Вступ:

Натхненний дискусією, яка триває вже багато років щодо виразу $6÷2(1+2)$ .

З виразом $6÷2(1+2)$ математики швидко побачать, що правильна відповідь - $1$ , тоді як люди з простим математичним фоном зі школи швидко побачать, що правильна відповідь - це $9$ . То звідки береться ця суперечка і тому різні відповіді? Існує два суперечливих правила в тому, як пишеться $6÷2(1+2)$ . Один через частину 2(, а один через символ поділу ÷.

Хоча і математики, і «звичайні люди» використовуватимуть PEMDAS ( парентез - експоненти - ділення / множення - додавання / віднімання), для математиків вираз оцінюється так, як нижче, тому що $2(3)$ само, як, наприклад, $2x^2$ одночлен aka " єдиний доданок через мається на увазі множення на співставлення " (і, отже, частина Pв PEMDAS), який буде оцінюватися інакше, ніж $2×(3)$ (двочлен як і два доданки):

$$6÷2(1+2) → \frac{6}{2(3)} → \frac{6}{6} → 1$$

$2(3)$$2×(3)$MDPEMDAS

$$6÷2(1+2) → 6/2×(1+2) → 6/2×3 → 3×3 → 9$$

However, even if we would have written the original expression as $6÷2×(1+2)$, there can still be some controversy due to the use of the division symbol ÷. In modern mathematics, the / and ÷ symbols have the exact same meaning: divide. Some rules pre-1918^† regarding the division symbol ÷^†† state that it had a different meaning than the division symbol /. This is because ÷ used to mean "divide the number/expression on the left with the number/expression on the right"^†††. So $a ÷ b$ then, would be $(a) / (b)$ or $\frac{a}{b}$ now. In which case $6÷2×(1+2)$ would be evaluated like this by people pre-1918:

^{†: Although I have found multiple sources explaining how ÷ was used in the past (see ††† below), I haven't been able to find definitive prove this changed somewhere around 1918. But for the sake of this challenge we assume 1918 was the turning point where ÷ and / starting to mean the same thing, where they differed in the past.}

^{††: Other symbols have also been used in the past for division, like : in 1633 (or now still in The Netherlands and other European non-English speaking countries, since this is what I've personally learned in primary school xD) or ) in the 1540s. But for this challenge we only focus on the pre-1918 meaning of the obelus symbol ÷.
†††: Sources: this article in general. And the pre-1918 rules regarding ÷ are mentioned in: this The American Mathematical Monthly article from February 1917; this German Teutsche Algebra book from 1659 page 9 and page 76; this A First Book in Algebra from 1895 page 46 [48/189].}

Slightly off-topic: regarding the actual discussion about this expression: It should never be written like this in the first place! The correct answer is irrelevant, if the question is unclear. *Clicks the "close because it's unclear what you're asking" button*.
And for the record, even different versions of Casio calculators don't know how to properly deal with this expression:

Challenge:

You are given two inputs:

• A (valid) mathematical expression consisting only of the symbols 0123456789+-×/÷()
• A year

And you output the result of the mathematical expression, based on the year (where ÷ is used differently when $year<1918$, but is used exactly the same as / when $year\ge1918$).

Challenge rules:

• You can assume the mathematical expression is valid and only uses the symbols 0123456789+-×/÷(). This also means you won't have to deal with exponentiation. (You are also allowed to use a different symbols for × or ÷ (i.e. * or %), if it helps the golfing or if your language only supports ASCII.)
• You are allowed to add space-delimiters to the input-expression if this helps the (perhaps manual) evaluation of the expression.
• I/O is flexible. Input can be as a string, character-array, etc. Year can be as an integer, date-object, string, etc. Output will be a decimal number.
• You can assume there won't be any division by 0 test cases.
• You can assume the numbers in the input-expression will be non-negative (so you won't have to deal with differentiating the - as negative symbol vs - as subtraction symbol). The output can however still be negative!
• You can assume N( will always be written as N×( instead. We'll only focus on the second controversy of the division symbols / vs ÷ in this challenge.
• Decimal output-values should have a precision of at least three decimal digits.
• If the input-expression contains multiple ÷ (i.e. $4÷2÷2$) with $year<1918$, they are evaluated like this: $4÷2÷2 → \frac{4}{\frac{2}{2}} → \frac{4}{1} → 4$. (Or in words: number $4$ is divided by expression $2 ÷2$, where expression $2 ÷2$ in turn means number $2$ is divided by number $2$.)
• Note that the way ÷ works implicitly means it has operator precedence over × and / (see test case $4÷2×2÷3$).
• You can assume the input-year is within the range $[0000, 9999]$.

