First Number to Contain Each Letter

Question

Given a single letter from A to Z (except J and K) as input, output the smallest non-negative integer containing that letter in its written form. Assume numbers never contain the word "and", so 101 is "one hundred one", not "one hundred and one". Assume American (short-scale) counting, so one million is 10^6 and one billion is 10^9.

a 1000                           one thousand
b 1000000000                     one billion
c 1000000000000000000000000000   one octillion
d 100                            one hundred
e 0                              zero
f 4                              four
g 8                              eight
h 3                              three
i 5                              five
j 
k
l 11                             eleven
m 1000000                        one million
n 1                              one
o 0                              zero
p 1000000000000000000000000      one septillion
q 1000000000000000               one quadrillion
r 0                              zero
s 6                              six
t 2                              two
u 4                              four
v 5                              five
w 2                              two
x 6                              six
y 20                             twenty
z 0                              zero

J and K are not part of the input specification, so your behavior is undefined for them. Given one of the above letters, output the (decimal) number next to it. You can take input in lowercase or uppercase, but you cannot require that some inputs are lowercase and others uppercase.

This is code-golf, so the shortest answer in bytes wins.

I'm not quite sure why this challenge has been downvoted so much? As far as I can see, it's clear and on-topic. Sure it's most likely to simply be encoding each letter to its corresponding number, but I don't think that justifies 3 downvotes? — caird coinheringaahin g, Sep 3, 2019 at 15:39
I reckon that a StackExchange user who thinks that k can't appear in a number's name is rather lakhing in imagination. — Andrew Grimm, Sep 4, 2019 at 5:01

Community · Accepted Answer · 2020-06-17 09:04:33Z

JavaScript (Node.js), 78 75 74 73 bytes

c=>(n=([x]=Buffer(c+'8>P7 $(#%  +;! MD &"$%"&4 '))[x-96]-53)<0?n+21:10**n

Try it online!

How?

Each value is encoded with a single printable character. We use the ASCII range \$[32..52]\$ to encode \$n-32\$ and the range \$[53..80]\$ to encode \$10^{n-53}\$.

Commented

c =>                                   // c = input character
  ( n =                                //
    ( [x] =                            // let x be the 1st byte of the
        Buffer(                        // buffer made of:
          c +                          //   c followed by
          '8>P7 $(#%  +;! MD &"$%"&4 ' //   the encoded values
        )                              //
    )[x - 96]                          // let n be the encoded value corresponding to c
    - 53                               // minus 53
  ) < 0 ?                              // if n is negative:
    n + 21                             //   return n + 21
  :                                    // else:
    10 ** n                            //   return 10 ** n

recursive · Accepted Answer · 2019-09-03 21:05:59Z

7

Stax, 33 bytes

º░¡µ?Äz*B╥╪╩ΓoΣ4ù↓|♂5%⌡ÿΩ²┼h{☻4O└

Run and debug it

Procedure:

Extract codepoint from input.
Index into constant array [3, 5, 7, 9, 11, -6, 1, 0, -24, -15, 0, 6, 2, 4, 5, 2, 6, 20, 0, -3, -9, -27, -2, 0, 4, 8] using codepoint. (with wrap-around)
If the result is negative, negate and raise 10 to that power, otherwise leave as-is.

answered Sep 3, 2019 at 21:05

recursive

10.4k19 silver badges36 bronze badges

Add a comment |

Level River St · Accepted Answer · 2019-09-04 18:12:46Z

7

Ruby, 65 bytes

->n{i="CI[B?;7<:??4F>?XO?9=;:=9+?"[n.ord-65].ord-64;i>0?10**i:~i}

Try it online!

Improvements inspired by GB's comment.

Ruby, 70 bytes

->n{i="ci}b@DHCE@@KfA@xo@FBDEBFT@"[n.ord-65].ord;i>96?10**(i-96):i-64}

Try it online!

edited Sep 4, 2019 at 18:12

answered Sep 3, 2019 at 19:34

Level River St

25.7k4 gold badges36 silver badges104 bronze badges

\$\begingroup\$ 67 bytes \$\endgroup\$
– G B
Sep 3, 2019 at 19:45
\$\begingroup\$ @GB thanks, the idea of having negative values for i was useful. Now down to 65. \$\endgroup\$
– Level River St
Sep 4, 2019 at 18:14

Add a comment |

negative seven · Accepted Answer · 2019-09-03 16:01:04Z

6

///, 125 bytes

/:/\/\///T/000:d/100:a/d0:m/aT:b/aTT:q/bTT:p/qTTT:c/pT:e/0:f/4:g/8:h/3:i/5:l/11:n/1:o/0:r/0:s/6:t/2:u/4:v/5:w/2:x/6:y/20:z/0/

Try it online!

