Roll Dungeons and Dragons dice

Question

I want to play Dungeons and Dragons, but I don't have any dice! Your challenge is to roll some D&D dice.

The input format specification in Backus-Naur form is:

<valid-input>  ::= <opt-integer> "d" <integer> <opt-modifier>
<opt-integer>  ::= | <integer>
<opt-modifier> ::= | "+" <integer>
<integer>      ::= "0" | "1" | "2" | "3" | "4" | "5" |
                   "6" | "7" | "8" | "9" | <integer> <integer>

The optional integer before the d is the number of dice to roll; it must be at least 1, and defaults to 1 if not supplied.

The required integer immediately after the d is the number of sides each die has; it must be at least 1. The sides of each die are distinct consecutive positive integers starting at 1.

The optional modifier may be +0, and it defaults to +0 if not specified.

For example, for the input 2d10+5, you generate two random numbers from 1 to 10 inclusive, add them together, and add 5. Then you will output the result.

If you receive invalid input, such as 2d, d20+, 0d4, 2d5+1+2, 2+2, or anything else that doesn't fit this format, you must output "Invalid input". Otherwise, you must output only a single random integer, weighted according to the input. For example, 3d6 should produce more 10s than 4s.

Test cases

Input      Minimum possible output    Maximum possible output
d1         1                          1
d6         1                          6
d1+3       4                          4
d20+3      4                          23
2d1        2                          2
2d6+2      4                          14
d01        1                          1
d01+0      1                          1
01d01+01   2                          2
3d20+10    13                         70

d          Invalid input
d0         Invalid input
d+0        Invalid input
d0+0       Invalid input
0d1        Invalid input
0d1+1      Invalid input
d1+        Invalid input
1d         Invalid input
1d1+       Invalid input
1d+1       Invalid input
2d+2d      Invalid input
d2+d2      Invalid input
d2+2+2     Invalid input
d2-1       Invalid input
-d2        Invalid input
-2d2       Invalid input
4*3        Invalid input
4*d2       Invalid input

This is code-golf, so the shortest code in bytes will win!

Answer 1

Perl, 109 95 93 96 89 bytes

s/^d/1d/;/^(\d+)d(\d+)(\+\d+)?$/;$d+=1+rand$2|0for
1..$1;$_=$1*$2?$d+$3:'Invalid input'

Requires the -p switch, which accounts for two of the bytes. Try it online on Ideone.

How it works

• Because of the -p switch, a line is read from STDIN and stored in $_.

• The command s/^d/1d/ prepends a 1 to $_ if it begins with a d, i. e., if the number of dice has not been specified.

• The regular expression /^(\d+)d(\d+)(\+\d+)?/ checks if the line consists of a number, a literal d, another number and, optionally, a third number preceded by a + sign.

  If there is a match, the numbers will be saved in $1, $2 and $3.

  In this case, the input will be valid if and only if $1 and $2 are both positive.

• $d += 1 + rand $2 | 0 adds a pseudo-randomly chosen integer from 1 to the specified number of sides to $d (initially treated as zero).

• for 1 .. $1 does the above once for every integer between 1 and the number of dice.

• The command $_ = $1 * $2 ? $d + $3 : 'Invalid input' does the following:

  • If $1 * $2 is zero, it sets $_ to Invalid input.

  • Otherwise, the input is valid and it sets $_ to the sum of the dice rolls and the modifier.

• Because of the -p switch, Perl prints the contents of $_.

• Since there a no further input lines, the script exits.

Answer 2

Fortran: 145

character(1)a;read(*,*)s,a,j,a,k;n=0;if(k<0.or.a=="-")then;print*,"error k<0";stop;endif;do l=1,int(s);n=n+int(s*rand(0)+1);enddo;print*,n+k;end;

Abuses implicit typing (i-n are all integers, everything else a real). Minor caveat: input must be space separated, so 2d10+5 must be entered as 2 d 10 + 5, otherwise you'll get an input conversion error.

