In dem Webcomic Darths & Droids behauptet Pete, der R2-D2 in der fiktiven Rollenspielkampagne spielt, auf der der Comic basiert, einmal (Warnung: Potenzielle Spoiler im verknüpften Comic), dass er sich mit dem Lost Orb von Phanastacoria auf seine Seite gestellt hat Schocksonde , er kann jetzt eine satte 1048576d4 austeilendes Schadens. (Der GM hat dies weder bestätigt noch geleugnet.) Da es einigermaßen offensichtlich sein sollte, dass fast niemand die Geduld haben wird, so viele Würfel zu würfeln, schreiben Sie ein Computerprogramm, um dies für ihn zu tun, und geben Sie den Gesamtwert aus, der in einem vernünftigen Rahmen gewürfelt wurde Format. Die Einträge werden nach Programmgröße (kürzestes Programm, Anzahl der Bytes, Anzahl der Siege), sowohl gesamt als auch pro Sprache, sortiert, wobei die Laufzeit unterbrochen wird. Die Antwort kann entweder ein vollständiges Programm oder eine Funktionsdefinition sein.

Scores pro Sprache

Pyth

Maltysen - 8 Bytes *

Jakube - 10 Bytes

APL

Alex A - 10 Bytes

CJam

Optimierer - 11 Bytes

J

ɐɔıɐɔuʇǝɥʇs - 12 Bytes **

Clip10

Ypnypn - 12 Bytes **

K

JohnE - 13 Bytes

Ti-84 BASIC

SuperJedi224 - 17 Bytes *

R

MickyT - 23 Bytes

OCTAVE / MATLAB

Oebele - 24 Bytes

PARI / GP

Charles - 25 Bytes **

Wolfram / Mathematica

LegionMammal978 - 27 Bytes

Perl

Nutki - 29 Bytes

AsciiThenAnsii - 34 Bytes

Rubin

Haegin - 32 Bytes **

ConfusedMr_C - 51 Bytes **

Commodore Basic

Markierung - 37 Bytes **

PHP

Ismael Miguel - 38 Bytes

VBA

Sean Cheshire - 40 Bytes **

Power Shell

Nacht - 41 Bytes **

Javascript

Ralph Marshall - 41 Bytes

edc65 - 54 bytes (Erfordert ES6-Funktionalität, die nicht in allen Browsern verfügbar ist.)

Lua

Kryptychon - 51 Bytes

Java

RobAu - 52 Bytes **

Geobits - 65 Bytes

C

Funktino - 57 Bytes

Python

CarpetPython - 58 Bytes

Postgre / SQL

Andrew - 59 Bytes **

Schnell

Skrundz - 69 Bytes

GoatInTheMachine - 81 Bytes

Haskell

Zeta - 73 Bytes **

ActionScript

Brian - 75 Bytes **

> <>

ConfusedMr_C - 76 Bytes

GEHEN

Kristoffer Sall-Storgaard - 78 Bytes

C #

Brandon - 91 Bytes **

Andrew - 105 Bytes

Ewan - 148 Bytes

Kratzen

SuperJedi224 - 102 Bytes

C ++

Michelfrancis Bustillos - 154 Bytes

Mehrsprachige

Ismael Miguel (Javascript / ActionScript2) - 67 Byte

Top 10 insgesamt

Maltysen
Alex A
Jakube
Optimierer
ɐɔıɐɔuʇǝɥʇs / Ypnypn (Bestellung ungewiss)
JohnE
SuperJedi224
MickyT
Oebele

Programmiert markiert ** Ich konnte noch nicht richtig testen

Pyth - 9 8 Bytes

Verwendet eine offensichtliche einfache Methode zur Summierung von Randint. Es dauerte eine Minute, bis mir klar 1048576wurde 2^20, dass ich mich jetzt wirklich dumm fühle. Vielen Dank an @Jakube, dass Sie mir ein Byte gespart haben, indem Sie darauf hingewiesen haben 2^20 = 4^10.

