Padovan sequence
You are encouraged to solve this task according to the task description, using any language you may know.
The Padovan sequence is similar to the Fibonacci sequence in several ways.
Some are given in the table below, and the referenced video shows some of the geometric
similarities.
Comment Padovan Fibonacci Named after. Richard Padovan Leonardo of Pisa: Fibonacci Recurrence initial values. P(0)=P(1)=P(2)=1 F(0)=0, F(1)=1 Recurrence relation. P(n)=P(n-2)+P(n-3) F(n)=F(n-1)+F(n-2) First 10 terms. 1,1,1,2,2,3,4,5,7,9 0,1,1,2,3,5,8,13,21,34 Ratio of successive terms... Plastic ratio, p Golden ratio, g 1.324717957244746025960908854… 1.6180339887498948482... Exact formula of ratios p and q. ((9+69**.5)/18)**(1/3) + ((9-69**.5)/18)**(1/3) (1+5**0.5)/2 Ratio is real root of polynomial. p: x**3-x-1 g: x**2-x-1 Spirally tiling the plane using. Equilateral triangles Squares Constants for ... s= 1.0453567932525329623 a=5**0.5 ... Computing by truncation. P(n)=floor(p**(n-1) / s + .5) F(n)=floor(g**n / a + .5) L-System Variables. A,B,C A,B L-System Start/Axiom. A A L-System Rules. A->B,B->C,C->AB A->B,B->AB
- Task
- Write a function/method/subroutine to compute successive members of the Padovan series using the recurrence relation.
- Write a function/method/subroutine to compute successive members of the Padovan series using the floor function.
- Show the first twenty terms of the sequence.
- Confirm that the recurrence and floor based functions give the same results for 64 terms,
- Write a function/method/... using the L-system to generate successive strings.
- Show the first 10 strings produced from the L-system
- Confirm that the length of the first 32 strings produced is the Padovan sequence.
Show output here, on this page.
- Ref
- The Plastic Ratio - Numberphile video.
11l
V pp = 1.324717957244746025960908854
V ss = 1.0453567932525329623
V Rules = [‘A’ = ‘B’, ‘B’ = ‘C’, ‘C’ = ‘AB’]
F padovan1(n)
V r = [1] * min(n, 3)
V (a, b, c) = (1, 1, 1)
V count = 3
L count < n
(a, b, c) = (b, c, a + b)
r [+]= c
count++
R r
F padovan2(n)
V r = [1] * (n > 1)
V p = 1.0
V count = 1
L count < n
r [+]= Int(round(p / :ss))
p *= :pp
count++
R r
F padovan3(n)
[String] r
V s = ‘A’
V count = 0
L count < n
r [+]= s
V next = ‘’
L(ch) s
next ‘’= Rules[ch]
s = next
count++
R r
print(‘First 20 terms of the Padovan sequence:’)
print(padovan1(20).join(‘ ’))
V list1 = padovan1(64)
V list2 = padovan2(64)
print(‘The first 64 iterative and calculated values ’(I list1 == list2 {‘are the same.’} E ‘differ.’))
print()
print(‘First 10 L-system strings:’)
print(padovan3(10).join(‘ ’))
print()
print(‘Lengths of the 32 first L-system strings:’)
V list3 = padovan3(32).map(x -> x.len)
print(list3.join(‘ ’))
print(‘These lengths are’(I list3 == list1[0.<32] {‘ ’} E ‘ not ’)‘the 32 first terms of the Padovan sequence.’)- Output:
First 20 terms of the Padovan sequence: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 The first 64 iterative and calculated values are the same. First 10 L-system strings: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB Lengths of the 32 first L-system strings: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 200 265 351 465 616 816 1081 1432 1897 2513 3329 4410 These lengths are the 32 first terms of the Padovan sequence.
ALGOL 68
BEGIN # show members of the Padovan Sequence calculated in various ways #
# returns the first n elements of the Padovan sequence by the #
# recurance relation: P(n)=P(n-2)+P(n-3) #
OP PADOVANI = ( INT n )[]INT:
BEGIN
[ 0 : n - 1 ]INT p; p[ 0 ] := p[ 1 ] := p[ 2 ] := 1;
FOR i FROM 3 TO UPB p DO
p[ i ] := p[ i - 2 ] + p[ i - 3 ]
OD;
p
END; # PADOVANI #
# returns the first n elements of the Padovan sequence by #
# computing by truncation P(n)=floor(p^(n-1) / s + .5) #
# where s = 1.0453567932525329623 #
# and p = the "plastic ratio" #
OP PADOVANC = ( INT n )[]INT:
BEGIN
LONG REAL s = 1.0453567932525329623;
LONG REAL p = 1.324717957244746025960908854;
LONG REAL pf := 1 / p;
[ 0 : n - 1 ]INT result;
FOR i FROM LWB result TO UPB result DO
result[ i ] := SHORTEN ENTIER ( pf / s + 0.5 );
pf *:= p
OD;
result
END; # PADOVANC #
# returns the first n L System strings of the Padovan sequence #
OP PADOVANL = ( INT n )[]STRING:
BEGIN
[ 0 : n - 1 ]STRING l; l[ 0 ] := "A"; l[ 1 ] := "B"; l[ 2 ] := "C";
FOR i FROM 3 TO UPB l DO
l[ i ] := l[ i - 3 ] + l[ i - 2 ]
OD;
l
END; # PADOVANC #
# returns TRUE if a and b have the same values, FALSE otherwise #
OP = = ( []INT a, b )BOOL:
IF LWB a /= LWB b OR UPB a /= UPB b
THEN # rows are not the same size # FALSE
ELSE
BOOL result := TRUE;
FOR i FROM LWB a TO UPB a WHILE result := a[ i ] = b[ i ] DO SKIP OD;
result
FI; # = #
# returns the number of elements in a #
OP LENGTH = ( []INT a )INT: ( UPB a - LWB a ) + 1;
# returns the number of characters in s #
OP LENGTH = ( STRING s )INT: ( UPB s - LWB s ) + 1;
# returns a string representation of n #
OP TOSTRING = ( INT n )STRING: whole( n, 0 );
# generate 64 elements of the sequence and 32 L System values #
[]INT iterative = PADOVANI 64;
[]INT calculated = PADOVANC 64;
[]STRING l system = PADOVANL 32;
[ LWB l system : UPB l system ]INT l length;
FOR i FROM LWB l length TO UPB l length DO l length[ i ] := LENGTH l system[ i ] OD;
# first 20 terms #
print( ( "First 20 terms of the Padovan Sequence", newline ) );
FOR i FROM LWB iterative TO 19 DO
print( ( " ", TOSTRING iterative[ i ] ) )
OD;
print( ( newline ) );
print( ( "The first "
, TOSTRING LENGTH iterative
, " iterative and calculated values "
, IF iterative = calculated THEN "are the same" ELSE "differ" FI
, newline
)
);
# print the first 10 values of the L System strings #
print( ( newline, "First 10 L System strings", newline ) );
FOR i FROM LWB l system TO 9 DO
print( ( " ", l system[ i ] ) )
OD;
print( ( newline ) );
print( ( "The first "
, TOSTRING LENGTH l length
, " iterative values and L System lengths "
, IF l length = iterative[ LWB l length : UPB l length @ LWB l length ] THEN "are the same" ELSE "differ" FI
, newline
)
)
END- Output:
First 20 terms of the Padovan Sequence 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 The first 64 iterative and calculated values are the same First 10 L System strings A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB The first 32 iterative values and L System lengths are the same
AppleScript
--------------------- PADOVAN NUMBERS --------------------
-- padovans :: [Int]
on padovans()
script f
on |λ|(abc)
set {a, b, c} to abc
{a, {b, c, a + b}}
end |λ|
end script
unfoldr(f, {1, 1, 1})
end padovans
-- padovanFloor :: [Int]
on padovanFloor()
script f
property p : 1.324717957245
property s : 1.045356793253
on |λ|(n)
{floor(0.5 + ((p ^ (n - 1)) / s)), 1 + n}
end |λ|
end script
unfoldr(f, 0)
end padovanFloor
-- padovanLSystem :: [String]
on padovanLSystem()
script rule
on |λ|(c)
if "A" = c then
"B"
else if "B" = c then
"C"
else
"AB"
end if
end |λ|
end script
script f
on |λ|(s)
{s, concatMap(rule, characters of s) as string}
end |λ|
end script
unfoldr(f, "A")
end padovanLSystem
--------------------------- TEST -------------------------
on run
unlines({"First 20 padovans:", ¬
showList(take(20, padovans())), ¬
"", ¬
"The recurrence and floor-based functions", ¬
"match over the first 64 terms:\n", ¬
prefixesMatch(padovans(), padovanFloor(), 64), ¬
"", ¬
"First 10 L-System strings:", ¬
showList(take(10, padovanLSystem())), ¬
"", ¬
"The lengths of the first 32 L-System", ¬
"strings match the Padovan sequence:\n", ¬
prefixesMatch(padovans(), fmap(|length|, padovanLSystem()), 32)})
