Zottffssentettffssen

Question

Print or return the following string:

zottffssentettffssenttttttttttttttttttttffffffffffffffffffffsssssssssssssssssssseeeeeeeeeennnnnnnnnno

This is the first letter of each of the numbers zero, one, two, all the way up to one hundred.

Standard I/O rules apply, but the output must be exactly as shown above, optionally with a single trailing newline. No other variation in output is permitted. This is code golf, shortest code in bytes wins.

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^{Note this has also been cross-posted to Anarchy Golf (by me)}

Answer 1

Python, 58 bytes

a='ttffssen'
print(f"zo{a}te{a+''.join(x*10for x in a)}o")

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Answer 2

x86-64 machine code, 41 bytes

66 B8 7A 6F 66 AB 48 B8 74 74 66 66 73 73 65 6E 48 AB AA C6 07 65 AE 48 AB 6A 0A 59 F3 AA 48 C1 E8 08 75 F5 B0 6F 66 AB C3

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Following the standard calling convention for Unix-like systems (from the System V AMD64 ABI), this takes in RDI a memory address at which to place the result, as a null-terminated byte string.

In assembly:

f:  mov ax, 'z' | 'o'<<8    # Put these two bytes in AX.
    stosw                   # Write them to the string, advancing the pointer.
    .byte 0x48, 0xB8        # Place the following eight bytes in RAX.
    .ascii "ttffssen"
    stosq                   # Write them to the string, advancing the pointer.
    stosb                   # Write the first byte 't' again, advancing the pointer.
    mov BYTE PTR [rdi], 'e' # Place the byte 'e' at the current output address.
    scasb                   # Advance RDI over the 'e' while comparing with AL.
    stosq                   # Write "ttffssen" again, advancing the pointer.
r:  push 10; pop rcx        # Set RCX to 10.
    rep stosb               # Write AL to the string 10 times, advancing the pointer.
    shr rax, 8              # Shift RAX right by 8 bits; AL (its low byte)
                            #  will be each of "ttffssen" in order.
    jnz r                   # Jump back, to repeat, if the result is nonzero.
    mov al, 'o'             # Set AL to 'o'.
    stosw                   # Write 'o' and a null byte, advancing the pointer.
    ret                     # Return.

Answer 3

Vyxal, 7 bytes

₁ʀ∆ċvhṅ

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₁ʀ∆ċvhṅ
₁ʀ      # Range [0, 100]
  ∆ċ    # Nth cardinal of each
    vh  # First char of each
      ṅ # Join together

Vyxal sMH, 4 bytes

ƛ∆ċh

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(thanks to lyxal)

sMH, why are there flags for everything

H flag  # Preset stack to 100
ƛ       # Map over 100
M flag  # Make that 100 be a range [0, 100] instead of [1, 100]
 ∆ċ     # Convert to cardinal
   h    # Get the first character
s flag  # Join together

Answer 4

Factor + math.text.english, 39 bytes

[ 100 [0,b] [ number>text first ] map ]

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Answer 5

JavaScript (ES6), 57 bytes

Builds the output recursively from right to left.

f=(k=8)=>k>0?f(k-.1)+`${s}o`[~~k]:`zo${s="ttffssen"}te`+s

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Note

Because of floating point errors, we don't reach exactly k = 0 on the last iteration. That's why we have to use the test k > 0 ? instead of just k ?.

This is still 1 byte shorter than the following integer version:

58 bytes

f=(k=81)=>k--?f(k)+`${s}o`[k/10|0]:`zo${s="ttffssen"}te`+s

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Answer 6

Wolfram Language (Mathematica), ^{49 48} 41 bytes

IntegerName@#~StringPart~1&~Array~100<>""

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-1 thanks to Steffan

-7 thanks to att!

My Wolfram is rusty and was never great, so probably golfable, but it seemed like the right tool.

TIO shows an error because it can't connect to the internet, but it's still prints the correct answer.

