Is it a Cyclops number? “Nobody” knows!Sum of Binary SubstringsIs this a Smith number?Is this number triangular?Test if two numbers are equalIs this number a repdigit?Am I divisible by double the sum of my digits?Is the number binary-heavy?Is it a whole number?Binary SubstringsDiluted Integer Sums
Is this nominative case or accusative case?
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Is it a Cyclops number? “Nobody” knows!
Sum of Binary SubstringsIs this a Smith number?Is this number triangular?Test if two numbers are equalIs this number a repdigit?Am I divisible by double the sum of my digits?Is the number binary-heavy?Is it a whole number?Binary SubstringsDiluted Integer Sums
$begingroup$
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 5 hours ago
TauTau
1365
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 4 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 5 hours ago
TauTau
1365
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
5 hours ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
5 hours ago
$begingroup$
@FryAmTheEggman thank you, I'll keep that in mind. The post has been edited.
$endgroup$
– Tau
5 hours ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
4 hours ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
4 hours ago
|
show 4 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 5 hours ago
TauTau
1365
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 5 hours ago
TauTau
1365
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 5 hours ago
TauTau
1365
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 4 hours ago
Tau
asked 5 hours ago
TauTau
1365
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 5 hours ago
TauTau
1365
asked 5 hours ago
TauTau
1365
1365
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
5 hours ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
5 hours ago
$begingroup$
@FryAmTheEggman thank you, I'll keep that in mind. The post has been edited.
$endgroup$
– Tau
5 hours ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
4 hours ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
4 hours ago
|
show 4 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
5 hours ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
5 hours ago
$begingroup$
@FryAmTheEggman thank you, I'll keep that in mind. The post has been edited.
$endgroup$
– Tau
5 hours ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
4 hours ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
4 hours 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
5 hours 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
5 hours ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
5 hours ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
5 hours ago
$begingroup$
@FryAmTheEggman thank you, I'll keep that in mind. The post has been edited.
$endgroup$
– Tau
5 hours ago
$begingroup$
@FryAmTheEggman thank you, I'll keep that in mind. The post has been edited.
$endgroup$
– Tau
5 hours ago
1
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
4 hours ago
$begingroup$
Note: This is A129868
$endgroup$
– tsh
4 hours ago
2
2
$begingroup$
Should
119 be true?$endgroup$
– Quintec
4 hours ago
$begingroup$
Should
119 be true?$endgroup$
– Quintec
4 hours ago
|
show 4 more comments
8 Answers
8
active
oldest
votes
$begingroup$
Python 2, 30 bytes
lambda n:(2*n^2*n+3)**2==8*n+9
Try it online!
answered 2 hours ago
xnorxnor
91.7k18187443
$endgroup$
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 4 hours ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
4 hours 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 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
4 hours ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
4 hours ago
add a comment |
$begingroup$
Japt -N, 6 bytes
Returns 1 for truthy, 0 or NaN for falsy.
¤q0 äè
Run it online
answered 2 hours ago
OliverOliver
5,2001832
$endgroup$
add a comment |
$begingroup$
Perl 6, 23 bytes
.base(2)~~/^(1*)0$0$/
Try it online!
Regex based solution
answered 4 hours 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 4 hours ago
tshtsh
9,32511652
$endgroup$
add a comment |
$begingroup$
Mathematica (Wolfram language), 32 bytes
OddQ@Log[2,#+Floor@Sqrt[#/2]+2]&
Try it online!
