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INTEGERS WITH PERFECT BINARY DIGITS

A perfect binary digit integer means that the count of the 1 digits in its binary representation add up to a number that is a square number.

Two examples:

The integer 23 is 10111 in binary. It has four 1 bits.
Four is a square number (sqrt(4) = 2), so 23 is a perfect binary digit integer.

The integer 7 is 111 in binary. It has three 1 bits. Three is not a square number. (sqrt(3) ~1.732), so seven is not a perfect binary digit integer.

This program finds the number of perfect binary digits between two positive integers.

./pft 1 7
3

How the program works

One way the program could work:

  1. get the next (or first) number in the given range.
  2. count the 1 bits in the number.
  3. See if the count is in one of (1, 4, 9, 16, 25, 36, 49, 64)
  4. If so, increment a counter.
  5. start over until we reach the end of our range.

Although easy to understand, the algorithm above is horribly slow.

To speed things up, the program relies on the nCr formula from discrete mathematics. As the program progesses through a given range of numbers, it finds opportunities to apply the formula. See counting problems in the wikipedia, and refer to this interactive calculator.

How Does the nCr Formula Help?

In order to be included in our count, a number's binary representation needs to have exactly 1, 4, 9, 16, 25, 36, 49, or 64 bits set to 1.

Consider the decimal number 64.

64 in binary is a 1 with 6 zeros behind it: 1 0 0 0 0 0 0. That's 7 bits total.

7 > 1, 7 > 4, but 7 < 9, so any numbers we care about will have either 4 bits set to 1 or just a single bit set to 1.

Set aside the single-bit case for the moment. Consider finding the numbers up to 64 with exactly 4 bits set.

Observe that the integers between 0 and 63 need at most 6 bits. They will all fit in the 6 trailing bits after the 1 in 64. This means that there are 6 slots for sets of 4 bits to slosh around in.

Apply the nCr combinations formula. We have 4 bits in 6 possible places.

C(n, r) = 
n! / (r! * (n - r)!)

6! / (4! * (6 - 4)!) = 15

There are 15 numbers with exactly 4 bits set to 1.

The case of numbers with only 1 bit set to 1 is straightforward.

Number  In Binary      1-bits
  64    1 0 0 0 0 0 0    1
  32      1 0 0 0 0 0    1
  16        1 0 0 0 0    1
   8          1 0 0 0    1
   4            1 0 0    1
   2              1 0    1
   1                1    1
 ---------------------------
                         7

There are 15 + 7 = 22 numbers with perfect binary digits between 0 and 64.

Larger numbers are broken up into an additive chain of decreasing powers of two and each link evaluated.

113 = 64 + 32 + 16 + 1.
127 = 64 + 32 + 16 + 8 + 4 + 2 + 1.
129 = 128 + 1.

Testing

Included are two tests. One in Go and the other in Bash. Both rely on an 'Oracle' (not the database) somewhere on the internet that knows the correct answer.

Run the Bash version like so:

make test

You should see something like this:

clang -Werror -g   -c -o pft.o pft.c
gcc -o pft pft.o -Werror -g 
chmod 755 pft
./testit.sh 10 100
10  oracle=4  pft=4   
11  oracle=4  pft=4   
12  oracle=4  pft=4   
13  oracle=4  pft=4   
14  oracle=4  pft=4   
15  oracle=5  pft=5   
16  oracle=6  pft=6   
17  oracle=6  pft=6   
...
...
...

The Go version will try 3,000,000,000 random numbers using 40 goroutines.

make gotest

The output:

clang -Werror -g   -c -o pft.o pft.c
gcc -o pft pft.o -Werror -g
chmod 755 pft
go run test-pft.go
11525030097865261574 oracle = 915096048791149969 beep = 915096048791149969 MATCH
3619520504219696927 oracle = 275987812508718171 beep = 275987812508718171 MATCH
12881502906308645028 oracle = 1024072386590392324 beep = 1024072386590392324 MATCH
18252189985895656162 oracle = 1499614885929439434 beep = 1499614885929439434 MATCH
7354949233259147073 oracle = 571788042953110264 beep = 571788042953110264 MATCH
15757828176933690593 oracle = 1269074278812468454 beep = 1269074278812468454 MATCH
...
...
...