# Mixed Fraction Equality

In elementary school, children learn about proper fractions, where the numerator is less than the denominator, and thus the value of the fraction is less than one. Later, they are taught about fractions where the value of the fraction is greater than one, and two different ways to express these fractions: mixed fractions and improper fractions.

Given a mixed fraction, determine if it is equivalent to the improper fraction where the integer value and the numerator are concatenated together. For example, for input 1 3/4, the improper fraction is 13/4.

## Test Cases

1 3/4        -> falsey
1 3/10       -> truthy
6 6/7        -> falsey
55 55/100    -> truthy
4 9/100      -> falsey
40 9/100     -> falsey
7 49/1000    -> falsey
9 1/2        -> falsey
999 999/1000 -> truthy
1 21/200     -> falsey
1 21/101     -> falsey


For input, you may take the integer part and the fraction part as separate inputs, but you may not take the fraction as input in two pieces, and you may not take it as a decimal value. You may drop the integer part (not take it as input) if you do not need to use it.

• Should or can the fraction be simplified? Like the fourth test case would be false as 54/100 simplifies to 27/50
– Jo King
Commented Jul 24, 2018 at 22:45
• Should output be two distinct, consistent values or any, possibly inconsistent, truthy/falsey values? Commented Jul 24, 2018 at 22:53
• By the way, fixing the 4th test case to have 55 wouldn't change the issue would it - 55/100 can also be simplified to 11/20, so the same question @JoKing raised arises there. Commented Jul 24, 2018 at 23:44
• "you may not take the fraction as input in two pieces" - err why? That is exactly what the / does :/ Commented Jul 24, 2018 at 23:54
• This seems to be equivalent to "given an input that doesn't matter and two numbers as a string separated by a slash, determine whether the second number equals 10 to the power of the length of the first number".
– xnor
Commented Jul 25, 2018 at 0:15

# MATL, 7 bytes

UsGXzU=


Input is a string. Output is 1 for truthy, 0 for falsey.

### Explanation

U     % Implicit input. Convert to number(s): gives a vector of two numbers
s     % Sum of that vector
G     % Push input again
Xz    % Remove spaces
U     % Convert to number
=     % Equal? Implicit display


# Perl 6, 16 12 bytes

{1+$_==1~$_}


Try it online!

Takes input as a string representing the fraction. Turns out Perl 6's dynamic typing can handle strings to rational fractions, who knew? So the string "1/10" when coerced to a number, returns 0.1

The anonymous code block simply checks if the fraction plus one equals one concatenated with the fraction. Thanks to xnor's Python answer for showing me that the integer part doesn't matter.

### Old solution, 27 26 bytes

{.nude[0]==.Int~[%] .nude}


Try it online!

Takes input as a rational mixed fraction, and returns true or false. Returns false for the fourth test case because it can be simplified.

### Explanation:

.nude returns a list of [numerator, denominator].

{                        } # Anonymous code block
.nude[0]    # Check if the numerator of the mixed fraction
==  # Is equal to
.Int  # The integer part of the fraction
~ # Concatenated to
[%] .nude  # The numerator modulo the denominator
# And return implicitly

• I'm guessing that .nude is named for numerator + denominator, but someone probably took great pleasure in being able to call it that. Commented Jul 24, 2018 at 23:42
• I was going to go with something that took it as a single string '1 3/10' {S/\s//==.words.sum} Commented Jul 25, 2018 at 0:20