Answer 3

Ruby, 116

Alternative Ruby version. I was trying to find a way to do it without the regular expressions, but the validation you have to do is a lot harder without them.

gets=~/^(\d+)?d(\d+)(\+\d+)?$/
a=$1||?1
puts$~&&a>?0?eval("r=#{$3||0};#{a}.times{r+=rand(#$2)+1};r"):'Invalid input'

This one is 112, using Dennis' clever Perl algorithm:

$p='(\d*[1-9]\d*)'
puts~/^#$p?d#$p(\+\d+)?$/?eval("r=#{$3||0};#{$1||1}.times{r+=rand(#$2)+1};r"):'Invalid input'

Answer 4

GolfScript (120 106 bytes)

.100?!1`*\+.43?)!'+0'*+.10,'d+':^*-!*.10,''*-^=*^1/{/n*}/~].,3=*3,or:x~;*{x~\{rand)+}+@*}'Invalid input'if

This is not only shorter than the first version, but also more elegant. The part which actually does the die rolling is

\{rand)+}+@*

The rest is mainly input validation, and a few characters for parsing.

# Start by converting valid inputs into valid inputs with all optional bits.
# Prepend a '1' if the string starts with 'd'.
.100?!1`*\+
# Append '+0' if there's no '+' in the string.
.43?)!'+0'*+
# Now we start knocking out the invalid inputs.
# If it contains a character other than [0-9d+], replace the string with ''.
.10,'d+':^*-!*
# If it doesn't contain exactly one 'd', exactly one '+', and the 'd' before the '+',
# replace the string with ''.
.10,''*-^=*
# Now we either have a valid string, an empty string, or a string which is almost valid
# but has some empty substrings which should be integers, or a forbidden 0 integer value.
# Replace the 'd' and '+' with newlines, eval the result, and gather into an array.
^1/{/n*}/~]
# If we had any empty parts, we'll have fewer than 3 items on the stack.
# In that case, replace with integer values which will fail the final validation step.
.,3=*3,or
# Final validation: number of dice * number of sides per die != 0.
:x~;*
# If we pass, do the actual die rolling. Otherwise give the error message.
{x~\{rand)+}+@*}'Invalid input'if

Online demo with test framework

Answer 5

Java, 378

Just wanted to try a solution with Java far from the best solution. But hey: Java isn't a golfing language in any case!

It gets the input from command line. First parameter args[0] is the input value.

class A{public static void main(String[]s){System.out.print(s[0].matches(
"(0+\\d+|[1-9]\\d*|)d(0+\\d+|[1-9]\\d*)(\\+\\d+)?")?z(s[0]):"Invalid input");}static int
z(String s){String[]a=s.split("d");String[]b=a[1].split("\\+");int c=a[0].isEmpty()?1:Byte.
decode(a[0]);int d=b.length<2?0:Byte.decode(b[1]);while(c-->0)d+=new java.util.Random().
nextInt(Byte.decode(b[0]))+1;return d;}}

Did you know, that decode is shorter than valueOf?

Answer 6

J - 130 (45?) char

This challenge seems to be a little biased towards regular expressions, especially with having to differentiate invalid input. J has a POSIX regex library, so it's not that bad, but it's not integrated as it is with Perl, so J fares no better than other languages.

+/@,`(1+?@#~)/`('Invalid input'"_)@.(0 e.$)0 1 1>.".>|.}.((,'?d','(\+[0-9]+)?$',~}.)'^([0-9]*[1-9][0-9]*)')(rxmatch rxfrom])1!:1]1

If you're just implementing the logic for valid expressions, like the Python/PHP solutions appear to, it's the more reasonable 45 chars:

+/,(1+[:?@#/1>.".;._2@,&'d')`".;._1'+',1!:1]1

Notable bits:

• 1!:1]1 is the input, and (rxmatch rxfrom]) is the logic that returns the subexpression matches.