smhO4^4T

Die Laufzeit ist schrecklich, es ist noch nicht auf meinem Computer zu beenden, so dass kein Punkt ist es online läuft so hier ist die 2^10eine: Probieren Sie es hier online .

s        Summation
 m       Map
  h      Incr (accounts for 0-indexed randint)
   O4    Randint 4
  ^4T    Four raised to ten

Perl - 48 44 37 39 34 Bytes

$-+=rand(4)+1for(1..2**20);print$-

Gibt die Summe ohne nachfolgende Zeile aus.
4 Bytes gespart durch Ersetzen von 2**20(danke Maltysen) und Entfernen von Anführungszeichen um Drucken.
Weitere 7 Bytes durch Umstellen des Codes gespart (Danke Thaylon!)
2 Bytes verloren, da mein alter Code 0-4 generiert hat (es sollte 1-4 sein).
Nochmals 5 Bytes gespart dank Caek und Nutki.

Ungolfed, richtig geschriebener Code:

my $s = 0
$s += int( rand(4) + 1 ) for (1 .. 2**20);
print "$s";

APL, 11-10 Bytes

+/?4⍴⍨2*20

Dies ergibt nur die Summe eines Arrays von 2 ²⁰ = 1048576 Zufallszahlen zwischen 1 und 4.

+/           ⍝ Reduce by summing a
  ?          ⍝ random integer
   4⍴⍨       ⍝ array with values between 1 and 4
      2*20   ⍝ of length 2^20

Sie können dies mit TryAPL vergleichen indem Sie den Zeitstempel davor und danach drucken. Es dauert ungefähr 0,02 Sekunden.

Dank marinus und FUZxxl ein Byte gespart!

Ti-84 Basic, 17 Bytes

Gesamter Footprint - Größe des Programmheaders = 17 Byte

Laufzeit: Unbekannt, geschätzt auf 5-6 Stunden basierend auf der Leistung für eine geringere Anzahl von Rollen (also im Grunde nicht sehr gut)

Σ (randInt (1,4), A, 1,2 ^ 20

R 32 24 23 21 Bytes

Bearbeiten: Wurde die as.integerund verwendete Ganzzahldivision los%/% . Beschleunigen Sie es leicht.

Vielen Dank an Alex A für den Probentipp ... und Giuseppe für das Entfernen des r=

sum(sample(4,2^20,T))

Getestet mit

i = s = 0
repeat {
i = i + 1
print(sum(sample(4,2^20,r=T)))
s = s + system.time(sum(sample(4,2^20,r=T)))[3]
if (i == 10) break
}
print (s/10)

Ausgänge

[1] 2621936
[1] 2620047
[1] 2621004
[1] 2621783
[1] 2621149
[1] 2619777
[1] 2620428
[1] 2621840
[1] 2621458
[1] 2620680
elapsed 
   0.029 

Für reine Geschwindigkeit wird das Folgende in Mikrosekunden ausgeführt. Ich bin mir jedoch nicht sicher, ob ich meine Logik richtig verstanden habe. Die Ergebnisse scheinen mit der Zufallsmethode übereinzustimmen. Schade, dass es eine längere Länge ist.

sum(rmultinom(1,2^20,rep(1,4))*1:4)

Hier ist ein Zeitablauf, den ich auf meiner Maschine gemacht habe

system.time(for(i in 1:1000000)sum(rmultinom(1,2^20,rep(1,4))*1:4))
                   user                  system                 elapsed 
7.330000000000040927262 0.000000000000000000000 7.370000000000345607987 

Python 2, 58 bytes

We get 1048576 random characters from the operating system, take 2 bits of each, and add them up. Using the os library seems to save a few characters over using the random library.

import os
print sum(1+ord(c)%4 for c in os.urandom(1<<20))

This takes about 0.2 seconds on my PC.

