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 1 hour 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 1 hour 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
1 hour 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
1 hour 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
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect/Perfectto make it clearer
$endgroup$
– Jo King
1 hour 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 1 hour 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 1 hour ago
Jo KingJo King
24.6k357126
asked 1 hour ago
Jo KingJo King
24.6k357126
edited 13 mins ago
Jo King
asked 1 hour ago
Jo KingJo King
24.6k357126
asked 1 hour ago
Jo KingJo King
24.6k357126
asked 1 hour 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
1 hour 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
1 hour 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
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect/Perfectto make it clearer
$endgroup$
– Jo King
1 hour 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
1 hour 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
1 hour 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
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases toImperfect/Perfectto make it clearer
$endgroup$
– Jo King
1 hour ago
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago
$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect/Perfect to make it clearer$endgroup$
– Jo King
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect/Perfect to make it clearer$endgroup$
– Jo King
1 hour ago
add a comment |
8 Answers
8
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 1 hour 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 1 hour 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
44 mins ago
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
$endgroup$
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 1 hour ago
zeveezevee
22016
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
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 24 mins ago
Neil A.Neil A.
1,278120
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
23 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
18 mins ago
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 28 mins 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 21 mins ago
Unrelated StringUnrelated String
91118
$endgroup$
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 26 secs ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
edited 1 hour ago
answered 1 hour ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
answered 1 hour ago
Luis felipe De jesus MunozLuis felipe De jesus Munoz
5,60821670
answered 1 hour 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 1 hour 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
44 mins 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 1 hour 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
44 mins 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 1 hour 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 1 hour ago
GiuseppeGiuseppe
16.8k31052
answered 1 hour ago
GiuseppeGiuseppe
16.8k31052
answered 1 hour ago
GiuseppeGiuseppe
16.8k31052
answered 1 hour 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
44 mins 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
44 mins 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
44 mins 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
44 mins ago
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
$endgroup$
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
$endgroup$
add a comment |
$begingroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
$endgroup$
CJam, 17 bytes
ri_,(;{1$%!},:+=
Try it online!
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
answered 1 hour ago
Esolanging FruitEsolanging Fruit
8,50932674
8,50932674
add a comment |
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 1 hour ago
zeveezevee
22016
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
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 1 hour ago
zeveezevee
22016
$endgroup$
$begingroup$
thanks! Added that in
$endgroup$
– zevee
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 1 hour ago
zeveezevee
22016
$endgroup$
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 1 hour ago
zeveezevee
22016
edited 1 hour ago
answered 1 hour ago
zeveezevee
22016
answered 1 hour ago
zeveezevee
22016
answered 1 hour ago
zeveezevee
22016
22016
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
add a comment |
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
$begingroup$
thanks! Added that in
$endgroup$
– zevee
1 hour ago
$begingroup$
thanks! Added that in
$endgroup$
– zevee
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 24 mins ago
Neil A.Neil A.
1,278120
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
23 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
18 mins 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 24 mins ago
Neil A.Neil A.
1,278120
$endgroup$
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
23 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
18 mins 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 24 mins ago
Neil A.Neil A.
1,278120
$endgroup$
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 24 mins ago
Neil A.Neil A.
1,278120
edited 23 mins ago
answered 24 mins ago
Neil A.Neil A.
1,278120
answered 24 mins ago
Neil A.Neil A.
1,278120
answered 24 mins ago
Neil A.Neil A.
1,278120
1,278120
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
23 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
18 mins ago
add a comment |
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
23 mins ago
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
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– Jonathan Frech
18 mins ago
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Yeah, that was a typo. I'll fix it now.
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– Neil A.
23 mins ago
$begingroup$
Yeah, that was a typo. I'll fix it now.
$endgroup$
– Neil A.
23 mins ago
1
1
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
18 mins ago
$begingroup$
Using the comprehension condition as a mask for your iteration variable would save a byte.
$endgroup$
– Jonathan Frech
18 mins ago
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 28 mins 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 28 mins 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 28 mins 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 28 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
edited 22 mins ago
answered 28 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
answered 28 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,588124
answered 28 mins 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 21 mins ago
Unrelated StringUnrelated String
91118
$endgroup$
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 21 mins ago
Unrelated StringUnrelated String
91118
$endgroup$
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 21 mins ago
Unrelated StringUnrelated String
91118
$endgroup$
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 21 mins ago
Unrelated StringUnrelated String
91118
answered 21 mins ago
Unrelated StringUnrelated String
91118
answered 21 mins ago
Unrelated StringUnrelated String
91118
answered 21 mins ago
Unrelated StringUnrelated String
91118
91118
add a comment |
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 26 secs ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 26 secs ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 26 secs ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 26 secs ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 26 secs ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 26 secs ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 26 secs ago
Mr. XcoderMr. Xcoder
32.1k759199
32.1k759199
add a comment |
add a comment |
If this is an answer to a challenge…
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…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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$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour 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
1 hour 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
1 hour ago
$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to
Imperfect/Perfectto make it clearer$endgroup$
– Jo King
1 hour ago