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Do I have a twin with permutated remainders?
Tips for golfing in 05AB1ESplit a word into parts with equal scoresChinese Remainder TheoremMaximise the squared differenceIs it a Mersenne Prime?Your Base to 1-2-3-Tribonacci to Binary back to Your BaseIs this number a factorial?Write numbers as a difference of Nth powersConjugate permutationsAm I a Pillai prime?Passwords Strong Against Bishops
$begingroup$
We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.
Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ such that $R_n+k$ is a permutation of $R_n$.
Examples
The criterion is met for $n=8$, because:
- we have $R_8=(0,2,3,1)$
- for $k=44$, we have $R_n+k=R_52=(0,1,2,3)$, which is a permutation of $R_8$
The criterion is not met for $n=48$, because:
- we have $R_48=(0,0,3,6)$
- the smallest integer $k>0$ such that $R_n+k$ is a permutation of $R_48$ is $k=210$ (leading to $R_258=(0,0,3,6)$ as well)
Rules
- You may either output a truthy value if $k$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
- This is code-golf.
Hint
Do you really need to compute $k$? Well, maybe. Or maybe not.
Test cases
Some values of $n$ for which $k$ exists:
3, 4, 5, 8, 30, 100, 200, 2019
Some values of $n$ for which $k$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
code-golf decision-problem number-theory
asked yesterday
ArnauldArnauld
80.4k797333
$endgroup$
add a comment |
$begingroup$
We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.
Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ such that $R_n+k$ is a permutation of $R_n$.
Examples
The criterion is met for $n=8$, because:
- we have $R_8=(0,2,3,1)$
- for $k=44$, we have $R_n+k=R_52=(0,1,2,3)$, which is a permutation of $R_8$
The criterion is not met for $n=48$, because:
- we have $R_48=(0,0,3,6)$
- the smallest integer $k>0$ such that $R_n+k$ is a permutation of $R_48$ is $k=210$ (leading to $R_258=(0,0,3,6)$ as well)
Rules
- You may either output a truthy value if $k$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
- This is code-golf.
Hint
Do you really need to compute $k$? Well, maybe. Or maybe not.
Test cases
Some values of $n$ for which $k$ exists:
3, 4, 5, 8, 30, 100, 200, 2019
Some values of $n$ for which $k$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
code-golf decision-problem number-theory
asked yesterday
ArnauldArnauld
80.4k797333
$endgroup$
add a comment |
$begingroup$
We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.
Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ such that $R_n+k$ is a permutation of $R_n$.
Examples
The criterion is met for $n=8$, because:
- we have $R_8=(0,2,3,1)$
- for $k=44$, we have $R_n+k=R_52=(0,1,2,3)$, which is a permutation of $R_8$
The criterion is not met for $n=48$, because:
- we have $R_48=(0,0,3,6)$
- the smallest integer $k>0$ such that $R_n+k$ is a permutation of $R_48$ is $k=210$ (leading to $R_258=(0,0,3,6)$ as well)
Rules
- You may either output a truthy value if $k$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
- This is code-golf.
Hint
Do you really need to compute $k$? Well, maybe. Or maybe not.
Test cases
Some values of $n$ for which $k$ exists:
3, 4, 5, 8, 30, 100, 200, 2019
Some values of $n$ for which $k$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
code-golf decision-problem number-theory
asked yesterday
ArnauldArnauld
80.4k797333
$endgroup$
We define $R_n$ as the list of remainders of the Euclidean division of $n$ by $2$, $3$, $5$ and $7$.
Given an integer $nge0$, you have to figure out if there exists an integer $0<k<210$ such that $R_n+k$ is a permutation of $R_n$.
