Is it a Cyclops number? “Nobody” knows!
$begingroup$
Task:
Given an integer input, figure out whether or not it is a Cyclops Number.
What is a Cyclops number, you may ask? Well, it's a number whose binary representation only has one 0 in the center!
Test Cases:
Input | Output
--------------
1 | falsy
5 | truthy
12 | falsy
27 | truthy
85 | falsy
101 | falsy
119 | truthy
Input: An integer or equivalent types. (int, long, decimal, etc.)
Output:
- Truthy or falsy (e.g.
trueorfalse,0or1)..
Challenge Rules:
Input that is less than 0 is assumed to be falsy.
If the length of the binary representation of the number is even, then the number cannot be a Cyclops number.
General Rules:
This is code-golf, so the shortest answers in bytes wins!.
Default loopholes are forbidden.
Standard rules apply for your answer with default I/O rules.
This is my first Programming Puzzles & Code Golf challenge, so any feedback on how I should improve would be much appreciated!
code-golf number decision-problem binary
asked 1 hour ago
TauTau
1165
New contributor
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 3 more comments
$begingroup$
Task:
Given an integer input, figure out whether or not it is a Cyclops Number.
What is a Cyclops number, you may ask? Well, it's a number whose binary representation only has one 0 in the center!
Test Cases:
Input | Output
--------------
1 | falsy
5 | truthy
12 | falsy
27 | truthy
85 | falsy
101 | falsy
119 | truthy
Input: An integer or equivalent types. (int, long, decimal, etc.)
Output:
- Truthy or falsy (e.g.
trueorfalse,0or1)..
Challenge Rules:
Input that is less than 0 is assumed to be falsy.
If the length of the binary representation of the number is even, then the number cannot be a Cyclops number.
General Rules:
This is code-golf, so the shortest answers in bytes wins!.
Default loopholes are forbidden.
Standard rules apply for your answer with default I/O rules.
This is my first Programming Puzzles & Code Golf challenge, so any feedback on how I should improve would be much appreciated!
code-golf number decision-problem binary
asked 1 hour ago
TauTau
1165
New contributor
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
$endgroup$
– FryAmTheEggman
1 hour ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
1 hour ago
1
$begingroup$
I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
$endgroup$
– FryAmTheEggman
1 hour ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
1 hour ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
52 mins ago
|
show 3 more comments
$begingroup$
Task:
Given an integer input, figure out whether or not it is a Cyclops Number.
What is a Cyclops number, you may ask? Well, it's a number whose binary representation only has one 0 in the center!
Test Cases:
Input | Output
--------------
1 | falsy
5 | truthy
12 | falsy
27 | truthy
85 | falsy
101 | falsy
119 | truthy
Input: An integer or equivalent types. (int, long, decimal, etc.)
Output:
- Truthy or falsy (e.g.
trueorfalse,0or1)..
Challenge Rules:
Input that is less than 0 is assumed to be falsy.
If the length of the binary representation of the number is even, then the number cannot be a Cyclops number.
General Rules:
This is code-golf, so the shortest answers in bytes wins!.
Default loopholes are forbidden.
Standard rules apply for your answer with default I/O rules.
This is my first Programming Puzzles & Code Golf challenge, so any feedback on how I should improve would be much appreciated!
code-golf number decision-problem binary
asked 1 hour ago
TauTau
1165
New contributor
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Task:
Given an integer input, figure out whether or not it is a Cyclops Number.
What is a Cyclops number, you may ask? Well, it's a number whose binary representation only has one 0 in the center!
Test Cases:
Input | Output
--------------
1 | falsy
5 | truthy
12 | falsy
27 | truthy
85 | falsy
101 | falsy
119 | truthy
Input: An integer or equivalent types. (int, long, decimal, etc.)
Output:
- Truthy or falsy (e.g.
trueorfalse,0or1)..
Challenge Rules:
Input that is less than 0 is assumed to be falsy.
If the length of the binary representation of the number is even, then the number cannot be a Cyclops number.
General Rules:
This is code-golf, so the shortest answers in bytes wins!.
Default loopholes are forbidden.
Standard rules apply for your answer with default I/O rules.
