1

I'm a programmer and I need to write an algebraic notation for a LOOP made in one of our programs. I don't have Mathematica software, but just MathType to write formulas and notations.

The program executes 2 kinds of operation with bytes and bits sets:

BYTE OPERATION related to BIT POSITION and BIT OPERATION related to BIT POSITION

The first operation provides a kind of TRANSPOSITION of the bytes in the set. The second operation provides a kind of SUBSTITUTION of the bytes in the same set.

Let me make a sample of these ops:

I have two groups of numbers, (A) and (B):

A = {1,0,0,1,1,0} and B = {14,45,98,23,99,09)

In the first step, I get all 'bits' from (A) and order (B) if the related position bit of (A) = 1. In the second step, I do the same but considering the 0 bits. At the end, I do an UNION of the both sets.

Ex.: First Step: A = {1,0,0,1,1,0} - B={14,45,98,23,99,09) Result = {14,23,99}

Second Step: A = {1,0,0,1,1,0} - B={14,45,98,23,99,09) Result = {45,98,09}

Result = {14,23,99,45,98,09}

In the second operation (bit by bit) the same algorithm is performed BUT instead to get the bytes of (B) I select bit-by-bit for the (B) set. But, it's fundamentaly the same operation.

My questions:

1- Is it a Permutation, an Arragement, a Simple Set?

2- How do I make a math/algebraic notation of these operations?

3- Can I point the total possibilities of these operations as (B!)? Is it really a factorial, even if I'm not exploring ALL arragements but just the achieved by the interaction of (A)?

Thanks in advance for ANY kind of help. Kind regards, David

David BS
  • 111

1 Answers1

0

The answer to question $1$ is that it is neither.

The answer to question $2$ is you can make any notation you want.

As for question three I will answer the question when all of the numbers in $B$ are different.

Notice if $A$ is made up of consecutive ones and then consecutive zeroes the result is always going to be the same as $B$. As an example when $A=\{1,2,3,4,5,6\}$ and $B=\{1,1,1,0,0,0\}$ you'll get back $B$.

Excluding those two cases if you take different values for $A$ you get different results.

The reason is that if $B$ is not of the type described above it contains a sub-string $01$, and because of that $B$ is uniquely determined. (Try to see why this happens).

So there are $2^n-n$ ways to arrange them (where $n$ is the number of elements in the set). In your case it is $2^6-6=58$

Asinomás
  • 105,651
  • Hi Jorge. That's the point! Sets (A) and (B) ALWAYS will be different in any cases. The algorithm provides TRANSPOSITION and SUBSTITUTION and it's effective - except if (A) or (B) is totally zero/one, but it does not occurr in any case. The fact is: the numbers ARE really different everytime and neither fullfiled of 1's or 0's. The permutation and substitution is done 100% right, but I need to make it in a algebraic notation. – David BS Dec 17 '14 at 17:50
  • I was estuding something about ARRANGEMENTS and I found that "an arragement is a combination where the order is important". So, if I consider as the order the set (A), couldn't I explain that utilizing the classical formula for arregements but like: A = B!/(B-A)! ?? – David BS Dec 17 '14 at 21:34