Fast way to place bits for puzzle

Question

There is a puzzle which I am writing code to solve that goes as follows.

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.

Say n = 2. Then wherever we set the bit we end up with exactly one bit set.

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.

For n = 4, whichever bit we set we end up with two set.

For n = 5, setting the first gives us 10101 with three bits set in the end.

For n = 7, we need to set the third bit to get 1010101 it seems.

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.

#!/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 > maxcount):
                maxcount = count
                id = i
            count = 0

#Deal with vector ending in 0s
    if (2*count >= maxcount and count >= v.index(1) and count >1):
        return n-1
#Deal with vector starting in 0s
    if (2*v.index(1) >= maxcount and v.index(1) > 1):
        return 0
    if (maxcount <=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 > max):
        max = score       
print max

Answer 1

Latest answer (O(log n) complexity)

If we believe the conjecture by templatetypedef and Aleksi Torhamo (update: proof at the end of this post), there is a closed form solution count(n) calculable in O(log n) (or O(1) if we assume logarithm and bit shifting is O(1)):

Python:

from math import log

def count(n): # The count, using position k conjectured by templatetypedef
    k = p(n-1)+1
    count_left = k/2
    count_right = f(n-k+1)
    return count_left + count_right

def f(n): # The f function calculated using Aleksi Torhamo conjecture
    return max(p(n-1)/2 + 1, n-p(n-1))

def p(n): # The largest power of 2 not exceeding n
    return 1 << int(log(n,2)) if n > 0 else 0

C++:

int log(int n){ // Integer logarithm, by counting the number of leading 0
    return 31-__builtin_clz(n);
}

int p(int n){ // The largest power of 2 not exceeding n
    if(n==0) return 0;
    return 1<<log(n);
}

int f(int n){ // The f function calculated using Aleksi Torhamo conjecture
    int val0 = p(n-1);
    int val1 = val0/2+1;
    int val2 = n-val0;
    return val1>val2 ? val1 : val2;
}

int count(int n){ // The count, using position k conjectured by templatetypedef
    int k = p(n-1)+1;
    int count_left = k/2;
    int count_right = f(n-k+1);
    return count_left + count_right;
}

This code can calculate the result for n=100,000,000 (and even n=1e24 in Python!) correctly in no time¹.

I have tested the codes with various values for n (using my O(n) solution as the standard, see Old Answer section below), and they still seem correct.

This code relies on the two conjectures by templatetypedef and Aleksi Torhamo². Anyone wants to proof those? =D (Update 2: PROVEN)

¹_{By no time, I meant almost instantly}
²_{The conjecture by Aleksi Torhamo on f function has been empirically proven for n<=100,000,000}

---

Old answer (O(n) complexity)

I can return the count of n=1,000,000 (the result is 475712) in 1.358s (in my iMac) using Python 2.7. Update: It's 0.198s for n=10,000,000 in C++. =)

Here is my idea, which achieves O(n) time complexity.

The Algorithm

Definition of f(n)

Define f(n) as the number of bits that will be set on bitvector of length n, assuming that the first and last bit are set (except for n=2, where only the first or last bit is set). So we know some values of f(n) as follows:

f(1) = 1
f(2) = 1
f(3) = 2
f(4) = 2
f(5) = 3

Note that this is different from the value that we are looking for, since the initial bit might not be at the first or last, as calculated by f(n). For example, we have f(7)=3 instead of 4.

Note that this can be calculated rather efficiently (amortized O(n) to calculate all values of f up to n) using the recurrence relation:

f(2n) = f(n)+f(n+1)-1
f(2n+1) = 2*f(n+1)-1

for n>=5, since the next bit set following the rule will be the middle bit, except for n=1,2,3,4. Then we can split the bitvector into two parts, each independent of each other, and so we can calculate the number of bits set using f( floor(n/2) ) + f( ceil(n/2) ) - 1, as illustrated below:

n=11              n=13
10000100001       1000001000001
<---->            <----->
 f(6)<---->        f(7) <----->
      f(6)               f(7)

n=12              n=14
100001000001      10000010000001
<---->            <----->
 f(6)<----->       f(7) <------>
      f(7)                f(8)

we have the -1 in the formula to exclude the double count of the middle bit.

Now we are ready to count the solution of original problem.

Definition of g(n,i)

Define g(n,i) as the number of bits that will be set on bitvector of length n, following the rules in the problem, where the initial bit is at the i-th bit (1-based). Note that by symmetry the initial bit can be anywhere from the first bit up to the ceil(n/2)-th bit. And for those cases, note that the first bit will be set before any bit in between the first and the initial, and so is the case for the last bit. Therefore the number of bit set in the first partition and the second partition is f(i) and f(n+1-i) respectively.

So the value of g(n,i) can be calculated as:

g(n,i) = f(i) + f(n+1-i) - 1

following the idea when calculating f(n).