CJam, 12 11 bytes

YK#_{4mr+}*

This is pretty straight foward:

YK                  e# Y is 2, K is 20
  #                 e# 2 to the power 20
   _                e# Copy this 2 to the power 20. The first one acts as a base value
    {    }*         e# Run this code block 2 to the power 20 times
     4mr            e# Get a random int from 0 to 3. 0 to 3 works because we already have
                    e# 2 to the power 20 as base value for summation.
        +           e# Add it to the current sum (initially 2 to the power 20)

But the beauty of this is that its really fast too! On my machine (and using the Java compiler) it takes on an average of 70 milliseconds.

The online version takes around 1.7 seconds on my machine.

Update: 1 byte saved thanks to DocMax

JavaScript (ES6), 54 bytes

Average time < 100 msec. Run snippet to test (in Firefox)

// This is the answer
f=t=>(i=>{for(t=i;i--;)t+=Math.random()*4|0})(1<<20)|t

// This is the test
test();

function test(){
  var time = ~new Date;
  var tot = f();
  time -= ~new Date;
  
  Out.innerHTML = "Tot: " + tot + " in msec: " + time + "\n" + Out.innerHTML;
}

<button onclick="test()">Repeat test</button><br>
<pre id=Out></pre>

Explanation

With no statistical package built-in, in Javascript the shortest way to obtain the sum of 1 million random number is to call random() for a million times. So simply

f=()=>{
   var t = 0, r, i
   for (i=1<<20; i--; ) 
   {
      r = Math.random()*4 // random number between 0 and 3.9999999
      r = r + 1 // range 1 ... 4.999999
      r = r | 0 // truncate to int, so range 1 ... 4
      t = t+r
   }
   return t
}

Now, adding 1 for a million times is exactly the same than adding 1 million, or even better, start the sum with 1 million and then add the rest:

f=()=>{
   var t, r, i
   for (t = i = 1<<20; i--; ) 
   {
      r = Math.random()*4 // random number between 0 and 3.9999999
      r = r | 0 // truncate to int, so range 0 ... 3
      t = t+r
   }
   return t
}

Then golf, drop the temp variable r and drop the declaration of local variables. t is a parameter, as one is needed to shorten the declaration of f. i is global (bad thing)

f=t=>{
   for(t=i=1<<20;i--;) 
      t+=Math.random()*4|0
   return t
}

Then find a way to avoid 'return' using a nameless inner function. As a side effect, we gain another parameter so no globals used

f=t=>(
  (i=>{ // start inner function body
     for(t=i;i--;)t=t+Math.random()*4|0 // assign t without returning it
   })(1<<20) // value assigned to parameter i
  | t // the inner function returns 'undefined', binary ored with t gives t again
) // and these open/close bracket can be removed too

Perl, 29

Generates a table of the required length.

print~~map{0..rand 4}1..2**20

J (12 bytes, about 9.8 milliseconds)

+/>:?4$~2^20

I suspect this is mostly memory bandwith-limited: I can't even get it to max out a single core...

You can test this with the following code:

   timeit =: 13 : '(1000 * >./ ($/x) 6!:2"0 1 y)'
   4 20 timeit '+/>:?4$~2^20'
9.90059

This runs it in 4 groups of 20 trails, and returns the number of milliseconds of the avarage time in the quickest group. An interpreter can be found here.

Pyth, 10 bytes

u+GhO4^4TZ

This has slightly more bytes than @Maltysen's Pyth solution. But it runs in 8.5 seconds on my laptop, while @Maltysen's solution produced no solution in 20 minutes running time.

But still a little bit too slow for the online compiler.

Explanation

u     ^4TZ   start with G = 0, for H in 0, ... 4^10-1:
                G = 
 +GhO4              G + (rand_int(4) + 1)
             result is printed implicitly 

Java, 65

Since we have scores listed by language, why not throw Java into the mix? There's not much to golf here, just a simple loop, but I was able to squeeze a couple out of my initial attempt:

int f(){int i=1<<20,s=i;while(i-->0)s+=Math.random()*4;return s;}

Matlab, 24

First submission ever!

sum(randi([1,4],1,2^20))

I had hoped to make use of randi([1,4],1024), which gives a matrix of 1048576 elements, but then I needed a double sum, which takes more characters than this.

