Am I not good enough for you?
$begingroup$
Background:
The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.
Challenge
Given a positive integer through any standard input format, distinguish between whether it is perfect or not.
A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.
Test Cases:
Imperfect:
1,12,13,18,20,1000,33550335
Perfect:
6,28,496,8128,33550336,8589869056
Rules
- Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
- Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
code-golf number decision-problem number-theory factoring
asked 2 hours ago
Jo KingJo King
24.6k357126
$endgroup$
add a comment |
$begingroup$
Background:
The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.
Challenge
Given a positive integer through any standard input format, distinguish between whether it is perfect or not.
A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.
Test Cases:
Imperfect:
1,12,13,18,20,1000,33550335
Perfect:
6,28,496,8128,33550336,8589869056
Rules
- Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
- Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
code-golf number decision-problem number-theory factoring
asked 2 hours ago
Jo KingJo King
24.6k357126
$endgroup$
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
2 hours ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect/Perfectto make it clearer
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
Background:
The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.
Challenge
Given a positive integer through any standard input format, distinguish between whether it is perfect or not.
A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.
Test Cases:
Imperfect:
1,12,13,18,20,1000,33550335
Perfect:
6,28,496,8128,33550336,8589869056
Rules
- Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
- Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
code-golf number decision-problem number-theory factoring
asked 2 hours ago
Jo KingJo King
24.6k357126
$endgroup$
Background:
The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely decision-problem repost of the challenge.
Challenge
Given a positive integer through any standard input format, distinguish between whether it is perfect or not.
A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, $6$ is a perfect number, since its divisors are $1,2,3$, which sum up to $6$, while $12$ is not a perfect number since its divisors ( $1,2,3,4,6$ ) sum up to $16$, not $12$.
Test Cases:
Imperfect:
1,12,13,18,20,1000,33550335
Perfect:
6,28,496,8128,33550336,8589869056
Rules
- Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
- Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
code-golf number decision-problem number-theory factoring
code-golf number decision-problem number-theory factoring
asked 2 hours ago
Jo KingJo King
24.6k357126
asked 2 hours ago
Jo KingJo King
24.6k357126
edited 1 hour ago
Jo King
asked 2 hours ago
Jo KingJo King
24.6k357126
asked 2 hours ago
Jo KingJo King
24.6k357126
asked 2 hours ago
Jo KingJo King
24.6k357126
24.6k357126
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
2 hours ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect/Perfectto make it clearer
$endgroup$
– Jo King
2 hours ago
add a comment |
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
2 hours ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect/Perfectto make it clearer
$endgroup$
– Jo King
2 hours ago
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
2 hours ago
$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
2 hours ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect/Perfect to make it clearer$endgroup$
– Jo King
2 hours ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect/Perfect to make it clearer$endgroup$
– Jo King
2 hours ago
add a comment |
9 Answers
9
active
oldest
votes
$begingroup$
Japt -!, 4 bytes
¥â¬x
For some reason ¦ doesnt work on tio so I need to use the -! flag and ¥ instead
Try it online!
answered 2 hours ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
$endgroup$
add a comment |
$begingroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE for perfect numbers ans FALSE for imperfect ones.
answered 2 hours ago
GiuseppeGiuseppe
16.8k31052
$endgroup$
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
1 hour ago
$begingroup$
@CTHall I misread the spec; they originally mapped0(perfect) toFALSEand nonzero toTRUEbut I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumerictological, often in conjunction withwhichor[.
$endgroup$
– Giuseppe
1 hour ago
add a comment |
$begingroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
answered 2 hours ago
zeveezevee
23016
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
2 hours ago
add a comment |
$begingroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
answered 1 hour ago
Neil A.Neil A.
1,288120
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
1 hour ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
1 hour ago
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 54 mins ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 2 hours ago
Esolanging FruitEsolanging Fruit
8,50932674
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
answered 1 hour ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
$endgroup$
add a comment |
$begingroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true. or false. if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
answered 1 hour ago
Unrelated StringUnrelated String
92118
$endgroup$
add a comment |
$begingroup$
Neim, 3 bytes
𝐕𝐬𝔼
Try it online!
