Equal, sum or difference!
$begingroup$
Write shortest possible code that will return true if the two given integer values are equal or their sum or difference is 5.
Example test cases:
4 1 => True
10 10 => True
1 3 => False
6 2 => False
1 6 => True
-256 -251 => True
6 1 => True
The shortest I could come up with in python2 is 56 characters long:
x=input();y=input();print all([x-y,x+y-5,abs(x-y)-5])<1
code-golf decision-problem
New contributor
$endgroup$
add a comment |
$begingroup$
Write shortest possible code that will return true if the two given integer values are equal or their sum or difference is 5.
Example test cases:
4 1 => True
10 10 => True
1 3 => False
6 2 => False
1 6 => True
-256 -251 => True
6 1 => True
The shortest I could come up with in python2 is 56 characters long:
x=input();y=input();print all([x-y,x+y-5,abs(x-y)-5])<1
code-golf decision-problem
New contributor
$endgroup$
1
$begingroup$
Just for info, you can reduce your 56 by 9 by replacingx=input();y=input()
withx,y=input()
and giving the input as 2 integers separated by a comma.
$endgroup$
– ElPedro
8 hours ago
4
$begingroup$
welcome to PPCG! This is a good first challenge -- the challenge is clearly defined, it has ample test cases, and uses our default I/O! If you stick around for a while and keep thinking up interesting challenges, I would recommend using The Sandbox to get feedback before posting them to this site. I hope you enjoy the time you spend here!
$endgroup$
– Giuseppe
8 hours ago
$begingroup$
This would have been very slightly more interesting if the question was whether the absolute value of the addition or subtraction was equal to five, as then the sign of the integers doesn't matter. As it is, if I wanted to write an answer in Retina I'd have to special-case for a sum of-5
. (And since Retina only really understands positive integers, this isn't easy.)
$endgroup$
– Neil
6 hours ago
1
$begingroup$
suggest a test case6 1 => True
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@Neil based on the recently added test case that is the case
$endgroup$
– Stephen
3 hours ago
add a comment |
$begingroup$
Write shortest possible code that will return true if the two given integer values are equal or their sum or difference is 5.
Example test cases:
4 1 => True
10 10 => True
1 3 => False
6 2 => False
1 6 => True
-256 -251 => True
6 1 => True
The shortest I could come up with in python2 is 56 characters long:
x=input();y=input();print all([x-y,x+y-5,abs(x-y)-5])<1
code-golf decision-problem
New contributor
$endgroup$
Write shortest possible code that will return true if the two given integer values are equal or their sum or difference is 5.
Example test cases:
4 1 => True
10 10 => True
1 3 => False
6 2 => False
1 6 => True
-256 -251 => True
6 1 => True
The shortest I could come up with in python2 is 56 characters long:
x=input();y=input();print all([x-y,x+y-5,abs(x-y)-5])<1
code-golf decision-problem
code-golf decision-problem
New contributor
New contributor
edited 5 hours ago
Vikrant Biswas
New contributor
asked 8 hours ago
Vikrant BiswasVikrant Biswas
964
964
New contributor
New contributor
1
$begingroup$
Just for info, you can reduce your 56 by 9 by replacingx=input();y=input()
withx,y=input()
and giving the input as 2 integers separated by a comma.
$endgroup$
– ElPedro
8 hours ago
4
$begingroup$
welcome to PPCG! This is a good first challenge -- the challenge is clearly defined, it has ample test cases, and uses our default I/O! If you stick around for a while and keep thinking up interesting challenges, I would recommend using The Sandbox to get feedback before posting them to this site. I hope you enjoy the time you spend here!
$endgroup$
– Giuseppe
8 hours ago
$begingroup$
This would have been very slightly more interesting if the question was whether the absolute value of the addition or subtraction was equal to five, as then the sign of the integers doesn't matter. As it is, if I wanted to write an answer in Retina I'd have to special-case for a sum of-5
. (And since Retina only really understands positive integers, this isn't easy.)
$endgroup$
– Neil
6 hours ago
1
$begingroup$
suggest a test case6 1 => True
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@Neil based on the recently added test case that is the case
$endgroup$
– Stephen
3 hours ago
add a comment |
1
$begingroup$
Just for info, you can reduce your 56 by 9 by replacingx=input();y=input()
withx,y=input()
and giving the input as 2 integers separated by a comma.
$endgroup$
– ElPedro
8 hours ago
4
$begingroup$
welcome to PPCG! This is a good first challenge -- the challenge is clearly defined, it has ample test cases, and uses our default I/O! If you stick around for a while and keep thinking up interesting challenges, I would recommend using The Sandbox to get feedback before posting them to this site. I hope you enjoy the time you spend here!
$endgroup$
– Giuseppe
8 hours ago
$begingroup$
This would have been very slightly more interesting if the question was whether the absolute value of the addition or subtraction was equal to five, as then the sign of the integers doesn't matter. As it is, if I wanted to write an answer in Retina I'd have to special-case for a sum of-5
. (And since Retina only really understands positive integers, this isn't easy.)
$endgroup$
– Neil
6 hours ago
1
$begingroup$
suggest a test case6 1 => True
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@Neil based on the recently added test case that is the case
$endgroup$
– Stephen
3 hours ago
1
1
$begingroup$
Just for info, you can reduce your 56 by 9 by replacing
x=input();y=input()
with x,y=input()
and giving the input as 2 integers separated by a comma.$endgroup$
– ElPedro
8 hours ago
$begingroup$
Just for info, you can reduce your 56 by 9 by replacing
x=input();y=input()
with x,y=input()
and giving the input as 2 integers separated by a comma.$endgroup$
– ElPedro
8 hours ago
4
4
$begingroup$
welcome to PPCG! This is a good first challenge -- the challenge is clearly defined, it has ample test cases, and uses our default I/O! If you stick around for a while and keep thinking up interesting challenges, I would recommend using The Sandbox to get feedback before posting them to this site. I hope you enjoy the time you spend here!
$endgroup$
– Giuseppe
8 hours ago
$begingroup$
welcome to PPCG! This is a good first challenge -- the challenge is clearly defined, it has ample test cases, and uses our default I/O! If you stick around for a while and keep thinking up interesting challenges, I would recommend using The Sandbox to get feedback before posting them to this site. I hope you enjoy the time you spend here!
$endgroup$
– Giuseppe
8 hours ago
$begingroup$
This would have been very slightly more interesting if the question was whether the absolute value of the addition or subtraction was equal to five, as then the sign of the integers doesn't matter. As it is, if I wanted to write an answer in Retina I'd have to special-case for a sum of
-5
. (And since Retina only really understands positive integers, this isn't easy.)$endgroup$
– Neil
6 hours ago
$begingroup$
This would have been very slightly more interesting if the question was whether the absolute value of the addition or subtraction was equal to five, as then the sign of the integers doesn't matter. As it is, if I wanted to write an answer in Retina I'd have to special-case for a sum of
-5
. (And since Retina only really understands positive integers, this isn't easy.)$endgroup$
– Neil
6 hours ago
1
1
$begingroup$
suggest a test case
6 1 => True
$endgroup$
– cleblanc
6 hours ago
$begingroup$
suggest a test case
6 1 => True
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@Neil based on the recently added test case that is the case
$endgroup$
– Stephen
3 hours ago
$begingroup$
@Neil based on the recently added test case that is the case
$endgroup$
– Stephen
3 hours ago
add a comment |
24 Answers
24
active
oldest
votes
$begingroup$
Python 2, 30 bytes
lambda a,b:a in(b,5-b,b-5,b+5)
Try it online!
One byte saved by Arnauld
Three bytes saved by alephalpha
$endgroup$
2
$begingroup$
lambda a,b:a in(b,5-b,b-5,b+5)
$endgroup$
– alephalpha
8 hours ago
$begingroup$
This is amazingly concise, thanks
$endgroup$
– Vikrant Biswas
7 hours ago
add a comment |
$begingroup$
JavaScript (ES6), 28 bytes
Takes input as (a)(b)
. Returns $0$ or $1$.
a=>b=>a+b==5|!(a-=b)|a*a==25
Try it online!
$endgroup$
$begingroup$
Damn, took me a long long time to figure out how this handling the difference part. :)
$endgroup$
– Vikrant Biswas
5 hours ago
add a comment |
$begingroup$
R, 40 bytes (or 34)
function(x,y)any((-1:1*5)%in%c(x+y,x-y))
Try it online!
For non-R users:
-1:1*5
expands to[-5, 0, 5]
- the
%in%
operator takes elements from the left and checks (element-wise) if they exist in the vector on the right
A direct port of @ArBo's solution has 35 34 bytes, so go upvote that answer if you like it:
function(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
$begingroup$
The 34 byte one can be reduced by 1 withfunction(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
– MickyT
3 hours ago
add a comment |
$begingroup$
J, 12 11 bytes
1 byte saved thanks to Adám
1#.=+5=|@-,+
Try it online!
Explanation
This is equivalent to:
1 #. = + 5 = |@- , +
This can be divided into the following fork chain:
(= + (5 e. (|@- , +)))
Or, visualized using 5!:4<'f'
:
┌─ =
├─ +
──┤ ┌─ 5
│ ├─ e.
└───┤ ┌─ |
│ ┌─ @ ─┴─ -
└────┼─ ,
└─ +
Annotated:
┌─ = equality
├─ + added to (boolean or)
──┤ ┌─ 5 noun 5
│ ├─ e. is an element of
└───┤ ┌─ | absolute value |
│ ┌─ @ ─┴─ - (of) subtraction |
└────┼─ , paired with |
└─ + addition | any of these?
$endgroup$
$begingroup$
Save a byte withe.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám How so? Shortest approach I got withe.
was=+.5 e.|@-,+
. Maybe you forget5e.
is an invalid token in J?
$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
Since two integers cannot simultaneously sum to 5 and be equal, you can use+
instead of+.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám Ah, I see, thank you.
$endgroup$
– Conor O'Brien
7 hours ago
add a comment |
$begingroup$
Dyalog APL, 9 bytes
=∨5∊+,∘|-
Try it online!
Spelled out:
= ∨ 5 ∊ + , ∘ | -
equal or 5 found in an array of sum and absolute of difference.
$endgroup$
add a comment |
$begingroup$
PowerShell, 48 44 bytes
param($a,$b)5-in($x=$b-$a),-$x,($a+$b)-or!$x
Try it online!
Takes input $a
and $b
. Checks if 5
is -in
the group $b-$a
, -$x
($a-$b
), or $a+$b
, stores the first into $x
, and -or
s the -in
check with !$x
to check equality.
-4 bytes thanks to mazzy.
$endgroup$
$begingroup$
($a-$b)
is-$x
:)
$endgroup$
– mazzy
7 hours ago
$begingroup$
@mazzy Ooo, good call.
