Hofstadter Figure-Figure sequences

From Rosetta Code
Task
Hofstadter Figure-Figure sequences
You are encouraged to solve this task according to the task description, using any language you may know.

These two sequences of positive integers are defined as:


The sequence is further defined as the sequence of positive integers not present in .

Sequence starts:

   1, 3, 7, 12, 18, ...

Sequence starts:

   2, 4, 5, 6, 8, ...


Task
  1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively.
    (Note that R(1) = 1 and S(1) = 2 to avoid off-by-one errors).
  2. No maximum value for n should be assumed.
  3. Calculate and show that the first ten values of R are:
    1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
  4. Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once.


References



Ada[edit]

Specifying a package providing the functions FFR and FFS:

package Hofstadter_Figure_Figure is
 
function FFR(P: Positive) return Positive;
 
function FFS(P: Positive) return Positive;
 
end Hofstadter_Figure_Figure;

The implementation of the package internally uses functions which generate an array of Figures or Spaces:

package body Hofstadter_Figure_Figure is
 
type Positive_Array is array (Positive range <>) of Positive;
 
function FFR(P: Positive) return Positive_Array is
Figures: Positive_Array(1 .. P+1);
Space: Positive := 2;
Space_Index: Positive := 2;
begin
Figures(1) := 1;
for I in 2 .. P loop
Figures(I) := Figures(I-1) + Space;
Space := Space+1;
while Space = Figures(Space_Index) loop
Space := Space + 1;
Space_Index := Space_Index + 1;
end loop;
end loop;
return Figures(1 .. P);
end FFR;
 
function FFR(P: Positive) return Positive is
Figures: Positive_Array(1 .. P) := FFR(P);
begin
return Figures(P);
end FFR;
 
function FFS(P: Positive) return Positive_Array is
Spaces: Positive_Array(1 .. P);
Figures: Positive_Array := FFR(P+1);
J: Positive := 1;
K: Positive := 1;
begin
for I in Spaces'Range loop
while J = Figures(K) loop
J := J + 1;
K := K + 1;
end loop;
Spaces(I) := J;
J := J + 1;
end loop;
return Spaces;
end FFS;
 
function FFS(P: Positive) return Positive is
Spaces: Positive_Array := FFS(P);
begin
return Spaces(P);
end FFS;
 
end Hofstadter_Figure_Figure;

Finally, a test program for the package, solving the task at hand:

with Ada.Text_IO, Hofstadter_Figure_Figure;
 
procedure Test_HSS is
 
use Hofstadter_Figure_Figure;
 
A: array(1 .. 1000) of Boolean := (others => False);
J: Positive;
 
begin
for I in 1 .. 10 loop
Ada.Text_IO.Put(Integer'Image(FFR(I)));
end loop;
Ada.Text_IO.New_Line;
 
for I in 1 .. 40 loop
J := FFR(I);
if A(J) then
raise Program_Error with Positive'Image(J) & " used twice";
end if;
A(J) := True;
end loop;
 
for I in 1 .. 960 loop
J := FFS(I);
if A(J) then
raise Program_Error with Positive'Image(J) & " used twice";
end if;
A(J) := True;
end loop;
 
for I in A'Range loop
if not A(I) then raise Program_Error with Positive'Image(I) & " unused";
end if;
end loop;
Ada.Text_IO.Put_Line("Test Passed: No overlap between FFR(I) and FFS(J)");
 
exception
when Program_Error => Ada.Text_IO.Put_Line("Test Failed"); raise;
end Test_HSS;

The output of the test program:

 1 3 7 12 18 26 35 45 56 69
Test Passed: No overlap between FFR(I) and FFS(J)

AutoHotkey[edit]

R(n){
if n=1
return 1
return R(n-1) + S(n-1)
}
 
S(n){
static ObjR:=[]
if n=1
return 2
ObjS:=[]
loop, % n
ObjR[R(A_Index)] := true
loop, % n-1
ObjS[S(A_Index)] := true
Loop
if !(ObjR[A_Index]||ObjS[A_Index])
return A_index
}
Examples:
Loop
MsgBox, 262144, , % "R(" A_Index ") = " R(A_Index) "`nS(" A_Index ") = " S(A_Index)
Outputs:
R(1) = 1, 3, 7, 12, 18, 26, 35,...
S(1) = 2, 4, 5,  6,  8,  9, 10,...

BBC BASIC[edit]

      PRINT "First 10 values of R:"
FOR i% = 1 TO 10 : PRINT ;FNffr(i%) " "; : NEXT : PRINT
PRINT "First 10 values of S:"
FOR i% = 1 TO 10 : PRINT ;FNffs(i%) " "; : NEXT : PRINT
PRINT "Checking for first 1000 integers:"
r% = 1 : s% = 1
ffr% = FNffr(r%)
ffs% = FNffs(s%)
FOR wanted% = 1 TO 1000
CASE TRUE OF
WHEN wanted% = ffr% : r% += 1 : ffr% = FNffr(r%)
WHEN wanted% = ffs% : s% += 1 : ffs% = FNffs(s%)
OTHERWISE: EXIT FOR
ENDCASE
NEXT
IF r% = 41 AND s% = 961 PRINT "Test passed" ELSE PRINT "Test failed"
END
 
DEF FNffr(N%)
LOCAL I%, J%, R%, S%, V%
DIM V% LOCAL 2*N%+1
V%?1 = 1
IF N% = 1 THEN = 1
R% = 1
S% = 2
FOR I% = 2 TO N%
FOR J% = S% TO 2*N%
IF V%?J% = 0 EXIT FOR
NEXT
V%?J% = 1
S% = J%
R% += S%
IF R% <= 2*N% V%?R% = 1
NEXT I%
= R%
 
DEF FNffs(N%)
LOCAL I%, J%, R%, S%, V%
DIM V% LOCAL 2*N%+1
V%?1 = 1
IF N% = 1 THEN = 2
R% = 1
S% = 2
FOR I% = 1 TO N%
FOR J% = S% TO 2*N%
IF V%?J% = 0 EXIT FOR
NEXT
V%?J% = 1
S% = J%
R% += S%
IF R% <= 2*N% V%?R% = 1
NEXT I%
= S%
First 10 values of R:
1 3 7 12 18 26 35 45 56 69
First 10 values of S:
2 4 5 6 8 9 10 11 13 14
Checking for first 1000 integers:
Test passed

C[edit]

#include <stdio.h>
#include <stdlib.h>
 
// simple extensible array stuff
typedef unsigned long long xint;
 
typedef struct {
size_t len, alloc;
xint *buf;
} xarray;
 
xarray rs, ss;
 
void setsize(xarray *a, size_t size)
{
size_t n = a->alloc;
if (!n) n = 1;
 
while (n < size) n <<= 1;
if (a->alloc < n) {
a->buf = realloc(a->buf, sizeof(xint) * n);
if (!a->buf) abort();
a->alloc = n;
}
}
 
void push(xarray *a, xint v)
{
while (a->alloc <= a->len)
setsize(a, a->alloc * 2);
 
a->buf[a->len++] = v;
}
 
 
// sequence stuff
void RS_append(void);
 
xint R(int n)
{
while (n > rs.len) RS_append();
return rs.buf[n - 1];
}
 
xint S(int n)
{
while (n > ss.len) RS_append();
return ss.buf[n - 1];
}
 
void RS_append()
{
int n = rs.len;
xint r = R(n) + S(n);
xint s = S(ss.len);
 
push(&rs, r);
while (++s < r) push(&ss, s);
push(&ss, r + 1); // pesky 3
}
 
int main(void)
{
push(&rs, 1);
push(&ss, 2);
 
int i;
printf("R(1 .. 10):");
for (i = 1; i <= 10; i++)
printf(" %llu", R(i));
 
char seen[1001] = { 0 };
for (i = 1; i <= 40; i++) seen[ R(i) ] = 1;
for (i = 1; i <= 960; i++) seen[ S(i) ] = 1;
for (i = 1; i <= 1000 && seen[i]; i++);
 
if (i <= 1000) {
fprintf(stderr, "%d not seen\n", i);
abort();
}
 
puts("\nfirst 1000 ok");
return 0;
}

C++[edit]

