# 2 factors factorization

Given a natural number n write a program or function to get a list of all the possible two factors multiplications that can be used to achieve n. To understand better what is pretended you can go to http://factornumber.com/?page=16777216 to see when n is 16777216 we get the following list:

   2 × 8388608
4 × 4194304
8 × 2097152
16 × 1048576
32 ×  524288
64 ×  262144
128 ×  131072
256 ×   65536
512 ×   32768
1024 ×   16384
2048 ×    8192
4096 ×    4096


No need to pretty print things like here. The requirement is that each entry (pair of factors) is well distinguished from each other and inside each pair, the first factor is also well distinguished from the other. If you choose to return a list/array, the inside element can be a list/array with two elements, or some structure of your language that supports a pair of things like C++ std::pair.

Do not print the multiplication by 1 entry, nor repeat entries with the first factor commuted by the second, as they are pretty useless.

No winner; it will be a per language basis code golf.

• Could you possibly add a smaller test case, such as 30? – caird coinheringaahing Dec 2 '17 at 22:27
• @cairdcoinheringaahing You can use factornumber.com to generate more test cases. – Jonathan Frech Dec 2 '17 at 22:39
• I've seen this "per language" competition recently. What's the point? Most Qs don't get more than 1 or 2 As per language, and you still can select just one A as correct. – fede s. Dec 2 '17 at 23:54
• @fedes. It's usually because there's no point in comparing between languages (i.e. Java vs. Jelly). – totallyhuman Dec 3 '17 at 0:06
• @totallyhuman yeah, I know. Most of my answers are in Factor, or even Smalltalk. No chance against the golfing languages. Maybe there could be some way of ranking languages by verbosity and boilerplatery – fede s. Dec 3 '17 at 0:28

# Java (OpenJDK 8), 8166 65 bytes

• -15 Bytes thanks to Olivier Grégoire.
• -1 Byte: ++j<=i/j -> j++<i/j.
i->{for(int j=1;j++<i/j;)if(i%j<1)System.out.println(j+" "+i/j);}


Try it online!

Old one (for reference)

# Java (OpenJDK 8), 126 bytes

i->{java.util.stream.IntStream.range(2,i).filter(d->d<=i/d&&i%d==0).forEach(e->System.out.println(""+e+"x"+i/e));return null;}


Try it online!

First codegolf submit and first lambda usage. Future self, please forgive me for the code.

• Nice first entry! Welcome to PPCG! Here is it golfed down to 66 bytes by removing all the superfluous: I couldn't golf your algorithm though. – Olivier Grégoire Dec 4 '17 at 14:03

# 05AB1E, 8 bytes

ÑÂ‚ø2äн¦


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• +1 from me we have nearly the same solutions. I thought of this 8-byter – Mr. Xcoder Dec 2 '17 at 23:21
• @Mr.Xcoder: Ah yes, nice :) It's too bad that the map is required there. – Emigna Dec 3 '17 at 8:41

# C (gcc), 5854 53 bytes

f(N,j){for(j=1;j++*j<N;)N%j||printf("|%d,%d",j,N/j);}


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# Python 2, 51 bytes

f=lambda n,k=2:n/k/k*[f]and[(k,n/k)][n%k:]+f(n,k+1)


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51 bytes (thanks to Luis Mendo for a byte)

lambda n:[(n/k,k)for k in range(1,n)if(k*k<=n)>n%k]


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51 bytes

lambda n:[(n/k,k)for k in range(1,n)if n/k/k>n%k*n]


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• I like the use of [f]. – Jonathan Frech Dec 2 '17 at 22:52
• You can save 1 byte in the second version with lambda n:[(n/k,k)for k in range(1,n)if(k*k<=n)>n%k] – Luis Mendo Dec 3 '17 at 2:49
• MemoryError on all approaches for 1512518520 – sergiol Dec 5 '17 at 0:37

f x=[(a,b)|a<-[2..x],b<-[2..a],a*b==x]


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• Time out for 1512518520 – sergiol Dec 5 '17 at 0:40

