Stern-Brocot sequence
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.
- Related tasks
- Ref
- Infinite Fractions - Numberphile (Video).
- Trees, Teeth, and Time: The mathematics of clock making.
- A002487 The On-Line Encyclopedia of Integer Sequences.
360 Assembly
<lang 360asm>* Stern-Brocot sequence - 12/03/2019 STERNBR CSECT
USING STERNBR,R13 base register
B 72(R15) skip savearea
DC 17F'0' savearea
SAVE (14,12) save previous context
ST R13,4(R15) link backward
ST R15,8(R13) link forward
LR R13,R15 set addressability
LA R4,SB+2 k=2; @sb(k)
LA R2,SB+2 i=1; @sb(k-i)
LA R3,SB+0 j=2; @sb(k-j)
LA R1,NN/2 loop counter
LOOP LA R4,2(R4) @sb(k)++
LH R0,0(R2) sb(k-i)
AH R0,0(R3) sb(k-i)+sb(k-j)
STH R0,0(R4) sb(k)=sb(k-i)+sb(k-j)
LA R3,2(R3) @sb(k-j)++
LA R4,2(R4) @sb(k)++
LH R0,0(R3) sb(k-j)
STH R0,0(R4) sb(k)=sb(k-j)
LA R2,2(R2) @sb(k-i)++
BCT R1,LOOP end loop
LA R9,15 n=15
MVC PG(5),=CL80'FIRST'
XDECO R9,XDEC edit n
MVC PG+5(3),XDEC+9 output n
XPRNT PG,L'PG print buffer
LA R10,PG @pg
LA R6,1 i=1
DO WHILE=(CR,R6,LE,R9) do i=1 to n
LR R1,R6 i
SLA R1,1 ~
LH R2,SB-2(R1) sb(i)
XDECO R2,XDEC edit sb(i)
MVC 0(4,R10),XDEC+8 output sb(i)
LA R10,4(R10) @pg+=4
LA R6,1(R6) i++
ENDDO , enddo i
XPRNT PG,L'PG print buffer
LA R7,1 j=1
DO WHILE=(C,R7,LE,=A(11)) do j=1 to 11
IF C,R7,EQ,=F'11' THEN if j=11 then
LA R7,100 j=100
ENDIF , endif
LA R6,1 i=1
DO WHILE=(C,R6,LE,=A(NN)) do i=1 to nn
LR R1,R6 i
SLA R1,1 ~
LH R2,SB-2(R1) sb(i)
CR R2,R7 if sb(i)=j
BE EXITI then leave i
LA R6,1(R6) i++
ENDDO , enddo i
EXITI MVC PG,=CL80'FIRST INSTANCE OF'
XDECO R7,XDEC edit j
MVC PG+17(4),XDEC+8 output j
MVC PG+21(7),=C' IS AT '
XDECO R6,XDEC edit i
MVC PG+28(4),XDEC+8 output i
XPRNT PG,L'PG print buffer
LA R7,1(R7) j++
ENDDO , enddo j
L R13,4(0,R13) restore previous savearea pointer
RETURN (14,12),RC=0 restore registers from calling sav
LTORG
NN EQU 2400 nn PG DC CL80' ' buffer XDEC DS CL12 temp for xdeco SB DC (NN)H'1' sb(nn)
REGEQU
END STERNBR</lang>
- Output:
FIRST 15 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 FIRST INSTANCE OF 1 IS AT 1 FIRST INSTANCE OF 2 IS AT 3 FIRST INSTANCE OF 3 IS AT 5 FIRST INSTANCE OF 4 IS AT 9 FIRST INSTANCE OF 5 IS AT 11 FIRST INSTANCE OF 6 IS AT 33 FIRST INSTANCE OF 7 IS AT 19 FIRST INSTANCE OF 8 IS AT 21 FIRST INSTANCE OF 9 IS AT 35 FIRST INSTANCE OF 10 IS AT 39 FIRST INSTANCE OF 100 IS AT 1179
The nice part is the coding of the sequense: <lang c> k=2; i=1; j=2;
while(k<nn);
k++; sb[k]=sb[k-i]+sb[k-j];
k++; sb[k]=sb[k-j];
i++; j++;
} </lang>
Only five registers are used. No Horner's rule to access sequence items. <lang 360asm> LA R4,SB+2 k=2; @sb(k)
LA R2,SB+2 i=1; @sb(k-i)
LA R3,SB+0 j=2; @sb(k-j)
LA R1,NN/2 k=nn/2 'loop counter
LOOP LA R4,2(R4) @sb(k)++
LH R0,0(R2) sb(k-i)
AH R0,0(R3) sb(k-i)+sb(k-j)
STH R0,0(R4) sb(k)=sb(k-i)+sb(k-j)
LA R3,2(R3) @sb(k-j)++
LA R4,2(R4) @sb(k)++
LH R0,0(R3) sb(k-j)
STH R0,0(R4) sb(k)=sb(k-j)
LA R2,2(R2) @sb(k-i)++
BCT R1,LOOP k--; if k>0 then goto loop </lang>
8080 Assembly
<lang 8080asm>puts: equ 9 ; CP/M syscall to print a string org 100h ;;; Generate the first 1200 elements of the Stern-Brocot sequence lxi b,600 ; 2 elements generated per loop lxi h,seq mov e,m ; Initialization inx h push h ; Save considered member pointer inx h ; Insertion pointer genseq: xthl ; Load considered member pointer mov d,e ; D = predecessor mov e,m ; E = considered member inx h ; Point at next considered member xthl ; Load insertion pointer mov a,d ; A = sum of both members add e mov m,a ; Append the sum inx h mov m,e ; Append the considered member inx h dcx b ; Decrement counter mov a,b ; Zero? ora c jnz genseq ; If not, generate next members of sequence pop h ; Remove pointer from stack ;;; Print first 15 members of sequence lxi d,seq mvi b,15 ; 15 members mvi h,0 p15: ldax d ; Get current member mov l,a call prhl ; Print member inx d ; Increment pointer dcr b ; Decrement counter jnz p15 ; If not zero, print another one lxi d,nl mvi c,puts call 5 ;;; Print indices of first occurrence of 1..10 lxi b,010Ah ; B = number, C = counter call fnext ;;; Print index of first occurrence of 100 lxi b,6401h call fnext ;;; Check if the GCD of first 1000 consecutive elements is 0 xra a ; Zero out 1001th element as end marker sta seq+1000 lxi h,seq ; Start of array mov e,m inx h gcdchk: mov d,e ; (D,E) = next pair mov e,m inx h mov a,e mov b,d ana a ; Reached the end? jz done call gcd ; If not, check GCD dcr a ; Check that it is 1 jz gcdchk ; If so, check next pair push h ; GCD not 1 - save pointer lxi d,gcdn ; Print message mvi c,puts call 5 pop h ; Calculate offset in array lxi d,-seq dad d jmp prhl ; Print offset of pair whose GCD is not 1 done: lxi d,gcdy ; Print OK message mvi c,puts jmp 5 ;;; GCD(A,B) gcd: cmp b rz ; If A=B, result = A jc b_le_a ; B>A? sub b ; If A>B, subtract B jmp gcd ; and loop b_le_a: mov c,a mov a,b sub c mov b,a mov a,c jmp gcd ;;; Print indices of occurrences of C numbers ;;; starting at B fnext: lxi d,seq fsrch: ldax d ; Get current member cmp b ; Is it the number we are looking for? inx d ; Increment number jnz fsrch ; If no match, check next number lxi h,-seq ; Match - subtract start of array dad d call prhl ; Print index inr b ; Look for next number dcr c ; If we need more numbers jnz fnext push d ; Save sequence pointer lxi d,nl ; Print newline mvi c,puts call 5 pop d ; Restore sequence pointer ret ;;; Print HL as ASCII number. prhl: push h ; Save all registers push d push b lxi b,pnum ; Store pointer to num string on stack push b lxi b,-10 ; Divisor prdgt: lxi d,-1 prdgtl: inx d ; Divide by 10 through trial subtraction dad b jc prdgtl mvi a,'0'+10 add l ; L = remainder - 10 pop h ; Get pointer from stack dcx h ; Store digit mov m,a push h ; Put pointer back on stack xchg ; Put quotient in HL mov a,h ; Check if zero ora l jnz prdgt ; If not, next digit pop d ; Get pointer and put in DE mvi c,9 ; CP/M print string call 5 pop b ; Restore registers pop d pop h ret db '*****' ; Placeholder for number pnum: db ' $' nl: db 13,10,'$' gcdn: db 'GCD not 1 at: $' gcdy: db 'GCD of all pairs of consecutive members is 1.$' seq: db 1,1 ; Sequence stored here</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 3 5 9 11 33 19 21 35 39 1179 GCD of all pairs of consecutive members is 1.