Works with: gcc
Works with: C++ version 11, 14, 17
#include <iomanip>
#include <iostream>
#include <set>
#include <vector>
 
using namespace std;
 
unsigned hofstadter(unsigned rlistSize, unsigned slistSize)
{
auto n = rlistSize > slistSize ? rlistSize : slistSize;
auto rlist = new vector<unsigned> { 1, 3, 7 };
auto slist = new vector<unsigned> { 2, 4, 5, 6 };
auto list = rlistSize > 0 ? rlist : slist;
auto target_size = rlistSize > 0 ? rlistSize : slistSize;
 
while (list->size() > target_size) list->pop_back();
 
while (list->size() < target_size)
{
auto lastIndex = rlist->size() - 1;
auto lastr = (*rlist)[lastIndex];
auto r = lastr + (*slist)[lastIndex];
rlist->push_back(r);
for (auto s = lastr + 1; s < r && list->size() < target_size;)
slist->push_back(s++);
}
 
auto v = (*list)[n - 1];
delete rlist;
delete slist;
return v;
}
 
ostream& operator<<(ostream& os, const set<unsigned>& s)
{
cout << '(' << s.size() << "):";
auto i = 0;
for (auto c = s.begin(); c != s.end();)
{
if (i++ % 20 == 0) os << endl;
os << setw(5) << *c++;
}
return os;
}
 
int main(int argc, const char* argv[])
{
const auto v1 = atoi(argv[1]);
const auto v2 = atoi(argv[2]);
set<unsigned> r, s;
for (auto n = 1; n <= v2; n++)
{
if (n <= v1)
r.insert(hofstadter(n, 0));
s.insert(hofstadter(0, n));
}
cout << "R" << r << endl;
cout << "S" << s << endl;
 
int m = max(*r.rbegin(), *s.rbegin());
for (auto n = 1; n <= m; n++)
if (r.count(n) == s.count(n))
clog << "integer " << n << " either in both or neither set" << endl;
 
return 0;
}
Output:
% ./hofstadter 40 100 2> /dev/null
R(40):
1 3 7 12 18 26 35 45 56 69 83 98 114 131 150 170 191 213 236 260
285 312 340 369 399 430 462 495 529 565 602 640 679 719 760 802 845 889 935 982
S(100):
2 4 5 6 8 9 10 11 13 14 15 16 17 19 20 21 22 23 24 25
27 28 29 30 31 32 33 34 36 37 38 39 40 41 42 43 44 46 47 48
49 50 51 52 53 54 55 57 58 59 60 61 62 63 64 65 66 67 68 70
71 72 73 74 75 76 77 78 79 80 81 82 84 85 86 87 88 89 90 91
92 93 94 95 96 97 99 100 101 102 103 104 105 106 107 108 109 110 111 112

C#[edit]

Creates an IEnumerable for R and S and uses those to complete the task

using System;
using System.Collections.Generic;
using System.Linq;
 
namespace HofstadterFigureFigure
{
class HofstadterFigureFigure
{
readonly List<int> _r = new List<int>() {1};
readonly List<int> _s = new List<int>();
 
public IEnumerable<int> R()
{
int iR = 0;
while (true)
{
if (iR >= _r.Count)
{
Advance();
}
yield return _r[iR++];
}
}
 
public IEnumerable<int> S()
{
int iS = 0;
while (true)
{
if (iS >= _s.Count)
{
Advance();
}
yield return _s[iS++];
}
}
 
private void Advance()
{
int rCount = _r.Count;
int oldR = _r[rCount - 1];
int sVal;
 
// Take care of first two cases specially since S won't be larger than R at that point
switch (rCount)
{
case 1:
sVal = 2;
break;
case 2:
sVal = 4;
break;
default:
sVal = _s[rCount - 1];
break;
}
_r.Add(_r[rCount - 1] + sVal);
int newR = _r[rCount];
for (int iS = oldR + 1; iS < newR; iS++)
{
_s.Add(iS);
}
}
}
 
class Program
{
static void Main()
{
var hff = new HofstadterFigureFigure();
var rs = hff.R();
var arr = rs.Take(40).ToList();
 
foreach(var v in arr.Take(10))
{
Console.WriteLine("{0}", v);
}
 
var hs = new HashSet<int>(arr);
hs.UnionWith(hff.S().Take(960));
Console.WriteLine(hs.Count == 1000 ? "Verified" : "Oops! Something's wrong!");
}
}
}
 

Output:

1
3
7
12
18
26
35
45
56
69
Verified

CoffeeScript[edit]

Translation of: Ruby
R = [ null, 1 ]
S = [ null, 2 ]
 
extend_sequences = (n) ->
current = Math.max(R[R.length - 1], S[S.length - 1])
i = undefined
while R.length <= n or S.length <= n
i = Math.min(R.length, S.length) - 1
current += 1
if current == R[i] + S[i]
R.push current
else
S.push current
 
ff = (X, n) ->
extend_sequences n
X[n]
 
console.log 'R(' + i + ') = ' + ff(R, i) for i in [1..10]
int_array = ([1..40].map (i) -> ff(R, i)).concat [1..960].map (i) -> ff(S, i)
int_array.sort (a, b) -> a - b
 
for i in [1..1000]
if int_array[i - 1] != i
throw 'Something\'s wrong!'
console.log '1000 integer check ok.'
Output:

As JavaScript.

Common Lisp[edit]

;;; equally doable with a list
(flet ((seq (i) (make-array 1 :element-type 'integer
:initial-element i
:fill-pointer 1
:adjustable t)))
(let ((rr (seq 1)) (ss (seq 2)))
(labels ((extend-r ()
(let* ((l (1- (length rr)))
(r (+ (aref rr l) (aref ss l)))
(s (elt ss (1- (length ss)))))
(vector-push-extend r rr)
(loop while (<= s r) do
(if (/= (incf s) r)
(vector-push-extend s ss))))))
(defun seq-r (n)
(loop while (> n (length rr)) do (extend-r))
(elt rr (1- n)))
 
(defun seq-s (n)
(loop while (> n (length ss)) do (extend-r))
(elt ss (1- n))))))
 
(defun take (f n)
(loop for x from 1 to n collect (funcall f x)))
 
(format t "First of R: ~a~%" (take #'seq-r 10))
 
(mapl (lambda (l) (if (and (cdr l)
(/= (1+ (car l)) (cadr l)))
(error "not in sequence")))
(sort (append (take #'seq-r 40)
(take #'seq-s 960))
#'<))
(princ "Ok")
Output:
First of R: (1 3 7 12 18 26 35 45 56 69)
Ok

D[edit]

Translation of: Go
int delegate(in int) nothrow ffr, ffs;
 
nothrow static this() {
auto r = [0, 1], s = [0, 2];
 
ffr = (in int n) nothrow {
while (r.length <= n) {
immutable int nrk = r.length - 1;
immutable int rNext = r[nrk] + s[nrk];
r ~= rNext;
foreach (immutable sn; r[nrk] + 2 .. rNext)
s ~= sn;
s ~= rNext + 1;
}
return r[n];
};
 
ffs = (in int n) nothrow {
while (s.length <= n)
ffr(r.length);
return s[n];
};
}
 
void main() {
import std.stdio, std.array, std.range, std.algorithm;
 
iota(1, 11).map!ffr.writeln;
auto t = iota(1, 41).map!ffr.chain(iota(1, 961).map!ffs);
t.array.sort().equal(iota(1, 1001)).writeln;
}
Output:
[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
true

Alternative version[edit]

Translation of: Python

(Same output)

import std.stdio, std.array, std.range, std.algorithm;
 
struct ffr {
static r = [int.min, 1];
 
static int opCall(in int n) nothrow {
assert(n > 0);
if (n < r.length) {
return r[n];
} else {
immutable int ffr_n_1 = ffr(n - 1);
immutable int lastr = r[$ - 1];
// Extend s up to, and one past, last r.
ffs.s ~= iota(ffs.s[$ - 1] + 1, lastr).array;
if (ffs.s[$ - 1] < lastr)
ffs.s ~= lastr + 1;
// Access s[n - 1] temporarily extending s if necessary.
immutable size_t len_s = ffs.s.length;
immutable int ffs_n_1 = (len_s > n) ?
ffs.s[n - 1] :
(n - len_s) + ffs.s[$ - 1];
immutable int ans = ffr_n_1 + ffs_n_1;
r ~= ans;
return ans;
}
}
}
 
struct ffs {
static s = [int.min, 2];
 
static int opCall(in int n) nothrow {
assert(n > 0);
if (n < s.length) {
return s[n];
} else {
foreach (immutable i; ffr.r.length .. n + 2) {
ffr(i);
if (s.length > n)
return s[n];
}
assert(false, "Whoops!");
}
}
}
 
void main() {
iota(1, 11).map!ffr.writeln;
auto t = iota(1, 41).map!ffr.chain(iota(1, 961).map!ffs);
t.array.sort().equal(iota(1, 1001)).writeln;
}

EchoLisp[edit]

(define (FFR n)
(+ (FFR (1- n)) (FFS (1- n))))
 
(define (FFS n)
(define next (1+ (FFS (1- n))))
(for ((k (in-naturals next)))
#:break (not (vector-search* k (cache 'FFR))) => k
))
 
(remember 'FFR #(0 1)) ;; init cache
(remember 'FFS #(0 2))
 
Output:
 
(define-macro m-range [a .. b] (range a (1+ b)))
 
(map FFR [1 .. 10])
(1 3 7 12 18 26 35 45 56 69)
 
;; checking
(equal? [1 .. 1000] (list-sort < (append (map FFR [1 .. 40]) (map FFS [1 .. 960]))))
→ #t

Euler Math Toolbox[edit]

 
>function RSstep (r,s) ...
$ n=cols(r);
$ r=r|(r[n]+s[n]);
$ s=s|(max(s[n]+1,r[n]+1):r[n+1]-1);
$ return {r,s};
$ endfunction
>function RS (n) ...
$ if n==1 then return {[1],[2]}; endif;
$ if n==2 then return {[1,3],[2]}; endif;
$ r=[1,3]; s=[2,4];
$ loop 3 to n; {r,s}=RSstep(r,s); end;
$ return {r,s};
$ endfunction
>{r,s}=RS(10);
>r
[ 1 3 7 12 18 26 35 45 56 69 ]
>{r,s}=RS(50);
>all(sort(r[1:40]|s[1:960])==(1:1000))
1
 

Factor[edit]

We keep lists S and R, and increment them when necessary.

