Write shortest possible code that will return true if the two given integer values are equal or their sum or absolute 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
-5 5 => False
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
-9, thanks @ElPedro. It takes input in format x,y:
x,y=input();print all([x-y,x+y-5,abs(x-y)-5])<1
lambda a,b:a in(b,5-b,b-5,b+5)
One byte saved by Arnauld
Three bytes saved by alephalpha
Takes input as (a)(b). Returns \$0\$ or \$1\$.
a=>b=>a+b==5|!(a-=b)|a*a==25
=∨5∊+,∘|-
Spelled out:
= ∨ 5 ∊ + , ∘ | -
equal or 5 found in an array of sum and absolute of difference.
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..
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
Edit: fixed bash stderr errors on TIO
Edit 2: improved formatting in TIO header and footer code
+,ạ5eo=
+,ạ5eo= Main link. Arguments: x, y (integers)
+ Yield x+y.
ạ Yield |x-y|.
, Pair; yield (x+y, |x-y|).
5e Test if 5 exists in the pair.
= Test x and y for equality.
o Logical OR.
function(x,y)any((-1:1*5)%in%c(x+y,x-y))
For non-R users:
-1:1*5 expands to [-5, 0, 5]%in% operator takes elements from the left and checks (element-wise) if they exist in the vector on the rightA 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)
function(x,y)x%in%c(y--1:1*5,5-y)
\$\endgroup\$
function(x,y)(x-y)%in%(-1:1*5), and drop it further to 24 bytes by dropping the function notation to scan() input: diff(scan())%in%(-1:1*5) Try it online!. Still very much the same method though.
\$\endgroup\$
Commented
Jan 17, 2019 at 14:38
pryr::f, which happens to work in both cases. Whether it can properly detect the arguments is entirely somewhat hit or miss but it seems to nail these two functions. e.g. pryr::f(x%in%c(y--1:1*5,5-y)) Try it online!. Gets you to 36 and 29 bytes respectively.
\$\endgroup\$
Commented
Jan 18, 2019 at 15:50
lambda a,b:a+b==5or`a-b`in"0-5"
Since I didn't manage to read the task carefully the first time, in order to fix it, I had to come up with a completely different approach, which is unfortunately not as concise.
1 byte saved thanks to Adám
1#.=+5=|@-,+
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?
e. was =+.5 e.|@-,+. Maybe you forget 5e. is an invalid token in J?
\$\endgroup\$
Commented
Jan 16, 2019 at 17:05
+ instead of +.
\$\endgroup\$
-2 bytes thanks to @DjMcMayhem
lambda a,b:a+b==5or abs(a-b)==5or a==b
5s and the ors
\$\endgroup\$
param($a,$b)$b-in($a-5),(5-$a),(5+$a),$a
Try it online! or Verify all Test Cases
Takes input $a and $b. Checks if $b is -in the group ($a-5, 5-$a 5+$a, or $a), which checks all possible combinations of $a,$b, and 5.
-4 bytes thanks to mazzy.
-4 bytes thanks to KGlasier.
5 and $b around you can cut off a couple bytes(ie param($a,$b)$b-in($a-5),(5-$a),($a+5),$a) Try it out here
\$\endgroup\$
$a+5 to 5+$a to get it to cast appropriately when taking command-line input, but otherwise awesome. Thanks!
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Commented
Jan 17, 2019 at 15:55
-1 byte thanks to Grimy
{$^a-$^b==5|0|-5|5-2*$b}
This uses the Any Junction but technically, ^ could work as well.
{ } # Anonymous code block
$^a-$^b== # Is the difference equal to
| | | # Any of
0
5
-5
5-2*$b
{$^a-$^b==5|0|-5|5-2*$b}
\$\endgroup\$
Edit: + input variables.
Procedure z(a,b:integer);begin Writeln((abs(a-b)in[0,5])or(a+b=5))end;
(abs(a-b)in[0,5])or(a+b=5)
I hope that my answer is according to all rules of code-golf. It was fun anyway.