8086 Assembly
<lang asm>puts: equ 9 cpu 8086 bits 16 org 100h section .text ;;; Generate the first 1200 elemets of the Stern-Brocot sequence mov cx,600 ; 2 elements generated per loop mov si,seq mov di,seq+2 lodsb mov ah,al ; AH = predecessor genseq: lodsb ; AL = considered member add ah,al ; AH = sum xchg ah,al ; Swap them (AL = sum, AH = member) stosw ; Write sum and current considered member loop genseq ; Loop 600 times ;;; Print first 15 members of sequence mov si,seq mov cx,15 p15: lodsb ; Get member cbw call prax ; Print member loop p15 call prnl ;;; Print first occurrences of [1..10] mov al,1 mov cx,10 call find call prnl ;;; Print first occurrence of 100 mov al,100 mov cx,1 call find call prnl ;;; Check pairs of GCDs mov cx,1000 ; 1000 times mov si,seq lodsb gcdchk: mov ah,al ; AH = last member, AL = current member lodsb mov dx,ax ; Calculate GCD call gcd dec dl ; See if it is 1 jnz .fail ; If not, failure loop gcdchk ; Otherwise, check next pair mov dx,gcdy ; GCD of all pairs is 0 jmp pstr .fail: mov dx,gcdn ; GCD of current pair is not 0 call pstr mov ax,si sub ax,seq+1 jmp prax ;;; DL = gcd(DL,DH) gcd: cmp dl,dh jl .lt ; DL < DH jg .gt ; DL > DH ret .lt: sub dh,dl ; DL < DH jmp gcd .gt: sub dl,dh ; DL > DH jmp gcd ;;; Print indices of CX consecutive numbers starting ;;; at AL. find: mov di,seq push cx ; Keep loop counter mov cx,-1 repne scasb ; Find AL starting at [DI] pop cx ; Restore loop counter xchg si,ax ; Keep AL in SI mov ax,di ; Calculate index in sequence sub ax,seq call prax ; Output xchg si,ax ; Restore AL inc ax ; Increment loop find ; Keep going CX times ret ;;; Print newline prnl: mov dx,nl jmp pstr ;;; Print number in AX ;;; Destroys AX, BX, DX, BP prax: mov bp,10 ; Divisor mov bx,numbuf .loop: xor dx,dx ; DX = 0 div bp ; Divide DX:AX by 10; DX = remainder dec bx ; Move string pointer back add dl,'0' ; Make ASCII digit mov [bx],dl ; Write digit test ax,ax ; Any digits left? jnz .loop mov dx,bx pstr: mov ah,puts ; Print number string int 21h ret section .data gcdn: db 'GCD not 1 at: $' gcdy: db 'GCD of all pairs of consecutive members is 1.$' db '*****' numbuf: db ' $' nl: db 13,10,'$' seq: db 1,1</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 1 3 5 9 11 33 19 21 35 39 1179 GCD of all pairs of consecutive members is 1.
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!
APL
<lang APL>task←{
stern←{⍵{
⍺←0 ⋄ ⍺⍺≤⍴⍵:⍺⍺↑⍵
(⍺+1)∇⍵,(+/,2⊃⊣)2↑⍺↓⍵
}1 1}
seq←stern 1200 ⍝ Cache 1200 elements
⎕←'First 15 elements:',15↑seq
⎕←'Locations of 1..10:',seq⍳⍳10
⎕←'Location of 100:',seq⍳100
⎕←'All GCDs 1:','no' 'yes'[1+1∧.=2∨/1000↑seq]
}</lang>
- Output:
First 15 elements: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 Locations of 1..10: 1 3 5 9 11 33 19 21 35 39 Location of 100: 1179 All GCDs 1: yes
AppleScript
<lang applescript>use AppleScript version "2.4" use framework "Foundation" use scripting additions
-- sternBrocot :: Generator [Int]
on sternBrocot()
script go
on |λ|(xs)
set x to snd(xs)
tail(xs) & {fst(xs) + x, x}
end |λ|
end script
fmapGen(my head, iterate(go, {1, 1}))
end sternBrocot
-- TEST ------------------------------------------------------------------
on run
set sbs to take(1200, sternBrocot())
set ixSB to zip(sbs, enumFrom(1))
script low
on |λ|(x)
12 ≠ fst(x)
end |λ|
end script
script sameFst
on |λ|(a, b)
fst(a) = fst(b)
end |λ|
end script
script asList
on |λ|(x)
{fst(x), snd(x)}
end |λ|
end script
script below100
on |λ|(x)
100 ≠ fst(x)
end |λ|
end script
script fullyReduced
on |λ|(ab)
1 = gcd(|1| of ab, |2| of ab)
end |λ|
end script
unlines(map(showJSON, ¬
{take(15, sbs), ¬
take(10, map(asList, ¬
nubBy(sameFst, ¬
sortBy(comparing(fst), ¬
takeWhile(low, ixSB))))), ¬
asList's |λ|(fst(take(1, dropWhile(below100, ixSB)))), ¬
all(fullyReduced, take(1000, zip(sbs, tail(sbs))))}))
end run
--> [1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] --> [[1,32],[2,24],[3,40],[4,36],[5,44],[6,33],[7,38],[8,42],[9,35],[10,39]] --> [100,1179] --> true
-- GENERIC ABSTRACTIONS -------------------------------------------------------
-- Absolute value.