SYMBOL: S  V{ 2 } S set
SYMBOL: R V{ 1 } R set
 
: next ( s r -- news newr )
2dup [ last ] bi@ + suffix
dup [
[ dup last 1 + dup ] dip member? [ 1 + ] when suffix
] dip ;
 
: inc-SR ( n -- )
dup 0 <=
[ drop ]
[ [ S get R get ] dip [ next ] times R set S set ]
if ;
 
: ffs ( n -- S(n) )
dup S get length - inc-SR
1 - S get nth ;
: ffr ( n -- R(n) )
dup R get length - inc-SR
1 - R get nth ;
( scratchpad ) 10 iota [ 1 + ffr ] map .
{ 1 3 7 12 18 26 35 45 56 69 }
( scratchpad ) 40 iota [ 1 + ffr ] map 960 iota [ 1 + ffs ] map append 1000 iota 1 v+n set= .
t

Go[edit]

package main
 
import "fmt"
 
var ffr, ffs func(int) int
 
// The point of the init function is to encapsulate r and s. If you are
// not concerned about that or do not want that, r and s can be variables at
// package level and ffr and ffs can be ordinary functions at package level.
func init() {
// task 1, 2
r := []int{0, 1}
s := []int{0, 2}
 
ffr = func(n int) int {
for len(r) <= n {
nrk := len(r) - 1 // last n for which r(n) is known
rNxt := r[nrk] + s[nrk] // next value of r: r(nrk+1)
r = append(r, rNxt) // extend sequence r by one element
for sn := r[nrk] + 2; sn < rNxt; sn++ {
s = append(s, sn) // extend sequence s up to rNext
}
s = append(s, rNxt+1) // extend sequence s one past rNext
}
return r[n]
}
 
ffs = func(n int) int {
for len(s) <= n {
ffr(len(r))
}
return s[n]
}
}
 
func main() {
// task 3
for n := 1; n <= 10; n++ {
fmt.Printf("r(%d): %d\n", n, ffr(n))
}
// task 4
var found [1001]int
for n := 1; n <= 40; n++ {
found[ffr(n)]++
}
for n := 1; n <= 960; n++ {
found[ffs(n)]++
}
for i := 1; i <= 1000; i++ {
if found[i] != 1 {
fmt.Println("task 4: FAIL")
return
}
}
fmt.Println("task 4: PASS")
}
Output:
r(1): 1
r(2): 3
r(3): 7
r(4): 12
r(5): 18
r(6): 26
r(7): 35
r(8): 45
r(9): 56
r(10): 69
task 4: PASS

The following defines two mutually recursive generators without caching results. Each generator will end up dragging a tree of closures behind it, but due to the odd nature of the two series' growth pattern, it's still a heck of a lot faster than the above method when producing either series in sequence.

package main
import "fmt"
 
type xint int64
func R() (func() (xint)) {
r, s := xint(0), func() (xint) (nil)
return func() (xint) {
switch {
case r < 1: r = 1
case r < 3: r = 3
default:
if s == nil {
s = S()
s()
}
r += s()
}
if r < 0 { panic("r overflow") }
return r
}
}
 
func S() (func() (xint)) {
s, r1, r := xint(0), xint(0), func() (xint) (nil)
return func() (xint) {
if s < 2 {
s = 2
} else {
if r == nil {
r = R()
r()
r1 = r()
}
s++
if s > r1 { r1 = r() }
if s == r1 { s++ }
}
if s < 0 { panic("s overflow") }
return s
}
}
 
func main() {
r, sum := R(), xint(0)
for i := 0; i < 10000000; i++ {
sum += r()
}
fmt.Println(sum)
}

Haskell[edit]

import Data.List (delete, sort)
 
-- Functions by Reinhard Zumkeller
ffr n = rl !! (n - 1) where
rl = 1 : fig 1 [2 ..]
fig n (x : xs) = n' : fig n' (delete n' xs) where n' = n + x
 
ffs n = rl !! n where
rl = 2 : figDiff 1 [2 ..]
figDiff n (x : xs) = x : figDiff n' (delete n' xs) where n' = n + x
 
main = do
print $ map ffr [1 .. 10]
let i1000 = sort (map ffr [1 .. 40] ++ map ffs [1 .. 960])
print (i1000 == [1 .. 1000])

Output:

[1,3,7,12,18,26,35,45,56,69]
True

Defining R and S literally:

import Data.List (sort)
 
r = scanl (+) 1 s
s = 2:4:tail (compliment (tail r)) where
compliment = concat.interval
interval x = zipWith (\x y -> [x+1..y-1]) x (tail x)
 
main = do
putStr "R: "; print (take 10 r)
putStr "S: "; print (take 10 s)
putStr "test 1000: ";
print ([1..1000] == sort ((take 40 r) ++ (take 960 s)))

output:

R: [1,3,7,12,18,26,35,45,56,69]
S: [2,4,5,6,8,9,10,11,13,14]
test 1000: True

Icon and Unicon[edit]

link printf,ximage 
 
procedure main()
printf("Hofstader ff sequences R(n:= 1 to %d)\n",N := 10)
every printf("R(%d)=%d\n",n := 1 to N,ffr(n))
 
L := list(N := 1000,0)
zero := dup := oob := 0
every n := 1 to (RN := 40) do
if not L[ffr(n)] +:= 1 then # count R occurrence
oob +:= 1 # count out of bounds
 
every n := 1 to (N-RN) do
if not L[ffs(n)] +:= 1 then # count S occurrence
oob +:= 1 # count out of bounds
 
every zero +:= (!L = 0) # count zeros / misses
every dup +:= (!L > 1) # count > 1's / duplicates
 
printf("Results of R(1 to %d) and S(1 to %d) coverage is ",RN,(N-RN))
if oob+zero+dup=0 then
printf("complete.\n")
else
printf("flawed\noob=%i,zero=%i,dup=%i\nL:\n%s\nR:\n%s\nS:\n%s\n",
oob,zero,dup,ximage(L),ximage(ffr(ffr)),ximage(ffs(ffs)))
end
 
procedure ffr(n)
static R,S
initial {
R := [1]
S := ffs(ffs) # get access to S in ffs
}
 
if n === ffr then return R # secret handshake to avoid globals :)
 
if integer(n) > 0 then
return R[n] | put(R,ffr(n-1) + ffs(n-1))[n]
end
 
procedure ffs(n)
static R,S
initial {
S := [2]
R := ffr(ffr) # get access to R in ffr
}
 
if n === ffs then return S # secret handshake to avoid globals :)
 
if integer(n) > 0 then {
if S[n] then return S[n]
else {
t := S[*S]
until *S = n do
if (t +:= 1) = !R then next # could be optimized with more code
else return put(S,t)[*S] # extend S
}
}
end

printf.icn provides formatting ximage.icn allows formatting entire structures

Output:
Hofstader ff sequences R(n:= 1 to 10)
R(1)=1
R(2)=3
R(3)=7
R(4)=12
R(5)=18
R(6)=26
R(7)=35
R(8)=45
R(9)=56
R(10)=69
Results of R(1 to 40) and S(1 to 960) coverage is complete.

