计算一排数字（见上下文的详细信息） [英] calculate a row of numbers(see context for details)

问题描述

有两排数字，第1行是从0开始的连续数字，现在要求你填写第2行，以确保行2中的数字是correspoding数第1行中出现的第2行的时间

there are two rows of numbers, row 1 is consecutive numbers starting from 0, now ask you to fill out row 2 to make sure the number in row 2 is the times of correspoding number in row 1 appearing in row 2.

例如：

0 1 1 2 3 4 5 6 7 8 9

_ _ _ _ _ _ _ _ _ _

具体而言，我们使用 ROW1 为行1和 ROW2 为行2，我们填写< $ C C> ROW2 $，以确保它satisies： ROW2 [I] =计数（ROW2，ROW1 [I]）。 计数（ROW2，ROW1 [I]）表示 ROW1 [I] 中的频率计数ROW2 。

To be more specific, we use row1 for row 1 and row2 for row 2, we fill out row2 to make sure it satisies: row2[i] = count(row2, row1[i]). count(row2, row1[i]) means frequency count of row1[i] among row2.

推荐答案

在对1000运行该解决方案必须运行循环中的平均3.608倍

Out of 1000 runs this solution had to run the loop an average of 3.608 times

import random

def f(x):
    l = []
    for i in range(10):
        l.append(x.count(i))
    return l

fast = list(range(10))

while f(fast) != fast:
    fast = []
    slow = []
    for i in range(10):
        r = random.randint(0,9)
        fast.append(r)
        slow.append(r)
    while True:
        fast = f(f(fast))
        slow = f(slow)
        if fast == slow:
            break

print(fast)

函数f（x）取猜测，x和返回计数。我们基本上是寻找一个解决方案，使得F（X）= X。

f(x) takes a guess, x, and returns the counts. We are essentially looking for a solution such that f(x) = x.

我们首先从0-9中选择10个随机整数，做一个清单。我们的目标是要反复设置该列表等于其自身，直到我们找到一条溶液或碰上一个周期。要检查周期，我们用乌龟和头发的算法，这在2的速度移动。速度快的两倍一样快的速度慢。如果这些都是平等的，我们遇到了一个周期，从一个新的随机场景开始。

We first choose 10 random integers from 0-9 and make a list. Our goal is to repeatedly set this list equal to itself until we either find a solution or run into a cycle. To check for cycles, we use the Tortoise and the Hair algorithm, which move at 2 speeds. A fast speed which is twice as quick as the slow speed. If these are equal, we have run into a cycle and start from a new random scenario.

我跑过了几次，发现对于n> 6的通解（其中，在这种情况下，N = 10）。它的形式为[正4,2,1,0 ...，0,1,0,0,0]

I ran through this a few times, and found the general solution for n>6 (where in this case n = 10). It is of the form [n-4,2,1,0...,0,1,0,0,0]

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