Codegolf Rainbow : Draw in Black-and-White

Question

Introduction:

A rainbow emerges from a forest.
_{Image credit: CC-By-SA 2.5 Wing‐Chi Poon via Wikimedia Commons}

When we look at a rainbow it will always have the colors from top to bottom:
Red; orange; yellow; green; blue; indigo; violet

If we look at these individual rings, the red ring is of course bigger than the violet ring.
In addition, it's also possible to have two or even three rainbow at the same time.

All this above combined will be used in this challenge:

Challenge:

Given an integer n, output that many rings of the (possibly more than one) 'rainbows', where we'll use the letters vibgyor for the colors.

See the test cases below to see how they are built upwards from n=1, and how the spacing should be handled (at n=8). As you can see, one space is added between two rainbows, including spacing at the top, before we add the ring of the next rainbow in line.

Challenge rules:

• You are allowed to use capital VIBGYOR instead of lowercase
• There should be a space between the individual rainbows
• Any amount of leading and/or trailing spaces/new-lines are allowed, as long as the actual rainbow (wherever it is placed on the screen) is correct
• The input will always be a positive integer (>= 1). The behavior when n=0 is therefore undefined, and the program/function can do whatever it wants (outputting nothing; outputting the rainbow of n=1; random output; fail with an error; etc.)
• You are allowed to output a list/array of strings or 2D array/list of characters if you want (you could add the actual pretty-printing code in the TIO footer).
• Ignore the fact that the outputs looks more like Mayan temples than rainbows.. xD

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code.
• Also, adding an explanation for your answer is highly recommended.

Test cases (first n=1 through n=10, and n=25):

1:
 vvv
v v
2:
 iii
 ivvvi
iv vi
3:
 bbb
 biiib
 bivvvib
biv vib
4:
 ggg
 gbbbg
 gbiiibg
 gbivvvibg
gbiv vibg
5:
 yyy
 ygggy
 ygbbbgy
 ygbiiibgy
 ygbivvvibgy
ygbiv vibgy
6:
 ooo
 oyyyo
 oygggyo
 oygbbbgyo
 oygbiiibgyo
 oygbivvvibgyo
oygbiv vibgyo
7:
 rrr
 rooor
 royyyor
 roygggyor
 roygbbbgyor
 roygbiiibgyor
 roygbivvvibgyor
roygbiv vibgyor
8:
 vvv
 v v
 v rrr v
 v rooor v
 v royyyor v
 v roygggyor v
 v roygbbbgyor v
 v roygbiiibgyor v
 v roygbivvvibgyor v
v roygbiv vibgyor v
9:
 iii
 ivvvi
 iv vi
 iv rrr vi
 iv rooor vi
 iv royyyor vi
 iv roygggyor vi
 iv roygbbbgyor vi
 iv roygbiiibgyor vi
 iv roygbivvvibgyor vi
iv roygbiv vibgyor vi
10:
 bbb
 biiib
 bivvvib
 biv vib
 biv rrr vib
 biv rooor vib
 biv royyyor vib
 biv roygggyor vib
 biv roygbbbgyor vib
 biv roygbiiibgyor vib
 biv roygbivvvibgyor vib
biv roygbiv vibgyor vib
25:
 ggg
 gbbbg
 gbiiibg
 gbivvvibg
 gbiv vibg
 gbiv rrr vibg
 gbiv rooor vibg
 gbiv royyyor vibg
 gbiv roygggyor vibg
 gbiv roygbbbgyor vibg
 gbiv roygbiiibgyor vibg
 gbiv roygbivvvibgyor vibg
 gbiv roygbiv vibgyor vibg
 gbiv roygbiv rrr vibgyor vibg
 gbiv roygbiv rooor vibgyor vibg
 gbiv roygbiv royyyor vibgyor vibg
 gbiv roygbiv roygggyor vibgyor vibg
 gbiv roygbiv roygbbbgyor vibgyor vibg
 gbiv roygbiv roygbiiibgyor vibgyor vibg
 gbiv roygbiv roygbivvvibgyor vibgyor vibg
 gbiv roygbiv roygbiv vibgyor vibgyor vibg
 gbiv roygbiv roygbiv rrr vibgyor vibgyor vibg
 gbiv roygbiv roygbiv rooor vibgyor vibgyor vibg
 gbiv roygbiv roygbiv royyyor vibgyor vibgyor vibg
 gbiv roygbiv roygbiv roygggyor vibgyor vibgyor vibg
 gbiv roygbiv roygbiv roygbbbgyor vibgyor vibgyor vibg
 gbiv roygbiv roygbiv roygbiiibgyor vibgyor vibgyor vibg
 gbiv roygbiv roygbiv roygbivvvibgyor vibgyor vibgyor vibg
gbiv roygbiv roygbiv roygbiv vibgyor vibgyor vibgyor vibg

