贪心 Makespan 算法答案

【问题标题】：Greedy Makespan algorithm贪心 Makespan 算法
【发布时间】：2022-12-07 06:11:17
【问题描述】：

我需要在 python 中实现这个贪心算法，但我无法理解如何找到 M[j] 最少的“处理器”。下面提供的算法...

greedy_min_make_span(T, m):
  # T is an array of n numbers, m >= 2
  A = [Nil, ... , Nil] # Initialize the assignments to nil (array size n)
  M = [ 0, 0, ...., 0] # initialize the current load of each processor to 0 (array size m)
  for i = 1 to n
    find processor j for which M[j] is the least.
    A[i] = j
    M[j] = M[j] + T[i]
 # Assignment achieves a makespan of max(M[1], .. M[m])
 return A


def greedy_makespan_min(times, m):
    # times is a list of n jobs.
    assert len(times) >= 1
    assert all(elt >= 0 for elt in times)
    assert m >= 2
    n = len(times)
    # please do not reorder the jobs in times or else tests will fail.
    # Return a tuple of two things: 
    #    - Assignment list of n numbers from 0 to m-1
    #    - The makespan of your assignment
    A = n*[0]
    M = m*[0]
    
    i = 1
    for i in range(i, n):
        j = M.index(min(M))
        A[i] = j
        M[j] = M[j] + times[i]
    return (A, M)

修复：当我尝试将 A[i] 分配给 j 时，我现在遇到的错误是“列表分配索引超出范围”。

实用功能：

def compute_makespan(times, m, assign):
    times_2 = m*[0]
    
    for i in range(len(times)):
        proc = assign[i]
        time = times[i]
        times_2[proc] = times_2[proc] + time
    return max(times_2)

我有的测试用例...

def do_test(times, m, expected):
    (a, makespan) = greedy_makespan_min(times,m )
    print('\t Assignment returned: ', a)
    print('\t Claimed makespan: ', makespan)
    assert compute_makespan(times, m, a) == makespan, 'Assignment returned is not consistent with the reported makespan'
    assert makespan == expected, f'Expected makespan should be {expected}, your core returned {makespan}'
    print('Passed')
print('Test 1:')
times = [2, 2, 2, 2, 2, 2, 2, 2, 3] 
m = 3
expected = 7
do_test(times, m, expected)

print('Test 2:')
times = [1]*20 + [5]
m = 5
expected =9
do_test(times, m, expected)

现在我没有通过测试用例。我返回的作业与报告的完工时间不一致。我返回的分配是：[0, 0, 1, 2, 0, 1, 2, 0, 1]，我声称的 makespan 是：[6, 7, 4]。当我期望 7 时，我的计算 makespan 返回 8。我在执行此算法时有什么想法是错误的吗？

【问题讨论】：

标签： python algorithm greedy

【解决方案1】：

将 A = n*[] 更改为 A = n*[0]。

A = n*[] 不会创建长度为n 的列表，而是创建一个空列表。由于您在每次迭代中分配 A[i] = j，因此更改在功能上不会对输出产生任何影响。

【讨论】：