Sort the unique numbers in a multiplication table

Question

Pretty simple challenge today:

Write a program or function that takes in a positive integer N and prints or returns a sorted list of the unique numbers that appear in the multiplication table whose row and column multiplicands both range from 1 to N inclusive.

The list may be sorted in ascending order (smallest to largest) or descending order (largest to smallest), and may be output in any reasonable format.

The shortest code in bytes wins!

Example

When N = 4, the multiplication table looks like:

 1 2 3 4
 -----------
1| 1 2 3 4
 |
2| 2 4 6 8
 |
3| 3 6 9 12
 |
4| 4 8 12 16

The unique numbers in the table are 1, 2, 3, 4, 6, 8, 9, 12, 16. These are already sorted, so

1, 2, 3, 4, 6, 8, 9, 12, 16

might be your exact output for N = 4. But since the sorting can be reversed and there's some leeway in the formatting, these would also be valid outputs:

[16,12,9,8,6,4,3,2,1]

1
2
3
4
6
8
9
12
16

16 12 9 8 4 3 2 1

Test Cases

N=1 -> [1]
N=2 -> [1, 2, 4]
N=3 -> [1, 2, 3, 4, 6, 9]
N=4 -> [1, 2, 3, 4, 6, 8, 9, 12, 16]
N=5 -> [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 20, 25]
N=6 -> [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36]
N=7 -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 28, 30, 35, 36, 42, 49]
N=8 -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 28, 30, 32, 35, 36, 40, 42, 48, 49, 56, 64]
N=9 -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 35, 36, 40, 42, 45, 48, 49, 54, 56, 63, 64, 72, 81]
N=10 -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 35, 36, 40, 42, 45, 48, 49, 50, 54, 56, 60, 63, 64, 70, 72, 80, 81, 90, 100]
N=11 -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 27, 28, 30, 32, 33, 35, 36, 40, 42, 44, 45, 48, 49, 50, 54, 55, 56, 60, 63, 64, 66, 70, 72, 77, 80, 81, 88, 90, 99, 100, 110, 121]
N=12 -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 27, 28, 30, 32, 33, 35, 36, 40, 42, 44, 45, 48, 49, 50, 54, 55, 56, 60, 63, 64, 66, 70, 72, 77, 80, 81, 84, 88, 90, 96, 99, 100, 108, 110, 120, 121, 132, 144]

• TanMath: So basically, the code returns a list of numbers in the multiplication table specified by N, except any number cannot be repeated?
• xsot: How big can N be?
• Calvin's Hobbies: @xsot You can assume N*N will be less than your language's maximum usual int value (probably 2^31-1)
• gregsdennis: So essentially this is 1-n and non primes up to n^2.
• Calvin's Hobbies: @gregsdennis No. There are plenty of composites not present. e.g. 91, 92, 93, 94, 95, 96 for N = 10.

Answer 1

Pyth, 8 bytes

S{*M^SQ2

Try it online.

Explanation: SQ takes the evaluated list input (Q) and makes a list [1, 2, ..., Q]. ^SQ2 takes the Cartesian product of that list with itself - all possible product combinations. *M multiplies all these pairs together to form all possible results in the multiplication table and S{ makes it unique and sorts it.

• Reto Koradi: @FryAmTheEggman Input 5 already needs sorting, otherwise the 10 and 9 in the output are out of order.
• Maltysen: darn keep on forgetting about that splatting on M. +1

Answer 2

Python 2, (削除) 61 (削除ここまで) 51 bytes

lambda n:sorted({~(i%n)*~(i/n)for i in range(n*n)})

Thanks to xnor for shortening some syntax.

• xnor: The set(...) can just be a set comp {...}. Also, functions are allowed by default here, so you can just write lambda n:....
• xsot: Thanks for reminding me about set comprehension, I completely forgot it existed.
• xnor: I can't see a better way to do this, best I see with recursion is 56: f=lambda n:n*[0]and sorted(set(range(n,n*n+n,n)+f(n-1))).

Answer 3

APL, (削除) 18 (削除ここまで) 16 bytes

×ばつ⍨⍳⍵]}

This is an unnamed monadic function. The output is in ascending order.

Explanation:

 ⍳⍵]} ⍝ Get the integers from 1 to the input
 ×ばつ⍨ ⍝ Compute the outer product of this with itself
 , ⍝ Flatten into an array
 ∪ ⍝ Select unique elements
 y← ⍝ Assign to y
 {y[⍋ ⍝ Sort ascending

Fixed an issue and saved 2 bytes thanks to Thomas Kwa!

