Am I a Rude Number?Print the fingering for the note on a saxophoneProgram touch-tone instructions for my...
Why didn't Tom Riddle take the presence of Fawkes and the Sorting Hat as more of a threat?
Explanation of a regular pattern only occuring for prime numbers
Existence of Riemann surface, holomorphic maps
Macro expansion inside href
Why do neural networks need so many training examples to perform?
Square Root Distance from Integers
What makes papers publishable in top-tier journals?
Looking for a specific 6502 Assembler
In Linux what happens if 1000 files in a directory are moved to another location while another 300 files were added to the source directory?
Globe trotting Grandpa. Where is he going next?
Why is Agricola named as such?
Is there a lava-breathing lizard creature (that could be worshipped by a cult) in 5e?
What happens when I Twin Life Transference?
Premature ending of generator in list comprehension
What is a DAG (Graph Theory)?
How can I play a serial killer in a party of good PCs?
Can 5 Aarakocra PCs summon an Air Elemental?
What language shall they sing in?
Why is it that Bernie Sanders is always called a "socialist"?
Potential client has a problematic employee I can't work with
Citing paywalled articles accessed via illegal web sharing
How do you catch Smeargle in Pokemon Go?
How do you funnel food off a cutting board?
How to not let the Identify spell spoil everything?
Am I a Rude Number?
Print the fingering for the note on a saxophoneProgram touch-tone instructions for my fingersIs this even or odd?Is this number a factorial?Is this number triangular?Am I a Cullen Number?Am I an Automorphic Number?Is this any number?Octal Calculator with Primality checkIs it a Sphenic Number?
$begingroup$
For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one hand), assume we continue with a third hand, with additional hands added as required.
Your Task:
Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.
Input:
An integer between 0 and 10^9 (inclusive).
Output:
A truthy/falsy value that indicates whether the input is a rude number.
Test Cases:
Input: Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy
Scoring:
This is code-golf, so the lowest score in bytes wins.
code-golf number decision-problem
asked 1 hour ago
GryphonGryphon
3,1891963
$endgroup$
add a comment |
$begingroup$
For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one hand), assume we continue with a third hand, with additional hands added as required.
Your Task:
Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.
Input:
An integer between 0 and 10^9 (inclusive).
Output:
A truthy/falsy value that indicates whether the input is a rude number.
Test Cases:
Input: Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy
Scoring:
This is code-golf, so the lowest score in bytes wins.
code-golf number decision-problem
asked 1 hour ago
GryphonGryphon
3,1891963
$endgroup$
2
$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
$endgroup$
– Veskah
1 hour ago
1
$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
15 mins ago
$begingroup$
BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
$endgroup$
– ASCII-only
3 mins ago
add a comment |
$begingroup$
For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one hand), assume we continue with a third hand, with additional hands added as required.
Your Task:
Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.
Input:
An integer between 0 and 10^9 (inclusive).
Output:
A truthy/falsy value that indicates whether the input is a rude number.
Test Cases:
Input: Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy
Scoring:
This is code-golf, so the lowest score in bytes wins.
code-golf number decision-problem
asked 1 hour ago
GryphonGryphon
3,1891963
$endgroup$
For a while now, I've been running into a problem when counting on my fingers, specifically, that I can only count to ten. My solution to that problem has been to count in binary on my fingers, putting up my thumb for one, my forefinger for two, both thumb and forefinger for three, etc. However, we run into a bit of a problem when we get to the number four. Specifically, it requires us to put up our middle finger, which results in a rather unfortunate gesture, which is not typically accepted in society. This type of number is a rude number. We come to the next rude number at 36, when we raise the thumb on our second hand and the middle finger of our first hand. The definition of a rude number is any number that, under this system of counting, results in us putting up only the middle finger of any hand. Once we pass 1023 (the maximum number reachable on one hand), assume we continue with a third hand, with additional hands added as required.
Your Task:
Write a program or function that receives an input and outputs a truthy/falsy value based on whether the input is a rude number.
Input:
An integer between 0 and 10^9 (inclusive).
Output:
A truthy/falsy value that indicates whether the input is a rude number.
