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Posted on Jun 11, 2020

Daily Challenge #257 - Halving Sum

#challenge

Given a positive integer n, calculate the following sum:
n + n/2 + n/4 + n/8 + ...

All elements of the sum are the results of integer division. Continue dividing the number until you reach a decimal (no decimals allowed in final answer).

Example
25 => 25 + 12 + 6 + 3 + 1 = 47

Tests
halvingSum(25)
halvingSum(127)

This challenge comes from Belia on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Top comments (16)

Anna • Jun 12 '20 • Edited

COBOL

Run it at jdoodle.com/a/280D

IDENTIFICATION DIVISION.
PROGRAM-ID. HALVING-SUM.
DATA DIVISION.
    WORKING-STORAGE SECTION.
        01 WS-NUMBER PIC 9(4).
        01 WS-SUM PIC 9(5) VALUE 0.
PROCEDURE DIVISION.
    DISPLAY 'Enter number from 1 to 9999:'
    ACCEPT WS-NUMBER FROM CONSOLE       
    PERFORM UNTIL WS-NUMBER < 1
        ADD WS-NUMBER TO WS-SUM
        DIVIDE WS-NUMBER BY 2 GIVING WS-NUMBER
    END-PERFORM     
    DISPLAY WS-SUM.     
STOP RUN.

Mike Talbot ⭐ • Jun 11 '20

const half = n => n > 0 ? n + half(n>>1) : n

Mihail Malo • Jun 13 '20

const half = n => n && half(n>>1) + n

Mihail Malo • Jun 13 '20 • Edited

Does "until you reach a decimal" mean "until the result of integer division is below 1"?
Because 25/2 is immediately 12.5, but the example doesn't stop there.
(I do see "integer division", but then there are no decimals)

Sam Ferree • Jun 12 '20 • Edited

Befunge-93

&>:02pv
      >2/:0`!#v_:02g+02p
              >02g.@

Edit:
Here is a version with comments

&>:02pv Read a value t from the user, store it at (0,2)
         divide t by 2, if t > 0 store (0,2)+t at (0,2) repeat
      >2/:0`!#v_:02g+02p
                if t <= 0 print (0,2) as integer, end program
              >02g.@

Mihail Malo • Jun 13 '20 • Edited

I like the word problem in the question, where decimals are mentioned, but not the example.
So the examples would be:

25 => 25 + ~~12.5~~ = 25
28 => 28 + 14 + 7 + ~~3.5~~ = 49

const half = n => n % 2 ? n : n + half(n / 2) // ew, not even tail recursion

1) However, we know that the infinite series n + n/2 + n/4 + n/8 + ... converges to 2n.
(because it is 2n * (1/2 + 1/4 + 1/8 + 1/16 + ⋯) link)
2) And we know that the highest power of 2 we can cleanly divide a number is the number of trailing zeros it has:

0b1100 / 2 = 0b110
0b110 / 2 = 0b11
0b11 / 2 = 'loss.jpg'

3) And we also know that the sum of the infinitely many items we don't get to add looks like, for example n/8 + n/16 + n/32 + ..., which is equivalent to 2n/8* (1/2 + 1/4 + 1/8 + 1/16 + ⋯) (and converges to (2n/8)).
4) So the sum without that tail is 2n - 2n/8, where 8 is 2 to the power of number of trailing zeros of n.
5) Well, turns out there is an intrinsic for trailing zeros.

Rust

fn half(x: u64) -> u64 {
    let highest = x.trailing_zeros();
    2 * x - (x >> highest)
}

which results in just this assembly

example::half:
        test    rdi, rdi
        je      .LBB0_1
        bsf     rcx, rdi
        jmp     .LBB0_3
.LBB0_1:
        mov     ecx, 64
.LBB0_3:
        mov     rdx, rdi
        shr     rdx, cl
        lea     rax, [rdi + rdi]
        sub     rax, rdx
        ret

Is that good, one might ask? No idea, IDK how to computers.

David Leger • Jun 12 '20

Looping

const halvingSum = (n: number): number => {
  let sum = 0;

  while (n !== 0) {
    sum += n;
    n = Math.floor(n / 2);
  }

  return sum;
}

Recursion

const halvingSum = (n: number): number => {
  if (n === 0) {
    return n;
  }

  return n + halvingSum(Math.floor(n / 2))
}

Tail Recursion

const halvingSum = (n: number, sum: number = 0): number => {
  if (n === 0) {
    return sum;
  }

  return halvingSum(Math.floor(n / 2), sum + n);
}

peter279k • Jun 18 '20

Here is the simple solution with Python:

def halving_sum(n): 
    # your code here
    exp = 0
    currentNumber = 0
    answer = 0
    while n >= int(pow(2, exp)):
        currentNumber = int(n / pow(2, exp))
        answer += currentNumber
        exp += 1

    return answer

Swarup Das • Jun 11 '20

/**
 * 
 * @param {Number} number 
 */
function halvingSum(number) {
    let total = 0;
    while (number > 0) {
        total += number;
        number = Math.floor(number / 2);
    }
    return total;
}

Vidit Sarkar • Jun 12 '20

Here is a Python solution,


halvingSum = lambda n : n and (n + halvingSum(n >> 1))

print(halvingSum(25)) # output -> 47
print(halvingSum(127)) # output -> 247

AmirabbasJ • Jun 13 '20

Javascript

function halvingSum(num){
    if(num === 1){
        return 1
    }
    return num+halvingSum(Math.floor(num/2))
}

View full discussion (16 comments)

DEV Community

Daily Challenge #257 - Halving Sum

Top comments (16)

COBOL

Rust

Looping

Recursion

Tail Recursion

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