Examples
The criterion is met for $n=8$, because:
- we have $R_8=(0,2,3,1)$
- for $k=44$, we have $R_n+k=R_52=(0,1,2,3)$, which is a permutation of $R_8$
The criterion is not met for $n=48$, because:
- we have $R_48=(0,0,3,6)$
- the smallest integer $k>0$ such that $R_n+k$ is a permutation of $R_48$ is $k=210$ (leading to $R_258=(0,0,3,6)$ as well)
Rules
- You may either output a truthy value if $k$ exists and a falsy value otherwise, or two distinct and consistent values of your choice.
- This is code-golf.
Hint
Do you really need to compute $k$? Well, maybe. Or maybe not.
Test cases
Some values of $n$ for which $k$ exists:
3, 4, 5, 8, 30, 100, 200, 2019
Some values of $n$ for which $k$ does not exist:
0, 1, 2, 13, 19, 48, 210, 1999
code-golf decision-problem number-theory
code-golf decision-problem number-theory
asked yesterday
ArnauldArnauld
80.4k797333
asked yesterday
ArnauldArnauld
80.4k797333
asked yesterday
ArnauldArnauld
80.4k797333
asked yesterday
ArnauldArnauld
80.4k797333
asked yesterday
ArnauldArnauld
80.4k797333
80.4k797333
add a comment |
add a comment |
16 Answers
16
active
oldest
votes
$begingroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s1<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s1
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered yesterday
Robin RyderRobin Ryder
5017
New contributor
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
10 hours ago
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
9 hours ago
add a comment |
$begingroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered yesterday
H.PWizH.PWiz
10.3k21351
$endgroup$
3
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
19 hours ago
add a comment |
$begingroup$
Perl 6, 64 61 59 43 bytes
map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!
Try it online!
-16 thanks to @Jo King
answered yesterday
VenVen
2,53511223
$endgroup$
1
$begingroup$
43 bytes
$endgroup$
– Jo King
yesterday
$begingroup$
Oh! Never thought to use$!that way. Using method version ofbagis nice, and using the sub version ofmapis obvious is hindsight... Thanks
$endgroup$
– Ven
21 hours ago
add a comment |
$begingroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered yesterday
xnorxnor
93.3k18190448
$endgroup$
add a comment |
$begingroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered yesterday
xnorxnor
93.3k18190448
$endgroup$
$begingroup$
Can you explain?
$endgroup$
– dfeuer
23 hours ago
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
11 hours ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 125 42 38 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,768126
$endgroup$
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
23 hours ago
add a comment |
$begingroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],k,209],Sort@R@#]&
R@n_:=n~Mod~2,3,5,7
Try it online!
answered yesterday
J42161217J42161217
13.8k21253
$endgroup$
add a comment |
$begingroup$
Ruby, 54 bytes
->n[2,3,5,7].each_cons(2).any?n%l!=n%h&&n%h<l
Try it online!
Uses Robin Ryder's clever solution.
answered yesterday
histocrathistocrat
19.2k43173
$endgroup$
add a comment |
$begingroup$
Java (JDK), 38 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered 20 hours ago
Olivier GrégoireOlivier Grégoire
9,39511944
$endgroup$
$begingroup$
Java and C# are the second shortest answers, second to only 05AB1E
$endgroup$
– Embodiment of Ignorance
13 hours ago
add a comment |
$begingroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered yesterday
Aaron HaymanAaron Hayman
3516
$endgroup$
add a comment |
$begingroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
answered yesterday
Nick KennedyNick Kennedy
1,31649
$endgroup$
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
yesterday
add a comment |
$begingroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n)while(++$k<210)if(($r=function($n)foreach([2,3,5,7]as$d)$o[]=$n%$d;sort($o);return$o;)($n+$k)==$r($n))return 1;
Try it online!
answered yesterday
gwaughgwaugh
2,0281517
$endgroup$
add a comment |
$begingroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered yesterday
attinatattinat
4697
$endgroup$
add a comment |
$begingroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
# Sort the list
Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered 22 hours ago
Kevin CruijssenKevin Cruijssen
42.3k570217
$endgroup$
add a comment |
$begingroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered 21 hours ago
Galen IvanovGalen Ivanov
7,38211034
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s=2,3,5,7])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below 2,3,5,7 in each coordinate. Note that Or@@ is False.
