Shortest Konami Code

Question

The Problem

You must write a program that, when the Konami Code is typed in during runtime, prints the string "+30 lives" and sounds a noise of your choice from the computer's speaker.

Definition

The "Konami Code" is defined as UUDDLRLRBA followed by pressing the enter key.

The Rules

• You may choose to use the up arrow for U, down for D, left for L, and right for R, as long as your code is consistent with either arrows or letters.

• Your input may be accepted from a controller or a keyboard, but does not need to support both.

• Existing answers may continue to use BABA instead of BA, but may also shorten it if they wish to do so. Future answers should all use BA for consistency.

• Empty input doesn't need to be supported.

Answer 1

05AB1E, 33 bytes (Fine on Empty Input)

Dg0Qiq}•ï“뙵yê!•36BQi7ç“+30™‹“J,

Try it online!

05AB1E, 30 26 bytes (Fails on Empty Input)

-4 thanks to Adnan

•ï“뙵yê!•36BQi7ç,“+30™‹“,

Try it online!

•ï“뙵yê!•36B                  # UUDDLRLRBABA in base-214, converted back to base-36.
             Qi                # If implicit input is equal to...
               7ç,             # Print 7 converted to a character to the console (bell sound)...
                  “+30™‹“, # Print +30 lives.

Disclaimer: Bell sound does not play on TryItOnline, maybe ask Dennis if one of his servers somewhere is pinging when you run it.

Answer 2

JavaScript (ES6), 202 200 198 bytes

Tested only on Chrome and Firefox. Uses UDLR rather than the arrow keys. Please make sure that CapsLock is off.

(o=(A=new AudioContext()).createOscillator()).connect(A.destination);s='';document.onkeyup=e=>{(s+=e.key[0]).slice(-11)=='uuddlrlrbaE'&&setTimeout('o.stop()',500,console.log('+30 lives'),o.start())}

Please click inside this area to make sure it gets focus.

Answer 3

Python 3,  55  53 bytes

-1 byte thanks to isaacg (use bitwise not to replace -10 with ~9).
-1 byte thanks to CalculatorFeline (it is possible to use a literal BEL byte).

while'UUDDLRLRBA'!=input()[~9:]:1
print('+30 lives7')

where the 7 shown above is a literal byte 0x07 (an unprintable in a code block).

Once enter is pressed the while loop condition is checked. If the last ~9 = 10 characters (at most) do not match the Contra command code then the no-op 1 is executed (replacement of pass for brevity); if they do match then the while loop ends and the print statement is executed, which writes the required text along with an ASCII bell character ("alarm"), 0x07, which produces a sound unless it has been explicitly disabled (or the terminal has no speaker!).

Answer 4

Jelly, 33 30 29 27 bytes

ɠ⁻“UUDDLRLRBA”$¿ṛ7Ọ“¡}ʠƈ[ỵ»

Turns out, the append (ṭ) is not needed, strings are automatically joined together both printed.

ɠ⁻“UUDDLRLRBABA”$¿ṛ7Ọṭ“¡}ʠƈ[ỵ»

---

Uncompressed

ɠ⁻“UUDDLRLRBABA”$¿ṛ7Ọṭ“+30 lives”

Try it online
If you remove everything after ¿, you can mess with the input and see that it only returns when the input string is correct.

ɠ⁻“UUDDLRLRBABA”$¿ṛ7Ọṭ“+30 lives” - main link
ɠ                                 - takes a line of input
 ⁻“UUDDLRLRBABA”$                 - $ groups it into a monadic !="UUDDLRLRLRBABA"
                 ¿                - while (the inequality, which takes from the getline)
                  ṛ               - take the right argument (to ignore the input string)
                   7Ọ             - chr(7), the bell character
                     ṭ“+30 lives” - append that bell character to the output

Thanks @JonathanAllan for helping me figure out string compression :)

Answer 5

Processing, 161 157 155 bytes

String k="";void keyTyped(){if(match(k+=key,"uuddlrlrba\\n$")!=null){print("+30 lives");new processing.sound.SoundFile(this,"a.mp3").play();}}void draw(){}

The audio file must be saved as sketchName/data/a.mp3. Note: I have only tested this program without the audio file because I am too lazy to download an mp3 file (since only limited extensions are supported from processing.sound.SoundFile).

The draw() function is needed to be there in order for keyTyped to work.

The reason we are using keyTyped is because Processing does not have STDIN, it can only listen for keys being pressed via the sketch being run.