Input is appended to the end of the code, as per I/O meta. The footer in the above TIO link tests all letters simultaneously, as a single newline-delimited string, but the code also works just fine when inputting a single character.

answered Sep 3, 2019 at 16:01

negative seven

2,1911 gold badge8 silver badges19 bronze badges

Add a comment |

the default. · Accepted Answer · 2019-09-03 17:46:13Z

6

Wolfram Language (Mathematica), 50 bytes

Letters b, c, m, p and q are skipped for performance reasons when testing on TIO.

(i=0;While[1>StringCount[IntegerName@i,#],i++];i)&

Try it online!

answered Sep 3, 2019 at 17:46

the default.

8,4151 gold badge34 silver badges61 bronze badges

\$\begingroup\$ 41 bytes \$\endgroup\$
– att
Sep 4, 2019 at 1:04

Add a comment |

Wernisch · Accepted Answer · 2019-09-04 09:43:55Z

6

Excel, 85 bytes

=CHOOSE(CODE(A1)-96,1E3,1E9,1E27,100,,4,8,3,5,,,11,1E6,1,,1E24,1E15,,6,2,4,5,2,6,20,)

2 Golfy bits:

Using exponents (e.g. 1E15) saves 26 bytes.
Default of CHOOSE when nothing is provided is 0, saves 4 bytes

answered Sep 4, 2019 at 9:43

Wernisch

3,0249 silver badges22 bronze badges

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

05AB1E, 36 bytes

•—ÃÅ¸usδtθ}™-5„©‘öæH•57в₆-sÇ`èD0‹iÄ°

Port of @recursive's Stax answer.
Input in lowercase.

Try it online or verify all test cases.

Explanation:

•—ÃÅ¸usδtθ}™-5„©‘öæH• # Push compressed integer 3133432551338094772548436198140408157771728287
 57в                  # Converted to base-57 as list: [39,41,43,45,47,30,37,36,12,21,36,42,38,40,41,38,42,56,36,33,27,9,34,36,40,44]
    ₆-                # Subtract 36 from each: [3,5,7,9,11,-6,1,0,-24,-15,0,6,2,4,5,2,6,20,0,-3,-9,-27,-2,0,4,8]
      sÇ`             # Swap to take the input, and convert it to its unicode value
         è            # Index it into the list (with automatic wraparound)
          D0‹i        # Create a copy, and if this is negative:
              Ä       #  Take the absolute value
               °      #  And then take 10 the power this value
                      # (implicitly output the top of the stack as result)

See this 05AB1E tip of mine (section How to compress large integers? and How to compress integer lists?) to understand why •—ÃÅ¸usδtθ}™-5„©‘öæH• is 3133432551338094772548436198140408157771728287 and •—ÃÅ¸usδtθ}™-5„©‘öæH•57в is [39,41,43,45,47,30,37,36,12,21,36,42,38,40,41,38,42,56,36,33,27,9,34,36,40,44].

\$\begingroup\$ 32. 05AB1E should not lose to Stax! \$\endgroup\$
– Grimmy
Sep 4, 2019 at 15:41 — Grimmy, Sep 4, 2019 at 15:41

Xcali · Accepted Answer · 2019-09-04 04:28:19Z

3

Perl 5 `-p`, 84 bytes

$_=(1e3,1e9,1e27,100,0,4,8,3,5,1,1,11,1e6,1,0,1e24,1e15,0,6,2,4,5,2,6,20,0)[-65+ord]

Try it online!

answered Sep 4, 2019 at 4:28

Xcali

13.4k2 gold badges14 silver badges37 bronze badges

\$\begingroup\$ 72 bytes using lookup table from @JoKing's Perl 6 solution \$\endgroup\$
– Nahuel Fouilleul
Sep 4, 2019 at 10:32

Add a comment |

Jitse · Accepted Answer · 2019-09-04 08:57:05Z

3

Python 3, 103 bytes

lambda x:(1000,10**9,10**27,100,0,4,8,3,5,0,0,11,10**6,1,0,10**24,10**15,0,6,2,4,5,2,6,20,0)[ord(x)-97]

Try it online!

answered Sep 4, 2019 at 8:57

Jitse

6,8763 gold badges18 silver badges40 bronze badges

Add a comment |

Gymhgy · Accepted Answer · 2019-09-05 03:16:03Z

C# (Visual C# Interactive Compiler), 77 74 68 bytes

x=>((x="1‘Ʊ!  aƁñ"[x-65])&15)*Math.Pow(10,x>>4)

Encodes each letter's output as \$a \times 10^b\$, where \$a\$ is represented by the lowest 4 bits of the character, while \$b\$ is represented by the next five bits after that. Uses capital letters.