Answer 4

Octave, 22 bytes

@(n)unique((a=1:n)'*a)

Answer 5

Julia, 24 bytes

n->sort(∪((x=1:n)*x'))

This is an anonymous function that accepts an integer and returns an integer array.

Ungolfed:

function f(n::Integer)
 # Construct a UnitRange from 1 to the input
 x = 1:n
 # Compute the outer product of x with itself
 o = x * transpose(x)
 # Get the unique elements, implicitly flattening
 # columnwise into an array
 u = unique(o)
 # Return the sorted output
 return sort(u)
end

Answer 6

CJam, (削除) 14 (削除ここまで) 12 bytes

Latest version with improvements proposed by @aditsu:

{)2m*::*0^$}

This is an anonymous function. Try it online, with input/output code needed for testing it.

@Martin proposed another very elegant solution ({,:)_ff*:|$}) with the same length. I used the one by aditsu because it was much more similar to my original solution.

The main difference to my original solution is that this keeps the 0 value in the original sequence, which saves 2 bytes at the start. You'd think that this would not help, because you have to remove the 0 value from the result. But the core of @aditsu's idea is the 0^ at the end, which is a set difference with 0. This removes the 0, and at the same time, since it's a set operation, eliminates duplicate elements from the solution set. Since I already needed 2 bytes to eliminate the duplicates before, removing the 0 is then essentially free.

Explanation:

{ Start anonymous function.
 ) Increment to get N+1.
 2m* Cartesian power, to get all pairs of numbers in range [0, N].
 ::* Reduce all pairs with multiplication.
 0^ Remove 0, and remove duplicates at the same time since this is a set operation.
 $ Sort the list.
} End anonymous function.

• Peter Taylor: For the same length, {2m*::)::*_&$}, {)2m*::*_&0ドル-}
• Martin Ender: How about this for two bytes less :) {,:)_ff*:|$}
• aditsu quit because SE is EVIL: Another way: {)2m*::*0^$}

Answer 7

R, 39 bytes

cat(unique(sort(outer(n<-1:scan(),n))))

This reads an integer from STDIN and writes a space delimited list to STDOUT.

We create the multiplication table as a matrix using outer, implicitly flatten into a vector and sort using sort, select unique elements using unique, and print space delimited using cat.

• Giuseppe: 35 bytes

Answer 8

MATLAB, 24 bytes

@(n)unique((1:n)'*(1:n))

• Luis Mendo: Good one! Soon it will be possible to do it in 7 or 8 bytes... :-)
• Stewie Griffin: Oooh, nice! :-)
• Stewie Griffin: @Luis have you tried this one in MATL?
• Luis Mendo: I don't have much time now to read the whole challenge, but seeing your Matlab code it looks like it could be done with MATL

Answer 9

zsh, (削除) 86 (削除ここまで) 56 bytes

thanks to @Dennis for saving 30(!) bytes

(for a in {1..1ドル};for b in {1..1ドル};echo $[a*b])|sort -nu

Explanation / ungolfed:

( # begin subshell
 for a in {1..1ドル} # loop through every pair of multiplicands
 for b in {1..1ドル}
 echo $[a*b] # calculate a * b, output to stdout
) | sort -nu # pipe output of subshell to `sort -nu', sorting
 # numerically (-n) and removing duplicates (-u for uniq)

This doesn't work in Bash because Bash doesn't expand {1..1ドル}—it just interprets it literally (so, a=5; echo {1..$a} outputs {1..5} instead of 1 2 3 4 5).

• Addison Crump: I was waiting for a *sh answer. :D
• Dennis: Relevant bash tip. Seems to apply to the Z shell as well.
• Digital Trauma: You can squeeze a bit more out of brace expansions

Answer 10

Ruby, (削除) 50 (削除ここまで) 48 bytes

->n{c=*r=1..n;r.map{|i|c|=r.map{|j|i*j}};c.sort}

Ungolfed:

->n {
 c=*r=1..n
 r.map { |i| c|=r.map{|j|i*j} }
 c.sort
}

Using nested loop to multiply each number with every other number upto n and then sorting the array.

