Replicate "dotnet ef" unicorn

Question

.Net Technology is an incredible framework with marvellous functionalities, and among all of them, the most important one obviously stands out:

Yep, your simple task is create a program that when it takes at input:

dotnet ef

It prints the following ascii art by output:

             /\__
       ---==/    \\
 ___  ___   |.    \|\
| __|| __|  |  )   \\\
| _| | _|   \_/ |  //|\\
|___||_|       /   \\\/\\

And you have to do it in the lowest number of bytes of course.

Rules

• If the program doesn't receive dotnet ef by input it doesn't output anything
• The program with the fewest bytes wins so this is a code golf challenge

Answer 1

Perl 5 + -p, 100 bytes

1/s/^dotnet ef$/oooo..oo.aJo..z..o8o..3.3.svo!.3.3.q.3.5.4.43oo)q!+./;s/./ord$&/ge;y;0-9;
 _\\/|.=)-

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Explanation

1/s/^dotnet ef$/... will exit with a division by zero unless the input is dotnet ef, then the data is packed into the 52-byte binary string. Since there were 10 distinct chars, it's possible to convert those original chars to digits 0-9, which were then packed (so 111111111111143 becomes chr(111) + chr(111) + chr(111) + chr(111) + chr(143)), This packed string is replaced into $_ (where the implicit input is stored) via s///ubstitution, then each char is s///ubstituted with its ordinal value and then transliterated (y///) back to the original chars.

---

Perl 5 + -pF/^(dotnet\x20ef)$/, 94 bytes

Same as above, except abusing flags for the early exit.

1/$#F;$_='oooo..oo.aJo..z..o8o..3.3.svo!.3.3.q.3.5.4.43oo)q!+.';s/./ord$&/ge;y;0-9;
 _\\/|.=)-

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Answer 2

Python 3, 129 bytes

lambda s:(s=='dotnet ef')*b'V`"D1PffayWPqfwgqUqqgc3uPvfwgU5P'.hex().translate('-/=\_|).\n '*9)

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Explanation

We use bytes.hex() to convert the byte-string to a string of hexadecimal digits, which looks like this:

11111111111113566011111112224431111550166611666111791111575071667716671171181115550716717167111563171133755076667767111111131115553550

From here, we perform a str.translate(), to map the digits '0123456789' to '-/=\_|).\n '. These are the ten characters that make up the entire unicorn drawing.

Special care should be taken to ensure that the compressed string won't contain any values exceeding the ASCII range. The program below finds such strings:

from itertools import *
ART = '             /\\__\n       ---==/    \\\\\n ___  ___   |.    \\|\\\n| __|| __|  |  )   \\\\\\\n| _| | _|   \\_/ |  //|\\\\\n|___||_|       /   \\\\\\/\\\\\n'
L = '\n )-./=\_|'
for p in permutations(range(10)):
    compressed = b''.fromhex(''.join('0123456789'[p[L.find(c)]] for c in ART))
    if max(compressed) < 128:
        trans = [None] * 10
        for i in range(10):
            trans[(p[i] + 48) % 10] = L[i]
        print('translation:', repr(''.join(trans)))
        print('compressed:', compressed.decode())
        break

Answer 3

JavaScript (Node.js), 152 bytes

Contains unprintable RS characters.

s=>Buffer(s=="dotnet ef"?'#:u"J`P 9&vvn139(v)&o(! 4e&((o sU*90v)o# 4R2':0).map(n=>o+=(c=` |\\)-/=
_.`)[n/10-3|0]+c[n%10],o='')&&o

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---

JavaScript (ES8), 152 bytes

s=>s=='dotnet ef'?`94/0__
7---==/400
1___2___3|.40|0
|1__||1__|2|2)3000
|1_|1|1_|30_/1|2//|00
|___||_|7/3000/00`.replace(/\d/g,n=>''.padEnd(n)||'\\'):''

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Answer 4

Charcoal, 79 bytes

¿⁼Ｓdotnet ef”{¶⊟u↗9≧fXj⪪⊙Ｏ|↔d<⭆)⟧ΦÞ‴ζＬＰ≧≦⁸1？_Ｕ~l7(Ｉ℅gºqu⪫ÀF◨Ｙ∧π⦄◧μ×⎚⁶↓sαbＧＡaF

Try it online! Link is to verbose version of code. Explanation: Just Charcoal's default string compression, conditional on the input line (which needs a trailing newline to prevent Charcoal's autosplit on spaces).