Answer 7

brainfuck, 145 bytes

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

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Commented code

--[----->+>+>+>+>+>+>+>+<<<<<<<<]    Set cell 0 to 254 and loop 102 times filling cells 1 to 8 with ascii f
++++++++[->++>+>>>++>++>>+<<<<<<<<]  Set cell 0 to 8 and loop 8 times boosting certain cells to give vnffvvfn 
>++++.                               Move to cell 1; boost and output z
>+.+++++..                           Move to cell 2; boost and output ott
>.>.>---.>---.>-.>.                  From cells 3 to 8 output ffssen modifying values as necessary
<<+.->.                              Move to cell 6; temporarily modify to output t; move to cell 7 to output e
[<]>------                           Move to cell 1 and modify from z to t; cells now contain ttffssen
[.>]                                 Move through ttffssen outputing each character once (12 to 19) until empty cell reached
<[<]>                                Move back to cell 1
[..........>]                        Move through ttffssen outputting each character 10 times (20 to 99) 
<+.                                  Move to cell 8; modify value to o and output 

Answer 8

Python 3, 68 bytes

print('zottffssentettffssen'+'t'*20+'f'*20+'s'*20+'e'*10+'n'*10+'o')

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I'm sure this isn't optimal. Idk.

Answer 9

sed, 59 bytes

s/^/ttffssen/
s/./&&&&&&&&&&/g
s/.*/zottffssentettffssen&o/

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Answer 10

R + english, 48 50

Edited per Dominic van Essen's comment

cat(substr(english::as.english(0:100),1,1),sep="")

Similar to chunes' answer, this solution uses a library / package to handle integer-to-English mapping, then parses and collapses the resulting vector.

Answer 11

Arturo, _{110 65} 63 bytes

• -45 bytes, -2 bytes thanks to Krenium

s:"ttffssen"print~{zo|s|te|s||join map split s=>[repeat&10]|o}

Readable version:

s: "ttffssen"
print ~{zo|s|te|s||join map split s => [repeat&10]|o}

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Answer 12

Pyth, 29 bytes

s["zo"J"ttffssen""te"Js*LTJ\o

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A near literal translation of 97.100.97.109's Python solution.

Answer 13

Ruby, 54 bytes

puts"zo#{a="ttffssen"}te#{a+a.chars.map{|c|c*10}*''}o"

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Answer 14

Python 2, 53 bytes

Identical to my (and @xnor's, presumably) Python solution to the cross-post on Anarchy Golf.

s='ttffssen'
print'zo%ste'%s+s+(('o'+s*10)*8)[-8::-8]

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Answer 15

Perl 5, 40 bytes

say+zo,$_=ttffssen,te.$_.s/./$&x10/gre.o

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Answer 16

C (clang), ^{106 \$\cdots\$ 77} 75 bytes

f(i){for(i=0;i<101;i++)putchar("zottffssentettffssen"[i>99?:i<20?i:i/10]);}

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Saved a whopping 22 28 30 bytes thanks to jdt!!!
Saved a byte thanks to ceilingcat!!!

Answer 17

Bash 5.2, 45 43 bytes

-2 bytes thanks to Sisyphus

ttffssen
echo zo$_\te$_${_//?/&&&&&&&&&&}o

Uses the brand new (set-by-default) shell option patsub_replacement! Look forward to seeing this in many a future Bash golf.

ATO is still on 5.1 at the time of writing, so if you want to try it out yourself, the easiest way is with alpine:edge:

#!/bin/sh
podman run --rm -i alpine:edge << 'EOF'
apk add bash >/dev/null
bash -c 'ttffssen;echo zo$_\te$_${_//?/&&&&&&&&&&}o'
EOF

Answer 18

Japt, 29 26 bytes

`zottffÑA`
+h`` +¢®pAÃ+'o

Test it

Note that there is an unprintable character between the `` on line 2

`zottffÑA`     # U = "zottffssen"
               # Start building the output with U implicitly
 h``           # Replace the first 2 characters of U with "te"
+              # Add that to the output
      ¢        # Remove the first 2 characters of U
       ®  Ã    # For each remaining character
        pA     #  Duplicate it 10 times
     +         # Add that to the output
           +'o # Add "o" to the output
               # Print implicitly

I also tried another version which does "U repeated 10 times, grouped by index % 8, then joined" for the 20-99 portion, but it didn't golf as well

Answer 19

Excel, 77 71 bytes

^{Saved 6 bytes thanks to JvdV}

=LET(a,"ttffssen",CONCAT("zo",a&"te"&a,REPT(MID(a,ROW(1:8),1),10),"o"))

This is only slightly more efficient than just typing the 101 characters manually.