Now, to calculate the final result is trivial.

Definition of g(n)

Define g(n) as the count being looked for in the original problem. We can then take the maximum of all possible i, the position of initial bit:

g(n) = maxi=1..ceil(n/2)(f(i) + f(n+1-i) - 1)

Python code:

import time
mem_f = [0,1,1,2,2]
mem_f.extend([-1]*(10**7)) # This will take around 40MB of memory
def f(n):
    global mem_f
    if mem_f[n]>-1:
        return mem_f[n]
    if n%2==1:
        mem_f[n] = 2*f((n+1)/2)-1
        return mem_f[n]
    else:
        half = n/2
        mem_f[n] = f(half)+f(half+1)-1
        return mem_f[n]

def g(n):
    return max(f(i)+f(n+1-i)-1 for i in range(1,(n+1)/2 + 1))

def main():
    while True:
        n = input('Enter n (1 <= n <= 10,000,000; 0 to stop): ')
        if n==0: break
        start_time = time.time()
        print 'g(%d) = %d, in %.3fs' % (n, g(n), time.time()-start_time)

if __name__=='__main__':
    main()

Complexity Analysis

Now, the interesting thing is, what is the complexity of calculating g(n) with the method described above?

We should first note that we iterate over n/2 values of i, the position of initial bit. And in each iteration we call f(i) and f(n+1-i). Naive analysis will lead to O(n * O(f(n))), but actually we used memoization on f, so it's much faster than that, since each value of f(i) is calculated only once, at most. So the complexity is actually added by the time required to calculate all values of f(n), which would be O(n + f(n)) instead.

So what's the complexity of initializing f(n)?

We can assume that we precompute every value of f(n) first before calculating g(n). Note that due to the recurrence relation and the memoization, generating the whole values of f(n) takes O(n) time. And the next call to f(n) will take O(1) time.

So, the overall complexity is O(n+n) = O(n), as evidenced by this running time in my iMac for n=1,000,000 and n=10,000,000:

> python max_vec_bit.py
Enter n (1 <= n <= 10,000,000; 0 to stop): 1000000
g(1000000) = 475712, in 1.358s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
>
> <restarted the program to remove the effect of memoization>
>
> python max_vec_bit.py
Enter n (1 <= n <= 10,000,000; 0 to stop): 10000000
g(10000000) = 4757120, in 13.484s
Enter n (1 <= n <= 10,000,000; 0 to stop): 6745231
g(6745231) = 3145729, in 3.072s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0

And as a by-product of memoization, the calculation of lesser value of n will be much faster after the first call to large n, as you can also see in the sample run. And with language better suited for number crunching such as C++, you might get significantly faster running time

I hope this helps. =)

The code using C++, for performance improvement

The result in C++ is about 68x faster (measured by clock()):

> ./a.out
Enter n (1 <= n <= 10,000,000; 0 to stop): 1000000
g(1000000) = 475712, in 0.020s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
>
> <restarted the program to remove the effect of memoization>
>
> ./a.out
Enter n (1 <= n <= 10,000,000; 0 to stop): 10000000
g(10000000) = 4757120, in 0.198s
Enter n (1 <= n <= 10,000,000; 0 to stop): 6745231
g(6745231) = 3145729, in 0.047s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0

Code in C++:

#include <cstdio>
#include <cstring>
#include <ctime>

int mem_f[10000001];
int f(int n){
    if(mem_f[n]>-1)
        return mem_f[n];
    if(n%2==1){
        mem_f[n] = 2*f((n+1)/2)-1;
        return mem_f[n];
    } else {
        int half = n/2;
        mem_f[n] = f(half)+f(half+1)-1;
        return mem_f[n];
    }
}

int g(int n){
    int result = 0;
    for(int i=1; i<=(n+1)/2; i++){
        int cnt = f(i)+f(n+1-i)-1;
        result = (cnt > result ? cnt : result);
    }
    return result;
}

int main(){
    memset(mem_f,-1,sizeof(mem_f));
    mem_f[0] = 0;
    mem_f[1] = mem_f[2] = 1;
    mem_f[3] = mem_f[4] = 2;
    clock_t start, end;
    while(true){
        int n;
        printf("Enter n (1 <= n <= 10,000,000; 0 to stop): ");
        scanf("%d",&n);
        if(n==0) break;
        start = clock();
        int result = g(n);
        end = clock();
        printf("g(%d) = %d, in %.3fs\n",n,result,((double)(end-start))/CLOCKS_PER_SEC);
    }
}

---

Proof

note that for the sake of keeping this answer (which is already very long) simple, I've skipped some steps in the proof