-- abs :: Num -> Num
on abs(x)
if 0 > x then
-x
else
x
end if
end abs
-- Applied to a predicate and a list, `all` determines if all elements -- of the list satisfy the predicate. -- all :: (a -> Bool) -> [a] -> Bool on all(p, xs)
tell mReturn(p)
set lng to length of xs
repeat with i from 1 to lng
if not |λ|(item i of xs, i, xs) then return false
end repeat
true
end tell
end all
-- comparing :: (a -> b) -> (a -> a -> Ordering) on comparing(f)
script
on |λ|(a, b)
tell mReturn(f)
set fa to |λ|(a)
set fb to |λ|(b)
if fa < fb then
-1
else if fa > fb then
1
else
0
end if
end tell
end |λ|
end script
end comparing
-- drop :: Int -> [a] -> [a]
-- drop :: Int -> String -> String
on drop(n, xs)
set c to class of xs
if c is not script then
if c is not string then
if n < length of xs then
items (1 + n) thru -1 of xs
else
{}
end if
else
if n < length of xs then
text (1 + n) thru -1 of xs
else
""
end if
end if
else
take(n, xs) -- consumed
return xs
end if
end drop
-- dropWhile :: (a -> Bool) -> [a] -> [a] -- dropWhile :: (Char -> Bool) -> String -> String on dropWhile(p, xs)
set lng to length of xs
set i to 1
tell mReturn(p)
repeat while i ≤ lng and |λ|(item i of xs)
set i to i + 1
end repeat
end tell
drop(i - 1, xs)
end dropWhile
-- enumFrom :: a -> [a] on enumFrom(x)
script
property v : missing value
property blnNum : class of x is not text
on |λ|()
if missing value is not v then
if blnNum then
set v to 1 + v
else
set v to succ(v)
end if
else
set v to x
end if
return v
end |λ|
end script
end enumFrom
-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
-- fmapGen <$> :: (a -> b) -> Gen [a] -> Gen [b] on fmapGen(f, gen)
script
property g : gen
property mf : mReturn(f)'s |λ|
on |λ|()
set v to g's |λ|()
if v is missing value then
v
else
mf(v)
end if
end |λ|
end script
end fmapGen
-- fst :: (a, b) -> a on fst(tpl)
if class of tpl is record then
|1| of tpl
else
item 1 of tpl
end if
end fst
-- gcd :: Int -> Int -> Int on gcd(a, b)
set x to abs(a)
set y to abs(b)
repeat until y = 0
if x > y then
set x to x - y
else
set y to y - x
end if
end repeat
return x
end gcd
-- head :: [a] -> a on head(xs)
if xs = {} then
missing value
else
item 1 of xs
end if
end head
-- iterate :: (a -> a) -> a -> Gen [a] on iterate(f, x)
script
property v : missing value
property g : mReturn(f)'s |λ|
on |λ|()
if missing value is v then
set v to x
else
set v to g(v)
end if
return v
end |λ|
end script
end iterate
-- length :: [a] -> Int
on |length|(xs)
set c to class of xs
if list is c or string is c then
length of xs
else
(2 ^ 29 - 1) -- (maxInt - simple proxy for non-finite)
end if
end |length|
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: First-class m => (a -> b) -> m (a -> b) on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- nubBy :: (a -> a -> Bool) -> [a] -> [a]
on nubBy(f, xs)
set g to mReturn(f)'s |λ|
script notEq
property fEq : g
on |λ|(a)
script
on |λ|(b)
not fEq(a, b)
end |λ|
end script
end |λ|
end script
script go
on |λ|(xs)
if (length of xs) > 1 then
set x to item 1 of xs
{x} & go's |λ|(filter(notEq's |λ|(x), items 2 thru -1 of xs))
else
xs
end if
end |λ|
end script
go's |λ|(xs)
end nubBy
-- partition :: predicate -> List -> (Matches, nonMatches) -- partition :: (a -> Bool) -> [a] -> ([a], [a]) on partition(f, xs)
tell mReturn(f)
set ys to {}
set zs to {}
repeat with x in xs
set v to contents of x
if |λ|(v) then
set end of ys to v
else
set end of zs to v
end if
end repeat
end tell
Tuple(ys, zs)
end partition
-- showJSON :: a -> String on showJSON(x)
set c to class of x
if (c is list) or (c is record) then
set ca to current application
set {json, e} to ca's NSJSONSerialization's dataWithJSONObject:x options:0 |error|:(reference)
if json is missing value then
e's localizedDescription() as text
else
(ca's NSString's alloc()'s initWithData:json encoding:(ca's NSUTF8StringEncoding)) as text
end if
else if c is date then
"\"" & ((x - (time to GMT)) as «class isot» as string) & ".000Z" & "\""
else if c is text then
"\"" & x & "\""
else if (c is integer or c is real) then
x as text
else if c is class then
"null"
else
try
x as text
on error
("«" & c as text) & "»"
end try
end if
end showJSON
-- snd :: (a, b) -> b on snd(tpl)
if class of tpl is record then
|2| of tpl
else
item 2 of tpl
end if
end snd
-- Enough for small scale sorts.
-- Use instead sortOn :: Ord b => (a -> b) -> [a] -> [a]
-- which is equivalent to the more flexible sortBy(comparing(f), xs)
-- and uses a much faster ObjC NSArray sort method
-- sortBy :: (a -> a -> Ordering) -> [a] -> [a]
on sortBy(f, xs)
if length of xs > 1 then
set h to item 1 of xs
set f to mReturn(f)
script
on |λ|(x)
f's |λ|(x, h) ≤ 0
end |λ|
end script
set lessMore to partition(result, rest of xs)
sortBy(f, |1| of lessMore) & {h} & ¬
sortBy(f, |2| of lessMore)
else
xs
end if
end sortBy
-- tail :: [a] -> [a] on tail(xs)
set blnText to text is class of xs
if blnText then
set unit to ""
else
set unit to {}
end if
set lng to length of xs
if 1 > lng then
missing value
else if 2 > lng then
unit
else
if blnText then
text 2 thru -1 of xs
else
rest of xs
end if
end if
end tail
-- take :: Int -> [a] -> [a] -- take :: Int -> String -> String on take(n, xs)
set c to class of xs
if list is c then
if 0 < n then
items 1 thru min(n, length of xs) of xs
else
{}
end if
else if string is c then
if 0 < n then
text 1 thru min(n, length of xs) of xs
else
""
end if
else if script is c then
set ys to {}
repeat with i from 1 to n
set v to xs's |λ|()
if missing value is v then
return ys
else
set end of ys to v
end if
end repeat
return ys
else
missing value
end if
end take
-- takeWhile :: (a -> Bool) -> [a] -> [a] -- takeWhile :: (Char -> Bool) -> String -> String on takeWhile(p, xs)
if script is class of xs then
takeWhileGen(p, xs)
else
tell mReturn(p)
repeat with i from 1 to length of xs
if not |λ|(item i of xs) then ¬
return take(i - 1, xs)
end repeat
end tell
return xs
end if
end takeWhile
-- takeWhileGen :: (a -> Bool) -> Gen [a] -> [a] on takeWhileGen(p, xs)
set ys to {}
set v to |λ|() of xs
tell mReturn(p)
repeat while (|λ|(v))
set end of ys to v
set v to xs's |λ|()
end repeat
end tell
return ys
end takeWhileGen
-- Tuple (,) :: a -> b -> (a, b) on Tuple(a, b)
{type:"Tuple", |1|:a, |2|:b, length:2}
end Tuple
-- unlines :: [String] -> String on unlines(xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set str to xs as text
set my text item delimiters to dlm
str
end unlines
-- zip :: [a] -> [b] -> [(a, b)] on zip(xs, ys)
zipWith(Tuple, xs, ys)
end zip
-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] on zipWith(f, xs, ys)
set lng to min(|length|(xs), |length|(ys))
if 1 > lng then return {}
set xs_ to take(lng, xs) -- Allow for non-finite
set ys_ to take(lng, ys) -- generators like cycle etc
set lst to {}
tell mReturn(f)
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs_, item i of ys_)
end repeat
return lst
end tell
end zipWith</lang>
- Output:
[1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] [[1,32],[2,24],[3,40],[4,36],[5,44],[6,33],[7,38],[8,42],[9,35],[10,39]] [100,1179] true
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.
