# Introduction

The four basic math operators (+, -, *, /) can be reduced to just two, due to the fact that:

x + y = x - (-y)
x * y = x / (1/y), y != 0
x * 0 = 0/x


# Challenge

The challenge is to take input as a "string" containing:

• Numbers
• Single character variables ("x", "y")
• The four basic math operators (+, -, *, /)
• Parenthesis

and output a string manipulated so that it would produce the same mathematical result as the input, but containing only the mathematical symbols '-' and '/'

# Specifics

• Input can be in any acceptable form (file, STDIN, etc.) and may be represented as a string or character array (but not an array of arrays)
• Output can be in any acceptable form (file, STDIN, etc.) and may be represented as a string or character array (but not an array of arrays)
• You must recognize and maintain balanced parenthesis
• Standard loopholes are disallowed
• It is your choice if you want to represent x + y as x - -y or x - (-y)
• You must maintain the order of operations
• You never have to handle invalid input
• Input can be empty or a single number/variable, in that case the program should output the input
• Note: You do not have to use the substitutions in the introduction, so long as input = output, your program could change 2 * 2 to 8/2, if you wanted
• You can assume that "0" is the only way a zero will appear in the equation (I.e. you don't have to handle 1 * (4 - 4))
• Suggestion: to test your program, go to this website type in input = output, where input is the input, and output is the output, and if the result is "true" your program handled that case successfully (example, example)

# Test Cases

Below are some test cases, input as a single string and output as a single string.

x + y
x - (-y)

x * y
x / (1/y)

x / y
x / y

x - y
x - y

1
1

5
5

-6
-6

+x
x

1 + (x * 4) - (512 * 3)
1 - (-(x / (1/4))) - (512 / (1/3))

1 - 3 / 4 + l / g
1 - 3/4 - (-(l / g))

5 * 0 / 2
0/5 / 2

(a + g) * 0
0/(a - (-g))


# Scoring

It's , so shortest answer in bytes wins. Ties are resolved by first-post.

• By the way x / 1/y = x/y because division isn't associative. I know what you're thinking, but even WolframAlpha doesn't recognize that you want spaces to change the order of operations.... so you probably should rethink this or not cite that as a valid way to check things. – Linus Sep 4 '16 at 20:32
• @Linus: It also isn't equivalent when y=0, but I'm guessing the challenge implicitly assumes that n/d => d != 0. – Tim Čas Sep 4 '16 at 20:59
• @TimČas ahh! Didn't think of that. I'll update the challenge, just know it will have to be handled properly – Socratic Phoenix Sep 4 '16 at 21:44
• As it stands, there's very little stopping us from simply evaluating the expression and returning the result (they're mathematically equal, after all). I'd recommend changing that, unless you want v to be a proper solution in Pyth. – Steven H. Sep 6 '16 at 3:29
• Wait, so if there are variables (like x and y), how can we divide without risking division by zero? 5 * (a - b) if a=b. And do we have to detect things like 5 * (a - a)? How about 5 * (4 - 4) and 5 * (a / a - 1) or 5 * (4 / 4 - 1)? – Adám Sep 6 '16 at 7:29

# Python 3, 267 bytes

Thanks to @ConorO'Brien

import re
q=re.sub
g=lambda m:'--'+m.group()[1:]
h=lambda m:'/(1/'+m.group()[1:]+')'
i=lambda m:'0/'+m.group()[:-2]
print(q(r'\*[^]+',h,q(r'[^]\*0',i,q(r'\+[^]+',g,q(r'\*$$[^$$]+\)',h,q(r'$$[^$$]+\)\*0',i,q(r'\+$$[^$$]+\)',g,input().replace(' ',''))))))))


Ideone it!

• Honestly, I've no idea what dark magic you've employed, but +1 for FGITW – Socratic Phoenix Sep 5 '16 at 19:51
• @SocraticPhoenix Haha the dark magic is called regex ;) – Beta Decay Sep 5 '16 at 19:54

# Dyalog APL, 42 bytes

This maintains APLs order of operations. Note that ÷x is 1÷x

'\+' '×'⎕R'--' '÷÷'('(.*)×0'⎕R'0÷\1'~∘' ')


TryAPL online!

( on the result of...

~∘' ' remove spaces

'(.*)×0'⎕R'0÷\1' replace anything followed by "×0" with "0÷" followed by it

) evaluate...

'\+' '×'⎕R'--' '÷÷' replace "+" with "--" and "×" with "÷÷"

# SED 272 246 239 213

s,^\+,,;s,[^+*/()-]\+,(&),g;t;:;s,)[^)]*)\*(0,&,;tr;s,$$(.*)$$\*(0),(0/\1),;ty;s,\*([^)]*(,&,;tr;s,\*$$([^)]*)$$,/(1/\1),;ty;s,\+([^)]*(,&,;tr;s,\+$$([^)]*)$$,-(-\1),;ty;p;q;:r;s,($$[^()]*$$),!\1@,;t;:y;y,!@,(),;b


Take input without spaces (e.g. x+y*2).
This is one of the few cases where it is actually shorter to just escape ( and ) for capture groups instead of using -r. I'm sure this can be golfed more, but I'm happy for now.

s,^\+,, #remove leading +
s,[^+*/()-]\+,(&),g #suround numbers/variables in ()
t start #reset the test because the line above will always match

:start
#------------- deal with *0 ------------------
s,)[^)]*)\*(0,&, #remove inner matching ()
t replace
s,$$(.*)$$\*(0),(0/\1),
t y
#------------- deal with generic * -----------
s,\*([^)]*(,&, #remove inner matching ()
t replace
s,\*$$([^)]*)$$,/(1/\1),
t y
#------------- deal with + -------------------
s,\+([^)]*(,&, #remove inner matching ()
t replace
s,\+$$([^)]*)$$,-(-\1),
t y

b end #all done, branch to the end

#------------- replace a set of () with !@ ---
#repeated application of this helps find the matching ( or )
:replace
s,($$[^()]*$$),!\1@,
t start

:y
y,!@,(),
b start

:end