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
2 hours 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 1 hour 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 1 hour 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
2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours ago
$begingroup$
What do the 2 !s in a row get you?
$endgroup$
– CT Hall
2 hours 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 1 hour ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 1 hour ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
add a comment |
$begingroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 1 hour ago
Mr. XcoderMr. Xcoder
32.1k759199
$endgroup$
Jelly, 3 bytes
Æṣ=
Try it online!
answered 1 hour ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 1 hour ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 1 hour ago
Mr. XcoderMr. Xcoder
32.1k759199
answered 1 hour 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 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 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 1 hour 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 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 |
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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$begingroup$
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
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– 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