## Challenge

Given a set of numbers with operators between them: "5 + 4 * 9 / 3 - 8", return all the possible results of the expression for every permutation of the order of basic operations: [/, *, +, -].

## Rules

• Standard loopholes forbidden
• I/O
• Input must be ordered with infix operations, but however that is easiest (string or array)
• You are not required to support unary operators (e.g. "-3 * 8 / +2")
• Integers can be replaced by floats for languages that implicitly parse type (e.g. 45 ⟶ 45.0)
• Output must be all of the possible results of the expression, no specified format or order
• All of the inputs are valid (e.g. do not need to deal with "7 / 3 + *"). This also means that you will never need to divide by zero.
• Operators are all left-associative so "20 / 4 / 2" = "(20 / 4) / 2"
• This is Code Golf so fewest number of bytes wins

## Test Cases (With explanation)

• "2 + 3 * 4" = [14, 20]
• 2 + (3 * 4) ⟶ 2 + (12) ⟶ 14
• (2 + 3) * 4 ⟶ (5) * 4 ⟶ 20
• "18 / 3 * 2 - 1" = [11, 2, 6]
• ((18 / 3) * 2) - 1 ⟶ ((6) * 2) - 1 ⟶ (12) - 1 ⟶ 11
• (18 / 3) * (2 - 1) ⟶ (6) * (1) ⟶ 6
• (18 / (3 * 2)) - 1 ⟶ (18 / (6)) - 1 ⟶ (3) - 1 ⟶ 2
• 18 / (3 * (2 - 1)) ⟶ 18 / (3 * (1)) ⟶ 6
• 18 / ((3 * 2) - 1) ⟶ 18 / 5 ⟶ 3.6

## Test Cases (Without explanation)

• "45 / 8 + 19 / 45 * 3" = [6.891666666666667, 18.141666666666666, 0.11111111111111113, 0.01234567901234568, 0.01234567901234568, 5.765740740740741]
• "2 + 6 * 7 * 2 + 6 / 4" = [112 196 23 87.5]
• Nice first challenge, though, by the way. – Shaggy Sep 8 '19 at 22:31
• Closely related – Arnauld Sep 9 '19 at 16:12
• Suggested test case 2 - 3 + 4 => [-5, 3] – Jo King Sep 10 '19 at 2:26
• Suggested test case: 2*3-6+2-9/6*8+5/2-9, giving 24 distinct results. – Arnauld Sep 10 '19 at 9:56

# JavaScript (V8),  118  112 bytes

Prints the results.

f=(s,x='',y=x,o='+-*/')=>[...o].map(v=>f(s.split(v).join(x+v+y),x+')',y+'(',o.replace(v,'')))|print(eval(y+s+x))


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# C# (Visual C# Interactive Compiler), 285 bytes

x=>{int c=0,j,t=1,i;for(;c++<25;t=c){var r="*+-/".ToList();for(i=j=1;j++<4;t=t/j+1)(r[j-1],r[t%j])=(r[t%j],r[j-1]);float k(float z,int p=4){char d;int l;float m;return i<x.Count&&(l=r.IndexOf(d=x[i]))<p?k((m=k(x[(i+=2)-1],l))*0+d<43?z*m:d<44?z+m:d<46?z-m:z/m,p):z;}Print(k(x));}}


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x=>{                                          //Lambda taking in a List<dynamic>
int c=0,j,t=1,i;                            //A bunch of delcarations jammed together to save bytes
for(;c++<25;t=c){                           //Loop 24 times (amount of permutations a set of length 4 can have)
var r="/+*-".ToList();                    //Initialize r as list of operators
for(i=j=1;j++<4;t=t/j+1)                    //Create the Tth permutation, saving result in r, also reset i to 1
(r[j-1],r[t%j])=(r[t%j],r[j-1]);
float k(float z,int p=4) {                //Define local function 'k', with z as current value accumalated and p as current precedence
char d;int l;float m;                   //Some helper variables
return i<x.Count                        //If this is not the last number
&&(l=r.IndexOf(d=x[i]))<p?         //  And the current operator's precedence is higher than the current precedence
k(                                      //  Recursive call with the accumalative value as
(m=k(x[(i+=2)-1],l))                  //    Another recursive call with the next number following the current operator as seed value,
//    And the next operator's precedence as the precedence value, and store that in variable 'm'
*0+d<43?z*m:d<44?z+m:d<46?z-m:z/m,    //    And doing the appropriate operation to m and current value ('z')
p)                                    //  Passing in the current precedence
:z;                                       //Else just return the current number
}
Print(k(x));                           //Print the result of calling k with the first number as starting value
}
}