end run
-- prefixesMatch :: [a] -> [a] -> Bool
on prefixesMatch(xs, ys, n)
take(n, xs) = take(n, ys)
end prefixesMatch
------------------------- GENERIC ------------------------
-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
set lng to length of xs
set acc to {}
tell mReturn(f)
repeat with i from 1 to lng
set acc to acc & (|λ|(item i of xs, i, xs))
end repeat
end tell
return acc
end concatMap
-- floor :: Num -> Int
on floor(x)
if class of x is record then
set nr to properFracRatio(x)
else
set nr to properFraction(x)
end if
set n to item 1 of nr
if 0 > item 2 of nr then
n - 1
else
n
end if
end floor
-- fmap <$> :: (a -> b) -> Gen [a] -> Gen [b]
on fmap(f, gen)
script
property g : mReturn(f)
on |λ|()
set v to gen's |λ|()
if v is missing value then
v
else
g's |λ|(v)
end if
end |λ|
end script
end fmap
-- intercalate :: String -> [String] -> String
on intercalate(delim, xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, delim}
set s to xs as text
set my text item delimiters to dlm
s
end intercalate
-- 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)
-- The list obtained by applying f
-- to each element of 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
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- properFraction :: Real -> (Int, Real)
on properFraction(n)
set i to (n div 1)
{i, n - i}
end properFraction
-- showList :: [a] -> String
on showList(xs)
"[" & intercalate(",", map(my str, xs)) & "]"
end showList
-- str :: a -> String
on str(x)
x as string
end str
-- 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 |λ|() of xs
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
-- unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
on unfoldr(f, v)
-- A lazy (generator) list unfolded from a seed value
-- by repeated application of f to a value until no
-- residue remains. Dual to fold/reduce.
-- f returns either nothing (missing value),
-- or just (value, residue).
script
property valueResidue : {v, v}
property g : mReturn(f)
on |λ|()
set valueResidue to g's |λ|(item 2 of (valueResidue))
if missing value ≠ valueResidue then
item 1 of (valueResidue)
else
missing value
end if
end |λ|
end script
end unfoldr
-- unlines :: [String] -> String
on unlines(xs)
-- A single string formed by the intercalation
-- of a list of strings with the newline character.
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set s to xs as text
set my text item delimiters to dlm
s
end unlines
- Output:
First 20 padovans: [1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151] The recurrence and floor-based functions match over the first 64 terms: true First 10 L-System strings: [A,B,C,AB,BC,CAB,ABBC,BCCAB,CABABBC,ABBCBCCAB] The lengths of the first 32 L-System strings match the Padovan sequence: true
C
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
/* Generate (and memoize) the Padovan sequence using
* the recurrence relationship */
int pRec(int n) {
static int *memo = NULL;
static size_t curSize = 0;
/* grow memoization array when necessary and fill with zeroes */
if (curSize <= (size_t) n) {
size_t lastSize = curSize;
while (curSize <= (size_t) n) curSize += 1024 * sizeof(int);
memo = realloc(memo, curSize * sizeof(int));
memset(memo + lastSize, 0, (curSize - lastSize) * sizeof(int));
}
/* if we don't have the value for N yet, calculate it */
if (memo[n] == 0) {
if (n<=2) memo[n] = 1;
else memo[n] = pRec(n-2) + pRec(n-3);
}
return memo[n];
}
/* Calculate the Nth value of the Padovan sequence
* using the floor function */
int pFloor(int n) {
long double p = 1.324717957244746025960908854;
long double s = 1.0453567932525329623;
return powl(p, n-1)/s + 0.5;
}
/* Given the previous value for the L-system, generate the
* next value */
void nextLSystem(const char *prev, char *buf) {
while (*prev) {
switch (*prev++) {
case 'A': *buf++ = 'B'; break;
case 'B': *buf++ = 'C'; break;
case 'C': *buf++ = 'A'; *buf++ = 'B'; break;
}
}
*buf = '\0';
}
int main() {
// 8192 is enough up to P_33.
#define BUFSZ 8192
char buf1[BUFSZ], buf2[BUFSZ];
int i;
/* Print P_0..P_19 */
printf("P_0 .. P_19: ");
for (i=0; i<20; i++) printf("%d ", pRec(i));
printf("\n");
/* Check that functions match up to P_63 */
printf("The floor- and recurrence-based functions ");
for (i=0; i<64; i++) {
if (pRec(i) != pFloor(i)) {
printf("do not match at %d: %d != %d.\n",
i, pRec(i), pFloor(i));
break;
}
}
if (i == 64) {
printf("match from P_0 to P_63.\n");
}
/* Show first 10 L-system strings */
printf("\nThe first 10 L-system strings are:\n");
for (strcpy(buf1, "A"), i=0; i<10; i++) {
printf("%s\n", buf1);
strcpy(buf2, buf1);
nextLSystem(buf2, buf1);
}
/* Check lengths of strings against pFloor up to P_31 */
printf("\nThe floor- and L-system-based functions ");
for (strcpy(buf1, "A"), i=0; i<32; i++) {
if ((int)strlen(buf1) != pFloor(i)) {
printf("do not match at %d: %d != %d\n",
i, (int)strlen(buf1), pFloor(i));
break;
}
strcpy(buf2, buf1);
nextLSystem(buf2, buf1);
}
if (i == 32) {
printf("match from P_0 to P_31.\n");
}
return 0;
}
- Output:
P_0 .. P_19: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 The floor- and recurrence-based functions match from P_0 to P_63. The first 10 L-system strings are: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB The floor- and L-system-based functions match from P_0 to P_31.
C++
#include <iostream>
#include <map>
#include <cmath>
// Generate the Padovan sequence using the recurrence
// relationship.
int pRec(int n) {
static std::map<int,int> memo;
auto it = memo.find(n);
if (it != memo.end()) return it->second;
if (n <= 2) memo[n] = 1;
else memo[n] = pRec(n-2) + pRec(n-3);
return memo[n];
}
// Calculate the N'th Padovan sequence using the
// floor function.
int pFloor(int n) {
long const double p = 1.324717957244746025960908854;
long const double s = 1.0453567932525329623;
return std::pow(p, n-1)/s + 0.5;
}
// Return the N'th L-system string
std::string& lSystem(int n) {
static std::map<int,std::string> memo;
auto it = memo.find(n);
if (it != memo.end()) return it->second;
if (n == 0) memo[n] = "A";
else {
memo[n] = "";
for (char ch : memo[n-1]) {
switch(ch) {
case 'A': memo[n].push_back('B'); break;
case 'B': memo[n].push_back('C'); break;
case 'C': memo[n].append("AB"); break;
}
}
}
return memo[n];
}
// Compare two functions up to p_N
using pFn = int(*)(int);
void compare(pFn f1, pFn f2, const char* descr, int stop) {
std::cout << "The " << descr << " functions ";
int i;
for (i=0; i<stop; i++) {
int n1 = f1(i);
int n2 = f2(i);
if (n1 != n2) {
std::cout << "do not match at " << i
<< ": " << n1 << " != " << n2 << ".\n";
break;
}
}
if (i == stop) {
std::cout << "match from P_0 to P_" << stop << ".\n";
}
}
int main() {
/* Print P_0 to P_19 */
std::cout << "P_0 .. P_19: ";
for (int i=0; i<20; i++) std::cout << pRec(i) << " ";
std::cout << "\n";
/* Check that floor and recurrence match up to P_64 */
compare(pFloor, pRec, "floor- and recurrence-based", 64);
/* Show first 10 L-system strings */
std::cout << "\nThe first 10 L-system strings are:\n";
for (int i=0; i<10; i++) std::cout << lSystem(i) << "\n";
std::cout << "\n";
/* Check lengths of strings against pFloor up to P_31 */
compare(pFloor, [](int n){return (int)lSystem(n).length();},
"floor- and L-system-based", 32);
return 0;
}
- Output:
P_0 .. P_19: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 The floor- and recurrence-based functions match from P_0 to P_64. The first 10 L-system strings are: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB The floor- and L-system-based functions match from P_0 to P_32.