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code (i.e. TIO).
• Also, adding an explanation for your answer is highly recommended.

Test cases:

Input-expression:   Input-year:   Output:      Expression interpretation with parenthesis:

6÷2×(1+2)           2018          9            (6/2)×(1+2)
6÷2×(1+2)           1917          1            6/(2×(1+2))
9+6÷3-3+15/3        2000          13           ((9+(6/3))-3)+(15/3)
9+6÷3-3+15/3        1800          3            (9+6)/((3-3)+(15/3))
4÷2÷2               1918          1            (4/2)/2
4÷2÷2               1900          4            4/(2/2)
(1÷6-3)×5÷2/2       2400          -3.541...    ((((1/6)-3)×5)/2)/2
(1÷6-3)×5÷2/2       1400          1.666...     ((1/(6-3))×5)/(2/2)
1×2÷5×5-15          2015          -13          (((1×2)/5)×5)-15
1×2÷5×5-15          1719          0.2          (1×2)/((5×5)-15)
10/2+3×7            1991          26           (10/2)+(3×7)
10/2+3×7            1911          26           (10/2)+(3×7)
10÷2+3×7            1991          26           (10/2)+(3×7)
10÷2+3×7            1911          0.434...     10/(2+(3×7))
4÷2+2÷2             2000          3            (4/2)+(2/2)
4÷2+2÷2             1900          2            4/((2+2)/2)
4÷2×2÷3             9999          1.333...     ((4/2)×2)/3
4÷2×2÷3             0000          3            4/((2×2)/3)
((10÷2)÷2)+3÷7      2000          2.928...     ((10/2)/2)+(3/7)
((10÷2)÷2)+3÷7      1900          0.785...     (((10/2)/2)+3)/7
(10÷(2÷2))+3×7+(10÷(2÷2))+3×7
                    1920          62           (10/(2/2))+(3×7)+(10/(2/2))+(3×7)
(10÷(2÷2))+3×7+(10÷(2÷2))+3×7
                    1750          62           (10/(2/2))+(3×7)+(10/(2/2))+(3×7)
10÷2/2+4            2000          6.5          ((10/2)/2)+4
10÷2/2+4            0100          2            10/((2/2)+4)
9+6÷3-3+15/3        9630          13           9+(6/3)-3+(15/3)
9+6÷3-3+15/3        0369          3            (9+6)/(3-3+(15/3))

R, 68 66 bytes

function(x,y,`=`=`/`)eval(parse(t=`if`(y<1918,x,gsub('=','/',x))))

Try it online!

Expects equality sign = instead of ÷ and * instead of ×.

The code makes use of some nasty operator overloading, making advantage of the fact that = is a right-to-left operator with very low precedence (the exact behavior that we want from pre-1918 ÷), and R retains its original precedence when it is overloaded. The rest is automatically done for us by eval.

As a bonus, here is the same exact approach implemented in terser syntax. This time our special division operator is tilde (~):

Julia 0.7, 51 bytes

~=/;f(x,y)=eval(parse(y<1918?x:replace(x,'~','/')))

Try it online!

JavaScript (ES6),  130 129  120 bytes

Saved 9 bytes thanks to @ScottHamper

Takes input as (year)(expr). Expects % and * instead of ÷ and ×.

y=>g=e=>(e!=(e=e.replace(/\([^()]*\)/,h=e=>eval(e.split`%`.reduceRight((a,c)=>y<1918?`(${c})/(${a})`:c+'/'+a))))?g:h)(e)

Try it online!

How?

Processing leaf expressions

The helper function $h$ expects a leaf expression $e$ as input, processes all % symbols according to the rules of the year $y$ (defined in the parent scope) and evaluates the resulting string.

If $y<1918$, we transform X%Y into (X)/(Y), to enforce low precedence and repeat this process for the entire string from right to left to enforce right-to-left associativity.