$endgroup$
– AdmBorkBork
7 hours ago
add a comment |
$begingroup$
Python 2, 38 bytes
-2 bytes thanks to @DjMcMayhem
lambda a,b:a+b==5or abs(a-b)==5or a==b
Try it online!
$endgroup$
$begingroup$
Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the5
s and theor
s
$endgroup$
– ElPedro
8 hours ago
3
$begingroup$
Actually, the TIO link could be 38 bytes
$endgroup$
– DJMcMayhem♦
8 hours ago
$begingroup$
@ElPedro the function itself was 40 bytes but I used f= in order to be able to call it
$endgroup$
– fəˈnɛtɪk
6 hours ago
1
$begingroup$
@DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example
$endgroup$
– fəˈnɛtɪk
6 hours ago
add a comment |
$begingroup$
C# (.NET Core), 43, 48, 47, 33 bytes
EDIT: Tried to use % and apparently forgot how to %. Thanks to Arnauld for pointing that out!
EDIT2: AdmBorkBork with a -1 byte golf rearranging the parentheses to sit next to the return so no additional space is needed!
EDIT3: Thanks to dana for -14 byte golf for the one-line return shortcut and currying the function (Ty Embodiment of Ignorance for linking to TIO).
C# (.NET Core), 33 bytes
a=>b=>a==b|a+b==5|(a-b)*(a-b)==25
Try it online!
$endgroup$
$begingroup$
Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D
$endgroup$
– Destroigo
8 hours ago
1
$begingroup$
You can get it down to 33 bytes applying dana's tips
$endgroup$
– Embodiment of Ignorance
6 hours ago
add a comment |
$begingroup$
x86 machine code, 39 bytes
00000000: 6a01 5e6a 055f 5251 31c0 39d1 0f44 c601 j.^j._RQ1.9..D..
00000010: d139 cf0f 44c6 595a 29d1 83f9 050f 44c6 .9..D.YZ).....D.
00000020: 83f9 fb0f 44c6 c3 ....D..
Assembly
section .text
global func
func: ;inputs int32_t ecx and edx
push 0x1
pop esi
push 0x5
pop edi
push edx
push ecx
xor eax, eax
;ecx==edx?
cmp ecx, edx
cmove eax, esi
;ecx+edx==5?
add ecx, edx
cmp edi, ecx
cmove eax, esi
;ecx-edx==5?
pop ecx
pop edx
sub ecx, edx
cmp ecx, 5
;ecx-edx==-5?
cmove eax, esi
cmp ecx, -5
cmove eax, esi
ret
Try it online!
$endgroup$
add a comment |
$begingroup$
Jelly, 7 bytes
+,ạ5eo=
Try it online!
How it works
+,ạ5eo= Main link. Arguments: x, y (integers)
+ Yield x+y.
ạ Yield |x-y|.
, Pair; yield (x+y, |x-y|).
5e Test fi 5 exists in the pair.
= Test x and y for equality.
o Logical OR.
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
!{#,5-#,#-5,#+5}~FreeQ~#2&
Try it online!
$endgroup$
add a comment |
$begingroup$
C (gcc), 41 34 bytes
f(a,b){a=5==abs(a-b)|a+b==5|a==b;}
Try it online!
$endgroup$
1
$begingroup$
Why doesf
returna
? Just some Undefined Behavior?
$endgroup$
– Tyilo
7 hours ago
$begingroup$
@Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value.
$endgroup$
– cleblanc
7 hours ago
$begingroup$
30 bytes Try it online!
$endgroup$
– Logern
6 hours ago
$begingroup$
@Logern Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@ceilingcat Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
|
show 1 more comment
$begingroup$
Scala, 45 bytes
def f(a:Int,b:Int)=a+b==5||(a-b).abs==5||a==b
Try it online!
$endgroup$
add a comment |
$begingroup$
Tcl, 53 bytes
proc P a b {expr abs($a-$b)==5|$a==$b|abs($a+$b)==5}
Try it online!
$endgroup$
$begingroup$
Same byte count: tio.run/##K0nO@f@/oCg/…
$endgroup$
– sergiol
7 hours ago
add a comment |
$begingroup$
Japt, 13 12 bytes
x ¥5|50ìøUra
Try it or run all test cases
x ¥5|50ìøUra
:Implicit input of array U
x :Reduce by addition
¥5 :Equal to 5?
| :Bitwise OR
50ì :Split 50 to an array of digits
ø :Contains?
Ur : Reduce U
a : By absolute difference
Alternative
50ìø[Ux Ura]
$endgroup$
add a comment |
$begingroup$
Japt, 14 13 bytes
¥VªaU ¥5ª5¥Nx
Try it online!
$endgroup$
add a comment |
$begingroup$
05AB1E, 13 12 bytes
ÐO5Qs`α5QrËO
Try it online!
Takes input as a list of integers, saving one byte. Thanks @Wisław!
Alternate 12 byte answer
Q¹²α5Q¹²+5QO
Try it online!
This one takes input on separate lines.
$endgroup$
1
$begingroup$
Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial|
?
$endgroup$
– Wisław
7 hours ago
$begingroup$
@Wisław Good point, I updated my answer. Thanks!
$endgroup$
– Cowabunghole
7 hours ago
$begingroup$
I found a 11 bytes alternative:OI`αª5¢IË~Ā
. Input is a list of integers.
$endgroup$
– Wisław
7 hours ago
$begingroup$
OIÆÄ)5QIËM
is 10.
$endgroup$
– Magic Octopus Urn
6 hours ago
1
$begingroup$
@MagicOctopusUrn I'm not sure exactly what the rules are but I think your solution is different enough from mine to submit your own answer, no? Also, unrelated but I've seen your username on this site for a long time but only after typing it out did I realize that it's "Urn", not "Um" :)
$endgroup$
– Cowabunghole
6 hours ago
|
show 2 more comments
$begingroup$
Batch, 81 bytes
@set/as=%1+%2,d=%1-%2
@if %d% neq 0 if %d:-=% neq 5 if %s% neq 5 exit/b
@echo 1
Takes input as command-line arguments and outputs 1 on success, nothing on failure. Batch can't easily do disjunctions so I use De Morgan's laws to turn it into a conjunction.
$endgroup$
add a comment |
$begingroup$
05AB1E, 10 bytes
OIÆ‚Ä50SåZ
Try it online!
O # Sum the input.
IÆ # Reduced subtraction of the input.
‚ # Wrap [sum,reduced_subtraction]
Ä # abs[sum,red_sub]
50S # [5,0]
å # [5,0] in abs[sum,red_sub]?
Z # Max of result, 0 is false, 1 is true.
Tried to do it using stack-only operations, but it was longer.
$endgroup$
$begingroup$
This will unfortunately return true if the sum is0
such as for[5, -5]
$endgroup$
– Emigna
4 hours ago
add a comment |
$begingroup$
Charcoal, 18 bytes
Nθ¿№⟦θ⁺⁵θ⁻⁵θ⁻θ⁵⟧N1
Try it online! Link is to verbose version of code. Port of @ArBo's Python 2 solution.
$endgroup$
add a comment |
$begingroup$
Runic Enchantments, 30 bytes
i::i::}3s=?!@-'|A"5"n:}=?!@+=@
Try it online!
With a pending update, 4 bytes can be saved, as presently integer inputs are treated as doubles and don't equal integers of the same value. Hence "5"n
instead of just 5
. This was an oversight and several factors are being adjusted to account for it (such as using Approximately(a, b)
instead of a.Equals(b)
).
Outputs 0
(exactly one zero) for false and any other output (literally whatever is left on the stack) for true.
$endgroup$
add a comment |
$begingroup$
Perl 5, 51 + 2 (-an) = 53 bytes
Pretty simple really. Outputs 0 for false, 1 for true. Not gonna lie, I don't know if bitwise OR is appropriate here, but is does work for all the test cases, so that's nice.
($a,$b)=@F;print($a==$b|$a+$b==5|$a-$b==5|$b-$a==5)
Try it online!
$endgroup$
add a comment |
$begingroup$
Java (JDK), 30 bytes
a->b->a+b==5|a==b|(b-=a)*b==25
Try it online!
$endgroup$
add a comment |
$begingroup$
Retina 0.8.2, 82 bytes
d+
$*
^(-?1*) 1$|^(-?1*)1{5} -?2$|^-?(-?1*) (3)1{5}$|^-?(1 ?){5}$|^(1 ?-?){5}$
Try it online! Link includes test cases. Explanation: The first two lines convert the inputs into unary. The final line then checks for any of the permitted matches:
^(-?1*) 1$ x==y
^(-?1*)1{5} -?2$ x>=0 y>=0 x=5+y i.e. x-y=5
x>=0 y<=0 x=5-y i.e. x+y=5
x<=0 y<=0 x=y-5 i.e. y-x=5
^-?(-?1*) (3)1{5}$ x<=0 y<=0 y=x-5 i.e. x-y=5
x<=0 y>=0 y=5-x i.e. x+y=5
x>=0 y>=0 y=5+x i.e. y-x=5
^-?(1 ?){5}$ x>=0 y>=0 y=5-x i.e. x+y=5
x<=0 y>=0 y=5+x i.e. y-x=5
^(1 ?-?){5}$ x>=0 y>=0 x=5-y i.e. x+y=5
x>=0 y<=0 x=5+y i.e. x-y=5
Pivoted by the last column we get:
x==y ^(-?1*) 1$
x+y=5 x>=0 y>=0 ^-?(1 ?){5}$
x>=0 y>=0 ^(1 ?-?){5}$
x>=0 y<=0 ^(-?1*)1{5} -?2$
x<=0 y>=0 ^-?(-?1*) (3)1{5}$
x<=0 y<=0 (impossible)
x-y=5 x>=0 y>=0 ^(-?1*)1{5} -?2$
x>=0 y<=0 ^(1 ?-?){5}$
x<=0 y>=0 (impossible)
x<=0 y<=0 ^-?(-?1*) (3)1{5}$
y-x=5 x>=0 y>=0 ^-?(-?1*) (3)1{5}$
x>=0 y<=0 (impossible)
x<=0 y>=0 ^-?(1 ?){5}$
x<=0 y<=0 ^(-?1*)1{5} -?2$
$endgroup$
add a comment |
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24 Answers
24
active
oldest
votes
24 Answers
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$begingroup$
Python 2, 30 bytes
lambda a,b:a in(b,5-b,b-5,b+5)
Try it online!