J[edit]

R=: 1 1 3
S=: 0 2 4
FF=: 3 :0
while. +./y>:R,&#S do.
R=: R,({:R)+(<:#R){S
S=: (i.<:+/_2{.R)-.R
end.
R;S
)
ffr=: { 0 {:: FF@(>./@,)
ffs=: { 1 {:: FF@(0,>./@,)

Required examples:

   ffr 1+i.10
1 3 7 12 18 26 35 45 56 69
(1+i.1000) -: /:~ (ffr 1+i.40), ffs 1+i.960
1

Java[edit]

Code:

import java.util.*;
 
class Hofstadter
{
private static List<Integer> getSequence(int rlistSize, int slistSize)
{
List<Integer> rlist = new ArrayList<Integer>();
List<Integer> slist = new ArrayList<Integer>();
Collections.addAll(rlist, 1, 3, 7);
Collections.addAll(slist, 2, 4, 5, 6);
List<Integer> list = (rlistSize > 0) ? rlist : slist;
int targetSize = (rlistSize > 0) ? rlistSize : slistSize;
while (list.size() > targetSize)
list.remove(list.size() - 1);
while (list.size() < targetSize)
{
int lastIndex = rlist.size() - 1;
int lastr = rlist.get(lastIndex).intValue();
int r = lastr + slist.get(lastIndex).intValue();
rlist.add(Integer.valueOf(r));
for (int s = lastr + 1; (s < r) && (list.size() < targetSize); s++)
slist.add(Integer.valueOf(s));
}
return list;
}
 
public static int ffr(int n)
{ return getSequence(n, 0).get(n - 1).intValue(); }
 
public static int ffs(int n)
{ return getSequence(0, n).get(n - 1).intValue(); }
 
public static void main(String[] args)
{
System.out.print("R():");
for (int n = 1; n <= 10; n++)
System.out.print(" " + ffr(n));
System.out.println();
 
Set<Integer> first40R = new HashSet<Integer>();
for (int n = 1; n <= 40; n++)
first40R.add(Integer.valueOf(ffr(n)));
 
Set<Integer> first960S = new HashSet<Integer>();
for (int n = 1; n <= 960; n++)
first960S.add(Integer.valueOf(ffs(n)));
 
for (int i = 1; i <= 1000; i++)
{
Integer n = Integer.valueOf(i);
if (first40R.contains(n) == first960S.contains(n))
System.out.println("Integer " + i + " either in both or neither set");
}
System.out.println("Done");
}
}

Output:

R(): 1 3 7 12 18 26 35 45 56 69
Done

JavaScript[edit]

Translation of: Ruby
var R = [null, 1];
var S = [null, 2];
 
var extend_sequences = function (n) {
var current = Math.max(R[R.length-1],S[S.length-1]);
var i;
while (R.length <= n || S.length <= n) {
i = Math.min(R.length, S.length) - 1;
current += 1;
if (current === R[i] + S[i]) {
R.push(current);
} else {
S.push(current);
}
}
}
 
var ffr = function(n) {
extend_sequences(n);
return R[n];
};
 
var ffs = function(n) {
extend_sequences(n);
return S[n];
};
 
for (var i = 1; i <=10; i += 1) {
console.log('R('+ i +') = ' + ffr(i));
}
 
var int_array = [];
 
for (var i = 1; i <= 40; i += 1) {
int_array.push(ffr(i));
}
for (var i = 1; i <= 960; i += 1) {
int_array.push(ffs(i));
}
 
int_array.sort(function(a,b){return a-b;});
 
for (var i = 1; i <= 1000; i += 1) {
if (int_array[i-1] !== i) {
throw "Something's wrong!"
} else { console.log("1000 integer check ok."); }
}

Output:

R(1) = 1
R(2) = 3
R(3) = 7
R(4) = 12
R(5) = 18
R(6) = 26
R(7) = 35
R(8) = 45
R(9) = 56
R(10) = 69
1000 integer check ok.

Julia[edit]

Much of this task would seem to lend itself to an iterator based solution. However, the first step calls for ffr(n) and ffs(n), which imply that the series values are to be "randomly" rather than "sequentially" accessed. Given this implied requirement, I chose to implement ffr and ffs as closures containing the type (data structure) FigureFigure, which are used to calculate their values as required. I address task requirement 2 (no maximum n) by having these functions extend this data structure as needed to accommodate values of n larger than those used for their creation.

Functions

 
type FigureFigure{T<:Integer}
r::Array{T,1}
rnmax::T
snmax::T
snext::T
end
 
function grow!{T<:Integer}(ff::FigureFigure{T}, rnmax::T=100)
ff.rnmax < rnmax || return nothing
append!(ff.r, zeros(T, (rnmax-ff.rnmax)))
snext = ff.snext
for i in (ff.rnmax+1):rnmax
ff.r[i] = ff.r[i-1] + snext
snext += 1
while snext in ff.r
snext += 1
end
end
ff.rnmax = rnmax
ff.snmax = ff.r[end] - rnmax
ff.snext = snext
return nothing
end
 
function FigureFigure{T<:Integer}(rnmax::T=10)
ff = FigureFigure([1], 1, 0, 2)
grow!(ff, rnmax)
return ff
end
 
function FigureFigure{T<:Integer}(rnmax::T, snmax::T)
ff = FigureFigure(rnmax)
while ff.snmax < snmax
grow!(ff, 2ff.rnmax)
end
return ff
end
 
function make_ffr{T<:Integer}(nmax::T=10)
ff = FigureFigure(nmax)
function ffr{T<:Integer}(n::T)
if n > ff.rnmax
grow!(ff, 2n)
end
ff.r[n]
end
end
 
function make_ffs{T<:Integer}(nmax::T=100)
ff = FigureFigure(13, nmax)
function ffs{T<:Integer}(n::T)
while ff.snmax < n
grow!(ff, 2ff.rnmax)
end
s = n
for r in ff.r
r <= s || return s
s += 1
end
end
end
 

Main

 
NR = 40
NS = 960
ffr = make_ffr(NR)
ffs = make_ffs(NS)
 
hi = 10
print("The first ", hi, " values of R are:\n ")
for i in 1:hi
print(ffr(i), " ")
end
println()
 
tally = falses(NR+NS)
iscontained = true
for i in 1:NR
try
tally[ffr(i)] = true
catch
iscontained = false
end
end
for i in 1:NS
try
tally[ffs(i)] = true
catch
iscontained = false
end
end
 
println()
print("The first ", NR, " values of R and ", NS, " of S are ")
if !iscontained
print("not ")
end
println("contained in the interval 1:", NR+NS, ".")
print("These values ")
if !all(tally)
print("do not ")
end
println("cover the entire interval.")
 
Output:
The first 10 values of R are:
    1  3  7  12  18  26  35  45  56  69  

The first 40 values of R and 960 of S are contained in the interval 1:1000.
These values cover the entire interval.

Kotlin[edit]

Translated from Java.

package hofstadter
 
fun ffr(n: Int) = get(n, 0)[n - 1]
 
fun ffs(n: Int) = get(0, n)[n - 1]
 
internal fun get(rSize: Int, sSize: Int): List<Int> {
val rlist = arrayListOf(1, 3, 7)
val slist = arrayListOf(2, 4, 5, 6)
val list = if (rSize > 0) rlist else slist
val targetSize = if (rSize > 0) rSize else sSize
 
while (list.size > targetSize)
list.removeAt(list.size - 1)
while (list.size < targetSize) {
val lastIndex = rlist.lastIndex
val lastr = rlist[lastIndex]
val r = lastr + slist[lastIndex]
rlist += r
var s = lastr + 1
while (s < r && list.size < targetSize)
slist += s++
}
return list
}
 
fun main(args: Array<String>) {
print("R():")
(1..10).forEach { print(" " + ffr(it)) }
println()
 
val first40R = (1..40).map { ffr(it) }
val first960S = (1..960).map { ffs(it) }
val indices = (1..1000).filter { it in first40R == it in first960S }
indices.forEach { println("Integer $it either in both or neither set") }
println("Done")
}

Mathematica / Wolfram Language[edit]

1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively.

   The instructions call for two functions.
   Because S[n] is generated while computing R[n], one would normally avoid redundancy by combining 
   R and S into a single function that returns both sequences.

2. No maximum value for n should be assumed.

 
ffr[j_] := Module[{R = {1}, S = 2, k = 1},
Do[While[Position[R, S] != {}, S++]; k = k + S; S++;
R = Append[R, k], {n, 1, j - 1}]; R]
 
ffs[j_] := Differences[ffr[j + 1]]
 

3. Calculate and show that the first ten values of R are: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69

 
ffr[10]
 
(* out *)
{1, 3, 7, 12, 18, 26, 35, 45, 56, 69}
 

4. Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once.

 
t = Sort[Join[ffr[40], ffs[960]]];
 
t == Range[1000]
 
(* out *)
True
 

MATLAB / Octave[edit]

1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively. 2. No maximum value for n should be assumed.

  function [R,S] = ffr_ffs(N) 
t = [1,0];
T = 1;
n = 1;
%while T<=1000,
while n<=N,
R = find(t,n);
S = find(~t,n);
T = R(n)+S(n);
 
% pre-allocate memory, this improves performance
if T > length(t), t = [t,zeros(size(t))]; end;
 
t(T) = 1;
n = n + 1;
end;
if nargout>0,
r = max(R);
s = max(S);
else
printf('Sequence R:\n'); disp(R);
printf('Sequence S:\n'); disp(S);
end;
end;

3. Calculate and show that the first ten values of R are: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69

>>ffr_ffs(10)
Sequence R:
    1    3    7   12   18   26   35   45   56   69
Sequence S:
    2    4    5    6    8    9   10   11   13   14

4. This is self-evident from the function definition, but also because R and S are complementary in t and ~t. However, one can also Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once. Modify the function above in such a way that, instead of r and s, R and S are returned, and run