00000000: 8bd0 2bc3 740d 7902 f7d8 3c05 7405 03d3 ..+.t.y...<.t...
00000010: 80fa 05c3 ....
Listing:
8B D0 MOV DX, AX ; Save AX to DX
2B C3 SUB AX, BX ; AX = AX - BX
74 0D JZ DONE ; if 0, they are equal (ZF=1)
79 02 JNS IS_POS ; if positive, check if result is 5
F7 D8 NEG AX ; is negative, negate the result to get abs value
IS_POS:
3C 05 CMP AL, 5 ; is result 5?
74 05 JZ DONE ; if so, exit with ZF=1
03 D3 ADD DX, BX ; DX = DX + BX
80 FA 05 CMP DL, 5 ; ZF = ( DX == 5 )
DONE:
C3 RET ; return to caller
Input numbers in AX and BX and returns Zero Flag (ZF) if result is truthy.
Try it online! (testing code borrowed and adapted from @Logem's answer - thanks!)
Explanation:
If the difference between the numbers is 0, they are equal. Otherwise if result is negative, first negate it (abs value) and check for 5. If still not true, add and check for 5.
Bonus:
If desired, you can also determine which condition was true with the following:
ZF = 1 and DX = 5 ; sum is 5ZF = 1 and AX = 5 ; diff is 5ZF = 1 and AX = 0 ; equalZF = 0 ; falsey
OIÆ‚Ä50SåZ
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.
0 such as for [5, -5]
\$\endgroup\$
OIÆÄ‚5QIËM) is correct for [5,-5].
\$\endgroup\$
Commented
Jan 17, 2019 at 9:31
OsÆÄ‚5åsË~. Almost identical to yours it seems. Try it online!
\$\endgroup\$
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).
a=>b=>a==b|a+b==5|(a-b)*(a-b)==25
f(a,b){a=!(a+b-5&&(a-=b)/6|a%5);}
Tried an approach I didn't see anyone else try using. The return expression is equivalent to a+b==5||((-6<a-b||a-b<6)&&(a-b)%5==0).
|| to |? I know it's possible in Java, C#, Python, or JavaScript, but not sure about Scala.
\$\endgroup\$
Commented
Jan 17, 2019 at 12:15
=|+5|-ȧ5
Takes input as a list of two numbers (use _ for negatives). Try it online!
Pretty much a direct translation of the spec:
= The two numbers are equal
| or
+ The sum of the two numbers
5 is 5
| or
- The difference of the two numbers
ȧ absolute value
5 is 5
f return a? Just some Undefined Behavior?
\$\endgroup\$
ÐO5Qs`α5QrËO
Takes input as a list of integers, saving one byte. Thanks @Wisław!
Q¹²α5Q¹²+5QO
This one takes input on separate lines.
|?
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OI`αª5¢IË~Ā. Input is a list of integers.
\$\endgroup\$
.nil? check to give output in the required format. ->(a,b){[a+5,a-5,5-a,a].index(b).nil?}, this is longer than the current one.
\$\endgroup\$
Commented
Jan 28, 2019 at 5:04
Saved 4 bytes thanks to @chunes!
[ { [ = ] [ - abs 5 = ] [ + 5 = ] } || ]
The repetition of = is annoying, but I couldn't find another way.
|| tries out each of the three quotations in {...} and checks if the result is true for any of them. The first, [ = ], checks if they're equal, the second finds the absolute value of their difference and compares that to 5, and the third compares their sum to 5.
||. Try it online!
\$\endgroup\$
@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.
Ｎθ¿№⟦θ⁺⁵θ⁻⁵θ⁻θ⁵⟧Ｎ1
Try it online! Link is to verbose version of code. Port of @ArBo's Python 2 solution.
\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$
x ¥5|50ìøUra
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
[-5,5] (should be falsey)
\$\endgroup\$
Commented
Jan 17, 2019 at 9:26