Clojure
(def padovan (map first (iterate (fn [[a b c]] [b c (+ a b)]) [1 1 1])))
(def pad-floor
(let [p 1.324717957244746025960908854
s 1.0453567932525329623]
(map (fn [n] (int (Math/floor (+ (/ (Math/pow p (dec n)) s) 0.5)))) (range))))
(def pad-l
(iterate (fn f [[c & s]]
(case c
\A (str "B" (f s))
\B (str "C" (f s))
\C (str "AB" (f s))
(str "")))
"A"))
(defn comp-seq [n seqa seqb]
(= (take n seqa) (take n seqb)))
(defn comp-all [n]
(= (map count (vec (take n pad-l)))
(take n padovan)
(take n pad-floor)))
(defn padovan-print [& args]
((print "The first 20 items with recursion relation are: ")
(println (take 20 padovan))
(println)
(println (str
"The recurrence and floor based algorithms "
(if (comp-seq 64 padovan pad-floor) "match" "not match")
" to n=64"))
(println)
(println "The first 10 L-system strings are:")
(println (take 10 pad-l))
(println)
(println (str
"The L-system, recurrence and floor based algorithms "
(if (comp-all 32) "match" "not match")
" to n=32"))))
- Output:
The first 20 items with recursion relation are: (1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151) The recurrence and floor based algorithms match to n=64 The first 10 L-system strings are: (A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB) The L-system, recurrence and floor based algorithms match to n=32
Delphi
Thanks Rudy Velthuis for the Velthuis.BigDecimals library.
Boost.Generics.Collection is part of DelphiBoostLib.
program Padovan_sequence;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
Velthuis.BigDecimals,
Boost.Generics.Collection;
type
TpFn = TFunc<Integer, Integer>;
var
RecMemo: TDictionary<Integer, Integer>;
lSystemMemo: TDictionary<Integer, string>;
function pRec(n: Integer): Integer;
begin
if RecMemo.HasKey(n) then
exit(RecMemo[n]);
if (n <= 2) then
RecMemo[n] := 1
else
RecMemo[n] := pRec(n - 2) + pRec(n - 3);
Result := RecMemo[n];
end;
function pFloor(n: Integer): Integer;
var
p, s, a: BigDecimal;
begin
p := '1.324717957244746025960908854';
s := '1.0453567932525329623';
a := p.IntPower(n - 1, 64);
Result := Round(BigDecimal.Divide(a, s));
end;
function lSystem(n: Integer): string;
begin
if n = 0 then
lSystemMemo[n] := 'A'
else
begin
lSystemMemo[n] := '';
for var ch in lSystemMemo[n - 1] do
begin
case ch of
'A':
lSystemMemo[n] := lSystemMemo[n] + 'B';
'B':
lSystemMemo[n] := lSystemMemo[n] + 'C';
'C':
lSystemMemo[n] := lSystemMemo[n] + 'AB';
end;
end;
end;
Result := lSystemMemo[n];
end;
procedure Compare(f1, f2: TpFn; descr: string; stop: Integer);
begin
write('The ', descr, ' functions ');
var i := 0;
while i < stop do
begin
var n1 := f1(i);
var n2 := f2(i);
if n1 <> n2 then
begin
write('do not match at ', i);
writeln(': ', n1, ' != ', n2, '.');
break;
end;
inc(i);
end;
if i = stop then
writeln('match from P_0 to P_', stop, '.');
end;
begin
RecMemo := TDictionary<Integer, Integer>.Create([], []);
lSystemMemo := TDictionary<Integer, string>.Create([], []);
write('P_0 .. P_19: ');
for var i := 0 to 19 do
write(pRec(i), ' ');
writeln;
Compare(pFloor, pRec, 'floor- and recurrence-based', 64);
writeln(#10'The first 10 L-system strings are:');
for var i := 0 to 9 do
writeln(lSystem(i));
writeln;
Compare(pFloor,
function(n: Integer): Integer
begin
Result := length(lSystem(n));
end, 'floor- and L-system-based', 32);
readln;
end.
Factor
USING: L-system accessors io kernel make math math.functions
memoize prettyprint qw sequences ;
CONSTANT: p 1.324717957244746025960908854
CONSTANT: s 1.0453567932525329623
: pfloor ( m -- n ) 1 - p swap ^ s /f .5 + >integer ;
MEMO: precur ( m -- n )
dup 3 < [ drop 1 ]
[ [ 2 - precur ] [ 3 - precur ] bi + ] if ;
: plsys, ( L-system -- )
[ iterate-L-system-string ] [ string>> , ] bi ;
: plsys ( n -- seq )
<L-system>
"A" >>axiom
{ qw{ A B } qw{ B C } qw{ C AB } } >>rules
swap 1 - '[ "A" , _ [ dup plsys, ] times ] { } make nip ;
"First 20 terms of the Padovan sequence:" print
20 [ pfloor pprint bl ] each-integer nl nl
64 [ [ pfloor ] [ precur ] bi assert= ] each-integer
"Recurrence and floor based algorithms match to n=63." print nl
"First 10 L-system strings:" print
10 plsys . nl
32 <iota> [ pfloor ] map 32 plsys [ length ] map assert=
"The L-system, recurrence and floor based algorithms match to n=31." print
- Output:
First 20 terms of the Padovan sequence:
1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151
Recurrence and floor based algorithms match to n=63.
First 10 L-system strings:
{
"A"
"B"
"C"
"AB"
"BC"
"CAB"
"ABBC"
"BCCAB"
"CABABBC"
"ABBCBCCAB"
}
The L-system, recurrence and floor based algorithms match to n=31.
Go
package main
import (
"fmt"
"math"
"math/big"
"strings"
)
func padovanRecur(n int) []int {
p := make([]int, n)
p[0], p[1], p[2] = 1, 1, 1
for i := 3; i < n; i++ {
p[i] = p[i-2] + p[i-3]
}
return p
}
func padovanFloor(n int) []int {
var p, s, t, u = new(big.Rat), new(big.Rat), new(big.Rat), new(big.Rat)
p, _ = p.SetString("1.324717957244746025960908854")
s, _ = s.SetString("1.0453567932525329623")
f := make([]int, n)
pow := new(big.Rat).SetInt64(1)
u = u.SetFrac64(1, 2)
t.Quo(pow, p)
t.Quo(t, s)
t.Add(t, u)
v, _ := t.Float64()
f[0] = int(math.Floor(v))
for i := 1; i < n; i++ {
t.Quo(pow, s)
t.Add(t, u)
v, _ = t.Float64()
f[i] = int(math.Floor(v))
pow.Mul(pow, p)
}
return f
}
type LSystem struct {
rules map[string]string
init, current string
}
func step(lsys *LSystem) string {
var sb strings.Builder
if lsys.current == "" {
lsys.current = lsys.init
} else {
for _, c := range lsys.current {
sb.WriteString(lsys.rules[string(c)])
}
lsys.current = sb.String()
}
return lsys.current
}
func padovanLSys(n int) []string {
rules := map[string]string{"A": "B", "B": "C", "C": "AB"}
lsys := &LSystem{rules, "A", ""}
p := make([]string, n)
for i := 0; i < n; i++ {
p[i] = step(lsys)
}
return p
}
// assumes lists are same length
func areSame(l1, l2 []int) bool {
for i := 0; i < len(l1); i++ {
if l1[i] != l2[i] {
return false
}
}
return true
}
func main() {
fmt.Println("First 20 members of the Padovan sequence:")
fmt.Println(padovanRecur(20))
recur := padovanRecur(64)
floor := padovanFloor(64)
same := areSame(recur, floor)
s := "give"
if !same {
s = "do not give"
}
fmt.Println("\nThe recurrence and floor based functions", s, "the same results for 64 terms.")