  [R1,S1] = ffr_ffs(40);	
  [R2,S2] = ffr_ffs(960);	
  all(sort([R1,S2])==1:1000) 
ans =  1

Nim[edit]

var cr = @[1]
var cs = @[2]
 
proc extendRS =
let x = cr[cr.high] + cs[cr.high]
cr.add x
for y in cs[cs.high] + 1 .. <x: cs.add y
cs.add x + 1
 
proc ffr(n): int =
assert n > 0
while n > cr.len: extendRS()
cr[n - 1]
 
proc ffs(n): int =
assert n > 0
while n > cs.len: extendRS()
cs[n - 1]
 
for i in 1..10: stdout.write ffr i," "
echo ""
 
var bin: array[1..1000, int]
for i in 1..40: inc bin[ffr i]
for i in 1..960: inc bin[ffs i]
var all = true
for x in bin:
if x != 1:
all = false
break
 
if all: echo "All Integers 1..1000 found OK"
else: echo "All Integers 1..1000 NOT found only once: ERROR"

Output:

/home/deen/git/nim-unsorted/hofstadter 
1 3 7 12 18 26 35 45 56 69 
All Integers 1..1000 found OK

Oforth[edit]

tvar: R
ListBuffer new 1 over add R put
 
tvar: S
ListBuffer new 2 over add S put
 
: buildnext
| r s current i |
R at ->r
S at ->s
r last r size s at + dup ->current r add
s last 1+ current 1- for: i [ i s add ]
current 1+ s add ;
 
: ffr(n)
while ( R at size n < ) [ buildnext ]
n R at at ;
 
: ffs(n)
while ( S at size n < ) [ buildnext ]
n S at at ;

Output :

>#[ ffr . ] 10 seqEach
1 3 7 12 18 26 35 45 56 69
ok
>#ffr 40 seq map  #ffs 960 seq map  + sort 1000 seq == .
1 ok

Perl[edit]

The program produces a table with the first 10 values of R and S. It also calculates R(40) which is 982, S(960) which is 1000, and R(41) which is 1030.

Then we go through the first 1000 outputs, mark those which are seen, then check if all values in the range one through one thousand were seen.

#!perl
use strict;
use warnings;
 
my @r = ( undef, 1 );
my @s = ( undef, 2 );
 
sub ffsr {
my $n = shift;
while( $#r < $n ) {
push @r, $s[$#r]+$r[-1];
push @s, grep { $s[-1]<$_ } $s[-1]+1..$r[-1]-1, $r[-1]+1;
}
return $n;
}
 
sub ffr { $r[ffsr shift] }
sub ffs { $s[ffsr shift] }
 
printf " i: R(i) S(i)\n";
printf "==============\n";
printf "%3d:  %3d  %3d\n", $_, ffr($_), ffs($_) for 1..10;
printf "\nR(40)=%3d S(960)=%3d R(41)=%3d\n", ffr(40), ffs(960), ffr(41);
 
my %seen;
$seen{ffr($_)}++ for 1 .. 40;
$seen{ffs($_)}++ for 1 .. 960;
if( 1000 == keys %seen and grep $seen{$_}, 1 .. 1000 ) {
print "All occured exactly once.\n";
} else {
my @missed = grep !$seen{$_}, 1 .. 1000;
my @dupped = sort { $a <=> $b} grep $seen{$_}>1, keys %seen;
print "These were missed: @missed\n";
print "These were duplicated: @dupped\n";
}
 

Perl 6[edit]

my @ffr;
my @ffs;
 
@ffr.plan: 0, 1, gather take @ffr[$_] + @ffs[$_] for 1..*;
@ffs.plan: 0, 2, 4..6, gather take @ffr[$_] ^..^ @ffr[$_+1] for 3..*;
 
say @ffr[1..10];
 
say "Rawks!" if (1...1000) eqv sort @ffr[1..40], @ffs[1..960];

Output:

1 3 7 12 18 26 35 45 56 69
Rawks!

PicoLisp[edit]

(setq *RNext 2)
 
(de ffr (N)
(cache '(NIL) N
(if (= 1 N)
1
(+ (ffr (dec N)) (ffs (dec N))) ) ) )
 
(de ffs (N)
(cache '(NIL) N
(if (= 1 N)
2
(let S (inc (ffs (dec N)))
(when (= S (ffr *RNext))
(inc 'S)
(inc '*RNext) )
S ) ) ) )

Test:

: (mapcar ffr (range 1 10))
-> (1 3 7 12 18 26 35 45 56 69)
 
: (=
(range 1 1000)
(sort (conc (mapcar ffr (range 1 40)) (mapcar ffs (range 1 960)))) )
-> T

PL/I[edit]

ffr: procedure (n) returns (fixed binary(31));
declare n fixed binary (31);
declare v(2*n+1) bit(1);
declare (i, j) fixed binary (31);
declare (r, s) fixed binary (31);
 
v = '0'b;
v(1) = '1'b;
 
if n = 1 then return (1);
 
r = 1;
do i = 2 to n;
do j = 2 to 2*n;
if v(j) = '0'b then leave;
end;
v(j) = '1'b;
s = j;
r = r + s;
if r <= 2*n then v(r) = '1'b;
end;
return (r);
end ffr;

Output:

Please type a value for n: 
    1    3    7   12   18   26   35   45   56   69   83   98  114  131  150
  170  191  213  236  260  285  312  340  369  399  430  462  495  529  565
  602  640  679  719  760  802  845  889  935  982
ffs: procedure (n) returns (fixed binary (31));
declare n fixed binary (31);
declare v(2*n+1) bit(1);
declare (i, j) fixed binary (31);
declare (r, s) fixed binary (31);
 
v = '0'b;
v(1) = '1'b;
 
if n = 1 then return (2);
 
r = 1;
do i = 1 to n;
do j = 2 to 2*n;
if v(j) = '0'b then leave;
end;
v(j) = '1'b;
s = j;
r = r + s;
if r <= 2*n then v(r) = '1'b;
end;
return (s);
end ffs;

Output of first 960 values:

Please type a value for n: 
    2    4    5    6    8    9   10   11   13   14   15   16   17   19   20
   21   22   23   24   25   27   28   29   30   31   32   33   34   36   37
  ...
  986  987  988  989  990  991  992  993  994  995  996  997  998  999 1000

Verification using the above procedures:

 
Dcl t(1000) Bit(1) Init((1000)(1)'0'b);
put skip list ('Verification that the first 40 FFR numbers and the first');
put skip list ('960 FFS numbers result in the integers 1 to 1000 only.');
do i = 1 to 40;
j = ffr(i);
if t(j) then put skip list ('error, duplicate value at ' || i);
else t(j) = '1'b;
end;
do i = 1 to 960;
j = ffs(i);
if t(j) then put skip list ('error, duplicate value at ' || i);
else t(j) = '1'b;
end;
if all(t = '1'b) then put skip list ('passed test');
 

Output:

Verification that the first 40 FFR numbers and the first 
960 FFS numbers result in the integers 1 to 1000 only. 
passed test 

Prolog[edit]

Constraint Handling Rules[edit]

CHR is a programming language created by Professor Thom Frühwirth.
Works with SWI-Prolog and module chr written by Tom Schrijvers and Jan Wielemaker

:- use_module(library(chr)).
 
:- chr_constraint ffr/2, ffs/2, hofstadter/1,hofstadter/2.
:- chr_option(debug, off).
:- chr_option(optimize, full).
 
% to remove duplicates
ffr(N, R1) \ ffr(N, R2) <=> R1 = R2 | true.
ffs(N, R1) \ ffs(N, R2) <=> R1 = R2 | true.
 
% compute ffr
ffr(N, R), ffr(N1, R1), ffs(N1,S1) ==>
N > 1, N1 is N - 1 |
R is R1 + S1.
 
% compute ffs
ffs(N, S), ffs(N1,S1) ==>
N > 1, N1 is N - 1 |
V is S1 + 1,
( find_chr_constraint(ffr(_, V)) -> S is V+1; S = V).
 
% init
hofstadter(N) ==> ffr(1,1), ffs(1,2).
% loop
hofstadter(N), ffr(N1, _R), ffs(N1, _S) ==> N1 < N, N2 is N1 +1 | ffr(N2,_), ffs(N2,_).
 
 

Output for first task :

 ?- hofstadter(10), bagof(ffr(X,Y), find_chr_constraint(ffr(X,Y)), L).
ffr(10,69)
ffr(9,56)
ffr(8,45)
ffr(7,35)
ffr(6,26)
ffr(5,18)
ffr(4,12)
ffr(3,7)
ffr(2,3)
ffr(1,1)
ffs(10,14)
ffs(9,13)
ffs(8,11)
ffs(7,10)
ffs(6,9)
ffs(5,8)
ffs(4,6)
ffs(3,5)
ffs(2,4)
ffs(1,2)
hofstadter(10)
L = [ffr(10,69),ffr(9,56),ffr(8,45),ffr(7,35),ffr(6,26),ffr(5,18),ffr(4,12),ffr(3,7),ffr(2,3),ffr(1,1)].

