Help! My calculator broke! (Turn integer expression into calculator keystrokes)

Question

Introduction

Help! I accidentally dropped my TI-84 calculator out my window (don't ask how) and it broke. I have a math test tomorrow and the only calculator I can find is one with these buttons:

7 8 9 +
4 5 6 -
1 2 3 *
0   = /

My math test is a review test on evaluating expressions. I need a program to take an expression such as 1+(5*4)/7 and convert it to the keystrokes needed to solve it on my spare calculator. (And in case you were wondering, this actually happened to me).

Challenge

Given a non-empty input string containing only the characters 0-9, (, ), +, -, *, and /, output the keystrokes in a space-separated string (eg. 1 + 3 / 3 =). There must always be an equal sign at the end of the output. Standard loopholes are not allowed.

Examples:

• Input: 1+(5*4)/7, Output: 5 * 4 / 7 + 1 =
• Input: 6*(2/3), Output: 2 / 3 * 6 =
• Input: (7-3)/2, Output: 7 - 3 / 2 =

To make this challenge easier:

• You may assume that the input has a series of keystrokes linked to it that does not require clearing the calculator (1-(7*3) is not valid since it would require you to find 7 * 3, then clear the calculator to do 1 - 21. All the above examples are valid since there is one, continuous output that does not require the user to clear the calculator and remember a number).
• You may assume that there will only be a single integer after a /, as having an input such as 21/(7*3) would not pass the first assumption either.
• You may assume that there will always be a * between an integer and a left parentheses (Valid: 6*(7), Invalid: 6(7)).
• You may assume the input always produces integer output.
• You may assume the input only has three levels of parentheses.

Non-examples

• 2-(14/2) as you would have to do 14 / 2, then clear, then 2 - 7.
• 36/(2*3) as you would have to do 2 * 3, then clear, then 36 / 6.
• 1024*4/(1*2+2) as you would have to do 1*2+2, then clear, then 1024 * 4 / 4.

Bonuses

• -5% if your program can recognize parentheses multiplication (it knows that 6(7)=6*(7)).
• -5% if your program can handle input with decimal numbers (3.4, 2.75, 7.8) and the output includes . (as there must be a . key on my spare calculator in this case).
• -5% if your program can handle unlimited levels of parentheses.

This is code-golf, shortest code in bytes (including the bonuses) wins!

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Answer 1

TI-BASIC, 605.2 bytes

Eligible for lirtosiast's TI-BASIC bounty under Fitting but Unsuitable Languages.

Qualifies for all 3 bonuses, 712 - 15% = 605.2. There are some golfing opportunities here and there, but I wanted to get this out first, as some of the potential golfs are non-trivial. Please note that TI-BASIC is a tokenized language, and the below is a textual representation of that program. Thus, the program is not 1182 bytes, since this program is not encoded under UTF-8. Note that ~ is equivalent to unary negation and -> to the STO> operator. Output is a string, retrievable from Ans or Str1.

The below program is the result of a few programmer hours thinking and programming, spread over the course of a few weeks.