answered 4 hours ago
Misha LavrovMisha Lavrov
4,251424
$endgroup$
add a comment |
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16 Answers
16
active
oldest
votes
16 Answers
16
active
oldest
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active
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active
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$begingroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s1<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s1
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered yesterday
Robin RyderRobin Ryder
5017
New contributor
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
10 hours ago
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
9 hours ago
add a comment |
$begingroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s1<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s1
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered yesterday
Robin RyderRobin Ryder
5017
New contributor
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
10 hours ago
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
9 hours ago
add a comment |
$begingroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s1<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s1
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered yesterday
Robin RyderRobin Ryder
5017
New contributor
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
R, 63 59 bytes
s=scan()%%c(2,3,5,7);i=which(s<c(0,2,3,5));any(s[i]-s[i-1])
Try it online!
-4 bytes thanks to Giuseppe
(The explanation contains a spoiler as to how to solve the problem without computing $k$.)
Explanation:
Let $s$ be the list of remainders. Note the constraint that s1<2, s[2]<3, s[3]<5 and s[4]<7. By the Chinese Remainder Theorem, there exists a $k$ iff there is a permutation of $s$, distinct from $s$, which verifies the constraint. In practice, this will be verified if one of the following conditions is verified:
- s[2]<2 and s[2]!=s1
- s[3]<3 and s[3]!=s[2]
- s[4]<5 and s[4]!=s[3]
The code can probably be golfed further.
answered yesterday
Robin RyderRobin Ryder
5017
New contributor
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 9 hours ago
answered yesterday
Robin RyderRobin Ryder
5017
New contributor
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Robin RyderRobin Ryder
5017
answered yesterday
Robin RyderRobin Ryder
5017
5017
New contributor
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Robin Ryder is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
10 hours ago
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
9 hours ago
add a comment |
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
10 hours ago
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
9 hours ago
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
10 hours ago
$begingroup$
Could you explain why the permutation is necessarily distinct from $s$?
$endgroup$
– dfeuer
10 hours ago
1
1
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
9 hours ago
$begingroup$
@dfeuer It is a consequence of the Chinese Remainder Theorem; I added a link. If two integers have the same remainders modulo 2, 3, 5 and 7 (without a permutation), then the two integers are equal modulo 2*3*5*7=210.
$endgroup$
– Robin Ryder
9 hours ago
add a comment |
$begingroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered yesterday
H.PWizH.PWiz
10.3k21351
$endgroup$
3
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
19 hours ago
add a comment |
$begingroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered yesterday
H.PWizH.PWiz
10.3k21351
$endgroup$
3
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
19 hours ago
add a comment |
$begingroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered yesterday
H.PWizH.PWiz
10.3k21351
$endgroup$
Haskell, 69 bytes
Based on the Chinese remainder theorem
m=[2,3,5,7]
f x|s<-mod x<$>m=or[m!!a>b|a<-[0..2],b<-drop a s,s!!a/=b]
Try it online!
answered yesterday
H.PWizH.PWiz
10.3k21351
answered yesterday
H.PWizH.PWiz
10.3k21351
answered yesterday
H.PWizH.PWiz
10.3k21351
answered yesterday
H.PWizH.PWiz
10.3k21351
10.3k21351
3
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
19 hours ago
add a comment |
3
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
19 hours ago
3
3
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
19 hours ago
$begingroup$
Actually, my working title for this challenge was "Do I have a Chinese twin?" :)
$endgroup$
– Arnauld
19 hours ago
add a comment |
$begingroup$
Perl 6, 64 61 59 43 bytes
map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!
Try it online!