Explanation

String k="";
void keyTyped(){
  if(match(k+=key,"uuddlrlrba\\n$")!=null){
    print("+30 lives");
    new processing.sound.SoundFile(this,"a.mp3").play();
  }
}
void draw(){
}

All of the user's keystrokes are stored as chars inside the String k. The keyTyped is an inbuilt function that is called whenever the user types a key. Simultaneously, we are checking if this String ends with the keystrokes. Then we print +30 lives and play the sound file. And the draw function is there to continuously update keyTyped. After the Konami code is entered, then nothing else will be outputted and no audio will be played.

Answer 6

><>, 132 69 64 62 bytes *><>, 84 bytes

i:1+?\0[
:1+?!\i
BA"a~/"UUDDLRLR
{v!?l/!?=
7/"+30 lives"
o<;!?l

Try it here!

This relys on the bell chime (ascii 7) for the noise at the end (not heard on the online interpreter).

Thanks to @TealPelican for saving another 15 bytes!

Saved two bytes checking for BA instead of BABA.

Answer 7

C 119 -1 - 2 = 116 bytes

Golfed

i;f(){char a[14];a[13]=0;while(strcmp(a,"UUDDLRLRBA"))for(i=0;i<12;a[i]=a[i+++1]);a[i]=getch();puts("+30 Lives\a");}  

Ungolfed

#include<stdlib.h>
#include<conio.h>

i;
f()
{
    char a[14];
    a[13]=0;
    while(strcmp(a,"UUDDLRLRBA"))
    {
        for(i=0;i<12;a[i]=a[i+++1]);
        a[i]=getch();
        puts(a); //This is to print every step in the runtime
    }
    puts("+30 Lives\a"); // '\a' is the acsii char that makes sound when passed to STDOUT
}

main()
{
    f();
}

Answer 8

Stacked, 41 bytes

['+30 lives'BEL+out]prompt'UUDDLRLRBA'=if

Reads input from keyboard.

['+30 lives'BEL+out]prompt'UUDDLRLRBA'=if
[                  ]prompt'UUDDLRLRBA'=if  if the input is the desired string,
 '+30 lives'BEL+out                           output '+30 lives' and the BEL character. 

Answer 9

AutoHotkey, 63 bytes

Input,K,L10
if(K="UUDDLRLRBA"){
MsgBox,"+30 lives" 
SoundBeep
}

After running this script, it will check if the next 10 keys are the Konami code and if it is, it will show a message box saying "+30 lives" and it should play a beep (I don't have speakers now to test).

Answer 10

GNU sed, 32 25 + 1 = 33 26 bytes

+1 byte for -n flag.

/UUDDLRLRBA$/a+30 Lives\a

Try it online!

Explanation

Every time enter is pressed ($), the program checks if the preceding 10 keystrokes were the code. If they were, the a command queues the text +30 Lives and the bell sound (\a) to be printed at the end of the current cycle (which, in this case, is immediately).

Answer 11

Mathematica, 50 bytes

While[x=!=UUDDLRLRBA,x=Input[]];Beep[];"+30 lives"

Initially, x is just a symbol and is thus not identical (=!=) to the symbol UUDDLRLRBA, so x=Input[] is evaluated. This will open a dialog box with the cursor already in the input field, so the user can immediately start typing on the keyboard.

If Enter is pressed or OK is clicked without typing anything, then InputField[] will return Null, which is not identical to UUDDLRLRBA, so the loop continues and another dialog box will be opened.

If the user clicks Cancel or otherwise exits the dialog box, then InputField will return $Canceled, which is also not identical to UUDDLRLRBA, so the loop will continue.

The user can type in the dialog box to their heart's desire. When they hit Enter, their input is interpreted as a Wolfram language expression (possibly just boxes). If that expression is anything other than the symbol UUDDLRLRBA, the loop will continue.

"Must loop until the Konami code is entered" is a little vague, but I believe this satisfies it. Once the While loop is completed, Beep[];"+30 lives".

Answer 12

Petit Computer BASIC, 91 bytes

@L
WAIT 1S$=S$+CHR$(B)*!!B
B=BTRIG()IF B<1024GOTO@L
IF"xxxxxxxxxx"==S$THEN BEEP?"+30 lives

Finally a solution that actually uses controller input!

I used Petit Computer rather than the newer version (SmileBASIC) because it has access to the Start button, and BTRIG() is shorter than BUTTON().