Answer 5

Jelly, 82 bytes

⁼“¤fÞ#³ıḄ×»“¡mƁµɼʋ"§ĿỵSėṭƊḲṂ§ÇḋỵL}Ṅnʠṫ2#ap{}fƓ*ƑṇḢ§ƓṾ®[~]sṘÐ/wỊḊ.÷⁹ß’ḣṃ“ \_¶-=|.)/

A full program accepting a string that prints the art when that string is "dotnet ef".

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How?

⁼“...»“...’ḣṃ“... - Main Link: string, S
 “...»            - dictionary compression of "dotnet ef"
⁼                 - S equals that? (1 if so; 0 otherwise)
      “...’       - base 250 integer (say, n) = 1111111111111023341111111555660111122413331133311178111127247133771337117119111222471371713711123017110072247333773711111110111222022
           ḣ      - head that (implicitly wrapped in a list) by (1 or 0)
                      -> [n] or []
             “... - the art's characters (¶ is a newline & no closing quote needed)
            ṃ     - decompress the integers (in either [n] or []) using the
                    characters as the digits [1,...,9,0] (as there are ten of them)
                  - implicit, smashing print

Answer 6

05AB1E, 85 bytes

’¥¥‡Ò ef’Qi•{Ã∍ûç=®¤[ˆ ôÆ-˜´≠pw:‡©[§∊è$#±8¡žR=α»ιœëā7¬ÄÒU₁ýmOÜÉt×›ʒ•"_ |\/
-=)."ÅвJëõ

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A compressed string .•1piN∍Ç• instead of dictionary string ’¥¥‡Ò ef’ would be the same byte-count.

Explanation:

’¥¥‡Ò ef’        # Push the dictionary string "dotnet ef"
         Qi      # If the (implicit) input is equal to this:
  •{Ã∍...×›ʒ•    #  Push compressed integer 1111111111111430051111111666774111133510001100011129111132352100221002112118111333521021210211130412114423352000220211111114111333433
   "_ |\/\n-=)." #  Push this string
    Åв           #  Convert the large integer to base-"_ |\/\n-=)."
                 #  which means base-length, and then indexing it into the string
      J          #  Join this list of characters together
          ë      # Else:
õ                #  Push an empty string instead
                 # (after which the top is output implicitly as result)

The compressed ASCII-art is generated using the generator program in this 05AB1E tip.
See this 05AB1E tip of mine (sections How to use the dictionary?, How to compress strings not part of the dictionary?, and How to compress large integers?) to understand why ’¥¥‡Ò ef’ is "dotnet ef"; .•1piN∍Ç• is "dotnet ef" as well; and •{Ã∍ûç=®¤[ˆ ôÆ-˜´≠pw:‡©[§∊è$#±8¡žR=α»ιœëā7¬ÄÒU₁ýmOÜÉt×›ʒ• is 1111111111111430051111111666774111133510001100011129111132352100221002112118111333521021210211130412114423352000220211111114111333433.

Answer 7

Python 3, 156 bytes

if"dotnet ef"==input():print(r"""		 /\__
	 ---==/    \\
 ___	___	|.	\|\
| __|| __|  |  )	 \\\
| _| | _|	\_/ |  //|\\
|___||_|	   /	 \\\/\\""".expandtabs(6))

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Credits to @AnttiP

Answer 8

[C (gcc)], ̶2̶1̶7̶ 212 202 bytes

#define S"\\\\\r"
main(c,v)int**v;{strcmp(v[1],"dotnet ef")||printf("%14c\\__\r%8c--==/%7s ___  ___   |.    \\|\\\r| __|| __|  |  )   \\"S"| _| | _|   \\_/ |  //|"S"|___||_|%8c   \\\\\\/"S,47,45,S,47);}