BASIC
<lang basic>10 DEFINT A,B,I,J,S: DIM S(1200) 20 S(1)=1: S(2)=1 30 FOR I=2 TO 600 40 S(I*2-1)=S(I)+S(I-1) 50 S(I*2)=S(I) 60 NEXT I 70 PRINT "First 15 elements: "; 80 FOR I=1 TO 15: PRINT USING"# ";S(I);: NEXT I 85 PRINT 90 FOR I=1 TO 10 100 FOR J=1 TO 1200: IF S(J)<>I THEN NEXT J 110 PRINT "First";I;"at";J 120 NEXT I 130 FOR J=1 TO 1200: IF S(J)<>100 THEN NEXT J 140 PRINT "First 100 at";J 150 FOR I=2 TO 1000 160 A=S(I): B=S(I-1) 170 J=A: A=B: B=J MOD A: IF B THEN 170 180 IF A<>1 THEN PRINT "GCD <> 1 at ";I: STOP 190 NEXT I 200 PRINT "All GCDs are 1." 210 END</lang>
- Output:
First 15 elements: 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 All GCDs are 1.
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 csharp>using System; using System.Collections.Generic; using System.Linq;
static class Program {
static List<int> l = new List<int>() { 1, 1 };
static int gcd(int a, int b) {
return a > 0 ? a < b ? gcd(b % a, a) : gcd(a % b, b) : b; }
static void Main(string[] args) {
int max = 1000; int take = 15; int i = 1;
int[] selection = new[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100 };
do { l.AddRange(new List<int>() { l[i] + l[i - 1], l[i] }); i += 1; }
while (l.Count < max || l[l.Count - 2] != selection.Last());
Console.Write("The first {0} items In the Stern-Brocot sequence: ", take);
Console.WriteLine("{0}\n", string.Join(", ", l.Take(take)));
Console.WriteLine("The locations of where the selected numbers (1-to-10, & 100) first appear:");
foreach (int ii in selection) {
int j = l.FindIndex(x => x == ii) + 1; Console.WriteLine("{0,3}: {1:n0}", ii, j); }
Console.WriteLine(); bool good = true;
for (i = 1; i <= max; i++) { if (gcd(l[i], l[i - 1]) != 1) { good = false; break; } }
Console.WriteLine("The greatest common divisor of all the two consecutive items of the" +
" series up to the {0}th item is {1}always one.", max, good ? "" : "not ");
}
}</lang>
- Output:
The first 15 items In the Stern-Brocot sequence: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 The locations of where the selected numbers (1-to-10, & 100) first appear: 1: 1 2: 3 3: 5 4: 9 5: 11 6: 33 7: 19 8: 21 9: 35 10: 39 100: 1,179 The greatest common divisor of all the two consecutive items of the series up to the 1000th item is always one.
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>;; each step adds two items (defn sb-step [v]
(let [i (quot (count v) 2)] (conj v (+ (v (dec i)) (v i)) (v i))))
- A lazy, infinite sequence -- `take` what you want.
(def all-sbs (sequence (map peek) (iterate sb-step [1 1])))
- zero-based
(defn first-appearance [n]
(first (keep-indexed (fn [i x] (when (= x n) i)) all-sbs)))
- inlined abs; rem is slightly faster than mod, and the same result for positive values
(defn gcd [a b]
(loop [a (if (neg? a) (- a) a)
b (if (neg? b) (- b) b)]
(if (zero? b)
a
(recur b (rem a b)))))
(defn check-pairwise-gcd [cnt]
(let [sbs (take (inc cnt) all-sbs)] (every? #(= 1 %) (map gcd sbs (rest sbs)))))
- one-based index required by problem statement
(defn report-sb []
(println "First 15 Stern-Brocot members:" (take 15 all-sbs)) (println "First appearance of N at 1-based index:") (doseq [n [1 2 3 4 5 6 7 8 9 10 100]] (println " first" n "at" (inc (first-appearance n)))) (println "Check pairwise GCDs = 1 ..." (check-pairwise-gcd 1000)) true)
(report-sb)</lang>
- Output:
First 15 Stern-Brocot members: (1 1 2 1 3 2 3 1 4 3 5 2 5 3 4) First appearance of N at 1-based index: 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 Check pairwise GCDs = 1 ... true true
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
F#
The function
<lang fsharp> // Generate Stern-Brocot Sequence. Nigel Galloway: October 11th., 2018 let sb=Seq.unfold(fun (n::g::t)->Some(n,[g]@t@[n+g;g]))[1;1] </lang>
The Task
Uses Greatest_common_divisor#F.23 <lang fsharp> sb |> Seq.take 15 |> Seq.iter(printf "%d ");printfn "" [1..10] |> List.map(fun n->(n,(sb|>Seq.findIndex(fun g->g=n))+1)) |> List.iter(printf "%A ");printfn "" printfn "%d" ((sb|>Seq.findIndex(fun g->g=100))+1) printfn "There are %d consecutive members, of the first thousand members, with GCD <> 1" (sb |> Seq.take 1000 |>Seq.pairwise |> Seq.filter(fun(n,g)->gcd n g <> 1) |> Seq.length) </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) 1179 There are 0 consecutive members, of the first thousand members, with GCD <> 1
Factor
Using the alternative function given in the C example for computing the Stern-Brocot sequence. <lang factor>USING: formatting io kernel lists lists.lazy locals math math.ranges prettyprint sequences ; IN: rosetta-code.stern-brocot
- fn ( n -- m )
[ 1 0 ] dip
[ dup zero? ] [
dup 1 bitand zero?
[ dupd [ + ] 2dip ]
[ [ dup ] [ + ] [ ] tri* ] if
-1 shift
] until drop nip ;
- search ( n -- m )
1 0 lfrom [ fn n = ] lfilter ltake list>array first ;
- first15 ( -- )
15 [1,b] [ fn pprint bl ] each "are the first fifteen." print ;
- first-appearances ( -- )
10 [1,b] 100 suffix [ dup search "First %3u at Stern #%u.\n" printf ] each ;
- gcd-test ( -- )
1,000 [1,b] [ dup 1 + [ fn ] bi@ gcd nip 1 = not ] filter empty? "" " not" ? "All GCDs are%s 1.\n" printf ;
- main ( -- ) first15 first-appearances gcd-test ;
MAIN: main</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 are the first fifteen. First 1 at Stern #1. First 2 at Stern #3. First 3 at Stern #5. First 4 at Stern #9. First 5 at Stern #11. First 6 at Stern #33. First 7 at Stern #19. First 8 at Stern #21. First 9 at Stern #35. First 10 at Stern #39. First 100 at Stern #1179. All GCDs are 1.