One byte saved by Arnauld
Three bytes saved by alephalpha
$endgroup$
2
$begingroup$
lambda a,b:a in(b,5-b,b-5,b+5)
$endgroup$
– alephalpha
8 hours ago
$begingroup$
This is amazingly concise, thanks
$endgroup$
– Vikrant Biswas
7 hours ago
add a comment |
$begingroup$
Python 2, 30 bytes
lambda a,b:a in(b,5-b,b-5,b+5)
Try it online!
One byte saved by Arnauld
Three bytes saved by alephalpha
$endgroup$
2
$begingroup$
lambda a,b:a in(b,5-b,b-5,b+5)
$endgroup$
– alephalpha
8 hours ago
$begingroup$
This is amazingly concise, thanks
$endgroup$
– Vikrant Biswas
7 hours ago
add a comment |
$begingroup$
Python 2, 30 bytes
lambda a,b:a in(b,5-b,b-5,b+5)
Try it online!
One byte saved by Arnauld
Three bytes saved by alephalpha
$endgroup$
Python 2, 30 bytes
lambda a,b:a in(b,5-b,b-5,b+5)
Try it online!
One byte saved by Arnauld
Three bytes saved by alephalpha
edited 8 hours ago
answered 8 hours ago
ArBoArBo
2115
2115
2
$begingroup$
lambda a,b:a in(b,5-b,b-5,b+5)
$endgroup$
– alephalpha
8 hours ago
$begingroup$
This is amazingly concise, thanks
$endgroup$
– Vikrant Biswas
7 hours ago
add a comment |
2
$begingroup$
lambda a,b:a in(b,5-b,b-5,b+5)
$endgroup$
– alephalpha
8 hours ago
$begingroup$
This is amazingly concise, thanks
$endgroup$
– Vikrant Biswas
7 hours ago
2
2
$begingroup$
lambda a,b:a in(b,5-b,b-5,b+5)
$endgroup$
– alephalpha
8 hours ago
$begingroup$
lambda a,b:a in(b,5-b,b-5,b+5)
$endgroup$
– alephalpha
8 hours ago
$begingroup$
This is amazingly concise, thanks
$endgroup$
– Vikrant Biswas
7 hours ago
$begingroup$
This is amazingly concise, thanks
$endgroup$
– Vikrant Biswas
7 hours ago
add a comment |
$begingroup$
JavaScript (ES6), 28 bytes
Takes input as (a)(b)
. Returns $0$ or $1$.
a=>b=>a+b==5|!(a-=b)|a*a==25
Try it online!
$endgroup$
$begingroup$
Damn, took me a long long time to figure out how this handling the difference part. :)
$endgroup$
– Vikrant Biswas
5 hours ago
add a comment |
$begingroup$
JavaScript (ES6), 28 bytes
Takes input as (a)(b)
. Returns $0$ or $1$.
a=>b=>a+b==5|!(a-=b)|a*a==25
Try it online!
$endgroup$
$begingroup$
Damn, took me a long long time to figure out how this handling the difference part. :)
$endgroup$
– Vikrant Biswas
5 hours ago
add a comment |
$begingroup$
JavaScript (ES6), 28 bytes
Takes input as (a)(b)
. Returns $0$ or $1$.
a=>b=>a+b==5|!(a-=b)|a*a==25
Try it online!
$endgroup$
JavaScript (ES6), 28 bytes
Takes input as (a)(b)
. Returns $0$ or $1$.
a=>b=>a+b==5|!(a-=b)|a*a==25
Try it online!
edited 8 hours ago
answered 8 hours ago
ArnauldArnauld
73.3k689307
73.3k689307
$begingroup$
Damn, took me a long long time to figure out how this handling the difference part. :)
$endgroup$
– Vikrant Biswas
5 hours ago
add a comment |
$begingroup$
Damn, took me a long long time to figure out how this handling the difference part. :)
$endgroup$
– Vikrant Biswas
5 hours ago
$begingroup$
Damn, took me a long long time to figure out how this handling the difference part. :)
$endgroup$
– Vikrant Biswas
5 hours ago
$begingroup$
Damn, took me a long long time to figure out how this handling the difference part. :)
$endgroup$
– Vikrant Biswas
5 hours ago
add a comment |
$begingroup$
R, 40 bytes (or 34)
function(x,y)any((-1:1*5)%in%c(x+y,x-y))
Try it online!
For non-R users:
-1:1*5
expands to[-5, 0, 5]
- the
%in%
operator takes elements from the left and checks (element-wise) if they exist in the vector on the right
A direct port of @ArBo's solution has 35 34 bytes, so go upvote that answer if you like it:
function(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
$begingroup$
The 34 byte one can be reduced by 1 withfunction(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
– MickyT
3 hours ago
add a comment |
$begingroup$
R, 40 bytes (or 34)
function(x,y)any((-1:1*5)%in%c(x+y,x-y))
Try it online!
For non-R users:
-1:1*5
expands to[-5, 0, 5]
- the
%in%
operator takes elements from the left and checks (element-wise) if they exist in the vector on the right
A direct port of @ArBo's solution has 35 34 bytes, so go upvote that answer if you like it:
function(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
$begingroup$
The 34 byte one can be reduced by 1 withfunction(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
– MickyT
3 hours ago
add a comment |
$begingroup$
R, 40 bytes (or 34)
function(x,y)any((-1:1*5)%in%c(x+y,x-y))
Try it online!
For non-R users:
-1:1*5
expands to[-5, 0, 5]
- the
%in%
operator takes elements from the left and checks (element-wise) if they exist in the vector on the right
A direct port of @ArBo's solution has 35 34 bytes, so go upvote that answer if you like it:
function(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
R, 40 bytes (or 34)
function(x,y)any((-1:1*5)%in%c(x+y,x-y))
Try it online!
For non-R users:
-1:1*5
expands to[-5, 0, 5]
- the
%in%
operator takes elements from the left and checks (element-wise) if they exist in the vector on the right
A direct port of @ArBo's solution has 35 34 bytes, so go upvote that answer if you like it:
function(x,y)x%in%c(y--1:1*5,5-y)
edited 3 hours ago
answered 6 hours ago
ngmngm
3,32924
3,32924
$begingroup$
The 34 byte one can be reduced by 1 withfunction(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
– MickyT
3 hours ago
add a comment |
$begingroup$
The 34 byte one can be reduced by 1 withfunction(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
– MickyT
3 hours ago
$begingroup$
The 34 byte one can be reduced by 1 with
function(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
– MickyT
3 hours ago
$begingroup$
The 34 byte one can be reduced by 1 with
function(x,y)x%in%c(y--1:1*5,5-y)
$endgroup$
– MickyT
3 hours ago
add a comment |
$begingroup$
J, 12 11 bytes
1 byte saved thanks to Adám
1#.=+5=|@-,+
Try it online!
Explanation
This is equivalent to:
1 #. = + 5 = |@- , +
This can be divided into the following fork chain:
(= + (5 e. (|@- , +)))
Or, visualized using 5!:4<'f'
:
┌─ =
├─ +
──┤ ┌─ 5
│ ├─ e.
└───┤ ┌─ |
│ ┌─ @ ─┴─ -
└────┼─ ,
└─ +
Annotated:
┌─ = equality
├─ + added to (boolean or)
──┤ ┌─ 5 noun 5
│ ├─ e. is an element of
└───┤ ┌─ | absolute value |
│ ┌─ @ ─┴─ - (of) subtraction |
└────┼─ , paired with |
└─ + addition | any of these?
$endgroup$
$begingroup$
Save a byte withe.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám How so? Shortest approach I got withe.
was=+.5 e.|@-,+
. Maybe you forget5e.
is an invalid token in J?
$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
Since two integers cannot simultaneously sum to 5 and be equal, you can use+
instead of+.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám Ah, I see, thank you.
$endgroup$
– Conor O'Brien
7 hours ago
add a comment |
$begingroup$
J, 12 11 bytes
1 byte saved thanks to Adám
1#.=+5=|@-,+
Try it online!
Explanation
This is equivalent to:
1 #. = + 5 = |@- , +
This can be divided into the following fork chain:
(= + (5 e. (|@- , +)))
Or, visualized using 5!:4<'f'
:
┌─ =
├─ +
──┤ ┌─ 5
│ ├─ e.
└───┤ ┌─ |
│ ┌─ @ ─┴─ -
└────┼─ ,
└─ +
Annotated:
┌─ = equality
├─ + added to (boolean or)
──┤ ┌─ 5 noun 5
│ ├─ e. is an element of
└───┤ ┌─ | absolute value |
│ ┌─ @ ─┴─ - (of) subtraction |
└────┼─ , paired with |
└─ + addition | any of these?
$endgroup$
$begingroup$
Save a byte withe.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám How so? Shortest approach I got withe.
was=+.5 e.|@-,+
. Maybe you forget5e.
is an invalid token in J?
$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
Since two integers cannot simultaneously sum to 5 and be equal, you can use+
instead of+.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám Ah, I see, thank you.
$endgroup$
– Conor O'Brien
7 hours ago
add a comment |
$begingroup$
J, 12 11 bytes
1 byte saved thanks to Adám
1#.=+5=|@-,+
Try it online!
Explanation
This is equivalent to:
1 #. = + 5 = |@- , +
This can be divided into the following fork chain:
(= + (5 e. (|@- , +)))
Or, visualized using 5!:4<'f'
:
┌─ =
├─ +
──┤ ┌─ 5
│ ├─ e.
└───┤ ┌─ |
│ ┌─ @ ─┴─ -
└────┼─ ,
└─ +
Annotated:
┌─ = equality
├─ + added to (boolean or)
──┤ ┌─ 5 noun 5
│ ├─ e. is an element of
└───┤ ┌─ | absolute value |
│ ┌─ @ ─┴─ - (of) subtraction |
└────┼─ , paired with |
└─ + addition | any of these?
$endgroup$
J, 12 11 bytes
1 byte saved thanks to Adám
1#.=+5=|@-,+
Try it online!
Explanation
This is equivalent to:
1 #. = + 5 = |@- , +
This can be divided into the following fork chain:
(= + (5 e. (|@- , +)))
Or, visualized using 5!:4<'f'
:
┌─ =
├─ +
──┤ ┌─ 5
│ ├─ e.
└───┤ ┌─ |
│ ┌─ @ ─┴─ -
└────┼─ ,
└─ +
Annotated:
┌─ = equality
├─ + added to (boolean or)
──┤ ┌─ 5 noun 5
│ ├─ e. is an element of
└───┤ ┌─ | absolute value |
│ ┌─ @ ─┴─ - (of) subtraction |
└────┼─ , paired with |
└─ + addition | any of these?
edited 7 hours ago
answered 8 hours ago
Conor O'BrienConor O'Brien
29.2k263162
29.2k263162
$begingroup$
Save a byte withe.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám How so? Shortest approach I got withe.
was=+.5 e.|@-,+
. Maybe you forget5e.
is an invalid token in J?