-16 thanks to @Jo King
answered yesterday
VenVen
2,53511223
$endgroup$
1
$begingroup$
43 bytes
$endgroup$
– Jo King
yesterday
$begingroup$
Oh! Never thought to use$!that way. Using method version ofbagis nice, and using the sub version ofmapis obvious is hindsight... Thanks
$endgroup$
– Ven
21 hours ago
add a comment |
$begingroup$
Perl 6, 64 61 59 43 bytes
map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!
Try it online!
-16 thanks to @Jo King
answered yesterday
VenVen
2,53511223
$endgroup$
1
$begingroup$
43 bytes
$endgroup$
– Jo King
yesterday
$begingroup$
Oh! Never thought to use$!that way. Using method version ofbagis nice, and using the sub version ofmapis obvious is hindsight... Thanks
$endgroup$
– Ven
21 hours ago
add a comment |
$begingroup$
Perl 6, 64 61 59 43 bytes
map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!
Try it online!
-16 thanks to @Jo King
answered yesterday
VenVen
2,53511223
$endgroup$
Perl 6, 64 61 59 43 bytes
map($!=(*X%2,3,5,7).Bag,^209+$_+1)∋.&$!
Try it online!
-16 thanks to @Jo King
answered yesterday
VenVen
2,53511223
edited 21 hours ago
answered yesterday
VenVen
2,53511223
answered yesterday
VenVen
2,53511223
answered yesterday
VenVen
2,53511223
2,53511223
1
$begingroup$
43 bytes
$endgroup$
– Jo King
yesterday
$begingroup$
Oh! Never thought to use$!that way. Using method version ofbagis nice, and using the sub version ofmapis obvious is hindsight... Thanks
$endgroup$
– Ven
21 hours ago
add a comment |
1
$begingroup$
43 bytes
$endgroup$
– Jo King
yesterday
$begingroup$
Oh! Never thought to use$!that way. Using method version ofbagis nice, and using the sub version ofmapis obvious is hindsight... Thanks
$endgroup$
– Ven
21 hours ago
1
1
$begingroup$
43 bytes
$endgroup$
– Jo King
yesterday
$begingroup$
43 bytes
$endgroup$
– Jo King
yesterday
$begingroup$
Oh! Never thought to use
$! that way. Using method version of bag is nice, and using the sub version of map is obvious is hindsight... Thanks$endgroup$
– Ven
21 hours ago
$begingroup$
Oh! Never thought to use
$! that way. Using method version of bag is nice, and using the sub version of map is obvious is hindsight... Thanks$endgroup$
– Ven
21 hours ago
add a comment |
$begingroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered yesterday
xnorxnor
93.3k18190448
$endgroup$
add a comment |
$begingroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered yesterday
xnorxnor
93.3k18190448
$endgroup$
add a comment |
$begingroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered yesterday
xnorxnor
93.3k18190448
$endgroup$
Python 2, 41 bytes
lambda n:n%5!=n%7<5or n%3!=n%5<3or-~n%6/4
Try it online!
Uses the same characterization as Robin Ryder. The check n%2!=n%3<2 is shortened to -~n%6/4. Writing out the three conditions turned out shorter than writing a general one:
46 bytes
lambda n:any(n%p!=n%(p+1|1)<p for p in[2,3,5])
Try it online!
answered yesterday
xnorxnor
93.3k18190448
edited yesterday
answered yesterday
xnorxnor
93.3k18190448
answered yesterday
xnorxnor
93.3k18190448
answered yesterday
xnorxnor
93.3k18190448
93.3k18190448
add a comment |
add a comment |
$begingroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered yesterday
xnorxnor
93.3k18190448
$endgroup$
$begingroup$
Can you explain?
$endgroup$
– dfeuer
23 hours ago
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
11 hours ago
add a comment |
$begingroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered yesterday
xnorxnor
93.3k18190448
$endgroup$
$begingroup$
Can you explain?
