Generalizing increasing number of nested loop algorithm

Question

Sorry for the terrible title, but I have no clue on how to generalize (or simplify) my loop case here.

I have a program that iterates to a sequence of integer, for example dimension=1 to 5. In each iteration, there will be a main loop, and inside the main loop, there will be a nested loop. The number of the nested loop will be [dimension]. For example, in dimension=1, there is a For loop. In dimension=2, there is a For loop inside a For loop. And so on.

Is there any possible way to simplify the algorithm? currently I'm manually write totally different code for each value of [dimension]. Imagine if dimension=1 to 100? I'll be dead.

Here's my piece of program (written in VB.NET)

for dimension=2

Dim result(2) As Integer
For i = 0 To 1
    For j = 0 To 1
        result(0)=i
        result(1)=j
    Next
Next

For dimension=3

 Dim result(3) As Integer
 For i = 0 To 1
     For j = 0 To 1
         For k = 0 To 1
             result(0)=i
             result(1)=j
             result(2)=k
         Next
     Next
 Next

For dimension=4

Dim result(4) As Integer
For i = 0 To 1
    For j = 0 To 1
        For k = 0 To 1
            For l = 0 To 1
                result(0)=i
                result(1)=j
                result(2)=k
                result(3)=l
            Next
        Next
   Next
Next

And so on..

Any suggestion? Thanks!

Related: Finding All Combinations (cartesian product) of list values in vb.net — Kevin
– Kevin, Commented Feb 17, 2016 at 18:12

user4668606 · Accepted Answer · 2016-02-17 18:28:19Z

2

There are plenty of solutions:

Recursion

Idk, if vb.net supports methods, but if it does, this would probably be the simplest:

void nestedLoop(int lower , int upper , int remaining_loops , int[] values)
    if(remaining_loops == 0)
        //process values list
    else
        for int i in [lower , upper)
            values[remaining_loops] = i
            nestedLoop(lower , upper , remaining_loops - 1)

Integer Transformation

In theory, a number can be represented by any radix:

d_i * radix ^ i + d_i-1 * radix ^ (i - 1) ... + d_0 * radix ^ 0

Consider each digit the value of one of the nested loops:

for int i in [0 , max)
    for int j in [0 , max)
        for int k in [0 , max)
            ...

Could be represented by a 3-digit number with radix max, where d_0 = i, d_1 = j, etc.. Basically how each digit is mapped to one of the values can be arbitrary and will only affect the order of the output.

void nestedLoops(int upper , int dimension)
    for int i in [0 , pow(upper , dimension))
        int[] values
        int digit_sub = 1

        int tmp = i
        for int j in [0 , dimension)
            values[j] = tmp % dimension
            tmp /= dimension

        //all values of the loops are now in values
        //process them here

There would be a few other options aswell, but these are the most common.

answered Feb 17, 2016 at 18:28

user4668606

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2 Comments

Mantas Anni Over a year ago

Thanks, working on the recursion methods adaptation for my program.

user4668606 Over a year ago

@MantasAnni glad to help you :)

Community · Accepted Answer · 2017-05-23 12:31:10Z

0

Please do note that when you do

Dim result(2) As Integer

You are actually declaring an array of 3 elements see this question for why. It's a subtle difference in VB.NET

That being said, I'll assume that you meant to declare an array of only 2 elements. If this is the case then you could build and call a recursive function like this

LoopOver(result) 

Sub LoopOver(ByRef array() As Integer, ByVal Optional level As Integer = 0)
    If array.Length = level Then
        Return
    Else
        array(level) = 1
        LoopOver(array, level + 1)
    End If
End Sub

This recursive function will call itself (i.e., it will loop) for as many times as the array's size.

edited May 23, 2017 at 12:31

CommunityBot

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answered Feb 17, 2016 at 18:56

Mr.Z

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2 Comments

Mantas Anni Over a year ago

Huh, that explains why my program return zero instead of crashing when I accidentally access the array(array.count). Thanks, working on the recursive subroutine.

Mr.Z Over a year ago

Best of luck! If you've found my answer somewhat useful please consider voting it up.

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