$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
Since two integers cannot simultaneously sum to 5 and be equal, you can use+
instead of+.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám Ah, I see, thank you.
$endgroup$
– Conor O'Brien
7 hours ago
add a comment |
$begingroup$
Save a byte withe.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám How so? Shortest approach I got withe.
was=+.5 e.|@-,+
. Maybe you forget5e.
is an invalid token in J?
$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
Since two integers cannot simultaneously sum to 5 and be equal, you can use+
instead of+.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám Ah, I see, thank you.
$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
Save a byte with
e.
$endgroup$
– Adám
7 hours ago
$begingroup$
Save a byte with
e.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám How so? Shortest approach I got with
e.
was =+.5 e.|@-,+
. Maybe you forget 5e.
is an invalid token in J?$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
@Adám How so? Shortest approach I got with
e.
was =+.5 e.|@-,+
. Maybe you forget 5e.
is an invalid token in J?$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
Since two integers cannot simultaneously sum to 5 and be equal, you can use
+
instead of +.
$endgroup$
– Adám
7 hours ago
$begingroup$
Since two integers cannot simultaneously sum to 5 and be equal, you can use
+
instead of +.
$endgroup$
– Adám
7 hours ago
$begingroup$
@Adám Ah, I see, thank you.
$endgroup$
– Conor O'Brien
7 hours ago
$begingroup$
@Adám Ah, I see, thank you.
$endgroup$
– Conor O'Brien
7 hours ago
add a comment |
$begingroup$
Dyalog APL, 9 bytes
=∨5∊+,∘|-
Try it online!
Spelled out:
= ∨ 5 ∊ + , ∘ | -
equal or 5 found in an array of sum and absolute of difference.
$endgroup$
add a comment |
$begingroup$
Dyalog APL, 9 bytes
=∨5∊+,∘|-
Try it online!
Spelled out:
= ∨ 5 ∊ + , ∘ | -
equal or 5 found in an array of sum and absolute of difference.
$endgroup$
add a comment |
$begingroup$
Dyalog APL, 9 bytes
=∨5∊+,∘|-
Try it online!
Spelled out:
= ∨ 5 ∊ + , ∘ | -
equal or 5 found in an array of sum and absolute of difference.
$endgroup$
Dyalog APL, 9 bytes
=∨5∊+,∘|-
Try it online!
Spelled out:
= ∨ 5 ∊ + , ∘ | -
equal or 5 found in an array of sum and absolute of difference.
edited 3 hours ago
answered 8 hours ago
dzaimadzaima
14.6k21755
14.6k21755
add a comment |
add a comment |
$begingroup$
PowerShell, 48 44 bytes
param($a,$b)5-in($x=$b-$a),-$x,($a+$b)-or!$x
Try it online!
Takes input $a
and $b
. Checks if 5
is -in
the group $b-$a
, -$x
($a-$b
), or $a+$b
, stores the first into $x
, and -or
s the -in
check with !$x
to check equality.
-4 bytes thanks to mazzy.
$endgroup$
$begingroup$
($a-$b)
is-$x
:)
$endgroup$
– mazzy
7 hours ago
$begingroup$
@mazzy Ooo, good call.
$endgroup$
– AdmBorkBork
7 hours ago
add a comment |
$begingroup$
PowerShell, 48 44 bytes
param($a,$b)5-in($x=$b-$a),-$x,($a+$b)-or!$x
Try it online!
Takes input $a
and $b
. Checks if 5
is -in
the group $b-$a
, -$x
($a-$b
), or $a+$b
, stores the first into $x
, and -or
s the -in
check with !$x
to check equality.
-4 bytes thanks to mazzy.
$endgroup$
$begingroup$
($a-$b)
is-$x
:)
$endgroup$
– mazzy
7 hours ago
$begingroup$
@mazzy Ooo, good call.
$endgroup$
– AdmBorkBork
7 hours ago
add a comment |
$begingroup$
PowerShell, 48 44 bytes
param($a,$b)5-in($x=$b-$a),-$x,($a+$b)-or!$x
Try it online!
Takes input $a
and $b
. Checks if 5
is -in
the group $b-$a
, -$x
($a-$b
), or $a+$b
, stores the first into $x
, and -or
s the -in
check with !$x
to check equality.
-4 bytes thanks to mazzy.
$endgroup$
PowerShell, 48 44 bytes
param($a,$b)5-in($x=$b-$a),-$x,($a+$b)-or!$x
Try it online!
Takes input $a
and $b
. Checks if 5
is -in
the group $b-$a
, -$x
($a-$b
), or $a+$b
, stores the first into $x
, and -or
s the -in
check with !$x
to check equality.
-4 bytes thanks to mazzy.
edited 7 hours ago
answered 8 hours ago
AdmBorkBorkAdmBorkBork
26.5k364229
26.5k364229
$begingroup$
($a-$b)
is-$x
:)
$endgroup$
– mazzy
7 hours ago
$begingroup$
@mazzy Ooo, good call.
$endgroup$
– AdmBorkBork
7 hours ago
add a comment |
$begingroup$
($a-$b)
is-$x
:)
$endgroup$
– mazzy
7 hours ago
$begingroup$
@mazzy Ooo, good call.
$endgroup$
– AdmBorkBork
7 hours ago
$begingroup$
($a-$b)
is -$x
:)$endgroup$
– mazzy
7 hours ago
$begingroup$
($a-$b)
is -$x
:)$endgroup$
– mazzy
7 hours ago
$begingroup$
@mazzy Ooo, good call.
$endgroup$
– AdmBorkBork
7 hours ago
$begingroup$
@mazzy Ooo, good call.
$endgroup$
– AdmBorkBork
7 hours ago
add a comment |
$begingroup$
Python 2, 38 bytes
-2 bytes thanks to @DjMcMayhem
lambda a,b:a+b==5or abs(a-b)==5or a==b
Try it online!
$endgroup$
$begingroup$
Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the5
s and theor
s
$endgroup$
– ElPedro
8 hours ago
3
$begingroup$
Actually, the TIO link could be 38 bytes
$endgroup$
– DJMcMayhem♦
8 hours ago
$begingroup$
@ElPedro the function itself was 40 bytes but I used f= in order to be able to call it
$endgroup$
– fəˈnɛtɪk
6 hours ago
1
$begingroup$
@DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example
$endgroup$
– fəˈnɛtɪk
6 hours ago
add a comment |
$begingroup$
Python 2, 38 bytes
-2 bytes thanks to @DjMcMayhem
lambda a,b:a+b==5or abs(a-b)==5or a==b
Try it online!
$endgroup$
$begingroup$
Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the5
s and theor
s
$endgroup$
– ElPedro
8 hours ago
3
$begingroup$
Actually, the TIO link could be 38 bytes
$endgroup$
– DJMcMayhem♦
8 hours ago
$begingroup$
@ElPedro the function itself was 40 bytes but I used f= in order to be able to call it
$endgroup$
– fəˈnɛtɪk
6 hours ago
1
$begingroup$
@DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example
$endgroup$
– fəˈnɛtɪk
6 hours ago
add a comment |
$begingroup$
Python 2, 38 bytes
-2 bytes thanks to @DjMcMayhem
lambda a,b:a+b==5or abs(a-b)==5or a==b
Try it online!
$endgroup$
Python 2, 38 bytes
-2 bytes thanks to @DjMcMayhem
lambda a,b:a+b==5or abs(a-b)==5or a==b
Try it online!
edited 6 hours ago
answered 8 hours ago
fəˈnɛtɪkfəˈnɛtɪk
3,6431637
3,6431637
$begingroup$
Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the5
s and theor
s
$endgroup$
– ElPedro
8 hours ago
3
$begingroup$
Actually, the TIO link could be 38 bytes
$endgroup$
– DJMcMayhem♦
8 hours ago
$begingroup$
@ElPedro the function itself was 40 bytes but I used f= in order to be able to call it
$endgroup$
– fəˈnɛtɪk
6 hours ago
1
$begingroup$
@DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example
$endgroup$
– fəˈnɛtɪk
6 hours ago
add a comment |
$begingroup$
Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the5
s and theor
s
$endgroup$
– ElPedro
8 hours ago
3
$begingroup$
Actually, the TIO link could be 38 bytes
$endgroup$
– DJMcMayhem♦
8 hours ago
$begingroup$
@ElPedro the function itself was 40 bytes but I used f= in order to be able to call it
$endgroup$
– fəˈnɛtɪk
6 hours ago
1
$begingroup$
@DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example
$endgroup$
– fəˈnɛtɪk
6 hours ago
$begingroup$
Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the
5
s and the or
s$endgroup$
– ElPedro
8 hours ago
$begingroup$
Your TIO is actually 42 bytes but you can fix it by deleting the spaces between the
5
s and the or
s$endgroup$
– ElPedro
8 hours ago
3
3
$begingroup$
Actually, the TIO link could be 38 bytes
$endgroup$
– DJMcMayhem♦
8 hours ago
$begingroup$
Actually, the TIO link could be 38 bytes
$endgroup$
– DJMcMayhem♦
8 hours ago
$begingroup$
@ElPedro the function itself was 40 bytes but I used f= in order to be able to call it
$endgroup$
– fəˈnɛtɪk
6 hours ago
$begingroup$
@ElPedro the function itself was 40 bytes but I used f= in order to be able to call it
$endgroup$
– fəˈnɛtɪk
6 hours ago
1
1
$begingroup$
@DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example
$endgroup$
– fəˈnɛtɪk
6 hours ago
$begingroup$
@DJMcMayhem I don't normally golf in python. I just did it because the question asker used python for their example
$endgroup$
– fəˈnɛtɪk
6 hours ago
add a comment |
$begingroup$
C# (.NET Core), 43, 48, 47, 33 bytes
EDIT: Tried to use % and apparently forgot how to %. Thanks to Arnauld for pointing that out!
EDIT2: AdmBorkBork with a -1 byte golf rearranging the parentheses to sit next to the return so no additional space is needed!
EDIT3: Thanks to dana for -14 byte golf for the one-line return shortcut and currying the function (Ty Embodiment of Ignorance for linking to TIO).
C# (.NET Core), 33 bytes
a=>b=>a==b|a+b==5|(a-b)*(a-b)==25
Try it online!
$endgroup$
$begingroup$
Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D
$endgroup$
– Destroigo
8 hours ago
1
$begingroup$
You can get it down to 33 bytes applying dana's tips
$endgroup$
– Embodiment of Ignorance
6 hours ago
add a comment |
$begingroup$
C# (.NET Core), 43, 48, 47, 33 bytes
EDIT: Tried to use % and apparently forgot how to %. Thanks to Arnauld for pointing that out!
