Stern-Brocot sequence: Difference between revisions
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=={{header|Perl 6}}== |
=={{header|Perl 6}}== |
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{{works with|rakudo|2017-03}} |
{{works with|rakudo|2017-03}} |
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<lang perl6>constant Stern-Brocot = |
<lang perl6>constant @Stern-Brocot = 1, 1, { |
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( |
|(@_[$_ - 1] + @_[$_], @_[$_]) given ++$ |
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| ⚫ | |||
|(@_[$_ - 1] + @_[$_], @_[$_]) given ++$ |
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| ⚫ | |||
| ⚫ | |||
| ⚫ | |||
for (flat 1..10, 100) -> $ix { |
for (flat 1..10, 100) -> $ix { |
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say "first occurrence of $ix is at index |
say "first occurrence of $ix is at index : ", 1 + @Stern-Brocot.first($ix, :k); |
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} |
} |
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say so 1 == all map ^1000: { [gcd] Stern-Brocot[$_, $_ + 1] }</lang> |
say so 1 == all map ^1000: { [gcd] @Stern-Brocot[$_, $_ + 1] }</lang> |
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{{out}} |
{{out}} |
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<pre>1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 |
<pre>1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 |
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Revision as of 02:01, 29 March 2017
You are encouraged to solve this task according to the task description, using any language you may know.
For this task, the Stern-Brocot sequence is to be generated by an algorithm similar to that employed in generating the Fibonacci sequence.
- The first and second members of the sequence are both 1:
- 1, 1
- Start by considering the second member of the sequence
- Sum the considered member of the sequence and its precedent, (1 + 1) = 2, and append it to the end of the sequence:
- 1, 1, 2
- Append the considered member of the sequence to the end of the sequence:
- 1, 1, 2, 1
- Consider the next member of the series, (the third member i.e. 2)
- GOTO 3
- ─── Expanding another loop we get: ───
- Sum the considered member of the sequence and its precedent, (2 + 1) = 3, and append it to the end of the sequence:
- 1, 1, 2, 1, 3
- Append the considered member of the sequence to the end of the sequence:
- 1, 1, 2, 1, 3, 2
- Consider the next member of the series, (the fourth member i.e. 1)
- The task is to
- Create a function/method/subroutine/procedure/... to generate the Stern-Brocot sequence of integers using the method outlined above.
- Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4)
- Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence.
- Show the (1-based) index of where the number 100 first appears in the sequence.
- Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.
Show your output on this page.
- Ref
- Infinite Fractions - Numberphile (Video).
- Trees, Teeth, and Time: The mathematics of clock making.
- A002487 The On-Line Encyclopedia of Integer Sequences.
- Related Tasks
Ada
<lang Ada>with Ada.Text_IO, Ada.Containers.Vectors;
procedure Sequence is
package Vectors is new
Ada.Containers.Vectors(Index_Type => Positive, Element_Type => Positive);
use type Vectors.Vector;
type Sequence is record
List: Vectors.Vector;
Index: Positive;
-- This implements some form of "lazy evaluation":
-- + List holds the elements computed, so far, it is extended
-- if the user tries to "Get" an element not yet computed;
-- + Index is the location of the next element under consideration
end record;
function Initialize return Sequence is
(List => (Vectors.Empty_Vector & 1 & 1), Index => 2);
function Get(Seq: in out Sequence; Location: Positive) return Positive is
-- returns the Location'th element of the sequence
-- extends Seq.List (and then increases Seq.Index) if neccessary
That: Positive := Seq.List.Element(Seq.Index);
This: Positive := That + Seq.List.Element(Seq.Index-1);
begin
while Seq.List.Last_Index < Location loop
Seq.List := Seq.List & This & That; Seq.Index := Seq.Index + 1;
end loop;
return Seq.List.Element(Location);
end Get;
S: Sequence := Initialize;
J: Positive;
use Ada.Text_IO;
begin
-- show first fifteen members
Put("First 15:");
for I in 1 .. 15 loop
Put(Integer'Image(Get(S, I)));
end loop;
New_Line;
-- show the index where 1, 2, 3, ... first appear in the sequence
for I in 1 .. 10 loop
J := 1;
while Get(S, J) /= I loop
J := J + 1;
end loop;
Put("First" & Integer'Image(I) & " at" & Integer'Image(J) & "; ");
if I mod 4 = 0 then
New_Line; -- otherwise, the output gets a bit too ugly
end if;
end loop;
-- show the index where 100 first appears in the sequence
J := 1;
while Get(S, J) /= 100 loop
J := J + 1;
end loop;
Put_Line("First 100 at" & Integer'Image(J) & ".");
-- check GCDs
declare
function GCD (A, B : Integer) return Integer is
M : Integer := A; N : Integer := B; T : Integer;
begin
while N /= 0 loop T := M; M := N; N := T mod N; end loop; return M;
end GCD;
A, B: Positive;
begin
for I in 1 .. 999 loop
A := Get(S, I); B := Get(S, I+1); if GCD(A, B) /= 1 then raise Constraint_Error; end if;
end loop;
Put_Line("Correct: The first 999 consecutive pairs are relative prime!");
exception
when Constraint_Error => Put_Line("Some GCD > 1; this is wrong!!!") ;
end;
end Sequence;</lang>
- Output:
First 15: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1; First 2 at 3; First 3 at 5; First 4 at 9; First 5 at 11; First 6 at 33; First 7 at 19; First 8 at 21; First 9 at 35; First 10 at 39; First 100 at 1179. Correct: The first 999 consecutive pairs are relative prime!
AutoHotkey
<lang AutoHotkey>Found := FindOneToX(100), FoundList := "" Loop, 10
FoundList .= "First " A_Index " found at " Found[A_Index] "`n"
MsgBox, 64, Stern-Brocot Sequence
, % "First 15: " FirstX(15) "`n" . FoundList . "First 100 found at " Found[100] "`n" . "GCDs of all two consecutive members are " (GCDsUpToXAreOne(1000) ? "" : "not ") "one."
return
class SternBrocot {
__New()
{
this[1] := 1
this[2] := 1
this.Consider := 2
}
InsertPair()
{
n := this.Consider
this.Push(this[n] + this[n - 1], this[n])
this.Consider++
}
}
- Show the first fifteen members of the sequence. (This should be
- 1, 1, 2, 1, 3, 2, 3, 1, 4, 3,
- 5, 2, 5, 3, 4)
FirstX(x) {
SB := new SternBrocot()
while SB.MaxIndex() < x
SB.InsertPair()
Loop, % x
Out .= SB[A_Index] ", "
return RTrim(Out, " ,")
}
- Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence.
- Show the (1-based) index of where the number 100 first appears in the sequence.
FindOneToX(x) {
SB := new SternBrocot(), xRequired := x, Found := []
while xRequired > 0 ; While the count of numbers yet to be found is > 0.
{
Loop, 2 ; Consider the second last member and then the last member.
{
n := SB[i := SB.MaxIndex() - 2 + A_Index]
; If number (n) has not been found yet, and it is less than the maximum number to
; find (x), record the index (i) and decrement the count of numbers yet to be found.
if (Found[n] = "" && n <= x)
Found[n] := i, xRequired--
}
SB.InsertPair() ; Insert the two members that will be checked next.
}
return Found
}
- Check that the greatest common divisor of all the two consecutive members of the series up to
- the 1000th member, is always one.
GCDsUpToXAreOne(x) {
SB := new SternBrocot()
while SB.MaxIndex() < x
SB.InsertPair()
Loop, % x - 1
if GCD(SB[A_Index], SB[A_Index + 1]) > 1
return 0
return 1
}
GCD(a, b) {
while b
b := Mod(a | 0x0, a := b)
return a
}</lang>
- Output:
First 15: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 First 1 found at 1 First 2 found at 3 First 3 found at 5 First 4 found at 9 First 5 found at 11 First 6 found at 33 First 7 found at 19 First 8 found at 21 First 9 found at 35 First 10 found at 39 First 100 found at 1179 GCDs of all two consecutive members are one.