$endgroup$
– dfeuer
23 hours ago
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
11 hours ago
add a comment |
$begingroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered yesterday
xnorxnor
93.3k18190448
$endgroup$
Haskell, 47 bytes
g.mod
g r|let p?q=r p/=r q&&r q<p=2?3||3?5||5?7
Try it online!
answered yesterday
xnorxnor
93.3k18190448
edited yesterday
answered yesterday
xnorxnor
93.3k18190448
answered yesterday
xnorxnor
93.3k18190448
answered yesterday
xnorxnor
93.3k18190448
93.3k18190448
$begingroup$
Can you explain?
$endgroup$
– dfeuer
23 hours ago
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
11 hours ago
add a comment |
$begingroup$
Can you explain?
$endgroup$
– dfeuer
23 hours ago
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
11 hours ago
$begingroup$
Can you explain?
$endgroup$
– dfeuer
23 hours ago
$begingroup$
Can you explain?
$endgroup$
– dfeuer
23 hours ago
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
11 hours ago
$begingroup$
@dfeuer It's using Robin Ryder's method.
$endgroup$
– Ørjan Johansen
11 hours ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 125 42 38 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,768126
$endgroup$
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
23 hours ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 125 42 38 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,768126
$endgroup$
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
23 hours ago
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 125 42 38 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,768126
$endgroup$
C# (Visual C# Interactive Compiler), 125 42 38 bytes
n=>n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Direct port of @xnor's answer, which is based off of @RobinRyder's solution.
Saved 4 bytes thanks to @Ørjan Johansen!
Try it online!
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,768126
edited 13 hours ago
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,768126
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,768126
answered yesterday
Embodiment of IgnoranceEmbodiment of Ignorance
2,768126
2,768126
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
23 hours ago
add a comment |
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
23 hours ago
1
1
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
23 hours ago
$begingroup$
I found a variation that only ties for xnor's languages but helps for this: 38 bytes
$endgroup$
– Ørjan Johansen
23 hours ago
add a comment |
$begingroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],k,209],Sort@R@#]&
R@n_:=n~Mod~2,3,5,7
Try it online!
answered yesterday
J42161217J42161217
13.8k21253
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],k,209],Sort@R@#]&
R@n_:=n~Mod~2,3,5,7
Try it online!
answered yesterday
J42161217J42161217
13.8k21253
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],k,209],Sort@R@#]&
R@n_:=n~Mod~2,3,5,7
Try it online!
answered yesterday
J42161217J42161217
13.8k21253
$endgroup$
Wolfram Language (Mathematica), 67 bytes
!FreeQ[Sort/@Table[R[#+k],k,209],Sort@R@#]&
R@n_:=n~Mod~2,3,5,7
Try it online!
answered yesterday
J42161217J42161217
13.8k21253
edited yesterday
answered yesterday
J42161217J42161217
13.8k21253
answered yesterday
J42161217J42161217
13.8k21253
answered yesterday
J42161217J42161217
13.8k21253
13.8k21253
add a comment |
add a comment |
$begingroup$
Ruby, 54 bytes
->n[2,3,5,7].each_cons(2).any?n%l!=n%h&&n%h<l
Try it online!
Uses Robin Ryder's clever solution.
answered yesterday
histocrathistocrat
19.2k43173
$endgroup$
add a comment |
$begingroup$
Ruby, 54 bytes
->n[2,3,5,7].each_cons(2).any?n%l!=n%h&&n%h<l
Try it online!
Uses Robin Ryder's clever solution.
answered yesterday
histocrathistocrat
19.2k43173
$endgroup$
add a comment |
$begingroup$
Ruby, 54 bytes
->n[2,3,5,7].each_cons(2).any?n%l!=n%h&&n%h<l
Try it online!
Uses Robin Ryder's clever solution.
answered yesterday
histocrathistocrat
19.2k43173
$endgroup$
Ruby, 54 bytes
->n[2,3,5,7].each_cons(2).any?n%l!=n%h&&n%h<l
Try it online!