# Retina 0.8.2, 17 16 bytes

(.)+/1(?<-1>0)*$ Try it online! Requires just the fraction part, so the linked test suite removes the integer from the test cases. Explanation: The improper concatenation is equal to the mixed number only if the denominator is a power of 10 and the numerator has one digit for every zero in the denominator. .NET's balancing groups are used to verify that sufficient digits exist. Edit: Saved 1 byte thanks to @sundar. • Doesn't work for 1 11/10. It seems to be a problem with your implementation, not the method Commented Jul 24, 2018 at 23:14 • Note that " You may drop the integer part (not take it as input) if you do not need to use it." - so the leading space may be unnecessary if you change the input to have only the fraction. Commented Jul 24, 2018 at 23:21 • @H.PWiz I don't think we have to deal with inputs where the numerator is greater than the denominator (since these are supposed to be mixed fractions with only the non-integral decimal part expressed as a fraction). But I'll ask the OP to confirm that. Commented Jul 24, 2018 at 23:23 • @sundar I'd have to change it to ^ instead, so it doesn't help. – Neil Commented Jul 24, 2018 at 23:54 • The / makes it unambiguous what you're matching, so I don't think you need the anchor there (going by usual regex matching rules, no Retina expertise here). Seems to work anyway: Try it online!. Commented Jul 24, 2018 at 23:56 # Husk, 8 bytes §·=r¤+r+  Try it online! ### Explanation §(·=r)(¤+r)(+) -- example arguments: "1" "3/10" § -- fork both arguments (¤ r) -- | read both: 1 3/10 ( + ) -- | and add them: 13/10 (+) -- | concatenate: "13/10" -- and do (· r) -- | read the second argument: 13/10 ( = ) -- | and compare: 13/10 == 13/10 -- : 1  # Python 2, 43 bytes lambda k:k.split('/')[1]==10**k.find('/')  Try it online! # R, 78 65 bytes function(n,e=function(a)eval(parse(t=sub(" ",a,n))))e("")==e("+")  Try it online! -13 bytes thanks to both Giuseppe and JayCe! • Just sub is fine here. Also, you can use t= instead of text= Commented Jul 25, 2018 at 12:04 • What can I say? Brilliant! it simplifies nicely to 65 bytes Commented Jul 25, 2018 at 13:48 • @JayCe Glad to see I was on the right page! Thanks! Commented Jul 25, 2018 at 14:41 • You can try porting xnor's Python 3 answer for probably 20 bytes... Commented Jul 26, 2018 at 15:09 • Commented Aug 4, 2018 at 20:16 # Stax, 5 bytes ╡ÄLσ  Run and debug it Explanation: +yj$e= Full program, implicit input
+      Add integer and fraction part
y     Push unparsed input
j$Split on spaces and flatten, i.e. Remove spaces e Evaluate = Check for equality  # Python 3, 26 bytes lambda k:eval(k+'+1==1'+k)  Try it online! For example, input 3/4 gives 3/4+1==13/4. Instead of taking the whole part of the fraction, we just set it to 1 to test the mixed fraction equality. Test cases from Chas Brown. # Brachylog, 15 bytes ḍ{lᵛ&ht¬ị&t↔ị1}  Try it online! Takes the fractional part alone as a string input. Indirectly uses the same idea as my Julia answer - "the denominator is 10^{length of numerator}" can be said as "the denominator is a power of ten, and the length of the denominator is equal to the length of the numerator + the length of "/" (i.e. 1). ḍ % split the input in half { } % and verify that lᵛ % each half has the same length (i.e. the string had even length) &ht¬ị % and the last character of the first half is % not a number (it should be "/") &t↔ị1 % and the second half when reversed is the number 1 % i.e. the denominator should be a power of 10  Older answer: ### 15 20 bytes a₀ᶠịˢtl;10↺^.&a₁ᶠịˢh  Try it online! (-1 byte thanks to @Fatalize, but unfortunately +6 bytes since I discovered bugs in the older method.) Same idea as my Julia answer. • You can shorten it by 1 byte by replacing variable A with the output variable . (and thus remove the last A because the output variable is implicitely there at the end) Commented Jul 26, 2018 at 6:52 • @Fatalize Thanks, I forget that the output is pretty much available as a free variable in these decision problems. Unfortunately I found bugs in the code I had: since it only asked for any numeric prefix and any numeric suffix, it was passing things like 61/10 (using just 6 as the numerator/prefix) 2/110 (using just 10 as the denominator/suffix). I've tried to fix it, not sure if this is the best way to do it though. Commented Jul 26, 2018 at 14:09 • I'm not sure I can help you because I don't understand the specs of this challenge at all, even after reading it 3 times. I don't know what "mixed fractions and improper fractions" are called in my country, or if it's even taught in elementary schools here. Commented Jul 26, 2018 at 14:41 • @Fatalize Fair enough. Do you have interest in reviving the chat room? I have a bunch of questions to pester you with, if you're interested and have the time. Commented Jul 27, 2018 at 17:34 • Sure, just ask a mod to revive the room and tag me when you ask questions Commented Jul 30, 2018 at 6:37 # Julia 0.6, 29 bytes r->10^ndigits(num(r))==den(r)  Try it online! Based on the idea that the output should be true only when the denominator is a power of ten with as many zeros as there are digits in the numerator. Takes input as a Rational type, checks that the denominator is equal to 10 raised to the number of digits in the numerator. # Clean, 57 bytes import StdEnv,Text$b#[u,v:_]=split"/"b
=v==""<+10^size u