p := padovanLSys(32)
lsyst := make([]int, 32)
for i := 0; i < 32; i++ {
lsyst[i] = len(p[i])
}
fmt.Println("\nFirst 10 members of the Padovan L-System:")
fmt.Println(p[:10])
fmt.Println("\nand their lengths:")
fmt.Println(lsyst[:10])
same = areSame(recur[:32], lsyst)
s = "give"
if !same {
s = "do not give"
}
fmt.Println("\nThe recurrence and L-system based functions", s, "the same results for 32 terms.")
- Output:
First 20 members of the Padovan sequence: [1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151] The recurrence and floor based functions give the same results for 64 terms. First 10 members of the Padovan L-System: [A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB] and their lengths: [1 1 1 2 2 3 4 5 7 9] The recurrence and L-system based functions give the same results for 32 terms.
Haskell
-- list of Padovan numbers using recurrence
pRec = map (\(a,_,_) -> a) $ iterate (\(a,b,c) -> (b,c,a+b)) (1,1,1)
-- list of Padovan numbers using self-referential lazy lists
pSelfRef = 1 : 1 : 1 : zipWith (+) pSelfRef (tail pSelfRef)
-- list of Padovan numbers generated from floor function
pFloor = map f [0..]
where f n = floor $ p**fromInteger (pred n) / s + 0.5
p = 1.324717957244746025960908854
s = 1.0453567932525329623
-- list of L-system strings
lSystem = iterate f "A"
where f [] = []
f ('A':s) = 'B':f s
f ('B':s) = 'C':f s
f ('C':s) = 'A':'B':f s
-- check if first N elements match
checkN n as bs = take n as == take n bs
main = do
putStr "P_0 .. P_19: "
putStrLn $ unwords $ map show $ take 20 pRec
putStr "The floor- and recurrence-based functions "
putStr $ if checkN 64 pRec pFloor then "match" else "do not match"
putStr " from P_0 to P_63.\n"
putStr "The self-referential- and recurrence-based functions "
putStr $ if checkN 64 pRec pSelfRef then "match" else "do not match"
putStr " from P_0 to P_63.\n\n"
putStr "The first 10 L-system strings are:\n"
putStrLn $ unwords $ take 10 lSystem
putStr "\nThe floor- and L-system-based functions "
putStr $ if checkN 32 pFloor (map length lSystem)
then "match" else "do not match"
putStr " from P_0 to P_31.\n"
- Output:
P_0 .. P_19: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 The floor- and recurrence-based functions match from P_0 to P_63. The self-referential- and recurrence-based functions match from P_0 to P_63. The first 10 L-system strings are: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB The floor- and L-system-based functions match from P_0 to P_31.
and a variant expressed in terms of unfoldr, which allows for a coherent treatment of these three cases – isolating the respects in which they differ – and also lends itself to a simple translation to the N-Step Padovan case, covered in another task.
import Data.List (unfoldr)
--------------------- PADOVAN NUMBERS --------------------
padovans :: [Integer]
padovans = unfoldr f (1, 1, 1)
where
f (a, b, c) = Just (a, (b, c, a + b))
padovanFloor :: [Integer]
padovanFloor = unfoldr f 0
where
f = Just . (((,) . g) <*> succ)
g = floor . (0.5 +) . (/ s) . (p **) . fromInteger . pred
p = 1.324717957244746025960908854
s = 1.0453567932525329623
padovanLSystem :: [String]
padovanLSystem = unfoldr f "A"
where
f = Just . ((,) <*> concatMap rule)
rule 'A' = "B"
rule 'B' = "C"
rule 'C' = "AB"
-------------------------- TESTS -------------------------
main :: IO ()
main =
mapM_
putStrLn
[ "First 20 padovans:\n",
show $ take 20 padovans,
[],
"The recurrence and floor based functions",
"match over 64 terms:\n",
show $ prefixesMatch padovans padovanFloor 64,
[],
"First 10 L-System strings:\n",
show $ take 10 padovanLSystem,
[],
"The length of the first 32 strings produced",
"is the Padovan sequence:\n",
show $
prefixesMatch
padovans
(fromIntegral . length <$> padovanLSystem)
32
]
prefixesMatch :: Eq a => [a] -> [a] -> Int -> Bool
prefixesMatch xs ys n = and (zipWith (==) (take n xs) ys)
- Output:
First 20 padovans: [1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151] The recurrence and floor based functions match over 64 terms: True First 10 L-System strings: ["A","B","C","AB","BC","CAB","ABBC","BCCAB","CABABBC","ABBCBCCAB"] The length of the first 32 strings produced is the Padovan sequence: True
J
Implementation:
padovanSeq=: (],+/@(_2 _3{]))^:([-3:)&1 1 1
NB. or, equivalently:
padovanSeq=: (, [: +/ _2 {. }:)@]^:([ - 2:)&1 1
realRoot=. {:@(#~ ]=|)@;@p.
padovanNth=: 0.5 <.@+ (realRoot _23 23 _2 1) %~ (realRoot _1 _1 0 1)^<:
padovanL=: rplc&('A';'B'; 'B';'C'; 'C';'AB')@]^:[&'A'
seqLen=. #@(-.&' ')"1
Typically, inductive sequences based on a function F with an initial value G can be expressed using an expression of the form F@]^:[@G or something similar. Here, [ represents the argument to the derived recurrence function. For padovanSeq, we are generating a sequence, so we want to retain the previous values. So instead of just using f=: +/@(_2 3{]) which adds up the second and third numbers from the end of the sequence, we also retain the previous values using F=: ],f
But, also, since our initial value is a sequence with three elements, we can make the argument to the recurrence function be the length of the desired sequence by replacing [ for the repetition count with [-3:
Task examples:
padovanSeq 20
1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151
(padovanSeq 64) -: padovanNth(i.64)
1
padovanL i.10
A
B
C
AB
BC
CAB
ABBC
BCCAB
CABABBC
ABBCBCCAB
(padovanSeq 32) -: seqLen padovanL i.32
1
JavaScript
(() => {
"use strict";
// ----------------- PADOVAN NUMBERS -----------------
// padovans :: [Int]
const padovans = () => {
// Non-finite series of Padovan numbers,
// defined in terms of recurrence relations.
const f = ([a, b, c]) => [
a,
[b, c, a + b]
];
return unfoldr(f)([1, 1, 1]);
};
// padovanFloor :: [Int]
const padovanFloor = () => {
// The Padovan series, defined in terms
// of a floor function.
const
// NB JavaScript loses some of this
// precision at run-time.
p = 1.324717957244746025960908854,
s = 1.0453567932525329623;
const f = n => [
Math.floor(((p ** (n - 1)) / s) + 0.5),
1 + n
];
return unfoldr(f)(0);
};
// padovanLSystem : [Int]
const padovanLSystem = () => {
// An L-system generating terms whose lengths
// are the values of the Padovan integer series.
const rule = c =>
"A" === c ? (
"B"
) : "B" === c ? (
"C"
) : "AB";
const f = s => [
s,
chars(s).flatMap(rule)
.join("")
];
return unfoldr(f)("A");
};
// ---------------------- TEST -----------------------
// main :: IO ()
const main = () => {
// prefixesMatch :: [a] -> [a] -> Bool
const prefixesMatch = xs =>
ys => n => and(
zipWith(a => b => a === b)(
take(n)(xs)
)(
take(n)(ys)
)
);
return [
"First 20 padovans:",
take(20)(padovans()),
"\nThe recurrence and floor-based functions",
"match over the first 64 terms:\n",
prefixesMatch(
padovans()
)(
padovanFloor()
)(64),
"\nFirst 10 L-System strings:",
take(10)(padovanLSystem()),
"\nThe lengths of the first 32 L-System",
"strings match the Padovan sequence:\n",
prefixesMatch(
padovans()
)(
fmap(length)(padovanLSystem())
)(32)
]
.map(str)
.join("\n");
};
// --------------------- GENERIC ---------------------
// and :: [Bool] -> Bool
const and = xs =>
// True unless any value in xs is false.
