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8
Introduction:
Inspired by a discussion that is already going on for many years regarding the expression \$6÷2(1+2)\$.
With the expression \$6÷2(1+2)\$, mathematicians will quickly see that the correct answer is \$1\$, whereas people with a simple math background from school will quickly see that the correct answer is \$9\$. So where does this controversy and therefore different answers come from? There are two conflicting rules in how \$6÷2(1+2)\$ is written. One due to the part
2(
, and one due to the division symbol÷
.Although both mathematicians and 'ordinary people' will use PEMDAS (Parenthesis - Exponents - Division/Multiplication - Addition/Subtraction), for mathematicians the expression is evaluated like this below, because \$2(3)\$ is just like for example \$2x^2\$ a monomial a.k.a. "a single term due to implied multiplication by juxtaposition" (and therefore part of the
P
inPEMDAS
), which will be evaluated differently than \$2×(3)\$ (a binomial a.k.a. two terms):$$6÷2(1+2) → \frac{6}{2(3)} → \frac{6}{6} → 1$$
Whereas for 'ordinary people', \$2(3)\$ and \$2×(3)\$ will be the same (and therefore part of the
MD
inPEMDAS
), so they'll use this instead:$$6÷2(1+2) → 6/2×(1+2) → 6/2×3 → 3×3 → 9$$
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:
$$6÷2×(1+2) → \frac{6}{2×(1+2)} → \frac{6}{2×3} → \frac{6}{6} → 1$$
†: 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 asN×(
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))
3
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=`=`/`` is diabolical! Great solution! – Gregor Thomas – 2019-02-28T14:02:40.707uuugggghhh I had thoughts on the same lines. Alas, you beat me to it by quite a lot. Try it online
– Giuseppe – 2019-03-01T17:17:44.190Although there are no answers in codegolf languages yet, I'm accepting your Julia answer as the shortest for now. It is possible this will change in the future of course if a shorter answer is posted. – Kevin Cruijssen – 2019-04-05T05:47:19.077