C
Recursive function. <lang c>#include <stdio.h>
typedef unsigned int uint;
/* the sequence, 0-th member is 0 */ uint f(uint n) { return n < 2 ? n : (n&1) ? f(n/2) + f(n/2 + 1) : f(n/2); }
uint gcd(uint a, uint b) { return a ? a < b ? gcd(b%a, a) : gcd(a%b, b) : b; }
void find(uint from, uint to) { do { uint n; for (n = 1; f(n) != from ; n++); printf("%3u at Stern #%u.\n", from, n); } while (++from <= to); }
int main(void) { uint n; for (n = 1; n < 16; n++) printf("%u ", f(n)); puts("are the first fifteen.");
find(1, 10); find(100, 0);
for (n = 1; n < 1000 && gcd(f(n), f(n+1)) == 1; n++); printf(n == 1000 ? "All GCDs are 1.\n" : "GCD of #%d and #%d is not 1", n, n+1);
return 0; }</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 are the first fifteen. 1 at Stern #1. 2 at Stern #3. 3 at Stern #5. 4 at Stern #9. 5 at Stern #11. 6 at Stern #33. 7 at Stern #19. 8 at Stern #21. 9 at Stern #35. 10 at Stern #39. 100 at Stern #1179. All GCDs are 1.
The above f() can be replaced by the following, which is much faster but probably less obvious as to how it arrives from the recurrence relations.
<lang c>uint f(uint n)
{
uint a = 1, b = 0;
while (n) {
if (n&1) b += a;
else a += b;
n >>= 1;
}
return b;
}</lang>
C++
<lang cpp>
- include <iostream>
- include <iomanip>
- include <algorithm>
- include <vector>
unsigned gcd( unsigned i, unsigned j ) {
return i ? i < j ? gcd( j % i, i ) : gcd( i % j, j ) : j;
} void createSequence( std::vector<unsigned>& seq, int c ) {
if( 1500 == seq.size() ) return; unsigned t = seq.at( c ) + seq.at( c + 1 ); seq.push_back( t ); seq.push_back( seq.at( c + 1 ) ); createSequence( seq, c + 1 );
} int main( int argc, char* argv[] ) {
std::vector<unsigned> seq( 2, 1 ); createSequence( seq, 0 );
std::cout << "First fifteen members of the sequence:\n ";
for( unsigned x = 0; x < 15; x++ ) {
std::cout << seq[x] << " ";
}
std::cout << "\n\n";
for( unsigned x = 1; x < 11; x++ ) {
std::vector<unsigned>::iterator i = std::find( seq.begin(), seq.end(), x );
if( i != seq.end() ) {
std::cout << std::setw( 3 ) << x << " is at pos. #" << 1 + distance( seq.begin(), i ) << "\n";
}
}
std::cout << "\n";
std::vector<unsigned>::iterator i = std::find( seq.begin(), seq.end(), 100 );
if( i != seq.end() ) {
std::cout << 100 << " is at pos. #" << 1 + distance( seq.begin(), i ) << "\n";
}
std::cout << "\n";
unsigned g;
bool f = false;
for( int x = 0, y = 1; x < 1000; x++, y++ ) {
g = gcd( seq[x], seq[y] );
if( g != 1 ) f = true;
std::cout << std::setw( 4 ) << x + 1 << ": GCD (" << seq[x] << ", "
<< seq[y] << ") = " << g << ( g != 1 ? " <-- ERROR\n" : "\n" );
}
std::cout << "\n" << ( f ? "THERE WERE ERRORS --- NOT ALL GCDs ARE '1'!" : "CORRECT: ALL GCDs ARE '1'!" ) << "\n\n";
return 0;
} </lang>
- Output:
First fifteen members of the sequence:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4
1 is at pos. #1
2 is at pos. #3
3 is at pos. #5
4 is at pos. #9
5 is at pos. #11
6 is at pos. #33
7 is at pos. #19
8 is at pos. #21
9 is at pos. #35
10 is at pos. #39
100 is at pos. #1179
1: GCD (1, 1) = 1
2: GCD (1, 2) = 1
3: GCD (2, 1) = 1
4: GCD (1, 3) = 1
5: GCD (3, 2) = 1
6: GCD (2, 3) = 1
7: GCD (3, 1) = 1
8: GCD (1, 4) = 1
[...]
993: GCD (26, 21) = 1
994: GCD (21, 37) = 1
995: GCD (37, 16) = 1
996: GCD (16, 43) = 1
997: GCD (43, 27) = 1
998: GCD (27, 38) = 1
999: GCD (38, 11) = 1
1000: GCD (11, 39) = 1
CORRECT: ALL GCDs ARE '1'!
Clojure
<lang clojure>;; compute the Nth (1-based) Stern-Brocot number directly (defn nth-stern-brocot [n]
(if (< n 2)
n
(let [h (quot n 2) h1 (inc h) hth (nth-stern-brocot h)]
(if (zero? (mod n 2)) hth (+ hth (nth-stern-brocot h1))))))
- return a lazy version of the entire Stern-Brocot sequence
(defn stern-brocot
([] (stern-brocot 1)) ([n] (cons (nth-stern-brocot n) (lazy-seq (stern-brocot (inc n))))))
(printf "Stern-Brocot numbers 1-15: %s%n"
(clojure.string/join ", " (take 15 (stern-brocot))))
(dorun (for [n (concat (range 1 11) [100])]
(printf "The first appearance of %3d is at index %4d.%n" n (inc (first (keep-indexed #(when (= %2 n) %1) (stern-brocot)))))))</lang>
- Output:
Stern-Brocot numbers 1-15: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 The first appearance of 1 is at index 1. The first appearance of 2 is at index 3. The first appearance of 3 is at index 5. The first appearance of 4 is at index 9. The first appearance of 5 is at index 11. The first appearance of 6 is at index 33. The first appearance of 7 is at index 19. The first appearance of 8 is at index 21. The first appearance of 9 is at index 35. The first appearance of 10 is at index 39. The first appearance of 100 is at index 1179.
Clojure: Using Lazy Sequences
<lang clojure>(ns test-p.core) (defn gcd
"(gcd a b) computes the greatest common divisor of a and b." [a b] (if (zero? b) a (recur b (mod a b))))
(defn stern-brocat-next [p]
" p is the block of the sequence we are using to compute the next block This routine computes the next block " (into [] (concat (rest p) [(+ (first p) (second p))] [(second p)])))
(defn seq-stern-brocat
([] (seq-stern-brocat [1 1]))
([p] (lazy-seq (cons (first p)
(seq-stern-brocat (stern-brocat-next p))))))
- First 15 elements
(println (take 15 (seq-stern-brocat)))
- Where numbers 1 to 10 first appear
(doseq [n (concat (range 1 11) [100])]
(println "The first appearnce of" n "is at index" (some (fn i k (when (= k n) (inc i))) (map-indexed vector (seq-stern-brocat)))))
- Check that gcd between 1st 1000 consecutive elements equals 1
- Create cosecutive pairs of 1st 1000 elements
(def one-thousand-pairs (take 1000 (partition 2 1 (seq-stern-brocat))))
- Check every pair has a gcd = 1
(println (every? (fn ith ith-plus-1 (= (gcd ith ith-plus-1) 1))
one-thousand-pairs))
</lang>
- Output:
(1 1 2 1 3 2 3 1 4 3 5 2 5 3 4) The first appearnce of 1 is at index 1 The first appearnce of 2 is at index 3 The first appearnce of 3 is at index 5 The first appearnce of 4 is at index 9 The first appearnce of 5 is at index 11 The first appearnce of 6 is at index 33 The first appearnce of 7 is at index 19 The first appearnce of 8 is at index 21 The first appearnce of 9 is at index 35 The first appearnce of 10 is at index 39 The first appearnce of 100 is at index 1179 true
Common Lisp
<lang lisp>(defun stern-brocot (numbers)
(declare ((or null (vector integer)) numbers))
(cond ((null numbers)
(setf numbers (make-array 2 :element-type 'integer :adjustable t :fill-pointer t
:initial-element 1)))
((zerop (length numbers))
(vector-push-extend 1 numbers)
(vector-push-extend 1 numbers))
(t
(assert (evenp (length numbers)))
(let* ((considered-index (/ (length numbers) 2))
(considered (aref numbers considered-index))
(precedent (aref numbers (1- considered-index))))
(vector-push-extend (+ considered precedent) numbers)
(vector-push-extend considered numbers))))
numbers)
(defun first-15 ()
(loop for input = nil then seq
for seq = (stern-brocot input)
while (< (length seq) 15)
finally (format t "First 15: ~{~A~^ ~}~%" (coerce (subseq seq 0 15) 'list))))
(defun first-1-to-10 ()
(loop with seq = (stern-brocot nil)
for i from 1 to 10
for index = (loop with start = 0
for pos = (position i seq :start start)
until pos
do (setf start (length seq)
seq (stern-brocot seq))
finally (return (1+ pos)))
do (format t "First ~D at ~D~%" i index)))
(defun first-100 ()
(loop for input = nil then seq
for start = (length input)
for seq = (stern-brocot input)
for pos = (position 100 seq :start start)
until pos
finally (format t "First 100 at ~D~%" (1+ pos))))
(defun check-gcd ()
(loop for input = nil then seq
for seq = (stern-brocot input)
while (< (length seq) 1000)
finally (if (loop for i from 0 below 999
always (= 1 (gcd (aref seq i) (aref seq (1+ i)))))
(write-line "Correct. The GCDs of all the two consecutive numbers are 1.")