EDIT2: AdmBorkBork with a -1 byte golf rearranging the parentheses to sit next to the return so no additional space is needed!
EDIT3: Thanks to dana for -14 byte golf for the one-line return shortcut and currying the function (Ty Embodiment of Ignorance for linking to TIO).
C# (.NET Core), 33 bytes
a=>b=>a==b|a+b==5|(a-b)*(a-b)==25
Try it online!
$endgroup$
$begingroup$
Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D
$endgroup$
– Destroigo
8 hours ago
1
$begingroup$
You can get it down to 33 bytes applying dana's tips
$endgroup$
– Embodiment of Ignorance
6 hours ago
add a comment |
$begingroup$
C# (.NET Core), 43, 48, 47, 33 bytes
EDIT: Tried to use % and apparently forgot how to %. Thanks to Arnauld for pointing that out!
EDIT2: AdmBorkBork with a -1 byte golf rearranging the parentheses to sit next to the return so no additional space is needed!
EDIT3: Thanks to dana for -14 byte golf for the one-line return shortcut and currying the function (Ty Embodiment of Ignorance for linking to TIO).
C# (.NET Core), 33 bytes
a=>b=>a==b|a+b==5|(a-b)*(a-b)==25
Try it online!
$endgroup$
C# (.NET Core), 43, 48, 47, 33 bytes
EDIT: Tried to use % and apparently forgot how to %. Thanks to Arnauld for pointing that out!
EDIT2: AdmBorkBork with a -1 byte golf rearranging the parentheses to sit next to the return so no additional space is needed!
EDIT3: Thanks to dana for -14 byte golf for the one-line return shortcut and currying the function (Ty Embodiment of Ignorance for linking to TIO).
C# (.NET Core), 33 bytes
a=>b=>a==b|a+b==5|(a-b)*(a-b)==25
Try it online!
edited 5 hours ago
answered 8 hours ago
DestroigoDestroigo
1815
1815
$begingroup$
Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D
$endgroup$
– Destroigo
8 hours ago
1
$begingroup$
You can get it down to 33 bytes applying dana's tips
$endgroup$
– Embodiment of Ignorance
6 hours ago
add a comment |
$begingroup$
Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D
$endgroup$
– Destroigo
8 hours ago
1
$begingroup$
You can get it down to 33 bytes applying dana's tips
$endgroup$
– Embodiment of Ignorance
6 hours ago
$begingroup$
Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D
$endgroup$
– Destroigo
8 hours ago
$begingroup$
Bah. Trying to avoid System.Math. Back to it! Thanks for pointing that out :D
$endgroup$
– Destroigo
8 hours ago
1
1
$begingroup$
You can get it down to 33 bytes applying dana's tips
$endgroup$
– Embodiment of Ignorance
6 hours ago
$begingroup$
You can get it down to 33 bytes applying dana's tips
$endgroup$
– Embodiment of Ignorance
6 hours ago
add a comment |
$begingroup$
x86 machine code, 39 bytes
00000000: 6a01 5e6a 055f 5251 31c0 39d1 0f44 c601 j.^j._RQ1.9..D..
00000010: d139 cf0f 44c6 595a 29d1 83f9 050f 44c6 .9..D.YZ).....D.
00000020: 83f9 fb0f 44c6 c3 ....D..
Assembly
section .text
global func
func: ;inputs int32_t ecx and edx
push 0x1
pop esi
push 0x5
pop edi
push edx
push ecx
xor eax, eax
;ecx==edx?
cmp ecx, edx
cmove eax, esi
;ecx+edx==5?
add ecx, edx
cmp edi, ecx
cmove eax, esi
;ecx-edx==5?
pop ecx
pop edx
sub ecx, edx
cmp ecx, 5
;ecx-edx==-5?
cmove eax, esi
cmp ecx, -5
cmove eax, esi
ret
Try it online!
$endgroup$
add a comment |
$begingroup$
x86 machine code, 39 bytes
00000000: 6a01 5e6a 055f 5251 31c0 39d1 0f44 c601 j.^j._RQ1.9..D..
00000010: d139 cf0f 44c6 595a 29d1 83f9 050f 44c6 .9..D.YZ).....D.
00000020: 83f9 fb0f 44c6 c3 ....D..
Assembly
section .text
global func
func: ;inputs int32_t ecx and edx
push 0x1
pop esi
push 0x5
pop edi
push edx
push ecx
xor eax, eax
;ecx==edx?
cmp ecx, edx
cmove eax, esi
;ecx+edx==5?
add ecx, edx
cmp edi, ecx
cmove eax, esi
;ecx-edx==5?
pop ecx
pop edx
sub ecx, edx
cmp ecx, 5
;ecx-edx==-5?
cmove eax, esi
cmp ecx, -5
cmove eax, esi
ret
Try it online!
$endgroup$
add a comment |
$begingroup$
x86 machine code, 39 bytes
00000000: 6a01 5e6a 055f 5251 31c0 39d1 0f44 c601 j.^j._RQ1.9..D..
00000010: d139 cf0f 44c6 595a 29d1 83f9 050f 44c6 .9..D.YZ).....D.
00000020: 83f9 fb0f 44c6 c3 ....D..
Assembly
section .text
global func
func: ;inputs int32_t ecx and edx
push 0x1
pop esi
push 0x5
pop edi
push edx
push ecx
xor eax, eax
;ecx==edx?
cmp ecx, edx
cmove eax, esi
;ecx+edx==5?
add ecx, edx
cmp edi, ecx
cmove eax, esi
;ecx-edx==5?
pop ecx
pop edx
sub ecx, edx
cmp ecx, 5
;ecx-edx==-5?
cmove eax, esi
cmp ecx, -5
cmove eax, esi
ret
Try it online!
$endgroup$
x86 machine code, 39 bytes
00000000: 6a01 5e6a 055f 5251 31c0 39d1 0f44 c601 j.^j._RQ1.9..D..
00000010: d139 cf0f 44c6 595a 29d1 83f9 050f 44c6 .9..D.YZ).....D.
00000020: 83f9 fb0f 44c6 c3 ....D..
Assembly
section .text
global func
func: ;inputs int32_t ecx and edx
push 0x1
pop esi
push 0x5
pop edi
push edx
push ecx
xor eax, eax
;ecx==edx?
cmp ecx, edx
cmove eax, esi
;ecx+edx==5?
add ecx, edx
cmp edi, ecx
cmove eax, esi
;ecx-edx==5?
pop ecx
pop edx
sub ecx, edx
cmp ecx, 5
;ecx-edx==-5?
cmove eax, esi
cmp ecx, -5
cmove eax, esi
ret
Try it online!
answered 5 hours ago
LogernLogern
75546
75546
add a comment |
add a comment |
$begingroup$
Jelly, 7 bytes
+,ạ5eo=
Try it online!
How it works
+,ạ5eo= Main link. Arguments: x, y (integers)
+ Yield x+y.
ạ Yield |x-y|.
, Pair; yield (x+y, |x-y|).
5e Test fi 5 exists in the pair.
= Test x and y for equality.
o Logical OR.
$endgroup$
add a comment |
$begingroup$
Jelly, 7 bytes
+,ạ5eo=
Try it online!
How it works
+,ạ5eo= Main link. Arguments: x, y (integers)
+ Yield x+y.
ạ Yield |x-y|.
, Pair; yield (x+y, |x-y|).
5e Test fi 5 exists in the pair.
= Test x and y for equality.
o Logical OR.
$endgroup$
add a comment |
$begingroup$
Jelly, 7 bytes
+,ạ5eo=
Try it online!
How it works
+,ạ5eo= Main link. Arguments: x, y (integers)
+ Yield x+y.
ạ Yield |x-y|.
, Pair; yield (x+y, |x-y|).
5e Test fi 5 exists in the pair.
= Test x and y for equality.
o Logical OR.
$endgroup$
Jelly, 7 bytes
+,ạ5eo=
Try it online!
How it works
+,ạ5eo= Main link. Arguments: x, y (integers)
+ Yield x+y.
ạ Yield |x-y|.
, Pair; yield (x+y, |x-y|).
5e Test fi 5 exists in the pair.
= Test x and y for equality.
o Logical OR.
answered 8 hours ago
Dennis♦Dennis
187k32297736
187k32297736
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
!{#,5-#,#-5,#+5}~FreeQ~#2&
Try it online!
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
!{#,5-#,#-5,#+5}~FreeQ~#2&
Try it online!
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 26 bytes
!{#,5-#,#-5,#+5}~FreeQ~#2&
Try it online!
$endgroup$
Wolfram Language (Mathematica), 26 bytes
!{#,5-#,#-5,#+5}~FreeQ~#2&
Try it online!
answered 8 hours ago
alephalphaalephalpha
21.2k32991
21.2k32991
add a comment |
add a comment |
$begingroup$
C (gcc), 41 34 bytes
f(a,b){a=5==abs(a-b)|a+b==5|a==b;}
Try it online!
$endgroup$
1
$begingroup$
Why doesf
returna
? Just some Undefined Behavior?
$endgroup$
– Tyilo
7 hours ago
$begingroup$
@Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value.
$endgroup$
– cleblanc
7 hours ago
$begingroup$
30 bytes Try it online!
$endgroup$
– Logern
6 hours ago
$begingroup$
@Logern Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@ceilingcat Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
|
show 1 more comment
$begingroup$
C (gcc), 41 34 bytes
f(a,b){a=5==abs(a-b)|a+b==5|a==b;}
Try it online!
$endgroup$
1
$begingroup$
Why doesf
returna
? Just some Undefined Behavior?
$endgroup$
– Tyilo
7 hours ago
$begingroup$
@Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value.
$endgroup$
– cleblanc
7 hours ago
$begingroup$
30 bytes Try it online!
$endgroup$
– Logern
6 hours ago
$begingroup$
@Logern Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@ceilingcat Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
|
show 1 more comment
$begingroup$
C (gcc), 41 34 bytes
f(a,b){a=5==abs(a-b)|a+b==5|a==b;}
Try it online!
$endgroup$
C (gcc), 41 34 bytes
f(a,b){a=5==abs(a-b)|a+b==5|a==b;}
Try it online!
edited 7 hours ago
answered 8 hours ago
cleblanccleblanc
3,200316
3,200316
1
$begingroup$
Why doesf
returna
? Just some Undefined Behavior?
$endgroup$
– Tyilo
7 hours ago
$begingroup$
@Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value.
$endgroup$
– cleblanc
7 hours ago
$begingroup$
30 bytes Try it online!
$endgroup$
– Logern
6 hours ago
$begingroup$
@Logern Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@ceilingcat Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
|
show 1 more comment
1
$begingroup$
Why doesf
returna
? Just some Undefined Behavior?
$endgroup$
– Tyilo
7 hours ago
$begingroup$
@Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value.