Fortran
Fortran IV
<lang fortran>* STERN-BROCOT SEQUENCE - FORTRAN IV
DIMENSION ISB(2400)
NN=2400
ISB(1)=1
ISB(2)=1
I=1
J=2
K=2
1 IF(K.GE.NN) GOTO 2
K=K+1
ISB(K)=ISB(K-I)+ISB(K-J)
K=K+1
ISB(K)=ISB(K-J)
I=I+1
J=J+1
GOTO 1
2 N=15
WRITE(*,101) N
101 FORMAT(1X,'FIRST',I4)
WRITE(*,102) (ISB(I),I=1,15)
102 FORMAT(15I4)
DO 5 J=1,11
JJ=J
IF(J.EQ.11) JJ=100
DO 3 I=1,K
IF(ISB(I).EQ.JJ) GOTO 4
3 CONTINUE
4 WRITE(*,103) JJ,I
103 FORMAT(1X,'FIRST',I4,' AT ',I4)
5 CONTINUE
END </lang>
- Output:
FIRST 15 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
Fortran 90
<lang fortran> ! Stern-Brocot sequence - Fortran 90
parameter (nn=2400)
dimension isb(nn)
isb(1)=1; isb(2)=1
i=1; j=2; k=2
do while(k.lt.nn)
k=k+1; isb(k)=isb(k-i)+isb(k-j)
k=k+1; isb(k)=isb(k-j)
i=i+1; j=j+1
end do
n=15
write(*,"(1x,'First',i4)") n
write(*,"(15i4)") (isb(i),i=1,15)
do j=1,11
jj=j
if(j==11) jj=100
do i=1,k
if(isb(i)==jj) exit
end do
write(*,"(1x,'First',i4,' at ',i4)") jj,i
end do
end </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
FreeBASIC
<lang freebasic>' version 02-03-2019 ' compile with: fbc -s console
- Define max 2000
Dim Shared As UInteger stern(max +2)
Sub stern_brocot
stern(1) = 1 stern(2) = 1
Dim As UInteger i = 2 , n = 2, ub = UBound(stern)
Do While i < ub
i += 1
stern(i) = stern(n) + stern(n -1)
i += 1
stern(i) = stern(n)
n += 1
Loop
End Sub
Function gcd(x As UInteger, y As UInteger) As UInteger
Dim As UInteger t
While y
t = y
y = x Mod y
x = t
Wend
Return x
End Function
' ------=< MAIN >=------
Dim As UInteger i
stern_brocot
Print "The first 15 are: " ; For i = 1 To 15
Print stern(i); " ";
Next
Print : Print Print " Index First nr." Dim As UInteger d = 1 For i = 1 To max
If stern(i) = d Then
Print Using " ######"; i; stern(i)
d += 1
If d = 11 Then d = 100
If d = 101 Then Exit For
i = 0
End If
Next
Print : Print d = 0 For i = 1 To 1000
If gcd(stern(i), stern(i +1)) <> 1 Then
d = gcd(stern(i), stern(i +1))
Exit For
End If
Next
If d = 0 Then
Print "GCD of two consecutive members of the series up to the 1000th member is 1"
Else
Print "The GCD for index "; i; " and "; i +1; " = "; d
End If
' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
The first 15 are: 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4
Index First nr.
1 1
3 2
5 3
9 4
11 5
33 6
19 7
21 8
35 9
39 10
1179 100
GCD of two consecutive members of the series up to the 1000th member is 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 (elemIndex)
sb :: [Int] sb = 1 : 1 : f (tail sb) sb
where f (a:aa) (b:bb) = a + b : a : f aa bb
main :: IO () 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
Or, expressed in terms of iterate:
<lang haskell>import Data.List (nubBy, sortBy) import Data.Ord (comparing) import Data.Monoid ((<>)) import Data.Function (on)
sternBrocot :: [Int] sternBrocot =
let go (a:b:xs) = (b : xs) <> [a + b, b] in head <$> iterate go [1, 1]
-- TEST ------------------------------------------------------------- main :: IO () main = do
print $ take 15 sternBrocot print $ take 10 $ nubBy (on (==) fst) $ sortBy (comparing fst) $ takeWhile ((110 >=) . fst) $ zip sternBrocot [1 ..] print $ take 1 $ dropWhile ((100 /=) . fst) $ zip sternBrocot [1 ..] print $ (all ((1 ==) . uncurry gcd) . (zip <*> tail)) $ take 1000 sternBrocot</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>
JavaScript
<lang JavaScript>(() => {
'use strict';
const main = () => {
// sternBrocot :: Generator [Int]
const sternBrocot = () => {
const go = xs => {
const x = snd(xs);
return tail(append(xs, [fst(xs) + x, x]));
};
return fmapGen(head, iterate(go, [1, 1]));
};
// TESTS ------------------------------------------
const
sbs = take(1200, sternBrocot()),
ixSB = zip(sbs, enumFrom(1));
return unlines(map(
JSON.stringify,
[
take(15, sbs),
take(10,
map(listFromTuple,
nubBy(
on(eq, fst),
sortBy(
comparing(fst),
takeWhile(x => 12 !== fst(x), ixSB)
)
)
)
),
listFromTuple(
take(1, dropWhile(x => 100 !== fst(x), ixSB))[0]
),
all(tpl => 1 === gcd(fst(tpl), snd(tpl)),
take(1000, zip(sbs, tail(sbs)))
)
]
));
};
// GENERIC ABSTRACTIONS -------------------------------
// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});
// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});
// Tuple (,) :: a -> b -> (a, b)
const Tuple = (a, b) => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});
// | Absolute value.
// abs :: Num -> Num const abs = Math.abs;
// Determines whether all elements of the structure // satisfy the predicate.