# APL (Dyalog), 28 bytes

{(⊢,⍵÷⊢)¨o/⍨0=⍵|⍨o←1↓⍳⌊⍵*.5}


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# Perl 6, 38 bytes

{map {$^a,$_/$a},grep$_%%*,2.. .sqrt}


Try it

## Expanded:

{   # bare block lambda with implicit parameter ｢$_｣ map {$^a, $_ /$a },  # map the number with the other factor

grep
$_ %% *, # is the input divisible by * 2 .. .sqrt # from 2 to the square root of the input }  # Brachylog, 8 bytes {~×≜Ċo}ᵘ  Try it online! ## Explanation {~×≜Ċo}ᵘ { }ᵘ List the unique outputs of this predicate. ~× Pick a list of integers whose product is the input. ≜ Force concrete values for its elements. Ċ Force its length to be 2. o Sort it and output the result.  The ~× part does not include 1s in its output, so for input N it gives [N] instead of [1,N], which is later culled by Ċ. I'm not entirely sure why ≜ is needed... • The ≜ is needed because otherwise there are no choice points for ᵘ: a length-2 list whose product is the input is the only answer if you don't actually ask for the values of the list. – Fatalize Dec 4 '17 at 7:01 # Japt, 9 bytes â¬Å£[XZo]  Test it online! Returns an array of arrays, with some nulls at the end; -R flag added to show output more clearly. • So I think the  -R should be considered for the byte count... – sergiol Dec 2 '17 at 23:41 • @sergiol, no, in this case it's just for formatting the output for better readability. – Shaggy Dec 3 '17 at 11:07 • Exactly the solution I had, except I filtered the nulls out at the end. – Shaggy Dec 3 '17 at 11:08 # Jelly, 8 bytes ½ḊpP⁼¥Ðf  A monadic link taking a number and returning a list of lists (pairs) of numbers. Try it online! (times out on TIO for the 16777216 example since it would create a list of 68.7 billion pairs and filter down to those with the correct product!) ### How? ½ḊpP⁼¥Ðf - Link: number, n e.g. 144 ½ - square root of n 12 Ḋ - dequeue* [2,3,4,5,6,7,8,9,10,11,12] p - Cartesian product** [[2,1],[2,2],...[2,144],[3,1],...,[3,144],...,[12,144] Ðf - filter keep if: ¥ - last two links as a dyad (n is on the right): P - product ⁼ - equals - [[2,72],[3,48],[4,36],[6,24],[8,18],[9,16],[12,12]]  * Ḋ, dequeue, implicitly makes a range of a numeric input prior to acting, and the range function implicitly floors its input, so with, say, n=24 the result of ½ is 4.898...; the range becomes [1,2,3,4]; and the dequeued result is [2,3,4] ** Similarly to above, p, Cartesian product, makes ranges for numeric input - here the right argument is n hence the right argument becomes [1,2,3,...,n] prior to the actual Cartisian product taking place. # Husk, 8 bytes tüOSze↔Ḋ  Try it online! ## Explanation tüOSze↔Ḋ Implicit input, say n=30. Ḋ List of divisors: [1,2,3,5,6,10,15,30] ↔ Reverse: [30,15,10,6,5,3,2,1] Sze Zip with original: [[1,30],[2,15],[3,10],[5,6],[6,5],[10,3],[15,2],[30,1]] üO Deduplicate by sort: [[1,30],[2,15],[3,10],[5,6]] t Drop first pair: [[2,15],[3,10],[5,6]]  # JavaScript (ES6), 55 bytes n=>eval('for(k=1,a=[];k*++k<n;n%k||a.push([k,n/k]));a')  ### Demo let f = n=>eval('for(k=1,a=[];k*++k<n;n%k||a.push([k,n/k]));a') console.log(JSON.stringify(f(6))) console.log(JSON.stringify(f(7))) console.log(JSON.stringify(f(16777216))) Try It Online! • Is it me or does this fail for 6? – Neil Dec 4 '17 at 10:33 • @Neil "We can fix it." (Thanks for reporting!) – Arnauld Dec 4 '17 at 10:41 • How can I supply a number to test? – sergiol Dec 5 '17 at 21:41 • You can Try it online! – Arnauld Dec 5 '17 at 21:47 # Python 2, 59 bytes lambda N:{(n,N/n,n)[n>N/n:][:2]for n in range(2,N)if N%n<1}  Try it online! • – sergiol Dec 5 '17 at 0:30 • @sergiol Yes, a MemoryError, since Python tries to evaluate range(2,N) and store it as a list, yet the allocated memory does not suffice. One could try replace range with xrange (Python 2's range generator), though this exceeds TIO's one minute of maximum runtime. On a machine with enough memory and time, this program should terminate and return the correct answer. – Jonathan Frech Dec 5 '17 at 16:01 # Jelly, 9 bytes ÆḌḊµżUṢ€Q  Try it online! # Octave, 42 bytes @(n)[y=find(~mod(n,x=2:n)&x.^2<=n)+1;n./y]  Try it online! # PHP, 70 bytes As string (70 bytes): $i++;while($i++<sqrt($a=$argv[1])){echo !($a%$i)?" {$i}x".($a/$i):'';}