Examples:

• 8%2 becomes (8)/(2), whose simplified form is 8/2
• 2+3%3+2 becomes (2+3)/(3+2)
• 8%2%2 becomes (8)/((2)/(2)), whose simplified form is 8/(2/2)

If $y\ge 1918$, each % is simply turned into a /.

h = e =>                    // e = input string
  eval(                     // evaluate as JS code:
    e.split`%`              //   split e on '%'
    .reduceRight((a, c) =>  //   for each element 'c', starting from the right and
                            //   using 'a' as the accumulator:
      y < 1918 ?            //     if y is less than 1918:
        `(${c})/(${a})`     //       transform 'X%Y' into '(X)/(Y)'
      :                     //     else:
        c + '/' + a         //       just replace '%' with '/'
    )                       //   end of reduceRight()
  )                         // end of eval()

Dealing with nested expressions

As mentioned above, the function $h$ is designed to operate on a leaf expression, i.e. an expression without any other sub-expression enclosed in parentheses.

That's why we use the helper function $g$ to recursively identify and process such leaf expressions.

g = e => (            // e = input
  e !=                // compare the current expression with
    ( e = e.replace(  // the updated expression where:
        /\([^()]*\)/, //   each leaf expression '(A)'
        h             //   is processed with h
      )               // end of replace()
    ) ?               // if the new expression is different from the original one:
      g               //   do a recursive call to g
    :                 // else:
      h               //   invoke h on the final string
)(e)                  // invoke either g(e) or h(e)

Python 3.8 (pre-release), 324 310 306 bytes

lambda s,y:eval((g(s*(y<1918))or s).replace('%','/'))
def g(s):
 if'%'not in s:return s
 l=r=j=J=i=s.find('%');x=y=0
 while j>-1and(x:=x+~-')('.find(s[j])%3-1)>-1:l=[l,j][x<1];j-=1
 while s[J:]and(y:=y+~-'()'.find(s[J])%3-1)>-1:r=[r,J+1][y<1];J+=1
 return g(s[:l]+'('+g(s[l:i])+')/('+g(s[i+1:r])+')'+s[r:])

Try it online!

Takes % instead of ÷ and * instead of ×

Perl 5, 47 97 95 bytes

/ /;$_="($`)";$'<1918?s-%-)/(-g:y-%-/-;$_=eval

$_="($F[0])";1while$F[1]<1918&&s-\([^()]+\)-local$_=$&;s,%,)/((,rg.")"x y,%,,-ee;y-%-/-;$_=eval

TIO

Rust - 1066 860 783 755 740 bytes

macro_rules! p{($x:expr)=>{$x.pop().unwrap()}}fn t(s:&str,n:i64)->f64{let (mut m,mut o)=(vec![],vec![]);let l=|v:&Vec<char>|*v.last().unwrap();let z=|s:&str|s.chars().nth(0).unwrap();let u=|c:char|->(i64,fn(f64,f64)->f64){match c{'÷'=>(if n<1918{-1}else{6},|x,y|y/x),'×'|'*'=>(4,|x,y|y*x),'-'=>(2,|x,y|y-x),'+'=>(2,|x,y|y+x),'/'=>(5,|x,y|y/x),_=>(0,|_,_|0.),}};macro_rules! c{($o:expr,$m:expr)=>{let x=(u(p!($o)).1)(p!($m),p!($m));$m.push(x);};};for k in s.split(" "){match z(k){'0'..='9'=>m.push(k.parse::<i64>().unwrap() as f64),'('=>o.push('('),')'=>{while l(&o)!='('{c!(o,m);}p!(o);}_=>{let j=u(z(k));while o.len()>0&&(u(l(&o)).0.abs()>=j.0.abs()){if j.0<0&&u(l(&o)).0<0{break;};c!(o,m);}o.push(z(k));}}}while o.len()>0{c!(o,m);}p!(m)}

Rust does not have anything like 'eval' so this is a bit tough. Basically, this is a bog-standard Djisktra shunting-yard infix evaluator with a minor modification. ÷ is an operator with a variable precedence: lower than everything else (but parenthesis) in <1918 mode, higher than everything else in >=1918 mode. It is also 'right associated' (or left?) for <1918 to meet the 4÷2÷2 specification, and association is 'faked' by making ÷ precedence negative, then during evaluation treating any precedence <0 as associated. There's more room for golfing but this is a good draft i think.

Ungolfed at play.rust-lang.org