My Works | Ackerman Function

Ackerman Function: Ackerman function is function with two arguments each of which can be assigned any nonnegative integer, such as 1,2,4,...... This function is defined as follows,
     (a) if m = 0, then A(m,n) = n + 1
     (b) if m > 0 but n = 0, then A(m,n) = A(m - 1,1)
     (c) if m > 0 and n > 0, then A(m,n) = A(m - 1,A(m,n - 1))

This function is recursive as it call itself in (b) and (c).

Before getting into code, let see an example using Ackerman Function. Supposing that in this example m = 1 and n = 2. So the function starts as follows,

(01) A(1,2) = A(0,A(1,1)) 'as it falls in condition (c) where m>0 and n>0. Here we got another call to 
                                          'A(1,1). Thus we will have to evaluate it first.
(02) A(1,1) = A(0,A(1,0))  'it also make another call to function A(1,0), so we now will evaluate A(1,0)
(03) A(1,0) = A(0,1)          'it falls in condition (b) and make a call to A(0,1)
(04) A(0,1) = 2                  'it falls in condition (a) and gives a result, 2. Now we can get back to step 
                                          'from where this A(0,1) was called, that is to step (03)
(05) A(1,0) = 2                  'Thus A(1,0) of step (03) gets a value so we van turn back to the step from 
                                          'where this A(1,0) was called that is step (02)
(06) A(1,1) = A(0,2)         'Well, unfortunately it results another call so we will have to do it first
(07) A(0,2) = 3                 'We get this value 3 from the call of step (06)
(08) A(1,1) = 3                 'thus we returned to step (06) and assigned 3 to the caller, that is A(1,1)
(09) A(1,2) = A(0,3)         'now we can move to the caller of A(1,1) at step (01) and it is producing 
                                         'another call to A(0,3), we will do it first
(10) A(0,3) = 4                 ' produces a value so we can get back to there from where A(0,3) was 
                                         'called, that is step (09)
(11) A(1,2) = 4                 'Here we assigned the value of A(0,3) to A(1,2) and we are done.

We can easily code this function. We just have to write a function inside which we will put
those above given conditions. Don't think too much and look at the following code,

The following downloadable program and code has been found error producing for dig entries, like when m=4 or when m is small but n is bigger. This error is because of Too Many Method Calls or Recursion. I am working on it and I hope that you, the readers will also try to modify it be error free.

Function ack(ByVal m As Integer, ByVal n As Integer) As Integer

        If m = 0 Then
            res = n + 1
        ElseIf n = 0 Then
            ack(m - 1, 1)
            ack(m - 1, ack(m, n - 1))
        End If

        Return res
End Function

Thus, in the above function we transfer our two arguments at m and n, and this function will return the result packed in the variable res. Isn't it so easy to implement !! Look, I haven't wrote a single line of code except the conditions those came with the definition of the Ackerman function.

You can download the project from here, but I hope that you will try it yourself first. Actually I have gave you 99% of the code already, you just have to write a statement which will pass the argument and will call the function and will keep the return to print it on the screen. However, you also download it. [Click Here to Download Ackerman Function]

Remember, I haven't done any kind of exception handling so just don't play with it, ok. Thanks for reading.

Recommended Recommends


Contact Us