Is it a Cyclops number? “Nobody” knows!#Hashtag_or_notIs it truthy or falsy?Is this a Smith number?Binary...
Sundering Titan and basic normal lands and snow lands
Is this nominative case or accusative case?
Why doesn't "adolescent" take any articles in "listen to adolescent agonising"?
Create chunks from an array
Named nets not connected in Eagle board design
Can a Mimic (container form) actually hold loot?
Can inspiration allow the Rogue to make a Sneak Attack?
What is the oldest European royal house?
Paper published similar to PhD thesis
I've given my players a lot of magic items. Is it reasonable for me to give them harder encounters?
Why do we call complex numbers “numbers” but we don’t consider 2 vectors numbers?
School performs periodic password audits. Is my password compromised?
Why are special aircraft used for the carriers in the United States Navy?
The Key to the Door
A bug in Excel? Conditional formatting for marking duplicates also highlights unique value
Practical reasons to have both a large police force and bounty hunting network?
When to use the term transposed instead of modulation?
Why do phishing e-mails use faked e-mail addresses instead of the real one?
How can I be pwned if I'm not registered on the compromised site?
An Undercover Army
The need of reserving one's ability in job interviews
Can a space-faring robot still function over a billion years?
Is there a way to find out the age of climbing ropes?
What does it mean when I add a new variable to my linear model and the R^2 stays the same?
Is it a Cyclops number? “Nobody” knows!
#Hashtag_or_notIs it truthy or falsy?Is this a Smith number?Binary to decimal converterIs this number triangular?Am I divisible by double the sum of my digits?Is the number binary-heavy?Nobody knows GAU NumbersIs it a whole number?How many consecutive descending numbers in my number?
$begingroup$
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
52 mins ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
49 mins ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
32 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 28 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
52 mins ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
49 mins ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
32 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
52 mins ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
49 mins ago
2
$begingroup$
Should119be true?
$endgroup$
– Quintec
32 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
52 mins 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
52 mins ago
1
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
49 mins ago
$begingroup$
Note: This is A129868
$endgroup$
– tsh
49 mins ago
2
2
$begingroup$
Should
119 be true?$endgroup$
– Quintec
32 mins ago
$begingroup$
Should
119 be true?$endgroup$
– Quintec
32 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 46 mins ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
2 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 30 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
28 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
18 mins ago
add a comment |
$begingroup$
Perl 6, 23 bytes
{.base(2)~~/^(1*)0$0$/}
Try it online!
Regex based solution
answered 23 mins ago
Jo KingJo King
24.4k357126
$endgroup$
add a comment |
$begingroup$
JavaScript (Node.js), 33 bytes
f=(p,q)=>p%2?f(p>>1,q+q|1):p==q+q
Try it online!
This one will report falsy for 0. But the RegExp based one works for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 21 mins ago
tshtsh
9,32511652
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "200"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Tau is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f181128%2fis-it-a-cyclops-number-nobody-knows%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 46 mins ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
2 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 46 mins ago
QuintecQuintec
1,9781726
$endgroup$
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
2 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 46 mins 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 46 mins ago
QuintecQuintec
1,9781726
edited 21 mins ago
answered 46 mins ago
QuintecQuintec
1,9781726
answered 46 mins ago
QuintecQuintec
1,9781726
answered 46 mins ago
QuintecQuintec
1,9781726
1,9781726
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
2 mins ago
add a comment |
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
2 mins ago
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
2 mins ago
$begingroup$
found an even shorter way, based on your solution :)
$endgroup$
– Embodiment of Ignorance
2 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 30 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
28 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
18 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 30 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
$endgroup$
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
28 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
18 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 30 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 30 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
edited 2 mins ago
answered 30 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
answered 30 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
answered 30 mins ago
Embodiment of IgnoranceEmbodiment of Ignorance
1,516123
1,516123
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
28 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
18 mins ago
add a comment |
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
28 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
18 mins ago
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
28 mins ago
$begingroup$
Well played, I should really learn regular expressions sometime. :) +1
$endgroup$
– Quintec
28 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
@Quintec Regex is awesome :)
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
18 mins ago
$begingroup$
Update: found shorter way :)
$endgroup$
– Quintec
18 mins ago
add a comment |
$begingroup$
Perl 6, 23 bytes
{.base(2)~~/^(1*)0$0$/}
Try it online!
Regex based solution
answered 23 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 23 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 23 mins ago
Jo KingJo King
24.4k357126
$endgroup$
Perl 6, 23 bytes
{.base(2)~~/^(1*)0$0$/}
Try it online!
Regex based solution
answered 23 mins ago
Jo KingJo King
24.4k357126
answered 23 mins ago
Jo KingJo King
24.4k357126
answered 23 mins ago
Jo KingJo King
24.4k357126
answered 23 mins ago
Jo KingJo King
24.4k357126
24.4k357126
add a comment |
add a comment |
$begingroup$
JavaScript (Node.js), 33 bytes
f=(p,q)=>p%2?f(p>>1,q+q|1):p==q+q
Try it online!
This one will report falsy for 0. But the RegExp based one works for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 21 mins ago
tshtsh
9,32511652
$endgroup$
add a comment |
$begingroup$
JavaScript (Node.js), 33 bytes
f=(p,q)=>p%2?f(p>>1,q+q|1):p==q+q
Try it online!
This one will report falsy for 0. But the RegExp based one works for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 21 mins ago
tshtsh
9,32511652
$endgroup$
add a comment |
$begingroup$
JavaScript (Node.js), 33 bytes
f=(p,q)=>p%2?f(p>>1,q+q|1):p==q+q
Try it online!
This one will report falsy for 0. But the RegExp based one works for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 21 mins ago
tshtsh
9,32511652
$endgroup$
JavaScript (Node.js), 33 bytes
f=(p,q)=>p%2?f(p>>1,q+q|1):p==q+q
Try it online!
This one will report falsy for 0. But the RegExp based one works for 0.
JavaScript (Node.js), 35 bytes
p=>p.toString(2).match(/^(1*)01$/)
Try it online!
answered 21 mins ago
tshtsh
9,32511652
edited 9 mins ago
answered 21 mins ago
tshtsh
9,32511652
answered 21 mins ago
tshtsh
9,32511652
answered 21 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…
…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.
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).
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f181128%2fis-it-a-cyclops-number-nobody-knows%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
52 mins ago
1
$begingroup$
Note: This is A129868
$endgroup$
– tsh
49 mins ago
2
$begingroup$
Should
119be true?$endgroup$
– Quintec
32 mins ago