Am I a Rude Number?
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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
$endgroup$
3
$begingroup$
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
31 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
19 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 58 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
3
$begingroup$
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
31 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
19 mins ago
add a comment |
3
$begingroup$
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
31 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
19 mins ago
3
3
$begingroup$
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
31 mins ago
$begingroup$
You can reach 1023 on one hand? O_o
$endgroup$
– ASCII-only
31 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
19 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
19 mins ago
add a comment |
6 Answers
6
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 36 mins ago
tshtsh
9,15511650
$endgroup$
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 32 mins ago
ASCII-onlyASCII-only
3,4801236
$endgroup$
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
43 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
37 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
33 mins ago
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@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
30 mins ago
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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
25 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 27 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 37 mins ago
DeadcodeDeadcode
1,7241419
$endgroup$
add a comment |
$begingroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered 8 mins ago
GrahamGraham
2,34678
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
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 36 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 36 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 36 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 36 mins ago
tshtsh
9,15511650
answered 36 mins ago
tshtsh
9,15511650
answered 36 mins ago
tshtsh
9,15511650
answered 36 mins ago
tshtsh
9,15511650
9,15511650
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 32 mins ago
ASCII-onlyASCII-only
3,4801236
$endgroup$
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 32 mins ago
ASCII-onlyASCII-only
3,4801236
$endgroup$
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 32 mins ago
ASCII-onlyASCII-only
3,4801236
$endgroup$
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 32 mins ago
ASCII-onlyASCII-only
3,4801236
answered 32 mins ago
ASCII-onlyASCII-only
3,4801236
answered 32 mins ago
ASCII-onlyASCII-only
3,4801236
answered 32 mins ago
ASCII-onlyASCII-only
3,4801236
3,4801236
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
43 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
37 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
33 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
30 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
25 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
43 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
37 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
33 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
30 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
25 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 30 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
43 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
37 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
33 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
30 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
25 mins ago
|
show 2 more comments
$begingroup$
This returns true for 2207, which has a binary representation of100010011111
$endgroup$
– Embodiment of Ignorance
43 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is00100.
$endgroup$
– Doorknob♦
37 mins ago
$begingroup$
I don't speak Ruby. But why not->n{n.to_s(32)=~/4/}?
$endgroup$
– tsh
33 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
30 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
25 mins ago
$begingroup$
This returns true for 2207, which has a binary representation of
100010011111$endgroup$
– Embodiment of Ignorance
43 mins ago
$begingroup$
This returns true for 2207, which has a binary representation of
100010011111$endgroup$
– Embodiment of Ignorance
43 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is
00100.$endgroup$
– Doorknob♦
37 mins ago
$begingroup$
@EmbodimentofIgnorance That is the correct result, is it not? The second hand is
00100.$endgroup$
– Doorknob♦
37 mins ago
$begingroup$
I don't speak Ruby. But why not
->n{n.to_s(32)=~/4/}?$endgroup$
– tsh
33 mins ago
$begingroup$
I don't speak Ruby. But why not
->n{n.to_s(32)=~/4/}?$endgroup$
– tsh
33 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
30 mins ago
$begingroup$
@tsh because I'm not as clever as you :)
$endgroup$
– Doorknob♦
30 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
25 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
25 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 27 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 27 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 27 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 27 mins ago
Jo KingJo King
23.7k257123
answered 27 mins ago
Jo KingJo King
23.7k257123
answered 27 mins ago
Jo KingJo King
23.7k257123
answered 27 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 37 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 37 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 37 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 37 mins ago
DeadcodeDeadcode
1,7241419
edited 30 mins ago
answered 37 mins ago
DeadcodeDeadcode
1,7241419
answered 37 mins ago
DeadcodeDeadcode
1,7241419
answered 37 mins ago
DeadcodeDeadcode
1,7241419
1,7241419
add a comment |
add a comment |
$begingroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered 8 mins ago
GrahamGraham
2,34678
$endgroup$
add a comment |
$begingroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered 8 mins ago
GrahamGraham
2,34678
$endgroup$
add a comment |
$begingroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered 8 mins ago
GrahamGraham
2,34678
$endgroup$
APL+WIN, 10 bytes
Prompts for input of integer
4∊(6⍴32)⊤⎕
Noting six hands are required to represent 10^9 converts to vector of 6 elements of the base 32 representation and checks if a 4 exists in any element.
answered 8 mins ago
GrahamGraham
2,34678
answered 8 mins ago
GrahamGraham
2,34678
answered 8 mins ago
GrahamGraham
2,34678
answered 8 mins ago
GrahamGraham
2,34678
2,34678
add a comment |
add a comment |
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3
$begingroup$
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
31 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
19 mins ago