChainLess(number):
    if (number % 2 == 0):
        return number / 2
    else:
        return ((number * 3) + 1)

collatzTree = [[1, 1]]

finallAns = "None"

def getAns(collatzTree, what):    
    awnser = "None"

    if (collatzTree[0][0] < limit and collatzTree[0][1] == A):
        awnser = collatzTree[0][0]

    while (len(collatzTree) > 250):
        currentHigh = collatzTree[0][0]
        highIndex = 0
        index = 0
        for x in collatzTree:
            if (x[0] > currentHigh):
                currentHigh = x[0]
                highIndex = index
            index += 1
        collatzTree.pop(highIndex)
            

    if (collatzTree[0][0] > 4):
        if (collatzTree[0][0] - 1) % 3 == 0:
            if (collatzTree[0][0] - 1) % 2 != 0:
                collatzTree += [[(collatzTree[0][0] - 1) / 3, int(collatzTree[0][1]) + 1]]
            collatzTree += [[collatzTree[0][0] * 2, int(collatzTree[0][1]) + 1]]
            collatzTree.pop(0)
        else:
            collatzTree += [[collatzTree[0][0] * 2, int(collatzTree[0][1]) + 1]]
            collatzTree.pop(0)

    else:
        collatzTree += [[collatzTree[0][0] * 2, int(collatzTree[0][1]) + 1]]
        collatzTree.pop(0)
    if (what == "C"):
        return collatzTree
    else:
        return awnser

while finallAns == "None":
    finallAns = getAns(collatzTree, "A")
    collatzTree = getAns(collatzTree, "C")
print(int(finallAns))

如果有人能帮忙，我真的会很感激。

Python 数学 collatz

'''
This code finds numbers that have the specified Collatz sequence length
(by brute force)
'''
import time
start = time.time()                      #start code timer

n = 1                                        #set low end of range
for number in range (10000000, n-1, -1):   #work down from max range
     num = number
     trial = 0
     while number > 1:                 #set up looping until 1 is reached
            
         if number % 2 == 0:         #set up normal Collatz sequence comp.
              number = number // 2             
                             
         else:
              number = (number * 3) + 1
                           
         trial = trial+1                  #update counter for each loop
         if number == 1 and trial == 500 or trial == 600:  #set target numbers
                  #print all numbers where specified target occured:
              print("Number with sequence "
                    "length of {0}: {1}".format(trial, num))
                                         
if number == n:
     print("end of range")                #tells user that code has ended

end = time.time()
print("elapsed time: ",end - start)

查找具有指定 collatz 序列长度的编号

Finding number with specified collatz sequence length

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