Lateral Thinking XVI

[Holmes]

Please ladies and gentlemen, give a big applause to ... ... ... Lateral Thinking XVI

A computer cable has seven connectors, arranged in a perfect circle -- so by rotating the plug, it can be connected to the outlet in any of seven different ways. Each of the connectors is numbered from one to seven, each number being used exactly once. The same is true for the holes in the outlet. The device that uses this cable only requires that one of the connectors match up to its corresponding hole in order to operate. How should you number the connectors on the plug and the holes in the outlet so that, no matter how the cable is rotated and plugged in, at least one connector matches up?

[bash7353]

Spoiler:

I would arrange the numbers in the outer circle exactly counterwise to the inner circle. In that case you can rotate one circle and always have one corresponding number.

[ayw]

@googleearth, yes that will work.

The following also works:

Spoiler:

Shift each id number on the plug clockwise by its value:
Plug          Socket
      1            4
   7     2      7     1
  6       3    3       5
    5   4        6   2

I wonder how many different solutions there are (other than those for which the labels are interchanged). The same tricks seem to work for 9 connectors or any number of odd connectors (proof?). What about eight connectors, or any number of even connectors?

[Holmes]

That was quick, nice work!

[Sayumi]

I wonder how many different solutions there are

Okay, I wrecked my mind over that one for the last couple of hours...

there are at least 5solutions (including ayw's and googleearth's ones)

original circle (stolen from ayw ;-) )
sockets (I will refer to it as the "original circle")       
      1           
   7     2     
  6       3   
    5   4   

All of the following is about the plugs, the sockets remain unchanged!
1 is always were it originally was, starting from there clockwise:

Spoiler:

a) 1,7,6,5,4,3,2 (googleearth)
b) 1,5,2,6,3,7,4 (ayw)
c) 1,4,7,3,6,2,5
d) 1,3,5,7,2,4,6
e) 1,6,4,2,7,5,3

Here a lot of little mathematical curiosities I noticed while thinking about whether there are more solutions than these... not really significant but maybe interesting...

Spoiler:

When writing them in circles (I couldn't be bothered doing that here)  b) - c)  and  d) - e) are mirror images of one another. The mirror image of a) would be the "original circle". When adding up the number opposite of one another (with 1 having no corresponding number) they will always add up to 9.

When writing down how far each number was shifted from the "original circle" (1 always being shifted by 7 or 0), you can see that all of these are different possibilities (not just different labels).
(All numbers from 1-7 being used is another way of saying no matter how you rotate the circle one of the plugs will always match the socket)

A) 7,5,3,1,6,4,2 (1 was shifted by 7, 2 was shifted by 5, 3 was shifted by 3...) -clockwise of course) - = 1,6,4,2,7,5,3,1 = e (same sequence)
B) 7,1,2,3,4,5,6 -  same sequence as the original circle
C) 7,4,1,5,2,6,3 - same sequence as b
D) 7,3,6,2,5,1,4 - same sequence as c
E) 7,2,4,6,1,3,5 - same sequence as d

When writing these sequences in a circle, with 7 always being at the top, the numbers which are oppsite of one another will always add up
to 7. Here (written in a circle with 7 being on top) A) - E) and C) - D) are mirror images!

All of this is totally unsignificant to the answer but since there seem to be so many mathematical similarities and realtions between these 5 solutions I assume that these are the only ones-but I'm not sure 

Few.. all of this took me about 2 and a half hours and 4pages of notes in 3 different colors... but at least the downloads for the two new DC episodes have finished now so I'll go  and watch those   

[ayw]

Sayumi, thanks for sharing your notes and thoughts. Having read them with a blank sheet in my mind, it was immediately obvious that you had found all possible combinations.

Spoiler:

Taking our original circle and going around clockwise in steps of 1, 2, 3, ..., 6 etc we get the following six sequences, which you've already written down
steps of 1: 1 2 3 4 5 6 7
steps of 2: 1 3 5 7 2 4 6
steps of 3: 1 4 7 3 6 2 5
steps of 4: 1 5 2 6 3 7 4
steps of 5: 1 6 4 2 7 5 3
steps of 6: 1 7 6 5 4 3 2
(steps of 7 = steps of 1)
Note that each row contains the numbers 1..7 exactly once, and that each column beyond the first contains the numbers 2..7 exactly once. There are no other combination of numbers starting with 1 that one could add to these series without getting at least one duplicate in any of the columns beyond the first.

Any arbitrary shift of a row would lead to a match of exactly one number with every other row, and the matching numbers are different for each row. This proves that Sayumi's solutions are unique and complete 8).  The plug and socket can be any pair of series above, i.e. there are 6×5 possibilities .

For the proof, it's not important that the above series was generated with the "steps-of-n" method. For a plug with N connectors the only thing one needs to do is create a N-1×N-1 square matrix and fill it with numbers 2, 3, 4...N so that each number only occurs once per row and once per column. Attach a 1 in front of each row, and you have the series for all N-1×N-2 possible plug socket pairings. The reason the "steps-of-n" method worked for 7 connectors is because 7 is a prime number.

[Holmes]

To be honest I really didn´t think you´d be able to find the six possible combinations. I gave me a big surprise, I bow before you.