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You will be given 3 integers as input. The inputs may or may not be different from each other. You have to output 1 if all three inputs are different from each other, and 0 if any input is repeated more than once.
This is code-golf, so make your code as short as possible!
ɠḲQL=3
From 5 to 6 bytes because this is my first time and I messed up (whoops) fixed it now
ɠḲQL=3
^^^^^
||||Is it equal to three?
|||How many unique numbers do we have? (length of unique numbers array)
||Sort By Unique
|Split by Spaces
Read Input
3 integers, or is it only functional for three digits?
\$\endgroup\$
Sep 7, 2018 at 16:58
echo(4==count(array_unique($argv)))*1;
Called with e.g. php codegolf.php 2 3 4. $argv always contains the filename of the script, hence 3 unique numbers make 4 elements.
~x=x==x∪x
-7 bytes (!!) thanks to MarcMush: replace length(x∪x)>2 with x==x∪x
..i..
.MiC.
}u)3=
Excluding the mandatory triangular padding, we're left with:
iMiC}u)3=
Which can minimally be explained as: Push all STDIN as an array of strings (representing each line of input), count the occurrences of all of STDIN's elements in itself (which, in case all are distinct, should yield [1, 1, 1]), sum, then check whether the sum is equal to 3. There are 2 tricks here (which together save no less than 14 bytes):
i and treating the inputs as strings rather than integers – which is quite uncommon, as )IE is frequently used instead, taking input as a list.
≔Ｅ³ＮθＩ¬⊖⌈Ｅθ№θι
Try it online! Link is to verbose version of code. 5 bytes could be saved by using a slightly non-standard input format [[x, y, z]]. Explanation:
≔Ｅ³Ｎθ
Input 3 numbers and make a list.
Ｉ¬⊖⌈Ｅθ№θι
Count the number of occurrences of each number in the list. Take the maximum, subtract 1, take the logical not, then cast to string for implicit print.
<?=array_unique($argv)==$argv;
takes input from command line arguments; prints 1 for truthy, empty string for falsy
-9 bytes thanks to GB !
m(int*a){a=*a==a[2]^*a==*++a^*a==*++a^1;}
No built-in sets in C, and iterating is too long, but at least i can shave off a few bytes by taking advantage of the fact that *a == a[0] ...
SWIFT 37 Bytes
Set(Array(readLine()!)).count==3 ?1:0
ä▼Z
Explanation
ux= #Full program, unpacked, implicit input as array
u #Keep only unique elements in array, maintaining first order of appearance
x= #Copy original input onto stack, is it equal to unique-ified array.
Short and sweet. Wish Stax would implicitly check against "X" register if there was nothing to check against, but it works out regardless.
g a b c=and$map(==)[a,b,c]
?>a
?>b
?>c
(a-b)&(b-c)&(a-c)
Using bitwise, comparison and logical operators Try it online!
(4 bytes saved courtesy of Jonathan Frech)
(a,b,c)=>a!=b&a!=c&b!=c
1^ inside the parentheses, changing == to != and | to & and then remove said parentheses to save four bytes.
\$\endgroup\$
Sep 23, 2018 at 7:26
y==z (returns true, should be false). You'll need to add a third test to check for that possibility.
\$\endgroup\$
0, 1, 2 are distinct, yet 0&... = 0.
\$\endgroup\$
Sep 22, 2018 at 14:43
rEeEREeErrEEEEEeRrEeErrEEEEEerEEEEEEERerEEEEEerRrEEEEEEeErREEREEEEErEeReeReEE
Try it online! Input is a,b,c. Change the "separated by" box to , to make it work!
s<2l{
Looks like someone already did pyth but I didn't know about the I for testing invariants.
Anything XORd with itself yields zero. So. Any language with multiplication and XOR can do this:
(a^b)*(b^c)*(a^c)
Since input specifics were rather vague I chose a liberal interpretation. Or if you want to be more strict, we can define it as a closure (kind of like a lambda function),
24 bytes
|a,b,c|(a^b)*(b^c)*(a^c)
$(($1-$2&&$2-$3&&$3-$1))
Sorry for a boring answer; at first if bashism would work, but I could not get out of POSIXism.
The code above evaluates to a string of either 0 or 1.
._._._.e.z.c.c.z1.o.d.a.a0.oTry it online! (with pxem.posixism)
XX.z
# for three times; do push( read integer); done
.a._._._XX.z
# call subroutine
## NOTE. When DOT-e is performed when stack is a,b,c,bottom,
## then subroutine begins with same content of stack
## And when subroutine ends when stack is d,e,f,bottom,
## then previous stack would be d,e,f,a,b,c,bottom
## to continue origimal procedure
## In this program content is empty, so is subroutine
## thus x,y,z,bottom to x,y,z,x,y,z,bottom,
## which duplicates entire stack
.a.eXX.z
# while size < 2 && pop != pop; do
.a.zXX.z
# dup; dup; while size<2 && pop!=pop; do
## this is equal to while stack is empty; do
.a.c.c.zXX.z
# push '1'; putc pop; exit
.a1.o.dXX.z
# done
.a.aXX.z
# done
.a.aXX.z
# push '0'; putc pop; (implicity) exit
.a0.oXX.z
.a
{(#x)=#?x}
A little bit more interesting answer in my opinion, but still boring. Takes a list of integers as input.
Explanations:
{(#x)=#?x} Main function. x is input
? Unique (Remove all duplicated elements)
# Length
= Is it equal to
(#x) The length of x itself
This is a complete program conforming to ISO standard 10206 “Extended Pascal” and at 91 bytes certainly more readable than brain****.
program p(input,output);var x,y,z:integer;begin read(x,y,z);write(ord(card([x,y,z])=3))end.
The [x,y,z] denotes a set, thus duplicate values get collapsed.
This means, the set literal [4, 4, 4] is identical to [4].
The card function returns the number of distinct elements in the set.
Here this will be anything between 1 and 3 inclusive.
We test for equality to 3 (… = 3) which evaluates to a Boolean expression (denoted by the built-in constants false and true) and convert this enumeration value into an integer using ord.
