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POFast way to place bits for puzzle
 

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  1. POFast way to place bits for puzzle
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    <p>There is a puzzle which I am writing code to solve that goes as follows.</p> <p>Consider a binary vector of length n that is initially all zeros. You choose a bit of the vector and set it to 1. Now a process starts that sets the bit that is the greatest distance from any 1 bit to $1$ (or an arbitrary choice of furthest bit if there is more than one). This happens repeatedly with the rule that no two 1 bits can be next to each other. It terminates when there is no more space to place a 1 bit. The goal is to place the initial 1 bit so that as many bits as possible are set to 1 on termination.</p> <p>Say n = 2. Then wherever we set the bit we end up with exactly one bit set.</p> <p>For n = 3, if we set the first bit we get 101 in the end. But if we set the middle bit we get 010 which is not optimal.</p> <p>For n = 4, whichever bit we set we end up with two set.</p> <p>For n = 5, setting the first gives us 10101 with three bits set in the end.</p> <p>For n = 7, we need to set the third bit to get 1010101 it seems. </p> <p>I have written code to find the optimal value but it does not scale well to large n. My code starts to get slow around n = 1000 but I would like to solve the problem for n around 1 million.</p> <pre><code>#!/usr/bin/python from __future__ import division from math import * def findloc(v): count = 0 maxcount = 0 id = -1 for i in xrange(n): if (v[i] == 0): count += 1 if (v[i] == 1): if (count &gt; maxcount): maxcount = count id = i count = 0 #Deal with vector ending in 0s if (2*count &gt;= maxcount and count &gt;= v.index(1) and count &gt;1): return n-1 #Deal with vector starting in 0s if (2*v.index(1) &gt;= maxcount and v.index(1) &gt; 1): return 0 if (maxcount &lt;=2): return -1 return id-int(ceil(maxcount/2)) def addbits(v): id = findloc(v) if (id == -1): return v v[id] = 1 return addbits(v) #Set vector length n=21 max = 0 for i in xrange(n): v = [0]*n v[i] = 1 v = addbits(v) score = sum([1 for j in xrange(n) if v[j] ==1]) # print i, sum([1 for j in xrange(n) if v[j] ==1]), v if (score &gt; max): max = score print max </code></pre>
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    1. COYour example for n = 7 is strange to me. You said you would set the 3rd bit but wouldn't you rather do 0000000 -> 1000000 -> 1000001 -> 1010001 -> 1010101 ?
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    2. CO@ShashankGupta The problem is that once you get to 1000001 the position that is furthest from any 1 is right in the middle so you end up with 1001001 which is not optimal. I don't believe there is any easy closed form solution sadly which is why I would like to speed up my code.
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    3. CO@user2179021, useful clarification! Thanks.
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      1. USTim Peters
 

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