Commented

x=>                                                 //Lambda taking in a char
  (x=                                         )     //Re-assign x to
     "1‘Ʊ!   aƁñ"[x-65]      //The character's value at index x-65
 (                                             &15) //Bitwise AND by 15                                
  * Math.Pow(10,x>>4)                               // Multiplied by 10**(floor(x/16))

Try it online!

Jo King · Accepted Answer · 2019-09-04 06:48:47Z

2

Perl 6, 67 bytes

{/\-/??10**-$_!!$_}o{'`ZHacgkfhccn]dcKTciegheiwc'.ords[.ord-97]-99}

Try it online!

Uses a lookup table where a negative number means that is the negative of the exponent, otherwise it is the number itself.

answered Sep 4, 2019 at 6:48

Jo King♦

47.1k5 gold badges124 silver badges184 bronze badges

Add a comment |

Grimmy · Accepted Answer · 2019-09-04 15:46:09Z

2

05AB1E, 32 bytes

•н“вüQ;æ¡ζæÀÛß%aÜ×₃t•56вsCè2‰`i°

Try it online!

•н“вüQ;æ¡ζæÀÛß%aÜ×₃t•56в  # compressed list:
# [31, 0, 12, 4, 8, 10, 4, 12, 40, 0, 7, 19, 55, 5, 0, 8, 16, 6, 10, 1, 1, 22, 13, 2, 0, 49]

s                         # swap so the input is at the top
 C                        # parse input as "binary" (a -> 36, b -> 37, ...)
  è                       # index (wraps around)
   2‰                     # divmod 2: [n / 2, n % 2]
     `                    # dump both on the stack
      i                   # if the modulo is 1:
       °                  #  10 ** the quotient
                          # implicit output

edited Sep 4, 2019 at 15:46

answered Sep 4, 2019 at 15:36

Grimmy

15.6k1 gold badge33 silver badges63 bronze badges

\$\begingroup\$ I still don't get that 'binary' builtin, haha. xD But I guess it can be useful like this sometimes. ;) \$\endgroup\$
– Kevin Cruijssen
Sep 4, 2019 at 17:00
\$\begingroup\$ @KevinCruijssen It appears to simply convert from "base 255" when it's not valid binary, but still raises digits to powers of 2. \$\endgroup\$
– Makonede
Mar 16, 2021 at 17:04

Add a comment |

spuck · Accepted Answer · 2019-09-04 19:48:47Z

Bash, 129 100 bytes

A=xDVw04835zzbA10SJ0624526k0
c=$[64#${A:$[64#$1-10]:1}]
[ $c -gt 30 ]&&printf 1%0$[c-30].0f||echo $c

Try it online!

~~Try it online!~~

How it works:

A=xDVw04835zzbA10SJ0624526k0

$A: Base64 encoded "a"-"z": numbers less than 100 are stored directly. Larger numbers are encoded as the number of zeros +30. (ex: 1,000 = 33, 100 = 32, etc.)

c=$[64#${A:$[64#$1-10]:1}]

Extract one letter from $A at position specified in argument $1 (base64 decoded, -10 to account for offset of 'a'). Base64 decode that character and store in c.

[ $c -gt 30 ]&&printf 1%0$[c-30].0f||echo $c

If $c is greater than 30, print "1" padded with $c-30 zeros. Else, print $c.

the default. · Accepted Answer · 2019-09-05 07:22:44Z

Sledgehammer, 17 bytes

Technically, this is 133 bits long, but that doesn't really make this 16.625 bytes, as the compressor claims.

⣜⢍⢞⢹⡱⡋⣽⡱⡆⢺⢦⡽⡐⡌⢗⠈⣵

This probably decodes to

x1 = Input[]; x2 = 0; While[StringFreeQ[IntegerName[x2], x1], x2++]; Print[x2]

(approximately the same as my Mathematica answer), although I barely got it encoded (it appears as though my PC has compatibility issues with everything), so good luck decoding it again to check. I might have made some mistakes while using the encoder, so be careful.

lyxal · Accepted Answer · 2022-09-09 01:48:08Z

2

Vyxal, 11 bytes

λ∆ċ‛λ¬-?c;ṅ

Try it Online!

Very literal implementation of the title.