$endgroup$
– cleblanc
7 hours ago
$begingroup$
30 bytes Try it online!
$endgroup$
– Logern
6 hours ago
$begingroup$
@Logern Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@ceilingcat Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
1
1
$begingroup$
Why does
f
return a
? Just some Undefined Behavior?$endgroup$
– Tyilo
7 hours ago
$begingroup$
Why does
f
return a
? Just some Undefined Behavior?$endgroup$
– Tyilo
7 hours ago
$begingroup$
@Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value.
$endgroup$
– cleblanc
7 hours ago
$begingroup$
@Tyilo Yes, it's implementation specific. So happens the first parameter is stored in the same register as the return value.
$endgroup$
– cleblanc
7 hours ago
$begingroup$
30 bytes Try it online!
$endgroup$
– Logern
6 hours ago
$begingroup$
30 bytes Try it online!
$endgroup$
– Logern
6 hours ago
$begingroup$
@Logern Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@Logern Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@ceilingcat Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@ceilingcat Doesn't work for f(6,1)
$endgroup$
– cleblanc
6 hours ago
|
show 1 more comment
$begingroup$
Scala, 45 bytes
def f(a:Int,b:Int)=a+b==5||(a-b).abs==5||a==b
Try it online!
$endgroup$
add a comment |
$begingroup$
Scala, 45 bytes
def f(a:Int,b:Int)=a+b==5||(a-b).abs==5||a==b
Try it online!
$endgroup$
add a comment |
$begingroup$
Scala, 45 bytes
def f(a:Int,b:Int)=a+b==5||(a-b).abs==5||a==b
Try it online!
$endgroup$
Scala, 45 bytes
def f(a:Int,b:Int)=a+b==5||(a-b).abs==5||a==b
Try it online!
answered 8 hours ago
Xavier GuihotXavier Guihot
2037
2037
add a comment |
add a comment |
$begingroup$
Tcl, 53 bytes
proc P a b {expr abs($a-$b)==5|$a==$b|abs($a+$b)==5}
Try it online!
$endgroup$
$begingroup$
Same byte count: tio.run/##K0nO@f@/oCg/…
$endgroup$
– sergiol
7 hours ago
add a comment |
$begingroup$
Tcl, 53 bytes
proc P a b {expr abs($a-$b)==5|$a==$b|abs($a+$b)==5}
Try it online!
$endgroup$
$begingroup$
Same byte count: tio.run/##K0nO@f@/oCg/…
$endgroup$
– sergiol
7 hours ago
add a comment |
$begingroup$
Tcl, 53 bytes
proc P a b {expr abs($a-$b)==5|$a==$b|abs($a+$b)==5}
Try it online!
$endgroup$
Tcl, 53 bytes
proc P a b {expr abs($a-$b)==5|$a==$b|abs($a+$b)==5}
Try it online!
edited 7 hours ago
answered 8 hours ago
sergiolsergiol
2,5271925
2,5271925
$begingroup$
Same byte count: tio.run/##K0nO@f@/oCg/…
$endgroup$
– sergiol
7 hours ago
add a comment |
$begingroup$
Same byte count: tio.run/##K0nO@f@/oCg/…
$endgroup$
– sergiol
7 hours ago
$begingroup$
Same byte count: tio.run/##K0nO@f@/oCg/…
$endgroup$
– sergiol
7 hours ago
$begingroup$
Same byte count: tio.run/##K0nO@f@/oCg/…
$endgroup$
– sergiol
7 hours ago
add a comment |
$begingroup$
Japt, 13 12 bytes
x ¥5|50ìøUra
Try it or run all test cases
x ¥5|50ìøUra
:Implicit input of array U
x :Reduce by addition
¥5 :Equal to 5?
| :Bitwise OR
50ì :Split 50 to an array of digits
ø :Contains?
Ur : Reduce U
a : By absolute difference
Alternative
50ìø[Ux Ura]
$endgroup$
add a comment |
$begingroup$
Japt, 13 12 bytes
x ¥5|50ìøUra
Try it or run all test cases
x ¥5|50ìøUra
:Implicit input of array U
x :Reduce by addition
¥5 :Equal to 5?
| :Bitwise OR
50ì :Split 50 to an array of digits
ø :Contains?
Ur : Reduce U
a : By absolute difference
Alternative
50ìø[Ux Ura]
$endgroup$
add a comment |
$begingroup$
Japt, 13 12 bytes
x ¥5|50ìøUra
Try it or run all test cases
x ¥5|50ìøUra
:Implicit input of array U
x :Reduce by addition
¥5 :Equal to 5?
| :Bitwise OR
50ì :Split 50 to an array of digits
ø :Contains?
Ur : Reduce U
a : By absolute difference
Alternative
50ìø[Ux Ura]
$endgroup$
Japt, 13 12 bytes
x ¥5|50ìøUra
Try it or run all test cases
x ¥5|50ìøUra
:Implicit input of array U
x :Reduce by addition
¥5 :Equal to 5?
| :Bitwise OR
50ì :Split 50 to an array of digits
ø :Contains?
Ur : Reduce U
a : By absolute difference
Alternative
50ìø[Ux Ura]
edited 7 hours ago
answered 8 hours ago
ShaggyShaggy
19.2k21666
19.2k21666
add a comment |
add a comment |
$begingroup$
Japt, 14 13 bytes
¥VªaU ¥5ª5¥Nx
Try it online!
$endgroup$
add a comment |
$begingroup$
Japt, 14 13 bytes
¥VªaU ¥5ª5¥Nx
Try it online!
$endgroup$
add a comment |
$begingroup$
Japt, 14 13 bytes
¥VªaU ¥5ª5¥Nx
Try it online!
$endgroup$
Japt, 14 13 bytes
¥VªaU ¥5ª5¥Nx
Try it online!
edited 7 hours ago
answered 8 hours ago
OliverOliver
4,7701831
4,7701831
add a comment |
add a comment |
$begingroup$
05AB1E, 13 12 bytes
ÐO5Qs`α5QrËO
Try it online!
Takes input as a list of integers, saving one byte. Thanks @Wisław!
Alternate 12 byte answer
Q¹²α5Q¹²+5QO
Try it online!
This one takes input on separate lines.
$endgroup$
1
$begingroup$
Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial|
?
$endgroup$
– Wisław
7 hours ago
$begingroup$
@Wisław Good point, I updated my answer. Thanks!
$endgroup$
– Cowabunghole
7 hours ago
$begingroup$
I found a 11 bytes alternative:OI`αª5¢IË~Ā
. Input is a list of integers.
$endgroup$
– Wisław
7 hours ago
$begingroup$
OIÆÄ)5QIËM
is 10.
$endgroup$
– Magic Octopus Urn
6 hours ago
1
$begingroup$
@MagicOctopusUrn I'm not sure exactly what the rules are but I think your solution is different enough from mine to submit your own answer, no? Also, unrelated but I've seen your username on this site for a long time but only after typing it out did I realize that it's "Urn", not "Um" :)
$endgroup$
– Cowabunghole
6 hours ago
|
show 2 more comments
$begingroup$
05AB1E, 13 12 bytes
ÐO5Qs`α5QrËO
Try it online!
Takes input as a list of integers, saving one byte. Thanks @Wisław!
Alternate 12 byte answer
Q¹²α5Q¹²+5QO
Try it online!
This one takes input on separate lines.
$endgroup$
1
$begingroup$
Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial|
?
$endgroup$
– Wisław
7 hours ago
$begingroup$
@Wisław Good point, I updated my answer. Thanks!
$endgroup$
– Cowabunghole
7 hours ago
$begingroup$
I found a 11 bytes alternative:OI`αª5¢IË~Ā
. Input is a list of integers.
$endgroup$
– Wisław
7 hours ago
$begingroup$
OIÆÄ)5QIËM
is 10.
$endgroup$
– Magic Octopus Urn
6 hours ago
1
$begingroup$
@MagicOctopusUrn I'm not sure exactly what the rules are but I think your solution is different enough from mine to submit your own answer, no? Also, unrelated but I've seen your username on this site for a long time but only after typing it out did I realize that it's "Urn", not "Um" :)
$endgroup$
– Cowabunghole
6 hours ago
|
show 2 more comments
$begingroup$
05AB1E, 13 12 bytes
ÐO5Qs`α5QrËO
Try it online!
Takes input as a list of integers, saving one byte. Thanks @Wisław!
Alternate 12 byte answer
Q¹²α5Q¹²+5QO
Try it online!
This one takes input on separate lines.
$endgroup$
05AB1E, 13 12 bytes
ÐO5Qs`α5QrËO
Try it online!
Takes input as a list of integers, saving one byte. Thanks @Wisław!
Alternate 12 byte answer
Q¹²α5Q¹²+5QO
Try it online!
This one takes input on separate lines.
edited 7 hours ago
answered 7 hours ago
CowabungholeCowabunghole
1,075419
1,075419
1
$begingroup$
Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial|
?
$endgroup$
– Wisław
7 hours ago
$begingroup$
@Wisław Good point, I updated my answer. Thanks!
$endgroup$
– Cowabunghole
7 hours ago
$begingroup$
I found a 11 bytes alternative:OI`αª5¢IË~Ā
. Input is a list of integers.
$endgroup$
– Wisław
7 hours ago
$begingroup$
OIÆÄ)5QIËM
is 10.
$endgroup$
– Magic Octopus Urn
6 hours ago
1
$begingroup$
@MagicOctopusUrn I'm not sure exactly what the rules are but I think your solution is different enough from mine to submit your own answer, no? Also, unrelated but I've seen your username on this site for a long time but only after typing it out did I realize that it's "Urn", not "Um" :)
$endgroup$
– Cowabunghole
6 hours ago
|
show 2 more comments
1
$begingroup$
Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial|
?
$endgroup$
– Wisław
7 hours ago
$begingroup$
@Wisław Good point, I updated my answer. Thanks!
$endgroup$
– Cowabunghole
7 hours ago
$begingroup$
I found a 11 bytes alternative:OI`αª5¢IË~Ā
. Input is a list of integers.
$endgroup$
– Wisław
7 hours ago
$begingroup$
OIÆÄ)5QIËM
is 10.
$endgroup$
– Magic Octopus Urn
6 hours ago
1
$begingroup$
@MagicOctopusUrn I'm not sure exactly what the rules are but I think your solution is different enough from mine to submit your own answer, no? Also, unrelated but I've seen your username on this site for a long time but only after typing it out did I realize that it's "Urn", not "Um" :)
$endgroup$
– Cowabunghole
6 hours ago
1
1
$begingroup$
Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial
|
?$endgroup$
– Wisław
7 hours ago
$begingroup$
Since it is not very clearly specified, can you not assume the input is a list of integers, thus eliminating the initial
|
?$endgroup$
– Wisław
7 hours ago
$begingroup$
@Wisław Good point, I updated my answer. Thanks!
