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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 , 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
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    \$\begingroup\$ 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? \$\endgroup\$ Commented Aug 14, 2018 at 12:42
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    \$\begingroup\$ @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. :) \$\endgroup\$ Commented Aug 14, 2018 at 12:51
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    \$\begingroup\$ @ChrisM Added a third related challenge implementing your comment to the Sandbox. \$\endgroup\$ Commented Aug 16, 2018 at 15:09
  • \$\begingroup\$ Oh cool, nice one :¬) \$\endgroup\$ Commented Aug 16, 2018 at 16:04

18 Answers 18

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Python 2, 84 82 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!

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4
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Dyalog APL, 41 39 38 bytes

↑{⌽(⌽,⊃,A↑⊢)⍵↑A⍴' vibgyor'}¨-⍳A←⌈⎕×8÷7

Try it online!

A similar approach to others: A←⌈⎕×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

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  • \$\begingroup\$ You can replace ⍕⍪ with ↑ - outputting a 2D array of characters is allowed \$\endgroup\$ Commented Aug 16, 2018 at 3:47
  • \$\begingroup\$ 39 bytes \$\endgroup\$ Commented Aug 16, 2018 at 4:14
  • \$\begingroup\$ 1+÷7 -> 8÷7 \$\endgroup\$ Commented Aug 25, 2018 at 15:04
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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)

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05AB1E, 32 31 23 bytes

.•VvÈ©•¹∍¬„ v:Rηε¬ý}.c

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-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.
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    \$\begingroup\$ "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. \$\endgroup\$ Commented Aug 14, 2018 at 14:51
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    \$\begingroup\$ @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 \$\endgroup\$ Commented Aug 14, 2018 at 15:14
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    \$\begingroup\$ For 23 bytes: .•VvÈ©•¹∍¬„ v:Rηε¬ý}.c \$\endgroup\$ Commented Aug 14, 2018 at 16:30
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    \$\begingroup\$ @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). \$\endgroup\$ Commented Aug 14, 2018 at 16:47
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    \$\begingroup\$ @Adnan also ¬ý is genius... Jeez... You think so differently and it's awesome. \$\endgroup\$ Commented Aug 15, 2018 at 14:53
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Canvas, 29 28 26 bytes

７÷Ｕ＋｛ <ibgyor＠¹×／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 ¹× 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"

25 24 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

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  • \$\begingroup\$ 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? \$\endgroup\$ Commented Aug 14, 2018 at 12:11
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    \$\begingroup\$ 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 \$\endgroup\$ Commented Aug 14, 2018 at 12:13
  • \$\begingroup\$ Ah ok, so if you would have used v here, and you then rotate it 90 degrees counterclockwise, it becomes > instead. I see. :D \$\endgroup\$ Commented Aug 14, 2018 at 12:15
  • \$\begingroup\$ @KevinCruijssen That'd have other consequences too \$\endgroup\$ Commented Aug 14, 2018 at 12:15
  • \$\begingroup\$ 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. ;) \$\endgroup\$ Commented Aug 14, 2018 at 12:17
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Haskell, 114 110 101 bytes

Thanks to [nimi][1] for -4 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)

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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)

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Charcoal, 30 bytes

↶≔… vibgyor⁺²÷×⁸⊖Ｎ⁷θθ⸿Ｅθ✂θκ‖Ｏ←

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

↶

Change the drawing direction to upwards.

≔… vibgyor⁺²÷×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.

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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...

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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)
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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
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  • \$\begingroup\$ You can save 6 bytes. Also make you to upvote R's nomination for language of the month :) \$\endgroup\$ Commented Aug 16, 2018 at 19:41
  • \$\begingroup\$ @JayCe: thanks ! updated and upvoted ! \$\endgroup\$ Commented Aug 16, 2018 at 20:32
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Haskell, 106 113 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)

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  • \$\begingroup\$ 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. \$\endgroup\$ Commented Aug 17, 2018 at 13:51
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    \$\begingroup\$ 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. \$\endgroup\$ Commented Aug 17, 2018 at 15:35
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PowerShell, 108 98 89 85 bytes

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

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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
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  • \$\begingroup\$ [int]$x+=$x/7 ? \$\endgroup\$ Commented Aug 17, 2018 at 4:21
  • \$\begingroup\$ @mazzy That fails for x=25. You have to truncate but casting to int rounds \$\endgroup\$ Commented Aug 17, 2018 at 20:05
  • \$\begingroup\$ Yes. And truncate is works \$\endgroup\$ Commented Aug 18, 2018 at 5:37
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    \$\begingroup\$ @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. \$\endgroup\$ Commented Aug 20, 2018 at 20:01
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Python 2, 132 131 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)

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Saved:

  • -1 byte, thanks to Jonathan Frech
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  • \$\begingroup\$ Why // in Python 2? \$\endgroup\$ Commented Aug 14, 2018 at 12:03
  • \$\begingroup\$ @JonathanFrech Because I'm dumb :P \$\endgroup\$ Commented Aug 14, 2018 at 12:06
  • \$\begingroup\$ @dzaima, Yeah. Should be fixed now \$\endgroup\$ Commented Aug 14, 2018 at 12:22
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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]]]

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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 ] ] ]
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1
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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;}

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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

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  • \$\begingroup\$ You can save 2 bytes by changing ,w=-~h*2+1,i=h+1 to ,i=h+1,w=i*2+1 \$\endgroup\$ Commented Aug 16, 2018 at 6:29
  • \$\begingroup\$ Wow, I should really not golf passed midnight! Thanks for this, @KevinCruijssen! :) \$\endgroup\$ Commented Aug 16, 2018 at 7:27
  • \$\begingroup\$ Suggest i-~i instead of i*2+1 \$\endgroup\$ Commented Sep 25, 2019 at 20:28
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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

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1
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V (vim), 105 bytes

"aDiaibgyor<esc>0vg_y$@ap0l@alhd$:s/a/ v/g qqYP0xjYP!!rev gJkhq2@a@qylpp02ld$0VG: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.

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