Input "",Str1
0->dim(L1
0->dim(L2
1->I
DelVar S
While I<length(Str1
DelVar T
".->Str3
1->C
While C and I<length(Str1
sub(Str1,I,1->Str2
inString("0123456789.",Str2->C
If C:Then
I+1->I
1->T
Str3+Str2->Str3
End
End
If T=0:Str3+Str2->Str3
sub(Str3,2,length(Str3)-1->Str3
If T=1:Then
expr(Str3->L2(1+dim(L2
End
inString("*+/-",Str3->M
If M:Then
I+1->I
2->T
1->C
While C
dim(L1->C
If C:Then
2fPart(L1(dim(L1))/2)>2fPart(M/2->C
If C:Then
~L1(dim(L1->L2(1+dim(L2
dim(L1)-1->dim(L1
End
End
End
M->L1(1+dim(L1
End
If Str3="(":Then
If S=1:Then
sub(Str1,1,I-1)+"*"+sub(Str1,I,length(Str1)-I+1)->Str1
Else
I+1->I
0->L1(dim(L1)+1
End
End
If Str3=")":Then
I+1->I
While L1(dim(L1
~L1(dim(L1->L2(1+dim(L2
dim(L1)-1->dim(L1
End
dim(L1)-1->dim(L1
End
T->S
End
augment(L2,-seq(L1(X),X,dim(L1),1,-1)->L2
0->dim(L1
1->C
{0,1->L3
For(I,1,dim(L2
L2(I->M
If M>=0:Then
M->L1(dim(L1)+1
Else
If C:Then
L1(dim(L1)-1
Goto S
Lbl A
Ans->Str1
L1(dim(L1->L1(dim(L1)-1
dim(L1)-1->dim(L1
0->C
End
Str1+" "+sub("*+/-",-M,1)+" ->Str1
L1(dim(L1
Goto S
Lbl B
Str1+Ans->Str1
dim(L1)-1->dim(L1
End
End
Goto F
Lbl S
L3Ans->L4
LinReg(ax+b) L3,L4,Y1
Equ►String(Y1,Str2
sub(Str2,1,length(Str2)-3
If C:Goto A
Goto B
Lbl F
Str1

General explanation

Here's the key I worked with while developing the program:

Str1        The input string.
Str2        Counter variable (e.g. current character)
Str3        The current token being built.
L1          The operator stack
L2          The output queue
L3          Temporary list
L4          Temporary list
I           Iterator index
T           Type of token (enum)
S           Type of previous token (enum)
M           Temporary variable
C           Conditional variable


Token Types (T)
0           Undefined
1           Number
2           Operator
3           Open Parenthesis

Operator Elements
0           left parenthesis ("(")
1           multiplication ("*")
2           addition ("+")
3           division ("/")
4           subtraction ("-")
5           right parenthesis (")")

Precedence Rule: Remainder(prec, levelno)
0 - add, sub
1 - mul, div
(levelno = 2)

And here is the equivalent, hand-written JavaScript code I used to test and develop this program.

let Str1, Str2, Str3, L1, L2, I, S, T, M, C;

let error = (type, ...args) => {
    let message = "ERR:" + type + " (" + args.join("; ") + ")";
    throw new Error(message);
};

let isInteger = (n) => n == Math.floor(n);

let inString = (haystack, needle, start=1) => {
    if(start < 1 || !isInteger(start)) {
        error("DOMAIN", haystacak, needle, start);
    }
    let index = haystack.indexOf(needle, start - 1);
    return index + 1;
};

let sub = (string, start, length) => {
    if(start < 1 || length < 1 || !isInteger(start) || !isInteger(length)) {
        error("DOMAIN", string, start, length);
    }
    if(start + length > string.length + 1) {
        error("INVALID DIM", string, start, length);
    }
    return string.substr(start - 1, length);
}

let fPart = (value) => value - Math.floor(value);

// Input "", Str1
Str1 = process.argv[2];
// 0->dim(L1
L1 = [];
// 0->dim(L2
L2 = [];
// 1->I
I = 1;
// DelVar S
S = 0;
// While I<=length(Str1
while(I <= Str1.length) {
    // DelVar T
    T = 0;
    // Disp "Starting",I
    console.log("Starting, I =", I);

    // " read token
    // ".->Str3
    Str3 = ".";
    // 1->C
    C = 1;
    // While C and I<=length(Str1
    while(C && I <= Str1.length) {
        // sub(Str1,I,1->Str2
        Str2 = sub(Str1, I, 1);
        // inString("0123456789",Str2->C
        C = inString("0123456789", Str2);
        // If C:Then
        if(C) {
            // I+1->I
            I++;
            // 1->T
            T = 1;
            // Str3+Str2->Str3
            Str3 += Str2;
        }
    }
    // If T=0:
    if(T == 0) {
        // console.log("Huh?T=0?", Str3, Str2);
        // Str3+Str2->Str3
        Str3 += Str2;
    }

    // " remove placeholder character
    // sub(Str3,2,length(Str3)-1->Str3
    Str3 = sub(Str3, 2, Str3.length - 1);

    // " number
    // If T=1:Then
    if(T == 1) {
        // expr(Str3->L2(1+dim(L2
        L2[L2.length] = eval(Str3);
    }