• Chris M: Maybe you know this and it's by design (I know that rainbows are not rhombuses or ascii, too, and higher orders' positions get more complicated), but aren't the colours reversed in the 2nd rainbow?
• Kevin Cruijssen: @ChrisM Ah, you're indeed right. With two rainbows the second is indeed reversed, but with three only the outer is reversed, and with four both outer are reversed. Ah well, bit too late to change it now. Maybe I'll think of a third related challenges with this later on. :)
• Kevin Cruijssen: @ChrisM Added a third related challenge implementing your comment to the Sandbox.
• Chris M: Oh cool, nice one :¬)

Answer 1

Python 2, (削除) 84 (削除ここまで) (削除) 82 (削除ここまで) 81 bytes

-2 bytes thanks to ElPedro.

n=input();n+=n/7;w=''
while-~n:w+=" vibgyor"[n%8];print' '*n+w+w[-1]+w[::-1];n-=1

Try it online!

Answer 2

Dyalog APL, (削除) 41 (削除ここまで) (削除) 39 (削除ここまで) 38 bytes

↑{⌽(⌽,⊃,A↑⊢)⍵↑A⍴' vibgyor'} ×ばつ8÷7

Try it online!

A similar approach to others: ×ばつ8÷7 finds the height of the rainbow (also the width of the longest 'half row' to the left/right of the centre) and assigns it to A for later use, while ̈-⍳ iterates through the values 1..A, negating them to select on the correct side when used with ↑.

A⍴' vibgyor' generates a 'half row' and ⍵↑ selects the correct length substring. (⌽,⊃,A↑⊢) generates the full row in reverse (which takes fewer characters to do), starting with a reversed half row (⌽), then the centre character taken from the beginning of the half row string (⊃) and finally a right padded version of the half row (A↑⊢). The final ⌽ reverses the row into the correct orientation and ↑ turns the vector of rows into a 2D array.

Edit: -2 thanks to dzaima

Edit: -1 thanks to ngn

• dzaima: You can replace ⍕⍪ with ↑ - outputting a 2D array of characters is allowed
• dzaima: 39 bytes
• ngn: 1+÷7 -> 8÷7

Answer 3

JavaScript (ES6), 100 bytes

Returns an array of strings.

f=(n,a=[i=' '])=>++i<n+n/7?f(n,[c=' vibgyor'[i&7],...a].map(s=>c+s+c)):a.map(s=>' '.repeat(--i)+s)

Try it online!

Answer 4

05AB1E, (削除) 32 (削除ここまで) (削除) 31 (削除ここまで) 23 bytes

.•VvÈ©•1∍¬„ v:Rηε¬ý}.c

Try it online!

-1 thanks to Kevin Cruijssen and -8 thanks to Adnan

---

Explanation (Stack example w/ input of 3):

.•VvÈ©• # Push 'aibgyor' | ['aibgyor']
 1∍ # Extend to input length. | ['aib']
 ¬ # Push head. | ['aib','a']
 „ v: # Replace with ' v'. | [' vib']
 R # Reverse. | ['biv ']
 η # Prefixes. | ['b', 'bi', 'biv', 'biv ']
 ε } # For each.... | []
 ¬ý # Bifurcate, join by head. | ['b','b'] -> ['bbb']
 | ['bi','ib'] -> ['biiib']
 | ['biv','vib'] -> ['bivvvib']
 | ['biv ',' vib'] -> ['biv vib']
 .c # Center the result. | Expected output.

