Next Identical Popcount, Revisited
June 26, 2018
In a comment on the prior exercise, Richard A. O’Keefe said:
This is just the “next k-subset of 1..n” problem. It is possible to go fairly directly from one popc-k integer to the next. Let the least significant 1 bit be at offset p. If bit(p+1) is 0, set it to 1 and clear bit(p). If bit(p+1) is 1, clear bit(p), set bit(0), advance bits (p+1..msb) to the next integer with k-1 bits.
The algorithm at the top of the page is spectacularly inefficient. Consider the case of 32-bit integers, bit(30) is on, and all the others are off. (k=1) The (next n) algorithm loops about 2**30 times.
Dr. O’Keefe is correct. My solution isn’t very good. In fact, it’s abominable.
Your task is to write a program to solve the next-identical-popcount problem in the manner Professor O’Keefe suggests. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
racket/scheme:
(define (pop-count num) (define (bin-weight n) (foldl (λ (c acc) (+ acc (if (equal? c #\1) 1 0))) 0 (string->list(number->string n 2)))) (let ((target (bin-weight num))) (let get-pop-count ((n (add1 num))) (if (= (bin-weight n) target) n (get-pop-count (add1 n))))))testing:
I believe that there is a problem with the described algorithm.
For example, for 0b111000, the next number should be 0b1000011, but the algorithm above would incorrectly return 0b1010001.
The problem is in the recursive call: “advance bits (p+1..msb) to the next integer with k-1 bits”
That should seemingly be “advance bits (1..msb) to the next integer with k-1 bits”
Where “p+1” was modified to “1”, and indexing is relative to the recursive call input, not the original bit string.
Here’s a solution in Python with the modification.
def nip(x): bits = list(reversed(bin(x)[2:])) + ["0"] p = bits.index("1") z = 0 while True: bits[p] = "0" if bits[p+1] == "0": bits[p+1] = "1" break else: bits[z] = "1" p += 1 z += 1 return int("".join(reversed(bits)), 2) print(nip(15)) print(nip(23))Output: