Letters challenge

Question

Using the word “conemon” print each character 2 times more than the last. The first letter, “c”, should be printed 10 times, the second 12 and so on. Each repeated-letter-string should be printed on a new line.

So the end-result becomes:

cccccccccc
oooooooooooo
nnnnnnnnnnnnnn
eeeeeeeeeeeeeeee
mmmmmmmmmmmmmmmmmm
oooooooooooooooooooo
nnnnnnnnnnnnnnnnnnnnnn

Answer 1

Python 2, 36 bytes

i=8
for c in'conemon':i+=2;print c*i

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Answer 2

R, 50 47 45 bytes

This is some serious R abuse. I create a quote R expression C(O,N,E,M,O,N) and feed it into the string repeat function strrep, which kindly coerces each name inside the quote into a string. Shorter than scan !

write(strrep(quote(C(O,N,E,M,O,N)),5:11*2),1)

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Answer 3

K (ngn/k), 20 19 bytes

(2*5+!7)#'"conemon"

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Answer 4

05AB1E, 16 14 bytes

.•Ω‡h₅•Sā·8+×»

-2 bytes thanks to @Emigna.

Try it online.

Explanation:

.•Ω‡h₅•           # Push "conemon"
       S          # Convert it to a list of character: ["c","o","n","e","m","o","n"]
        ā         # Push a list in the range [1, length]: [1,2,3,4,5,6,7]
         ·        # Double each: [2,4,6,8,10,12,14]
          8+      # Add 8: [10,12,14,16,18,20,22]
            ×     # Repeat the characters that many times
             »    # Join the list by newlines (and output implicitly)

See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•Ω‡h₅• is "conemon".

Answer 5

MathGolf, 15 bytes

╕│Pùmon+ô8î∞+*p

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Explanation

╕│P                 Compression of 'cone'
   ùmon             Push 'mon'
       +            Concatenate together
        ô          Foreach over each letter
         8         Push 8
          î∞       Push the index of the loop (1 based) and double it
            +      Add the 8
             *p    Repeat the letter that many times and print with a newline

Answer 6

Java 11, 78 bytes

v->{int i=8;for(var c:"conemon".split(""))System.out.println(c.repeat(i+=2));}

Try it online. (NOTE: String.repeat(int) is emulated as repeat(String,int) for the same byte-count, because Java 11 isn't on TIO yet.)

Explanation:

v->{                     // Method with empty unused parameter and no return-type
  int i=8;               //  Integer `i`, starting at 8
  for(var c:"conemon".split(""))
                         //  Loop over the characters (as Strings) of "codemon"
    System.out.println(  //   Print with trailing new-line:
      c.repeat(          //    The current character repeated
        i+=2));}         //    `i` amount of times, after we've first increased `i` by 2

Answer 7

Javascript, 55 52 bytes

saved 3 bytes thanks to @Arnauld

s=>[...'conemon'].map(x=>x.repeat(i+=2),i=8).join`
`

f=s=>[...'conemon'].map(x=>x.repeat(i+=2),i=8).join`
`
console.log(f());

Answer 8

Japt -R, 13 bytes

`¬¶n`¬Ë²p5+E

Try it

---

Explanation

`¬¶n`¬Ë²p5+E
`¦n`            :Compressed string "conemon"
     ¬           :Split
      Ë          :Map each character at 0-based index E
       ²         :  Repeat twice
        p5+E     :  Repeat 5+E times
                 :Implicitly join with new lines and output

Answer 9

PowerShell, 35 34 bytes

'conemon'|% t*y|%{"$_$_"*(++$j+4)}

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-1 byte thanks to mazzy

Literal string 'conemon' is converted toCharArray, then for each character we multiply it out by the appropriate length. This is handled by doubling up the character $_$_ then multiplying by $j+4 with $j pre-incremented each time (i.e., so it'll start at 1+4 = 5, which gets us 10 characters).

Each newly formed string is left on the pipeline, and implicit Write-Output gives us newlines for free.

Answer 10

Tcl, 78 bytes

set i 8
lmap x {c o n e m o n} {set s ""
time {set s $s$x} [incr i 2]
puts $s}

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Answer 11

K (oK), 29 bytes

{i::8;{i+::2;i#x}'x}"conemon"

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I'm not the biggest fan of this solution as it's bad practice (global variables), but it's short. May work on a better solution.

Answer 12

Husk, 11 bytes

zR↓4İ0¨cΦ◄‰

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Explanation

zR↓4İ0¨cΦ◄‰
      ¨cΦ◄‰        A = The compressed string "conemon"
    İ0             B = The infinite list of even positive numbers
  ↓4                             without its first four elements: [10,12,14,16...]
zR                 Create a list of strings replicating each character in A as many times
                   as the corresponding number in B

A list of strings in Husk is printed by joining them with newlines.

Answer 13

Perl 6, 36 34 bytes

-2 bytes thanks to nwellnhof

("conemon".comb Zx(5..*X*2))>>.say

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Explanation:

 "conemon".comb                     # Get conemon as a list of characters
                   5..*             # A lazy list from 5 to infinity
                       X*2          # Multiplied by 2
                                    # This results in the list 10,12,14 etc.
                Zx                  # Zip the two together with the string multiplication operator
                                    # This multiplies each character by 10, 12 etc.
                            >>.say  # Print each line

Answer 14

Jelly,  14  13 bytes

-1 Thanks to Erik the Outgolfer (pointing me to the real optimal string compressor)
...and, of course, to user202729 (for creating it)!

“¦[Þ⁷ƥ»J+4×ḤY

A full program which prints the output required.

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How?

“¦[Þ⁷ƥ»J+4×ḤY - Main Link: no arguments
“¦[Þ⁷ƥ»       - compressed string as a list of characters -> ['c','o','n','e','m','o','n']
              - (...due to this being a leading constant this is now the argument too)
       J      - range of length  -> [ 1, 2, 3, 4, 5, 6, 7]
         4    - literal four
        +     - add (vectorises) -> [ 5, 6, 7, 8, 9,10,11]
          ×   - multiply by the argument -> ['ccccc','oooooo',...,'nnnnnnnnnnn']
           Ḥ  - multiply by 2 (vectorises) -> ['cccccccccc','oooooooooooo',...,'nnnnnnnnnnnnnnnnnnnnnn']
            Y - join with newline characters
              - implicit print

Answer 15

Red, 53 bytes

n: 8 foreach c"conemon"[loop n: n + 2[prin c]print""]

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Answer 16

J, 25 24 bytes

 echo'conemon'#"0~2*5+i.7

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Note: There are trailing spaces on all lines except the last one.

Answer 17

Charcoal, 14 bytes

Ｅconemon×ι⁺χ⊗κ

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

 conemon        Literal string
Ｅ               Map over characters
             κ  Current index
            ⊗   Doubled
           χ    Predefined variable 10
          ⁺     Add
         ι      Current character
        ×       Repeat
                Implicitly print each string on separate lines

Answer 18

APL (Dyalog Classic), 21 bytes

⎕←↑'conemon'⍴¨⍨8+2×⍳7

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Answer 19

Japt, 17 bytes

"conemon"¬Ëp°EÑ+8

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---

Japt, 14 bytes

`¬¶n`¬Ëp°EÑ+8

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---

Explanation

"conemon"¬Ëp°EÑ+8           Full Program
"conemon"¬                  Split each letter       
          Ëp                Map and repeat  
            °EÑ+8           ((1 + Index) * 2) + 8

Answer 20

Haskell, 44 bytes

i=[0..]
g=["conemon"!!n<$take(10+n*2)i|n<-i]

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We generate a list of lists of lengths 10, 12 etc. (by taking the appropriate amount of elements from an infinite list) and then replace each element in each such list with corresponding character from the required string.

Answer 21

Perl 5, 33 bytes

conemon=~s/./say$&x(8+2*++$i)/reg

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Explanation

         s/ /                /     # Replace
                                g  # each
           .                       # character
       =~                          # in
conemon                            # string "conemon"
                              r    # non-destructively (source is read-only)
                               e   # only for the side effect of
             say                   # printing with newline
                $&                 # the full match (character)
                  x                # repeated
                        ++$i       # 1,2,3,...
                      2*           # 2,4,6,...
                    8+             # 10,12,14,...
                   (        )      # times

Answer 22

Canvas, 14 bytes

conemon｛²«８＋×］

Try it here!

Answer 23

SOGL V0.12, 13 bytes

╝⌠Ei№‘{ē«L+*P

Try it Here!

Explanation:

╝⌠Ei№‘       push "conemon" - 2 words "cone" and "mon" compressed
      {      for each
       ē       push the value of e (default 0), and increment it (aka e++)
        «      multiply that by 2
         L+    add 10 to that
           *   repeat the character that many times
            P  and print it

Answer 24

F#, 71 bytes

let d=Seq.iteri(fun i c->System.String(c,10+i*2)|>printfn"%s")"conemon"

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Seq.iteri iterates through the sequence and applies the function to the index of the item i and the item itself c. In this case the string, every character in the string conemon.

System.String is a shortform of new System.String, taking the current letter c and repeating it 10+i*2 times, where i is the index of the letter. It then prints the string to the output with a new line.

You can omit the string and shorten it to:

let d=Seq.iteri(fun i c->System.String(c,10+i*2)|>printfn"%s")

And this will work with every string. But given this challenge is specifically for conemon the string is hard-coded.

Answer 25

C# (Visual C# Interactive Compiler) 65 64 bytes

for(int i=0;i<7;){WriteLine(new string("conemon"[i],i++*2+10));}

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thanks to @Jonathan Frech saving 1 byte

new string(char, int) repeats the char as often as the int value.

C#, 72 bytes without "Console" as static using (example outside interactive compiler):

for(int i=0;i<7;){Console.WriteLine(new string("conemon"[i],i++*2+10));}

Answer 26

Pyth, 17 16 bytes

V"conemon"*N~+T2

Try it online here.

V"conemon"*N~+T2   Implicit: T=10
V"conemon"         For each character in "conemon", as N:
          *N  T      Repeat N T times, implicit print with newline
            ~+T2     T += 2

Answer 27

PowerShell, 46 bytes

$i=8;'c','o','n','e','m','o','n'|%{$_*($i+=2)}

# $i=8;foreach($c in [char[]]'conemon'){[string]$c*($i+=2)}

Answer 28

jq, 56 characters

(52 characters code + 4 characters command line options)

[[range(5;12)],"conemon"/""]|transpose[]|.[1]*.[0]*2

Sample run:

bash-4.4$ jq -nr '[[range(5;12)],"conemon"/""]|transpose[]|.[1]*.[0]*2'
cccccccccc
oooooooooooo
nnnnnnnnnnnnnn
eeeeeeeeeeeeeeee
mmmmmmmmmmmmmmmmmm
oooooooooooooooooooo
nnnnnnnnnnnnnnnnnnnnnn

Try it online!

Answer 29

Ruby, 48 bytes

->{a=4;'conemon'.each_char{|s|puts s*(a=1+a)*2}}

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Answer 30

Pyth, 17 bytes

j.e*b+Tyk"conemon

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Explanation

j.e*b+Tyk"conemon
 .e      "conemon    For each character 'b' and index 'k' in "conemon"...
   *b+Tyk            ... get 2k + 10 copies of b.
j                    Join the result with newlines.

Answer 31

MBASIC, 80 73 bytes

1 S$="conemon":FOR I=1 TO LEN(S$):PRINT STRING$(8+I*2,MID$(S$,I,1)):NEXT

Saved a loop by using the STRING$ function to generate the string of characters. Saved another 7 bytes by computing the length relative to the loop index.

Output:

cccccccccc
oooooooooooo
nnnnnnnnnnnnnn
eeeeeeeeeeeeeeee
mmmmmmmmmmmmmmmmmm
oooooooooooooooooooo
nnnnnnnnnnnnnnnnnnnnnn

Answer 32

D, 48 bytes

foreach(i,c;"conemon")c.repeat(10+i*2).writeln;

Explanation:

foreach(i,c;"conemon")                           // Loop over string with index i, char c
                      c.repeat(10+i*2)           // Repeat character 10 + (i*2) times
                                      .writeln;  // Write with newline

Run with rdmd:

$ rdmd --eval='foreach(i,c;"conemon")c.repeat(10+i*2).writeln;'
cccccccccc
oooooooooooo
nnnnnnnnnnnnnn
eeeeeeeeeeeeeeee
mmmmmmmmmmmmmmmmmm
oooooooooooooooooooo
nnnnnnnnnnnnnnnnnnnnnn

Answer 33

V, 27 26 bytes

Iµc
¶o
·n
¸e
¹m
±o
±±n<esc>Îä$

<esc> represents the escape character (ascii 27)

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Explanation

I                Enter insert mode
µc                Write 5 c and a newline
¶o                Write 6 o and a newline
·n                Write 7 n and a newline
¸e                Write 8 e and a newline
¹m                Write 9 m and a newline
±o                Write 10 o and a newline
±±n               Write 11 n
<esc>            End insert mode
Îä$              Duplicate every line