• Kevin Cruijssen: "vibgyor" can be golfed by 1 byte to .•2Bãθ(•. (Here the explanation for it at the "How to compress strings not part of the dictionary?" section.) In addition, „vr...v r:ð« can be golfed to 'v„v .:. So 28 bytes. Nice answer though, +1 from me.
• Magic Octopus Urn: @KevinCruijssen oh I know all about string compression-- I just figured (as is usually the case with super small strings) that the 3-byte bloat of .•• wouldn't save a byte. Figures the one time I don't actually check is the time it would save bytes xD. Good catch
• Adnan: For 23 bytes: .•VvÈ©•¹∍¬„ v:Rηε¬ý}.c
• Magic Octopus Urn: @Adnan fair enough, the lack of the mirror makes it a significant refactor in my eyes though (that and I doubt I'd've got there on my own :P).
• Magic Octopus Urn: @Adnan also ¬ý is genius... Jeez... You think so differently and it's awesome.

Answer 5

Canvas, (削除) 29 (削除ここまで) (削除) 28 (削除ここまで) 26 bytes

7÷U+{ <ibg×ばつ/n}⇵K2*∔─↶

Try it here!

Explanation:

7÷U+ ceil(input/7) + input
 { } for n in 1..the above
 <ibgyor@ in the string " <ibgyor", pick the nth character
 ×ばつ repeat n times
 / create a diagonal of that
 n and overlap the top 2 stack items (the 1st time around this does nothing, leaving an item for the next iterations)
 ⇵ reverse the result vertically
 K take off the last line (e.g. " <ibgyor <ib")
 2* repeat that vertically twice
 ∔ and append that back to the diagonals
 ─ palindromize vertically
 ↶ and rotate 90° anti-clockwise. This rotates "<" to "v"

(削除) 25 (削除ここまで) (削除) 24 (削除ここまで) 22 bytes after fixing that mold should cycle if the wanted length is bigger than the inputs length and fixing / for like the 10th time

• Kevin Cruijssen: Out of curiosity, why is the v sideways (<)? Is v already a reserved keyword in Canvas, if so, why not use < as that reversed keyword instead?
• dzaima: the characters are used vertically and then rotated, and Canvas is smart enough to figure out that < rotated 90° anticlockwise equals v :P All ASCII are part of strings in Canvas btw
• Kevin Cruijssen: Ah ok, so if you would have used v here, and you then rotate it 90 degrees counterclockwise, it becomes > instead. I see. :D
• dzaima: @KevinCruijssen That'd have other consequences too
• Kevin Cruijssen: Ah, you also have a mirror there somewhere? I can't read Canvas tbh, so I'm looking forward to the explanation of the code. ;)

Answer 6

Haskell, (削除) 114 (削除ここまで) (削除) 110 (削除ここまで) 101 bytes

Thanks to [nimi][1] for -(削除) 4 (削除ここまで) 13 bytes!

f n=""#(n+1+div n 7)
w#0=[]
w#n|x<-cycle"r vibgyo"!!n=((' '<$[2..n])++reverse w++x:x:x:w):(x:w)#(n-1)

Try it online!

Answer 7

Python 2, 108 bytes

n=input()-1
n+=n/7+2
o=[]
for s in(' vibgyor'*n)[:n]:o=[s+l+s for l in[s]+o]
for l in o:print l.center(n-~n)

Try it online!

Answer 8

Charcoal, 30 bytes

↶≔... vibg×ばつ8⊖N7θθ⸿Eθ✂θκ‖O←

Try it online! Link is to verbose version of code. Explanation:

↶

Change the drawing direction to upwards.

≔... vibg×ばつ8⊖N7θ

Calculate the height of the rainbow and repeat the literal string to that length.

θ⸿

Print the central line of the rainbow.

Eθ✂θκ

Print the right half of the rainbow by taking successive slices and printing each on its own "line".