Try it online!

This one is a little shorter but breaks for large numerator/denominators.

# Clean, 776160 58 bytes

-1 thanks to OMᗺ's tip on my other answer

import StdEnv,Text
$b#[u,v:_]=split"/"b =1<+[48\\_<-:u]==v  Try it online! This uses Neil's method, it's a bit shorter than doing it directly. There's some trickery with conversion overloading, where 1<+[48\\_<-:u] converts [Int] to [Char] and then to {#Char} (:== String), but Int directly to String. # Clean, 91 89 bytes import StdEnv,Text t=toInt$a b#[b,c:_]=map t(split"/"b)
#d=t(a<+b)
=d/c==t a&&d-d/c*c==b


Try it online!

Defines a function $:: String String -> Bool which extracts the numerator and denominator, string-concatenates the integer part and the numerator, and checks equivalence. # 05AB1E, 7 bytes '/¡ćg°Q  Only takes the fractions as input. Explanation: '/¡ # Split the input by '/' # i.e. '3/10' → ['3', '10'] ć # Head extracted: # i.e. ['3', '10'] → 10 and 3 g # Take the length of the numerator # i.e. '3' → 1 ° # Take 10 to the power of this length # 1 → 10**1 → 10 Q # Check if that value equals the denominator # 10 and 10 → 1 (truthy)  Or a more general explanation: We have to validate two things: • Is the denominator a factor 10 (1, 10, 100, 1000, etc.)? • Does the length of the numerator + 1 equal the length of the denominator? • This second part is done by checking if the denominator as is, is equal to 10 to the power of the length of the numerator, which saves 2 bytes PS: If we could take the numerator and denominator as separated inputs, just 3 bytes would have been enough: g°Q. # JavaScript, 26 bytes Takes input in currying syntax (f(x)(y)) where x is the integer and y is the fraction as a string. x=>y=>x==eval(x+y)-eval(y)  Try it online # Java 10, 1077067 57 bytes f->new Long(f.split("/")[1])==Math.pow(10,f.indexOf("/"))  Welcome to the world without eval.. -40 bytes by creating a port of @ChasBrown's Python 2 answer. -10 bytes thanks to @Shaggy (I should have read @ChasBrown's answer better and his use of find (indexOf)..) Try it online. Explanation: f-> // Method with String parameter and boolean return-type new Long(f.split("/")[1]) // Take the denominator as integer ==Math.pow(10, // And check whether it is equal to 10 to the power of: f.indexOf("/")) // the length of the numerator-String  • 57 bytes Commented Jul 25, 2018 at 12:26 • @Shaggy Ah, Chas Brown even has the same in the Python 2 answer that I linked.. Not sure why I didn't already use that.. Thanks! Commented Jul 25, 2018 at 12:29 # Haskell, 47 40 bytes -7 thanks to OMᗺ f b|(u,v)<-span('/'<)b="/1"++('0'<$u)==v


Try it online!

A port of my Clean answer using Neil's method.

# Perl 5 -p, 23 bytes

$_=eval=~s/..//r eq$_+0


Try it online!

Takes the fractional part alone as input (as allowed by OP), outputs 1 for true and nothing for false.

$_= # assign to be printed by -p eval # evaluate fraction to get 0.something, for eg. 0.54 =~s/..//r # remove the 0. part, giving 54 eq # string equality check$_+0      # after first coercing input to number to remove / and denominator


The decimal part taken by itself would be exactly equal to the numerator only when the denominator is the immediate next power of ten greater than the numerator, which is the condition we need to check for.