(write-line "Wrong."))))
(defun main ()
(first-15) (first-1-to-10) (first-100) (check-gcd))</lang>
- Output:
First 15: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First 1 at 1 First 2 at 3 First 3 at 5 First 4 at 9 First 5 at 11 First 6 at 33 First 7 at 19 First 8 at 21 First 9 at 35 First 10 at 39 First 100 at 1179 Correct. The GCDs of all the two consecutive numbers are 1.
D
<lang d>import std.stdio, std.numeric, std.range, std.algorithm;
/// Generates members of the stern-brocot series, in order, /// returning them when the predicate becomes false. uint[] sternBrocot(bool delegate(in uint[]) pure nothrow @safe @nogc pred=seq => seq.length < 20) pure nothrow @safe {
typeof(return) sb = [1, 1];
size_t i = 0;
while (pred(sb)) {
sb ~= [sb[i .. i + 2].sum, sb[i + 1]];
i++;
}
return sb;
}
void main() {
enum nFirst = 15;
writefln("The first %d values:\n%s\n", nFirst,
sternBrocot(seq => seq.length < nFirst).take(nFirst));
foreach (immutable nOccur; iota(1, 10 + 1).chain(100.only))
writefln("1-based index of the first occurrence of %3d in the series: %d",
nOccur, sternBrocot(seq => nOccur != seq[$ - 2]).length - 1);
enum nGcd = 1_000;
auto s = sternBrocot(seq => seq.length < nGcd).take(nGcd);
assert(zip(s, s.dropOne).all!(ss => ss[].gcd == 1),
"A fraction from adjacent terms is reducible.");
}</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179
This uses a queue from the Queue/usage Task: <lang d>import std.stdio, std.algorithm, std.range, std.numeric, queue_usage2;
struct SternBrocot {
private auto sb = GrowableCircularQueue!uint(1, 1);
enum empty = false;
@property uint front() pure nothrow @safe @nogc {
return sb.front;
}
uint popFront() pure nothrow @safe {
sb.push(sb.front + sb[1]);
sb.push(sb[1]);
return sb.pop;
}
}
void main() {
SternBrocot().drop(50_000_000).front.writeln;
}</lang>
- Output:
7004
Direct Version:
<lang d>void main() {
import std.stdio, std.numeric, std.range, std.algorithm, std.bigint, std.conv;
/// Stern-Brocot sequence, 0-th member is 0.
T sternBrocot(T)(T n) pure nothrow /*safe*/ {
T a = 1, b = 0;
while (n) {
if (n & 1) b += a;
else a += b;
n >>= 1;
}
return b;
}
alias sb = sternBrocot!uint;
enum nFirst = 15;
writefln("The first %d values:\n%s\n", nFirst, iota(1, nFirst + 1).map!sb);
foreach (immutable nOccur; iota(1, 10 + 1).chain(100.only))
writefln("1-based index of the first occurrence of %3d in the series: %d",
nOccur, sequence!q{n}.until!(n => sb(n) == nOccur).walkLength);
auto s = iota(1, 1_001).map!sb;
assert(s.zip(s.dropOne).all!(ss => ss[].gcd == 1),
"A fraction from adjacent terms is reducible.");
sternBrocot(10.BigInt ^^ 20_000).text.length.writeln;
}</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179 7984
EchoLisp
Function
<lang lisp>
- stern (2n ) = stern (n)
- stern(2n+1) = stern(n) + stern(n+1)
(define (stern n) (cond (( < n 3) 1) ((even? n) (stern (/ n 2))) (else (let ((m (/ (1- n) 2))) (+ (stern m) (stern (1+ m))))))) (remember 'stern) </lang>
- Output:
<lang lisp>
- generate the sequence and check GCD
(for ((n 10000))
(unless (= (gcd (stern n) (stern (1+ n))) 1) (error "BAD GCD" n))) → #t
- first items
(define sterns (cache 'stern)) (subvector sterns 1 16)
→ #( 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4)
- first occurences index
(for ((i (in-range 1 11))) (write (vector-index i sterns))) → 0 3 5 9 11 33 19 21 35 39
- 100
(writeln (vector-index 100 sterns)) → 1179
(stern 900000) → 446 (stern 900001) → 2479 </lang>
Stream
From A002487, if we group the elements by two, we get (uniquely) all the rationals. Another way to generate the rationals, hence the stern sequence, is to iterate the function f(x) = floor(x) + 1 - fract(x).
<lang lisp>
- grouping
(for ((i (in-range 2 40 2))) (write (/ (stern i)(stern (1+ i))))) → 1/2 1/3 2/3 1/4 3/5 2/5 3/4 1/5 4/7 3/8 5/7 2/7 5/8 3/7 4/5 1/6 5/9 4/11 7/10
- computing f(1), f(f(1)), etc.
(define (f x)
(let [(a (/ (- (floor x) -1 (fract x))))] (if (> a 1) (f a) (cons a a))))
(define T (make-stream f 1)) (for((i 19)) (write (stream-iterate T)))
→ 1/2 1/3 2/3 1/4 3/5 2/5 3/4 1/5 4/7 3/8 5/7 2/7 5/8 3/7 4/5 1/6 5/9 4/11 7/10 </lang>
Elixir
<lang elixir>defmodule SternBrocot do
def sequence do
Stream.unfold({0,{1,1}}, fn {i,acc} ->
a = elem(acc, i)
b = elem(acc, i+1)
{a, {i+1, Tuple.append(acc, a+b) |> Tuple.append(b)}}
end)
end
def task do
IO.write "First fifteen members of the sequence:\n "
IO.inspect Enum.take(sequence, 15)
Enum.each(Enum.concat(1..10, [100]), fn n ->
i = Enum.find_index(sequence, &(&1==n)) + 1
IO.puts "#{n} first appears at #{i}"
end)
Enum.take(sequence, 1000)
|> Enum.chunk(2,1)
|> Enum.all?(fn [a,b] -> gcd(a,b) == 1 end)
|> if(do: "All GCD's are 1", else: "Whoops, not all GCD's are 1!")
|> IO.puts
end
defp gcd(a,0), do: abs(a)
defp gcd(a,b), do: gcd(b, rem(a,b))
end
SternBrocot.task</lang>
- Output:
First fifteen members of the sequence: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1 first appears at 1 2 first appears at 3 3 first appears at 5 4 first appears at 9 5 first appears at 11 6 first appears at 33 7 first appears at 19 8 first appears at 21 9 first appears at 35 10 first appears at 39 100 first appears at 1179 All GCD's are 1
Go
<lang go>package main
import (
"fmt"
"sternbrocot"
)
func main() {
// Task 1, using the conventional sort of generator that generates // terms endlessly. g := sb.Generator()
// Task 2, demonstrating the generator.
fmt.Println("First 15:")
for i := 1; i <= 15; i++ {
fmt.Printf("%2d: %d\n", i, g())
}
// Task 2 again, showing a simpler technique that might or might not be
// considered to "generate" terms.
s := sb.New()
fmt.Println("First 15:", s.FirstN(15))
// Tasks 3 and 4.
for _, x := range []int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100} {
fmt.Printf("%3d at 1-based index %d\n", x, 1+s.Find(x))
}
// Task 5.
fmt.Println("1-based indexes: gcd")
for n, f := range s.FirstN(1000)[:999] {
g := gcd(f, (*s)[n+1])
fmt.Printf("%d,%d: gcd(%d, %d) = %d\n", n+1, n+2, f, (*s)[n+1], g)
if g != 1 {
panic("oh no!")
return
}
}
}
// gcd copied from greatest common divisor task func gcd(x, y int) int {
for y != 0 {
x, y = y, x%y
}
return x
}</lang> <lang go>// SB implements the Stern-Brocot sequence. // // Generator() satisfies RC Task 1. For remaining tasks, Generator could be // used but FirstN(), and Find() are simpler methods for specific stopping // criteria. FirstN and Find might also be considered to satisfy Task 1, // in which case Generator would not really be needed. Anyway, there it is. package sb
// Seq represents an even number of terms of a Stern-Brocot sequence. // // Terms are stored in a slice. Terms start with 1. // (Specifically, the zeroth term, 0, given in OEIS A002487 is not represented.) // Term 1 (== 1) is stored at slice index 0. // // Methods on Seq rely on Seq always containing an even number of terms. type Seq []int
// New returns a Seq with the two base terms. func New() *Seq {
return &Seq{1, 1} // Step 1 of the RC task.