‖O←

Reflect to complete the rainbow.

Answer 9

Jelly, 31 bytes

:7+‘μ" vibgyor"ṁṚ,Ṛjṛ/ƲƤṭ"ḶṚ6ẋƲ

Try it online!

Check out a test suite!

ಠ_ಠ This is overly complicated because Jelly doesn't have a centralize function...

Answer 10

Jelly, 28 bytes

:7+‘" vibgyor"ṁμṫJZz6U;"8ŒBṚ

A monadic link accepting an integer which yields a list of lists of characters.

Try it online! (footer joins with newline characters)

Or see the test-suite.

How?

:7+‘" vibgyor"ṁμṫJZz6U;"8ŒBṚ - Link: integer
:7 - integer divide by seven (number of full rainbows)
 ‘ - increment (the input integer)
 + - add (gets the number bands)
 " vibgyor" - list of characters = " vibgyor"
 ṁ - mould like the result above (as a range)
 μ - start a new monadic chain
 J - range of length
 ṫ - tail (vectorises) (gets the suffixes)
 Z - transpose
 z6 - transpose with filler space character
 - (together these pad with spaces to the right)
 U - reverse each
 - (now we have the left side of the rainbow upside down)
 8 - chain's left argument, as right argument of...
 " - zip with:
 ; - concatenation
 - (adds the central character)
 ŒB - bounce (vectorises at depth 1)
 - (reflects each row like [1,2,3,4] -> [1,2,3,4,3,2,1])
 Ṛ - reverse (turn the rainbow up the right way)

Answer 11

R, 130 bytes

function(n,k=n%/%7*8+1+n%%7,a=el(strsplit(' vibgyor'/k,'')))for(i in k:1)cat(d<-' '/(i-1),a[c(k:i,i,i:k)],d,sep='','
')
"/"=strrep

Try it online!

• -6 bytes thanks to @JayCe

• JayCe: You can save 6 bytes. Also make you to upvote R's nomination for language of the month :)
• digEmAll: @JayCe: thanks ! updated and upvoted !

Answer 12

Haskell, 106 (削除) 113 (削除ここまで) bytes

I can't yet comment other posts (namely this) so I have to post the solution as a separate answer.

Golfed away 7 bytes by ovs

p x=reverse x++x!!0:x
u m|n<-m+div(m-1)7=[(' '<$[z..n])++p(drop(n-z)$take(n+1)$cycle" vibgyor")|z<-[0..n]]

Try it online!

(Old version, 113 bytes)

• Kevin Cruijssen: Nice answer. I don't know Haskell, but the code seems rather different from the other Haskell answer. PS: That other Haskell answer is actually 110 bytes after the golf-tips in @nimi's comment. Regardless, this is a nice alternative Haskell answer, so +1 from me.
• ovs: It seems like list comprehensions were the better tool for this challenge. I was able to golf your solution to 106 bytes by reducing the total number of functions. Feel free to adapt these changes.

Answer 13

PowerShell, (削除) 108 (削除ここまで) (削除) 98 (削除ここまで) (削除) 89 (削除ここまで) 85 bytes

param($x)($x+=$x/7-replace'\..*')..0|%{' '*$_+-join(" vibgyor"*$x)[$x..$_+$_+$_..$x]}

Try it online!

This one feels pretty alright now. Banker's rounding is still the devil and I figured out how to make a non-dumb join. I tried monkeying with $ofs to not much success. Speaking of, the results without joins look pretty good, a bit melty:

 vvv
 v v
 v rrr v
 v r ooo r v
 v r o yyy o r v
 v r o y ggg y o r v
 v r o y g bbb g y o r v
 v r o y g b iii b g y o r v
 v r o y g b i vvv i b g y o r v
v r o y g b i v v i b g y o r v

• mazzy: [int]$x+=$x/7 ?
• Veskah: @mazzy That fails for x=25. You have to truncate but casting to int rounds
• mazzy: Yes. And truncate is works
• Veskah: @mazzy I know, the only ways I know to truncate is either [math]::truncate() or the regex trick used above. [int]$x rounds the number. If you know a better way, I'm all ears.

