Visualize the greatest common divisor

Question

Background

The greatest common divisor (gcd for short) is a convenient mathematical function, since it has many useful properties. One of them is Bézout's identity: if d = gcd(a, b), then there exist integers x and y such that d = x*a + y*b. In this challenge, your task is to visualize this property with simple ASCII art.

Input

Your inputs are two positive integers a and b, given in any reasonable format. You may also take unary inputs (repetitions of a single printable ASCII character of your choice), but you must be consistent and use the same format for both inputs. The inputs may be in any order, and they may be equal.

Output

Your output is a string s of length lcm(a, b) + 1 (lcm stands for lowest common multiple). The characters of s represent integers from 0 to lcm(a, b). The character s[i] is a lowercase o if i is a multiple of a or b, and a period . otherwise. Note that zero is a multiple of every number. Now, because of Bézout's identity, there will be at least one pair of characters o in s whose distance is exactly gcd(a, b). The leftmost such pair is to be replaced by uppercase Os; this is the final output.

Example

Consider the inputs a = 4 and b = 6. Then we have gcd(a, b) = 2 and lcm(a, b) = 12, so the length of s will be 13. The multiples of a and b are highlighted as follows:

0 1 2 3 4 5 6 7 8 9 10 11 12
o . . . o . o . o . . . o

There are two pairs of os with distance two, but we will only replace the leftmost ones with Os, so the final output is

o...O.O.o...o

Rules and scoring

You can write a full program or a function. The lowest byte count wins, and standard loopholes are disallowed.

Test cases

 1 1 -> OO
 2 2 -> O.O
 1 3 -> OOoo
 4 1 -> OOooo
 2 6 -> O.O.o.o
 2 3 -> o.OOo.o
10 2 -> O.O.o.o.o.o
 4 5 -> o...OO..o.o.o..oo...o
 8 6 -> o.....O.O...o...o.o.....o
12 15 -> o...........O..O........o.....o.....o........o..o...........o
19 15 -> o..............o...o..........o.......o......o...........o..o..............OO.............o....o.........o........o.....o............o.o..............o.o............o.....o........o.........o....o.............oo..............o..o...........o......o.......o..........o...o..............o

• Martin Ender: When taking unary input, can we choose any character? (In particular, how about ., o or O.) Or does it have to be 1? Or 0?
• Zgarb: @MartinBüttner It can be any character, as long as you're consistent and use the same format for both inputs.
• Neil: I'm surprised you didn't use 3 and 5 as one of your test cases.
• Xwtek: Can I use buildin?
• Zgarb: @ChristianIrwan Yes, all built-ins are allowed.

Answer 1

Julia, (削除) 111 (削除ここまで) (削除) 110 (削除ここまで) (削除) 107 (削除ここまで) (削除) 103 (削除ここまで) 96 bytes

f(a,b)=replace(join([i%a*(i%b)<1?"o":"."for i=0:lcm(a,b)]),"o$(d="."^(gcd(a,b)-1))o","O$(d)O",1)

This is a function that accepts two integers and returns a string.

Ungolfed:

function f(a::Int, b::Int)
 # Construct an array of dots and o's
 x = [i % a * (i % b) < 1 ? "o" : "." for i = 0:lcm(a, b)]
 # Join it into a string
 j = join(x)
 # Replace the first pair with distance gcd(a, b) - 1
 replace(j, "o$(d = "."^(gcd(a, b) - 1))o", "O$(d)O", 1) 
end

Saved a byte thanks to nimi!

Answer 2

Retina, (削除) 112 (削除ここまで) (削除) 109 (削除ここまで) (削除) 99 (削除ここまで) (削除) 94 (削除ここまで) 91 bytes

^
. 
+r`(?<!^1円+). (.+) 
$'0ドル
.(?=.* (.+) (.+))(?=1円* |2円* )
o
o(\.*)o((1円\.*o)*) .*
O1ドルO2ドル

Not very competitive, I think, but number theory in Retina is always quite fun. :)

Takes input as unary numbers using . as the unary digit.

Try it online.

Explanation

^
. 

This inserts a . and a space in front of the input. This will ultimately become the output.

+r`(?<!^1円+). (.+) 
$'0ドル

This prepends the LCM of a and b to the string. Since we already have a . there, we'll end up with lcm(a,b)+1. This is accomplished by repeatedly prepending b as long as a does not divide this new prefix. We capture a into a group one and then check if we can reach the beginning of the string by matching that capture at least once. b is then inserted into the string via the rarely used $' which inserts everything after the match into the substitution.