// all :: (a -> Bool) -> [a] -> Bool const all = (p, xs) => xs.every(p);
// append (++) :: [a] -> [a] -> [a] // append (++) :: String -> String -> String const append = (xs, ys) => xs.concat(ys);
// chr :: Int -> Char const chr = String.fromCodePoint;
// comparing :: (a -> b) -> (a -> a -> Ordering)
const comparing = f =>
(x, y) => {
const
a = f(x),
b = f(y);
return a < b ? -1 : (a > b ? 1 : 0);
};
// dropWhile :: (a -> Bool) -> [a] -> [a]
// dropWhile :: (Char -> Bool) -> String -> String
const dropWhile = (p, xs) => {
const lng = xs.length;
return 0 < lng ? xs.slice(
until(
i => i === lng || !p(xs[i]),
i => 1 + i,
0
)
) : [];
};
// enumFrom :: a -> [a]
function* enumFrom(x) {
let v = x;
while (true) {
yield v;
v = succ(v);
}
}
// eq (==) :: Eq a => a -> a -> Bool
const eq = (a, b) => {
const t = typeof a;
return t !== typeof b ? (
false
) : 'object' !== t ? (
'function' !== t ? (
a === b
) : a.toString() === b.toString()
) : (() => {
const aks = Object.keys(a);
return aks.length !== Object.keys(b).length ? (
false
) : aks.every(k => eq(a[k], b[k]));
})();
};
// fmapGen <$> :: (a -> b) -> Gen [a] -> Gen [b]
function* fmapGen(f, gen) {
const g = gen;
let v = take(1, g)[0];
while (0 < v.length) {
yield(f(v))
v = take(1, g)[0]
}
}
// fst :: (a, b) -> a const fst = tpl => tpl[0];
// gcd :: Int -> Int -> Int
const gcd = (x, y) => {
const
_gcd = (a, b) => (0 === b ? a : _gcd(b, a % b)),
abs = Math.abs;
return _gcd(abs(x), abs(y));
};
// head :: [a] -> a const head = xs => xs.length ? xs[0] : undefined;
// isChar :: a -> Bool
const isChar = x =>
('string' === typeof x) && (1 === x.length);
// iterate :: (a -> a) -> a -> Gen [a]
function* iterate(f, x) {
let v = x;
while (true) {
yield(v);
v = f(v);
}
}
// Returns Infinity over objects without finite length // this enables zip and zipWith to choose the shorter // argument when one is non-finite, like cycle, repeat etc
// length :: [a] -> Int const length = xs => xs.length || Infinity;
// listFromTuple :: (a, a ...) -> [a]
const listFromTuple = tpl =>
Array.from(tpl);
// map :: (a -> b) -> [a] -> [b] const map = (f, xs) => xs.map(f);
// nubBy :: (a -> a -> Bool) -> [a] -> [a]
const nubBy = (p, xs) => {
const go = xs => 0 < xs.length ? (() => {
const x = xs[0];
return [x].concat(
go(xs.slice(1)
.filter(y => !p(x, y))
)
)
})() : [];
return go(xs);
};
// e.g. sortBy(on(compare,length), xs)
// on :: (b -> b -> c) -> (a -> b) -> a -> a -> c const on = (f, g) => (a, b) => f(g(a), g(b));
// ord :: Char -> Int const ord = c => c.codePointAt(0);
// snd :: (a, b) -> b const snd = tpl => tpl[1];
// sortBy :: (a -> a -> Ordering) -> [a] -> [a]
const sortBy = (f, xs) =>
xs.slice()
.sort(f);
// succ :: Enum a => a -> a
const succ = x =>
isChar(x) ? (
chr(1 + ord(x))
) : isNaN(x) ? (
undefined
) : 1 + x;
// tail :: [a] -> [a] const tail = xs => 0 < xs.length ? xs.slice(1) : [];
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = (n, xs) =>
xs.constructor.constructor.name !== 'GeneratorFunction' ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));
// takeWhile :: (a -> Bool) -> [a] -> [a]
// takeWhile :: (Char -> Bool) -> String -> String
const takeWhile = (p, xs) =>
xs.constructor.constructor.name !==
'GeneratorFunction' ? (() => {
const lng = xs.length;
return 0 < lng ? xs.slice(
0,
until(
i => lng === i || !p(xs[i]),
i => 1 + i,
0
)
) : [];
})() : takeWhileGen(p, xs);
// takeWhileGen :: (a -> Bool) -> Gen [a] -> [a]
const takeWhileGen = (p, xs) => {
const ys = [];
let
nxt = xs.next(),
v = nxt.value;
while (!nxt.done && p(v)) {
ys.push(v);
nxt = xs.next();
v = nxt.value
}
return ys;
};
// uncons :: [a] -> Maybe (a, [a])
const uncons = xs => {
const lng = length(xs);
return (0 < lng) ? (
lng < Infinity ? (
Just(Tuple(xs[0], xs.slice(1))) // Finite list
) : (() => {
const nxt = take(1, xs);
return 0 < nxt.length ? (
Just(Tuple(nxt[0], xs))
) : Nothing();
})() // Lazy generator
) : Nothing();
};
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
let v = x;
while (!p(v)) v = f(v);
return v;
};
// Use of `take` and `length` here allows for zipping with non-finite // lists - i.e. generators like cycle, repeat, iterate.
// zip :: [a] -> [b] -> [(a, b)]
const zip = (xs, ys) => {
const lng = Math.min(length(xs), length(ys));
return Infinity !== lng ? (() => {
const bs = take(lng, ys);
return take(lng, xs).map((x, i) => Tuple(x, bs[i]));
})() : zipGen(xs, ys);
};
// zipGen :: Gen [a] -> Gen [b] -> Gen [(a, b)]
const zipGen = (ga, gb) => {
function* go(ma, mb) {
let
a = ma,
b = mb;
while (!a.Nothing && !b.Nothing) {
let
ta = a.Just,
tb = b.Just
yield(Tuple(fst(ta), fst(tb)));
a = uncons(snd(ta));
b = uncons(snd(tb));
}
}
return go(uncons(ga), uncons(gb));
};
// MAIN --- return main();
})();</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
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>
Julia
<lang julia>using Printf
function sternbrocot(f::Function=(x) -> length(x) ≥ 20)::Vector{Int}
rst = Int[1, 1]
i = 3
while !f(rst)
append!(rst, Int[rst[i-1] + rst[i-2], rst[i-2]])
i += 1
end
return rst
end
println("First 15 elements of Stern-Brocot series:\n", sternbrocot(x -> length(x) ≥ 15)[1:15], "\n")
for i in (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100)
occurr = findfirst(x -> x == i, sternbrocot(x -> i ∈ x))
@printf("Index of first occurrence of %3i in the series: %4i\n", i, occurr)
end
print("\nAssertion: the greatest common divisor of all the two\nconsecutive members of the series up to the 1000th member, is always one: ") sb = sternbrocot(x -> length(x) > 1000) if all(gcd(prev, this) == 1 for (prev, this) in zip(sb[1:1000], sb[2:1000]))
println("Confirmed.")
else
println("Rejected.")
end</lang>
- Output:
First 15 elements of Stern-Brocot series: [1, 1, 2, 1, 3, 1, 3, 2, 4, 1, 4, 3, 4, 1, 5] Index of first occurrence of 1 in the series: 1 Index of first occurrence of 2 in the series: 3 Index of first occurrence of 3 in the series: 5 Index of first occurrence of 4 in the series: 9 Index of first occurrence of 5 in the series: 15 Index of first occurrence of 6 in the series: 17 Index of first occurrence of 7 in the series: 23 Index of first occurrence of 8 in the series: 31 Index of first occurrence of 9 in the series: 33 Index of first occurrence of 10 in the series: 51 Index of first occurrence of 100 in the series: 1855 Assertion: the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one: Rejected.
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, $generator->());
}
}</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 first occurrence of 1 is at index 1 first occurrence of 2 is at index 3 first occurrence of 3 is at index 5 first occurrence of 4 is at index 9 first occurrence of 5 is at index 11 first occurrence of 6 is at index 33 first occurrence of 7 is at index 19 first occurrence of 8 is at index 21 first occurrence of 9 is at index 35 first occurrence of 10 is at index 39 first occurrence of 100 is at index 1179
A slightly different method:
<lang perl>use ntheory qw/gcd vecsum vecfirst/;
sub stern_diatomic {
my ($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 $n=$_; vecfirst { stern_diatomic($_) == $n } 1..10000 } 1..10; print "Index of 1-10 first occurrence: [@s]\n"; print "Index of 100 first occurrence: ", (vecfirst { stern_diatomic($_) == 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.
