Replace twos with threes

Question

Given a positive integer n write some code to take its prime factorization and replace all of its factors of 2 with 3.

For example

12 = 2 * 2 * 3 -> 3 * 3 * 3 = 27

This is code-golf so the goal is to minimize the byte count of your answer.

Test cases

1 -> 1
2 -> 3
3 -> 3
4 -> 9
5 -> 5
6 -> 9
7 -> 7
8 -> 27
9 -> 9
10 -> 15
11 -> 11
12 -> 27
13 -> 13
14 -> 21
15 -> 15
16 -> 81
17 -> 17
18 -> 27
19 -> 19
20 -> 45
21 -> 21
22 -> 33
23 -> 23
24 -> 81
25 -> 25
26 -> 39
27 -> 27
28 -> 63
29 -> 29

Answer 1

Fractran, 3 bytes

3/2

Fractran literally only has one builtin, but it happens to do exactly what this task is asking for. (It's also Turing-complete, by itself.)

The language doesn't really have a standardised syntax or interpreter. This interpreter (in a comment to a blog post – it's a very simple language) will accept the syntax shown here. (There are other Fractran interpreters with other syntaxes, e.g. some would write this program as 3 2, or even using 3 and 2 as command line arguments, which would lead to a score of 0+3 bytes. I doubt it's possible to do better than 3 in a pre-existing interpreter, though.)

Explanation

3/2
 /   Replace a factor of
  2  2
3    with 3
     {implicit: repeat until no more replacements are possible}

Answer 2

Python 2, 28 bytes

f=lambda n:n%2*n or 3*f(n/2)

Try it online!

Recursively divide the number by 2 and multiplies the result by 3, as long as the number is even. Odd numbers return themselves.

32 byte alt:

lambda n:n*(n&-n)**0.58496250072

Try it online. Has some float error. The constant is log_2(3)-1.

Uses (n&-n) to find the greatest power-of-2 factor of n, the converts 3**k to 2**k by raising it to the power of log_2(3)-1.

Answer 3

05AB1E, 4 bytes

Ò1~P

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How it works

Ò     Compute the prime factorization of the input.
 1~   Perform bitwise OR with 1, making the only even prime (2) odd (3).
   P  Take the product of the result.

Answer 4

Haskell, 24 23 bytes

until odd$(*3).(`div`2)

The divide by two and multiply by 3 until odd trick in Haskell.

Try it online!

Alternative with a lambda instead of a pointfree function and with the same byte count:

odd`until`\x->div(x*3)2

Edit: @ais523 saved a byte in the original version and @Ørjan Johansen one in the alternative version, so both version have still the same length. Thanks!

Answer 5

JavaScript (ES6), 19 bytes

f=x=>x%2?x:f(x*1.5)

While the input is divisible by two, multiplies it by 1.5, which is equivalent to dividing by 2 and multiplying by 3.

Answer 6

Brain-Flak, 76 bytes

{{({}[()]<([({})]()<({}{})>)>)}{}([{}]()){{}((({})){}{})<>}<>}<>({}({}){}())

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Explanation

This program works by dividing the number by two and tripling until it gets a remainder of one from the division. Then it stops looping and doubles and adds one to the final number.

More detailed explanation eventually...

Answer 7

Mathematica, 22 19 bytes

Thanks to lanlock4 for saving 3 bytes!

#//.x_?EvenQ:>3x/2&

Pure function that does the replacement repeatedly, one factor of 2 at a time. Works on all positive integers less than 2⁶⁵⁵³⁷.

Answer 8

Perl 6, 14 bytes

{1.5**.lsb*$_}

lsb returns the position of the least-significant bit, counted from the right. That is, how many trailing zeroes in the binary representation, which is the same as the number of factors of 2. So raise 3/2 to that power and we're done.

say {$_*1.5**.lsb}(24);
> 81

Answer 9

Lean, 139 82 71 bytes

11 bytes saved by ovs.

import data.nat.prime
open list
λx,prod$map(λp,p+1-p%2)$nat.factors x

Try it online!

I'm really new to lean so, this answer is probably really inefficient in ways I don't yet know. But it works.

We get a list of factors with nat.factors replace 2s with threes using a map and roll it back up with list.prod.

Answer 10

MATL, 7, 6 bytes

Yf1Z|p

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1 byte saved thanks to Dennis's genius observation

---

The best way to explain this is to show the stack at various points.

Yf  % Prime factors

[2 2 2 3]

1Z| % Bitwise OR with 1

[3 3 3 3]

p   % Product

81

Alternate solution:

Yfto~+p

Answer 11

05AB1E, 6 5 bytes

Saved a byte thanks to Adnan.

