Cheat at Rock-Paper-Scissors-Lizard-Spock

Question

When playing Rock-Paper-Scissors-Lizard-Spock, your robot reflexes let you see your opponent's throw before they make it. You just need to choose a throw that beats theirs, in as few bytes as possible of course.

There's always two winning throws, and either one will work.

Input    -> Possible outputs
------------------------------
Rock     -> Spock or Paper
Spock    -> Paper or Lizard
Paper    -> Lizard or Scissors
Lizard   -> Scissors or Rock
Scissors -> Rock or Spock

Take a string as input, and output one of the strings that beats it. Strings must be exactly as given, with capitalization. List of characters may substitute for strings.

Answer 1

Perl 6, 36 30 bytes

{<Spock Lizard Rock>[.comb%4]}

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Gets each output based on the length of the input modulo 4. I think this is probably the optimal strategy.

Answer 2

Perl 5 -p, 28 bytes

$_=/k/?Paper:/i/?Rock:Lizard

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Tried @JoKing's length mod 4 method, but this turned out to be even shorter.

Answer 3

PHP, 39 bytes

Shorter md5 variants by Christoph and Benoit Esnard.

<?=[Lizard,Spock,Rock][md5(y.$argn)%3];

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<?=[Spock,Rock,Lizard][md5($argn.m)%7];

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PHP, 40 bytes

<?=[Rock,Spock,Lizard][md5($argn)[5]%3];

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Takes sixth character of md5 of input, which gives unique number of 1 for "Rock" and "Scissors", unique number of 2 for "Spock" and "Paper" and unique number of 9 for "Lizard". A mod 3 on those numbers and we have 0 or 1 or 2 indexes.

| Input    | MD5                              | 6th char | %3 | Output |
|----------|----------------------------------|----------|----|--------|
| Rock     | 4cfbb125e9878528bab91d12421134d8 |        1 |  1 | Spock  |
| Spock    | 5769c28299713c949cd59d5469e40ace |        2 |  2 | Lizard |
| Paper    | d0a662a5235ecde30739fe50cf0de830 |        2 |  2 | Lizard |
| Lizard   | 2f7569e00c4d97aef5f2c2b3a4d2213f |        9 |  0 | Rock   |
| Scissors | 28204140cee6e34a9843b64e5a490b08 |        1 |  1 | Spock  |

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PHP, 40 bytes

Alternative 40 bytes, port of Jo King's answer.

<?=[Spock,Lizard,Rock][strlen($argn)%4];

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PHP, 41 bytes

This is based on 79037662's answer.

<?=[Lizard,Rock,Paper][ord($argn[-1])%3];

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PHP (7.4), 42 bytes

Uses fifth and sixth characters of input to get one of 0, 1 or 2.

fn($s)=>[Rock,Lizard,Paper][!$s[5]+!$s[4]]

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PHP, 43 bytes

Uses CRC32 of input and mod to get 0, 1 or 2.

<?=[Rock,Lizard,Paper][crc32($argn)%866%3];

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PHP, 43 bytes

Uses second to the last character of input ("Rock" and "Spock" = "c", "Paper" = "e", "Lizard" and "Scissors" = "r").

<?=[c=>Paper,e=>Lizard,r=>Rock][$argn[-2]];

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Answer 4

APL (Dyalog Unicode), 28 bytes

'Spock' 'Lizard' 'Rock'⊃⍨4|≢

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Uses the simple strategy based on input length modulo 4 (idea from Jo King's answer). ⎕IO←0.

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APL (Dyalog Unicode), 34 bytes

'Lizard' 'Rock' 'Paper'⊃⍨'Xbj'⍸2∘⊃

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Uses the shortest choice for all cases. ⎕IO←1.

How it works

'Lizard' 'Rock' 'Paper'⊃⍨'Xbj'⍸2∘⊃
2∘⊃     Take the second character
'Xbj'⍸  Classify it by intervals:
        'X'≤c<'b': 1
        'b'≤c<'j': 2
        'j'≤c    : 3
...⊃⍨   Give the nth string

Answer 5

Jelly, 16 14 bytes

Lị“ŀ¡ḋ¤:1KỌ»Ḳ¤

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-2 bytes thanks to Jonathan Allan and Nick Kennedy both independently coming up with splitting on spaces instead of using a list literal.

A translation of Jo King's Perl 6 solution, using Jelly's compressed strings and modular indexing.

 ị                The element of
  “ŀ¡ḋ¤:1KỌ»      "Lizard Rock  Spock"
            Ḳ¤    split on spaces
 ị                at (1-based) index
L                 length of the input
  “ŀ¡ḋ¤:1KỌ»Ḳ¤    mod 4.

