Project 1: The Game of Hog--Problem 9 (2 pt)问题

合集 - cs61a(5)

1.Project 1: The Game of Hog--Problem 8 (2 pt)问题2024-02-19

2.Project 1: The Game of Hog--Problem 9 (2 pt)问题2024-02-20

3.cs61a回顾2024-02-204.cs61a-How to draw an Environment Diagram（1.6）2024-02-215.CS61A阅读指南2024-03-01

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问题9

Implement max_scoring_num_rolls, which runs an experiment to determine the number of rolls (from 1 to 10) that gives the maximum average score for a turn. Your implementation should use make_averaged and roll_dice.

If two numbers of rolls are tied for the maximum average score, return the lower number. For example, if both 3 and 6 achieve a maximum average score, return 3.

You might find it useful to read the doctest and the example shown in the doctest for this problem before doing the unlocking test.

Important: In order to pass all of our tests, please make sure that you are testing dice rolls starting from 1 going up to 10, rather than from 10 to 1.

函数原型描述：

def max_scoring_num_rolls(dice=six_sided, samples_count=1000):
    """Return the number of dice (1 to 10) that gives the highest average turn score
    by calling roll_dice with the provided DICE a total of SAMPLES_COUNT times.
    Assume that the dice always return positive outcomes.
 
    >>> dice = make_test_dice(1, 6)
    >>> max_scoring_num_rolls(dice)
    1
    """
    # BEGIN PROBLEM 9
    "*** YOUR CODE HERE ***"
    # END PROBLEM 9

分析

这个函数想实现：给定一个dice()，然后确定这个骰子扔多少个（num_rolls）才能在函数make_averaged()中获取最大平均值。

在这举个例子，看下函数roll_dice(num_rolls, dice)以相同的点数策略dice()，但是在不同的骰子数量num_rolls的影响下，产生的结果。这里先固定下dice()函数的策略：

>>> dice = make_test_dice(3, 1, 5, 6)

然后看下roll_dice(num_rolls, dice)在不同的骰子数量num_rolls的影响下（这里仅举1、2、3数量，其他数量可以自行推算💦），产生的结果：

1. num_rolls = 1：反复调用roll_dice()函数，省略各个骰子的结果，可以得到分数分别为：3、1、5、6。各个结果的概率均为1/4。
2. num_rolls = 2：反复调用roll_dice()函数，省略各个骰子的结果，分数分别为1、11。得到每个结果的概率为1/2
3. num_rolls = 3：反复调用roll_dice()函数，骰子的结果分别为：3,1,5、6,3,1、5,6,3、1,5,6、3,1,5...这是4个得分循环，每次得分分别为：1、1、14、1。各个结果的概率均为1/4。

这个例子说明，函数roll_dice(num_rolls, dice)在相同的点数策略dice()，而在不同的骰子数量num_rolls的影响下，重复调用会出现完全不一致的结果。这样就会导致函数make_averaged()在选用相同点数策略但是不同骰子数量的roll_dice()时候，得到的结果不一致。

所以要确定：函数roll_dice(num_rolls, dice)以相同的点数策略dice()，但是以多少骰子数量num_rolls才能在函数make_averaged()获取到较大的平均值。

实现

def max_scoring_num_rolls(dice=six_sided, samples_count=1000):
    """Return the number of dice (1 to 10) that gives the highest average turn score
    by calling roll_dice with the provided DICE a total of SAMPLES_COUNT times.
    Assume that the dice always return positive outcomes.
 
    >>> dice = make_test_dice(1, 6)
    >>> max_scoring_num_rolls(dice)
    1
    """
    # BEGIN PROBLEM 9
    "*** YOUR CODE HERE ***"
    i = num_rolls = 1
    higher_avg_score = 0
    while i <= 10:
        avg_score = make_averaged(roll_dice, samples_count)(i, dice)
        if avg_score > higher_avg_score:
            higher_avg_score = avg_score
            num_rolls = i
        i += 1
    return num_rolls
    # END PROBLEM 9

这样便通过了。