| Bytes | Lang | Time | Link |
|---|---|---|---|
| 111 | Perl 5 | 241227T113955Z | Donat |
| 026 | K ngn/k | 180922T212741Z | ngn |
| 024 | APLNARS | 241222T163013Z | Rosario |
| 019 | Uiua | 241221T025840Z | nyxbird |
| 019 | Japt g | 180921T162147Z | Shaggy |
| 024 | APL Dyalog Unicode | 180925T175438Z | J. Sall& |
| 186 | C gcc | 180921T232818Z | Jonathan |
| 088 | Ruby | 180925T112701Z | G B |
| 064 | Wolfram Language Mathematica | 180925T080456Z | alephalp |
| 097 | Python 2 | 180922T204536Z | Chas Bro |
| 013 | 05AB1E | 180921T170608Z | Kevin Cr |
| 010 | Jelly | 180921T172812Z | Jonathan |
| 014 | MATL | 180921T155225Z | Luis Men |
Perl 5, 111 bytes
sub{%r=%d=();map$r{$_*$_%$n}++,1..($n=$_[0]);//,map$d{($_-$'+$n)%$n}++,@/for@/=keys%r;(sort{$a-$b}values%d)[0]}
K (ngn/k), 26 bytes
{&/#'=,/x!r-\:r:?x!i*i:!x}
{ } function with argument x
!x integers from 0 to x-1
i: assign to i
x! mod x
? unique
r: assign to r
-\: subtract from each left
r-\:r matrix of all differences
x! mod x
,/ concatenate the rows of the matrix
= group, returns a dictionary from unique values to lists of occurrence indices
#' length of each value in the dictionary
&/ minimum
APL(NARS), 24 chars
{⌊/+/∘.=⍨,⍵∣∘.-⍨∪⍵∣×⍨⍳⍵}
I see the other APL solution and i find "∘.-⍨" is shorter than "-/¨∘.,⍨"..., test
{⌊/+/∘.=⍨,⍵∣∘.-⍨∪⍵∣×⍨⍳⍵}¨1..50
1 2 1 1 1 2 2 1 1 2 3 1 3 4 1 1 4 2 5 1 2 6 6 1 2 6 2 2 7 2 8 1 3 8 2 1 9 10 3 1 10 4 11 3 1 12 12 1 8 4
in the test it seems ok even i not sure... the size of intervall is 2x and not has the 0 (but result until 50 seems ok)...
Uiua, 19 bytes
/↧°⊚⊛♭◿:⊞-.◴◿:°√⇡..
/↧°⊚⊛♭◿:⊞-.◴◿:°√⇡..
°√⇡ # the squared range from 0 to n-1
◴◿: . # mod n, deduplicated
◿:⊞-. . # subtraction table, mod n
°⊚⊛♭ # the count of each value
/↧ # minimum
The .s and ◿:s feel golfable, but I'm not really sure how.
Japt -g, 22 20 19 bytes
Outputs the nth term in the sequence. Starts struggling when input >900.
õȲuUÃâ ïÍmuU ü mÊÍ
APL (Dyalog Unicode), 28 24 bytes
{⌊/⊢∘≢⌸∊⍵|∘.-⍨∪⍵|×⍨⍳⍵+1}
Prefix direct function. Uses ⎕IO←0.
Thanks to Cows quack for 4 bytes!
How:
{⌊/⊢∘≢⌸∊⍵|∘.-⍨∪⍵|×⍨⍳⍵+1} ⍝ Dfn, argument ⍵
⍳⍵+1 ⍝ Range [0..⍵]
×⍨ ⍝ Squared
⍵| ⍝ Modulo ⍵
∪ ⍝ Unique
∘.-⍨ ⍝ Pairwise subtraction table
∊⍵| ⍝ Modulo ⍵, flattened
⌸ ⍝ Key; groups indices (in its ⍵) of values (in its ⍺).
⊢∘≢ ⍝ Tally (≢) the indices. This returns the number of occurrences of each element.
⌊/ ⍝ Floor reduction; returns the smallest number.
C (gcc), 202 200 190 188 187 186 bytes
- Saved
twotwelvefourteenfifteen bytes thanks to ceilingcat. - Saved a byte.
