Graphical to Mathematical representation of changing the order of some elements

AI Thread Summary
The discussion focuses on the mathematical representation of permutations through operations that change the order of elements in a set. It begins with a graphical approach to identify operations needed to transform an original container of numbers into a target configuration. The conversation highlights the importance of the order of operations, as they are not commutable, and introduces the concept of cycle notation for expressing permutations mathematically. Participants discuss the possibility of deriving a general expression for any N number group based on the original and final states. The dialogue concludes with a request for resources on a more efficient method that requires fewer swapping operations.
arsenal_51
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I have a question that is a little hard to explain, since i don't know the name of this method, but I'll try my best, if anyone knows the name please do tell me.

So let's say we have three numbers, 1 2 3 (in this order)
and we have a container for this numbers: C123
and we have some operations: O12, O13 and O23
each of these operations act on those numbers changing their positions.

For example O12 will change the position of the first and second elements.
So let's say: O12 . C123 will equal: C213

And if we want to find out what operations to use when we have the original Container and the target Container we can do it easily graphically.
For example:
Original: C123
Target: C231

This can be done graphically:
example1.jpg


The point where the lines intercept represent the operation between those two numbers. And the order is important, since these operations are not commutable.

So that's the same as: O12 . O23 . C123 = C231


One last example:
The container doesn't need to hold all of the numbers of the three number-space
Original: C12
Target: C31

example2.jpg


Or: O23 . O12 . C12 = C31


So graphically its easy to find out the operations of any N number-space.
But how do we express that in a mathematical general expression?
 
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I'm not exactly sure what you're asking. These are just basic permutations, so cycle notation should communicate everything that you need.
 
Thanks for your reply, I'm going to read about that.

I'm trying to find out the permutations needed to do mathematically for any N number group, knowing only the original and the final state.
Ideally something of the format: O1i . O2j . C12 = Cij
But for a N number group instead of just a this small example that might not even be correct.
 
Last edited:
arsenal_51 said:
Thanks for your reply, I'm going to read about that.

I'm trying to find out the permutations needed to do mathematically for any N number group, knowing only the original and the final state.
Ideally something of the format: O1i . O2j . C12 = Cij
But for a N number group instead of just a this small example that might not even be correct.

Your notation is somewhat unconventional, but I think you mean the following: Given an ordered set of the first N counting numbers, i.e. ##S = (s_1\,s_2\,s_3\,\ldots\,s_N)##, you want to find ##X## "swappings" of the form ##(f_1\,t_1),\ldots,(f_X\,t_X)## such that their product will take the ordered set ##(1\,2\,3\,\ldots\,N)## to ##S##, i.e.
$$
(1\,2\,3\,\ldots\,N)\cdot(f_1\,t_1)\cdot\ldots \cdot (f_X\,t_X)=(s_1\,s_2\,s_3\,\ldots\,i_N).
$$
This is possible, and you can construct the pairs ##(f_k\,t_k)## quite easily. I'll just hint by saying this much: choose ##(f_1\,t_1)## so that it swaps the elements ##1## and ##s_N##, and then let ##(f_2\,t_2)=(1\,N)##. This means that those two swapping put ##s_N## at position ##N##. In the next step you place ##s_{N-1}## into position ##N-1## etc. till you end up with the ordered set ##S## you wanted.

This way you may need ##X=2N-1## swappings, and it's actually possible to get from ##(1\,2\,3\,\ldots\,N)## to any ##(s_1\,s_2\,s_3\,\ldots\,s_N)## with only ##N-1## swappings, but not as easily as by my method (one element at a time goes into position ##1## and then to its proper place).
 
Thank you, i will try out your method, it seems pretty clear.

After i try that out i would like to check that other more efficient method you were talking about where you only need N - 1 swapping operations. Do you know where i can read more about that other method or the name of it?

Thanks again.
 
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