Regarding the running speed mentioned in the question, timeit tells me the runtime is about 0.031 seconds. So, pretty much instant.

Haskell, 73 bytes

import System.Random
f=fmap sum.(sequence.replicate(2^20))$randomRIO(1,4)

Usage:

$ ghci sourcefile.hs
ghci> f
2622130

C#: 105 bytes

using System.Linq;class C{int D(){var a=new System.Random();return new int[1<<20].Sum(i=>a.Next(1,5));}}

PHP, 38 37 bytes

This uses a very simple idea: sum them all!

Also, I've noticed that 1048576 is 10000000000000000000 in binary, equivalent to 1<<20.

Here's the code:

while($i++<1<<20)$v+=rand(1,4);echo$v

Test in your browser (with VERY LITTLE changes):

$i=$v=0;while($i++<1<<20)$v+=rand(1,4);printf($v);

//'patch' to make a TRUE polyglot, to work in JS

if('\0'=="\0")//this will be false in PHP, true in JS
{
  //rand function, takes 2 parameters
  function rand($m,$n){
    return ((Math.random()*($n-$m+1))+$m)>>0;
    
    /*
     *returns an integer number between $m and $n
     *example run, with rand(4,9):
     *(Math.random()*(9-4+1))+4
     *
     *if you run Math.random()*(9-4+1),
     *it will returns numbers between 0 and 5 (9-4+1=6, but Math.random() never returns 1)
     *but the minimum is 4, so, we add it in the end
     *this results in numbers between 0+4 and 4+5
     *
     */
    
  }
  
  //for this purpose, this is enough
  function printf($s){
    document.write($s);
  }
}


/*
 *Changes:
 *
 *- instead of echo, use printf
 *    printf outputs a formatted string in php
 *    if I used echo instead, PHP would complain
 *- set an initial value on $i and $v
 *    this avoids errors in Javascript
 *
 */


$i=$v=0;while($i++<1<<20)$v+=rand(1,4);printf($v);

All the changes in the code are explained in comments.

Mathematica, 30 27 bytes

Tr[RandomInteger[3,2^20]+1]

Mathematica has quite long function names...

C, 57 bytes

main(a,b){for(b=a=1<<20;a--;b+=rand()%4);printf("%d",b);}

This code works... once. If you ever need to roll those dice again, you'll need to put srand(time(0)) in there, adding 14 bytes.

PostgreSQL, 59 bytes

select sum(ceil(random()*4)) from generate_series(1,1<<20);

I'll admit to the slight problem that random() could, in theory, produce exactly zero, in which case the die roll would be zero.

Ruby, 32 bytes

(1..2**20).inject{|x|x-~rand(4)}

In a more readable form:

(1..2**20).inject(0) do |x|
  x + rand(4) + 1
end

It creates a range from 1 to 1048576 and then iterates over the block that many times. Each time the block is executed the value from the previous iteration is passed in as x (initially 0, the default for inject). Each iteration it calculates a random number between 0 and 3 (inclusive), adds one so it simulates rolling a d4 and adds that to the total.

On my machine it's pretty fast to run (0.25 real, 0.22 user, 0.02 sys).

If you've got Ruby installed you can run it with ruby -e 'p (1..2**20).inject{|x|x+rand(4)+1}' (the p is necessary to see the output when run in this manner, omit it if you don't care for that or just run it inside IRB where the result is printed to the screen for you). I've tested it on Ruby 2.1.6.

Thanks to histocrat for the bit twiddling hack that replaces x + rand(4) + 1 with x-~rand(4).