Uses Robin Ryder's clever solution.
answered yesterday
histocrathistocrat
19.2k43173
answered yesterday
histocrathistocrat
19.2k43173
answered yesterday
histocrathistocrat
19.2k43173
answered yesterday
histocrathistocrat
19.2k43173
19.2k43173
add a comment |
add a comment |
$begingroup$
Java (JDK), 38 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered 20 hours ago
Olivier GrégoireOlivier Grégoire
9,39511944
$endgroup$
$begingroup$
Java and C# are the second shortest answers, second to only 05AB1E
$endgroup$
– Embodiment of Ignorance
13 hours ago
add a comment |
$begingroup$
Java (JDK), 38 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered 20 hours ago
Olivier GrégoireOlivier Grégoire
9,39511944
$endgroup$
$begingroup$
Java and C# are the second shortest answers, second to only 05AB1E
$endgroup$
– Embodiment of Ignorance
13 hours ago
add a comment |
$begingroup$
Java (JDK), 38 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered 20 hours ago
Olivier GrégoireOlivier Grégoire
9,39511944
$endgroup$
Java (JDK), 38 bytes
n->n%7<5&5<n%35|n%5<3&3<n%15|-~n%6/4>0
Try it online!
Credits
- Port of xnor's solution, improved by Ørjan Johansen.
answered 20 hours ago
Olivier GrégoireOlivier Grégoire
9,39511944
answered 20 hours ago
Olivier GrégoireOlivier Grégoire
9,39511944
answered 20 hours ago
Olivier GrégoireOlivier Grégoire
9,39511944
answered 20 hours ago
Olivier GrégoireOlivier Grégoire
9,39511944
9,39511944
$begingroup$
Java and C# are the second shortest answers, second to only 05AB1E
$endgroup$
– Embodiment of Ignorance
13 hours ago
add a comment |
$begingroup$
Java and C# are the second shortest answers, second to only 05AB1E
$endgroup$
– Embodiment of Ignorance
13 hours ago
$begingroup$
Java and C# are the second shortest answers, second to only 05AB1E
$endgroup$
– Embodiment of Ignorance
13 hours ago
$begingroup$
Java and C# are the second shortest answers, second to only 05AB1E
$endgroup$
– Embodiment of Ignorance
13 hours ago
add a comment |
$begingroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered yesterday
Aaron HaymanAaron Hayman
3516
$endgroup$
add a comment |
$begingroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered yesterday
Aaron HaymanAaron Hayman
3516
$endgroup$
add a comment |
$begingroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered yesterday
Aaron HaymanAaron Hayman
3516
$endgroup$
R, 72 bytes
n=scan();b=c(2,3,5,7);for(i in n+1:209)F=F|all(sort(n%%b)==sort(i%%b));F
Try it online!
answered yesterday
Aaron HaymanAaron Hayman
3516
answered yesterday
Aaron HaymanAaron Hayman
3516
answered yesterday
Aaron HaymanAaron Hayman
3516
answered yesterday
Aaron HaymanAaron Hayman
3516
3516
add a comment |
add a comment |
$begingroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
answered yesterday
Nick KennedyNick Kennedy
1,31649
$endgroup$
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
yesterday
add a comment |
$begingroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
answered yesterday
Nick KennedyNick Kennedy
1,31649
$endgroup$
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
yesterday
add a comment |
$begingroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
answered yesterday
Nick KennedyNick Kennedy
1,31649
$endgroup$
Jelly, 15 bytes
8ÆR©PḶ+%Ṣ¥€®ċḢ$
Try it online!
I’m sure there’s a golfier answer. I’ve interpreted a truthy value as being anything that isn’t zero, so here it’s the number of possible values of k. If it needs to be two distinct values that costs me a further byte.
answered yesterday
Nick KennedyNick Kennedy
1,31649
answered yesterday
Nick KennedyNick Kennedy
1,31649
answered yesterday
Nick KennedyNick Kennedy
1,31649
answered yesterday
Nick KennedyNick Kennedy
1,31649
1,31649
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
yesterday
add a comment |
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (?is the if-else construct in Jelly; for some languages it's a harder question).
