Round up my number

Question

^Sandbox

Given two positive integers a and b, return the smallest multiple of b that is greater than or equal to a.

Examples

Definitions: true = is a multiple of b, false = is not a multiple of b.

• f(1,3) = 3 because 1 is false, 1+1 is false, but 1+1+1 is true. Therefore 3 is returned.

• f(5,3) = 6 because 5 is false, but 5+1 is true. Therefore 6 is returned.

• f(9,3) = 9 because 9 is already a multiple of 3.

• f(12,5) = 15 because 12 is false, 12+1 is false, 12+2 is false, but 12+3 is true. Therefore 15 is returned.

Rules

• Input/output can be given by any convenient method.
• You can print it to STDOUT or return it as a function result.
• Either a full program or a function are acceptable.
• Any amount of extraneous whitespace is permitted, provided the characters line up appropriately.
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

The Catalogue

The Stack Snippet at the bottom of this post generates the catalogue from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](https://esolangs.org/wiki/Fish), 121 bytes

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• flawr: I think from your wording it is not so clear what actually should be returned. I'd just rephrase the challenge to something like "Given two positive integers a and b, return the smallest number that is at least as big as a which is also a multiple of b."

Answer 1

Python, 17 bytes

lambda a,b:a+-a%b

Try it online!

Since this doesn't use division, it works for both Python 2 and 3. Note that we can't cut out the + sign because a-a%b does a-(a%b), whereas we want to do(-a)%b and add that to a. We could also write it as -a%b+a.

Answer 2

Japt, 2 bytes

cV

Try it

cV :Implicit input of integers U & V
cV :Round U up to the nearest multiple of V

With input as an array:

rc

Try it

rc :Implicit input of array
r :Reduce by
 c :Rounding the first element up to the nearest multiple of the second

Less Trivial!, 6 bytes

_vV}aU

Try it

_vV}aU :Implicit input of integers U & V
_ :Function taking an integer Z as input
 vV : Is Z divisible by V?
 } :End function
 aU :Starting from U, return the first integer that returns true when passed through that function

ôV ævV

Try it

ôV ævV :Implicit input of integers U & V
ôV :Range [U,U+V]
 æ :Get first element
 vV : Divisible by V

• findusl: Your description bascially repeats the task. I don't know japt, is it possible to describe what each character does?
• Shaggy: @findusl, "full" explanation added!

Answer 3

Intcode (day 5), (削除) 70 (削除ここまで) (削除) 66 (削除ここまで) 64 bytes

3,0,3,2,102,-1,2,4,1,0,4,0,1,1,8,1,7,9,0,5,5,5,14,2,1,2,0,4,0,99

Try it online!

Intcode is a programming language invented for Advent of Code 2019. It was first introduced on day 2, then further expanded upon on day 5, which is the version used here.

Since on PPCG, languages are defined by their interpreters, I am using the interpreter that I wrote for my day 5 solution (which was written before this challenge).

Explanation:

3,0, store first input in cell 0
3,2, store second input in cell 2
 (cell 1 is a counter, because it's already initialized to zero)
102,-1,2,4, negate the second input and store in cell 4
 (begin loop)
1,0,4,0, subtract the second input from the first input
1,1,8,1, increment the counter (by adding cell 8)
7,9,0,5, if the first input is greater than zero (cell 9)...
5,5,14, ... jump to the beginning of the loop (reusing the cell 8 reference)
2,1,2,0, multiply the counter by the second input
4,0,99 output and quit

Answer 4

x86 Machine Code, 7 bytes

Since x86 integer division instructions do only floor rounding, this uses the formula: f(a,b) = floor((a+b-1)/b)*b to accomplish ceiling rounding.

02 C3 ADD AL, BL ; a = a + b
48 DEC AX ; a = a - 1
F6 F3 DIV BL ; a = a / b
F6 E3 MUL BL ; a = a * b

Input is in AL and BL, output is AL.

Testing

Here is output from a test program for PC DOS that takes interactive input.

enter image description here

Download and test ROUNDMY.COM.

Answer 5

MATL, 4 bytes

:ien

Try it online!

Explanation

Consider inputs 12, 5.