.(?=.* (.+) (.+))(?=1円* |2円* )
o

This one matches characters at positions which are divided by a or b. It makes use of the fact that the result is symmetric: since lcm(a,b) is divided by both a and b going left by subtracting instances of a or b yields the same pattern as going right from 0 by adding them. The first lookahead simply captures a and b. The second lookahead checks that there is a multiple of each a or b characters before the first space.

o(\.*)o((1円\.*o)*) .*
O1ドルO2ドル

As stated on Wikipedia, in addition to Bézout's identity it is also true that

The greatest common divisor d is the smallest positive integer that can be written as ax + by.

This implies that the GCD will correspond to the shortest gap between two os in the output. So we don't have to bother finding the GCD at all. Instead we just look for first instance of the shortest gap. o(\.*)o matches a candidate gap and captures its width into group 1. Then we try to reach the first space by alternating between a backreference to group 1 and os (with optional additional .s). If there is a shorter gap further to the right, this will fail to match, because we cannot get past that gap with the backreference. As soon as all further gaps are at least as wide as the current one, this matches. We capture the end of the LCM-string into group 2 and match the remainder of the string with .*. We write back the uppercase Os (with the gap in between) as well as the remainder of the LCM string, but discard everything starting from the space, to remove a and b from final result.

• Conor O'Brien: I don't know much about Retina number theory, but wouldn't setting the input character to something that does not require escaping save bytes? I.e. (\.*) => (a*)
• Martin Ender: @CᴏɴᴏʀO'Bʀɪᴇɴ Yes, but then I'd have to replace it with . later, which costs four bytes (and getting rid of the escapes only saves 3).
• Conor O'Brien: Ohh. Cool! Very interesting answer.

Answer 3

Jolf, 52 bytes

on*'.wm9jJΡR m*Yhm8jJDN?<*%Sj%SJ1'o'.}"'o%o"n"O%O"n

I will split this code up into two parts.

on*'.wm9jJ
on set n
 *'. to a dot repeated
 m9jJ the gcd of two numeric inputs
ΡR m*Yhm8jJDN?<*%Sj%SJ1'o'.}"'o%o"n"O%O"n
 *Y multiply (repeat) Y (Y = [])
 hm8jJ by the lcm of two inputs + 1
 _m DN } and map the array of that length
 ?<*%Sj%SJ1'o'. "choose o if i%a*(i%b)<1; otherwise choose ."
 R "' join by empty string
Ρ 'o%o"n replace once (capital Rho, 2 bytes): "o"+n+"o"
 "O%O"n with "O"+n+"O"
 implicit printing

Try it here!

• Riker: Shorter than everything else so far. :P
• Conor O'Brien: @RikerW Yes! I'm hoping Jolf will finally win, once.

Answer 4

ESMin, 50 chars / 90 bytes

⩥Мū(îí+1)m$%î⅋$%í?⍘.:⍘o)⨝ċɼ(`o⦃⍘.ĘМũ(îí-1)}o",↪$ú⬮

Try it here (Firefox only).

There must be a way to golf this further!

Explanation

This is a basic two-phase algorithm. It's actually quite simple.

Phase 1

⩥Мū(îí+1)m$%î⅋$%í?⍘.:⍘o)⨝

First, we create a range from 0 to the LCM+1. Then we map over it, checking if either of the inputs is a factor of the current item in the range. If so, we replace that item with an o; otherwise, we replace it with a . . Joining it gives us a series of o's and dots that we can pass to phase two.

Phase 2

ċɼ(`o⦃⍘.ĘМũ(îí-1)}o",↪$ú⬮

This is just one big replace function. A regex is created as o[dots]o, where the amount of dots is determined by the GCD-1. Since this regex is not global, it will only match the first occurrence. After, the match is replaced by O[dots]O using a toUpperCase function.

Answer 5

MATL, 72 bytes

Uses version 6.0.0, which is earlier than this challenge. The code runs in Matlab and in Octave.

2$tZm1+:1-bbvtbw\A~otbZ}ZdXK1+ltb(3X53$X+1K2$lh*t2=f1)tK+hwg1+Ib('.oO'w)

Example

>> matl
 > 2$tZm1+:1-bbvtbw\A~otbZ}ZdXK1+ltb(3X53$X+1K2$lh*t2=f1)tK+hwg1+Ib('.oO'w)
 > 
> 1
> 1
OO
>> matl
 > 2$tZm1+:1-bbvtbw\A~otbZ}ZdXK1+ltb(3X53$X+1K2$lh*t2=f1)tK+hwg1+Ib('.oO'w)
 > 
> 2
> 3
o.OOo.o
>> matl
 > 2$tZm1+:1-bbvtbw\A~otbZ}ZdXK1+ltb(3X53$X+1K2$lh*t2=f1)tK+hwg1+Ib('.oO'w)
 > 
> 12
> 15
o...........O..O........o.....o.....o........o..o...........o

Explanation

I have no idea how it works. I just typed characters randomly. I think there is some convolution involved.

Edit: Try it online! The code in the link has been slightly modified to conform to changes in the language (as of June 2, 2016).