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Previous versions :

char*S="\\\\\r";main(c,v)char**v;{if(!strcmp(v[1],"dotnet ef"))printf("%14s\\__\r%8s--==/%7s ___  ___   |.    \\|\\\r| __|| __|  |  )   \\%s| _| | _|   \\_/ |  //|%s|___||_|%8s   \\\\\\/%s","/","-",S,S,S,"/",S);}

Contributions :

Makonede :

• remove the brace for the if statement
• change the define by a global variable

celingcat :

• replace the if by the OR test
• put the define S inside the string of the printf
• change some characters by theirs ascii values.

Answer 9

Julia, 177 163 bytes

f(i)=i=="dotnet ef"&&print(" "^13,"/\\__
"," "^7,raw"---==/    \\
 ___  ___   |.    \|\ 
| __|| __|  |  )   \\\
| _| | _|   \_/ |  //|\
|___||_|       /   \\/\\
")

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Thanks to MarcMush we can bring it down to 163 bytes by using raw which allows us to use \ without escaping!

Answer 10

Canvas, 75 bytes

dotnet ef≡？“ｃL⤢43¶h\＾»）√＠ｈ┘S┼N（fＶ／3ＸＶｋ＃６Ｋ}⁰⤢ＫＥ @＠sMｏ↷.(ＲｑｇａＦ╶u┬∑ｚ７F│┘@&k］⁰‟

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just canvas's string compression.

Answer 11

Vyxal O, 82 bytes

«ƛ4-ʀWṪ«=[»|ȮW¼∩r¶‛v←^ǔṠlȯ¦U4;o→₄v₂V⊍cbε×√ẋÞrAĖRṫ⇧ż÷₇Ṗ⌈¨YΠ1ʁ⋎Ẋ8…ṡ@»`_ |\/\n-=).`τ₴

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Explanation

A port of Kevin Cruijssen's O5AB1E answer

<O flag> (disable implicit output, so anything other than 'dotnet ef' doesnt print anything)
«ƛ4-ʀWṪ«=[                                - If the input equals 'dotnet ef':
          »|ȮW¼...8…ṡ@»                   - Push a big compressed integer
                       `_ |\/\n-=).`τ₴    - Replace the digits with symbols, and print.

Answer 12

C (gcc), 175 169 bytes

-6 bytes by reusing the input variable

i;f(char*p){if(!strcmp(p,"dotnet ef"))for(p="76QI:i7[bQ4Ji1;2;3Aq4IAIiA1:B1:A2A2y3KiA19A1A19A3I9Q1A2RAJiA;B9A7Q3KQJ";i=*p++;)while(i--&7)putchar(" _|\\/-=\n.)"[i/8-6]);}

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I used RLE compression, where the lower 3 bits are the length and the high bits encode the character index. By shifting the index by 6 we get an encoding that doesn't require escaping in a string literal. I used this function for encoding:

void encode() {
    char chars[] = " _|\\/-=\n.)";
    int cur, prev = ' ', count = 0;
    while ((cur = getchar()) != EOF) {
        if (cur == prev && count < 7) {
            count++;
        } else {
            putchar((strchr(chars, prev) - chars + 6) * 8 + count);
            count = 1;
        }
        prev = cur;
    }
}

Answer 13

05AB1E, 83 82 bytes

’¥¥‡Ò ef’Q•{Ã∍ûçαÖв₂þ₂ÕµŒHθǝ¤kªγVíαвÒf¾ΘMãwà5ÚX¿Q₆Sì=ЧòqxheΘ+ãkÛž•"/ \_
-=|.)"Åв×

Try it online! Outputs as a list of characters. Link includes a footer to format the output.

_{Kevin Cruijssen, why so bad?}

Answer 14

Python3, 177 bytes

if"dotnet ef"==input():print("             /\__\n       ---==/    \\\\\n ___  ___   |.    \|\ \n| __|| __|  |  )   \\\\\ \n| _| | _|   \_/ |  //|\\\n|___||_|       /   \\\/\\")

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