 % Implicit input
: % Range
 % STACK: [1 2 3 4 5 6 7 8 9 10 11 12]
i % Input
 % STACK: [1 2 3 4 5 6 7 8 9 10 11 12], 5
e % Reshape as a matrix with that number of rows, padding with zeros
 % STACK: [1 6 11;
 2 7 12;
 3 8 0;
 4 9 0;
 5 10 0]
n % Number of elements
 % STACK: 15
 % Implicit display

• Galen Ivanov: Shouldn't the stack after Range be [1 2 3 4 5 6 7 8 9 10 11 12] ?
• Luis Mendo: @Galen Yes, thank you! Corrected

Answer 6

Octave, 17 bytes

Same method as this JS answer:

@(a,b)ceil(a/b)*b

Try it online!

Answer 7

Python 2, (削除) 23 (削除ここまで) (削除) 20 (削除ここまで) 18 bytes

lambda a,b:-a/b*-b

Try it online!

2 bytes thanks to tsh

• Raserhin: My guess is that is not necessary as Python 2 division (/) actually returns an integer itself. And in Python 3 you can directly use // as a floor division. Both of wich are more compact expressions.
• Neil: a/-b*-b also works, and is slightly more readable, I think?
• Chas Brown: @Neil - Readability?!? Blasphemy! :)

Answer 8

Haskell, 14 bytes

a%b=a+mod(-a)b

Try it online!

Answer 9

C++ (gcc), (削除) 43 (削除ここまで) (削除) 41 (削除ここまで) (削除) 40 (削除ここまで) (削除) 36 (削除ここまで) 25 bytes

#define f(a,b)(b-a%b)%b+a

Try it online!

Saved 4 bytes thanks to @Neil!!!

Saved 11 bytes thanks to @AZTECCO!!!

• Neil: -4 bytes: (b-a%b)%b+a
• Noodle9: @Neil Thanks so much - absolutely brilliant!
• AZTECCO: 25 using #macro
• AZTECCO: @Neil the same solution I suggested here codegolf.stackexchange.com/a/196710/84844 in the last comment, which is basically a port of xnor solution
• Noodle9: @AZTECCO Wow! I was assuming we're stuck with the first run of bytes as the shortest possible in C++ but you came up with a way forwards!

Answer 10

JavaScript (Node.js), 17 bytes

a=>b=>a+(b-a%b)%b

Try it online!

• -1 byte by Neil

• Neil: a-(a%b||b)+b=b-(a%b||b)+a=(b-a%b)%b+a to save a byte.
• Neil: Wasn't enough though, someone's come up with a 15-byter!

Answer 11

Jelly, 3 bytes

×ばつ

Try it online!

Simply divide, ceiling, multiply.

Answer 12

C (gcc), (削除) 47 (削除ここまで) (削除) 31 (削除ここまで) 25 bytes

f(a,b){a=a%b?f(++a,b):a;}

Try it online!

-16 bytes thanks to A̲̲ (forgot that return types and parameters would default to int if not defined, whoops!)

-6 thanks to AZTECCO for using a ternary and making the function recursive

• AZTECCO: You can save 1 with ternary Try it online!
• AZTECCO: 25
• AZTECCO: Try it online! using Tcc and a different approach

Answer 13

Wolfram Language (Mathematica), 7 bytes

Ceiling

Just pointing out that Mathematica's builtin, which is most commonly called with one argument, actually does exactly the desired operation when called with two arguments.

Answer 14

Brainfuck, (削除) 52 (削除ここまで) 49 bytes

Thanks to Jo King for finding a bug, helping fix it with some extra golfing.

,>>,<<[>+>->+<[>]>[<+>-]<<[<]>-]>>>[<[-<+>]>,]<<.

Input and output is in character codes.

Try it online!

Explanation

,>>,<< [>a, 0, b, 0]
[>+>->+<[>]>[<+>-]<<[<]>-] [>0, a, b-a%b, a%b]
>>>[<[-<+>]>,] If a%b is nonzero, add b-a%b to a, and clear a%b
<<. Output a+(b-a%b)

Modulus algorithm is straight from the esolang list of brainfuck algorithms.

• Wheat Wizard: Welcome to the site!
• Exalted Toast: Yikes. Yeah, somehow it slipped my mind "outputting a null won't matter" when I'm using character codes. Will update the code to your version.
• Exalted Toast: Okay listen, just 'cause you're right...

Answer 15

JavaScript (V8), 22 bytes

This is the function \$(a,b) \mapsto \left\lceil \frac{a}{b} \right\rceil b\$.

a=>b=>Math.ceil(a/b)*b

Try it online!

Answer 16

R, 20 bytes

function(a,b)a+-a%%b

Try it online!

Answer 17

K (oK), 8 bytes

{x+y!-x}

Try it online!

Another port of xnor's Python answer.

Answer 18

Vyxal, 3 bytes

/⌈*

Try it Online!