[...xs].every(Boolean);
// chars :: String -> [Char]
const chars = s =>
s.split("");
// fmap <$> :: (a -> b) -> Gen [a] -> Gen [b]
const fmap = f =>
function* (gen) {
let v = take(1)(gen);
while (0 < v.length) {
yield f(v[0]);
v = take(1)(gen);
}
};
// length :: [a] -> Int
const length = xs =>
// 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
"GeneratorFunction" !== xs.constructor
.constructor.name ? (
xs.length
) : Infinity;
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => "GeneratorFunction" !== xs
.constructor.constructor.name ? (
xs.slice(0, n)
) : [].concat(...Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));
// str :: a -> String
const str = x =>
"string" !== typeof x ? (
JSON.stringify(x)
) : x;
// unfoldr :: (b -> Maybe (a, b)) -> b -> Gen [a]
const unfoldr = f =>
// A lazy (generator) list unfolded from a seed value
// by repeated application of f to a value until no
// residue remains. Dual to fold/reduce.
// f returns either Null or just (value, residue).
// For a strict output list,
// wrap with `list` or Array.from
x => (
function* () {
let valueResidue = f(x);
while (null !== valueResidue) {
yield valueResidue[0];
valueResidue = f(valueResidue[1]);
}
}()
);
// zipWithList :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = f =>
// A list constructed by zipping with a
// custom function, rather than with the
// default tuple constructor.
xs => ys => ((xs_, ys_) => {
const lng = Math.min(length(xs_), length(ys_));
return take(lng)(xs_).map(
(x, i) => f(x)(ys_[i])
);
})([...xs], [...ys]);
// MAIN ---
return main();
})();
- Output:
First 20 padovans: [1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151] The recurrence and floor-based functions match over the first 64 terms: true First 10 L-System strings: ["A","B","C","AB","BC","CAB","ABBC","BCCAB","CABABBC","ABBCBCCAB"] The lengths of the first 32 L-System strings match the Padovan sequence: true
jq
Works with gojq, the Go implementation of jq
This entry differs from others in that the L-system is used to define two functions - `padovanStrings` and `padovanNumbers` - that emit unbounded streams of the Padovan strings and numbers respectively.
# Output: first $n Padovans
def padovanRecur($n):
[range(0;$n) | 1] as $p
| if $n < 3 then $p
else reduce range(3;$n) as $i ($p; .[$i] = .[$i-2] + .[$i-3])
end;
# Output: first $n Padovans
def padovanFloor($n):
{ p: 1.324717957244746025960908854,
s: 1.0453567932525329623,
pow: 1 }
| reduce range (1;$n) as $i ( .f = [ ((.pow/.p/.s) + 0.5)|floor];
.f[$i] = (((.pow/.s) + 0.5)|floor)
| .pow *= .p)
| .f ;
# Output: a stream of the L-System Padovan strings
def padovanStrings:
{A: "B", B: "C", C: "AB", "": "A"} as $rules
| $rules[""]
| while(true;
ascii_downcase
| gsub("a"; $rules["A"]) | gsub("b"; $rules["B"]) | gsub("c"; $rules["C"]) ) ;
# Output: a stream of the Padovan numbers using the L-System strings
def padovanNumbers:
padovanStrings | length;
def task:
def s1($n):
if padovanFloor($n) == padovanRecur($n) then "give" else "do not give" end;
def s2($n):
if [limit($n; padovanNumbers)] == padovanRecur($n) then "give" else "do not give" end;
"The first 20 members of the Padovan sequence:", padovanRecur(20),
"",
"The recurrence and floor-based functions \(s1(64)) the same results for 64 terms.",
"",
([limit(10; padovanStrings)]
| "First 10 members of the Padovan L-System:", .,
"and their lengths:",
map(length)),
"",
"The recurrence and L-system based functions \(s2(32)) the same results for 32 terms."
;
task- Output:
The first 20 members of the Padovan sequence: [1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151] The recurrence and floor-based functions give the same results for 64 terms. First 10 members of the Padovan L-System: ["A","B","C","AB","BC","CAB","ABBC","BCCAB","CABABBC","ABBCBCCAB"] and their lengths: [1,1,1,2,2,3,4,5,7,9] The recurrence and L-system based functions give the same results for 32 terms.
Julia
""" Recursive Padovan """
rPadovan(n) = (n < 4) ? one(n) : rPadovan(n - 3) + rPadovan(n - 2)
""" Floor function calculation Padovan """
function fPadovan(n)::Int
p, s = big"1.324717957244746025960908854", big"1.0453567932525329623"
return Int(floor(p^(n-2) / s + .5))
end
""" LSystem Padovan """
function list_LsysPadovan(N)
rules = Dict("A" => "B", "B" => "C", "C" => "AB")
seq, lens = ["A"], [1]
for i in 1:N
str = prod([rules[string(c)] for c in seq[end]])
push!(seq, str)
push!(lens, length(str))
end
return seq, lens
end
const lr, lf = [rPadovan(i) for i in 1:64], [fPadovan(i) for i in 1:64]
const sL, lL = list_LsysPadovan(32)
println("N Recursive Floor LSystem String\n=============================================")
foreach(i -> println(rpad(i, 4), rpad(lr[i], 12), rpad(lf[i], 12),
rpad(i < 33 ? lL[i] : "", 12), (i < 11 ? sL[i] : "")), 1:64)
- Output:
N Recursive Floor LSystem String ============================================= 1 1 1 1 A 2 1 1 1 B 3 1 1 1 C 4 2 2 2 AB 5 2 2 2 BC 6 3 3 3 CAB 7 4 4 4 ABBC 8 5 5 5 BCCAB 9 7 7 7 CABABBC 10 9 9 9 ABBCBCCAB 11 12 12 12 12 16 16 16 13 21 21 21 14 28 28 28 15 37 37 37 16 49 49 49 17 65 65 65 18 86 86 86 19 114 114 114 20 151 151 151 21 200 200 200 22 265 265 265 23 351 351 351 24 465 465 465 25 616 616 616 26 816 816 816 27 1081 1081 1081 28 1432 1432 1432 29 1897 1897 1897 30 2513 2513 2513 31 3329 3329 3329 32 4410 4410 4410 33 5842 5842 34 7739 7739 35 10252 10252 36 13581 13581 37 17991 17991 38 23833 23833 39 31572 31572 40 41824 41824 41 55405 55405 42 73396 73396 43 97229 97229 44 128801 128801 45 170625 170625 46 226030 226030 47 299426 299426 48 396655 396655 49 525456 525456 50 696081 696081 51 922111 922111 52 1221537 1221537 53 1618192 1618192 54 2143648 2143648 55 2839729 2839729 56 3761840 3761840 57 4983377 4983377 58 6601569 6601569 59 8745217 8745217 60 11584946 11584946 61 15346786 15346786 62 20330163 20330163 63 26931732 26931732 64 35676949 35676949
Mathematica / Wolfram Language
ClearAll[Padovan1,a,p,s]
p=N[Surd[((9+Sqrt[69])/18),3]+Surd[((9-Sqrt[69])/18),3],200];
s=1.0453567932525329623;
Padovan1[nmax_Integer]:=RecurrenceTable[{a[n+1]==a[n-1]+a[n-2],a[0]==1,a[1]==1,a[2]==1},a,{n,0,nmax-1}]
Padovan2[nmax_Integer]:=With[{},Floor[p^Range[-1,nmax-2]/s+1/2]]
Padovan1[20]
Padovan2[20]
Padovan1[64]===Padovan2[64]
SubstitutionSystem[{"A"->"B","B"->"C","C"->"AB"},"A",10]//Column
(StringLength/@SubstitutionSystem[{"A"->"B","B"->"C","C"->"AB"},"A",31])==Padovan2[32]
- Output:
