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Challenges

Comments on Lone ​​​​​​ones

Parent

Lone ​​​​​​ones

+7
−0

Given an integer, output the number of lone 1s in its binary representation.

Input

  • An integer from $0$ to $4294967295$ (that is, $2^{32}-1$). In binary that's from 00000000000000000000000000000000 to 11111111111111111111111111111111.
  • You may take input in any form that does not contribute to solving the challenge. For example:
    • A signed or unsigned integer of $32$ bits or more.
    • A string of decimal digits.
    • A binary string.
    • A sequence of bits.

Output

  • The number of 1s in the binary representation that do not have a 1 to their left or right.
  • For avoidance of doubt:
    • A 1 at the left end of the binary representation does not have a 1 to its left.
    • A 1 at the right end of the binary representation does not have a 1 to its right.

Examples

Input $4$ corresponds to 100 in binary. The output is $1$ because there is $1$ 1 bit that is not adjacent to another 1 bit.

Input $5$ corresponds to 101 in binary. The output is $2$ because there are $2$ 1 bits that are not adjacent to another 1 bit.

Input $6$ corresponds to 110 in binary. The output is $0$ because each of the 1 bits is adjacent to another 1 bit.

Test cases

Test cases are in the format input : output.

Unsigned integer test cases

0 : 0
4294967295 : 0
2147483648 : 1
1 : 1
2147483649 : 2
98304 : 0
1310720 : 2
4294639615 : 1
4294967294 : 0
2147483647 : 0
3221225471 : 1
4294967293 : 1
2863311530 : 16
1431655765 : 16
1717986918 : 0

Binary test cases

00000000000000000000000000000000 : 0
11111111111111111111111111111111 : 0
10000000000000000000000000000000 : 1
00000000000000000000000000000001 : 1
10000000000000000000000000000001 : 2
00000000000000011000000000000000 : 0
00000000000101000000000000000000 : 2
11111111111110101111111111111111 : 1
11111111111111111111111111111110 : 0
01111111111111111111111111111111 : 0
10111111111111111111111111111111 : 1
11111111111111111111111111111101 : 1
10101010101010101010101010101010 : 16
01010101010101010101010101010101 : 16
01100110011001100110011001100110 : 0

Scoring

This is a code golf challenge. Your score is the number of bytes in your code. Lowest score for each language wins.

Explanations are optional, but I'm more likely to upvote answers that have one.

History

0 comment threads

Post
+2
−0

C (gcc), 67 bytes

a;r;i;f(long n){for(;i<33;)a=n&1l<<i++?a?2:1:a&1?r++,0:0;return a;}

Try it online!


General algorithm:

  • This is a little state machine where the variable 'alone' a can have the value 0=zero, 1=alone candidate, 2=not alone. Essentially a 'carry' but with several states.

    The non-golfed code would look something like if(!a) a=1; else a=2; which becomes a=a?2:1.

  • Then the outer if-else non-golfed would look something like

    if(n&1l<<i)
      a=a?2:1;
    else
    {
      if(a&1)
        r++;
      a=0;
    }
    

    ...which turns into the one-liner ?: mess.

  • When the state machine has found an alone candidate and transits from value 1 to zero, increase count r

  • Whenever there's a zero, reset the state machine.

  • To solve the special case at the end where there might be a lone 1, iterate one more than 32 times to get comparison with a zero in there. Which will only work if the function parameter is not sign extended! To cheat I used long which is 64 bit on Linux that tio.run uses - won't work on 32 bit long systems. Making the parameter unsigned would be portable but unsigned has more digits than long. Same thing with the 1l prefix, which needs to be 1ull to be portable.

C-specific stuff:

  • Plain int is 32 bits.
  • Operator precedence ++ over << over & over ?: over = over ,.
  • Shifting negative numbers is of course wildly poorly-specified behavior but the code gets away with it on gcc/Linux.

Test cases:

  • Taken from the challenge. An ugly macro takes a digit and expected output, then prints OK or FAIL.
  • As usual, the function-only solution relying on zero init of global variables need to be reset by the caller between tests.
  • Also print test cases as hex because the decimal numbers really don't tell me much :)
History

1 comment thread

Missing default rule for using globals (4 comments)
Missing default rule for using globals
trichoplax‭ wrote 7 months ago

This uses the Code Golf rule that a global may hold the result to save a return

Is this a rule established on SE's Meta? Currently it doesn't seem to be covered in Codidact's Default Rules: Code Golf I/O. I'm mentioning this in case you want to add an answer there to cover this case.

Lundin‭ wrote 7 months ago

trichoplax‭ Huh I might have mixed up several rules with each other. Well in that case I suppose I should add a return statement.

trichoplax‭ wrote 7 months ago · edited 7 months ago

You may modify the input in-place is perhaps the most similar existing rule. Not quite as generous but does still allow avoiding return.

Lundin‭ wrote 7 months ago

trichoplax‭ Ok in this case I must use another named parameter so I don't think that would improve golfing. I'll add the return.