As array dump (71 bytes):

$i++;while($i++<sqrt($a=$argv[1]))!($a%$i)?$b[$i]=$a/$i:'';print_r($b);  (im not sure if i can use return$b; instead of print_r since it no longer outputs the array, otherwise i can save 2 bytes here. )

The array gives the results like:

Array
(
[2] => 8388608
[4] => 4194304
[8] => 2097152
[16] => 1048576

• "If you choose to return a list/array" To me it means you can print or return as you see fit. – fede s. Dec 4 '17 at 6:17
• On second thought, returning should be valid for a function, and printing for a program. You seem to have a snippet/program, not a function, so I'd say in this case you should be printing. – fede s. Dec 4 '17 at 6:20

# Jelly, 12 bytes

ÆDµżUḣLHĊ$$Ḋ  Try it online! ## How it works ÆDµżUḣLHĊ$$Ḋ - Main monadic link;
- Argument: n (integer) e.g. 30
ÆD           - Divisors                   [1, 2, 3, 5, 6, 10, 15, 30]
U        - Reverse                    [30, 15, 10, 6, 5, 3, 2, 1]
ż         - Interleave                 [[1, 30], [2, 15], [3, 10], [5, 6], [6, 5], [10, 3], [15, 2], [30, 1]]
  - Last 3 links as a monad
L      -   Length                   8
H     -   Halve                    4
Ċ    -   Ceiling                  4
ḣ       - Take first elements        [[1, 30], [2, 15], [3, 10], [5, 6]]
Ḋ - Dequeue                    [[2, 15], [3, 10], [5, 6]]


## Wolfram Language (Mathematica), 41 bytes

nRest@Union[Sort@{#,n/#}&/@Divisors@n]


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 is the Function operator, which introduces an unnamed function with named parameter n.

# Factor, 58

Well, there has to be some Factor in this question!

[ divisors dup reverse zip dup length 1 + 2 /i head rest ]


It's a quotation. call it with the number on the stack, leaves an assoc (an array of pairs) on the stack.

I'm never sure if all the imports count or not, as they're part of the language. This one uses:

USING: math.prime.factors sequences assocs math ;


(If they count, I should look for a longer solution with shorter imports, which is kind of silly)

As a word:

: 2-factors ( x -- a ) divisors dup reverse zip dup length 1 + 2 /i head rest ;

50 2-factors .
--> { { 2 25 } { 5 10 } }


# Ruby, 43 bytes

->n{(2..n**0.5).map{|x|[[x,n/x]][n%x]}-[p]}


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### How it works:

For every number up to sqrt(n), generate the pair [[x, n/x]], then take the n%xth element of this array. If n%x==0 this is [x, n/x], otherwise it's nil. when done, remove all nil from the list.

# Pari/GP, 4934 38 bytes

n->[[d,n/d]|d<-divisors(n),d>1&d<=n/d]


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Set builder notation for all pairs [d, n/d] where d runs through all divisors d of n subject to d > 1 and d <= n/d.

Huge improvement by alephalpha.