}
// TwoMore appends two more terms to p. // It's the body of the loop in the RC algorithm. // Generate(), FirstN(), and Find() wrap this body in different ways. func (p *Seq) TwoMore() {
s := *p n := len(s) / 2 // Steps 2 and 5 of the RC task. c := s[n] *p = append(s, c+s[n-1], c) // Steps 3 and 4 of the RC task.
}
// Generator returns a generator function that returns successive terms // (until overflow.) func Generator() func() int {
n := 0
p := New()
return func() int {
if len(*p) == n {
p.TwoMore()
}
t := (*p)[n]
n++
return t
}
}
// FirstN lazily extends p as needed so that it has at least n terms. // FirstN then returns a list of the first n terms. func (p *Seq) FirstN(n int) []int {
for len(*p) < n {
p.TwoMore()
}
return []int((*p)[:n])
}
// Find lazily extends p as needed until it contains the value x // Find then returns the slice index of x in p. func (p *Seq) Find(x int) int {
for n, f := range *p {
if f == x {
return n
}
}
for {
p.TwoMore()
switch x {
case (*p)[len(*p)-2]:
return len(*p) - 2
case (*p)[len(*p)-1]:
return len(*p) - 1
}
}
}</lang>
- Output:
First 15: 1: 1 2: 1 3: 2 4: 1 5: 3 6: 2 7: 3 8: 1 9: 4 10: 3 11: 5 12: 2 13: 5 14: 3 15: 4 First 15: [1 1 2 1 3 2 3 1 4 3 5 2 5 3 4] 1 at 1-based index 1 2 at 1-based index 3 3 at 1-based index 5 4 at 1-based index 9 5 at 1-based index 11 6 at 1-based index 33 7 at 1-based index 19 8 at 1-based index 21 9 at 1-based index 35 10 at 1-based index 39 100 at 1-based index 1179 1-based indexes: gcd 1,2: gcd(1, 1) = 1 2,3: gcd(1, 2) = 1 3,4: gcd(2, 1) = 1 4,5: gcd(1, 3) = 1 ... 998,999: gcd(27, 38) = 1 999,1000: gcd(38, 11) = 1
Haskell
<lang haskell>import Data.List
sb = 1:1: f (tail sb) sb where f (a:aa) (b:bb) = a+b : a : f aa bb
main = do print $ take 15 sb print [(i,1 + (\(Just i)->i) (elemIndex i sb)) | i <- [1..10]++[100]] print $ all (\(a,b)->1 == gcd a b) $ take 1000 $ zip sb (tail sb)</lang>
- Output:
[1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] [(1,1),(2,3),(3,5),(4,9),(5,11),(6,33),(7,19),(8,21),(9,35),(10,39),(100,1179)] True
J
We have two different kinds of list specifications here (length of the sequence, and the presence of certain values in the sequence). Also the underlying list generation mechanism is somewhat arbitrary. So let's generate the sequence iteratively and provide a truth valued function of the intermediate sequences to determine when we have generated one which is adequately long:
<lang J>sternbrocot=:1 :0
ind=. 0
seq=. 1 1
while. -. u seq do.
ind=. ind+1
seq=. seq, +/\. seq {~ _1 0 +ind
end.
)</lang>
(Grammatical aside: this is an adverb which generates a noun without taking any x/y arguments. So usage is: u sternbrocot. J does have precedence rules, just not very many of them. Users of other languages can get a rough idea of the grammatical terms like this: adverb is approximately like a macro, verb approximately like a function and noun is approximately like a number. Also x and y are J's names for left and right noun arguments, and u and v are J's names for left and right verb arguments. An adverb has a left verb argument. There are some other important constraints but that's probably more than enough detail for this task.)
First fifteen members of the sequence:
<lang J> 15{.(15<:#) sternbrocot 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4</lang>
One based indices of where numbers 1-10 first appear in the sequence:
<lang J> 1+(10 e. ]) sternbrocot i.1+i.10 1 3 5 9 11 33 19 21 35 39</lang>
One based index of where the number 100 first appears:
<lang J> 1+(100 e. ]) sternbrocot i. 100 1179</lang>
List of distinct greatest common divisors of adjacent number pairs from a sternbrocot sequence which includes the first 1000 elements:
<lang J> ~.2 +./\ (1000<:#) sternbrocot 1</lang>
Java
This example generates the first 1200 members of the sequence since that is enough to cover all of the tests in the description. It borrows the gcd method from BigInteger rather than using its own.
<lang java5>import java.math.BigInteger;
import java.util.LinkedList;
public class SternBrocot { static LinkedList<Integer> sequence = new LinkedList<Integer>()Template:Add(1); add(1);;
private static void genSeq(int n){ for(int conIdx = 1; sequence.size() < n; conIdx++){ int consider = sequence.get(conIdx); int pre = sequence.get(conIdx - 1); sequence.add(consider + pre); sequence.add(consider); }
}
public static void main(String[] args){ genSeq(1200); System.out.println("The first 15 elements are: " + sequence.subList(0, 15)); for(int i = 1; i <= 10; i++){ System.out.println("First occurrence of " + i + " is at " + (sequence.indexOf(i) + 1)); }
System.out.println("First occurrence of 100 is at " + (sequence.indexOf(100) + 1));
boolean failure = false; for(int i = 0; i < 999; i++){ failure |= !BigInteger.valueOf(sequence.get(i)).gcd(BigInteger.valueOf(sequence.get(i + 1))).equals(BigInteger.ONE); } System.out.println("All GCDs are" + (failure ? " not" : "") + " 1"); } }</lang>
- Output:
The first 15 elements are: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] First occurrence of 1 is at 1 First occurrence of 2 is at 3 First occurrence of 3 is at 5 First occurrence of 4 is at 9 First occurrence of 5 is at 11 First occurrence of 6 is at 33 First occurrence of 7 is at 19 First occurrence of 8 is at 21 First occurrence of 9 is at 35 First occurrence of 10 is at 39 First occurrence of 100 is at 1179 All GCDs are 1
Stern-Brocot Tree
<lang java>import java.awt.*; import javax.swing.*;
public class SternBrocot extends JPanel {
public SternBrocot() {
setPreferredSize(new Dimension(800, 500));
setFont(new Font("Arial", Font.PLAIN, 18));
setBackground(Color.white);
}
private void drawTree(int n1, int d1, int n2, int d2,
int x, int y, int gap, int lvl, Graphics2D g) {
if (lvl == 0)
return;
// mediant
int numer = n1 + n2;
int denom = d1 + d2;
if (lvl > 1) {
g.drawLine(x + 5, y + 4, x - gap + 5, y + 124);
g.drawLine(x + 5, y + 4, x + gap + 5, y + 124);
}
g.setColor(getBackground());
g.fillRect(x - 10, y - 15, 35, 40);
g.setColor(getForeground());
g.drawString(String.valueOf(numer), x, y);
g.drawString("_", x, y + 2);
g.drawString(String.valueOf(denom), x, y + 22);
drawTree(n1, d1, numer, denom, x - gap, y + 120, gap / 2, lvl - 1, g);
drawTree(numer, denom, n2, d2, x + gap, y + 120, gap / 2, lvl - 1, g);
}
@Override
public void paintComponent(Graphics gg) {
super.paintComponent(gg);
Graphics2D g = (Graphics2D) gg;
g.setRenderingHint(RenderingHints.KEY_ANTIALIASING,
RenderingHints.VALUE_ANTIALIAS_ON);
int w = getWidth();
drawTree(0, 1, 1, 0, w / 2, 50, w / 4, 4, g); }
public static void main(String[] args) {
SwingUtilities.invokeLater(() -> {
JFrame f = new JFrame();
f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
f.setTitle("Stern-Brocot Tree");
f.setResizable(false);
f.add(new SternBrocot(), BorderLayout.CENTER);
f.pack();
f.setLocationRelativeTo(null);
f.setVisible(true);
});
}
}</lang>
jq
In jq 1.4, there is no equivalent of "yield" for unbounded streams, and so the following uses "until".