# Noether, 17 bytes

I#I"/"^WL1-%WL_=P


Try it online!

## Explanation

So how does this work? Well, if you look at the test cases, the only true cases are when the denominator is a power of ten, $10^a$, where $a$ is the length of the numerator plus one ($a = \lfloor \log_{10} n \rfloor + 1$, where $n$ is the numerator and $\lfloor x \rfloor$ represents the flooring function).

I#                - Push the first input then pop it off the stack
I"/"^           - Push the second input and split the string at "/"
W          - Convert the top (the denominator) of the stack from a string to a number
L1-       - Take the log 10 of the top of the stack and subtract 1 (a)
%      - Rotate the stack
WL_   - Convert the top of the stack (the numerator) to a number, take the log10 and floor the result (b)
=  - Check if a and b are equal
P - Print the top of the stack


# TeaScript, 25 bytes

First input is the fraction, second is the integer.

I just started in TeaScript so it might be golfed down a lot

s"/")▒⌐ep(xs"/")░.n


Try it online!

# R, 53 bytes

function(n,x=el(strsplit(n,"/")))x[2]==10^nchar(x[1])


Try it online!

Takes only the fractional part as input. As mentioned by xnor in a comment:

This seems to be equivalent to "given an input that doesn't matter and two numbers as a string separated by a slash, determine whether the second number equals 10 to the power of the length of the first number".

Robert S.'s answer is less golfy but much more interesting than mine.

# C (gcc), 6756 55 bytes

m;x(char*_){for(m=1;*_++;m*=10)*_*=*_^47;m=m==atoi(_);}


Try it online!

• 64 bytes: tio.run/… Commented Jul 25, 2018 at 12:49
• Actually, do you even need r? tio.run/… Commented Jul 25, 2018 at 12:51
• @O.O.Balance No I do not. Thank you. Commented Jul 25, 2018 at 14:31

# Excel, 52 bytes

=10^FIND("/",B1)/10-MID(B1,FIND("/",B1)+1,LEN(B1))=0


Ignores the Integer input. Basically: IS Denominator = 10^LEN(Numerator)

For denominators limited to <10^9: 48 bytes:

=10^FIND("/",B1)/10-MID(B1,FIND("/",B1)+1,1E9)=0


Bulk of logic is splitting on /. If input could be taken separately, 16 bytes:

=10^LEN(B1)-C1=0


# Elixir, 81 bytes

fn b->[n,d]=String.split b,"/";String.to_integer(d)==:math.pow 10,byte_size n end


Try it online!

Might be able to get somewhere with {n,"/"<>d}=Integer.parse b, but not sure how.

# 2DFuck, 86 bytes

..!x..!...,,,,[>,,,,,,,,],,,,,,,,,[v!],[v!],[v!],![v!],,,[,,[v!],[v!],[v!],[v!]<,,,]r.


Try it online!

Takes input without integer part.

# C (gcc/clang), 5949 47 bytes

f(a,b){a=atoi(b=strchr(a,47)+1)==pow(10,b+~a);}


Port of Chas Brown's Python 2 answer. Try it online here.

Ignores the integer part of the input. Thanks to Jonathan Frech for golfing 2 bytes.

Ungolfed:

f(a, b) { // function returning an int and taking a string as input; also declaring another string variable
// this abuses the parameters as much as possible, omitting the type int and implicitly converting it to char *
a =                             // return a truthy or falsey value based on
atoi(b = strchr(a, 47) + 1) // the denominator (as integer; 47 is the ASCII code for '/')
== pow(10, b + ~a);         // is equal to 10 to the power of the length of the numerator-string
}

• '/' can most likely be 47. Commented Jul 25, 2018 at 14:36
• Possible 47 bytes. Commented Jul 25, 2018 at 14:40
• You are welcome. I think you forgot to update your header to reflect the new byte count. Commented Jul 25, 2018 at 15:35
• Suggest index() instead of strchr() Commented Aug 21, 2021 at 7:29

# ForceLang, 86 78 bytes

set a io.readnum()
`