(I don't actually know how to run all of the test cases at once, since I started learning Neim about fifteen minutes ago, but I did check them individually.)
Prints 0 for imperfect, 1 for perfect.
𝐕 Pop an int from the stack and push its proper divisors,
implicitly reading the int from a line of input as the otherwise absent top of the stack.
𝐬 Pop a list from the stack and push the sum of the values it contains.
𝔼 Pop two ints from the stack and push 1 if they are equal, 0 if they are not;
implicitly reading the same line of input that was already read as the second int, I guess?
Implicitly print the contents of the stack, or something like that.
answered 48 mins ago
Unrelated StringUnrelated String
92118
$endgroup$
add a comment |
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9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Japt -!, 4 bytes
¥â¬x
For some reason ¦ doesnt work on tio so I need to use the -! flag and ¥ instead
Try it online!
answered 2 hours ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
$endgroup$
add a comment |
$begingroup$
Japt -!, 4 bytes
¥â¬x
For some reason ¦ doesnt work on tio so I need to use the -! flag and ¥ instead
Try it online!
answered 2 hours ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
$endgroup$
add a comment |
$begingroup$
Japt -!, 4 bytes
¥â¬x
For some reason ¦ doesnt work on tio so I need to use the -! flag and ¥ instead
Try it online!
answered 2 hours ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
$endgroup$
Japt -!, 4 bytes
¥â¬x
For some reason ¦ doesnt work on tio so I need to use the -! flag and ¥ instead
Try it online!
answered 2 hours ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
edited 2 hours ago
answered 2 hours ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
answered 2 hours ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
answered 2 hours ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
5,60821670
add a comment |
add a comment |
$begingroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE for perfect numbers ans FALSE for imperfect ones.
answered 2 hours ago
GiuseppeGiuseppe
16.8k31052
$endgroup$
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
1 hour ago
$begingroup$
@CTHall I misread the spec; they originally mapped0(perfect) toFALSEand nonzero toTRUEbut I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumerictological, often in conjunction withwhichor[.
$endgroup$
– Giuseppe
1 hour ago
add a comment |
$begingroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE for perfect numbers ans FALSE for imperfect ones.
answered 2 hours ago
GiuseppeGiuseppe
16.8k31052
$endgroup$
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
1 hour ago
$begingroup$
@CTHall I misread the spec; they originally mapped0(perfect) toFALSEand nonzero toTRUEbut I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumerictological, often in conjunction withwhichor[.
$endgroup$
– Giuseppe
1 hour ago
add a comment |
$begingroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE for perfect numbers ans FALSE for imperfect ones.
answered 2 hours ago
GiuseppeGiuseppe
16.8k31052
$endgroup$
R, 33 bytes
!2*(n=scan())-sum(which(!n%%1:n))
Try it online!
Returns TRUE for perfect numbers ans FALSE for imperfect ones.
answered 2 hours ago
GiuseppeGiuseppe
16.8k31052
answered 2 hours ago
GiuseppeGiuseppe
16.8k31052
answered 2 hours ago
GiuseppeGiuseppe
16.8k31052
answered 2 hours ago
GiuseppeGiuseppe
16.8k31052
16.8k31052
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
1 hour ago
$begingroup$
@CTHall I misread the spec; they originally mapped0(perfect) toFALSEand nonzero toTRUEbut I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumerictological, often in conjunction withwhichor[.
$endgroup$
– Giuseppe
1 hour ago
add a comment |
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
1 hour ago
$begingroup$
@CTHall I misread the spec; they originally mapped0(perfect) toFALSEand nonzero toTRUEbut I removed one of them to reverse the mapping. It's a useful golfing trick to cast fromnumerictological, often in conjunction withwhichor[.