• Whether or not the input was legal is handled by the regex matching, so we can set the defaults for n and p with 0 1 1>.. It looks backwards (n is 1 by default and p is 0) because we had to reverse (|.) the list earlier, so that the logic at the end executes in the right order.

• @. is the Agenda conjunction, essentially a J-ish switch statement. If the matches are empty (if 0 is an e.lement of the $hape: 0 e.$), we emit the error message, else we go through with rolling the dice: #~ to set out the dice, 1+? to roll, and +/@, to add the modifier p and sum.

Answer 7

TinyMUSH, 239

@dig/t +
@op d=+
@lo d=d
@fail d=Invalid input
@cr .
@set .=com
&d .=$*:\ifelse(regmatch(%0,^(\\\\d+)?d(\\\\d+)(\\\\+\\\\d+)?$,0 1 2 3),ifzero(and(or(not(strlen(%q1)),%q1),%q2),Invalid input,add(die(usetrue(%q1,1),%q2),%q3)),Invalid input)

The first four lines deal with the fact that "d" is an alias for the universal "down" exit with a built-in failure message when it doesn't exist; exits are scanned before user-defined commands. The remaining lines create an object with a user-defined command making use of the built-in die() function.

Answer 8

PHP, 129

<?eval(preg_filter(~Сף›ÔÖÀ›×£›Ö×£Ô£›ÔÖÀÛÐ,~ÛžÂÝÛÎÝÀÅÎÄ™×ÄÛ–ÔÔÃÛžÄیԞ‘›×ÎÓÛÍÖÖÄšœ—ÛŒÛÌÄ,$_GET[0])?:~šœ—ݶ‘‰ž“–›ß–‘Š‹ÝÄ);

Uses a regex to create a expression which PHP then evaluates. Input is fed in via url: ?0=argument. Make sure you urlencode the + to %2b. Here's what it looks like in a more readable form:

eval(preg_filter('/^(\\d)?d(\\d)(\\+\\d)?$/','$a="$1"?:1;for(;$i++<$a;$s+=rand(1,$2));echo$s$3;',$_GET[0])?:'echo"Invalid input";');

Bitwise inverting the strings using ~ not only saves a character because you don't need quotes (PHP assumes they are strings) but also saves characters because you don't have to escape the backslashes in the regular expression.

The ?: operator is a special form of the ternary operator. $foo = $a ? $a : $b is the same as $foo = $a ?: $b.

Answer 9

Ruby, 167 147

/^(\d+)?d(\d+)(\+\d+)?$/.match gets
abort'Invalid input'if !$~||$1==?0||!$2||$2==?0
p eval(([0]*($1||1).to_i).map{rand($2.to_i)+1}*?+)+($3||0).to_i

Uses a regexp to do all the work. Since I'm using \d+, the only things I need to check for invalid input are that there was a match, that neither n nor m were 0, and that there was an m. If any of those are found, it aborts with a message ('Invalid input'). Then it just prints the result, since it would have aborted by now if the input was invalid.

The result-printing isn't that interesting, but...

([0]*($1||1).to_i)    # create an array of n elements (1 if there is no n)
.map{rand($2.to_i)+1} # fill it up with random numbers, where the number x is 1 < x < m+1
.inject(:+)           # add them all up
+($3||0).to_i         # and finally add the modifier (0 if there is none)

I later changed .inject(:+) to eval(...*?+), but the idea is the same.

Answer 10

Python 3, 184 bytes

import random,re
try:a,b,c=re.findall("^(\d*)d(\d+)(\+\d+)?$",input())[0];t=int(c or 0)+(sum(random.randint(1,int(b))for i in range(int(a or 1)))or q)
except:t="Invalid input"
print(t)

Passes all tests. If zero dice were allowed, it would be 6 bytes shorter by leaving out (or q).

Answer 11

Vyxal V, 89 bytes

h\d=[1p|]⁽±Ḋ→a⟨←aL5=|←a⌊±T⟨0|2|4⟩=|←a1i\d=|←a3i\+=⟩A[←a0i⌊ʁ(←a2i⌊℅⅛)¾∑←a4i⌊+,|«¢q⁰∷Ṗ-←¥«,

Try it Online!