$endgroup$
– Cowabunghole
7 hours ago
$begingroup$
@Wisław Good point, I updated my answer. Thanks!
$endgroup$
– Cowabunghole
7 hours ago
$begingroup$
I found a 11 bytes alternative:
OI`αª5¢IË~Ā
. Input is a list of integers.$endgroup$
– Wisław
7 hours ago
$begingroup$
I found a 11 bytes alternative:
OI`αª5¢IË~Ā
. Input is a list of integers.$endgroup$
– Wisław
7 hours ago
$begingroup$
OIÆÄ)5QIËM
is 10.$endgroup$
– Magic Octopus Urn
6 hours ago
$begingroup$
OIÆÄ)5QIËM
is 10.$endgroup$
– Magic Octopus Urn
6 hours ago
1
1
$begingroup$
@MagicOctopusUrn I'm not sure exactly what the rules are but I think your solution is different enough from mine to submit your own answer, no? Also, unrelated but I've seen your username on this site for a long time but only after typing it out did I realize that it's "Urn", not "Um" :)
$endgroup$
– Cowabunghole
6 hours ago
$begingroup$
@MagicOctopusUrn I'm not sure exactly what the rules are but I think your solution is different enough from mine to submit your own answer, no? Also, unrelated but I've seen your username on this site for a long time but only after typing it out did I realize that it's "Urn", not "Um" :)
$endgroup$
– Cowabunghole
6 hours ago
|
show 2 more comments
$begingroup$
Batch, 81 bytes
@set/as=%1+%2,d=%1-%2
@if %d% neq 0 if %d:-=% neq 5 if %s% neq 5 exit/b
@echo 1
Takes input as command-line arguments and outputs 1 on success, nothing on failure. Batch can't easily do disjunctions so I use De Morgan's laws to turn it into a conjunction.
$endgroup$
add a comment |
$begingroup$
Batch, 81 bytes
@set/as=%1+%2,d=%1-%2
@if %d% neq 0 if %d:-=% neq 5 if %s% neq 5 exit/b
@echo 1
Takes input as command-line arguments and outputs 1 on success, nothing on failure. Batch can't easily do disjunctions so I use De Morgan's laws to turn it into a conjunction.
$endgroup$
add a comment |
$begingroup$
Batch, 81 bytes
@set/as=%1+%2,d=%1-%2
@if %d% neq 0 if %d:-=% neq 5 if %s% neq 5 exit/b
@echo 1
Takes input as command-line arguments and outputs 1 on success, nothing on failure. Batch can't easily do disjunctions so I use De Morgan's laws to turn it into a conjunction.
$endgroup$
Batch, 81 bytes
@set/as=%1+%2,d=%1-%2
@if %d% neq 0 if %d:-=% neq 5 if %s% neq 5 exit/b
@echo 1
Takes input as command-line arguments and outputs 1 on success, nothing on failure. Batch can't easily do disjunctions so I use De Morgan's laws to turn it into a conjunction.
answered 6 hours ago
NeilNeil
79.8k744177
79.8k744177
add a comment |
add a comment |
$begingroup$
05AB1E, 10 bytes
OIÆ‚Ä50SåZ
Try it online!
O # Sum the input.
IÆ # Reduced subtraction of the input.
‚ # Wrap [sum,reduced_subtraction]
Ä # abs[sum,red_sub]
50S # [5,0]
å # [5,0] in abs[sum,red_sub]?
Z # Max of result, 0 is false, 1 is true.
Tried to do it using stack-only operations, but it was longer.
$endgroup$
$begingroup$
This will unfortunately return true if the sum is0
such as for[5, -5]
$endgroup$
– Emigna
4 hours ago
add a comment |
$begingroup$
05AB1E, 10 bytes
OIÆ‚Ä50SåZ
Try it online!
O # Sum the input.
IÆ # Reduced subtraction of the input.
‚ # Wrap [sum,reduced_subtraction]
Ä # abs[sum,red_sub]
50S # [5,0]
å # [5,0] in abs[sum,red_sub]?
Z # Max of result, 0 is false, 1 is true.
Tried to do it using stack-only operations, but it was longer.
$endgroup$
$begingroup$
This will unfortunately return true if the sum is0
such as for[5, -5]
$endgroup$
– Emigna
4 hours ago
add a comment |
$begingroup$
05AB1E, 10 bytes
OIÆ‚Ä50SåZ
Try it online!
O # Sum the input.
IÆ # Reduced subtraction of the input.
‚ # Wrap [sum,reduced_subtraction]
Ä # abs[sum,red_sub]
50S # [5,0]
å # [5,0] in abs[sum,red_sub]?
Z # Max of result, 0 is false, 1 is true.
Tried to do it using stack-only operations, but it was longer.
$endgroup$
05AB1E, 10 bytes
OIÆ‚Ä50SåZ
Try it online!
O # Sum the input.
IÆ # Reduced subtraction of the input.
‚ # Wrap [sum,reduced_subtraction]
Ä # abs[sum,red_sub]
50S # [5,0]
å # [5,0] in abs[sum,red_sub]?
Z # Max of result, 0 is false, 1 is true.
Tried to do it using stack-only operations, but it was longer.
answered 6 hours ago
Magic Octopus UrnMagic Octopus Urn
12.5k444125
12.5k444125
$begingroup$
This will unfortunately return true if the sum is0
such as for[5, -5]
$endgroup$
– Emigna
4 hours ago
add a comment |
$begingroup$
This will unfortunately return true if the sum is0
such as for[5, -5]
$endgroup$
– Emigna
4 hours ago
$begingroup$
This will unfortunately return true if the sum is
0
such as for [5, -5]
$endgroup$
– Emigna
4 hours ago
$begingroup$
This will unfortunately return true if the sum is
0
such as for [5, -5]
$endgroup$
– Emigna
4 hours ago
add a comment |
$begingroup$
Charcoal, 18 bytes
Nθ¿№⟦θ⁺⁵θ⁻⁵θ⁻θ⁵⟧N1
Try it online! Link is to verbose version of code. Port of @ArBo's Python 2 solution.
$endgroup$
add a comment |
$begingroup$
Charcoal, 18 bytes
Nθ¿№⟦θ⁺⁵θ⁻⁵θ⁻θ⁵⟧N1
Try it online! Link is to verbose version of code. Port of @ArBo's Python 2 solution.
$endgroup$
add a comment |
$begingroup$
Charcoal, 18 bytes
Nθ¿№⟦θ⁺⁵θ⁻⁵θ⁻θ⁵⟧N1
Try it online! Link is to verbose version of code. Port of @ArBo's Python 2 solution.
$endgroup$
Charcoal, 18 bytes
Nθ¿№⟦θ⁺⁵θ⁻⁵θ⁻θ⁵⟧N1
Try it online! Link is to verbose version of code. Port of @ArBo's Python 2 solution.
answered 6 hours ago
NeilNeil
79.8k744177
79.8k744177
add a comment |
add a comment |
$begingroup$
Runic Enchantments, 30 bytes
i::i::}3s=?!@-'|A"5"n:}=?!@+=@
Try it online!
With a pending update, 4 bytes can be saved, as presently integer inputs are treated as doubles and don't equal integers of the same value. Hence "5"n
instead of just 5
. This was an oversight and several factors are being adjusted to account for it (such as using Approximately(a, b)
instead of a.Equals(b)
).
Outputs 0
(exactly one zero) for false and any other output (literally whatever is left on the stack) for true.
$endgroup$
add a comment |
$begingroup$
Runic Enchantments, 30 bytes
i::i::}3s=?!@-'|A"5"n:}=?!@+=@
Try it online!
With a pending update, 4 bytes can be saved, as presently integer inputs are treated as doubles and don't equal integers of the same value. Hence "5"n
instead of just 5
. This was an oversight and several factors are being adjusted to account for it (such as using Approximately(a, b)
instead of a.Equals(b)
).
Outputs 0
(exactly one zero) for false and any other output (literally whatever is left on the stack) for true.
$endgroup$
add a comment |
$begingroup$
Runic Enchantments, 30 bytes
i::i::}3s=?!@-'|A"5"n:}=?!@+=@
Try it online!
With a pending update, 4 bytes can be saved, as presently integer inputs are treated as doubles and don't equal integers of the same value. Hence "5"n
instead of just 5
. This was an oversight and several factors are being adjusted to account for it (such as using Approximately(a, b)
instead of a.Equals(b)
).
Outputs 0
(exactly one zero) for false and any other output (literally whatever is left on the stack) for true.
$endgroup$
Runic Enchantments, 30 bytes
i::i::}3s=?!@-'|A"5"n:}=?!@+=@
Try it online!
With a pending update, 4 bytes can be saved, as presently integer inputs are treated as doubles and don't equal integers of the same value. Hence "5"n
instead of just 5
. This was an oversight and several factors are being adjusted to account for it (such as using Approximately(a, b)
instead of a.Equals(b)
).
Outputs 0
(exactly one zero) for false and any other output (literally whatever is left on the stack) for true.
answered 5 hours ago
Draco18sDraco18s
1,261619
1,261619
add a comment |
add a comment |
$begingroup$
Perl 5, 51 + 2 (-an) = 53 bytes
Pretty simple really. Outputs 0 for false, 1 for true. Not gonna lie, I don't know if bitwise OR is appropriate here, but is does work for all the test cases, so that's nice.
($a,$b)=@F;print($a==$b|$a+$b==5|$a-$b==5|$b-$a==5)
Try it online!
$endgroup$
add a comment |
$begingroup$
Perl 5, 51 + 2 (-an) = 53 bytes
Pretty simple really. Outputs 0 for false, 1 for true. Not gonna lie, I don't know if bitwise OR is appropriate here, but is does work for all the test cases, so that's nice.
($a,$b)=@F;print($a==$b|$a+$b==5|$a-$b==5|$b-$a==5)
Try it online!
$endgroup$
add a comment |
$begingroup$
Perl 5, 51 + 2 (-an) = 53 bytes
Pretty simple really. Outputs 0 for false, 1 for true. Not gonna lie, I don't know if bitwise OR is appropriate here, but is does work for all the test cases, so that's nice.
($a,$b)=@F;print($a==$b|$a+$b==5|$a-$b==5|$b-$a==5)
Try it online!
$endgroup$
Perl 5, 51 + 2 (-an) = 53 bytes
Pretty simple really. Outputs 0 for false, 1 for true. Not gonna lie, I don't know if bitwise OR is appropriate here, but is does work for all the test cases, so that's nice.
($a,$b)=@F;print($a==$b|$a+$b==5|$a-$b==5|$b-$a==5)
Try it online!
edited 4 hours ago
answered 5 hours ago
Geoffrey H.Geoffrey H.