Explained

λ∆ċ‛λ¬-?c;ṅ
λ        ;ṅ  # The first number
 ∆ċ‛λ¬-      #   (converted to words, with any "and"s removed)
       ?c    # to contain the input

answered Sep 9, 2022 at 1:48

lyxal

32.1k2 gold badges56 silver badges138 bronze badges

Add a comment |

Nick Kennedy · Accepted Answer · 2019-09-03 19:20:59Z

Jelly, 36 bytes

Oị“[@ịẆþĊ`o&÷ḲḞṘḂỊP¥t’b48¤_⁹⁵*ɗ¹>?20

Try it online!

A monadic link taking a lower case letter as its argument and returning an integer. Returns 0 for j and k.

Explanation

O                       | Convert to code point
 ị          ¤           | Index into following as a nilad (wraps around):
  “[...t’               | - Integer 5370441668223940717846370165240010583188867 (stored base 250)
         b48            | - Convert to base 48
                 ɗ >?20 | If >20, following as a dyad using 20 as right argument:
             _⁹         | - Subtract right argument (20)
               ⁵*       | - 10 to the power of this
                  ¹     | Else: leave unchanged (identity function)

Neil · Accepted Answer · 2019-09-03 19:39:47Z

Retina 0.8.2, 89 bytes

^
$'
T`l`111104835__111011062452620`^.
T`abcm\pq`139285
\d$
$*0$&$*0$&$*0
d
00
T`\lyl`10_

Try it online! Link includes test cases. Explanation:

^
$'

Duplicate the input.

T`l`111104835__111011062452620`^.

Change the first copy to the (first) digit of the relevant result.

T`abcm\pq`139285

If the number has a multiple of 3 trailing zeros, get that multiple now.

\d$
$*0$&$*0$&$*0

And actually convert it into the relevant number of trailing zeros. (Note that this would simplify to *3*0 in Retina 1.)

d
00

Fix up d.

T`\lyl`10_

Fix up l and y and remove any remaining letters.

Night2 · Accepted Answer · 2019-09-04 12:37:19Z

PHP, 104 bytes

<?=A<($a='^FX]0483500GC10UL0624526P0'[ord($argn)-97])?20==($b=ord($a)%30)||11==$b?$b:str_pad(1,$b,0):$a;

Try it online!

I have a string ^FX]0483500GC10UL0624526P0 which holds a single character for every input letter from "a" to "z". I extract this character based on input and store it in $a. If the character is not a digit, its ASCII code mod 30 will be stored in $b.

If $a is a digit, same digit is printed, this is used for any input that needs an output between 0 and 9 (like "e", "f", etc).

Else if $b is 20 or 11, same number is printed, this is used for "l" and "y".

Else, digit "1" padded to $b with "0"s is printed. For example for input of "a", the character is "^" which has an ASCII code of 94. 94 % 30 = 4 and "1" padded to 4 with "0"s will be "1000".

Stack Exchange Network

First Number to Contain Each Letter

18 Answers 18

JavaScript (Node.js), 78 75 74 73 bytes

How?

Commented

Stax, 33 bytes

Procedure:

Ruby, 65 bytes

Ruby, 70 bytes

///, 125 bytes

Wolfram Language (Mathematica), 50 bytes

Excel, 85 bytes

05AB1E, 36 bytes

Perl 5 `-p`, 84 bytes

Python 3, 103 bytes

C# (Visual C# Interactive Compiler), 77 74 68 bytes

Perl 6, 67 bytes

05AB1E, 32 bytes

Bash, 129 100 bytes

How it works:

Sledgehammer, 17 bytes

Vyxal, 11 bytes

Explained

Jelly, 36 bytes

Explanation

Retina 0.8.2, 89 bytes

PHP, 104 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
number
kolmogorov-complexity
or ask your own question.

Linked

Hot Network Questions

First Number to Contain Each Letter

18 Answers 18

JavaScript (Node.js), 78 75 74 73 bytes

How?

Commented

Stax, 33 bytes

Procedure:

Ruby, 65 bytes

Ruby, 70 bytes

///, 125 bytes

Wolfram Language (Mathematica), 50 bytes

Excel, 85 bytes

05AB1E, 36 bytes

Perl 5 -p, 84 bytes

Python 3, 103 bytes

C# (Visual C# Interactive Compiler), 77 74 68 bytes

Perl 6, 67 bytes

05AB1E, 32 bytes

Bash, 129 100 bytes

How it works:

Sledgehammer, 17 bytes

Vyxal, 11 bytes

Explained

Jelly, 36 bytes

Explanation

Retina 0.8.2, 89 bytes

PHP, 104 bytes

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged code-golfnumberkolmogorov-complexity or ask your own question.

Linked

Related

Hot Network Questions

Perl 5 `-p`, 84 bytes

Not the answer you're looking for? Browse other questions tagged
code-golf
number
kolmogorov-complexity
or ask your own question.