{1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151}
{1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151}
True
A
B
C
AB
BC
CAB
ABBC
BCCAB
CABABBC
ABBCBCCAB
BCCABCABABBC
True
Nim
import sequtils, strutils, tables
const
P = 1.324717957244746025960908854
S = 1.0453567932525329623
Rules = {'A': "B", 'B': "C", 'C': "AB"}.toTable
iterator padovan1(n: Natural): int {.closure.} =
## Yield the first "n" Padovan values using recurrence relation.
for _ in 1..min(n, 3): yield 1
var a, b, c = 1
var count = 3
while count < n:
(a, b, c) = (b, c, a + b)
yield c
inc count
iterator padovan2(n: Natural): int {.closure.} =
## Yield the first "n" Padovan values using formula.
if n > 1: yield 1
var p = 1.0
var count = 1
while count < n:
yield (p / S).toInt
p *= P
inc count
iterator padovan3(n: Natural): string {.closure.} =
## Yield the strings produced by the L-system.
var s = "A"
var count = 0
while count < n:
yield s
var next: string
for ch in s:
next.add Rules[ch]
s = move(next)
inc count
echo "First 20 terms of the Padovan sequence:"
echo toSeq(padovan1(20)).join(" ")
let list1 = toSeq(padovan1(64))
let list2 = toSeq(padovan2(64))
echo "The first 64 iterative and calculated values ",
if list1 == list2: "are the same." else: "differ."
echo ""
echo "First 10 L-system strings:"
echo toSeq(padovan3(10)).join(" ")
echo ""
echo "Lengths of the 32 first L-system strings:"
let list3 = toSeq(padovan3(32)).mapIt(it.len)
echo list3.join(" ")
echo "These lengths are",
if list3 == list1[0..31]: " " else: " not ",
"the 32 first terms of the Padovan sequence."
- Output:
First 20 terms of the Padovan sequence: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 The first 64 iterative and calculated values are the same. First 10 L-system strings: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB Lengths of the 32 first L-system strings: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 200 265 351 465 616 816 1081 1432 1897 2513 3329 4410 These lengths are the 32 first terms of the Padovan sequence.
Perl
use strict;
use warnings;
use feature <state say>;
use List::Lazy 'lazy_list';
my $p = 1.32471795724474602596;
my $s = 1.0453567932525329623;
my %rules = (A => 'B', B => 'C', C => 'AB');
my $pad_recur = lazy_list { state @p = (1, 1, 1, 2); push @p, $p[1]+$p[2]; shift @p };
sub pad_floor { int 1/2 + $p**($_<3 ? 1 : $_-2) / $s }
my($l, $m, $n) = (10, 20, 32);
my(@pr, @pf);
push @pr, $pad_recur->next() for 1 .. $n; say join ' ', @pr[0 .. $m-1];
push @pf, pad_floor($_) for 1 .. $n; say join ' ', @pf[0 .. $m-1];
my @L = 'A';
push @L, join '', @rules{split '', $L[-1]} for 1 .. $n;
say join ' ', @L[0 .. $l-1];
$pr[$_] == $pf[$_] and $pr[$_] == length $L[$_] or die "Uh oh, n=$_: $pr[$_] vs $pf[$_] vs " . length $L[$_] for 0 .. $n-1;
say '100% agreement among all 3 methods.';
- Output:
1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB 100% agreement among all 3 methods.
Phix
with javascript_semantics sequence padovan = {1,1,1} function padovanr(integer n) while length(padovan)<n do padovan &= padovan[$-2]+padovan[$-1] end while return padovan[n] end function constant p = 1.324717957244746025960908854, s = 1.0453567932525329623 function padovana(integer n) return floor(power(p,n-2)/s + 0.5) end function constant l = {"B","C","AB"} function padovanl(string prev) string res = "" for i=1 to length(prev) do res &= l[prev[i]-64] end for return res end function sequence pl = "A", l10 = {} for n=1 to 64 do integer pn = padovanr(n) if padovana(n)!=pn or length(pl)!=pn then crash("oops") end if if n<=10 then l10 = append(l10,pl) end if pl = padovanl(pl) end for printf(1,"The first 20 terms of the Padovan sequence: %v\n\n",{padovan[1..20]}) printf(1,"The first 10 L-system strings: %v\n\n",{l10}) printf(1,"recursive, algorithmic, and l-system agree to n=64\n")
- Output:
The first 20 terms of the Padovan sequence: {1,1,1,2,2,3,4,5,7,9,12,16,21,28,37,49,65,86,114,151}
The first 10 L-system strings: {"A","B","C","AB","BC","CAB","ABBC","BCCAB","CABABBC","ABBCBCCAB"}
recursive, algorithmic, and l-system agree to n=64
Python
Python: Idiomatic
from math import floor
from collections import deque
from typing import Dict, Generator
def padovan_r() -> Generator[int, None, None]:
last = deque([1, 1, 1], 4)
while True:
last.append(last[-2] + last[-3])
yield last.popleft()
_p, _s = 1.324717957244746025960908854, 1.0453567932525329623
def padovan_f(n: int) -> int:
return floor(_p**(n-1) / _s + .5)
def padovan_l(start: str='A',
rules: Dict[str, str]=dict(A='B', B='C', C='AB')
) -> Generator[str, None, None]:
axiom = start
while True:
yield axiom
axiom = ''.join(rules[ch] for ch in axiom)
if __name__ == "__main__":
from itertools import islice
print("The first twenty terms of the sequence.")
print(str([padovan_f(n) for n in range(20)])[1:-1])
r_generator = padovan_r()
if all(next(r_generator) == padovan_f(n) for n in range(64)):
print("\nThe recurrence and floor based algorithms match to n=63 .")
else:
print("\nThe recurrence and floor based algorithms DIFFER!")
print("\nThe first 10 L-system string-lengths and strings")
l_generator = padovan_l(start='A', rules=dict(A='B', B='C', C='AB'))
print('\n'.join(f" {len(string):3} {repr(string)}"
for string in islice(l_generator, 10)))
r_generator = padovan_r()
l_generator = padovan_l(start='A', rules=dict(A='B', B='C', C='AB'))
if all(len(next(l_generator)) == padovan_f(n) == next(r_generator)
for n in range(32)):
print("\nThe L-system, recurrence and floor based algorithms match to n=31 .")
else:
print("\nThe L-system, recurrence and floor based algorithms DIFFER!")
- Output:
The first twenty terms of the sequence.
1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151
The recurrence and floor based algorithms match to n=63 .
The first 10 L-system string-lengths and strings
1 'A'
1 'B'
1 'C'
2 'AB'
2 'BC'
3 'CAB'
4 'ABBC'
5 'BCCAB'
7 'CABABBC'
9 'ABBCBCCAB'
The L-system, recurrence and floor based algorithms match to n=31 .
Python: Expressed in terms of a generic anamorphism (unfoldr)
'''Padovan series'''
from itertools import chain, islice
from math import floor
from operator import eq
# padovans :: [Int]
def padovans():
'''Non-finite series of Padovan numbers,
defined in terms of recurrence relations.
'''
def recurrence(abc):
a, b, c = abc
return a, (b, c, a + b)
return unfoldr(recurrence)(
(1, 1, 1)
)
# padovanFloor :: [Int]
def padovanFloor():
'''The Padovan series, defined in terms
of a floor function.