• LET(a,"ttffssen" stores the string ttffssen for later repeated use.
• CONCAT("zo",a&"te"&a combines strings to make the first part. This is only 2 bytes more efficient than hard-coding it.
• MID(a,ROW(1:8),1) turns the 8 character string into an array of 8 characters.
• REPT(MID(~),10) repeats each of those characters 10 times. These last two lines are the big byte-saver I missed the first time. Storing the string as a is not efficient by itself but it is if you later re-use it in the REPT() function.
• CONCAT(~,~,~,"o") add the last letter to the rest of the bits.

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Original 77 byte version:

="zottffssentettffssen"&CONCAT(REPT({"t","f","s","e","n"},5*{4,4,4,2,2}))&"o"

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This 78 byte alternative looked like it'd be shorter since each character is repeated the same number of times (allowing you to replace an array with a single integer) but it's only a close second:

="zottffssentettffssen"&CONCAT(REPT({"t","f","s"},20))&"eeeeeeeeeennnnnnnnnno"

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This 79 byte alternative was an attempt at compressing that array of characters but the savings where offset by the bytes required to convert them to strings:

="zottffssentettffssen"&CONCAT(REPT(CHAR(101+{15,1,14,0,9}),5*{4,4,4,2,2}))&"o"

Answer 20

Lua, 68 bytes

a='ttffssen'print('zo'..a..'te'..a..a:gsub('.',('%1'):rep(10))..'o')

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Answer 21

Tcl, 301 57 bytes

saved bytes thanks to the comment of @Sisyphus

Golfed version provided by me, try it online!

puts zo[set x ttffssen]te$x[regsub -all . $x &&&&&&&&&&]o

Original ungolfed version

set a "ttffssen"

proc repeat_string {str count} {
    set result ""
    for {set i 0} {$i < $count} {incr i} {
        append result $str
    }
    return $result
}

proc generate_string {a} {
    set result ""
    foreach char [split $a ""] {
        append result [repeat_string $char 10]
    }
    return $result
}

puts "zo${a}te${a}[generate_string $a]o"
puts "zottffssentettffssenttttttttttttttttttttffffffffffffffffffffsssssssssssssssssssseeeeeeeeeennnnnnnnnno"

Answer 22

05AB1E, 23 bytes

.•p≠²δK•ÐST×JŠ"zoÿteÿÿo

Port of @97.100.97.109's Python answer, so make sure to upvote him/her as well!

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Explanation:

.•p≠²δK•           # Push compressed string "ttffssen"
 Ð                 # Triplicate it
  S                # Convert the top copy to a list of characters
   T×              # Repeat each character 10 times as string
     J             # Join these back together to a single string
      Š            # Triple-swap so the other two strings are on top
       "zoÿteÿÿo  "# Push string "zoÿteÿÿo", where the `ÿ` are automatically filled with
                   # the three strings
                   # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress strings not part of the dictionary?) to understand why .•p≠²δK• is "ttffssen".

Answer 23

Retina 0.8.2, 44 bytes

zo_te_20$*t20$*f20$*s10$*e10$*no
_
ttffssen

Try it online! Link is to verbose version of code. Explanation: Mostly run-length compression via $*, with a final substitution to save a further 3 bytes.

Answer 24

Charcoal, 22 bytes

≔ttffssenθzoθteθＦθ×ιχo

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

≔ttffssenθ

Save the first letters of two/twelve/twenty, three/thirteen/thirty, four(teen)/forty, five/fifteen/fifty, six(teen/ty), seven(teen/ty), eight(een/y) and nine(teen/ty) in a variable.

zo

Output the first letters of zero and one.

θ

Output the first letters of two to nine.

te

Output the first letters of ten and eleven.