Code for the second task

hofstadter :-
hofstadter(960),
% fetch the values of ffr
bagof(Y, X^find_chr_constraint(ffs(X,Y)), L1),
% fetch the values of ffs
bagof(Y, X^(find_chr_constraint(ffr(X,Y)), X < 41), L2),
% concatenate then
append(L1, L2, L3),
% sort removing duplicates
sort(L3, L4),
% check the correctness of the list
( (L4 = [1|_], last(L4, 1000), length(L4, 1000)) -> writeln(ok); writeln(ko)),
% to remove all pending constraints
fail.
 

Output for second task

 ?- hofstadter.
ok
false.

Python[edit]

def ffr(n):
if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1")
try:
return ffr.r[n]
except IndexError:
r, s = ffr.r, ffs.s
ffr_n_1 = ffr(n-1)
lastr = r[-1]
# extend s up to, and one past, last r
s += list(range(s[-1] + 1, lastr))
if s[-1] < lastr: s += [lastr + 1]
# access s[n-1] temporarily extending s if necessary
len_s = len(s)
ffs_n_1 = s[n-1] if len_s > n else (n - len_s) + s[-1]
ans = ffr_n_1 + ffs_n_1
r.append(ans)
return ans
ffr.r = [None, 1]
 
def ffs(n):
if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1")
try:
return ffs.s[n]
except IndexError:
r, s = ffr.r, ffs.s
for i in range(len(r), n+2):
ffr(i)
if len(s) > n:
return s[n]
raise Exception("Whoops!")
ffs.s = [None, 2]
 
if __name__ == '__main__':
first10 = [ffr(i) for i in range(1,11)]
assert first10 == [1, 3, 7, 12, 18, 26, 35, 45, 56, 69], "ffr() value error(s)"
print("ffr(n) for n = [1..10] is", first10)
#
bin = [None] + [0]*1000
for i in range(40, 0, -1):
bin[ffr(i)] += 1
for i in range(960, 0, -1):
bin[ffs(i)] += 1
if all(b == 1 for b in bin[1:1000]):
print("All Integers 1..1000 found OK")
else:
print("All Integers 1..1000 NOT found only once: ERROR")
Output
ffr(n) for n = [1..10] is [1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
All Integers 1..1000 found OK

Alternative[edit]

cR = [1]
cS = [2]
 
def extend_RS():
x = cR[len(cR) - 1] + cS[len(cR) - 1]
cR.append(x)
cS += range(cS[-1] + 1, x)
cS.append(x + 1)
 
def ff_R(n):
assert(n > 0)
while n > len(cR): extend_RS()
return cR[n - 1]
 
def ff_S(n):
assert(n > 0)
while n > len(cS): extend_RS()
return cS[n - 1]
 
# tests
print([ ff_R(i) for i in range(1, 11) ])
 
s = {}
for i in range(1, 1001): s[i] = 0
for i in range(1, 41): del s[ff_R(i)]
for i in range(1, 961): del s[ff_S(i)]
 
# the fact that we got here without a key error
print("Ok")
output
[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
Ok

Using cyclic iterators[edit]

Translation of: Haskell

Defining R and S as mutually recursive generators. Follows directly from the definition of the R and S sequences.

from itertools import islice
 
def R():
n = 1
yield n
for s in S():
n += s
yield n;
 
def S():
yield 2
yield 4
u = 5
for r in R():
if r <= u: continue;
for x in range(u, r): yield x
u = r + 1
 
def lst(s, n): return list(islice(s(), n))
 
print "R:", lst(R, 10)
print "S:", lst(S, 10)
print sorted(lst(R, 40) + lst(S, 960)) == list(range(1,1001))
 
# perf test case
# print sum(lst(R, 10000000))
Output:
R: [1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
S: [2, 4, 5, 6, 8, 9, 10, 11, 13, 14]
True

Racket[edit]

Translation of: Java

We store the values of r and s in hash-tables. The first values are added by hand. The procedure extend-r-s! adds more values.

#lang racket/base
 
(define r-cache (make-hash '((1 . 1) (2 . 3) (3 . 7))))
(define s-cache (make-hash '((1 . 2) (2 . 4) (3 . 5) (4 . 6))))
 
(define (extend-r-s!)
(define r-count (hash-count r-cache))
(define s-count (hash-count s-cache))
(define last-r (ffr r-count))
(define new-r (+ (ffr r-count) (ffs r-count)))
(hash-set! r-cache (add1 r-count) new-r)
(define offset (- s-count last-r))
(for ([val (in-range (add1 last-r) new-r)])
(hash-set! s-cache (+ val offset) val)))

The functions ffr and ffs simply retrieve the value from the hash table if it exist, or call extend-r-s until they are long enought.

(define (ffr n)
(hash-ref r-cache n (lambda () (extend-r-s!) (ffr n))))
 
(define (ffs n)
(hash-ref s-cache n (lambda () (extend-r-s!) (ffs n))))

Tests:

(displayln (map ffr (list 1 2 3 4 5 6 7 8 9 10)))
(displayln (map ffs (list 1 2 3 4 5 6 7 8 9 10)))
 
(displayln "Checking for first 1000 integers:")
(displayln (if (equal? (sort (append (for/list ([i (in-range 1 41)])
(ffr i))
(for/list ([i (in-range 1 961)])
(ffs i)))
<)
(for/list ([i (in-range 1 1001)])
i))
"Test passed"
"Test failed"))

Sample Output:

(1 3 7 12 18 26 35 45 56 69)
(2 4 5 6 8 9 10 11 13 14)
Checking for first 1000 integers: Test passed

REXX[edit]

version 1[edit]

This REXX example makes use of sparse arrays.

Over a third of the program was for verification of the first thousand numbers in the Hofstadter Figure-Figure sequences.

/*REXX program  calculates and verifies  the  Hofstadter Figure─Figure sequences.       */
parse arg x top bot . /*obtain optional arguments from the CL*/
if x=='' | x=="," then x= 10 /*Not specified? Then use the default.*/
if top=='' | top=="," then top=1000 /* " " " " " " */
if bot=='' | bot=="," then bot= 40 /* " " " " " " */
low=1; if x<0 then low=abs(x) /*only display a single │X│ value? */
r.=0; r.1=1; rr.=r.; rr.1=1; s.=r.; s.1=2 /*initialize the R, RR, and S arrays.*/
errs=0 /*the number of errors found (so far).*/
do i=low to abs(x) /*display the 1st X values of R & S.*/
say right('R('i") =",20) right(FFR(i),7) right('S('i") =",20) right(FFS(i),7)
end /*i*/
/* [↑] list the 1st X Fig─Fig numbers.*/
if x<1 then exit /*if X isn't positive, then we're done.*/
$.=0 /*initialize the memoization ($) array.*/
do m=1 for bot; r=FFR(m); $.r=1 /*calculate the first forty R values.*/
end /*m*/ /* [↑] ($.) is used for memoization. */
/* [↓] check for duplicate #s in R & S*/
do n=1 for top-bot; s=FFS(n) /*calculate the value of FFS(n). */
if $.s then call ser 'duplicate number in R and S lists:' s; $.s=1
end /*n*/ /* [↑] calculate the 1st 960 S values.*/
/* [↓] check for missing values in R│S*/
do v=1 for top; if \$.v then call ser 'missing R │ S:' v
end /*v*/ /* [↑] are all 1≤ numbers ≤1k present?*/
say
if errs==0 then say 'verification completed for all numbers from 1 ──►' top " [inclusive]."
else say 'verification failed with' errs "errors."
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
FFR: procedure expose r. rr. s.; parse arg n /*obtain the number from the arguments.*/
if r.n\==0 then return r.n /*R.n defined? Then return the value.*/
_=FFR(n-1) + FFS(n-1) /*calculate the FFR and FFS values.*/
r.n=_; rr._=1; return _ /*assign the value to R & RR; return.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
FFS: procedure expose r. s. rr.; parse arg n /*search for not null R or S number. */
if s.n==0 then do k=1 for n /* [↓] 1st IF is a SHORT CIRCUIT. */
if s.k\==0 then if r.k\==0 then iterate /*are both defined?*/
call FFR k /*define R.k via the FFR subroutine*/
km=k-1; _=s.km+1 /*calc. the next S number, possibly.*/
_=_+rr._; s.k=_ /*define an element of the S array. */
end /*k*/
return s.n /*return S.n value to the invoker. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
ser: errs=errs+1; say '***error***' arg(1); return

To see the talk section about this REXX program's timings, click here:     timings for the REXX solutions.

output   when using the default inputs:

              R(1) =       1               S(1) =       2
              R(2) =       3               S(2) =       4
              R(3) =       7               S(3) =       5
              R(4) =      12               S(4) =       6
              R(5) =      18               S(5) =       8
              R(6) =      26               S(6) =       9
              R(7) =      35               S(7) =      10
              R(8) =      45               S(8) =      11
              R(9) =      56               S(9) =      13
             R(10) =      69              S(10) =      14

verification completed for all numbers from  1 ──► 1000   [inclusive].