Ties Perl

h\d=[1p|]         # Checks if the first character is a d, and prepends a 1 
        ⁽±Ḋ→a     # Splits the string on non-numeric values, so consecutive digits stay grouped, then casts the list to variable "a"

     ⟨←aL5=|           # is it Length 5 (3 numbers, the "+" and the "d")
       ←a⌊±T⟨0|2|4⟩=|  # convert the numbers-as-strings to numbers and check if the numbers are in positions 1 3 and 5 (0 indexed)
         ←a1i\d=|     # is the second item in the list  "d"
           ←a3i\+=⟩    # is the fourth item "+"
                A     # are all of these checks true?

  [←a0i⌊ʁ               # If all the checks are true, create a range from 0-a[0]
        (←a2i⌊℅⅛)       #  for i in said range, generate a random number 0-a[2] and push to global array
             ¾∑←a4i⌊+,  # push global array, sum the values, and add a[4] 

            |«¢q⁰∷Ṗ-←¥«, # If any of the checks are false, print "invalid input"
```

Answer 12

Zsh, 205 223 bytes

set -e;e=Invalid\ input;trap '<<<$e' ERR
echo "^0d\n+$\nd+\n+d\n+.+\n[-|*]">p;[[ `grep -cfp<<<$1`>0 ]]&&<<<$e&&exit
q=(${(s: :)${1//(#s)d/1d}//[^0-9]/ });for i ({1..$q[1]})((s+=RANDOM%q[2]+1))
<<<$[s+q[3]]

Try it Online (with all test cases)

Writing the validation code was a chore. Without this guff, the solution would be a svelte 89 bytes:

q=(${(s: :)${1//(#s)d/1d}//[^0-9]/ });for i ({1..$q[1]})((s+=RANDOM%q[2]+1))
<<<$[s+q[3]]

Answer 13

Python3, 204B

Mine beats the existing Python answer by adding in the required error handling and reading d20 as 1d20 rather than 0d20 :)

import random,re
try:a,b,c=re.findall('^([1-9]\d*)?d(\d+)(\+\d+)?$',input())[0];I=int;R=sum(random.randrange(I(b))+1for x in[0]*(1if a==''else I(a)))+(0if c==''else I(c))
except:R='Invalid input'
print(R)

edited to fix 2 typos: I(x) => I(c), Invalid Input => Invalid input

edited to fix regex: \+?(\d*) => (\+\d+)?

Answer 14

Go, unseeded, 366 bytes

import(."regexp";."strconv";."math/rand")
func f(s string)int{k:=MustCompile(`^(\d*)d(\d+)(\+\d+)?$`).FindAllStringSubmatch(s,-1)
if len(k)<1||k[0][2]==""{return-1}
r:=k[0]
var n,l,f,m,t int
l=len(r)
n,_=Atoi(r[1])
if n<1&&r[1]==""{n=1}else if n<1{return-1}
if l>2{f,_=Atoi(r[2])}
if f<1{return-1}
if l>3{m,_=Atoi(r[3])}
for i:=0;i<n;i++{t+=Intn(f)+1}
t+=m
return t}

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Go, seeded, 393 bytes

import(."regexp";."strconv";."math/rand";."time")
func f(s string)int{k:=MustCompile(`^(\d*)d(\d+)(\+\d+)?$`).FindAllStringSubmatch(s,-1)
if len(k)<1||k[0][2]==""{return-1}
r:=k[0]
var n,l,f,m,t int
l=len(r)
n,_=Atoi(r[1])
if n<1&&r[1]==""{n=1}else if n<1{return-1}
if l>2{f,_=Atoi(r[2])}
if f<1{return-1}
if l>3{m,_=Atoi(r[3])}
Seed(Now().Unix())
for i:=0;i<n;i++{t+=Intn(f)+1}
t+=m
return t}

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