50 bytes

->n{r=1..n;r.flat_map{|i|r.map{|j|i*j}}.uniq.sort}

Usage:

->n{c=*r=1..n;r.map{|i|c|=r.map{|j|i*j}};c.sort}[4]
=> [1, 2, 3, 4, 6, 8, 9, 12, 16]

Answer 11

Shell + common utilities, 41

seq -f"seq -f%g*%%g 1ドル" 1ドル|sh|bc|sort -nu

---

Or alternatively:

Bash + coreutils, 48

eval printf '%s\\n' \$[{1..1ドル}*{1..1ドル}]|sort -nu

Constructs a brace expansion inside an arithmetic expansion:

\$[{1..n}*{1..n}] expands to the arithmetic expansions $[1*1] $[1*2] ... $[1*n] ... $[n*n] which are evaluated and passed to printf, which prints one per line, which is piped to sort.

Careful use of quotes, escapes and eval ensure the expansions happen in the required order.

---

Or alternatively:

Pure Bash, 60

eval a=($(eval echo [\$[{1..1ドル}*{1..1ドル}\]]=1))
echo ${!a[@]}

Answer 12

Husk, 6 bytes

OumΠπ2

Try it online!

Explanation

OumΠπ2
 π2 cartesian square
 mΠ map product
 u unique
O sort

Answer 13

Haskell, (削除) 55 (削除ここまで) 54 bytes

import Data.List
f n=sort$nub[x*y|x<-[1..n],y<-[1..x]]

Usage example: f 4 -> [1,2,3,4,6,8,9,12,16].

nub removes duplicate elements from a list.

Edit: @Zgarb found a superfluous $.

Answer 14

JavaScript (Node.js), 69 bytes

n=>[...a=Array(i=n*n)].map(x=>a[q=~(--i%n)*~(i/n)]=q)&&a.filter(x=>x)

Try it online!

JavaScript (Node.js), 76 bytes

n=>[...Array(n*n)].flatMap((_,i,x)=>x.every((_,j)=>~(j/n)*~(j%n)+~i)?[]:i+1)

Try it online!

No recursion like other answers

Answer 15

Mathematica, 25 bytes

Union@@Array[1##&,{#,#}]&

Answer 16

K, 17 bytes

t@<t:?,/t*\:t:1+!

Not much to say here. Sort (t@<t:) the unique items (?) of the flattened (,/) multiplied cartesian self-product (t*\:t:) of 1 up to and including N (1+!).

In action:

 t@<t:?,/t*\:t:1+!5
1 2 3 4 5 6 8 9 10 12 15 16 20 25

Answer 17

J, (削除) 21 (削除ここまで) 20 bytes

Thanks to @Zgarb for -1 byte!

/:~@~.@,@(1*/~@:+i.)

My first J answer! Golfing tips are appreciated, if there is something to golf.

This is a monadic function; we take the outer product by multiplication of the list 1..input with itself, flatten, take unique elements, and sort.

Answer 18

Kotlin, 70 bytes

val a={i:Int->(1..i).flatMap{(1..i).map{j->it*j}}.distinct().sorted()}

Ungolfed version:

val a: (Int) -> List<Int> = { 
 i -> (1..i).flatMap{ j -> (1..i).map{ k -> j * k } }.distinct().sorted()
}

Test it with:

fun main(args: Array<String>) {
 for(i in 1..12) {
 println(a(i))
 }
}

Answer 19

Pyt, 7 bytes

řĐɐ*ƑỤş

Try it online!

ř implicit input (n); push [1,2,...,n]
 Đ duplicate
 ɐ* generate multiplication table
 Ƒ flatten array
 Ụ get unique elements
 ş sort in ascending order; implicit print

Answer 20

Thunno, \$ 12 \log_{256}(96) \approx \$ 9.88 bytes

R1+DZ!.PZU

Explanation

R1+DZ!.PZUz: # Implicit input Example (n=5)
R1+ # [1..input] [1, 2, 3, 4, 5]
 DZ! # Cartesian product with itself [[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [5, 1], [5, 2], [5, 3], [5, 4], [5, 5]]
 .P # Product of each inner list [1, 2, 3, 4, 5, 2, 4, 6, 8, 10, 3, 6, 9, 12, 15, 4, 8, 12, 16, 20, 5, 10, 15, 20, 25]
 ZU # Uniquified [1, 2, 3, 4, 5, 6, 8, 10, 9, 12, 15, 16, 20, 25]
 z: # And sorted [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 20, 25]
 # Implicit output

Screenshot

Screenshot

Answer 21

Vyxal, 7 bytes

ɾẊƛΠ;Us

Try it Online!

• lyxal: Try it Online! for 5 or 4.25 bytes

Answer 22

Jelly, 5 bytes

Ṭẋ)ST

Try it online!