Answer 34

T-SQL, 84 bytes

DECLARE @ INT=1a:PRINT REPLICATE(SUBSTRING('conemon',@,1),2*@+8)SET @+=1IF @<8GOTO a

The variable/loop approach turned out shorter than the best set-based variation I came up with (91 bytes):

SELECT REPLICATE(SUBSTRING('conemon',n,1),2*n+8)FROM(VALUES(1),(2),(3),(4),(5),(6),(7))a(n)

I don't know what it is, but I found this question particularly annoying. Which probably means it's a good question.

Answer 35

Z80Golf, 35 31 bytes

00000000: 2118 0016 0a42 7efe 0028 0cff 10f8 3e0a  !....B~..(....>.
00000010: ff23 1414 4218 ef76 636f 6e65 6d6f 6e    .#..B..vconemon

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Assembly:

ld hl,str
ld d, 10
ld b, d
loop:
	ld a,(hl)
	cp 0
	jr z, hlt
	rst 38h; putchar
	djnz loop
ld a, 10; newline
rst 38h
inc hl
inc d
inc d
ld b, d
jr loop
hlt:
	halt
str:
	db 'conemon'

-4 bytes changed call 8000h to rst 38h

Answer 36

Stax, 12 bytes

ü¡│\½◙S·«═▓╞

Run and debug it

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

`Mwys-` compressed literal "conemon"
m       for each character output the result after running the rest of the program
 ]      wrap in singleton array
 iHA+   `2 * i + 10` where i is the 0-based iteration index
 *      repeat array

Run this one

Answer 37

APL(NARS), 19 chars, 38 bytes

⊃'conemon'/¨⍨2×4+⍳7

test:

  ⊃'conemon'/¨⍨2×4+⍳7
cccccccccc            
oooooooooooo          
nnnnnnnnnnnnnn        
eeeeeeeeeeeeeeee      
mmmmmmmmmmmmmmmmmm    
oooooooooooooooooooo  
nnnnnnnnnnnnnnnnnnnnnn

Answer 38

[PHP], 76 bytes

$i=8;foreach(str_split("conemon") as $k){$i+=2;echo str_repeat($k,$i)."\n";}

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Answer 39

Z80Golf, 23 bytes

00000000: 0907 060d 0507 0676 1a3c 473c 127e ee68  .......v.<G<.~.h
00000010: ff10 fd3e 0a23 e5                        ...>.#.

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Disassembly

start:
  add hl, bc ; db ('c'^$68)-2
  rlca       ; db 'o'^$68
  ld b, $0d  ; db 'n'^$68
             ; db 'e'^$68
  dec b      ; db 'm'^$68
  rlca       ; db 'o'^$68
  ld b, $76  ; db 'n'^$68
             ; halt

  ld a, (de) ; manipulate the byte at address 0
  inc a
  ld b, a    ; loop count
  inc a
  ld (de), a ; save +2 of previous

  ld a, (hl)
  xor $68    ; the char to print
loop:
  rst $38    ; print it `b` times
  djnz loop

  ld a, $0a  ; newline to print
  inc hl     ; string index & return address
  push hl

The Hello World trick strikes again. The first byte is also used to track the loop count.

Answer 40

GolfScript, 26 bytes

7:a;"conemon"{{.}a)):a*n}%

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I'm surprised nobody else has done a solution in this language by now. However, I won't be surprised when this submission is eventually beaten.

Answer 41

JavaScript (Node.js), 56 bytes

[..."conemon"].map((e,i)=>console.log(e.repeat(i*2+10)))

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Answer 42

///, 74 bytes

/E/eeee//M/mmm//N/nnnn//O/oooo//;/OOO
NNNnn/cccccccccc
;
EEEE
MMMMMM
OO;NN

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Answer 43

Retina, 29 bytes

K`conemon
L$`.
$.(5*_$`)*2*$&

Try it online! Explanation:

K`conemon

Initialise the buffer with the text.

L$`.

Loop over each character and output each substituion on its own line.

$.(5*_$`)*2*$&

Repeat the match (5 + index) * 2 times.

Answer 44

LUA, 67 66 bytes

s="conemon"
l=10
for i in s:gmatch'.'do print(i.rep(i,l))l=l+2 end

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