Code Golf: Mathematical expression evaluator (that respects PEMDAS)

Answer 1

I know, I know..this code-golf seems to be closed. Still, I felt the urge to code this stuff in erlang __, so here is an erlang version (didn't found the will to golf-format it, so these are 58 lines, about 1400 chars)

-module (math_eval).
-export ([eval/1]).
eval( Str ) ->
  ev(number, Str,[]).
ev( _, [], Stack ) -> [Num] = do(Stack), Num;
ev( State, [$ |Str], Stack ) ->
  ev( State,Str,Stack );
ev( number, [$(|Str], Stack ) ->
  ev( number,Str,[$(|Stack] );
ev( number, Str, Stack ) ->
  {Num,Str1} = r(Str),
  ev( operator,Str1,[Num|Stack] );
ev( operator, [$)|Str], Stack) ->
  ev( operator, Str, do(Stack) );
ev( operator, [Op2|Str], [N2,Op,N1|T]=Stack ) when is_float(N1) andalso is_float(N2) ->
  case p(Op2,Op) of
    true -> ev( number, Str, [Op2|Stack]);
    false -> ev( operator, [Op2|Str], [c(Op,N1,N2)|T] )
  end;
ev( operator, [Op|Str], Stack ) ->
  ev( number,Str,[Op|Stack] ).
do(Stack) ->
  do(Stack,0).
do([],V) -> [V];
  do([$(|Stack],V) -> [V|Stack];
do([N2,Op,N1|Stack],0) ->
  do(Stack,c(Op,N1,N2));
do([Op,N1|Stack],V) ->
  do(Stack,c(Op,N1,V)).
p(O1,O2) -> op(O1) < op(O2).
op(O) ->
  case O of
    $) -> 0; $( -> 0;
    $^ -> 1;
    $* -> 2; $/ -> 2;
    $+ -> 3; $- -> 3;
    $  -> 4; _ -> -1
  end.
r(L) ->
  r(L,[]).
r([], Out) ->
  {f( lists:reverse(Out) ),[]};
r([$-|R],[]) ->
  r(R,[$-]);
r([C|T]=R,O) ->
  if (C =< $9 andalso C >= $0) orelse C =:= $. -> r(T,[C|O]);
    true -> {f(lists:reverse(O)),R}
  end.
f(L) ->
  case lists:any(fun(C) -> C =:= $. end,L) of
    true -> list_to_float(L);
    false -> list_to_float(L++".0")
  end.
c($+,A,B) -> A+B;
c($-,A,B) -> A-B;
c($*,A,B) -> A*B;
c($/,A,B) -> A/B;
c($^,A,B) -> math:pow(A,B).

Answer 2

F#, 589 chars

I golf-compressed my prior solution into this gem:

let rec D a=function|c::s when System.Char.IsDigit c->D(c::a)s|s->a,s
and L p o s=
 let rec K(a,s)=match o s with|None->a,s|Some(o,t)->let q,t=p t in K(o a q,t)
 K(p s)
and E=L(L F (function|'*'::s->Some((*),s)|'/'::s->Some((/),s)|_->None))(
function|'+'::s->Some((+),s)|'-'::s->Some((-),s)|_->None)
and F s=match P s with|x,'^'::s->let y,s=F s in x**y,s|r->r
and P=function|'('::s->let r,_::s=E s in r,s|s->(
let a,s=match(match s with|'-'::t->D['-']t|_->D[]s)with|a,'.'::t->D('.'::a)t|r->r
float(new string(Seq.to_array(List.rev a))),s)
and G s=string(fst(E(Seq.to_list s)))

Tests:

printfn "%s" (G "1.1+2.2+10^2^3") // 100000003.3
printfn "%s" (G "10+3.14/2")      // 11.57
printfn "%s" (G "(10+3.14)/2")    // 6.57
printfn "%s" (G "-1^(-3*4/-6)")   // 1
printfn "%s" (G "-2^(2^(4-1))")   // 256 
printfn "%s" (G "2*6/4^2*4/3")    // 1
printfn "%s" (G "3-2-1")          // 0

Answer 3

I wrote an attp at http://www.sumtree.com as an educational tool for teachers that does this.

Used bison for the parsing and wxwidgets for the GUI.

Answer 4

F#, 70 lines

Ok, I implement a monadic parser combinator library, and then use that library to solve this problem. All told it's still just 70 lines of readable code.

I assume exponentiation associates to the right, and the other operators associate to the left. Everything works on floats (System.Doubles). I did not do any error handling for bad inputs or divide-by-zero.