Pure function taking an integer as input and returning True or False. Based on the fact (fun to prove!) that a number n is Cyclops if and only if n plus the square root of n/2 plus 2 rounds down to an odd power of 2. (One can replace Floor by either Ceiling or Round as long as one also replaces +2 by +1.) Returns True on input 0.
answered 2 hours ago
Greg MartinGreg Martin
12.3k21059
$endgroup$
add a comment |
$begingroup$
Jelly, 9 bytes
BŒḂaB¬S=1
Try it online!
answered 1 hour ago
Nick KennedyNick Kennedy
52127
$endgroup$
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
$begingroup$
Python 2, 30 bytes
lambda n:(2*n^2*n+3)**2==8*n+9
Try it online!
answered 2 hours ago
xnorxnor
91.7k18187443
$endgroup$
add a comment |
$begingroup$
Python 2, 30 bytes
lambda n:(2*n^2*n+3)**2==8*n+9
Try it online!
answered 2 hours ago
xnorxnor
91.7k18187443
$endgroup$
add a comment |
$begingroup$
Python 2, 30 bytes
lambda n:(2*n^2*n+3)**2==8*n+9
Try it online!
answered 2 hours ago
xnorxnor
91.7k18187443
$endgroup$
Python 2, 30 bytes
lambda n:(2*n^2*n+3)**2==8*n+9
Try it online!
answered 2 hours ago
xnorxnor
91.7k18187443
answered 2 hours ago
xnorxnor
91.7k18187443
answered 2 hours ago
xnorxnor
91.7k18187443
answered 2 hours ago
xnorxnor
91.7k18187443
91.7k18187443
add a comment |
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 4 hours ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
4 hours 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 4 hours ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
4 hours 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 4 hours 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 4 hours ago
QuintecQuintec
1,9781726
edited 4 hours ago
answered 4 hours ago
QuintecQuintec
1,9781726
answered 4 hours ago
QuintecQuintec
1,9781726
answered 4 hours ago
QuintecQuintec
1,9781726
1,9781726
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
add a comment |
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
4 hours 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 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
4 hours ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
4 hours 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 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
4 hours ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
4 hours 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 4 hours 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 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
edited 4 hours ago
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
answered 4 hours ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
1,516123
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
4 hours ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
4 hours ago
add a comment |
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
4 hours ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
4 hours ago
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
4 hours ago
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
4 hours ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
4 hours ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
4 hours ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
4 hours ago
add a comment |
$begingroup$
Japt -N, 6 bytes
Returns 1 for truthy, 0 or NaN for falsy.
¤q0 äè
Run it online
answered 2 hours ago
OliverOliver
5,2001832
$endgroup$
add a comment |
$begingroup$
Japt -N, 6 bytes
Returns 1 for truthy, 0 or NaN for falsy.
¤q0 äè
Run it online
answered 2 hours ago
OliverOliver
5,2001832
$endgroup$
add a comment |
$begingroup$
Japt -N, 6 bytes
Returns 1 for truthy, 0 or NaN for falsy.
¤q0 äè
Run it online
answered 2 hours ago
OliverOliver
5,2001832
$endgroup$
Japt -N, 6 bytes
Returns 1 for truthy, 0 or NaN for falsy.
¤q0 äè
Run it online
answered 2 hours ago
OliverOliver
5,2001832
answered 2 hours ago
OliverOliver
5,2001832
answered 2 hours ago
OliverOliver
5,2001832
answered 2 hours ago
OliverOliver
5,2001832
5,2001832
add a comment |
add a comment |
$begingroup$
Perl 6, 23 bytes
.base(2)~~/^(1*)0$0$/
Try it online!
Regex based solution
answered 4 hours 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 4 hours 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 4 hours ago
Jo KingJo King
24.4k357126
$endgroup$
Perl 6, 23 bytes
.base(2)~~/^(1*)0$0$/
Try it online!
Regex based solution
answered 4 hours ago
Jo KingJo King
24.4k357126
answered 4 hours ago
Jo KingJo King
24.4k357126
answered 4 hours ago
Jo KingJo King
24.4k357126
answered 4 hours 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 4 hours 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 4 hours 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 4 hours 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 4 hours ago
tshtsh
9,32511652
edited 3 hours ago
answered 4 hours ago
tshtsh
9,32511652
answered 4 hours ago
tshtsh
9,32511652
answered 4 hours ago
tshtsh
9,32511652
9,32511652
add a comment |
add a comment |
$begingroup$
Mathematica (Wolfram language), 32 bytes
OddQ@Log[2,#+Floor@Sqrt[#/2]+2]&
Try it online!
