Next Identical Popcount, Revisited

June 26, 2018

We solved the next-permutation problem in a previous exercise, and it is easy to adapt that solution to the current task:

(define (next n)
  (undigits (reverse (next-perm < (reverse (cons 0 (digits n 2))))) 2))

Working from the inside out: we convert decimal to binary, cons a 0 onto the front (so there is a next permutation when all bits are 1), reverse (because that’s the order next-perm expects its arguments), call next-perm to advance to the next permutation, reverse (to put it back in the correct order, and convert back from a list of binary digits to a decimal number. Here are some examples:

> (next 15)
23
> (next 23)
27

And here’s the example O’Keefe gave:

> (time (next (expt 2 30)))
(time (next (expt 2 ...)))
    no collections
    0.000008361s elapsed cpu time
    0.000006637s elapsed real time
    2624 bytes allocated
2147483648

You can run the program at https://ideone.com/OcRmSv.

Posted by programmingpraxis

Filed in Exercises

2 Comments »

2 Responses to “Next Identical Popcount, Revisited”

Kevin said

June 26, 2018 at 1:14 PM

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:

(pop-count     1) ==> 2
(pop-count     7) ==> 11
(pop-count    10) ==> 12
(pop-count   100) ==> 104
(pop-count 65535) ==> 98303

Daniel said
June 29, 2018 at 7:49 PM
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:
```
23
27
```

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