 ) For each x in 1 .. n:
Ṭ Create a list containing 1 at index x and 0s before it,
 ẋ and repeat it x times.
 S Sum the results element-wise,
 T and yield the indices of non-zeros in the result.

Answer 23

Brachylog, 9 bytes

×ばつ∋⟩uo

Try it online!

Explanation

×ばつ∋⟩uo
⟦1 Range from 1 to input number, inclusive
 u Get all unique outputs of this predicate:
 ∋ Pick a number from that range
 ∋ Pick another number from that range
 ⟨ ×ばつ ⟩ Multiply them together
 o Sort the resulting list

Answer 24

Pyth, 10

S{m*Fd^SQ2

Try it Online or run a Test Suite

Answer 25

Minkolang 0.14, (削除) 25 (削除ここまで) (削除) 22 (削除ここまで) 18 bytes

I remembered that I very conveniently implemented Cartesian products before this question was posted!

1nLI20P[x*1R]sS$N.

Try it here. (Outputs in reverse order.)

Explanation

1 Push a 1 onto the stack
 n Take number from input (n)
 L Pushes 1,2,...,n onto the stack
 I Pushes length of stack so 0P knows how many items to pop
 2 Pushes 2 (the number of repeats)
 0P Essentially does itertools.product(range(1,n+1), 2)
 [ Open for loop that repeats n^2 times (0P puts this on the stack)
 x Dump (I know each product has exactly two numbers
 * Multiply
 1R Rotate 1 step to the right
 ] Close for loop
 s Sort
 S Remove duplicates ("set")
 $N. Output whole stack as numbers and stop.

Answer 26

C, 96 bytes

i,a[1<<16];main(n){for(scanf("%d",&n);i<n*n;a[~(i%n)*~(i++/n)]="%d ");while(i)printf(a[i--],i);}

This prints the numbers in descending order. Suggestions are welcomed as this looks far from optimal.

Answer 27

JavaScript (ES6), (削除) 92 (削除ここまで) 90 bytes

n=>eval(`for(r=[],a=n;a;a--)for(b=n;b;)~r.indexOf(x=a*b--)||r.push(x);r.sort((a,b)=>a-b)`)

Explanation

n=>eval(` // use eval to remove need for return keyword
 for(r=[],a=n;a;a--) // iterate for each number a
 for(b=n;b;) // iterate for each number b
 ~r.indexOf(x=a*b--) // check if it is already in the list, x = value
 ||r.push(x); // add the result
 r.sort((a,b)=>a-b) // sort the results by ascending value
 // implicit: return r
`)

Test

N = <input type="number" oninput="result.innerHTML=(
n=>eval(`for(r=[],a=n;a;a--)for(b=n;b;)~r.indexOf(x=a*b--)||r.push(x);r.sort((a,b)=>a-b)`)
)(+this.value)" /><pre id="result"></pre>

Answer 28

Perl 6, 27 bytes

{squish sort 1..$_ X*1..$_} # 27
{unique sort 1..$_ X*1..$_} # 27
{sort unique 1..$_ X*1..$_} # 27

Example usage:

say {squish sort 1..$_ X*1..$_}(3); # (1 2 3 4 6 9)␤
my $code = {squish sort 1..$_ X*1..$_}
for 1..100 -> \N { say $code(N) }
my &code = $code;
say code 4; # (1 2 3 4 6 8 9 12 16)␤

Answer 29

Haskell, 51 bytes

f n=[i|i<-[1..n*n],elem i[a*b|a<-[1..n],b<-[1..n]]]

Pretty boring. Just filters the list [1..n*n] to the elements of the form a*b with a and b in [1..n]. Using filter gives the same length

f n=filter(`elem`[a*b|a<-[1..n],b<-[1..n]])[1..n*n]

I tried for a while to generate the list of products with something more clever like concatMap or mapM, but only got longer results. A more sophisticated check of membership came in at 52 bytes, 1 byte longer, but can perhaps be shortened.

f n=[k|k<-[1..n*n],any(\a->k`mod`a<1&&k<=n*a)[1..n]]

• ბიმო: You can save 3 bytes by using (*)<$>..<*>.. like this

Answer 30

JAVA - 86 Bytes

Set a(int a){Set s=new TreeSet();for(;a>0;a--)for(int b=a;b>0;)s.add(a*b--);return s;}

---

Ungolfed

Set a(int a){
 Set s = new TreeSet();
 for (;a>0;a--){
 for(int b = a;b>0;){
 s.add(a*b--);
 }
 }
 return s;
}