Conjecture of Aleksi Torhamo on the value of f

For `n>=1`, prove that:  
f(2n+k) = 2n-1+1 for k=1,2,…,2n-1  ...(1)  
f(2n+k) = k     for k=2n-1+1,…,2n  ...(2)
given f(0)=f(1)=f(2)=1

The result above can be easily proven using induction on the recurrence relation, by considering the four cases:

• Case 1: (1) for even k
• Case 2: (1) for odd k
• Case 3: (2) for even k
• Case 4: (2) for odd k

Suppose we have the four cases proven for n. Now consider n+1.
Case 1:
f(2n+1+2i) = f(2n+i) + f(2n+i+1) - 1, for i=1,…,2n-1
          = 2n-1+1 + 2n-1+1 - 1
          = 2n+1

Case 2:
f(2n+1+2i+1) = 2*f(2n+i+1) - 1, for i=0,…,2n-1-1
            = 2*(2n-1+1) - 1
            = 2n+1

Case 3:
f(2n+1+2i) = f(2n+i) + f(2n+i+1) - 1, for i=2n-1+1,…,2n
          = i + (i+1) - 1
          = 2i

Case 4:
f(2n+1+2i+1) = 2*f(2n+i+1) - 1, for i=2n-1+1,…,2n-1
            = 2*(i+1) - 1
            = 2i+1

So by induction the conjecture is proven.

Conjecture of templatetypedef on the best position

For n>=1 and k=1,…,2n, prove that g(2n+k) = g(2n+k, 2n+1)
That is, prove that placing the first bit on the 2n+1-th position gives maximum number of bits set.

The proof:

First, we have
g(2n+k,2n+1) = f(2n+1) + f(k-1) - 1

Next, by the formula of f, we have the following equalities:
f(2n+1-i) = f(2n+1), for i=-2n-1,…,-1
f(2n+1-i) = f(2n+1)-i, for i=1,…,2n-2-1
f(2n+1-i) = f(2n+1)-2n-2, for i=2n-2,…,2n-1

and also the following inequality:
f(k-1+i) <= f(k-1), for i=-2n-1,…,-1
f(k-1+i) <= f(k-1)+i , for i=1,…,2n-2-1
f(k-1+i) <= f(k-1)+2n-2, for i=2n-2,…,2n-1

and so we have:
f(2n+1-i)+f(k-1+i) <= f(2n+1)+f(k-1), for i=-2n-1,…,2n-1

Now, note that we have:
g(2n+k) = maxi=1..ceil(2n-1+1-k/2)(f(i) + f(2n+k+1-i) - 1)
       <= f(2n+1) + f(k-1) - 1
        = g(2n+k,2n+1)

And so the conjecture is proven.

Answer 2

So in a break with my normal tradition of not posting algorithms I don't have a proof for, I think I should mention that there's an algorithm that appears to be correct for numbers up to 50,000+ and runs in O(log n) time. This is due to Sophia Westwood, who I worked on this problem with for about three hours today. All credit for this is due to her. Empirically it seems to work beautifully, and it's much, much faster than the O(n) solutions.

One observation about the structure of this problem is that if n is sufficiently large (n ≥ 5), then if you put a 1 anywhere, the problem splits into two subproblems, one to the left of the 1 and one to the right. Although the 1s might be placed in the different halves at different times, the eventual placement is the same as if you solved each half separately and combined them back together.

The next observation is this: suppose you have an array of size 2^k + 1 for some k. In that case, suppose that you put a 1 on either side of the array. Then:

• The next 1 is placed on the other side of the array.
• The next 1 is placed in the middle.
• You now have two smaller subproblems of size 2^k-1 + 1.

The important part about this is that the resulting bit pattern is an alternating series of 1s and 0s. For example:

• For 5 = 4 + 1, we get 10101
• For 9 = 8 + 1, we get 101010101
• For 17 = 16 + 1, we get 10101010101010101

The reason this matters is the following: suppose you have n total elements in the array and let k be the largest possible value for which 2^k + 1 ≤ n. If you place the 1 at position 2^k + 1, then the left part of the array up to that position will end up getting tiled with alternating 1s and 0s, which puts a lot of 1s into the array.

What's not obvious is that placing the 1 bit there, for all numbers up to 50,000, appears to yield an optimal solution! I've written a Python script that checks this (using a recurrence relation similar to the one @justhalf) and it seems to work well. The reason that this fact is so useful is that it's really easy to compute this index. In particular, if 2^k + 1 ≤ n, then 2^k ≤ n - 1, so k ≤ lg (n - 1). Choosing the value ⌊lg (n - 1) ⌋ as your choice of k then lets you compute the bit index by computing 2^k + 1. This value of k can be computed in O(log n) time and the exponentiation can be done in O(log n) time as well, so the total runtime is Θ(log n).

The only issue is that I haven't formally proven that this works. All I know is that it's right for the first 50,000 values we've tried. :-)

Hope this helps!