# Husk, 14 12 bytes

tumoOSe/⁰Ḋ⁰


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### Explanation

tum(OSe/⁰)Ḋ⁰  -- input ⁰, eg. 30
Ḋ⁰  -- divisors [1..⁰]: [1,2,3,5,6,10,15,30]
m(      )    -- map the following function (example on 10):
Se        --   create list with 10 and ..
/⁰     --   .. flipped division by ⁰ (30/10): [10,3]
O          --   sort: [3,10]
-- [[1,30],[2,15],[3,10],[5,6],[5,6],[3,10],[2,15],[1,30]]
u             -- remove duplicates: [[1,30],[2,15],[3,10],[5,6]]
t              -- tail: [[2,15],[3,10],[5,6]]


# APL+WIN, 32 bytes

m,[.1]n÷m←(0=m|n)/m←1↓⍳⌊(n←⎕)*.5


Explanation:

(n←⎕) Prompts for screen input

m←(0=m|n)/m←1↓⍳⌊(n←⎕)*.5 Calculates the factors dropping the first

m,[.1]n÷ Identifies the pairs and concatenates into a list.


L,F@pB]dBRBcE#S


Try it online!

## How it works

L,   - Create a lambda function
- Example argument:     30
F  - Factors;     STACK = [1 2 3 5 6 10 15]
@  - Reverse;     STACK = [15 10 6 5 3 2 1]
p  - Pop;         STACK = [15 10 6 5 3 2]
B] - Wrap;        STACK = [[15 10 6 5 3 2]]
d  - Duplicate;   STACK = [[15 10 6 5 3 2] [15 10 6 5 3 2]]
BR - Reverse;     STACK = [[15 10 6 5 3 2] [2 3 5 6 10 15]]
Bc - Zip;         STACK = [[15 2] [10 3] [6 5] [5 6] [3 10] [2 15]]
E# - Sort each;   STACK = [[2 15] [3 10] [5 6] [5 6] [3 10] [2 15]]
S  - Deduplicate; STACK = [[[2 15] [3 10] [5 6]]]


# Mathematica, 53 bytes

Array[s[[{#,-#}]]&,⌈Length[s=Divisors@#]/2⌉-1,2]&


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# Befunge-93, 56 bytes

&1vg00,+55./.:   <
+1< v\g00/g00:p00
_ ^@_::00g%!00g\#v


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# Julia 0.6, 41 bytes

~x=[(y,div(x,y))for y=2:x if x%y<1>y^2-x]


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Redefines the inbuild unary operator ~ and uses an array comprehension to build the output.

• div(x,y) is neccessary for integer division. x/y saves 5 bytes but the output is ~4=(2,2.0).
• Julia allows chaining the comparisons, saving one byte.
• Looping all the way to x avoids Int(floor(√x)).

# APL NARS 99 chars

r←f w;i;h
r←⍬⋄i←1⋄→0×⍳0≠⍴⍴w⋄→0×⍳''≡0↑w⋄→0×⍳w≠⌊w⋄→0×⍳w≠+w
A:i+←1⋄→A×⍳∼0=i∣w⋄→0×⍳i>h←w÷i⋄r←r,⊂i h⋄→A


9+46+41+3=99 Test: (where not print nothing, it return something it return ⍬ the list null one has to consider as "no solution")

  f 101

f 1 2 3

f '1'

f '123'

f 33 1.23

f 1.23

⎕←⊃f 16777216
2 8388608
4 4194304
8 2097152
16 1048576
32  524288
64  262144
128  131072
256   65536
512   32768
1024   16384
2048    8192
4096    4096
f 123
3 41


# Pyt, 67 65 bytes

←ĐðĐ0↔/⅟ƖŽĐŁ₂20ŕ3ȘĐ05Ș↔ŕ↔Đ4Ș⇹3Ș⦋ƥ⇹⁺Ɩ3ȘĐ05Ș↔ŕ↔Đ4Ș⇹3Ș⦋ƤĐ3Ș⁺ƖĐ3Ș<łĉ


I'm pretty sure this can be golfed.

Basically, the algorithm generates a list of all of the divisors of the input (let's call it n), makes the same list, but flipped, interleaves the two (e.g., if n=24, then, at this point, it has [1,24,2,12,3,8,4,6,6,4,8,3,12,2,24,1]), and prints out the elements from index 2 until half the array length, printing each number on a new line, and with an extra new line in between every pair.

Most of the work is done in actually managing the stack.

Saved 2 bytes by using increment function.

# Perl 5, 50 bytes

sub{map[$_,$_[0]/$_],grep!($_[0]%$_),2..sqrt$_[0]}


Ungolfed:

sub {
return map  { [$_,$_[0] / $_] } grep { !($_[0] % $_) } (2 .. sqrt($_[0]));
}
`