Test Cases:
Input: Output:
0 ---> falsy
3 ---> falsy
4 ---> truthy
25 ---> falsy
36 ---> truthy
127 ---> falsy
131 ---> truthy
Scoring:
This is code-golf, so the lowest score in bytes wins.
code-golf number decision-problem
code-golf number decision-problem
asked 1 hour ago
GryphonGryphon
3,1891963
asked 1 hour ago
GryphonGryphon
3,1891963
edited 42 mins ago
Gryphon
asked 1 hour ago
GryphonGryphon
3,1891963
asked 1 hour ago
GryphonGryphon
3,1891963
asked 1 hour ago
GryphonGryphon
3,1891963
3,1891963
2
$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
$endgroup$
– Veskah
1 hour ago
1
$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
15 mins ago
$begingroup$
BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
$endgroup$
– ASCII-only
3 mins ago
add a comment |
2
$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.
$endgroup$
– Veskah
1 hour ago
1
$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
15 mins ago
$begingroup$
BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
$endgroup$
– ASCII-only
3 mins ago
2
2
$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.$endgroup$
– Veskah
1 hour ago
$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.$endgroup$
– Veskah
1 hour ago
1
1
$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
15 mins ago
$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
15 mins ago
$begingroup$
BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
$endgroup$
– ASCII-only
3 mins ago
$begingroup$
BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
$endgroup$
– ASCII-only
3 mins ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
JavaScript (SpiderMonkey), 27 bytes
x=>/4/.test(x.toString(32))
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
answered 20 mins ago
tshtsh
9,15511650
$endgroup$
add a comment |
$begingroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered 1 hour ago
Doorknob♦Doorknob
54.9k17115352
$endgroup$
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
21 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
17 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
14 mins ago
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
9 mins ago
|
show 2 more comments
$begingroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered 11 mins ago
Jo KingJo King
23.7k257123
$endgroup$
add a comment |
$begingroup$
Regex (ECMAScript), 37 bytes
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered 21 mins ago
DeadcodeDeadcode
1,7241419
$endgroup$
add a comment |
$begingroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered 16 mins ago
ASCII-onlyASCII-only
3,4601236
$endgroup$
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
});
}
});
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%2f180481%2fam-i-a-rude-number%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
JavaScript (SpiderMonkey), 27 bytes
x=>/4/.test(x.toString(32))
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
answered 20 mins ago
tshtsh
9,15511650
$endgroup$
add a comment |
$begingroup$
JavaScript (SpiderMonkey), 27 bytes
x=>/4/.test(x.toString(32))
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
answered 20 mins ago
tshtsh
9,15511650
$endgroup$
add a comment |
$begingroup$
JavaScript (SpiderMonkey), 27 bytes
x=>/4/.test(x.toString(32))
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
answered 20 mins ago
tshtsh
9,15511650
$endgroup$
JavaScript (SpiderMonkey), 27 bytes
x=>/4/.test(x.toString(32))
Try it online!
This is a trivial solution, you just want to convert to base 32 and check if there is a 4 in it.
answered 20 mins ago
tshtsh
9,15511650
answered 20 mins ago
tshtsh
9,15511650
answered 20 mins ago
tshtsh
9,15511650
answered 20 mins ago
tshtsh
9,15511650
9,15511650
add a comment |
add a comment |
$begingroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered 1 hour ago
Doorknob♦Doorknob
54.9k17115352
$endgroup$
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
21 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
17 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
14 mins ago
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
9 mins ago
|
show 2 more comments
$begingroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered 1 hour ago
Doorknob♦Doorknob
54.9k17115352
$endgroup$
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
21 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
17 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
14 mins ago
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
9 mins ago
|
show 2 more comments
$begingroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered 1 hour ago
Doorknob♦Doorknob
54.9k17115352
$endgroup$
Ruby, 36 19 bytes
->n{n.to_s(32)[?4]}
Try it online!
Saved 17 bytes with @tsh's method.