Version 2 from PL/I[edit]

/* REXX **************************************************************
* 21.11.2012 Walter Pachl transcribed from PL/I
**********************************************************************/

Call time 'R'
Say 'Verification that the first 40 FFR numbers and the first'
Say '960 FFS numbers result in the integers 1 to 1000 only.'
t.=0
num.=''
do i = 1 to 40
j = ffr(i)
if t.j then Say 'error, duplicate value at ' || i
else t.j = 1
num.i=j
end
nn=0
Say time('E') 'seconds elapsed'
Do i=1 To 3
ol=''
Do j=1 To 15
nn=nn+1
ol=ol right(num.nn,3)
End
Say ol
End
do i = 1 to 960
j = ffs(i)
if t.j then
Say 'error, duplicate value at ' || i
else t.j = 1
end
Do i=1 To 1000
if t.i=0 Then
Say i 'was not set'
End
If i>1000 Then
Say 'passed test'
Say time('E') 'seconds elapsed'
Exit
 
ffr: procedure Expose v.
Parse Arg n
v.= 0
v.1 = 1
if n = 1 then return 1
r = 1
do i = 2 to n
do j = 2 to 2*n
if v.j = 0 then leave
end
v.j = 1
s = j
r = r + s
if r <= 2*n then v.r = 1
end
return r
 
ffs: procedure Expose v.
Parse Arg n
v.= 0
v.1 = 1
if n = 1 then return 2
r = 1
do i = 1 to n
do j = 2 to 2*n
if v.j = 0 then leave
end
v.j = 1
s = j
r = r + s
if r <= 2*n then v.r = 1
end
return s
Output:
Verification that the first 40 FFR numbers and the first
960 FFS numbers result in the integers 1 to 1000 only.
0.011000 seconds elapsed
   1   3   7  12  18  26  35  45  56  69  83  98 114 131 150
 170 191 213 236 260 285 312 340 369 399 430 462 495 529 565
 602 640 679 719 760 802 845 889 935 982                    
passed test
Windows (ooRexx)  33.183000 seconds elapsed
Windows (Regina)  22.627000 seconds elapsed
TSO interpreted: 139.699246 seconds elapsed
TSO compiled:      9.749457 seconds elapsed

Ruby[edit]

Translation of: Tcl
$r = [nil, 1]
$s = [nil, 2]
 
def buildSeq(n)
current = [ $r[-1], $s[-1] ].max
while $r.length <= n || $s.length <= n
idx = [ $r.length, $s.length ].min - 1
current += 1
if current == $r[idx] + $s[idx]
$r << current
else
$s << current
end
end
end
 
def ffr(n)
buildSeq(n)
$r[n]
end
 
def ffs(n)
buildSeq(n)
$s[n]
end
 
require 'set'
require 'test/unit'
 
class TestHofstadterFigureFigure < Test::Unit::TestCase
def test_first_ten_R_values
r10 = 1.upto(10).map {|n| ffr(n)}
assert_equal(r10, [1, 3, 7, 12, 18, 26, 35, 45, 56, 69])
end
 
def test_40_R_and_960_S_are_1_to_1000
rs_values = Set.new
rs_values.merge( 1.upto(40).collect {|n| ffr(n)} )
rs_values.merge( 1.upto(960).collect {|n| ffs(n)} )
assert_equal(rs_values, Set.new( 1..1000 ))
end
end

outputs

Loaded suite hofstadter.figurefigure
Started
..
Finished in 0.511000 seconds.

2 tests, 2 assertions, 0 failures, 0 errors, 0 skips

Using cyclic iterators[edit]

Translation of: Python
R = Enumerator.new do |y|
y << n = 1
S.each{|s_val| y << n += s_val}
end
 
S = Enumerator.new do |y|
y << 2
y << 4
u = 5
R.each do |r_val|
next if u > r_val
(u...r_val).each{|r| y << r}
u = r_val+1
end
end
 
p R.take(10)
p S.take(10)
p (R.take(40)+ S.take(960)).sort == (1..1000).to_a
 
Output:
[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
[2, 4, 5, 6, 8, 9, 10, 11, 13, 14]
true

Scala[edit]

Translation of: Go
object HofstadterFigFigSeq extends App {
import scala.collection.mutable.ListBuffer
 
val r = ListBuffer(0, 1)
val s = ListBuffer(0, 2)
 
def ffr(n: Int): Int = {
val ffri: Int => Unit = i => {
val nrk = r.size - 1
val rNext = r(nrk)+s(nrk)
r += rNext
(r(nrk)+2 to rNext-1).foreach{s += _}
s += rNext+1
}
 
(r.size to n).foreach(ffri(_))
r(n)
}
 
def ffs(n:Int): Int = {
while (s.size <= n) ffr(r.size)
s(n)
}
 
(1 to 10).map(i=>(i,ffr(i))).foreach(t=>println("r("+t._1+"): "+t._2))
println((1 to 1000).toList.filterNot(((1 to 40).map(ffr(_))++(1 to 960).map(ffs(_))).contains)==List())
}

Output:

r(1): 1
r(2): 3
r(3): 7
r(4): 12
r(5): 18
r(6): 26
r(7): 35
r(8): 45
r(9): 56
r(10): 69
true

Sidef[edit]

Translation of: Perl
var r = [nil, 1]
var s = [nil, 2]
 
func ffsr(n) {
while(r.end < n) {
r << s[r.end]+r[-1]
s << [(s[-1]+1 .. r[-1]-1)..., r[-1]+1].grep{ s[-1] < _ }...
}
return n
}
 
func ffr(n) { r[ffsr(n)] }
func ffs(n) { s[ffsr(n)] }
 
printf(" i: R(i) S(i)\n")
printf("==============\n")
{ |i|
printf("%3d:  %3d  %3d\n", i, ffr(i), ffs(i))
} << 1..10
printf("\nR(40)=%3d S(960)=%3d R(41)=%3d\n", ffr(40), ffs(960), ffr(41))
 
var seen = Hash()
 
{|i| seen{ffr(i)} := 0 ++ } << 1..40
{|i| seen{ffs(i)} := 0 ++ } << 1..960
 
if (seen.count {|k,v| (k.to_i >= 1) && (k.to_i <= 1000) && (v == 1) } == 1000) {
say "All occured exactly once."
}
else {
var missed = { !seen.has_key(_) }.grep(1..1000)
var dupped = seen.grep { |_, v| v > 1 }.keys.sort
say "These were missed: #{missed}"
say "These were duplicated: #{dupped}"
}
Output:
  i: R(i) S(i)
==============
  1:    1    2
  2:    3    4
  3:    7    5
  4:   12    6
  5:   18    8
  6:   26    9
  7:   35   10
  8:   45   11
  9:   56   13
 10:   69   14

R(40)=982 S(960)=1000 R(41)=1030
All occured exactly once.

Tcl[edit]

Library: Tcllib (Package: struct::set)
package require Tcl 8.5
package require struct::set
 
# Core sequence generator engine; stores in $R and $S globals
set R {R:-> 1}
set S {S:-> 2}
proc buildSeq {n} {
global R S
set ctr [expr {max([lindex $R end],[lindex $S end])}]
while {[llength $R] <= $n || [llength $S] <= $n} {
set idx [expr {min([llength $R],[llength $S]) - 1}]
if {[incr ctr] == [lindex $R $idx]+[lindex $S $idx]} {
lappend R $ctr
} else {
lappend S $ctr
}
}
}
 
# Accessor procedures
proc ffr {n} {
buildSeq $n
lindex $::R $n
}
proc ffs {n} {
buildSeq $n
lindex $::S $n
}
 
# Show some things about the sequence
for {set i 1} {$i <= 10} {incr i} {
puts "R($i) = [ffr $i]"
}
puts "Considering {1..1000} vs {R(i)|i\u2208\[1,40\]}\u222a{S(i)|i\u2208\[1,960\]}"
for {set i 1} {$i <= 1000} {incr i} {lappend numsInSeq $i}
for {set i 1} {$i <= 40} {incr i} {
lappend numsRS [ffr $i]
}
for {set i 1} {$i <= 960} {incr i} {
lappend numsRS [ffs $i]
}
puts "set sizes: [struct::set size $numsInSeq] vs [struct::set size $numsRS]"
puts "set equality: [expr {[struct::set equal $numsInSeq $numsRS]?{yes}:{no}}]"

Output:

R(1) = 1
R(2) = 3
R(3) = 7
R(4) = 12
R(5) = 18
R(6) = 26
R(7) = 35
R(8) = 45
R(9) = 56
R(10) = 69
Considering {1..1000} vs {R(i)|i∈[1,40]}∪{S(i)|i∈[1,960]}
set sizes: 1000 vs 1000
set equality: yes

uBasic/4tH[edit]

Note that uBasic/4tH has no dynamic memory facilities and only one single array of 256 elements. So the only way to cram over a 1000 values there is to use a bitmap. This bitmap consists of an R range and an S range. In each range, a bit represents a positional value (bit 0 = "1", bit 1 = "2", etc.). The R(x) and S(x) functions simply count the number of bits set they encountered. To determine whether all integers between 1 and 1000 are complementary, both ranges are XORed, which would result in a value other than 231-1 if there were any discrepancies present. An extra check determines if there are exactly 40 R values.