ÒDÈ+P

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Explanation

Ò       # push list of prime factors of input
 D      # duplicate
  È     # check each factor for evenness (1 if true, else 0)
   +    # add list of factors and list of comparison results
    P   # product

Answer 12

Java, 38 bytes

int f(int n){return n%2>0?n:f(n/2*3);}

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---

Previous 43-byte solution:

int f(int n){for(;n%2<1;)n=n/2*3;return n;}

Try it online!

Answer 13

Alice, 9 bytes

2/S 3
o@i

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Alice has a built-in to replace a divisor of a number with another. I didn't think I'd actually get to make use of it so soon...

Using the code points of characters for I/O, this becomes 6 bytes: I23SO@.

Explanation

2   Push 2 (irrelevant).
/   Reflect to SE. Switch to Ordinal.
i   Read all input as a string.
    The IP bounces up and down, hits the bottom right corner and turns around,
    bounces down again.
i   Try to read more input, but we're at EOF so this pushes an empty string.
/   Reflect to W. Switch to Cardinal.
2   Push 2.
    The IP wraps around to the last column.
3   Push 3.
S   Implicitly discard the empty string and convert the input string to the integer
    value it contains. Then replace the divisor 2 with the divisor 3 in the input.
    This works by multiplying the value by 3/2 as long as it's divisible by 2.
/   Reflect to NW. Switch to Ordinal.
    Immediately bounce off the top boundary. Move SW.   
o   Implicitly convert the result to a string and print it.
    Bounce off the bottom left corner. Move NE.
/   Reflect to S. Switch to Cardinal.
@   Terminate the program.

Answer 14

Hexagony, 112 91 bytes

Grid size 6 (91 bytes)

      ? { 2 . . <
     / = { * = \ "
    . & . . . { _ '
   . . { . . } ' * 2
  _ } { / . } & . ! "
 . > % . . < | . . @ |
  \ . . \ . | ~ . . 3
   . . " . } . " . "
    . & . \ = & / 1
     \ = { : = } .
      [ = { { { <

Compact version

?{2..</={*=\".&...{_'..{..}'*2_}{/.}&.!".>%..<|..@|\..\.|~..3..".}.".".&.\=&/1\={:=}.[={{{<

Grid size 7 (112 bytes)

       ? { 2 " ' 2 <
      / { * \ / : \ "
     . = . = . = | . 3
    / & / { . . } { . "
   . > $ } { { } / = . 1
  _ . . } . . _ . . & . {
 . > % < . . ~ & . . " . |
  | . . } - " . ' . @ | {
   . . . = & . . * ! . {
    . . . _ . . . _ . =
     > 1 . . . . . < [
      . . . . . . . .
       . . . . . . .

Try it online!

Compact Version:

?{2"'2</{*\/:\".=.=.=|.3/&/{..}{.".>$}{{}/=.1_..}.._..&.{.>%<..~&..".||..}-".'.@|{...=&..*!.{..._..._.=>1.....<[

Ungolfed Version for better readability:

Approximate Memory Layout

Grey Path (Memory initialization)

?     Read input as integer (Input)
{     Move to memory edge "Divisor left"
2     Set current memory edge to 2 
" '   Move to memory edge "Divisor right" 
2     Set current memory edge to 2
"     Move to memory edge "Multiplier" 
3     Set current memory edge to 3
"     Move to memory edge "Temp 2" 
1     Set current memory edge to 1 
{ { { Move to memory edge "Modulo"
=     Turn memory pointer around 
[     Continue with next instruction pointer

Loop entry

%     Set "Modulo" to Input mod Divisor
<     Branch depending on result

Green Path (value is still divisible by 2)

} } {     Move to memory edge "Result"
=         Turn memory pointer around 
*         Set "Result" to "Temp 2" times "Multiplier" (3) 
{ = &     Copy "Result" into "Temp2" 
{ { } } = Move to "Temp"
:         Set "Temp" to Input / Divisor (2)
{ = &     Copy "Temp" back to "Input"
"         Move back to "Modulo"

Red Path (value is no longer divisible by 2)

} = & " ~ & ' Drag what's left of "Input" along to "Multiplier"
*             Multiply "Multiplier" with "Temp 2"
! @           Output result, exit program

Answer 15

Jelly, 8 5 bytes

Æf3»P

Æf3»P  Main Link, argument is z
Æf     Prime factors
  3»   Takes maximum of 3 and the value for each value in the array
    P  Takes the product of the whole thing

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-3 bytes thanks to a hint from @Dennis!