Foundations: <lang jq>def until(cond; update):
def _until: if cond then . else (update | _until) end; try _until catch if .== "break" then empty else . end ;
def gcd(a; b):
# subfunction expects [a,b] as input
# i.e. a ~ .[0] and b ~ .[1]
def rgcd: if .[1] == 0 then .[0]
else [.[1], .[0] % .[1]] | rgcd
end;
[a,b] | rgcd ;</lang>
The A002487 integer sequence:
The following definition is in strict accordance with https://oeis.org/A002487: i.e. a(0) = 0, a(1) = 1; for n > 0: a(2*n) = a(n), a(2*n+1) = a(n) + a(n+1). The n-th element of the Rosetta Code sequence (counting from 1) is thus a[n], which accords with the fact that jq arrays have an index origin of 0. <lang jq># If n is non-negative, then A002487(n)
- generates an array with at least n elements of
- the A002487 sequence;
- if n is negative, elements are added until (-n)
- is found.
def A002487(n):
[0,1]
| until(
length as $l
| if n >= 0 then $l >= n
else .[$l-1] == -n
end;
length as $l
| ($l / 2) as $n
| .[$l] = .[$n]
| if (.[$l-2] == -n) then .
else .[$l + 1] = .[$n] + .[$n+1]
end ) ;</lang> The tasks: <lang jq># Generate a stream of n integers beginning with 1,1... def stern_brocot(n): A002487(n+1) | .[1:n+1][];
- Return the index (counting from 1) of n in the
- sequence starting with 1,1,...
def stern_brocot_index(n):
A002487(-n) | length -1 ;
def index_task:
(range(1;11), 100) as $i | "index of \($i) is \(stern_brocot_index($i))" ;
def gcd_task:
A002487(1000)
| . as $A
| reduce range(0; length-1) as $i
( [];
gcd( $A[$i]; $A[$i+1] ) as $gcd
| if $gcd == 1 then . else . + [$gcd] end)
| if length == 0 then "GCDs are all 1"
else "GCDs include \(.)" end ;
"First 15 integers of the Stern-Brocot sequence",
"as defined in the task description are:",
stern_brocot(15),
"",
"Using an index origin of 1:",
index_task,
"",
gcd_task
</lang>
- Output:
<lang sh>$ jq -r -n -f stern_brocot.jq First 15 integers of the Stern-Brocot sequence as defined in the task description are: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4
Using an index origin of 1: index of 1 is 1 index of 2 is 3 index of 3 is 5 index of 4 is 9 index of 5 is 11 index of 6 is 33 index of 7 is 19 index of 8 is 21 index of 9 is 35 index of 10 is 39 index of 100 is 1179
GCDs are all 1</lang>
Kotlin
<lang scala>// version 1.1.0
val sbs = mutableListOf(1, 1)
fun sternBrocot(n: Int, fromStart: Boolean = true) {
if (n < 4 || (n % 2 != 0)) throw IllegalArgumentException("n must be >= 4 and even")
var consider = if (fromStart) 1 else n / 2 - 1
while (true) {
val sum = sbs[consider] + sbs[consider - 1]
sbs.add(sum)
sbs.add(sbs[consider])
if (sbs.size == n) break
consider++
}
}
fun gcd(a: Int, b: Int): Int = if (b == 0) a else gcd(b, a % b)
fun main(args: Array<String>) {
var n = 16 // needs to be even to ensure 'considered' number is added
println("First 15 members of the Stern-Brocot sequence")
sternBrocot(n)
println(sbs.take(15))
val firstFind = IntArray(11) // all zero by default
firstFind[0] = -1 // needs to be non-zero for subsequent test
for ((i, v) in sbs.withIndex())
if (v <= 10 && firstFind[v] == 0) firstFind[v] = i + 1
loop@ while (true) {
n += 2
sternBrocot(n, false)
val vv = sbs.takeLast(2)
var m = n - 1
for (v in vv) {
if (v <= 10 && firstFind[v] == 0) firstFind[v] = m
if (firstFind.all { it != 0 }) break@loop
m++
}
}
println("\nThe numbers 1 to 10 first appear at the following indices:")
for (i in 1..10) println("${"%2d".format(i)} -> ${firstFind[i]}")
print("\n100 first appears at index ")
while (true) {
n += 2
sternBrocot(n, false)
val vv = sbs.takeLast(2)
if (vv[0] == 100) {
println(n - 1); break
}
if (vv[1] == 100) {
println(n); break
}
}
print("\nThe GCDs of each pair of the series up to the 1000th member are ")
for (p in 0..998 step 2) {
if (gcd(sbs[p], sbs[p + 1]) != 1) {
println("not all one")
return
}
}
println("all one")
}</lang>
- Output:
First 15 members of the Stern-Brocot sequence [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] The numbers 1 to 10 first appear at the following indices: 1 -> 1 2 -> 3 3 -> 5 4 -> 9 5 -> 11 6 -> 33 7 -> 19 8 -> 21 9 -> 35 10 -> 39 100 first appears at index 1179 The GCDs of each pair of the series up to the 1000th member are all one
Lua
<lang Lua>-- Task 1 function sternBrocot (n)
local sbList, pos, c = {1, 1}, 2
repeat
c = sbList[pos]
table.insert(sbList, c + sbList[pos - 1])
table.insert(sbList, c)
pos = pos + 1
until #sbList >= n
return sbList
end
-- Return index in table 't' of first value matching 'v' function findFirst (t, v)
for key, value in pairs(t) do
if v then
if value == v then return key end
else
if value ~= 0 then return key end
end
end
return nil
end
-- Return greatest common divisor of 'x' and 'y' function gcd (x, y)
if y == 0 then
return math.abs(x)
else
return gcd(y, x % y)
end
end
-- Check GCD of adjacent values in 't' up to 1000 is always 1 function task5 (t)
for pos = 1, 1000 do
if gcd(t[pos], t[pos + 1]) ~= 1 then return "FAIL" end
end
return "PASS"
end
-- Main procedure local sb = sternBrocot(10000) io.write("Task 2: ") for n = 1, 15 do io.write(sb[n] .. " ") end print("\n\nTask 3:") for i = 1, 10 do print("\t" .. i, findFirst(sb, i)) end print("\nTask 4: " .. findFirst(sb, 100)) print("\nTask 5: " .. task5(sb))</lang>
- Output:
Task 2: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4
Task 3:
1 1
2 3
3 5
4 9
5 11
6 33
7 19
8 21
9 35
10 39
Task 4: 1179
Task 5: PASS
Oforth
<lang Oforth>: stern(n) | l i |
ListBuffer new dup add(1) dup add(1) dup ->l n 1- 2 / loop: i [ l at(i) l at(i 1+) tuck + l add l add ] n 2 mod ifFalse: [ dup removeLast drop ] dup freeze ;
stern(10000) Constant new: Sterns</lang>
- Output:
>Sterns left(15) . [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] ok >10 seq apply(#[ dup . Sterns indexOf . printcr ]) 1 1 2 3 3 5 4 9 5 11 6 33 7 19 8 21 9 35 10 39 ok >Sterns indexOf(100) . 1179 ok >999 seq map(#[ dup Sterns at swap 1 + Sterns at gcd ]) conform(#[ 1 == ]) . 1 ok >
PARI/GP
<lang parigp> \\ Stern-Brocot sequence \\ 5/27/16 aev SternBrocot(n)={ my(L=List([1,1]),k=2); if(n<3,return(L)); for(i=2,n, listput(L,L[i]+L[i-1]); if(k++>=n, break); listput(L,L[i]);if(k++>=n, break)); return(Vec(L)); } \\ Find the first item in any list starting with sind index (return 0 or index). \\ 9/11/2015 aev findinlist(list, item, sind=1)={ my(idx=0, ln=#list); if(ln==0 || sind<1 || sind>ln, return(0)); for(i=sind, ln, if(list[i]==item, idx=i; break;)); return(idx); } { \\ Required tests: my(v,j); v=SternBrocot(15); print1("The first 15: "); print(v); v=SternBrocot(1200); print1("The first i@n: "); \\print(v); for(i=1,10, if(j=findinlist(v,i), print1(i,"@",j,", "))); if(j=findinlist(v,100), print(100,"@",j)); v=SternBrocot(10000); print1("All GCDs=1?: "); j=1; for(i=2,10000, j*=gcd(v[i-1],v[i])); if(j==1, print("Yes"), print("No")); } </lang>
- Output:
The first 15: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] The first i@n: 1@1, 2@3, 3@5, 4@9, 5@11, 6@33, 7@19, 8@21, 9@35, 10@39, 100@1179 All GCDs=1?: Yes
Pascal
<lang pascal>program StrnBrCt; {$IFDEF FPC}
{$MODE DELPHI}
{$ENDIF} const
MaxCnt = 10835282;{ seq[i] < 65536 = high(Word) }
//MaxCnt = 500*1000*1000;{ 2Gbyte -> real 0.85 s user 0.31 } type
tSeqdata = word;//cardinal LongWord pSeqdata = pWord;//pcardinal pLongWord tseq = array of tSeqdata;
function SternBrocotCreate(size:NativeInt):tseq; var
pSeq,pIns : pSeqdata; PosIns : NativeInt; sum : tSeqdata;
Begin
setlength(result,Size+1); dec(Size); //== High(result) pIns := @result[size];// set at end PosIns := -size+2; // negative index campare to 0 pSeq := @result[0];
sum := 1; pSeq[0]:= sum;pSeq[1]:= sum; repeat pIns[PosIns+1] := sum;//append copy of considered inc(sum,pSeq[0]); pIns[PosIns ] := sum; inc(pSeq); inc(PosIns,2);sum := pSeq[1];//aka considered until PosIns>= 0; setlength(result,length(result)-1);
end;
function FindIndex(const s:tSeq;value:tSeqdata):NativeInt; Begin
result := 0;
while result <= High(s) do
Begin
if s[result] = value then
EXIT(result+1);
inc(result);
end;
end;
function gcd_iterative(u, v: NativeInt): NativeInt; //http://rosettacode.org/wiki/Greatest_common_divisor#Pascal_.2F_Delphi_.2F_Free_Pascal var
t: NativeInt;
begin
while v <> 0 do begin t := u;u := v;v := t mod v; end; gcd_iterative := abs(u);
end;
var
seq : tSeq; i : nativeInt;
Begin
seq:= SternBrocotCreate(MaxCnt);
// Show the first fifteen members of the sequence.