Answer 19

Retina 0.8.2, 28 bytes

\d+
$*
r`1+(2円*) (1+)
1ドル2ドル
1

Try it online! Link includes test cases. Explanation:

\d+
$*

Convert the inputs to unary.

r`

Process this stage from right-to-left, so that 2円 gets to match \$ b \$ first, allowing us to refer to it when matching within \$ a \$.

1+(2円*) (1+)

Match \$ b \$ in 2円, then as many copies of \$ b \$ as we can, without completely exhausting \$ a \$. 1円 thus matches \$ b \lfloor \frac {a - 1} b \rfloor = b \lceil \frac {a - b} b \rceil \$

1ドル2ドル

Add that to \$ b \$ to give \$ b \lceil \frac a b \rceil \$.

1

Convert to decimal.

Answer 20

Ruby, 13 bytes

->a,b{-a%b+a}

Try it online!

Answer 21

Red, 31 bytes

func[a b][round/to/ceiling a b]

Try it online!

Just a built-in round with refinements to, which scales the first argument according to the second, and ceiling, which indicates the direction.

Red, 25 bytes

func[a b][0 - a // b + a]

Try it online!

A port of xnor's Python answer. Don't forget to upvote his answer!

Answer 22

Prolog (SWI) + CLPFD, 30 bytes

f(R,X,Y):-R#=_*Y,R-Y#<X,R#>=X.

Try it online!

f(R,X,Y)

f is a predicate taking in a variable R, and two integers X and Y.

The CLPFD constraints are expressed using the #-prefixed operators. , means and, so f is true iff all these constraints are satisfied, i.e.,

1) R#=_*Y R is a multiple of Y , and

2) R-Y#<X R-Y < X , and

3) R#>=X R>= X .

Answer 23

SimpleTemplate, 61 bytes

This was quite fun! Sadly, I had to do the dumb way :(

{@fn F A,B}{@whileA is notmultipleB}{@incA}{@/}{@returnA}{@/}

Creates a function that iterates until it finds a number that's multiple of the 2nd number.

This was actually way smaller than doing it without a function.

---

Ungolfed:

{@fn F numberA, numberB}
 {@while numberA is not multiple of numberB}
 {@inc by 1 numberA}
 {@/}
 {@return numberA}
{@/}

Should be mostly straightforward.

You can see it running on https://ideone.com/MR7t1E (golfed version only).

• user85052: Glad you enjoyed it! I find simple challenges seem to capture folks' imaginations and are more likely to take off than something overcomplicated. --AJFaraday

Answer 24

Charcoal, 10 bytes

×ばつθ÷Nθ

Try it online! Link is to verbose version of code. Takes \$ b \$ first and \$ a \$ second. Explanation:

≔±Nθ

Input \$ b \$, negate it, and save the result.

×ばつθ÷Nθ

Input \$ a \$ and compute \$ -b \lfloor \frac a {-b} \rfloor \$, then cast to string for implicit print.

• Neil: For those of you counting, this was my 3000th answer!

Answer 25

GolfScript, 6 bytes

.~)@%+

Try it online!

Function which consumes A and B from the stack, assuming B is below A.

Explanation:

 #Example input: A = 12, B = 5 stack is 5 12
.~) #Duplicate A and negate the duplicate stack is 5 12 -12
 @ #Pull 3rd from top to top stack is 12 -12 5
 % #Second to top mod top stack is 12 3
 + #Add remaining values stack is 15

Answer 26

Javascript (ES6), 19 bytes

a=>b=>a%b?a-a%b+b:a

Naive attempt, pretty sure this can be golfed further.

Answer 27

Java, 99 bytes

static int r(int a,int b){int t=a%b<1||b%a<1?a:r(a+1,b);return t;}

Answer 28

JavaScript, 15 bytes

a=>b=>--a+b-a%b

More expressively, the answer is a => b => a + b-((a-1)%b) - 1, but by decrementing a we can handle both the "subtract one" steps at once, saving a couple of bytes. Then we just get lucky with operator presidencies.

Answer 29

Perl 5, (111 bytes with footer) 52 bytes

sub f{($a,$b)=@_;$c=$b;$c++until!($c%$b)&&$c>=$a;$c}

Try it online!

Takes a and b as arguments to a subroutine named f and increases $c until it is both a multiple of $b and greater than or equal to $a. Then returns $c (the last value evaluated in a subroutine is the return value).

Answer 30

Java (JDK), 15 bytes

b->a->--a+b-a%b

Try it online!