// Core Parser Library
open System
let Fail() = fun i -> None
type ParseMonad() =
    member p.Return x = fun i -> Some(x,i)
    member p.Bind(m,f) = fun i -> 
        match m i with
        | Some(x,i2) -> f x i2
        | None -> None
let parse = ParseMonad()
let (<|>) p1 p2 = fun i -> 
    match p1 i with
    | Some r -> Some(r)
    | None -> p2 i
let Sat pred = fun i -> 
    match i with
    | [] -> None
    | c::cs -> if pred c then Some(c, cs) else None
// Auxiliary Parser Library
let Digit = Sat Char.IsDigit
let Lit (c : char) r = 
    parse { let! _ = Sat ((=) c)
            return r }
let Opt p = p <|> parse { return [] }
let rec Many p = Opt (Many1 p)
and Many1 p = parse { let! x = p
                      let! xs = Many p
                      return x :: xs }
let Num = parse {
    let! sign = Opt(Lit '-' ['-'])
    let! beforeDec = Many Digit
    let! rest = parse { let! dec = Lit '.' '.'
                        let! afterDec = Many Digit
                        return dec :: afterDec } |> Opt
    let s = new string(List.concat([sign;beforeDec;rest])
                       |> List.to_array) 
    match(try Some(float s) with e -> None)with
    | Some(r) -> return r
    | None -> return! Fail() }
let Chainl1 p op = 
    let rec Help x = parse { let! f = op
                             let! y = p
                             return! Help (f x y) } 
                     <|> parse { return x }
    parse { let! x = p
            return! Help x }
let rec Chainr1 p op =
    parse { let! x = p
            return! parse { let! f = op
                            let! y = Chainr1 p op
                            return f x y }
                    <|> parse { return x } }
// Expression grammar of this code-golf question
let AddOp = Lit '+' (fun x y -> 0. + x + y) 
        <|> Lit '-' (fun x y -> 0. + x - y)
let MulOp = Lit '*' (fun x y -> 0. + x * y) 
        <|> Lit '/' (fun x y -> 0. + x / y)
let ExpOp = Lit '^' (fun x y -> Math.Pow(x,y))
let rec Expr = Chainl1 Term AddOp
and Term = Chainl1 Factor MulOp
and Factor = Chainr1 Part ExpOp
and Part = Num <|> Paren
and Paren = parse { do! Lit '(' ()
                    let! e = Expr
                    do! Lit ')' ()
                    return e }
let CodeGolf (s:string) =
    match Expr(Seq.to_list(s.ToCharArray())) with
    | None -> "bad input"
    | Some(r,_) -> r.ToString()
// Examples
printfn "%s" (CodeGolf "1.1+2.2+10^2^3") // 100000003.3
printfn "%s" (CodeGolf "10+3.14/2")      // 11.57
printfn "%s" (CodeGolf "(10+3.14)/2")    // 6.57
printfn "%s" (CodeGolf "-1^(-3*4/-6)")   // 1
printfn "%s" (CodeGolf "-2^(2^(4-1))")   // 256 
printfn "%s" (CodeGolf "2*6/4^2*4/3")    // 1

The parser representation type is

type P<'a> = char list -> option<'a * char list>

by the way, a common one for non-error-handling parsers.

Answer 5

C, 609 characters

(625 including formatted as below to avoid horizontal scrolling, 42 lines if I make it readable.)

double x(char*e,int*p);
D(char c){return c>=48&&c<=57;}
S(char c){return c==43||c==45;}
double h(char*e,int*p){double r=0,s=1,f=0,m=1;int P=*p;if(e[P]==40){
 P++;r=x(e,&P);P++;}else if(D(e[P])||S(e[P])){s=S(e[P])?44-e[P++]:s;
 while(D(e[P]))r=r*10+e[P++]-48;if(e[P]==46)while(D(e[++P])){f=f*10+e[P]-48;
 m*=10;}r=s*(r+f/m);}*p=P;return r;}
double x(char*e,int*p){double r=0,t,d,x,s=1;do{char o=42;t=1;do{d=h(e,p);
 while(e[*p]==94){(*p)++;x=h(e,p);d=pow(d,x);}t=o==42?t*d:t/d;o=e[*p];
 if(o==42||o==47)(*p)++;else o=0;}while(o);r+=s*t;s=S(e[*p])?44-e[(*p)++]:0;
}while(s);return r;}
double X(char*e){int p=0;return x(e,&p);}

It's my first code golf.

I'm parsing floats myself and the only library function I use is pow.

I corrected errors with multiple elevations to a power and handling of parentheses. I also made the main function X() that takes just a string as argument. It still returns a double, though.

Expanded

42 non-blank lines

double x(char*e, int*p);

D(char c) { return c>=48 && c<=57; }
S(char c) { return c==43 || c==45; }

double h(char*e, int*p) {
    double r=0, s=1, f=0, m=1;
    int P=*p;
    if(e[P]==40) {
        P++;
        r=x(e, &P);
        P++; }
    else if(D(e[P]) || S(e[P])) {
        s=S(e[P]) ? 44-e[P++] : s;
        while(D(e[P]))
            r=r*10+e[P++]-48;
        if(e[P]==46)
            while(D(e[++P])) {
                f=f*10+e[P]-48;
                m*=10; }
        r=s*(r+f/m); }
        *p=P;
    return r; }

double x(char*e, int*p) {
    double r=0, t, d, x, s=1;
    do {
        char o=42;
        t=1;
        do {
            d=h(e, p);
            while(e[*p]==94) {
                (*p)++;
                x=h(e, p);
                d=pow(d, x); }
            t=o==42 ? t*d : t/d;
            o=e[*p];
            if(o==42 || o==47) (*p)++;
            else o=0;
        } while(o);
        r+=s*t;
        s=S(e[*p]) ? 44-e[(*p)++] : 0;
    } while(s);
    return r; }

double X(char*e) {int p=0; return x(e, &p);}

Answer 6

J

:[[/%^(:[[+-/^,&i|:[$[' ']^j+0__:k<3:]]