answered 1 hour ago
Doorknob♦Doorknob
54.9k17115352
edited 14 mins ago
answered 1 hour ago
Doorknob♦Doorknob
54.9k17115352
answered 1 hour ago
Doorknob♦Doorknob
54.9k17115352
answered 1 hour ago
Doorknob♦Doorknob
54.9k17115352
54.9k17115352
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
21 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
17 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
14 mins ago
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
9 mins ago
|
show 2 more comments
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
21 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
17 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
14 mins ago
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 220710001, the second00111, and the third11? None of them have their middle finger only up
$endgroup$
– Embodiment of Ignorance
9 mins ago
$begingroup$
This returns true for 2207, which has a binary representation of
100010011111$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
This returns true for 2207, which has a binary representation of
100010011111$endgroup$
– Embodiment of Ignorance
27 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is
00100.$endgroup$
– Doorknob♦
21 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is
00100.$endgroup$
– Doorknob♦
21 mins ago
$begingroup$
I don't speak Ruby. But why not
->n{n.to_s(32)=~/4/}?$endgroup$
– tsh
17 mins ago
$begingroup$
I don't speak Ruby. But why not
->n{n.to_s(32)=~/4/}?$endgroup$
– tsh
17 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
14 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
14 mins ago
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 2207
10001, the second 00111, and the third 11? None of them have their middle finger only up$endgroup$
– Embodiment of Ignorance
9 mins ago
$begingroup$
Forgive me if I'm not understanding the question, but isn't the first hand of 2207
10001, the second 00111, and the third 11? None of them have their middle finger only up$endgroup$
– Embodiment of Ignorance
9 mins ago
|
show 2 more comments
$begingroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered 11 mins ago
Jo KingJo King
23.7k257123
$endgroup$
add a comment |
$begingroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered 11 mins ago
Jo KingJo King
23.7k257123
$endgroup$
add a comment |
$begingroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered 11 mins ago
Jo KingJo King
23.7k257123
$endgroup$
Perl 6, 16 bytes
{.base(32)~~/4/}
Try it online!
Checks if there is a 4 in the base 32 representation of the number. Returns either Nil as false or a Match containing a 4.
You can prove this by the fact that $2^5 = 32$ so each digit is the state of each hand.
answered 11 mins ago
Jo KingJo King
23.7k257123
answered 11 mins ago
Jo KingJo King
23.7k257123
answered 11 mins ago
Jo KingJo King
23.7k257123
answered 11 mins ago
Jo KingJo King
23.7k257123
23.7k257123
add a comment |
add a comment |
$begingroup$
Regex (ECMAScript), 37 bytes
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered 21 mins ago
DeadcodeDeadcode
1,7241419
$endgroup$
add a comment |
$begingroup$
Regex (ECMAScript), 37 bytes
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered 21 mins ago
DeadcodeDeadcode
1,7241419
$endgroup$
add a comment |
$begingroup$
Regex (ECMAScript), 37 bytes
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered 21 mins ago
DeadcodeDeadcode
1,7241419
$endgroup$
Regex (ECMAScript), 37 bytes
^((?=(x+)(2{31}x*))3)*(x{32})*x{4}$
Try it online!
^
(
(?=(x+)(2{31}x*)) # 2 = floor(tail / 32); 3 = tool to make tail = 2
3 # tail = 2
)* # Loop the above as many times as necessary to make
# the below match
(x{32})*x{4}$ # Assert that tail % 32 == 4
answered 21 mins ago
DeadcodeDeadcode
1,7241419
edited 14 mins ago
answered 21 mins ago
DeadcodeDeadcode
1,7241419
answered 21 mins ago
DeadcodeDeadcode
1,7241419
answered 21 mins ago
DeadcodeDeadcode
1,7241419
1,7241419
add a comment |
add a comment |
$begingroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered 16 mins ago
ASCII-onlyASCII-only
3,4601236
$endgroup$
add a comment |
$begingroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered 16 mins ago
ASCII-onlyASCII-only
3,4601236
$endgroup$
add a comment |
$begingroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered 16 mins ago
ASCII-onlyASCII-only
3,4601236
$endgroup$
Japt, 5 bytes
sH ø4
Try it online!
Explanation
// Implicit input
sH // To a base-H (=32) string
ø // Contains
4 // 4 (JavaScript interprets this as a string)
answered 16 mins ago
ASCII-onlyASCII-only
3,4601236
answered 16 mins ago
ASCII-onlyASCII-only
3,4601236
answered 16 mins ago
ASCII-onlyASCII-only
3,4601236
answered 16 mins ago
ASCII-onlyASCII-only
3,4601236
3,4601236
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.
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%2f180481%2fam-i-a-rude-number%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
2
$begingroup$
assume we continue with a third hand, When it comes to being rude, teamwork makes the dream work.$endgroup$
– Veskah
1 hour ago
1
$begingroup$
@Veskah turns out that for the bounds of the question, you only need 3 people to make any given number. Sure beats the old kind of counting on fingers.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
Also, I don't have the time at the moment, but if anyone could figure out an equation for this sequence, that'd be great.
$endgroup$
– Gryphon
1 hour ago
$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
15 mins ago
$begingroup$
BTW, pretty sure there is no short equation for this sequence, but you could (kinda) make a separate equation for each hand
$endgroup$
– ASCII-only
3 mins ago