• I have fixed it so you do not need to omit duplicates as it is not a fundamental part of the problem as pointed out. – Freddie R Sep 9 '19 at 13:05
• @Arnauld Fixed at the cost of 4 bytes, it was because my permutations algorithm was a bit wrong – Embodiment of Ignorance Sep 11 '19 at 4:56

a=>(w=[],F=(b,a)=>b?[...b].map(q=>F(b.replace(q,""),a.replace(eval(/[\\d.-]+( \${q} [\\d.-]+)+/g),eval))):w.push(a))("+-*/",a)&&w  Try it online! This allows duplicated outputs. # JavaScript (Node.js), 165161155153152 137 bytes a=>Object.keys((F=(b,a)=>b?[...b].map(q=>F(b.replace(q,""),a.replace(eval(/[\\d.-]+( \${q} [\\d.-]+)+/g),eval))):F[a]=1)("+-*/",a)&&F)


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Takes a string with spaces between operators and numbers.

a=>                                             // Main function:
Object.keys(                                   //  Return the keys of the -
(
F=(                                          //   Index container (helper function):
b,                                          //    Operators
a                                           //    The expression
)=>
b                                           //    If there are operators left:
?[...b].map(                                //     For each operator:
q=>
F(                                        //      Recur the helper function -
b.replace(q,""),                         //       With the operator deleted
a.replace(                               //       And all -
eval(/[\\d.-]+( \\${q} [\\d.-]+)+/g), // Expressions using the operator eval // Replaced with the evaluated result ) ) ) :F[a]=1 // Otherwise - set the result flag. )( "+-*/", // Starting with the four operators a // And the expression ) &&F )  • @JoKing Implemented the fix I've stated before, should output [3, -5] now. – Shieru Asakoto Sep 10 '19 at 3:42 # Perl 6, 92 90 88 bytes {map {[o](@_)($_)},<* / + ->>>.&{$^a;&{S:g{[\-?<[\d.]>+]+%"$a "}=$/.EVAL}}.permutations}  Try it online! Takes a string with a space after any operators and returns a set of numbers. This mostly works by substituting all instances of n op n with the eval'ed result for all permutations of the operators. ### Explanation: { } # Anonymous code block <* / + ->>>.&{ } # Map the operators to:$^a;&{                             }  # Functions that:
S:g{                }      # Substitute all matches of:
\-?<[\d.]>+]+        # Numbers
%$a # Joined by the operator =$/.EVAL   # With the match EVAL'd
map {           },                                                    .permutations   # Map each of the permutations of these operators
[o](@_)        # Join the functions
($_) # And apply it to the input  • You can remove the set, as the condition to eliminate duplicates has been removed. Nice code. – Freddie R Sep 9 '19 at 13:12 # Python 3, 108 bytes f=lambda e,s={*"+-*/"}:[str(eval(p.join(g)))for p in s for g in zip(*map(f,e.split(p),[s-{p}]*len(e)))]or[e]  Try it online! The function takes a single string as input and returns a list of possible results. # Ungolfed def get_all_eval_results(expr, operators={*"+-*/"}): results = [] for operator in operators: remaining_operators = operators - {operator} # Split expression with the current operator and recursively evaluate each subexpression with remaining operators sub_expr_results = (get_all_eval_results(sub_expr, remaining_operators) for sub_expr in expr.split(operator)) for result_group in zip(*sub_expr_results): # Iterate over each group of subexpression evaluation outcomes expr_to_eval = operator.join(result_group) # Join subexpression outcomes with current operator results.append(str(eval(expr_to_eval))) # Evaluate and append outcome to result list of expr return results or [expr] # If results is empty (no operators), return [expr]  Try it online! # Jelly, 30 bytes œṡ⁹¹jṪḢƭ€jŒVɗßʋFL’$?
Ḋm2QŒ!çƒ€