θ

Output the first letters of twelve to nineteen.

Ｆθ

Loop over the first letters of twenty to ninety.

×ιχ

Output each letter repeated ten times.

o

Output the first letter of one [hundred].

Answer 25

simply, 64 bytes

Just uses the same method as plenty of answers.

out"zo"$A="ttffssen""te"$A;each$A as$v;for$_ in0..9out$v;out'o'

Outputs the expected string, without any trailing whitespace.

Ungolfed

Extremely verbose, almost looking like pseudo-code:

Set $initials to the value "ttffssen".
Show the values "zo", $initials, "te", $initials.

Loop through $initials as $letter.
Begin.
    Loop from 0 to 9 as $i.
    Begin.
        Show the contents of $letter.
    End.
End.

Show the value "o".

More code-y looking:

$initials = "ttffssen";
echo "zo", $initials, "te", $initials;

each $initials as $letter {
    for $i in 0..9 {
        echo $letter;
    }
}

echo "o";

All versions do exactly the same.

Answer 26

MathGolf, 19 18 bytes

╢├╕◄ε╕f♣+∙╢►@♂m*'o

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Explanation:

╢├        # Push compressed "zo"
╕◄ε       # Push compressed "ttff"
   ╕f♣    # Push compressed "ssen"
      +   # Append the top two together: "ttffssen"
       ∙  # Triplicate it
╢►        # Push compressed "te"
  @       # Triple-swap the top three values (a,b,c to c,a,b)
 m        # Map over the characters of the top "ttffssen":
♂ *       #  Repeat the character 10 times
'o       '# Push character "o"
          # (after which the entire stack is joined and output implicitly as result)

Answer 27

PowerShell, 56 bytes

"zo$(($a='ttffssen'))te$a$(-join($a|% T*y|%{"$_"*10}))o"

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"zo                                                    " # start s new string with "zo"
   $(               )                                    # evaluate the subexpression $(...), which ...
     ($a='ttffssen')                                     # ... sets $a to "ttffssen", and outputs it straight back into the string (because of the additional brackets)
                     te                                  # literal "te"
                       $a                                # insert $a ("ttffssen")
                         $(                          )   # insert the results of another subexpression $(...)
                                 $a|% T*y                # pipe $a to % (alias for ForEach-Object) and call the member ToCharArray()
                                         |%{       }     # pipe the single characters to % and process them in the scriptblock {...}; loop variable is $_
                                            "$_"*10      # turn the char back into a string, and "multiply" it with 10
                           -join(                   )    # join those 10-char-long strings together
                                                      o  # Add the final "one hundred"

Output is implicit.

Answer 28

Java (JDK), 92 bytes

Interesting challenge!

()->{var s="ottffssente";for(int i=0;i<90;)s+=++i==9?"":s.charAt(i<9?i:i/10%9);return"z"+s;}

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I'm really glad that it was possible to beat the trivial solution :') (95 bytes) :

()->"zottffssentettffssen"+"t".repeat(20)+"f".repeat(20)+"s".repeat(20)+"eeeeeeeeeennnnnnnnnno"

Answer 29

Japt, 22 bytes

`zo{=`ttffÑA`}{+mpA}o

Test it

`zo{=`ttffÑA`}\x92{+mpA}o
`zo                     o     :Compressed string, starting with "zo" and ending with "o"
   {                          :Interpolate
    =                         :  Assign to variable U
     `ttffÑA`                 :  Compressed string "ttffssen"
             }                :End interpolation
              \x92            :Compressed "te"
                  {    }      :Interpolate
                   +          :  Append to U
                    m         :  Map each character in U
                     p        :    Repeat
                      A       :    Ten times

Answer 30

Oracle SQL, 97 bytes

select 'z'||listagg(substr(to_char(to_date(level,'J'),'jsp'),1,1))from"DUAL"connect by level<=100

Or, a bit more readable:

select
   'z' ||
   listagg(
      substr(
         to_char(to_date(level, 'J'), 'jsp')
         ,1,1)
      )
from
  "DUAL"
connect by
   level<=100;