$endgroup$
– Giuseppe
1 hour ago
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
1 hour ago
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
1 hour ago
$begingroup$
@CTHall I misread the spec; they originally mapped
0 (perfect) to FALSE and nonzero to TRUE but I removed one of them to reverse the mapping. It's a useful golfing trick to cast from numeric to logical, often in conjunction with which or [.$endgroup$
– Giuseppe
1 hour ago
$begingroup$
@CTHall I misread the spec; they originally mapped
0 (perfect) to FALSE and nonzero to TRUE but I removed one of them to reverse the mapping. It's a useful golfing trick to cast from numeric to logical, often in conjunction with which or [.$endgroup$
– Giuseppe
1 hour ago
add a comment |
$begingroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
answered 2 hours ago
zeveezevee
23016
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
2 hours ago
add a comment |
$begingroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
answered 2 hours ago
zeveezevee
23016
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
2 hours ago
add a comment |
$begingroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
answered 2 hours ago
zeveezevee
23016
$endgroup$
Javascript, 62
n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)
Explanation (although it's pretty simple)
n=> //return function that takes n
n== //and returns if n is equal to
[...Array(n).keys()] //an array [0..(n-1)]...
.filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
.reduce((a,b)=>a+b) //summed up
Thanks to Jo King for the improvement!
answered 2 hours ago
zeveezevee
23016
edited 2 hours ago
answered 2 hours ago
zeveezevee
23016
answered 2 hours ago
zeveezevee
23016
answered 2 hours ago
zeveezevee
23016
23016
$begingroup$
thanks! Added that in
$endgroup$
– zevee
2 hours ago
add a comment |
$begingroup$
thanks! Added that in
$endgroup$
– zevee
2 hours ago
$begingroup$
thanks! Added that in
$endgroup$
– zevee
2 hours ago
$begingroup$
thanks! Added that in
$endgroup$
– zevee
2 hours ago
add a comment |
$begingroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
answered 1 hour ago
Neil A.Neil A.
1,288120
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
1 hour ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
1 hour ago
add a comment |
$begingroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
answered 1 hour ago
Neil A.Neil A.
1,288120
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
1 hour ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
1 hour ago
add a comment |
$begingroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
answered 1 hour ago
Neil A.Neil A.
1,288120
$endgroup$
Python 3, 46 bytes
lambda x:sum(i for i in range(1,x)if x%i<1)==x
Try it online!
Brute force, sums the factors and checks for equality.
answered 1 hour ago
Neil A.Neil A.
1,288120
edited 1 hour ago
answered 1 hour ago
Neil A.Neil A.
1,288120
answered 1 hour ago
Neil A.Neil A.
1,288120
answered 1 hour ago
Neil A.Neil A.
1,288120
1,288120
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
1 hour ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
1 hour ago
add a comment |
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
1 hour ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
1 hour ago
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
1 hour ago
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
1 hour ago
1
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
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– Jonathan Frech
1 hour ago
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
1 hour ago
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 54 mins ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 54 mins ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 54 mins ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 54 mins ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 54 mins ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 54 mins ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 54 mins ago
Mr. XcoderMr. Xcoder
32.1k759199
32.1k759199
add a comment |
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 2 hours ago
Esolanging FruitEsolanging Fruit
8,50932674
$endgroup$
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 2 hours ago
Esolanging FruitEsolanging Fruit
8,50932674
$endgroup$
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 2 hours ago
Esolanging FruitEsolanging Fruit
8,50932674
$endgroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 2 hours ago
Esolanging FruitEsolanging Fruit
8,50932674
answered 2 hours ago
Esolanging FruitEsolanging Fruit
8,50932674
answered 2 hours ago
Esolanging FruitEsolanging Fruit
8,50932674
answered 2 hours ago
Esolanging FruitEsolanging Fruit
8,50932674
8,50932674
add a comment |
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
answered 1 hour ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
answered 1 hour ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
answered 1 hour ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
$endgroup$
C# (Visual C# Interactive Compiler), 49 47 bytes
n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2
Try it online!
answered 1 hour ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
edited 1 hour ago
answered 1 hour ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
answered 1 hour ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
answered 1 hour ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
1,588124
add a comment |
add a comment |
$begingroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true. or false. if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
answered 1 hour ago
Unrelated StringUnrelated String
92118
$endgroup$
add a comment |
$begingroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true. or false. if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
answered 1 hour ago
Unrelated StringUnrelated String
92118
$endgroup$
add a comment |
$begingroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true. or false. if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
answered 1 hour ago
Unrelated StringUnrelated String
92118
$endgroup$
Brachylog, 4 bytes
fk+?
