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Place numbers around a circle to make all length 3 binary strings. Question
There are 8 binary strings of length 3.
They are:
000
001
010
011
100
101
110
111
The goal of this puzzle is to arrange some 0’s and some 1’s around a circle (using as few 0’s and 1’s as possible) so that all 8 binary strings listed above appear at least once around the circle.
A binary string is considered to appear around the circle if the 3 numbers of the binary string are located in three consecutive clockwise positions around the circle.
In the example below, the binary string 110 appears starting at the 1 marked with a star. But the binary string 101 does not appear anywhere.
2 answers
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| will.octagon.gibson | (no comment) | Mar 5, 2026 at 05:43 |
There is already an answer, so I'll show a way the answer can be derived with simple logic without requiring fancy math most here have probably never heard of.
Since there are 8 unique patterns, there must be at least 8 positions in the circle. Each position is then the starting point for one of the 3-digit patterns. At this point we don't know that the problem can be solved with only 8 digits (circle positions), but we know trying for less is pointless. We will therefore try with 8 digits, at least to start.
Since the 8 required patterns include 000 and 111, those must each appear in the final 8-digit sequences. This leaves us with only three options:
000111-- 000-111- 000--111
Of these, we can rule out the middle one. No matter what we set the remaining unknown digits to, we won't be able to get 101 and 010. That leaves:
000111-- 000--111
Given these sequences, there is only one way each can be filled in so that the pattern 101 exists:
00011101 00010111
If 8 digits is enough, then at least one of these sequences must work.
After checking all 8 patterns against these sequences, we find that they both work.
Note that since the digits are supposed to be arranged in a circle, which has no start or end, any rotation of a sequence is really the same sequence. For example, the following are all the same:
00011101 10001110 01000111 10100011 11010001 11101000 01110100 00111010
Of course this also applies to the other sequence. We can see that the two sequences are therefore the same as in Moshi's answer.
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The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| will.octagon.gibson | (no comment) | Mar 3, 2026 at 21:21 |
This can be considered a de Bruijn sequence, specifically $B(2,3)$, which has two solutions: $00010111$ and $11101000$.
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