Proc _SetBitR(1)                       ' Set the first R value
Proc _SetBitS(2) ' Set the first S value
 
Print "Creating bitmap, wait.." ' Create the bitmap
Proc _MakeBitMap
Print
 
Print "R(1 .. 10):"; ' Print first 10 R-values
 
For x = 1 To 10
Print " ";FUNC(_Rx(x));
Next
 
Print : Print "S(1 .. 10):"; ' Print first 10 S-values
 
For x = 1 To 10
Print " ";FUNC(_Sx(x));
Next
 
Print : Print ' Terminate and skip line
 
For x = 0 To (1000/31) ' Check the first 1000 values
Print "Checking ";(x*31)+1;" to ";(x*31)+31;":\t";
If XOR(@(x), @(x+64)) = 2147483647 Then
Print "OK" ' XOR R() and S() ranges
Else ' should deliver MAX-N
Print "Fail!" ' or we did have an error
EndIf
Next
 
For x = 1 to 40 ' Prove there are only 40 R(x) values
If FUNC(_Rx(x)) > 1000 Then Print "R(";x;") value greater than 1000"
Next ' below 1000
 
If FUNC(_Rx(x)) < 1001 Then Print "R(";x;") value also below 1000"
End
 
 
_MakeBitMap ' Create the bitmap
Local (4)
 
a@ = 1 ' Previous R(x) level
b@ = 1 ' Previous R(x) value
 
Do Until b@ > (1000/31)*32 ' Fill up an entire array element
' calculate R(x+1) level
c@ = FUNC(_Rx(a@)) + FUNC(_Sx(a@))
Proc _SetBitR (c@) ' Set R(x+1) in the bitmap
 
For d@ = b@ + 1 To c@ - 1 ' Set all intermediate S() values
Proc _SetBitS (d@) ' between R(x) and R(x+1)
Next
 
Proc _SetBitS (c@+1) ' Number after R(x) is always S()
b@ = c@ ' R(x+1) now becomes R(x)
a@ = a@ + 1 ' Increment level
Loop ' Now do it again
Return
 
 
_Rx Param(1) ' Return value R(x)
Local(2)
 
b@ = 0 ' No value found so far
 
For c@ = 1 To (64*31)-1 ' Check the entire bitmap
If (FUNC(_GetBitR(c@))) Then b@ = b@ + 1
Until b@ = a@ ' If a value found, increment counter
Next ' Until the required level is reached
Return (c@) ' Return position in bitmap
 
 
_Sx Param(1) ' Return value S(x)
Local(2)
 
b@ = 0 ' No value found so far
 
For c@ = 1 To (64*31)-1 ' Check the entire bitmap
If (FUNC(_GetBitS(c@))) Then b@ = b@ + 1
Until b@ = a@ ' If a value found, increment counter
Next ' Until the required level is reached
Return (c@) ' Return position in bitmap
 
 
_SetBitR Param(1) ' Set bit n-1 in R-bitmap
a@ = a@ - 1
@(a@/31) = OR(@(a@/31), SHL(1,a@%31))
Return
 
_GetBitR Param(1) ' Return bit n-1 in R-bitmap
a@ = a@ - 1
Return (AND(@(a@/31), SHL(1,a@%31))#0)
 
_SetBitS Param(1) ' Set bit n-1 in S-bitmap
a@ = a@ - 1
@(64+a@/31) = OR(@(64+a@/31), SHL(1,a@%31))
Return
 
_GetBitS Param(1) ' Return bit n-1 in S-bitmap
a@ = a@ - 1
Return (AND(@(64+a@/31), SHL(1,a@%31))#0)
Output:
Creating bitmap, wait..

R(1 .. 10): 1 3 7 12 18 26 35 45 56 69
S(1 .. 10): 2 4 5 6 8 9 10 11 13 14

Checking 1 to 31:       OK
Checking 32 to 62:      OK
Checking 63 to 93:      OK
Checking 94 to 124:     OK
Checking 125 to 155:    OK
Checking 156 to 186:    OK
Checking 187 to 217:    OK
Checking 218 to 248:    OK
Checking 249 to 279:    OK
Checking 280 to 310:    OK
Checking 311 to 341:    OK
Checking 342 to 372:    OK
Checking 373 to 403:    OK
Checking 404 to 434:    OK
Checking 435 to 465:    OK
Checking 466 to 496:    OK
Checking 497 to 527:    OK
Checking 528 to 558:    OK
Checking 559 to 589:    OK
Checking 590 to 620:    OK
Checking 621 to 651:    OK
Checking 652 to 682:    OK
Checking 683 to 713:    OK
Checking 714 to 744:    OK
Checking 745 to 775:    OK
Checking 776 to 806:    OK
Checking 807 to 837:    OK
Checking 838 to 868:    OK
Checking 869 to 899:    OK
Checking 900 to 930:    OK
Checking 931 to 961:    OK
Checking 962 to 992:    OK
Checking 993 to 1023:   OK

0 OK, 0:875

VBScript[edit]

 
'Initialize the r and the s arrays.
Set r = CreateObject("System.Collections.ArrayList")
Set s = CreateObject("System.Collections.ArrayList")
 
'Set initial values of r.
r.Add ""  : r.Add 1
 
'Set initial values of s.
s.Add "" : s.Add 2
 
'Populate the r and the s arrays.
For i = 2 To 1000
ffr(i)
ffs(i)
Next
 
'r function
Function ffr(n)
r.Add r(n-1)+s(n-1)
End Function
 
's function
Function ffs(n)
'index is the value of the last element of the s array.
index = s(n-1)+1
Do
'Add to s if the current index is not in the r array.
If r.IndexOf(index,0) = -1 Then
s.Add index
Exit Do
Else
index = index + 1
End If
Loop
End Function
 
'Display the first 10 values of r.
WScript.StdOut.Write "First 10 Values of R:"
WScript.StdOut.WriteLine
For j = 1 To 10
If j = 10 Then
WScript.StdOut.Write "and " & r(j)
Else
WScript.StdOut.Write r(j) & ", "
End If
Next
WScript.StdOut.WriteBlankLines(2)
 
'Show that the first 40 values of r plus the first 960 values of s include all the integers from 1 to 1000 exactly once.
'The idea here is to create another array(integer) with 1000 elements valuing from 1 to 1000. Go through the first 40 values
'of the r array and remove the corresponding element in the integer array. Do the same thing with the first 960 values of
'the s array. If the resultant count of the integer array is 0 then it is a pass.
Set integers = CreateObject("System.Collections.ArrayList")
For k = 1 To 1000
integers.Add k
Next
For l = 1 To 960
If l <= 40 Then
integers.Remove(r(l))
End If
integers.Remove(s(l))
Next
WScript.StdOut.Write "Test for the first 1000 integers: "
If integers.Count = 0 Then
WScript.StdOut.Write "Passed!!!"
WScript.StdOut.WriteLine
Else
WScript.StdOut.Write "Miserably Failed!!!"
WScript.StdOut.WriteLine
End If
 
Output:
First 10 Values of R:
1, 3, 7, 12, 18, 26, 35, 45, 56, and 69

Test for the first 1000 integers: Passed!!!

zkl[edit]

fcn genRS(reset=False){ //-->(n,R,S)
var n=0, Rs=L(0,1), S=2;
if(True==reset){ n=0; Rs=L(0,1); S=2; return(); }
 
if (n==0) return(n=1,1,2);
R:=Rs[-1] + S; Rs.append(R);
foreach s in ([S+1..]){
if(not Rs.holds(s)) { S=s; break; } // trimming Rs doesn't save space
}
return(n+=1,R,S);
}
fcn ffrs(n) { genRS(True); do(n){ n=genRS() } n[1,2] } //-->( R(n),S(n) )
Output:
(0).pump(10,List,genRS).apply("get",1).println();
L(1,3,7,12,18,26,35,45,56,69)
genRS(True);  // reset
sink:=(0).pump(40,List, 'wrap(ns){ T(Void.Write,Void.Write,genRS()[1,*]) });
sink= (0).pump(960-40,sink,'wrap(ns){ T(Void.Write,genRS()[2]) });
(sink.sort()==[1..1000].walk()).println("<-- should be True");
Output:
True<-- should be True