For i := 0 to 13 do write(seq[i],',');writeln(seq[14]);
//Show the (1-based) index of where the numbers 1-to-10 first appears in the
For i := 1 to 10 do write(i,' @ ',FindIndex(seq,i),','); writeln(#8#32);
//Show the (1-based) index of where the number 100 first appears in the sequence.
writeln(100,' @ ',FindIndex(seq,100));
//Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.
i := 999;
if i > High(seq) then
i := High(seq);
Repeat
IF gcd_iterative(seq[i],seq[i+1]) <>1 then
Begin
writeln(' failure at ',i+1,' ',seq[i],' ',seq[i+1]);
BREAK;
end;
dec(i);
until i <0;
IF i< 0 then
writeln('GCD-test is O.K.');
setlength(seq,0);
end.</lang>
- Output:
1,1,2,1,3,2,3,1,4,3,5,2,5,3,41 @ 1,2 @ 3,3 @ 5,4 @ 9,5 @ 11,6 @ 33,7 @ 38,8 @ 42,9 @ 47,10 @ 57 100 @ 1179
GCD-test is O.K.
Perl
<lang perl>use strict; use warnings;
sub stern_brocot {
my @list = (1, 1);
sub {
push @list, $list[0] + $list[1], $list[1]; shift @list;
}
}
{
my $generator = stern_brocot; print join ' ', map &$generator, 1 .. 15; print "\n";
}
for (1 .. 10, 100) {
my $index = 1; my $generator = stern_brocot; $index++ until $generator->() == $_; print "first occurrence of $_ is at index $index\n";
}
{
sub gcd {
my ($u, $v) = @_; $v ? gcd($v, $u % $v) : abs($u);
}
my $generator = stern_brocot;
my ($a, $b) = ($generator->(), $generator->());
for (1 .. 1000) {
die "unexpected GCD for $a and $b" unless gcd($a, $b) == 1; ($a, $b) = ($b, $generator->());
}
}</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 first occurrence of 1 is at index 1 first occurrence of 2 is at index 3 first occurrence of 3 is at index 5 first occurrence of 4 is at index 9 first occurrence of 5 is at index 11 first occurrence of 6 is at index 33 first occurrence of 7 is at index 19 first occurrence of 8 is at index 21 first occurrence of 9 is at index 35 first occurrence of 10 is at index 39 first occurrence of 100 is at index 1179
A slightly different method: <lang perl>use List::Util qw/first/; use ntheory qw/gcd vecsum/;
sub stern_diatomic {
my ($p,$q,$i) = (0,1,shift);
while ($i) {
if ($i & 1) { $p += $q; } else { $q += $p; }
$i >>= 1;
}
$p;
}
my @s = map { stern_diatomic($_) } 1..15; print "First fifteen: [@s]\n"; @s = map { my $n=$_; first { stern_diatomic($_) == $n } 1..10000 } 1..10; print "Index of 1-10 first occurrence: [@s]\n"; print "Index of 100 first occurrence: ", (first { stern_diatomic($_) == 100 } 1..10000), "\n"; print "The first 999 consecutive pairs are ",
(vecsum( map { gcd(stern_diatomic($_),stern_diatomic($_+1)) } 1..999 ) == 999)
? "all coprime.\n" : "NOT all coprime!\n";</lang>
- Output:
First fifteen: [1 1 2 1 3 2 3 1 4 3 5 2 5 3 4] Index of 1-10 first occurrence: [1 3 5 9 11 33 19 21 35 39] Index of 100 first occurrence: 1179 The first 999 consecutive pairs are all coprime.
Perl 6
<lang perl6>constant @Stern-Brocot = 1, 1, {
|(@_[$_ - 1] + @_[$_], @_[$_]) given ++$
} ... *; say @Stern-Brocot[^15]; for (flat 1..10, 100) -> $ix {
say "first occurrence of $ix is at index : ", 1 + @Stern-Brocot.first($ix, :k);
} say so 1 == all map ^1000: { [gcd] @Stern-Brocot[$_, $_ + 1] }</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 first occurrence of 1 is at index : 1 first occurrence of 2 is at index : 3 first occurrence of 3 is at index : 5 first occurrence of 4 is at index : 9 first occurrence of 5 is at index : 11 first occurrence of 6 is at index : 33 first occurrence of 7 is at index : 19 first occurrence of 8 is at index : 21 first occurrence of 9 is at index : 35 first occurrence of 10 is at index : 39 first occurrence of 100 is at index : 1179 True
PowerShell
<lang PowerShell>
- An iterative approach
function iter_sb($count = 2000) {
# Taken from RosettaCode GCD challenge
function Get-GCD ($x, $y)
{
if ($y -eq 0) { $x } else { Get-GCD $y ($x%$y) }
}
$answer = @(1,1)
$index = 1
while ($answer.Length -le $count)
{
$answer += $answer[$index] + $answer[$index - 1]
$answer += $answer[$index]
$index++
}
0..14 | foreach {$answer[$_]}
1..10 | foreach {'Index of {0}: {1}' -f $_, ($answer.IndexOf($_) + 1)}
'Index of 100: {0}' -f ($answer.IndexOf(100) + 1)
[bool] $gcd = $true
1..999 | foreach {$gcd = $gcd -and ((Get-GCD $answer[$_] $answer[$_ - 1]) -eq 1)}
'GCD = 1 for first 1000 members: {0}' -f $gcd
} </lang>
- Output:
PS C:\> iter_sb 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 Index of 1: 1 Index of 2: 3 Index of 3: 5 Index of 4: 9 Index of 5: 11 Index of 6: 33 Index of 7: 19 Index of 8: 21 Index of 9: 35 Index of 10: 39 Index of 100: 1179 GCD = 1 for first 1000 members: True
Python
Python: procedural
<lang python>def stern_brocot(predicate=lambda series: len(series) < 20):
"""\ Generates members of the stern-brocot series, in order, returning them when the predicate becomes false
>>> print('The first 10 values:',
stern_brocot(lambda series: len(series) < 10)[:10])
The first 10 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3]
>>>
"""
sb, i = [1, 1], 0
while predicate(sb):
sb += [sum(sb[i:i + 2]), sb[i + 1]]
i += 1
return sb
if __name__ == '__main__':
from fractions import gcd
n_first = 15
print('The first %i values:\n ' % n_first,
stern_brocot(lambda series: len(series) < n_first)[:n_first])
print()
n_max = 10
for n_occur in list(range(1, n_max + 1)) + [100]:
print('1-based index of the first occurrence of %3i in the series:' % n_occur,
stern_brocot(lambda series: n_occur not in series).index(n_occur) + 1)
# The following would be much faster. Note that new values always occur at odd indices
# len(stern_brocot(lambda series: n_occur != series[-2])) - 1)
print()
n_gcd = 1000
s = stern_brocot(lambda series: len(series) < n_gcd)[:n_gcd]
assert all(gcd(prev, this) == 1
for prev, this in zip(s, s[1:])), 'A fraction from adjacent terms is reducible'</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179
Python: More functional
An iterator is used to produce successive members of the sequence. (its sb variable stores less compared to the procedural version above by popping the last element every time around the while loop.