    // Disp "Str3",Str3
    console.log("post processing, Str3 = \"" + Str3 + "\"");

    // inString("*+/-",Str3->M
    M = inString("*+/-", Str3);
    // " operator
    // If M:Then
    if(M) {
        // I+1->I
        I++;
        // 2->T
        T = 2;
        // Disp "op",M,dim(L1
        console.log("op", M, L1.length);
        // " parse previous operators
        // 1->C
        C = 1;
        // While C
        while(C) {
            // dim(L1->C
            C = L1.length;
            // If C:Then
            if(C) {
                // 2fPart(L1(dim(L1))/2)>2fPart(M/2->C
                C = 2 * fPart(L1[L1.length - 1] / 2) > 2 * fPart(M / 2);
                // If C:Then
                if(C) {
                    // ~L1(dim(L1->L2(1+dim(L2
                    L2[L2.length] = -L1[L1.length - 1];
                    // dim(L1)-1->dim(L1
                    L1.length--;
                }
            }
        }
        // " push current operator
        // M->L1(1+dim(L1
        L1[L1.length] = M;
    }
    // If Str3="(":Then
    if(Str3 == "(") {
        // 3->T
        T = 3;
        // If S=1:Then
        if(S == 1) {
            // sub(Str1,1,I-1)+"*"+sub(Str1,I,length(Str1)-I+1)->Str1
            Str1 = sub(Str1, 1, I - 1) + "*" + sub(Str1, I, Str1.length - I + 1);
        }
        // Else
        else {
            // I+1->I
            I++;
            // 0->L1(dim(L1)+1
            L1[L1.length] = 0;
        }
        // End
    }
    // If Str3=")":Then
    if(Str3 == ")") {
        // I+1->I
        I++;
        // While L1(dim(L1
        while(L1[L1.length - 1]) {
            // ~L1(dim(L1->L2(1+dim(L2
            L2[L2.length] = -L1[L1.length - 1];
            // dim(L1)-1->dim(L1
            L1.length--;
        }
        // End
        // dim(L1)-1->dim(L1
        L1.length--;
    }
    // Disp "Ending",I
    console.log("Ending", I);
    // T->S
    S = T;
    // Pause
    console.log("-".repeat(40));
}

// augment(L2,-seq(L1(X),X,dim(L1),1,-1)->L2
L2 = L2.concat(L1.map(e => -e).reverse());

// Disp L1, L2
console.log("L1", L1);
console.log("..", "[ " + L1.map(e=>"*+/-"[e-1]).join`, ` + " ]");
console.log("L2", L2);
console.log("..", "[ " + L2.map(e=>e<0?"*+/-"[~e]:e).join`, ` + " ]");

// post-processing
let res = "";
// 0->dim(L1
L1.length = 0;
// 1->C
C = 1;
// For(I,1,dim(L2
for(I = 1; I <= L2.length; I++) {
    // L2(I->M
    M = L2[I - 1];
    // If M>=0:Then
    if(M >= 0) {
        // M->L1(dim(L1)+1
        L1[L1.length] = M;
    }
    // Else
    else {
        // If C:Then
        if(C) {
            // L1(dim(L1)-1
            // Goto ST
            // Lbl A0
            // Ans->Str1
            res += L1[L1.length - 2];
            // L1(dim(L1->L1(dim(L1)-1
            L1[L1.length - 2] = L1[L1.length - 1];
            // dim(L1)-1->dim(L1
            L1.length--;
            // 0->C
            C = 0;
        }
        // End
        // Str1+" "+sub("*+/-",-M,1)+" ->Str1
        res += " " + "*+/-"[-M - 1] + " ";
        // L1(dim(L1
        // Goto ST
        // Lbl A1
        // Str1+Ans->Str1
        res += L1[L1.length - 1];
        // dim(L1)-1->dim(L1
        L1.length--;
    }
}
// Goto EF
// Lbl ST
// L3Ans->L4
// LinReg(ax+b) L3,L4,Y1
// Equ►String(Y1,Str2
// sub(Str2,1,length(Str2)-3
// If C:Goto A0
// Goto A1
// Lbl EF
// Str1
console.log(res);

I will provide a more in-depth explanation once I'm sure I'm done golfing, but in the meantime, this might help provide a cursory understanding of the code.