PARI/GP, 25 bytes

Really, no need for golfing here -- this is the straightforward way of doing the calculation in GP. It runs in 90 milliseconds on my machine. Hoisting the +1 saves about 20 milliseconds.

sum(i=1,2^20,random(4)+1)

Just for fun: if one were optimizing for performance in PARI,

inline long sum32d4(void) {
  long n = rand64();
  // Note: __builtin_popcountll could replace hamming_word if using gcc
  return hamming_word(n) + hamming_word(n & 0xAAAAAAAAAAAAAAAALL);
}

long sum1048576d4(void) {
  long total = 0;
  int i;
  for(i=0; i<32768; i++) total += sum32d4();
  return total;
}

has a very small total operation count -- if xorgens needs ~27 cycles per 64-bit word (can anyone verify this?), then a processor with POPCNT should take only about 0.5 cycle/bit, or a few hundred microseconds for the final number.

This should have close-to-optimal worst-case performance among methods using random numbers of similar or higher quality. It should be possible to greatly increase average speed by combining cases -- maybe a million rolls at a time -- and selecting with (essentially) arithmetic coding.

Javascript, 55 53 50 47 41 bytes

for(a=i=1<<20;i--;)a+=(Math.random()*4)|0

I didn't realize that non-random numbers were a known irritant, so I figure that I ought to post a real solution. Meant no disrespect.

Commentary: as noted by others above you can skip the +1 to each roll by starting off with the number of rolls in your answer, and by not having to write a=0,i=1<<20 you save two bytes, and another 2 because you don't add +1 to each roll. The parseInt function does the same thing as Math.floor but is 2 characters shorter.

Clip 10, 12 bytes

r+`m[)r4}#WT

         #4T    .- 4^10 = 1048576             -.
   m[   }       .- that many...               -.
     )r4        .-          ...random numbers -.
r+`             .- sum                        -.

It takes approximately 0.6 seconds to run on my machine.

Go, 78 bytes

Golfed

import."math/rand";func r()(o int){for i:=2<<19;i>=0;i--{o+=Intn(4)+1};return}

Still working on it

Run online here http://play.golang.org/p/pCliUpu9Eq

Go, 87 bytes

Naive solution

import"math/rand";func r(){o,n:=0,2<<19;for i:=0;i<n;i++{o+=rand.Intn(4)};println(o+n)}

Run online here: http://play.golang.org/p/gwP5Os7_Sq

Due to the way the Go playground works you have to manually change the seed (time is always the same)

Commodore Basic, 37 bytes

1F┌I=1TO2↑20:C=C+INT(R/(1)*4+1):N─:?C

PETSCII substitutions: ─ = SHIFT+E, / = SHIFT+N, ┌ = SHIFT+O

Estimated runtime based on runs with lower dice counts: 4.25 hours.

It's tempting to try to golf off two bytes by making C an integer, getting implicit rounding of the random numbers. However, the range on integers in Commodore Basic is -32678 to 32767 -- not enough, when the median answer is 2621440.

PowerShell, 41 37 bytes

1..1mb|%{(get-random)%4+1}|measure -s

Took my machine 2 minutes 40 seconds

Ruby, 51 47 chars

x=[];(2**20).times{x<<rand(4)+1};p x.inject(:+)

I looked at all of the answers before I did this, and the sum(2**20 times {randInt(4)}) strategy really stuck out, so I used that.

><>, 76 chars

012a*&>2*&1v
|.!33&^?&:-<
3.v < >-:v >
   vxv1v^<;3
  1234    n+
  >>>> >?!^^

I'm not sure if this one works, because my browser crashed when I tried to test it, but here's the online interpreter.

Swift, 64 bytes

Nothing clever, golfing in Swift is hard...

func r()->Int{var x=0;for _ in 0..<(2<<19) {x+=Int(arc4random()%4)+1;};return x;}

Version 2 (too late)

var x=0;for _ in 0..<(2<<19){x+=Int(arc4random()%4)+1;};print(x)

Java (Java 8) - 52

int f(){return new Random().ints(1<<20,1,5).sum();}