# The Challenge

You are the owner of an amazing service called Coyote Beta, which magically answers math questions its users send to it over the internet.

But it turns out, bandwidth is expensive. You have two choices, either create a "Coyote Beta Pro" or find some way to solve this. Just recently, someone queried (x + 2). Couldn't the client send x+2, and the user would see no difference?

Your task is to "minify" math expressions. Given an input expression, you must get rid of whitespace and parentheses until it gives a minimal representation of the same input. The parentheses around associative operations need not be preserved.

The only operators given here are +, -, *, /, and ^ (exponentiation), with standard mathematical associativity and precedence. The only whitespace given in the input will be actual space characters.

# Sample Input/Output

Input       | Output
------------|--------------
(2+x) + 3   | 2+x+3
((4+5))*x   | (4+5)*x
z^(x+42)    | z^(x+42)
x - ((y)+2) | x-(y+2)
(z - y) - x | z-y-x
x^(y^2)     | x^y^2
x^2 / z     | x^2/z
- (x + 5)+3 | -(x+5)+3


# Scoring

Input/output can use any preferred method. The smallest program in bytes wins.

Exact bits

Exponentiation is right associative and also follows standard math precedence (being the highest). A valid numeric literal is /[0-9]+/, and a valid variable literal is /[a-z]+/. A single variable literal represents a single value even when its character length is longer than 1.

What is meant by "the parentheses around associative operations need not be preserved" is that the output should consist of an expression that results in an identical parse tree, with the exception that associative operations can be rearranged.

• The idea is to create a minimal equivalent statement that results in the same parse tree. This is so that Coyote Beta can display it visually when the user makes a query.
– TND
Oct 17, 2015 at 17:55
• If a valid variable is /[a-z]+/, that means multiplication by juxtaposition like ab is disallowed? Oct 17, 2015 at 18:01
• You do want 2+(3+4) to be changed to 2+3+4, right? This does change the parse tree. Oct 17, 2015 at 18:26
• I take issue with the claim that x^(y/2)=x^y/2; exponentiation has a higher order precedence, ergo, x^y/2=(x^y)/2. Oct 17, 2015 at 18:50
• Aww man, I was going to submit Prompt X:expr(X) in TI-BASIC but you can't simplify :( Oct 17, 2015 at 23:49

# C#, 523519 504 bytes

Check the in-code comments to see how it works!

## Golfed

using System;using System.Collections.Generic;namespace n{class p{static void Main(string[]a){foreach(String s in a){String r=s.Replace(" ","");List<int>l=new List<int>();for(int i=0;i<r.Length;i++){if(r[i]=='('){l.Add(i);continue;}if(r[i]==')'){switch(r[Math.Max(l[l.Count-1]-1,0)]){case'+':case'(':switch(r[Math.Min(i+1,r.Length-1)]){case'+':case'-':case')':r=r.Remove(Math.Max(l[l.Count-1],0),1);r=r.Remove(Math.Min(i,r.Length)-1,1);i-=2;break;}break;}l.RemoveAt(l.Count-1);}}Console.WriteLine(r);}}}}


## Ungolfed

using System;
using System.Collections.Generic;

namespace n {
class p {
static void Main( string[] a ) {
// Loop every String given for the program
foreach (String s in a) {
// Get rid of the spaces
String r = s.Replace( " ", "" );

// A little helper that will have the indexes of the '('
List<int> l = new List<int>();

// Begin the optimizatio process
for (int i = 0; i < r.Length; i++) {
// If char is an '(', add the index to the helper list and continue
if (r[ i ] == '(') {
continue;
}

// If the char is an ')', validate the group
if (r[ i ] == ')') {
// If the char before the last '(' is an '+' or '(' ...
switch (r[ Math.Max( l[ l.Count - 1 ] - 1, 0 ) ]) {
case '+':
case '(':
// ... and the char after the ')' we're checking now is an '+', '-' or ')' ...
switch (r[ Math.Min( i + 1, r.Length - 1 ) ]) {
case '+':
case '-':
case ')':
// Remove the '()' since they're most likely desnecessary.
r = r.Remove( Math.Max( l[ l.Count - 1 ], 0 ), 1 );
r = r.Remove( Math.Min( i, r.Length ) - 1, 1 );

// Go two steps back in the loop since we removed 2 chars from the String,
//   otherwise we would miss some invalid inputs
i -= 2;
break;
}

break;
}

// Remove the last inserted index of '(' from the list,
//   since we matched an ')' for it.
l.RemoveAt( l.Count - 1 );
}
}

// Print the result
Console.WriteLine( r );
}
}
}
}


## Side notes

1. Fixed some typos and renamed some vars.
2. Nested an switch to get rid of an unnecessary variable. Also, fixed a bug that would render some solutions invalid, reported by Anders Kaseorg.

P.S.: If you have a tip or found a bug, please let me know in the comments and I'll try to fix it ( I'll then add a note about the bug fix with your name ;) )

• Nice answer! :D substantial answers here are generally better recieved if you include an explanation :P
– cat
Apr 15, 2016 at 14:04
• Can I do it in the form of code comments? Apr 15, 2016 at 14:24
• Sure, whatever works c:
– cat
Apr 15, 2016 at 14:25
• Then I'll do that! I'll also try to add a summary somewhere. Apr 15, 2016 at 14:28
• Welcome to Programming Puzzles and Code Golf, by the way! (even though it's not your first answer)
– cat
Apr 15, 2016 at 14:30

## C++, 284 bytes

Golfed

#include<iostream>
#include<algorithm>
int main(){std::string e;std::getline(std::cin,e);e.erase(std::remove_if(e.begin(),e.end(),isspace),e.end());for(int x=0;x<e.length();x++){if(e[x]=='('&&e[x+1]=='('){e.erase(x,1);}if(e[x]==')'&&e[x+1]==')'){e.erase(x,1);}}std::cout<<e;return 0;}


Ungolfed

#include<iostream>
#include<algorithm>

int main()
{
std::string e;
std::getline(std::cin, e);
e.erase(std::remove_if(e.begin(), e.end(), isspace), e.end());
for(int x = 0; x < e.length(); x++) {
if (e[x] == '(' && e[x+1] == '('){
e.erase(x, 1);
}
if (e[x] == ')' && e[x+1] == ')'){
e.erase(x, 1);
}
}
std::cout<<e;
return 0;
}

• This doesn’t have any precedence logic and fails many of the given test cases. Apr 15, 2016 at 18:01