Phix
<lang Phix>sequence sb = {1,1} integer c = 2
function stern_brocot(integer n)
while length(sb)<n do
sb &= sb[c]+sb[c-1] & sb[c]
c += 1
end while
return sb[1..n]
end function
sequence s = stern_brocot(15) puts(1,"first 15:") ?s integer n = 16, k sequence idx = tagset(10) for i=1 to length(idx) do
while 1 do
k = find(idx[i],s)
if k!=0 then exit end if
n *= 2
s = stern_brocot(n)
end while
idx[i] = k
end for puts(1,"indexes of 1..10:") ?idx puts(1,"index of 100:") while 1 do
k = find(100,s) if k!=0 then exit end if n *= 2 s = stern_brocot(n)
end while ?k s = stern_brocot(1000) integer maxgcd = 1 for i=1 to 999 do
maxgcd = max(gcd(s[i],s[i+1]),maxgcd)
end for printf(1,"max gcd:%d\n",{maxgcd})</lang>
- Output:
first 15:{1,1,2,1,3,2,3,1,4,3,5,2,5,3,4}
indexes of 1..10:{1,3,5,9,11,33,19,21,35,39}
index of 100:1179
max gcd:1
PicoLisp
Using the gcd function defined at Greatest_common_divisor#PicoLisp: <lang PicoLisp>(de nmbr (N)
(let (A 1 B 0)
(while (gt0 N)
(if (bit? 1 N)
(inc 'B A)
(inc 'A B) )
(setq N (>> 1 N)) )
B ) )
(let Lst (mapcar nmbr (range 1 2000))
(println 'First-15: (head 15 Lst))
(for N 10
(println 'First N 'found 'at: (index N Lst)) )
(println 'First 100 'found 'at: (index 100 Lst))
(for (L Lst (cdr L) (cddr L))
(test 1 (gcd (car L) (cadr L))) )
(prinl "All consecutive pairs are relative prime!") )</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 All consecutive pairs are relative prime!
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>
Python: Composing pure (curried) functions
Composing and testing a Stern-Brocot function by composition of generic and reusable functional abstractions (curried for more flexible nesting and rearrangement).
<lang python>Stern-Brocot sequence
from itertools import (count, dropwhile, islice, takewhile) import operator import math
- sternBrocot :: Generator [Int]
def sternBrocot():
Non-finite list of the Stern-Brocot
sequence of integers.
def go(xs):
x = xs[1]
return (tail(xs) + [x + head(xs), x])
return fmapGen(head)(
iterate(go)([1, 1])
)
- TESTS ---------------------------------------------------
- main :: IO ()
def main():
Various tests
[eq, ne, gcd] = map(
curry,
[operator.eq, operator.ne, math.gcd]
)
sbs = take(1200)(sternBrocot()) ixSB = zip(sbs, enumFrom(1))
print(unlines(map(str, [
# First 15 members of the sequence.
take(15)(sbs),
# Indices of where the numbers [1..10] first appear.
take(10)(
nubBy(on(eq)(fst))(
sorted(
takewhile(
compose(ne(12))(fst),
ixSB
),
key=fst
)
)
),
# Index of where the number 100 first appears.
take(1)(dropwhile(compose(ne(100))(fst), ixSB)),
# Is the gcd of any two consecutive members,
# up to the 1000th member, always one ?
every(compose(eq(1)(gcd)))(
take(1000)(zip(sbs, tail(sbs)))
)
])))
- GENERIC ABSTRACTIONS ------------------------------------
- compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
Right to left function composition. return lambda f: lambda x: g(f(x))
- curry :: ((a, b) -> c) -> a -> b -> c
def curry(f):
A curried function derived
from an uncurried function.
return lambda a: lambda b: f(a, b)
- enumFrom :: Enum a => a -> [a]
def enumFrom(x):
A non-finite stream of enumerable values,
starting from the given value.
return count(x) if isinstance(x, int) else (
map(chr, count(ord(x)))
)
- every :: (a -> Bool) -> [a] -> Bool
def every(p):
True if p(x) holds for every x in xs return lambda xs: all(map(p, xs))
- fmapGen <$> :: (a -> b) -> Gen [a] -> Gen [b]
def fmapGen(f):
A function f mapped over a
non finite stream of values.
def go(g):
while True:
v = next(g, None)
if None is not v:
yield f(v)
else:
return
return lambda gen: go(gen)
- fst :: (a, b) -> a
def fst(tpl):
First member of a pair. return tpl[0]
- head :: [a] -> a
def head(xs):
The first element of a non-empty list. return xs[0]
- iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
An infinite list of repeated
applications of f to x.
def go(x):
v = x
while True:
yield v
v = f(v)
return lambda x: go(x)
- nubBy :: (a -> a -> Bool) -> [a] -> [a]
def nubBy(p):
A sublist of xs from which all duplicates,
(as defined by the equality predicate p)
are excluded.
def go(xs):
if not xs:
return []
x = xs[0]
return [x] + go(
list(filter(
lambda y: not p(x)(y),
xs[1:]
))
)
return lambda xs: go(xs)
- on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
def on(f):
A function returning the value of applying
the binary f to g(a) g(b)
return lambda g: lambda a: lambda b: f(g(a))(g(b))
- tail :: [a] -> [a]
- tail :: Gen [a] -> [a]
def tail(xs):
The elements following the head of a
(non-empty) list or generator stream.
if isinstance(xs, list):
return xs[1:]
else:
list(islice(xs, 1)) # First item dropped.
return xs
- take :: Int -> [a] -> [a]
- take :: Int -> String -> String
def take(n):
The prefix of xs of length n,
or xs itself if n > length xs.
return lambda xs: (
xs[0:n]
if isinstance(xs, list)
else list(islice(xs, n))
)
- unlines :: [String] -> String
def unlines(xs):
A single string derived by the intercalation
of a list of strings with the newline character.
return '\n'.join(xs)
- MAIN ---
if __name__ == '__main__':
main()</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
R
For loop
<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! >
While loop
Tasks like this are R's bread and butter. The previous solution uses smart mathematical tricks to generate the desired sequence with a for loop and uses a similarly clever for loop for the task involving gcds. However, R is smart enough to let us avoid this work by writing some much more idiomatic code.
As with the previous solution, we have used a library for our gcd function. In this case, we have used gmp. <lang r>genNStern<-function(n) {
sternNums<-c(1L,1L)
i<-2
while((endIndex<-length(sternNums))<n)
{
#To show off R's vectorization, the following line is deliberately terse.
#It assigns sternNums[i-1]+sternNums[i] to sternNums[endIndex+1]
#and it assigns sternNums[i], the "considered" number, to sternNums[endIndex+2], now the end of the sequence.
#Note that we do not have to initialize a big sternNums array to do this.