Answer 6

JavaScript, 40 bytes

s=>['Spock','Lizard','Rock'][s.length%4]

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Answer 7

Runic Enchantments, 42 bytes

il͍4%:0)9*?~"Spock"@1)9*?"Lizard"@"Rock"@

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Footer runs all five input possibilities (the y sequences make them print in the correct order, no I'm not sure why the last one needed so many) and presets the stack with an input value (the input command then reads no input) and a newline. Remove the footer to use input instead.

Answer 8

Python 3, 44 bytes

lambda s:["Spock","Lizard","Rock"][len(s)%4]

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Uses the input-length-modulo-4 strategy in Jo King's answer.

Answer 9

C (gcc), 51 49 bytes

Thanks to Unrelated String for the suggestion.

The second byte of the string has sufficient uniqueness that I can take its two least-significant bits and use them to then index into a string (with the longest-length result at the end if I rotate the result, fortunately!) The separator between the strings should be a literal NUL character, but I put a space in between them below so that you can see the gaps.

f(char*s){s="Spock Paper Scissors"+(s[1]-3)%4*6;}

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Answer 10

Haskell, 43 bytes

(k!!).length
k="Spock":"Lizard":"Rock":"":k

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The mod-4 approach. Longer approaches:

f t|t<"Q"="Scissors"|t<"Sd"="Spock"|1<2="Paper"   (47)
f.last;f 'k'="Paper";f 'r'="Lizard";f _="Rock"    (46)
(["Spock","Lizard","Rock"]!!).(`mod`4).length     (45)
(words"Spock Lizard Rock"!!).(`mod`4).length      (44)
(cycle["Spock","Lizard","Rock",""]!!).length      (44)
(cycle(words"Spock Lizard Rock .")!!).length      (44)

Answer 11

Python 2, 73 69 bytes

lambda i:l[l.index(i)-1]
l="Rock Lizard Spock Scissors Paper".split()

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Since there are already plenty of ports of @JoKing's excellent answer, here is a different way of doing it.

Answer 12

Retina, 30 27 bytes

'R(`R
Sp
'i(K`Rock
K`Lizard

Try it online! Link includes all possible test cases. @BusinessCat's answer works perfectly well in Retina 0.8.2 but for Retina 1 we can do a little better by using a conditional stage. Explanation:

'R(`

If the input contains R...

R
Sp

... then replace the R with Sp, thus turning Rock into Spock...

'i(

... otherwise if the input contains i...

K`Rock

... then overwrite it with the string Rock, which beats Lizard and Scissors...

K`Lizard

... otherwise overwrite the input with Lizard, which beats Spock and Paper.

Answer 13

Ruby, 42 36 bytes

->c{%w(Spock Lizard Rock)[c.size%4]}

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Ruby, 43 39 bytes

->c{%w(Lizard Rock Paper)[c[-1].ord%3]}

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The second one is an alternative I found to Jo's method, using the ASCII value of the last character, but alas is 3 bytes longer in Ruby.

It is conceivable that my method is better in some language where finding the ASCII value of the last character is shorter than finding the string's length, but I don't know of any.

EDIT: Saved several bytes thanks to @IMP1

Answer 14

F# (.NET Core), 53 49 46 bytes

 ["Spock";"Lizard";"Rock"].[String.length n%4]

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I am using the input-length-modulo-4 strategy from Jo King's answer, so please check his answer.
And I know it's not in competition but I wanted to try something new…

Answer 15

V (vim), 28 bytes

2ORock
Lizard
SpockLÝkYHVGp

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Hexdump:

00000000: 324f 526f 636b 0d4c 697a 6172 640d 5370  2ORock.Lizard.Sp
00000010: 6f63 6b1b 4cdd 6b59 4856 4770            ock.L.kYHVGp

Answer 16

Japt, 24 bytes

`Spock Lizd Rock`¸gUÊu4

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"Spock Lizard Rock" //=>compressed
.q(S) // split on " "
.g(U.l().u(4)) // return element at index ( length of input % 4 )

Thanks to @Shaggy for the help

Answer 17

Python 3, 69 64 61 57 bytes

lambda s:[["Scissors","Paper"]['k'in s],"Rock"]['i'in s] 

Not as short as the mod 4 solution, but not a bad alternative.

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Answer 18

Pyth, 24 bytes

@c."Lzž›^ºéëEL”‡"\`lz

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Port of Jo King's mod-4 answer. Without the compressed string, it's 2 bytes longer: @c"Spock`Lizard`Rock`"\`lz

Answer 19

Excel, 49 bytes

Mod4 approach:

=CHOOSE(MOD(LEN(A1),4)+1,"Spock","Lizard","Rock")

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Switching on last character: 57 bytes

=CHOOSE(MOD(CODE(RIGHT(A1)),3)+1,"Lizard","Rock","Paper")

Answer 20

R, 52 bytes

function(x)c("Spock","Lizard","Rock")[nchar(x)%%4+1]

Answer 21

C# dotnet core, 47 bytes

h=>(new[]{"Spock","Lizard","Rock"})[h.Length%4]

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Uses the input-length-modulo-4 strategy in Jo King's answer.