I've replaced the data string with x's, it should contain characters with ascii codes 1,1,2,2,4,8,4,8,32,16

Boring version, 46 bytes

INPUT S$IF"UUDDLRLRBA"==S$THEN BEEP?"+30 lives

Answer 13

Vyxal, 21 bytes

«_sa}]∷«⇧=`+30 æ§\7`*

Based on @Makonede's 05AB1E answer.

Try it Online!

Explained:

«_sa}]∷«⇧=`+30 æ§\7`*    
                    *    # Repeat...
          `+30 æ§\7`     # the string '+30 lives␇'... (escape code \7 is BEL)
                         # once if ...
         =               # the (implicit) input equals...
«_sa}]∷«                 # the compressed code...
        ⇧                # in uppercase...
                         # otherwise repeat zero times

Answer 14

Powershell, 89 86 Bytes

if((Read-Host)-eq"UUDDLRLRBABA"){"+30 Lives";[System.Media.SystemSounds]::Beep.Play()}

enter submits the string to the Read-Host, so it should.. work?

Answer 15

C# (the boring version), 80 bytes

void n(){while(Console.ReadLine()!="UUDDLRLRBA"){}Console.Write("+30 Lives\a");}

C# (The interesting one), 202 bytes

void n(){t();Console.Write("+30 Lives\a");}void t(){var k="UUDDLRLRBA";var s="";while(s!=k){s+=Console.ReadKey().Key.ToString();s=k.StartsWith(s)?s:"";}if(Console.ReadKey().Key!=ConsoleKey.Enter){t();}}

I think this works, sadly online testers do not support the existence of a console input, as such I will have to go on the tests I have done

Can most likely be golfed hugely - I'm not great at this! :)

Ungolfed:

void n()
{
    t();
    Console.WriteLine("+30 Lives\a");
}

void t()
{
    var k = "UUDDLRLRBA";
    var s = "";
    while(s != k)
    {
        s += Console.ReadKey().Key.ToString();
        s = k.StartsWith(s) ? s : "";
    }
    if(Console.ReadKey().Key != ConsoleKey.Enter)
    {
        t();
    }
}

Answer 16

Wonder, 50 bytes

ol"•";f\@[="UUDDLRLRBA"rl0?ol"+30 lives•"?f0];f0

Replace • with the character of code \x07 (the BEL control char). Takes input through STDIN.

Answer 17

flex, 37 + 5 = 42 bytes

%%
UUDDLRLRBA puts("+30 lives\a");

The code itself is 37 bytes, compile it with the "-main" option which adds 5 bytes. Naturally, you have to compile the resulting C file with your favorite C compiler, but I don't think that step should count toward the byte count.

I could save a byte by using a literal BEL character instead of \a, but I'd rather be able to read my own code.

Answer 18

C, 87 85 81 bytes

s[13];f(){for(;strcmp(s,"UUDDLRLRBA\n");fgets(s,12,stdin));puts("+30 Lives!\a");}

Answer 19

shortC, 52 bytes

s[13];AO;Ms,"UUDDLRLRBA\n");Ys,12,N));J"+30 Lives!\a

Explanation:

s[13];                                                // declare array with 13 members
      A                                               // main function
       O;Ms,"UUDDLRLRBA\n");                          // for loop while input is not equal to the Konami Code
                            Ys,12,N));                // read 12 characters of input into array
                                      J"+30 Lives!\a  // print the string

Try it online!

Answer 20

C, 114 bytes

#define X "UUDDLRLRBA"
main(i,j){for(i=j=0;!j;)(getchar()==X[i++])&&((j=1?!X[i]|puts("+30 lives\a"):0)|1)||(i=0);}

Answer 21

Stax, 21 bytes

é>Δ┼3Σöîc9qz╔°→cz╣╖↑└

Run and debug it

Explanation

`!j.o6F>"`=D"+30 lives"7+
`!j.o6F>"`=               input equals sequence?
           D              if so,
            "+30 lives"   push +30 lives
                       7+ add a bell to it           

Answer 22

05AB1E, 22 bytes

.•›Gβ…Ýò•uQ7ç“+30™‹ÿ“×

Try it online!

.•...•uQ7ç“...“×  # trimmed program
          “...“   # push "+30 livesÿ"...
                  # (implicit) with ÿ replaced by...
         ç        # unicode character with codepoint...
        7         # literal...
               ×  # repeated...
       Q          # is...
                  # implicit input...
       Q          # equal to...
.•...•            # "uuddlrlrba"...
      u           # in uppercase...
               ×  # times
                  # implicit output