414
414
add a comment |
add a comment |
$begingroup$
Java (JDK), 30 bytes
a->b->a+b==5|a==b|(b-=a)*b==25
Try it online!
$endgroup$
add a comment |
$begingroup$
Java (JDK), 30 bytes
a->b->a+b==5|a==b|(b-=a)*b==25
Try it online!
$endgroup$
add a comment |
$begingroup$
Java (JDK), 30 bytes
a->b->a+b==5|a==b|(b-=a)*b==25
Try it online!
$endgroup$
Java (JDK), 30 bytes
a->b->a+b==5|a==b|(b-=a)*b==25
Try it online!
answered 3 hours ago
Olivier GrégoireOlivier Grégoire
8,88511843
8,88511843
add a comment |
add a comment |
$begingroup$
Retina 0.8.2, 82 bytes
d+
$*
^(-?1*) 1$|^(-?1*)1{5} -?2$|^-?(-?1*) (3)1{5}$|^-?(1 ?){5}$|^(1 ?-?){5}$
Try it online! Link includes test cases. Explanation: The first two lines convert the inputs into unary. The final line then checks for any of the permitted matches:
^(-?1*) 1$ x==y
^(-?1*)1{5} -?2$ x>=0 y>=0 x=5+y i.e. x-y=5
x>=0 y<=0 x=5-y i.e. x+y=5
x<=0 y<=0 x=y-5 i.e. y-x=5
^-?(-?1*) (3)1{5}$ x<=0 y<=0 y=x-5 i.e. x-y=5
x<=0 y>=0 y=5-x i.e. x+y=5
x>=0 y>=0 y=5+x i.e. y-x=5
^-?(1 ?){5}$ x>=0 y>=0 y=5-x i.e. x+y=5
x<=0 y>=0 y=5+x i.e. y-x=5
^(1 ?-?){5}$ x>=0 y>=0 x=5-y i.e. x+y=5
x>=0 y<=0 x=5+y i.e. x-y=5
Pivoted by the last column we get:
x==y ^(-?1*) 1$
x+y=5 x>=0 y>=0 ^-?(1 ?){5}$
x>=0 y>=0 ^(1 ?-?){5}$
x>=0 y<=0 ^(-?1*)1{5} -?2$
x<=0 y>=0 ^-?(-?1*) (3)1{5}$
x<=0 y<=0 (impossible)
x-y=5 x>=0 y>=0 ^(-?1*)1{5} -?2$
x>=0 y<=0 ^(1 ?-?){5}$
x<=0 y>=0 (impossible)
x<=0 y<=0 ^-?(-?1*) (3)1{5}$
y-x=5 x>=0 y>=0 ^-?(-?1*) (3)1{5}$
x>=0 y<=0 (impossible)
x<=0 y>=0 ^-?(1 ?){5}$
x<=0 y<=0 ^(-?1*)1{5} -?2$
$endgroup$
add a comment |
$begingroup$
Retina 0.8.2, 82 bytes
d+
$*
^(-?1*) 1$|^(-?1*)1{5} -?2$|^-?(-?1*) (3)1{5}$|^-?(1 ?){5}$|^(1 ?-?){5}$
Try it online! Link includes test cases. Explanation: The first two lines convert the inputs into unary. The final line then checks for any of the permitted matches:
^(-?1*) 1$ x==y
^(-?1*)1{5} -?2$ x>=0 y>=0 x=5+y i.e. x-y=5
x>=0 y<=0 x=5-y i.e. x+y=5
x<=0 y<=0 x=y-5 i.e. y-x=5
^-?(-?1*) (3)1{5}$ x<=0 y<=0 y=x-5 i.e. x-y=5
x<=0 y>=0 y=5-x i.e. x+y=5
x>=0 y>=0 y=5+x i.e. y-x=5
^-?(1 ?){5}$ x>=0 y>=0 y=5-x i.e. x+y=5
x<=0 y>=0 y=5+x i.e. y-x=5
^(1 ?-?){5}$ x>=0 y>=0 x=5-y i.e. x+y=5
x>=0 y<=0 x=5+y i.e. x-y=5
Pivoted by the last column we get:
x==y ^(-?1*) 1$
x+y=5 x>=0 y>=0 ^-?(1 ?){5}$
x>=0 y>=0 ^(1 ?-?){5}$
x>=0 y<=0 ^(-?1*)1{5} -?2$
x<=0 y>=0 ^-?(-?1*) (3)1{5}$
x<=0 y<=0 (impossible)
x-y=5 x>=0 y>=0 ^(-?1*)1{5} -?2$
x>=0 y<=0 ^(1 ?-?){5}$
x<=0 y>=0 (impossible)
x<=0 y<=0 ^-?(-?1*) (3)1{5}$
y-x=5 x>=0 y>=0 ^-?(-?1*) (3)1{5}$
x>=0 y<=0 (impossible)
x<=0 y>=0 ^-?(1 ?){5}$
x<=0 y<=0 ^(-?1*)1{5} -?2$
$endgroup$
add a comment |
$begingroup$
Retina 0.8.2, 82 bytes
d+
$*
^(-?1*) 1$|^(-?1*)1{5} -?2$|^-?(-?1*) (3)1{5}$|^-?(1 ?){5}$|^(1 ?-?){5}$
Try it online! Link includes test cases. Explanation: The first two lines convert the inputs into unary. The final line then checks for any of the permitted matches:
^(-?1*) 1$ x==y
^(-?1*)1{5} -?2$ x>=0 y>=0 x=5+y i.e. x-y=5
x>=0 y<=0 x=5-y i.e. x+y=5
x<=0 y<=0 x=y-5 i.e. y-x=5
^-?(-?1*) (3)1{5}$ x<=0 y<=0 y=x-5 i.e. x-y=5
x<=0 y>=0 y=5-x i.e. x+y=5
x>=0 y>=0 y=5+x i.e. y-x=5
^-?(1 ?){5}$ x>=0 y>=0 y=5-x i.e. x+y=5
x<=0 y>=0 y=5+x i.e. y-x=5
^(1 ?-?){5}$ x>=0 y>=0 x=5-y i.e. x+y=5
x>=0 y<=0 x=5+y i.e. x-y=5
Pivoted by the last column we get:
x==y ^(-?1*) 1$
x+y=5 x>=0 y>=0 ^-?(1 ?){5}$
x>=0 y>=0 ^(1 ?-?){5}$
x>=0 y<=0 ^(-?1*)1{5} -?2$
x<=0 y>=0 ^-?(-?1*) (3)1{5}$
x<=0 y<=0 (impossible)
x-y=5 x>=0 y>=0 ^(-?1*)1{5} -?2$
x>=0 y<=0 ^(1 ?-?){5}$
x<=0 y>=0 (impossible)
x<=0 y<=0 ^-?(-?1*) (3)1{5}$
y-x=5 x>=0 y>=0 ^-?(-?1*) (3)1{5}$
x>=0 y<=0 (impossible)
x<=0 y>=0 ^-?(1 ?){5}$
x<=0 y<=0 ^(-?1*)1{5} -?2$
$endgroup$
Retina 0.8.2, 82 bytes
d+
$*
^(-?1*) 1$|^(-?1*)1{5} -?2$|^-?(-?1*) (3)1{5}$|^-?(1 ?){5}$|^(1 ?-?){5}$
Try it online! Link includes test cases. Explanation: The first two lines convert the inputs into unary. The final line then checks for any of the permitted matches:
^(-?1*) 1$ x==y
^(-?1*)1{5} -?2$ x>=0 y>=0 x=5+y i.e. x-y=5
x>=0 y<=0 x=5-y i.e. x+y=5
x<=0 y<=0 x=y-5 i.e. y-x=5
^-?(-?1*) (3)1{5}$ x<=0 y<=0 y=x-5 i.e. x-y=5
x<=0 y>=0 y=5-x i.e. x+y=5
x>=0 y>=0 y=5+x i.e. y-x=5
^-?(1 ?){5}$ x>=0 y>=0 y=5-x i.e. x+y=5
x<=0 y>=0 y=5+x i.e. y-x=5
^(1 ?-?){5}$ x>=0 y>=0 x=5-y i.e. x+y=5
x>=0 y<=0 x=5+y i.e. x-y=5
Pivoted by the last column we get:
x==y ^(-?1*) 1$
x+y=5 x>=0 y>=0 ^-?(1 ?){5}$
x>=0 y>=0 ^(1 ?-?){5}$
x>=0 y<=0 ^(-?1*)1{5} -?2$
x<=0 y>=0 ^-?(-?1*) (3)1{5}$
x<=0 y<=0 (impossible)
x-y=5 x>=0 y>=0 ^(-?1*)1{5} -?2$
x>=0 y<=0 ^(1 ?-?){5}$
x<=0 y>=0 (impossible)
x<=0 y<=0 ^-?(-?1*) (3)1{5}$
y-x=5 x>=0 y>=0 ^-?(-?1*) (3)1{5}$
x>=0 y<=0 (impossible)
x<=0 y>=0 ^-?(1 ?){5}$
x<=0 y<=0 ^(-?1*)1{5} -?2$
answered 14 mins ago
NeilNeil
79.8k744177
79.8k744177
add a comment |
add a comment |
Vikrant Biswas is a new contributor. Be nice, and check out our Code of Conduct.
Vikrant Biswas is a new contributor. Be nice, and check out our Code of Conduct.
Vikrant Biswas is a new contributor. Be nice, and check out our Code of Conduct.
Vikrant Biswas is a new contributor. Be nice, and check out our Code of Conduct.
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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1
$begingroup$
Just for info, you can reduce your 56 by 9 by replacing
x=input();y=input()
withx,y=input()
and giving the input as 2 integers separated by a comma.$endgroup$
– ElPedro
8 hours ago
4
$begingroup$
welcome to PPCG! This is a good first challenge -- the challenge is clearly defined, it has ample test cases, and uses our default I/O! If you stick around for a while and keep thinking up interesting challenges, I would recommend using The Sandbox to get feedback before posting them to this site. I hope you enjoy the time you spend here!
$endgroup$
– Giuseppe
8 hours ago
$begingroup$
This would have been very slightly more interesting if the question was whether the absolute value of the addition or subtraction was equal to five, as then the sign of the integers doesn't matter. As it is, if I wanted to write an answer in Retina I'd have to special-case for a sum of
-5
. (And since Retina only really understands positive integers, this isn't easy.)$endgroup$
– Neil
6 hours ago
1
$begingroup$
suggest a test case
6 1 => True
$endgroup$
– cleblanc
6 hours ago
$begingroup$
@Neil based on the recently added test case that is the case
$endgroup$
– Stephen
3 hours ago