'''
p = 1.324717957244746025960908854
s = 1.0453567932525329623
def f(n):
return floor(p ** (n - 1) / s + 0.5), 1 + n
return unfoldr(f)(0)
# padovanLSystem : [Int]
def padovanLSystem():
'''An L-system generating terms whose lengths
are the values of the Padovan integer series.
'''
def rule(c):
return 'B' if 'A' == c else (
'C' if 'B' == c else 'AB'
)
def f(s):
return s, ''.join(list(concatMap(rule)(s)))
return unfoldr(f)('A')
# ------------------------- TEST -------------------------
# prefixesMatch :: [a] -> [a] -> Bool
def prefixesMatch(xs, ys, n):
'''True if the first n items of each
series are the same.
'''
return all(map(eq, take(n)(xs), ys))
# main :: IO ()
def main():
'''Test three Padovan functions for
equivalence and expected results.
'''
print('\n'.join([
"First 20 padovans:\n",
repr(take(20)(padovans())),
"\nThe recurrence and floor-based functions" + (
" match over 64 terms:\n"
),
repr(prefixesMatch(
padovans(),
padovanFloor(),
64
)),
"\nFirst 10 L-System strings:\n",
repr(take(10)(padovanLSystem())),
"\nThe lengths of the first 32 L-System strings",
"match the Padovan sequence:\n",
repr(prefixesMatch(
padovans(),
(len(x) for x in padovanLSystem()),
32
))
]))
# ----------------------- GENERIC ------------------------
# concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
'''A concatenated map'''
def go(xs):
return chain.from_iterable(map(f, xs))
return go
# unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
def unfoldr(f):
'''A lazy (generator) list unfolded from a seed value
by repeated application of f until no residue remains.
Dual to fold/reduce.
f returns either None, or just (value, residue).
For a strict output list, wrap the result with list()
'''
def go(x):
valueResidue = f(x)
while None is not valueResidue:
yield valueResidue[0]
valueResidue = f(valueResidue[1])
return go
# 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.
'''
def go(xs):
return (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)
return go
# MAIN ---
if __name__ == '__main__':
main()
- Output:
First 20 padovans: [1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151] The recurrence and floor-based functions match over 64 terms: True First 10 L-System strings: ['A', 'B', 'C', 'AB', 'BC', 'CAB', 'ABBC', 'BCCAB', 'CABABBC', 'ABBCBCCAB'] The lengths of the first 32 L-System strings match the Padovan sequence: True
Quackery
v** is defined at Exponentiation operator#Quackery.
( --------------------- Recurrence -------------------- )
[ dup 0 = iff
[ drop ' [ ] ] done
dup 1 = iff
[ drop ' [ 1 ] ] done
dip [ [] 0 1 1 ]
2 - times
[ dip [ 2dup + ] swap
3 pack dip join
unpack ]
3 times join behead drop ] is padovan1 ( n --> [ )
say "With recurrence: " 20 padovan1 echo cr cr
( ------------------- Floor Function ------------------ )
$ "bigrat.qky" loadfile
[ [ $ "1.324717957244746025960908854"
$->v drop join ] constant
do ] is p ( --> n/d )
[ [ $ "1.0453567932525329623"
$->v drop join ] constant
do ] is s ( --> n/d )
[ 1 -
p rot v** s v/ 1 2 v+ / ] is padovan2 ( n --> n )
say "With floor function: "
[]
20 times [ i^ padovan2 join ]
echo cr cr
( ---------------------- L-System --------------------- )
[ $ "" swap witheach
[ nested quackery join ] ] is expand ( $ --> $ )
[ $ "B" ] is A ( $ --> $ )
[ $ "C" ] is B ( $ --> $ )
[ $ "AB" ] is C ( $ --> $ )
$ "A"
say "First 10 L System strings: "
9 times
[ dup echo$ sp
expand ]
echo$ cr cr
[] $ "A"
31 times
[ dup size
swap dip join
expand ]
size join
32 padovan1 = iff
[ say "The first 32 recurrence terms and L System lengths are the same." ]
else [ say "Oh no! It's all gone pear-shaped!" ]- Output:
With recurrence: [ 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 ] With floor function: [ 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 ] First 10 L System strings: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB The first 32 recurrence terms and L System lengths are the same.
Raku
constant p = 1.32471795724474602596;
constant s = 1.0453567932525329623;
constant %rules = A => 'B', B => 'C', C => 'AB';
my @pad-recur = 1, 1, 1, -> $c, $b, $ { $b + $c } … *;
my @pad-floor = { floor 1/2 + p ** ($++ - 1) / s } … *;
my @pad-L-sys = 'A', { %rules{$^axiom.comb}.join } … *;
my @pad-L-len = @pad-L-sys.map: *.chars;
say @pad-recur.head(20);
say @pad-L-sys.head(10);
say "Recurrence == Floor to N=64" if (@pad-recur Z== @pad-floor).head(64).all;
say "Recurrence == L-len to N=32" if (@pad-recur Z== @pad-L-len).head(32).all;
- Output:
(1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151) (A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB) Recurrence == Floor to N=64 Recurrence == L-len to N=32
REXX
/*REXX pgm computes the Padovan seq. (using 2 methods), and also computes the L─strings.*/
numeric digits 40 /*better precision for Plastic ratio. */
parse arg n nF Ln cL . /*obtain optional arguments from the CL*/
if n=='' | n=="," then n= 20 /*Not specified? Then use the default.*/
if nF=='' | nF=="," then nF= 64 /* " " " " " " */
if Ln=='' | Ln=="," then Ln= 10 /* " " " " " " */
if cL=='' | cL=="," then cL= 32 /* " " " " " " */
PR= 1.324717957244746025960908854 /*the plastic ratio (constant). */
s= 1.0453567932525329623 /*tge "s" constant. */
@.= .; @.0= 1; @.1= 1; @.2= 1 /*initialize 3 terms of the Padovan seq*/
!.= .; !.0= 1; !.1= 1; !.2= 1 /* " " " " " " " */
call req1; call req2; call req3; call req4 /*invoke the four task's requirements. */
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
floor: procedure; parse arg x; t= trunc(x); return t - (x<0) * (x\=t)
pF: procedure expose !. PR s; parse arg x; !.x= floor(PR**(x-1)/s + .5); return !.x
th: parse arg th; return th||word('th st nd rd',1+(th//10)*(th//100%10\==1)*(th//10<4))
/*──────────────────────────────────────────────────────────────────────────────────────*/
L_sys: procedure: arg x; q=; a.A= 'B'; a.B= 'C'; a.C= 'AB'; if x=='' then return 'A'
do k=1 for length(x); _= substr(x, k, 1); q= q || a._
end /*k*/; return q
/*──────────────────────────────────────────────────────────────────────────────────────*/
p: procedure expose @.; parse arg x; if @.x\==. then return @.x /*@.X defined?*/
xm2= x - 2; xm3= x - 3; @.x= @.xm2 + @.xm3; return @.x
/*──────────────────────────────────────────────────────────────────────────────────────*/
req1: say 'The first ' n " terms of the Pandovan sequence:";
$= @.0; do j=1 for n-1; $= $ p(j)
end /*j*/
say $; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
req2: ok= 1; what= ' terms match for recurrence and floor─based functions.'
do j=0 for nF; if p(j)==pF(j) then iterate
say 'the ' th(j) " terms don't match:" p(j) pF(j); ok= 0
end /*j*/
say
if ok then say 'all ' nF what; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
req3: y=; $= 'A'
do j=1 for Ln-1; y= L_sys(y); $= $ L_sys(y)
end /*j*/
say
say 'L_sys:' $; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
req4: y=; what=' terms match for Padovan terms and lengths of L_sys terms.'
ok= 1; do j=1 for cL; y= L_sys(y); L= length(y)
if L==p(j-1) then iterate
say 'the ' th(j) " Padovan term doesn't match the length of the",
'L_sys term:' p(j-1) L; ok= 0
end /*j*/
say
if ok then say 'all ' cL what; return
- output when using the default inputs:
The first 20 terms of the Padovan sequence: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 all 64 terms match for recurrence and floor─based functions. L_sys: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB all 32 terms match for Padovan terms and lengths of L_sys terms.