This is my first Programming Puzzles & Code Golf challenge, so any feedback on how I should improve would be much appreciated!
code-golf number decision-problem binary
code-golf number decision-problem binary
asked 1 hour ago
TauTau
1165
New contributor
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
TauTau
1165
New contributor
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 48 mins ago
Tau
asked 1 hour ago
TauTau
1165
New contributor
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
TauTau
1165
asked 1 hour ago
TauTau
1165
1165
New contributor
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tau is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
$endgroup$
– FryAmTheEggman
1 hour ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
1 hour ago
1
$begingroup$
I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
$endgroup$
– FryAmTheEggman
1 hour ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
1 hour ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
52 mins ago
|
show 3 more comments
1
$begingroup$
You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
$endgroup$
– FryAmTheEggman
1 hour ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
1 hour ago
1
$begingroup$
I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
$endgroup$
– FryAmTheEggman
1 hour ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
1 hour ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
52 mins ago
1
1
$begingroup$
You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
$endgroup$
– FryAmTheEggman
1 hour ago
$begingroup$
You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
$endgroup$
– FryAmTheEggman
1 hour ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
1 hour ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
1 hour ago
1
1
$begingroup$
I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
$endgroup$
– FryAmTheEggman
1 hour ago
$begingroup$
I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
$endgroup$
– FryAmTheEggman
1 hour ago
1
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
1 hour ago
$begingroup$
Note: This is A129868
$endgroup$
– tsh
1 hour ago
2
2
$begingroup$
Should
119 be true?$endgroup$
– Quintec
52 mins ago
$begingroup$
Should
119 be true?$endgroup$
– Quintec
52 mins ago
|
show 3 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Japt, 25 19 10 bytes
¢èT ¶1©¢ês
That's more like it...
Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.
Try it online!
answered 1 hour ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
22 mins ago
add a comment |
$begingroup$
Japt, 8 bytes
¤øT ©¢ês
Based off of Quintec's solution.
Try it Online!
Old regex-based solution, 15 bytes
¤f/^(1*)01$/ l
Returns 1 for true, 0 for false.
Try it Online!
answered 50 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
48 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
47 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
38 mins ago
add a comment |
$begingroup$
Perl 6, 23 bytes
{.base(2)~~/^(1*)0$0$/}
Try it online!
Regex based solution
answered 43 mins ago
Jo KingJo King
24.4k357126
$endgroup$
add a comment |
$begingroup$
JavaScript (Node.js), 32 bytes
f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)
Try it online!
31 bytes version
f=(p,q)=>p&1?f(p>>1,q+q|2):p==q
This one will report falsy for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 41 mins ago
tshtsh
9,32511652
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Japt, 25 19 10 bytes
¢èT ¶1©¢ês
That's more like it...
Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.
Try it online!
answered 1 hour ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
22 mins ago
add a comment |
$begingroup$
Japt, 25 19 10 bytes
¢èT ¶1©¢ês
That's more like it...
Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.
Try it online!
answered 1 hour ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
22 mins ago
add a comment |
$begingroup$
Japt, 25 19 10 bytes
¢èT ¶1©¢ês
That's more like it...
Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.
Try it online!
answered 1 hour ago
QuintecQuintec
1,9781726
$endgroup$
Japt, 25 19 10 bytes
¢èT ¶1©¢ês
That's more like it...
Checks if there is only one 0 in the binary representation and that the binary representation is a palindrome.
Try it online!
answered 1 hour ago
QuintecQuintec
1,9781726
edited 41 mins ago
answered 1 hour ago
QuintecQuintec
1,9781726
answered 1 hour ago
QuintecQuintec
1,9781726
answered 1 hour ago
QuintecQuintec
1,9781726
1,9781726
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
22 mins ago
add a comment |
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
22 mins ago
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
22 mins ago
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
22 mins ago
add a comment |
$begingroup$
Japt, 8 bytes
¤øT ©¢ês
Based off of Quintec's solution.
Try it Online!
Old regex-based solution, 15 bytes
¤f/^(1*)01$/ l
Returns 1 for true, 0 for false.
Try it Online!
answered 50 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
48 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
47 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
38 mins ago
add a comment |
$begingroup$
Japt, 8 bytes
¤øT ©¢ês
Based off of Quintec's solution.
Try it Online!
Old regex-based solution, 15 bytes
¤f/^(1*)01$/ l
Returns 1 for true, 0 for false.
Try it Online!
answered 50 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
48 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
47 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
38 mins ago
add a comment |
$begingroup$
Japt, 8 bytes
¤øT ©¢ês
Based off of Quintec's solution.