• CalculatorFeline: You can't type a 72 byte program randomly. Will calculate probability later (after sleeping and ACTing for a while)

Answer 6

Vyxal, 25 bytes

ġ/ʀṘ*fDÞǔ‛.oİ‟Ẋ'?ġ+↔;h(ɾ¢

Try it Online! Input as pair of numbers, output as char list.

ġ/ # Divide both by gcd
 ʀ # Range from 0 to each inclusive
 Ṙ* # Multiply by each other
 f # And flatten, resulting in all divisors of the inputs up to lcm(a,b)
 Þǔ # Untruth; create a list with 1s at indices in list
 ‛.oİ # Index into ".o"
 D ‟Ẋ # All pairs of the previous list
 '----;h # Find the first one where
 ↔ # One element is equal to
 ?ġ+ # The other + gcd(input)
 ¢ # At those two indices
 (ɾ # Uppercase

Answer 7

Japt, 83 bytes

'.pD=U*V/(C=(G=@Y?G$($YX%Y :X} $($UV)+1 £Y%U©Y%V?".:o"} $.replace($E=`o{'.pC-1}o`Eu

Not fully golfed yet... And doesn't want to be golfed :/

• ETHproductions: Can you not use r in place of $.replace($?
• nicael: @Eth I haven't figured out how to replace without g flag, so no, I can't.

Answer 8

Javascript, (削除) 170 (削除ここまで) (削除) 164 (削除ここまで) (削除) 161 (削除ここまで) (削除) 153 (削除ここまで) (削除) 145 (削除ここまで) (削除) 141 (削除ここまで) 136 bytes

(a,b)=>[...Array(a*b/(c=(g=(a,b)=>b?g(b,a%b):a)(a,b))+1)].map((x,i)=>i%a&&i%b?'.':'o').join``.replace(`o${e='.'.repeat(c-1)}o`,`O${e}O`)

That's quite lonnnggggg....

Demo, explicitly defined variables because the interpreter uses strict mode.

• Mama Fun Roll: Try replacing i%a<1||i%b<1?'o':'.' with i%a&&i%b?'.':'o'
• Mama Fun Roll: Oh yeah, I think you can alias join.
• nicael: @ןnɟuɐɯɹɐןoɯ thanks, also replacing arrays with simple repeat.
• Mama Fun Roll: Oh, then in that case, you probably shouldn't alias join unless you have 3 occurrences of it.
• Neil: [...Array((d=a*b/(c=(g=(a,b)=>b?g(b,a%b):a)(a,b)))+1).keys()].map(i=>i%a&&i%b?'.':'o') saves you two bytes. (I also tried to use string indexing to create the '.' and 'o' but that actually costs two bytes.)

Answer 9

Python 2, (削除) 217 (削除ここまで) (削除) 200 (削除ここまで) 191 bytes

This is a little blunt, but it works. Any golfing tips are appreciated, (削除) especially if you know how to fix that s[i] = s[v] = "o" problem I encountered, where that would overwrite "O"s (削除ここまで) Got it!

g=lambda a,b:b and g(b,a%b)or a
def f(a,b):
 h=g(a,b);x=1+a*b/h;s=["."]*x;v=k=0
 for i in range(x):
 if(i%a)*(i%b)<1:
 if k:s[i]="o"
 else:k=i==h+v;s[i]=s[v]="oO"[k]
 v=i
 return''.join(s)

Ungolfed:

def gcd(a,b): # recursive gcd function
 if b:
 return g(b,a%b)
 else:
 return a
def f(a,b):
 h = gcd(a,b)
 x = 1 + a*b/h # 1 + lcm(a,b)
 s = ["."] * x
 v = 0
 k = 0
 for i in range(x):
 if i%a == 0 and i % b == 0:
 if k == 0:
 k = (i == h+v) # correct distance apart?
 if k: # if "O" just found
 s[i] = s[v] = "O"
 else:
 s[i] = s[v] = "o"
 else:
 s[i] = "o" # if "O" already found, always "o"
 v = i # If we found an "o" or an "O", i is the new v
 return ''.join(s)

Answer 10

Python 3, 160 bytes

def f(a,b):*o,l=111,lcm(a,b);y='.'*(a*b//l-1);x=[46]*a*b;x[::b]=o*a;x[::a]=o*b;return''.join(map(chr,x[:l])).replace(f'o{y}o',f'O{y}O',1)+'o'
from numpy import*

Try it online!

Uses slice replacement: x[::b]=o*a; x[::a]=o*b

Uses lcm from numpy

Answer 11

Raku (Perl 6) (rakudo), 82 bytes

->@a {{S/o($('.'x+^-[gcd] @a))o/O0ドルO/}([~] map {$[e-mail]??'.'!!'o'},0..[lcm] @a)}

Attempt This Online!