In checking the gcd's, two iterators are tee'd off from the one stream with the second advanced by one value with its call to next().
See the talk page for how a deque was selected over the use of a straightforward list' <lang python>>>> from itertools import takewhile, tee, islice >>> from collections import deque >>> from fractions import gcd >>> >>> def stern_brocot():
sb = deque([1, 1])
while True:
sb += [sb[0] + sb[1], sb[1]]
yield sb.popleft()
>>> [s for _, s in zip(range(15), stern_brocot())]
[1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4]
>>> [1 + sum(1 for i in takewhile(lambda x: x != occur, stern_brocot()))
for occur in (list(range(1, 11)) + [100])]
[1, 3, 5, 9, 11, 33, 19, 21, 35, 39, 1179] >>> prev, this = tee(stern_brocot(), 2) >>> next(this) 1 >>> all(gcd(p, t) == 1 for p, t in islice(zip(prev, this), 1000)) True >>> </lang>
R
<lang r>
- Stern-Brocot sequence
- 12/19/16 aev
SternBrocot <- function(n){
V <- 1; k <- n/2;
for (i in 1:k)
{ V[2*i] = V[i]; V[2*i+1] = V[i] + V[i+1];}
return(V);
}
- Required tests:
require(pracma); { cat(" *** The first 15:",SternBrocot(15),"\n"); cat(" *** The first i@n:","\n"); V=SternBrocot(40); for (i in 1:10) {j=match(i,V); cat(i,"@",j,",")} V=SternBrocot(1200); i=100; j=match(i,V); cat(i,"@",j,"\n"); V=SternBrocot(1000); j=1; for (i in 2:1000) {j=j*gcd(V[i-1],V[i])} if(j==1) {cat(" *** All GCDs=1!\n")} else {cat(" *** All GCDs!=1??\n")} } </lang>
- Output:
> require(pracma) Loading required package: pracma *** The first 15: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 *** The first i@n: 1 @ 1 ,2 @ 3 ,3 @ 5 ,4 @ 9 ,5 @ 11 ,6 @ 33 ,7 @ 19 ,8 @ 21 ,9 @ 35 ,10 @ 39 ,100 @ 1179 *** All GCDs=1! >
Racket
<lang racket>#lang racket
- OEIS Definition
- A002487
- Stern's diatomic series
- (or Stern-Brocot sequence)
- a(0) = 0, a(1) = 1;
- for n > 0
- a(2*n) = a(n),
- a(2*n+1) = a(n) + a(n+1).
(define A002487
(let ((memo (make-hash '((0 . 0) (1 . 1)))))
(lambda (n)
(hash-ref! memo n
(lambda ()
(define n/2 (quotient n 2))
(+ (A002487 n/2) (if (even? n) 0 (A002487 (add1 n/2)))))))))
(define Stern-Brocot A002487)
(displayln "Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4)") (for/list ((i (in-range 1 (add1 15)))) (Stern-Brocot i))
(displayln "Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence.") (for ((n (in-range 1 (add1 10))))
(for/first ((i (in-naturals 1))
#:when (= n (Stern-Brocot i)))
(printf "~a first found at a(~a)~%" n i)))
(displayln "Show the (1-based) index of where the number 100 first appears in the sequence.") (for/first ((i (in-naturals 1)) #:when (= 100 (Stern-Brocot i))) i)
(displayln "Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.") (unless
(for/first ((i (in-range 1 1000))
#:unless (= 1 (gcd (Stern-Brocot i) (Stern-Brocot (add1 i))))) #t)
(display "\tdidn't find gcd > (or otherwise ≠) 1"))</lang>
- Output:
Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4) (1 1 2 1 3 2 3 1 4 3 5 2 5 3 4) Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence. 1 first found at a(1) 2 first found at a(3) 3 first found at a(5) 4 first found at a(9) 5 first found at a(11) 6 first found at a(33) 7 first found at a(19) 8 first found at a(21) 9 first found at a(35) 10 first found at a(39) Show the (1-based) index of where the number 100 first appears in the sequence. 1179 Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one. didn't find gcd > (or otherwise ≠) 1
REXX
<lang rexx>/*REXX program generates & displays a Stern─Brocot sequence; finds 1─based indices; GCDs*/ parse arg N idx fix chk . /*get optional arguments from the C.L. */ if N== | N=="," then N= 15 /* N not defined? Then use default.*/ if idx== | idx=="," then idx= 10 /*IDX " " " " " */ if fix== | fix=="," then fix= 100 /*FIX " " " " " */ if chk== | chk=="," then chk=1000 /*CHK " " " " " */
say center('the first' N "numbers in the Stern─Brocot sequence", 70, '═') a=Stern_Brocot(N) /*invoke function to generate sequence.*/ say a /*display the sequence to the terminal.*/ say say center('the 1-based index for the first' idx "integers", 70, '═') a=Stern_Brocot(-idx) /*invoke function to generate sequence.*/
do i=1 for idx
say 'for ' right(i,length(idx))", the index is: " wordpos(i,a)
end /*i*/
say say center('the 1-based index for' fix, 70, "═") a=Stern_Brocot(-fix) /*invoke function to generate sequence.*/ say 'for ' fix", the index is: " wordpos(fix, a) say say center('checking if all two consecutive members have a GCD=1', 70, '═') a=Stern_Brocot(chk) /*invoke function to generate sequence.*/
do c=1 for chk-1; if gcd(subword(a,c,2))==1 then iterate
say 'GCD check failed at member' c"."; exit 13
end /*c*/
say '───── All ' chk " two consecutive members have a GCD of unity." exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ gcd: procedure; $=; do i=1 for arg(); $=$ arg(i); end /*i*/ /*arg list. */
parse var $ x z .; if x=0 then x=z; x=abs(x) /*zero case?*/
do j=2 to words($); y=abs(word($,j)); if y=0 then iterate /*ignore 0's*/
do until y==0; parse value x//y y with y x; end /*heavy work*/
end /*j*/
return x /*return GCD*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ Stern_Brocot: parse arg h 1 f; $=1 1; if h<0 then h=1e9
else f=0; f=abs(f)
do k=2 until words($)>=h | wordpos(f,$)\==0
_=word($,k); $=$ (_+word($,k-1)) _; if f==0 then iterate
end /*until*/
if f==0 then return subword($,1,h)
return $</lang>
output when using the default inputs:
══════════the first 15 numbers in the Stern─Brocot sequence═══════════ 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 ═════════════the 1-based index for the first 10 integers══════════════ for 1, the index is: 1 for 2, the index is: 3 for 3, the index is: 5 for 4, the index is: 9 for 5, the index is: 11 for 6, the index is: 33 for 7, the index is: 19 for 8, the index is: 21 for 9, the index is: 35 for 10, the index is: 39 ══════════════════════the 1-based index for 100═══════════════════════ for 100, the index is: 1179 ═════════checking if all two consecutive members have a GCD=1═════════ ───── All 1000 two consecutive members have a GCD of unity.