Try it online!
The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true. or false. if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
answered 1 hour ago
Unrelated StringUnrelated String
92118
answered 1 hour ago
Unrelated StringUnrelated String
92118
answered 1 hour ago
Unrelated StringUnrelated String
92118
answered 1 hour ago
Unrelated StringUnrelated String
92118
92118
add a comment |
add a comment |
$begingroup$
Neim, 3 bytes
𝐕𝐬𝔼
Try it online!
(I don't actually know how to run all of the test cases at once, since I started learning Neim about fifteen minutes ago, but I did check them individually.)
Prints 0 for imperfect, 1 for perfect.
𝐕 Pop an int from the stack and push its proper divisors,
implicitly reading the int from a line of input as the otherwise absent top of the stack.
𝐬 Pop a list from the stack and push the sum of the values it contains.
𝔼 Pop two ints from the stack and push 1 if they are equal, 0 if they are not;
implicitly reading the same line of input that was already read as the second int, I guess?
Implicitly print the contents of the stack, or something like that.
answered 48 mins ago
Unrelated StringUnrelated String
92118
$endgroup$
add a comment |
$begingroup$
Neim, 3 bytes
𝐕𝐬𝔼
Try it online!
(I don't actually know how to run all of the test cases at once, since I started learning Neim about fifteen minutes ago, but I did check them individually.)
Prints 0 for imperfect, 1 for perfect.
𝐕 Pop an int from the stack and push its proper divisors,
implicitly reading the int from a line of input as the otherwise absent top of the stack.
𝐬 Pop a list from the stack and push the sum of the values it contains.
𝔼 Pop two ints from the stack and push 1 if they are equal, 0 if they are not;
implicitly reading the same line of input that was already read as the second int, I guess?
Implicitly print the contents of the stack, or something like that.
answered 48 mins ago
Unrelated StringUnrelated String
92118
$endgroup$
add a comment |
$begingroup$
Neim, 3 bytes
𝐕𝐬𝔼
Try it online!
(I don't actually know how to run all of the test cases at once, since I started learning Neim about fifteen minutes ago, but I did check them individually.)
Prints 0 for imperfect, 1 for perfect.
𝐕 Pop an int from the stack and push its proper divisors,
implicitly reading the int from a line of input as the otherwise absent top of the stack.
𝐬 Pop a list from the stack and push the sum of the values it contains.
𝔼 Pop two ints from the stack and push 1 if they are equal, 0 if they are not;
implicitly reading the same line of input that was already read as the second int, I guess?
Implicitly print the contents of the stack, or something like that.
answered 48 mins ago
Unrelated StringUnrelated String
92118
$endgroup$
Neim, 3 bytes
𝐕𝐬𝔼
Try it online!
(I don't actually know how to run all of the test cases at once, since I started learning Neim about fifteen minutes ago, but I did check them individually.)
Prints 0 for imperfect, 1 for perfect.
𝐕 Pop an int from the stack and push its proper divisors,
implicitly reading the int from a line of input as the otherwise absent top of the stack.
𝐬 Pop a list from the stack and push the sum of the values it contains.
𝔼 Pop two ints from the stack and push 1 if they are equal, 0 if they are not;
implicitly reading the same line of input that was already read as the second int, I guess?
Implicitly print the contents of the stack, or something like that.
answered 48 mins ago
Unrelated StringUnrelated String
92118
answered 48 mins ago
Unrelated StringUnrelated String
92118
answered 48 mins ago
Unrelated StringUnrelated String
92118
answered 48 mins ago
Unrelated StringUnrelated String
92118
92118
add a comment |
add a comment |
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.
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…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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$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
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– Esolanging Fruit
2 hours ago
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@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
2 hours ago
$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
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– Esolanging Fruit
2 hours ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect/Perfectto make it clearer$endgroup$
– Jo King
2 hours ago