Try it Online!
Old regex-based solution, 15 bytes
¤f/^(1*)01$/ l
Returns 1 for true, 0 for false.
Try it Online!
answered 50 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
Japt, 8 bytes
¤øT ©¢ês
Based off of Quintec's solution.
Try it Online!
Old regex-based solution, 15 bytes
¤f/^(1*)01$/ l
Returns 1 for true, 0 for false.
Try it Online!
answered 50 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
edited 22 mins ago
answered 50 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
answered 50 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
answered 50 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
1,516123
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
48 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
47 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
38 mins ago
add a comment |
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
48 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
47 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
38 mins ago
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
48 mins ago
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
48 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
47 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
47 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
38 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
38 mins ago
add a comment |
$begingroup$
Perl 6, 23 bytes
{.base(2)~~/^(1*)0$0$/}
Try it online!
Regex based solution
answered 43 mins ago
Jo KingJo King
24.4k357126
$endgroup$
add a comment |
$begingroup$
Perl 6, 23 bytes
{.base(2)~~/^(1*)0$0$/}
Try it online!
Regex based solution
answered 43 mins ago
Jo KingJo King
24.4k357126
$endgroup$
add a comment |
$begingroup$
Perl 6, 23 bytes
{.base(2)~~/^(1*)0$0$/}
Try it online!
Regex based solution
answered 43 mins ago
Jo KingJo King
24.4k357126
$endgroup$
Perl 6, 23 bytes
{.base(2)~~/^(1*)0$0$/}
Try it online!
Regex based solution
answered 43 mins ago
Jo KingJo King
24.4k357126
answered 43 mins ago
Jo KingJo King
24.4k357126
answered 43 mins ago
Jo KingJo King
24.4k357126
answered 43 mins ago
Jo KingJo King
24.4k357126
24.4k357126
add a comment |
add a comment |
$begingroup$
JavaScript (Node.js), 32 bytes
f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)
Try it online!
31 bytes version
f=(p,q)=>p&1?f(p>>1,q+q|2):p==q
This one will report falsy for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 41 mins ago
tshtsh
9,32511652
$endgroup$
add a comment |
$begingroup$
JavaScript (Node.js), 32 bytes
f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)
Try it online!
31 bytes version
f=(p,q)=>p&1?f(p>>1,q+q|2):p==q
This one will report falsy for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 41 mins ago
tshtsh
9,32511652
$endgroup$
add a comment |
$begingroup$
JavaScript (Node.js), 32 bytes
f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)
Try it online!
31 bytes version
f=(p,q)=>p&1?f(p>>1,q+q|2):p==q
This one will report falsy for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 41 mins ago
tshtsh
9,32511652
$endgroup$
JavaScript (Node.js), 32 bytes
f=(p,q)=>p&1?f(p/2,q+q|2):!(p^q)
Try it online!
31 bytes version
f=(p,q)=>p&1?f(p>>1,q+q|2):p==q
This one will report falsy for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 41 mins ago
tshtsh
9,32511652
edited 9 mins ago
answered 41 mins ago
tshtsh
9,32511652
answered 41 mins ago
tshtsh
9,32511652
answered 41 mins ago
tshtsh
9,32511652
9,32511652
add a comment |
add a comment |
Tau is a new contributor. Be nice, and check out our Code of Conduct.
Tau is a new contributor. Be nice, and check out our Code of Conduct.
Tau is a new contributor. Be nice, and check out our Code of Conduct.
Tau is a new contributor. Be nice, and check out our Code of Conduct.
If this is an answer to a challenge…
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1
$begingroup$
You've essentially made two separate challenges, one a decision problem and one about outputting the next number of a sequence. This will not do what you want, which is invite more answers, but instead put off users who now need to consider three options about what to program before posting. I'd recommend removing the option, and in the future you can try posting to our sandbox first where hopefully you will get helpful feedback before posting. Good luck!
$endgroup$
– FryAmTheEggman
1 hour ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
1 hour ago
1
$begingroup$
I'd also recommend removing the hard coding restriction. By default solutions would need to handle all cyclops numbers in theory, and there are clearly infinitely many of them, so they couldn't all be hardcoded.
$endgroup$
– FryAmTheEggman
1 hour ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
1 hour ago
2
$begingroup$
Should
119be true?$endgroup$
– Quintec
52 mins ago