Ruby
padovan = Enumerator.new do |y|
ar = [1, 1, 1]
loop do
ar << ar.first(2).sum
y << ar.shift
end
end
P, S = 1.324717957244746025960908854, 1.0453567932525329623
def padovan_f(n) = (P**(n-1) / S + 0.5).floor
puts "Recurrence Padovan: #{padovan.take(20)}"
puts "Floor function: #{(0...20).map{|n| padovan_f(n)}}"
n = 63
bool = (0...n).map{|n| padovan_f(n)} == padovan.take(n)
puts "Recurrence and floor function are equal upto #{n}: #{bool}."
puts
def l_system(axiom = "A", rules = {"A" => "B", "B" => "C", "C" => "AB"} )
return enum_for(__method__, axiom, rules) unless block_given?
loop do
yield axiom
axiom = axiom.chars.map{|c| rules[c] }.join
end
end
puts "First 10 elements of L-system: #{l_system.take(10).join(", ")} "
n = 32
bool = l_system.take(n).map(&:size) == padovan.take(n)
puts "Sizes of first #{n} l_system strings equal to recurrence padovan? #{bool}."
- Output:
Recurrence Padovan: [1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151] Floor function: [1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151] Recurrence and floor function are equal upto 63: true. First 10 elements of L-system: A, B, C, AB, BC, CAB, ABBC, BCCAB, CABABBC, ABBCBCCAB Sizes of first 32 l_system strings equal to recurrence padovan? true.
Rust
fn padovan_recur() -> impl std::iter::Iterator<Item = usize> {
let mut p = vec![1, 1, 1];
let mut n = 0;
std::iter::from_fn(move || {
let pn = if n < 3 { p[n] } else { p[0] + p[1] };
p[0] = p[1];
p[1] = p[2];
p[2] = pn;
n += 1;
Some(pn)
})
}
fn padovan_floor() -> impl std::iter::Iterator<Item = usize> {
const P: f64 = 1.324717957244746025960908854;
const S: f64 = 1.0453567932525329623;
(0..).map(|x| (P.powf((x - 1) as f64) / S + 0.5).floor() as usize)
}
fn padovan_lsystem() -> impl std::iter::Iterator<Item = String> {
let mut str = String::from("A");
std::iter::from_fn(move || {
let result = str.clone();
let mut next = String::new();
for ch in str.chars() {
match ch {
'A' => next.push('B'),
'B' => next.push('C'),
_ => next.push_str("AB"),
}
}
str = next;
Some(result)
})
}
fn main() {
println!("First 20 terms of the Padovan sequence:");
for p in padovan_recur().take(20) {
print!("{} ", p);
}
println!();
println!(
"\nRecurrence and floor functions agree for first 64 terms? {}",
padovan_recur().take(64).eq(padovan_floor().take(64))
);
println!("\nFirst 10 strings produced from the L-system:");
for p in padovan_lsystem().take(10) {
print!("{} ", p);
}
println!();
println!(
"\nLength of first 32 strings produced from the L-system = Padovan sequence? {}",
padovan_lsystem()
.map(|x| x.len())
.take(32)
.eq(padovan_recur().take(32))
);
}
- Output:
First 20 terms of the Padovan sequence: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 Recurrence and floor functions agree for first 64 terms? true First 10 strings produced from the L-system: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB Length of first 32 strings produced from the L-system = Padovan sequence? true
Swift
import Foundation
class PadovanRecurrence: Sequence, IteratorProtocol {
private var p = [1, 1, 1]
private var n = 0
func next() -> Int? {
let pn = n < 3 ? p[n] : p[0] + p[1]
p[0] = p[1]
p[1] = p[2]
p[2] = pn
n += 1
return pn
}
}
class PadovanFloor: Sequence, IteratorProtocol {
private let P = 1.324717957244746025960908854
private let S = 1.0453567932525329623
private var n = 0
func next() -> Int? {
let p = Int(floor(pow(P, Double(n - 1)) / S + 0.5))
n += 1
return p
}
}
class PadovanLSystem: Sequence, IteratorProtocol {
private var str = "A"
func next() -> String? {
let result = str
var next = ""
for ch in str {
switch (ch) {
case "A": next.append("B")
case "B": next.append("C")
default: next.append("AB")
}
}
str = next
return result
}
}
print("First 20 terms of the Padovan sequence:")
for p in PadovanRecurrence().prefix(20) {
print("\(p)", terminator: " ")
}
print()
var b = PadovanRecurrence().prefix(64)
.elementsEqual(PadovanFloor().prefix(64))
print("\nRecurrence and floor functions agree for first 64 terms? \(b)")
print("\nFirst 10 strings produced from the L-system:");
for p in PadovanLSystem().prefix(10) {
print(p, terminator: " ")
}
print()
b = PadovanLSystem().prefix(32).map{$0.count}
.elementsEqual(PadovanRecurrence().prefix(32))
print("\nLength of first 32 strings produced from the L-system = Padovan sequence? \(b)")
- Output:
First 20 terms of the Padovan sequence: 1 1 1 2 2 3 4 5 7 9 12 16 21 28 37 49 65 86 114 151 Recurrence and floor functions agree for first 64 terms? true First 10 strings produced from the L-system: A B C AB BC CAB ABBC BCCAB CABABBC ABBCBCCAB Length of first 32 strings produced from the L-system = Padovan sequence? true
Wren
L-System stuff is based on the Julia implementation.
import "/big" for BigRat
import "/dynamic" for Struct
var padovanRecur = Fn.new { |n|
var p = List.filled(n, 1)
if (n < 3) return p
for (i in 3...n) p[i] = p[i-2] + p[i-3]
return p
}
var padovanFloor = Fn.new { |n|
var p = BigRat.fromDecimal("1.324717957244746025960908854")
var s = BigRat.fromDecimal("1.0453567932525329623")
var f = List.filled(n, 0)
var pow = BigRat.one
f[0] = (pow/p/s + 0.5).floor.toInt
for (i in 1...n) {
f[i] = (pow/s + 0.5).floor.toInt
pow = pow * p
}
return f
}
var LSystem = Struct.create("LSystem", ["rules", "init", "current"])
var step = Fn.new { |lsys|
var s = ""
if (lsys.current == "") {
lsys.current = lsys.init
} else {
for (c in lsys.current) s = s + lsys.rules[c]
lsys.current = s
}
return lsys.current
}
var padovanLSys = Fn.new { |n|
var rules = {"A": "B", "B": "C", "C": "AB"}
var lsys = LSystem.new(rules, "A", "")
var p = List.filled(n, null)
for (i in 0...n) p[i] = step.call(lsys)
return p
}
System.print("First 20 members of the Padovan sequence:")
System.print(padovanRecur.call(20))
var recur = padovanRecur.call(64)
var floor = padovanFloor.call(64)
var areSame = (0...64).all { |i| recur[i] == floor[i] }
var s = areSame ? "give" : "do not give"
System.print("\nThe recurrence and floor based functions %(s) the same results for 64 terms.")
var p = padovanLSys.call(32)
var lsyst = p.map { |e| e.count }.toList
System.print("\nFirst 10 members of the Padovan L-System:")
System.print(p.take(10).toList)
System.print("\nand their lengths:")
System.print(lsyst.take(10).toList)
recur = recur.take(32).toList
areSame = (0...32).all { |i| recur[i] == lsyst[i] }
s = areSame ? "give" : "do not give"
System.print("\nThe recurrence and L-system based functions %(s) the same results for 32 terms.")
- Output:
First 20 members of the Padovan sequence: [1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151] The recurrence and floor based functions give the same results for 64 terms. First 10 members of the Padovan L-System: [A, B, C, AB, BC, CAB, ABBC, BCCAB, CABABBC, ABBCBCCAB] and their lengths: [1, 1, 1, 2, 2, 3, 4, 5, 7, 9] The recurrence and L-system based functions give the same results for 32 terms.