Answer 16

Brachylog, 7 bytes

~×₂×₃↰|

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How it works

~×₂×₃↰|      original program
?~×₂×₃↰.|?.  with implicit input (?) and output (.) added

?~×₂         input "un-multiplied" by 2
    ×₃       multiplied by 3
      ↰      recursion
       .     is the output
        |    or (in case the above fails, meaning that the input
                 cannot be "un-multiplied" by 2)
         ?.  the input is the output

Answer 17

Hexagony, 28 27 26 bytes

?'2{{(\}}>='!:@=$\%<..>)"*

Try it online!

Laid out:

    ? ' 2 {
   { ( \ } }
  > = ' ! : @
 = $ \ % < . .
  > ) " * . .
   . . . . .
    . . . .

This basically runs:

num = input()
while num%2 == 0:
    num = (num/2)*3
print num

At this point it's an exercise on how torturous I can get the loop path to minimise bytes.

Answer 18

Retina, 23 bytes

.+
$*
+`^(1+)\1$
$&$1
1

Try it online! Link is to test suite that runs automatically runs all inputs from 1 to 29. Explanation:

.+
$*

Convert to unary.

+`^(1+)\1$
$&$1

While the value is even, multiply it by 3/2.

1

Convert to decimal.

Note that any FRACTRAN program is trivially convertible to a Retina program with unary I/O. The fraction itself is converted into a match for a string that appears an exact number of times given by the denominator, and the substitution is then $1 repeated a number of times given by the numerator, although if this is greater than the denominator then $& can be used to golf down the code length. It then remains to loop each fraction back to the start of the script, so that the earliest possible fraction is always the next to be substituted. In the case of a single fraction no group is required of course.

Answer 19

Pyth - 14 10 9 bytes

*^1.5/PQ2

Counts the number of 2s in the prime factorization (/PQ2). Multiplies the input by 1.5^(# of 2s)

Try it

Answer 20

Risky, 9 bytes

*\?+0+0+0__?\?+?:2

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Answer 21

BQN (CBQN), 30 bytes

{0=2|𝕩?3×𝕊𝕩÷2;𝕩}

Attempt This Online!

Answer 22

J, 11 bytes

[:*/q:+2=q:

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[: cap (placeholder to call the next verb monadically)

*/ the product of

q: the prime factors

+ plus (i.e. with one added where)

2 two

= is equal to (each of)

q: the prime factors

Answer 23

J, 15 12 10 bytes

(+2&=)&.q:

Try it online! Works similar to below, just has different logic concerning replacement of 2 with 3.

15 bytes

(2&={,&3)"+&.q:

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Explanation

(2&={,&3)"+&.q:
           &.    "under"; applies the verb on the right, then the left,
                 then the inverse of the right
             q:  prime factors
(       )"+      apply inside on each factor
     ,&3         pair with three: [factor, 3]
 2&=             equality with two: factor == 2
    {            list selection: [factor, 3][factor == 2]
                 gives us 3 for 2 and factor for anything else
           &.q:  under prime factor

Answer 24

PHP, 36 Bytes

for($a=$argn;$a%2<1;)$a*=3/2;echo$a;

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Answer 25

CJam, 10 9 bytes

rimf1f|:*

Really simple.

Explanation:

ri  e# Read integer:         | 28
mf  e# Prime factors:        | [2 2 7]
1   e# Push 1:               | [2 2 7] 1
f|  e# Bitwise OR with each: | [3 3 7]
:*  e# Product:              | 63

Answer 26

Japt, 7 bytes

k mw3 ×

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Explanation

k mw3 ×

k        // Factorize the input.
  mw3    // Map each item X by taking max(X, 3).
      ×  // Take the product of the resulting array.
         // Implicit: output result of last expression

Answer 27

Pyth, 9 bytes

Integer output \o/

*F+1.|R1P

Test suite.

How it works

*F+1.|R1P
        P   prime factorization
    .|R1    bitwise OR each element with 1
*F+1        product

Answer 28

APL (Dyalog Unicode), 26 bytes

{⍵(⊣×(3*⊢)×(÷2*⊢))2⍟⍵∨2*⍵}

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This is too verbose, I must be doing something wrong...

Answer 29

R, 42 bytes

The only right amount of bytes in an answer.

x=gmp::factorize(scan());x[x==2]=3;prod(x)

Pretty straightforward, uses the gmp package to factorize x, replaces 2s by 3s, and returns the product.

Answer 30

Japt, 19 16 10 9 7 bytes

k ®w3Ã×

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Explanation

 k ®   w3Ã ×
Uk mZ{Zw3} r*1
U              # (input)
 k m           # for every prime factor
    Z{Zw3}     # replace it by the maximum of itself and 3
           r*1 # output the product