TL;DR

Where we solve Perl Weekly Challenge #079 task #1.

Let’s try to adopt a systematic approach!

# The challenge

It’s stated in a simple yet somehow vague form, like always:

# The solution

There is a quick and dirty solution that addresses the problem with brute force:

sub count_bits_brute_force ($N,$m = 1000000007) {
my $n_bits = 0;$n_bits = ($n_bits + (sprintf('%b',$_) =~ tr/1/1/)) % $m for 1 ..$N;
return $n_bits; }  It’s a pretty straightforward translation of the input challenge: we iterate over all integers from 1 to the input $N and count the bits at each iteration, accumulating their count in $n_bits making sure that we perform the modulus operation at each iteration. This is necessary because the count might well exceed the maximum integer value we can represent (whatever it is), simply because the count of set bits for the maximum integer value will be way above it (hence, it’s not representable directly). OK, that works but it’s not that good: the iteration goes with$O(N)$and each step goes with something that might arguably be about$O(log(N))$, because the number of bits we get transforming $N in binary representation is proportional to $log(N)$. Urgh.

As it is, anyway, it’s a pretty decent solution to start with:

• it’s so simple that it’s quickly coded and hardly bugged
• it provides us a reference for checking more sophisticated solutions

so it does its job! Additionally, on my virtual machine it behaves in human time for inputs up about ten million, so depending on the boundaries of the problem (see the questions) it might solve the problem and let us move ahead.

But… let’s assume that we want to actually have a function that works in reasonable time for any integer input that our programming language can support natively. It’s time to call the math to the rescue.

One insight is that all numbers of the type $2^k-1$ are very easily calculated. Their binary representation is a sequence of $k$ 1s, and below them they have all possible sequences of $k$ bits. Yes, we can also count the sequence of $k 0s, because it does not alter our calculation (as it does not contain any 1). Why are$2^k$numbers easy? Simply because in this case we have to consider the whole grid of possible arrangements of$k$bits, which gives us a total of$k \cdot 2^k$bits. It’s easy to see that half of them will be 0 and the other half will be 1, so the number we are after in this case will be$k \cdot 2^{k-1}$. Easy! Let’s see this at work: 3 --> 11 --> 4 bits 10 01 00  In this case, we have$k = 2$and$k \cdot 2^{k-1} = 2 \cdot 2^1 = 4$. Another example: 7 --> 111 --> 12 bits 110 101 100 011 010 001 000  In this case we have$k = 3$and$k \cdot 2^{k-1} = 3 \cdot 2^2 = 12$. What happens in general? The representation of a number in binary form will have the most significant bit set to 1, followed by a sequence of$k$bits that can be either 0 or 1: 1xxx...xxx  When we count backwards from $N, we will hit a point where this most significant bit will turn to 0 and all the remaining bits will be 1:

1xxx...xxx )
...         > section A
1000...000 )

0111...111 )
...         > section B
0000...000 )


It’s easy to see that section B is the same as what we calculated before, i.e. $2^k-1$, so we know how many 1s we have (i.e. $k \cdot 2^{k-1}$).

Let’s move to section A. The first observation is that the most significant bit is always 1, and it repeats for a number of times that is $X = N - (2^k - 1) = (N - 2^k) + 1$.

At this point, we have to evaluate how many 1s are present in section A but excluding the leading 1. It’s easy to see that it’s the same number of 1 that appear in the solution to the problem for $X - 1$, so we can just repeat the process!

This leads us to the following code:

 1 sub count_bits ($n,$m = 1000000007) {
2    my $mask = 1; 3 my$mask_bit = 0;
4    while (($n & ~$mask) > $mask) { # scan for highest set bit 5$mask <<= 1;
6       $mask_bit++; 7 } 8 my$n_bits = 0;
9    while ($n) { 10 while (($n & $mask) == 0) { # scan for next high bit 11$mask_bit--;
12          $mask >>= 1; 13 } 14$n &= ~$mask; # this makes$n less than half of itself
15       $n_bits = ($n_bits + 1 + $n +$mask_bit * ($mask >> 1) %$m) % $m; 16 } ## end while ($n)
17    return $n_bits; 18 } ## end sub count_bits  At the beginning (lines 2 to 7) we find out the position of the most significant bit in $n (that is our input $N$) both as a one-bit mask $mask and as a bit position $mask_bit, which represents our $k$.

When ready, we initialize our count to 0 (line 8) and start looping until $n becomes 0, which mean that we don’t have to look for further 1s. Lines 10 through 13 take care to move the $mask/$mask_bit pair onto the most significant bit of the current value of $n, going down because $n is decremented at each loop. As a matter of fact, the operation in line 14 is the same as doing$N = N - 2^k$, preparing for the following iteration. Line 15 does the calculation of the bits in this iteration: the 1 +$n part represents the leading 1s and the $mask_bit * ($mask >> 1) represents the $k \cdot 2^{k - 1}$ part. The most critical operations are performed modulo \$m as requested to avoid overflows.

I think this is it!