Q(u,a){int*d,*r,A[u],t,i[a=u],*c=i,k;for(;a--;k||(*c++=a*a%u))for(k=a[A]=0,r=i;r<c;)k+=a*a%u==*r++;for(r=c;i-r--;)for(d=i;d<c;++A[(u+*r-*d++)%u]);for(t=*A;++a<u;t=k&&k<t?k:t)k=A[a];u=t;}
Ruby, 88 bytes
->n{(z=(w=(0..n).map{|x|x*x%n}|[]).product(w).map{|a,b|(a-b)%n}).map{|c|z.count(c)}.min}
Wolfram Language (Mathematica), 64 bytes
Min[Last/@Tally@Mod[Tuples[Union@Mod[Range@#^2,#],2].{1,-1},#]]&
Python 2, 97 bytes
def f(n):r={i*i%n for i in range(n)};r=[(s-t)%n for s in r for t in r];return min(map(r.count,r))
05AB1E, 22 20 15 13 bytes
LnI%êãÆI%D.m¢
-2 bytes thanks to @Mr. Xcoder.
Try it online or verify the first 99 test cases (in about 3 seconds). (NOTE: The Python legacy version is used on TIO instead of the new Elixir rewrite. It's about 10x faster, but requires a trailing ¬ (head) because .m returns a list instead of a single item, which I've added to the footer.)
Explanation:
L # Create a list in the range [1, (implicit) input]
n # Square each
I% # And then modulo each with the input
ê # Sort and uniquify the result (faster than just uniquify apparently)
ã # Create pairs (cartesian product with itself)
Æ # Get the differences between each pair
I% # And then modulo each with the input
D.m # Take the least frequent number (numbers in the legacy version)
¢ # Take the count it (or all the numbers in the legacy version, which are all the same)
# (and output it implicitly)
Jelly, 13 10 bytes
-1 thanks to Dennis (forcing dyadic interpretation with a leading ð)
-2 more also thanks to Dennis (since the pairs may be de-duplicated we can avoid an R and a 2)
ðp²%QI%ĠẈṂ
A monadic link accepting a positive integer which yields a non-negative integer.
Try it online! Or see the first 50 terms.
How?
ðp²%QI%ĠẈṂ - Link: integer, n e.g. 6
ð - start a new dyadic chain - i.e. f(Left=n, Right=n)
p - Cartesian product of (implicit ranges) [[1,1],[1,2],[1,3],[1,4],[1,5],[1,6],[2,1],[2,2],[2,3],[2,4],[2,5],[2,6],[3,1],[3,2],[3,3],[3,4],[3,5],[3,6],[4,1],[4,2],[4,3],[4,4],[4,5],[4,6],[5,1],[5,2],[5,3],[5,4],[5,5],[5,6],[6,1],[6,2],[6,3],[6,4],[6,5],[6,6]]
² - square (vectorises) [[1,1],[1,4],[1,9],[1,16],[1,25],[1,36],[4,1],[4,4],[4,9],[4,16],[4,25],[4,36],[9,1],[9,4],[9,9],[9,16],[9,25],[9,36],[16,1],[16,4],[16,9],[16,16],[16,25],[16,36],[25,1],[25,4],[25,9],[25,16],[25,25],[25,36],[36,1],[36,4],[36,9],[36,16],[36,25],[36,36]]
% - modulo (by Right) (vectorises) [[1,1],[1,4],[1,3],[1,4],[1,1],[1,0],[4,1],[4,4],[4,3],[4,4],[4,1],[4,0],[3,1],[3,4],[3,3],[3,4],[3,1],[3,0],[4,1],[4,4],[4,3],[4,4],[4,1],[4,0],[1,1],[1,4],[1,3],[1,4],[1,1],[1,0],[0,1],[0,4],[0,3],[0,4],[0,1],[0,0]]
Q - de-duplicate [[1,1],[1,4],[1,3],[1,0],[4,1],[4,4],[4,3],[4,0],[3,1],[3,4],[3,3],[3,0],[0,1],[0,4],[0,3],[0,0]]
I - incremental differences (vectorises) [0,3,2,-1,-3,0,-1,-4,-2,1,0,-3,1,4,3,0]
% - modulo (by Right) (vectorises) [0,3,2,5,3,0,5,2,4,1,0,3,1,4,3,0]
Ġ - group indices by value [[1,6,11,16],[10,13],[3,8],[2,5,12,15],[9,14],[4,7]]
Ẉ - length of each [3,2,2,4,2,2]
Ṃ - minimum 2
MATL, 14 bytes
:UG\u&-G\8#uX<
Try it online! Or verify the first 30 values.
Explanation
: % Implicit input. Range
U % Square, element-wise
G % Push input again
\ % Modulo, element-wise
u % Unique elements
&- % Table of pair-wise differences
G % Push input
\ % Modulo, element-wise
8#u % Number of occurrences of each element
X< % Minimum. Implicit display