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# /sci/ - Science & Math

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No.14596124 [Reply] [Original] [archived.moe]

Hi, Kim. I've got some mathematics questions for you.

I was looking at a clock face and it got me wondering about something. What are the minimum amount of numbers necessary in a set of numbers that would allow them to be scrambled so that:

A) No two consecutive numbers remain as such, including the first and last numbers (as though they were the ends of a line conjoined as a circle).
B) No number can remain in or in immediate proximity to the same place it was initially.

For example, five wouldn't work for several reasons:
1.) Simply reversing the order of the numbers 1-5 still leaves 3 as the third number in the list, 5&4, 4&3, 3&2, 2&1 are still consecutive, and 1&5 would still be able to loop back to each other (fails A and B).
2.) 31524 fails because 1 is second and 4 is fifth in this arrangement (fails B), and even though 3 and 4 are not at the ends of the original list, they could be considered "consecutive" in this context as 1 and 5 are normally (fails A).t

I came up with eight as the minimum number, with 35827146 as a satisfactory example (if not the only one). I almost thought seven would work as well. This brings me to my actual questions:

1.) Does eight being the minimum imply that any greater amount of numbers will also satisfy the conditions?
2.) Is this already a thing that mathematicians have worked out?
3.) If the answer is yes, what is it called?

 >> Anonymous Thu Jun 23 22:37:39 2022 No.14597542 >>14596124>Hi, Kim.Explain yourself.>rest of the postI have no idea what you are doing
 >> Kim Thu Jun 23 23:48:56 2022 No.14597669 >>14596124>Does eight being the minimum imply that any greater amount of numbers will also satisfy the conditions?Bare minimum it works for 19 and higher. Clearly it works for 16 (just do the 8 ordering on the first 8 and the last 8). Then for 17, put 17 in the 7th position and order the first 8 like we did before. Then we have to order 9-16 in the positions 10-17, which shouldn't be a problem, we're basically doing an 8 ordering with a free space.For n>18 we do the same process. Put 17-n into the 9-(n-16) spots, ordering to satisfy condition (A), as we get (B) for free. Then all that remains is to do the regular 8 ordering on 1-8 in their normal spots and 9-16 in the (n-8)-n spots.This means that we'd only need to check n=9-15,18 by hand
 >> Anonymous Thu Jun 23 23:50:20 2022 No.14597671 >>14597669Not OP, but good to see you Kim. How you been doing?
 >> Anonymous Thu Jun 23 23:50:51 2022 No.14597672 >>14597669Hey Kim! Long time.
 >> Kim Thu Jun 23 23:51:52 2022 No.14597673 >>14597671>>14597672aw fuck i didn't realize kim was an actual person someone call the fbi i did a identity theft
 >> Anonymous Fri Jun 24 01:32:19 2022 No.14597869 >>14597669https://pastebin.com/UtzVqU57Quick and dirty Python script confirms 9-15 and 18 all have permutations.
 >> Anonymous Fri Jun 24 02:03:02 2022 No.14597919 File: 2.34 MB, 3120x4160, IMG_20220624_115608.jpg [View same] [iqdb] [saucenao] [google] Here's some simple tree graphs showing why 7 isn't possibleThe key in the top-left shows all possible numbers for each position, and the trees show how each potential combination will eventually fail (either by reusing a number or running out of valid numbers to use)The closest combinations are 351624(7) and 571624(3), but they obviously don't fulfill the requirements of both A and B. I'm fairly confident 6 also fails, which means 8 is indeed the minimum
>>
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