Ruby
<lang ruby>def sb
return enum_for :sb unless block_given? a=[1,1] 0.step do |i| yield a[i] a << a[i]+a[i+1] << a[i+1] end
end
puts "First 15: #{sb.first(15)}"
[*1..10,100].each do |n|
puts "#{n} first appears at #{sb.find_index(n)+1}."
end
if sb.take(1000).each_cons(2).map { |a,b| a.gcd(b) }.all? { |n| n==1 }
puts "All GCD's are 1"
else
puts "Whoops, not all GCD's are 1!"
end</lang>
- Output:
First 15: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1 first appears at 1. 2 first appears at 3. 3 first appears at 5. 4 first appears at 9. 5 first appears at 11. 6 first appears at 33. 7 first appears at 19. 8 first appears at 21. 9 first appears at 35. 10 first appears at 39. 100 first appears at 1179. All GCD's are 1
Scala
<lang scala>lazy val sbSeq: Stream[BigInt] = {
BigInt("1") #::
BigInt("1") #::
(sbSeq zip sbSeq.tail zip sbSeq.tail).
flatMap{ case ((a,b),c) => List(a+b,c) }
}
// Show the results { println( s"First 15 members: ${(for( n <- 0 until 15 ) yield sbSeq(n)) mkString( "," )}" ) println for( n <- 1 to 10; pos = sbSeq.indexOf(n) + 1 ) println( s"Position of first $n is at $pos" ) println println( s"Position of first 100 is at ${sbSeq.indexOf(100) + 1}" ) println println( s"Greatest Common Divisor for first 1000 members is 1: " +
(sbSeq zip sbSeq.tail).take(1000).forall{ case (a,b) => a.gcd(b) == 1 } )
} </lang>
- Output:
First 15 members: 1,1,2,1,3,2,3,1,4,3,5,2,5,3,4 Position of first 1 is at 1 Position of first 2 is at 3 Position of first 3 is at 5 Position of first 4 is at 9 Position of first 5 is at 11 Position of first 6 is at 33 Position of first 7 is at 19 Position of first 8 is at 21 Position of first 9 is at 35 Position of first 10 is at 39 Position of first 100 is at 1179 Greatest Common Divisor for first 1000 members is 1: true
Sidef
<lang ruby># Declare a function to generate the Stern-Brocot sequence func stern_brocot {
var list = [1, 1];
func {
list.append(list[0]+list[1], list[1]);
list.shift;
}
}
- Show the first fifteen members of the sequence.
1..15 -> map(stern_brocot()).join(' ').say;
- Show the (1-based) index of where the numbers 1-to-10 first appears
- in the sequence, and where the number 100 first appears in the sequence.
[(1..10)..., 100].each { |i|
var index = 1;
var generator = stern_brocot();
while (generator() != i) {
++index;
}
say "First occurrence of #{i} is at index #{index}";
}
- Check that the greatest common divisor of all the two consecutive
- members of the series up to the 1000th member, is always one.
var generator = stern_brocot(); var (a, b) = (generator(), generator()); {
assert_eq(Math.gcd(a, b), 1); a = b; b = generator();
} * 1000;
say "All GCD's are 1";</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 First occurrence of 1 is at index 1 First occurrence of 2 is at index 3 First occurrence of 3 is at index 5 First occurrence of 4 is at index 9 First occurrence of 5 is at index 11 First occurrence of 6 is at index 33 First occurrence of 7 is at index 19 First occurrence of 8 is at index 21 First occurrence of 9 is at index 35 First occurrence of 10 is at index 39 First occurrence of 100 is at index 1179 All GCD's are 1
VBScript
<lang VBScript>sb = Array(1,1) i = 1 'considered j = 2 'precedent n = 0 'loop counter Do ReDim Preserve sb(UBound(sb) + 1) sb(UBound(sb)) = sb(UBound(sb) - i) + sb(UBound(sb) - j) ReDim Preserve sb(UBound(sb) + 1) sb(UBound(sb)) = sb(UBound(sb) - j) i = i + 1 j = j + 1 n = n + 1 Loop Until n = 2000
WScript.Echo "First 15: " & DisplayElements(15)
For k = 1 To 10 WScript.Echo "The first instance of " & k & " is in #" & ShowFirstInstance(k) & "." Next
WScript.Echo "The first instance of " & 100 & " is in #" & ShowFirstInstance(100) & "."
Function DisplayElements(n) For i = 0 To n - 1 If i < n - 1 Then DisplayElements = DisplayElements & sb(i) & ", " Else DisplayElements = DisplayElements & sb(i) End If Next End Function
Function ShowFirstInstance(n) For i = 0 To UBound(sb) If sb(i) = n Then ShowFirstInstance = i + 1 Exit For End If Next End Function</lang>
- Output:
First 15: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 The first instance of 1 is in #1. The first instance of 2 is in #3. The first instance of 3 is in #5. The first instance of 4 is in #9. The first instance of 5 is in #11. The first instance of 6 is in #33. The first instance of 7 is in #19. The first instance of 8 is in #21. The first instance of 9 is in #35. The first instance of 10 is in #39. The first instance of 100 is in #1179.
Tcl
<lang tcl>
- !/usr/bin/env tclsh
package require generator ;# from tcllib
namespace eval stern-brocot {
proc generate Template:Count 100 { set seq {1 1} set n 0 while {[llength $seq] < $count} { lassign [lrange $seq $n $n+1] a b lappend seq [expr {$a + $b}] $b incr n } return $seq }
proc genr {} {
yield [info coroutine]
set seq {1 1}
while {1} {
set seq [lassign $seq a]
set b [lindex $seq 0]
set c [expr {$a + $b}]
lappend seq $c $b
yield $a
}
}
proc Step {a b args} {
set c [expr {$a + $b}]
list $a [list $b {*}$args $c $b]
}
generator define gen {} {
set cmd [list 1 1]
while {1} {
lassign [Step {*}$cmd] a cmd
generator yield $a
}
}
namespace export {[a-z]*}
namespace ensemble create
}
interp alias {} sb {} stern-brocot
- a simple adaptation of gcd from http://wiki.tcl.tk/2891
proc coprime {a args} {
set gcd $a
foreach arg $args {
while {$arg != 0} {
set t $arg
set arg [expr {$gcd % $arg}]
set gcd $t
if {$gcd == 1} {return true}
}
}
return false
}
proc main {} {
puts "#1. First 15 members of the Stern-Brocot sequence:" puts \t[generator to list [generator take 16 [sb gen]]]
puts "#2. First occurrences of 1 through 10:"
set first {}
set got 0
set i 0
generator foreach x [sb gen] {
incr i
if {$x>10} continue
if {[dict exists $first $x]} continue
dict set first $x $i
if {[incr got] >= 10} break
}
foreach {a b} [lsort -integer -stride 2 $first] {
puts "\tFirst $a at $b"
}
puts "#3. First occurrence of 100:"
set i 0
generator foreach x [sb gen] {
incr i
if {$x eq 100} break
}
puts "\tFirst $x at $i"
puts "#4. Check first 1k elements for common divisors:"
set prev [expr {2*3*5*7*11*13*17*19+1}] ;# a handy prime
set i 0
generator foreach x [sb gen] {
if {[incr i] >= 1000} break
if {![coprime $x $prev]} {
error "Element $i, $x is not coprime with $prev!"
}
set prev $x
}
puts "\tFirst $i elements are all pairwise coprime"
}
main </lang>
- Output:
#1. First 15 members of the Stern-Brocot sequence:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4
#2. First occurrences of 1 through 10:
First 1 at 1
First 2 at 3
First 3 at 5
First 4 at 9
First 5 at 11
First 6 at 33
First 7 at 19
First 8 at 21
First 9 at 35
First 10 at 39
#3. First occurrence of 100:
First 100 at 1179
#4. Check first 1k elements for common divisors:
First 1000 elements are all pairwise coprime
zkl
<lang zkl>fcn SB // Stern-Brocot sequence factory --> Walker
{ Walker(fcn(sb,n){ a,b:=sb; sb.append(a+b,b); sb.del(0); a }.fp(L(1,1))) }
SB().walk(15).println();
[1..10].zipWith('wrap(n){ [1..].zip(SB())
.filter(1,fcn(n,sb){ n==sb[1] }.fp(n)) })
.walk().println();
[1..].zip(SB()).filter1(fcn(sb){ 100==sb[1] }).println();
sb:=SB(); do(500){ if(sb.next().gcd(sb.next())!=1) println("Oops") }</lang>
- Output:
L(1,1,2,1,3,2,3,1,4,3,5,2,5,3,4) L(L(L(1,1)),L(L(3,2)),L(L(5,3)),L(L(9,4)),L(L(11,5)),L(L(33,6)),L(L(19,7)),L(L(21,8)),L(L(35,9)),L(L(39,10))) L(1179,100)