Thread: Proof of permutation in sequential sets, also question on opposite ability...

1. Proof of permutation in sequential sets, also question on opposite ability...

I have two things here. First, I was looking for a way to arrive at a set of increasing numbers, like a permutation, where the set never runs into a situation which the permutation of any part of the set can collide...

I haven't gone far enough to figure out if there is a proof to do that, but I have an interesting proof in permutations that you might enjoy if none of you have stumbled on it. Basically, when I was trying to find the above, and stopped for a while, I found this which is Not what I want...

Take the following examples and you will see what I mean...

(a) in a descending set, pos 1&5 sum collides with 2&4. There is no way to distinguish the set no matter the starting integer, or any type of inversion...
(- is the first sum, + is the second sum, for brevity...)

Code:
(a1)		(a2)		(a3)		(a4)		(a5)		(a6)
-1	1+	+1	1-	-1	2+	+1	2-	-1	2-	+1	2+
+2	2-	-2	2+	+2	3-	-2	3+	+2	3+	-2	3-
+3	3-	+3	3-	+3	4-	+3	4-	+3	4+	+3	4+
+4	4-	-4	4+	+4	5-	-4	5+	+4	5+	-4	5-
-5	5+	+5	5-	-5	6+	+5	6-	-5	6-	+5	6+
-----	-----	-----	----	----	----	----	----	----	----	----	----
6	9	6	9	6	12	6	12	6	8	6	8
9	6	9	6	9	8	9	8	9	12	9	12

So we have found this:
(a1,1,2) = (a2,1,2); (a3,1,2 = a4,1,2); (a5,1,2 = a6,1,2); ....

So then we can conclude (am I using the right terms?) that any sequential set that has a permutation collision, assuming the same positions are used in the set, is congruent - a collision will always exist between those sets no matter the starting integer or whether you do an intersection or inversion of the permutation (all based on positions of the set).

Now, I was wondering if you all have ever come across this before, and if so, what is the mathematical proof going to be in mathematical terms?

Then, notice this as well:

(b) even using odd/even progression, permutation collisions of sets results in the same dillemma...
Code:
(b1)		(b2)
-1	2+	+1	2+
+3	4-	-3	4-
+5	6-	-5	6-
-7	8+	+7	8+
----	----	----	----
8	10	8	10
8	10	8	10
Here we still note (b1,1 = b1,2), (b2,1 = b2,2), (b1,1 = b2,1), (b1,2 = b2,2).. etc... There is no distinction no matter how you arrange this...

So what have we just concluded, does this form a mathematical proof for permutations about colliding permutations... i.e. a permutation result has numerous possibilities, that given any set are indistinguishable.

Then we have the question I was going to see about, to see if you all already have the answer:

(c) question domain: Is there an ordered set so that no permutation result of any members of the set collide? i.e. Pa = Pb?

Thanks, this was kind of interesting when I worked it out on paper....

2. Re: Proof of permutation in sequential sets, also question on opposite ability...

(sorry about how sloppy the tables turned out, I couldn't copy/paste from excel, or export to an image, or upload attachments, etc. so the best I could do was wrap the tables in code tags for approximate monospace...)

First, I was attempting to explain that given two ordered integral sets, a partition collision is indistinguishable regardless of the method used (intersection or inversion) and regardless of the starting integer of the set. The positions in the set gives away its collision result in a permutation for any ascending integer set....

Then, we showed that this holds true for competing odd/even sets - the permutation collision there is indistinguishable given the positions in the set that collide and a properly progressive set.

Finally, the question I was trying to solve was - is there a way to generate an integral set so that all possible permutations are unique and calculatable? If so, what is that method?

If none of you have seen this before, well, I guess we may have a new proof in permutations of sets that needs to be mathematically represented...

Regards, and thanks for taking the time to read and provide any input...