#True to the algorithm, the new entries are appended to the end of the old sequence.
sternNums[endIndex+c(1,2)]<-c(sum(sternNums[c(i-1,i)]),sternNums[i])
i<-i+1
}
sternNums[1:n]
}
- N=5000 was picked arbitrarily. The code runs very fast regardless of this number being much more than we need.
firstFiveThousandTerms<-genNStern(5000) match(1:10, firstFiveThousandTerms) match(100, firstFiveThousandTerms) all(sapply(1:999, function(i) gmp::gcd(firstFiveThousandTerms[i],firstFiveThousandTerms[i+1]))==1)</lang>
- Output:
> firstFiveThousandTerms<-genNStern(5000) > match(1:10, firstFiveThousandTerms) [1] 1 3 5 9 11 33 19 21 35 39 > match(100, firstFiveThousandTerms) [1] 1179 > all(sapply(1:999, function(i) gmp::gcd(firstFiveThousandTerms[i],firstFiveThousandTerms[i+1]))==1) [1] TRUE
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
Raku
(formerly 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
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 /*Not specified? Then use the default.*/ if idx== | idx=="," then idx= 10 /* " " " " " " */ if fix== | fix=="," then fix= 100 /* " " " " " " */ if chk== | chk=="," then chk=1000 /* " " " " " " */
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.*/ w=length(idx); do i=1 for idx
say 'for ' right(i, w)", 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 index' 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) /*get arg list. */
end /*i*/
parse var $ x z .; if x=0 then x=z /*is zero case? */
x=abs(x) /*use absolute x*/
do j=2 to words($); y=abs( word($, j) )
if y=0 then iterate /*ignore zeros. */
do until y==0; parse value x//y y with y x /* ◄──heavy work*/
end /*until*/
end /*j*/
return x /*return the 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($, k-1) ) _
end /*k*/
if f==0 then return subword($, 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.
Ring
<lang ring>
- Project : Stern-Brocot sequence
limit = 1200 item = list(limit+1) item[1] = 1 item[2] = 1 nr = 2 gcd = 1 gcdall = 1 for num = 3 to limit
item[num] = item[nr] + item[nr-1]
item[num+1] = item[nr]
nr = nr + 1
num = num + 1
next showarray(item,15)
for x = 1 to 100
if x < 11 or x = 100
totalitem(x)
ok
next
for n = 1 to len(item) - 1
if gcd(item[n],item[n+1]) != 1
gcdall = gcd
ok
next
if gcdall = 1
see "Correct: The first 999 consecutive pairs are relative prime!" + nl
ok
func totalitem(p)
pos = find(item, p)
see string(x) + " at Stern #" + pos + "." + nl
func showarray(vect,ln)
svect = ""
for n = 1 to ln
svect = svect + vect[n] + ", "
next
svect = left(svect, len(svect) - 2)
see svect
see nl
func gcd(gcd,b)
while b
c = gcd
gcd = b
b = c % b
end
return gcd
</lang> Output:
1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 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. Correct: The first 999 consecutive pairs are relative prime!
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).all? { |a,b| a.gcd(b) == 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]
{
list.append(list[0]+list[1], list[1])
list.shift
}
}
- Show the first fifteen members of the sequence.
say 15.of(stern_brocot()).join(' ')
- 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.
for i (1..10, 100) {
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(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
Snobol
<lang snobol>* GCD function
DEFINE('GCD(A,B)') :(GCD_END)
GCD GCD = A
EQ(B,0) :S(RETURN)
A = B
B = REMDR(GCD,B) :(GCD)
GCD_END
- Find first occurrence of element in array
DEFINE('IDX(ARR,ELM)') :(IDX_END)
IDX IDX = 1 ITEST EQ(ARR<IDX>,ELM) :S(RETURN)
IDX = IDX + 1 :(ITEST)
IDX_END
- Declare array
SEQ = ARRAY(1200,1)
- Fill array with Stern-Brocot sequence
IX = 1
FILL IX = IX + 1
SEQ<IX * 2 - 1> = SEQ<IX> + SEQ<IX - 1>
SEQ<IX * 2> = SEQ<IX> :S(FILL)
- Print first 15 elements
DONE IX = 1
S = "First 15 elements:"
P15 S = S " " SEQ<IX>
IX = IX + 1 LT(IX,15) :S(P15)
OUTPUT = S
- Print first occurrence of 1..10 and 100
N = 1
FIRSTN OUTPUT = "First " N " at " IDX(SEQ,N)
N = N + 1 LT(N,10) :S(FIRSTN)
OUTPUT = "First 100 at " IDX(SEQ,100)
- Test GCD between 1000 consecutive members
IX = 2
GCDTEST EQ(GCD(SEQ<IX - 1>,SEQ<IX>),1) :F(GCDFAIL)
IX = IX + 1 LT(IX,1000) :S(GCDTEST)
OUTPUT = "All GCDs are 1." :(END)
GCDFAIL OUTPUT = "GCD is not 1 at " IX "."
END</lang>
- Output:
First 15 elements: 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 All GCDs are 1.
Swift
<lang swift>struct SternBrocot: Sequence, IteratorProtocol {
private var seq = [1, 1]
mutating func next() -> Int? {
seq += [seq[0] + seq[1], seq[1]]
return seq.removeFirst() }
}
func gcd<T: BinaryInteger>(_ a: T, _ b: T) -> T {
guard a != 0 else {
return b
}
return a < b ? gcd(b % a, a) : gcd(a % b, b)
}
print("First 15: \(Array(SternBrocot().prefix(15)))")
var found = Set<Int>()
for (i, val) in SternBrocot().enumerated() {
switch val {
case 1...10 where !found.contains(val), 100 where !found.contains(val):
print("First \(val) at \(i + 1)")
found.insert(val)
case _:
continue
}
if found.count == 11 {
break
}
}
let firstThousand = SternBrocot().prefix(1000) let gcdIsOne = zip(firstThousand, firstThousand.dropFirst()).allSatisfy({ gcd($0.0, $0.1) == 1 })
print("GCDs of all two consecutive members are \(gcdIsOne ? "" : "not")one")</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 7 at 19 First 8 at 21 First 6 at 33 First 9 at 35 First 10 at 39 First 100 at 1179 GCDs of all two consecutive members are one
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
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.
Visual Basic .NET
<lang vbnet>Imports System Imports System.Collections.Generic Imports System.Linq
Module Module1
Dim l As List(Of Integer) = {1, 1}.ToList()
Function gcd(ByVal a As Integer, ByVal b As Integer) As Integer
Return If(a > 0, If(a < b, gcd(b Mod a, a), gcd(a Mod b, b)), b)
End Function
Sub Main(ByVal args As String())
Dim max As Integer = 1000, take As Integer = 15, i As Integer = 1,
selection As Integer() = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 100}
Do : l.AddRange({l(i) + l(i - 1), l(i)}.ToList) : i += 1
Loop While l.Count < max OrElse l(l.Count - 2) <> selection.Last()
Console.Write("The first {0} items In the Stern-Brocot sequence: ", take)
Console.WriteLine("{0}" & vbLf, String.Join(", ", l.Take(take)))
Console.WriteLine("The locations of where the selected numbers (1-to-10, & 100) first appear:")
For Each ii As Integer In selection
Dim j As Integer = l.FindIndex(Function(x) x = ii) + 1
Console.WriteLine("{0,3}: {1:n0}", ii, j)
Next : Console.WriteLine() : Dim good As Boolean = True : For i = 1 To max
If gcd(l(i), l(i - 1)) <> 1 Then good = False : Exit For
Next
Console.WriteLine("The greatest common divisor of all the two consecutive items of the" &
" series up to the {0}th item is {1}always one.", max, If(good, "", "not "))
End Sub
End Module</lang>
- Output:
The first 15 items In the Stern-Brocot sequence: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4 The locations of where the selected numbers (1-to-10, & 100) first appear: 1: 1 2: 3 3: 5 4: 9 5: 11 6: 33 7: 19 8: 21 9: 35 10: 39 100: 1,179 The greatest common divisor of all the two consecutive items of the series up to the 1000th item is always one.
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)
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