Issue in not getting the expected output while iterating through array

Question

I have the following code. The output currently I am receiving is not the expected output. The pseudocode I am trying to solve is described below.

   for each i in 1 · · · N do
         TEi = fmob(Li)
         TCi= fc(Li)
         TUi =fd(Li) 
         return

enter image description here

Python code

def optimal_partition():                   
 TE=[10,1,3]
 TC=[2,3,1]
 TU=[2,3,1]

 N = len(TU)-1
 SUMS = [0] * N
 for j in range(N):
    for i in range(1, j + 1):
        SUMS[j] += TE[i]
    for k in range(j - 1, N + 1):
        SUMS[j] += TC[k]
    SUMS[j] += TU[j]
 return SUMS.index(min(SUMS))

For the above code, I need the expected output to be [16,15]. Thanks, help is highly appreciated.

What is the actual output? I'm guessing an error, so please include the full text of the error traceback. — quamrana
– quamrana, Commented Sep 14, 2020 at 13:35
@quamrana, Thanks, The expected output is [9, 10], not retrieving error but the expected output is not what want. — emaan daavis
– emaan daavis, Commented Sep 14, 2020 at 13:38
I'm guessing that if I copy and run the code you supply I will get an error. If you don't get an error, then you must be running different code. Please update the question with the actual code you have. — quamrana
– quamrana, Commented Sep 14, 2020 at 13:40
Hi @quamrana I have provided the complete code. please check. Thanks — emaan daavis
– emaan daavis, Commented Sep 14, 2020 at 13:44
Following your algorithm, SUMS should have 3 values in it (j = 1..3, or when we use zero-based indexing, j = 0..2), not 2 values as you currently have. — Stuart
– Stuart, Commented Sep 14, 2020 at 14:13

Stuart · Accepted Answer · 2020-09-14 14:20:00Z

1

This is a better implementation of the algorithm using list slices instead of nested loops:

def part_sum(TE, TC, TU, j):
    return sum(TE[:j+1]) + sum(TC[j+1:]) + TU[j]

def optimal_partition(TE, TC, TU):
    return min(range(len(TE)), key=lambda j: part_sum(TE, TC, TU, j)) 

TE = [10,1,3]
TC = [2,3,1]
TU = [2,3,1]
print("The sums are: ", [part_sum(TE, TC, TU, j) for j in range(3)])
print("The optimal partition is at:", optimal_partition(TE, TC, TU))

Note that there are 3 sums, not 2, and that the returned value for j uses zero-based indexing. If you want to return a one-based index then just add 1 to the optimal partition result.

answered Sep 14, 2020 at 14:20

Stuart

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Comments

quamrana · Accepted Answer · 2020-09-14 14:05:37Z

0

You are suffering from the 0-based indexing syndrome of computing.

Computer Science sometimes uses 0-based indexing and maths, it seems, uses 1-based indexing.

This program seems to give you the expected output:

def optimal_partition():                   
    TE=[10,1,3]
    TC=[2,3,1]
    TU=[2,3,1]

    N = len(TU) - 1
    SUMS = [0] * N
    for j in range(N):
        for i in range(j + 1):
            SUMS[j] += TE[i]
        for k in range(j + 1, N + 1):
            SUMS[j] += TC[k]
        SUMS[j] += TU[j]
    return SUMS

SUMS = optimal_partition()
print(SUMS)
print(SUMS.index(min(SUMS)))

Output:

[16, 15]
1

answered Sep 14, 2020 at 14:05

quamrana

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1 Comment

emaan daavis Over a year ago

thank you for your wonderful help @quamrana, this is exactly what I want has been struggling for a long time on this.

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