Answer 2

Python 3. 337 327 - 10% = 295 bytes

Supports floating-point and unlimited levels of parentheses, so qualifies for bonus -10%.

import re
def v(s):
 if s[:1]=='(':l,s=e(s[1:]);return l,s[1:]
 m=re.match(r"\d+\.?\d*|\.\d+",s)
 if m:return[m.group()],s[m.end():]
def b(s,f,*O):
 l,s=f(s)
 while s[:1]in O:
  o=s[0]
  r,s=f(s[1:])
  if len(r)>1:l,r=r,l
  l+=[o]+r
 return l,s
m=lambda s:b(s,v,'*','/')
e=lambda s:b(s,m,'+','-')
print(' '.join(e(input())[0]))

Answer 3

Javascript (ES6), 535 - 80 (15% bonus) = 455 bytes

f=z=>{a=[],i=0;c=n=>{while(n[M='match'](/\(/)){n=n[R='replace'](/\(([^()]+)\)/g,(q,p)=>{m=++i;a[m]=c(p);return'['+m+']'})}n=n[R](/(\](?=\[|[\d])|[\d](?=\[))/g,'$1*');n=n[R](/\[?[\d\.]+\]?[*/]\[?[\d\.]+\]?/g,q=>{a[++i]=q;return'['+i+']'});n=n[R](/([\d.]+)\+(\[[\d]+\])/g,'$2+$1');while(n[M](/\[/)){n=n[R](/(\[?[\d\.]+\]?)\*(\[?[\d\.]+\]?)/g,(q,p,r)=>{t=a[r[R](/[\[\]]/g,'')];return r[M](/\[/)?(t&&t[M](/\+|\-/)?(r+'*'+p):q):q});n=n[R](/\[(\d+)\]/g,(q,p)=>{return a[p]+(a[p][M](/\+|-/)?'=':'')})}return n};return c(z)[R](/./g,'$& ')+'='}

Not a minimal solution I'm sure, but pretty complete, allowing for all three bonuses. Some instances require multiple pressings of equals key, but do not require clearing of calculator contents. (ex. 3,5,6 & 7 in fiddle)

Link to JSFiddle with some tests: https://jsfiddle.net/2v8rkysp/3/

Here's some unfolded, semi-unobfuscated code with a few comments for good measure.

function f(z) {
var a=[],i=0;
function c(n) {
    //// Tokenize parentheses groups recursively
    while (n.match(/\(/)) {
    n = n.replace(/\(([^()]+)\)/g, function(q,p) {
      m = ++i;
      a[m]=c(p);
      return '['+m+']';
    });
    }

    //// Allow implied multiplication with parentheses
    n = n.replace(/(\](?=\[|[\d])|[\d](?=\[))/g, '$1*');

    //// Tokenize mult/division
    n = n.replace(/\[?[\d\.]+\]?[*\/]\[?[\d\.]+\]?/g, function(q) {
      a[++i]=q;
      return '['+i+']';
    });

    //// Move addition tokens to the front
    n = n.replace(/([\d.]+)\+(\[[\d]+\])/g,'$2+$1');

    //// Detokenize
    while (n.match(/\[/)) {
        //// If a token includes addition or subtraction,
        ////   move it to the front of other tokens 
        n = n.replace(/(\[?[\d\.]+\]?)\*(\[?[\d\.]+\]?)/g,function(q,p,r) {
          t=a[r.replace(/[\[\]]/g,'')];
          return r.match(/\[/)?(t&&t.match(/\+|\-/)?(r+'*'+p):q):q;
        });
        //// If a token includes addition or subtraction,
        ////   add the equals keypress
        n = n.replace(/\[(\d+)\]/g, function(q,p) {
           return a[p]+(a[p].match(/\+|-/)?'=':'');
        });
    }
    return n;
}
//// Add spaces and final equals keypress
return c(z).replace(/./g, '$& ')+'=';
}