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
Oh, and you could get distinct values for no cost withḢe$if you wanted :)
$endgroup$
– Jonathan Allan
yesterday
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
yesterday
1
1
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (
? is the if-else construct in Jelly; for some languages it's a harder question).$endgroup$
– Jonathan Allan
yesterday
$begingroup$
All good regarding truthy vs falsey. Using the meta agreed definition of truthy and falsey (effectively "what does the language's if-else construct do if there is one) zero is falsey and non-zeros are truthy (
? is the if-else construct in Jelly; for some languages it's a harder question).$endgroup$
– Jonathan Allan
yesterday
$begingroup$
Oh, and you could get distinct values for no cost with
Ḣe$ if you wanted :)$endgroup$
– Jonathan Allan
yesterday
$begingroup$
Oh, and you could get distinct values for no cost with
Ḣe$ if you wanted :)$endgroup$
– Jonathan Allan
yesterday
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
yesterday
$begingroup$
@JonathanAllan yes of course, thanks. :)
$endgroup$
– Nick Kennedy
yesterday
add a comment |
$begingroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n)while(++$k<210)if(($r=function($n)foreach([2,3,5,7]as$d)$o[]=$n%$d;sort($o);return$o;)($n+$k)==$r($n))return 1;
Try it online!
answered yesterday
gwaughgwaugh
2,0281517
$endgroup$
add a comment |
$begingroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n)while(++$k<210)if(($r=function($n)foreach([2,3,5,7]as$d)$o[]=$n%$d;sort($o);return$o;)($n+$k)==$r($n))return 1;
Try it online!
answered yesterday
gwaughgwaugh
2,0281517
$endgroup$
add a comment |
$begingroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n)while(++$k<210)if(($r=function($n)foreach([2,3,5,7]as$d)$o[]=$n%$d;sort($o);return$o;)($n+$k)==$r($n))return 1;
Try it online!
answered yesterday
gwaughgwaugh
2,0281517
$endgroup$
PHP, 81 78 72 bytes
while($y<3)if($argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u)die(T);
A riff on @Robin Ryder's answer. Input is via STDIN, output is 'T' if truthy, and empty '' if falsy.
$ echo 3|php -nF euc.php
T
$ echo 5|php -nF euc.php
T
$ echo 2019|php -nF euc.php
T
$ echo 0|php -nF euc.php
$ echo 2|php -nF euc.php
$ echo 1999|php -nF euc.php
Try it online!
Or 73 bytes with 1 or 0 response
while($y<3)$r|=$argn%($u='235'[$y])!=($b=$argn%'357'[$y++])&$b<$u;echo$r;
$ echo 2019|php -nF euc.php
1
$ echo 1999|php -nF euc.php
0
Try it online (all test cases)!
Original answer, 133 127 bytes
function($n)while(++$k<210)if(($r=function($n)foreach([2,3,5,7]as$d)$o[]=$n%$d;sort($o);return$o;)($n+$k)==$r($n))return 1;
Try it online!
answered yesterday
gwaughgwaugh
2,0281517
edited yesterday
answered yesterday
gwaughgwaugh
2,0281517
answered yesterday
gwaughgwaugh
2,0281517
answered yesterday
gwaughgwaugh
2,0281517
2,0281517
add a comment |
add a comment |
$begingroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered yesterday
attinatattinat
4697
$endgroup$
add a comment |
$begingroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered yesterday
attinatattinat
4697
$endgroup$
add a comment |
$begingroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered yesterday
attinatattinat
4697
$endgroup$
Python 3, 69 bytes
lambda x:int('4LR2991ODO5GS2974QWH22YTLL3E3I6TDADQG87I0',36)&1<<x%210
Try it online!
