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MIT PRIMES/Art of Problem Solving

CROWDMATH 2018: Neural Codes

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Major Breakthrough (from #of intervals to #of codewords - one dimensional)
lilcritters   12
N Nov 23, 2018 by haha0201
I hope this has not been posted by anyone else...because then I'd be violating rules of CrowdMath...


First, define $P\prime$ to be the number of codewords possible with $P$ endpoints. For example, $2\prime=2 $ as in the diagram (assuming the segment is open, not closed):

IMAGE

giving the neural code [0,1].
However, $2\prime$ could also equal 3, as in this diagram (still assuming the endpoints are open, not closed):

IMAGE

giving the neural code [10, 00, 01]. Thus, $2\prime$=2 and 3. I found out that
$$  P\prime = \begin{cases} [\frac{P+1}{2}+1,P+2]\in \mathbb{N} & \text{if } P \equiv 1 (mod 2) \\ [\frac{P}{2},P+2]\in \mathbb{N} & \text{if } P \equiv 0 (mod 2) \end{cases}  $$for all $P$>0$\in\mathbb{N}$.

*$\mathbb{N}=1,2,3, \cdots$
12 replies
lilcritters
Jun 15, 2018
haha0201
Nov 23, 2018
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