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A pair of links. The second one is the main link, and takes as its argument a Jelly list of floats/integers interspersed with the operators as characters. This is a flattened version of the way Jelly takes its input when run as a full program with command line arguments. The return value of the link is a list of list of single member lists, each of which is a possible value for the expression.

## Explanation

Takes a list of floats/integers alternating with operators (as characters) as its left argument and an operator as a character as its right argument; returns the input list after evaluating numbers separated by the relevant operator, working left to right.

œṡ⁹                  | Split once by the right argument (the operator currently being processed)
? | If:
$| - Following as a monad L | - Length ’ | - Decremented by 1 ʋ | Then, following as a dyad: ¹ | - Identity function (used because of Jelly’s ordering of dyadic links at the start of a dyadic chain) j ɗ | - Join with the following as a dyad, using the original left and right arguments for this chain: ṪḢƭ€ | - Tail of first item (popping from list) and head from second item (again popping from list); extracts the numbers that were either side of the operator, while removing them from the split list j | - Joined with the operator ŒV | - Evaluate as Python (rather than V because of Jelly’s handling of decimals with a leading zero) ß | - Recursive call to this helper link (in case there are further of the same operator) F | Else: Flatten  ### Main link Takes a list of floats/integers alternating with operators (as characters) Ḋ | Remove first item (which will be a number) m2 | Every 2nd item, starting with the first (i.e. the operators) Q | Uniquify Œ! | Permutations çƒ€ | For each permuted list of operators, reduce using the helper link and with the input list as the starting point  # Python 2, 182 172 bytes import re def f(s,P=set('+-/*')): S=[eval(s)] for p in P: t=s while p+' 'in t:t=re.sub(r'[-\d.]+ \%s [-\d.]+'%p,lambda m:eval(m.group()),t,1) S+=f(t,P-{p}) return S  Try it online! Takes input with ints formatted as floats, as per "Integers can be replaced by floats for languages that implicitly parse type". # Julia 1.2, 88 (82) bytes f(t)=get(t,(),[f.([t[1:i-1];t[i+1](t[i],t[i+2]);t[i+3:end]] for i=1:2:length(t)-2)...;])  julia> f([2, +, 3, *, 4]) 2-element Array{Int64,1}: 20 14 julia> f([18, /, 3, *, 2, -, 1]) 6-element Array{Float64,1}: 11.0 6.0 2.0 3.6 6.0 6.0  Takes a tape in the form of a vector of numbers and infix functions, evaluates each single function call and recursively passes each resulting tape back to itself until only a single number is left. Unfortunately, get(t, (), ...) doesn't work properly in Julia 1.0, so a newer version is needed. Six bytes can be saved, if a bunch of nested arrays is acceptable as an output: f(t)=get(t,(),f.([t[1:i-1];t[i+1](t[i],t[i+2]);t[i+3:end]] for i=1:2:length(t)-2))  Output: julia> f([18, /, 3, *, 2, -, 1]) 3-element Array{Array{Array{Float64,1},1},1}: [[11.0], [6.0]] [[2.0], [3.6]] [[6.0], [6.0]]  # Perl 5 (-alp), 89 bytes my$x;map{$x.=$.(eval$&.$1).$2.$"while/\d+[-+*\/](?=(\d+)(.*))/g}@F;$_=$x;/[-+*\/]/&&redo


TIO

or unique values, 99 bytes

my%H;map{$H{$.(eval$&.$1).$2}++while/\d+[-+*\/](?=(\d+)(.*))/g}@F;$_=join\$",keys%H;/[-+*\/]/&&redo