Answer 22

Python 2, 50 bytes

lambda s:["Lizard","Rock","Paper"][cmp(s[-2],"e")]

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Answer 23

Wren, 47 bytes

Fn.new{|s|["Spock","Lizard","Rock"][s.count%4]}

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Uses the input-length-modulo-4 strategy in Jo King's answer.

Answer 24

05AB1E, 16 bytes

”‰«†ÕâŸard”#I3öè

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Answer 25

Retina 0.8.2, 36 33 31 bytes

-3 thanks to Neil

R
s
.{6,}
Rock
.{5}
Lizard
s
Sp

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Answer 26

Batch, 87 bytes

@set/ps=
@for %%a in (Rock.d Rock.s Paper.k Lizard.r)do @if %%~xa==.%s:~-1% echo %%~na

Takes input on STDIN. Works by switching on the last character of the input, which saves a case.

Answer 27

K, 26 bytes

`Paper`Rock`Lizard@&/"ki"?

Answer 28

///, 81 70 bytes

/~/\/\/.//.Sp/.~R/.~ock/Paper~Paper/Lizard~Lizard/Rock~Scissors/Rock/.

Because there is no other way to take input in ///, it goes after the . at the end of the code. For an input of Spock:

/~/\/\/.//.Sp/.~R/.~ock/Paper~Paper/Lizard~Lizard/Rock~Scissors/Rock/.Spock

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Test suite

Explanation:

/// works by replacing strings with other strings, within the code.

The first chunk of code, /~/\/\/./, works as a macro replacing ~ with //.. After this, the code expands to:

/.Sp/.//.R/.//.ock/Paper//.Paper/Lizard//.Lizard/Rock//.Scissors/Rock/

/.Sp/./ and /.R/./ change spock and rock to ock, allowing me to just replace ock with Paper.

The rest of the cases are just explicitly handled.

The period before the input is to make sure that none of the code itself is changed, just the input.

Answer 29

Shakespeare Programming Language, 1359 bytes

(Whitespace added for readability)

R.Ajax,.Puck,.Act I:.Scene I:.[Enter Ajax and Puck]Ajax:
Open mind.Open mind.You is the remainder of the quotient betweenyou a big big cat.
Be you nicer a cat?If solet usScene V.
Be you nicer I?If solet usScene X.
You is twice twice twice twice the sum ofa big big cat a cat.Speak thy.You is the sum ofthe sum ofyou a big big big big cat a cat.Speak thy.You is the sum ofthe sum ofyou a big big big big cat a pig.Speak thy.You is the sum ofthe square oftwice the sum ofa big big cat a cat a cat.Speak thy.You is the sum ofyou the sum ofa cat twice twice the sum ofa big cat a cat.Let usScene L.
Scene V:.Ajax:
You is the sum ofa big cat the square ofthe sum ofa big big big cat a cat.Speak thy.You is twice twice twice twice the sum ofa big big big cat a pig.Speak thy.You is the sum ofyou a pig.Speak thy.You is the sum ofyou twice twice the sum ofa big pig a pig.Speak thy.You is the sum ofyou a big big big cat.Let usScene L.
Scene X:.Ajax:
You is the sum ofa big cat the square ofthe sum ofa big big big cat a cat.Speak thy.You is the sum ofyou a big big big big cat.Speak thy.You is the sum ofyou twice the sum ofa big cat a cat.Speak thy.You is the sum ofyou twice the sum ofa big big cat a cat.Remember you.Speak thy.Speak thy.You is the sum ofyou a big big pig.Speak thy.You is the sum ofyou the sum ofa big cat a cat.Speak thy.Recall.
Scene L:.Ajax:Speak thy.

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Explanation:

I get the second character of the input mod 4:

• This is 0 when the input is Spock, at which point Puck is neither nicer than a cat nor nicer than Ajax. Then, Scene I prints Paper.
• This is 1 when the input is Paper or Lizard, at which point Puck is not nicer than a cat, but he is nicer than Ajax. Then, Scene X prints Scissors.
• This is 3 when the input is Rock or Scissors, at which point Puck is nicer than a cat. Then, Scene L prints Spock.

Possible improvements:

• Better math when printing to get between the letters, including switching characters and using Recall more
• Better modulus because some of the things are longer than others, and I could reuse the ock
• Possibly better ways to write the existing numbers. I actually am using a brute forcer that I wrote to do this, but it doesn’t yet account for getting between the letters (using ‘you’). GitHub link coming soon.

This was really fun and I love this language and I have no regrets.