Pure function taking an integer as input and returning True or False. Based on the fact (fun to prove!) that a number n is Cyclops if and only if n plus the square root of n/2 plus 2 rounds down to an odd power of 2. (One can replace Floor by either Ceiling or Round as long as one also replaces +2 by +1.) Returns True on input 0.
answered 2 hours ago
Greg MartinGreg Martin
12.3k21059
$endgroup$
add a comment |
$begingroup$
Mathematica (Wolfram language), 32 bytes
OddQ@Log[2,#+Floor@Sqrt[#/2]+2]&
Try it online!
Pure function taking an integer as input and returning True or False. Based on the fact (fun to prove!) that a number n is Cyclops if and only if n plus the square root of n/2 plus 2 rounds down to an odd power of 2. (One can replace Floor by either Ceiling or Round as long as one also replaces +2 by +1.) Returns True on input 0.
answered 2 hours ago
Greg MartinGreg Martin
12.3k21059
$endgroup$
add a comment |
$begingroup$
Mathematica (Wolfram language), 32 bytes
OddQ@Log[2,#+Floor@Sqrt[#/2]+2]&
Try it online!
Pure function taking an integer as input and returning True or False. Based on the fact (fun to prove!) that a number n is Cyclops if and only if n plus the square root of n/2 plus 2 rounds down to an odd power of 2. (One can replace Floor by either Ceiling or Round as long as one also replaces +2 by +1.) Returns True on input 0.
answered 2 hours ago
Greg MartinGreg Martin
12.3k21059
$endgroup$
Mathematica (Wolfram language), 32 bytes
OddQ@Log[2,#+Floor@Sqrt[#/2]+2]&
Try it online!
Pure function taking an integer as input and returning True or False. Based on the fact (fun to prove!) that a number n is Cyclops if and only if n plus the square root of n/2 plus 2 rounds down to an odd power of 2. (One can replace Floor by either Ceiling or Round as long as one also replaces +2 by +1.) Returns True on input 0.
answered 2 hours ago
Greg MartinGreg Martin
12.3k21059
answered 2 hours ago
Greg MartinGreg Martin
12.3k21059
answered 2 hours ago
Greg MartinGreg Martin
12.3k21059
answered 2 hours ago
Greg MartinGreg Martin
12.3k21059
12.3k21059
add a comment |
add a comment |
$begingroup$
Jelly, 9 bytes
BŒḂaB¬S=1
Try it online!
answered 1 hour ago
Nick KennedyNick Kennedy
52127
$endgroup$
add a comment |
$begingroup$
Jelly, 9 bytes
BŒḂaB¬S=1
Try it online!
answered 1 hour ago
Nick KennedyNick Kennedy
52127
$endgroup$
add a comment |
$begingroup$
Jelly, 9 bytes
BŒḂaB¬S=1
Try it online!
answered 1 hour ago
Nick KennedyNick Kennedy
52127
$endgroup$
Jelly, 9 bytes
BŒḂaB¬S=1
Try it online!
answered 1 hour ago
Nick KennedyNick Kennedy
52127
answered 1 hour ago
Nick KennedyNick Kennedy
52127
answered 1 hour ago
Nick KennedyNick Kennedy
52127
answered 1 hour ago
Nick KennedyNick Kennedy
52127
52127
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…
…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.
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…Avoid asking for help, clarification or responding to other answers (use comments instead).
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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
5 hours ago
$begingroup$
@SriotchilismO'Zaic thanks for the feedback! I have edited the post.
$endgroup$
– Tau
5 hours ago
$begingroup$
@FryAmTheEggman thank you, I'll keep that in mind. The post has been edited.
$endgroup$
– Tau
5 hours ago
1
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Note: This is A129868
$endgroup$
– tsh
4 hours ago
2
$begingroup$
Should
119be true?$endgroup$
– Quintec
4 hours ago