Answer 14

Python 2, (削除) 132 (削除ここまで) 131 bytes

def f(n):
 t=n+n/7;s=('vibgyor '*n)[:t];r=[s[~i:]+t*' 'for i in range(t)]
 for l in zip(*r+3*[' '+s]+r[::-1])[::-1]:print''.join(l)

Try it online!

---

Saved:

• -1 byte, thanks to Jonathan Frech

• Jonathan Frech: Why // in Python 2?
• TFeld: @JonathanFrech Because I'm dumb :P
• TFeld: @dzaima, Yeah. Should be fixed now

Answer 15

Red, 153 bytes

func[n][r: take/last/part append/dup copy"""roygbiv "n l: 9 * n + 8 / 8
repeat i l[print rejoin[t: pad/left take/part copy r i l last t reverse copy t]]]

Try it online!

Slightly more readable:

f: func[ n ] [
 r: copy ""
 append/dup r "roygbiv " n
 r: take/last/part r l: 9 * n + 8 / 8
 repeat i l [
 print rejoin [ t: pad/left take/part copy r i l
 last t 
 reverse copy t ]
 ]
]

Answer 16

Java (JDK 10), 184 bytes

n->{int h=n+n/7,i=h+1,w=i*2+1,j,k=0;var o=new char[i][w];for(;i-->0;o[i][w/2]=o[i][w/2+1])for(j=w/2;j-->0;)o[i][j]=o[i][w+~j]=i<h?j<1?32:o[i+1][j-1]:" vibgyor".charAt(k++%8);return o;}

Try it online!

Prints an extra leading and trailing space for each multiple of 7.

Explanation

n->{ // IntFunction
 int h=n+n/7, // Declare that height = n + n/7
 i=h+1, // that index = h + 1
 w=i*2+1, // that width = (h+1)*2+1
 j, // j
 k=0; // that k = 0
 var o=new char[i][w]; // Declare a 2D char array
 for(; // Loop
 i-->0; // Until i is 0
 o[i][w/2]=o[i][w/2+1] // After each run, copy the middle letter.
 )
 for(j=w/2; // Loop on j = w/2
 j-->0; // Until j = 0
 ) //
 o[i][j] // copy letters to the left side,
 =o[i][w+~j] // and the right side
 =i<h // if it's not the last line
 ?j<1 // if it's the first (and last) character
 ?32 // set it to a space.
 :o[i+1][j-1] // else set it to the previous character on the next line.
 :" vibgyor".charAt(k++%8); // else assign the next letter.
 return o; // return everything
}

Credits

• -2 bytes thanks to Kevin Cruijssen

• Kevin Cruijssen: You can save 2 bytes by changing ,w=-~h*2+1,i=h+1 to ,i=h+1,w=i*2+1
• Olivier Grégoire: Wow, I should really not golf passed midnight! Thanks for this, @KevinCruijssen! :)
• ceilingcat: Suggest i-~i instead of i*2+1

Answer 17

Stax, 23 bytes

⌡G'5h!M╩EV[Ez ▼>≈<S⌡⌡0`

Run and debug it

Unpacked, ungolfed, and commented, it looks like this.

" vibgyor" string literal
,8*7/^ input * 8 / 7 + 1
:m repeat literal to that length
|] get all prefixes
Mr rectangularize, transpose array of arrays, then reverse
 this is the same as rotating counter-clockwise
m map over each row with the rest of the program, then implicitly output
 the stack starts with just the row itself
 _h push the first character of the row
 _r push the reversed row
 L wrap the entire stack in a single array

Run this one

Answer 18

V (vim), 105 bytes

"aDiaibgyor<esc>0vg_y$@ap0l@alhd$:s/a/ v/g
qqYP0xjYP!!rev
gJkhq2@a@qylpp02ld0ドルVG:center<c-r>=2*(<c-r>a+ceil(1+<c-r>a/7))

Try it online!

There's many possible saves here, but for now I'm just happy I got it working.