Hardcoded
answered yesterday
attinatattinat
4697
answered yesterday
attinatattinat
4697
answered yesterday
attinatattinat
4697
answered yesterday
attinatattinat
4697
4697
add a comment |
add a comment |
$begingroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
# Sort the list
Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered 22 hours ago
Kevin CruijssenKevin Cruijssen
42.3k570217
$endgroup$
add a comment |
$begingroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
# Sort the list
Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered 22 hours ago
Kevin CruijssenKevin Cruijssen
42.3k570217
$endgroup$
add a comment |
$begingroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
# Sort the list
Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered 22 hours ago
Kevin CruijssenKevin Cruijssen
42.3k570217
$endgroup$
05AB1E, 16 bytes
Ƶ.L+ε‚ε4Åp%Ë}à
Try it online or verify all test cases.
Explanation:
Ƶ.L # Create a list in the range [1,209] (which is k)
+ # Add the (implicit) input to each (which is n+k)
ε # Map each value to:
‚ # Pair it with the (implicit) input
ε # Map both to:
4Åp # Get the first 4 primes: [2,3,5,7]
% # Modulo the current number by each of these four (now we have R_n and R_n+k)
# Sort the list
Ë # After the inner map: check if both sorted lists are equal
}à # After the outer map: check if any are truthy by taking the maximum
# (which is output implicitly as result)
See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ. is 209.
answered 22 hours ago
Kevin CruijssenKevin Cruijssen
42.3k570217
answered 22 hours ago
Kevin CruijssenKevin Cruijssen
42.3k570217
answered 22 hours ago
Kevin CruijssenKevin Cruijssen
42.3k570217
answered 22 hours ago
Kevin CruijssenKevin Cruijssen
42.3k570217
42.3k570217
add a comment |
add a comment |
$begingroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered 21 hours ago
Galen IvanovGalen Ivanov
7,38211034
$endgroup$
add a comment |
$begingroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered 21 hours ago
Galen IvanovGalen Ivanov
7,38211034
$endgroup$
add a comment |
$begingroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered 21 hours ago
Galen IvanovGalen Ivanov
7,38211034
$endgroup$
J, 40 bytes
1 e.(>:+i.@209)-:&(/:~)&(2 3 5 7&|"1 0)]
Try it online!
Brute force...
answered 21 hours ago
Galen IvanovGalen Ivanov
7,38211034
answered 21 hours ago
Galen IvanovGalen Ivanov
7,38211034
answered 21 hours ago
Galen IvanovGalen Ivanov
7,38211034
answered 21 hours ago
Galen IvanovGalen Ivanov
7,38211034
7,38211034
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s=2,3,5,7])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below 2,3,5,7 in each coordinate. Note that Or@@ is False.
answered 4 hours ago
Misha LavrovMisha Lavrov
4,251424
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s=2,3,5,7])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below 2,3,5,7 in each coordinate. Note that Or@@ is False.
answered 4 hours ago
Misha LavrovMisha Lavrov
4,251424
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s=2,3,5,7])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below 2,3,5,7 in each coordinate. Note that Or@@ is False.
answered 4 hours ago
Misha LavrovMisha Lavrov
4,251424
$endgroup$
Wolfram Language (Mathematica), 56 bytes
Or@@(Min[s-#]>0&/@Rest@Permutations@Mod[#,s=2,3,5,7])&
Try it online!
Finds all non-identity permutations of the remainders of the input modulo 2, 3, 5, 7, and checks if any of them are below 2,3,5,7 in each coordinate. Note that Or@@ is False.
answered 4 hours ago
Misha LavrovMisha Lavrov
4,251424
answered 4 hours ago
Misha LavrovMisha Lavrov
4,251424
answered 4 hours ago
Misha LavrovMisha Lavrov
4,251424
answered 4 hours ago
Misha LavrovMisha Lavrov
4,251424
4,251424
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).
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