Hofstadter Figure-Figure sequences: Difference between revisions

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Content added Content deleted
(Updated both D entries)
m (→‎version 1: used a short-circuit type IF statement for faster execution. -- ~~~~)
Line 1,461: Line 1,461:
if s.n\==0 then return s.n /*Defined? Then return the value.*/
if s.n\==0 then return s.n /*Defined? Then return the value.*/
do k=1 for n while s.n==0 /*search for not null R │ S num.*/
do k=1 for n while s.n==0 /*search for not null R │ S num.*/
if s.k\==0 & ffr(k)\==0 then iterate
if s.k\==0 then if ffr(k)\==0 then iterate /*short circuit*/
km=k-1; _=s.km+1 /*the next SS number, possibly.*/
km=k-1; _=s.km+1 /*the next SS number, possibly.*/
_=_+rr._ /*maybe adjust for the FRR num.*/
_=_+rr._ /*maybe adjust for the FRR num.*/

Revision as of 20:35, 20 July 2013

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 R starts: 1, 3, 7, 12, 18, ...
Sequence S 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

Specifying a package providing the functions FFR and FFS: <lang Ada>package Hofstadter_Figure_Figure is 

  function FFR(P: Positive) return Positive;

  function FFS(P: Positive) return Positive;

end Hofstadter_Figure_Figure;</lang>

The implementation of the package internally uses functions which generate an array of Figures or Spaces: <lang Ada>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;</lang>

Finally, a test program for the package, solving the task at hand: <lang Ada>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;</lang>

The output of the test program: <lang> 1 3 7 12 18 26 35 45 56 69 Test Passed: No overlap between FFR(I) and FFS(J)</lang>

BBC BASIC

Works with: BBC BASIC for Windows

<lang bbcbasic> 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%</lang>

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

Common Lisp

<lang lisp>;;; 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")</lang>output<lang>First of R: (1 3 7 12 18 26 35 45 56 69) Ok</lang>

C#

Creates an IEnumerable for R and S and uses those to complete the task <lang Csharp>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!"); } } } </lang> Output: 

1
3
7
12
18
26
35
45
56
69
Verified

C

<lang c>#include <stdio.h>

  1. 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; }</lang>

D

Translation of: Go

<lang d>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;

}</lang>

Output:
[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
true

Alternative version

Translation of: Python

(Same output) <lang d>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() { 

   map!ffr(iota(1, 11)).writeln;
   auto t = iota(1, 41).map!ffr.chain(iota(1, 961).map!ffs);
   t.array().sort().equal(iota(1, 1001)).writeln;

}</lang>

Euler Math Toolbox

<lang Euler Math Toolbox> >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

</lang>

Factor

We keep lists S and R, and increment them when necessary. <lang factor>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 ;</lang>

<lang factor>( 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</lang>

Go

<lang go>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")

}</lang> 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. <lang go>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) }</lang>

Haskell

<lang haskell>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])</lang>

Output: 

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

Defining R and S literally: <lang haskell>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)))</lang> 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

<lang Icon>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</lang>

Library: Icon Programming Library

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

<lang j>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,>./@,)</lang>

Required examples:

<lang j> 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</lang>

Java

Code:

<lang java>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");
 }

}</lang>

Output: 

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

JavaScript

Translated from Ruby. <lang JavaScript>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."); } }</lang> 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.


Mathematica

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.

<lang Mathematica> 

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]]

</lang>

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

<lang Mathematica> 

ffr[10]

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

</lang>

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.

<lang Mathematica> 

t = Sort[Join[ffr[40], ffs[960]]];

t == Range[1000]

(* out *)
True

</lang>

MATLAB / Octave

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.

<lang MATLAB> 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; </lang>

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

PicoLisp

<lang PicoLisp>(setq *RNext 2)

(de ffr (N) 

  (cache '(NIL) (pack (char (hash N)) N)
     (if (= 1 N)
        1
        (+ (ffr (dec N)) (ffs (dec N))) ) ) )

(de ffs (N) 

  (cache '(NIL) (pack (char (hash N)) N)
     (if (= 1 N)
        2
        (let S (inc (ffs (dec N)))
           (when (= S (ffr *RNext))
              (inc 'S)
              (inc '*RNext) )
           S ) ) ) )</lang>

Test: <lang PicoLisp>: (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</lang>

Perl 6

<lang perl6>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];</lang> Output: 

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

PL/I

<lang PL/I>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;</lang> 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

<lang>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;</lang> 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: <lang> 

  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');

</lang> 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

Constraint Handling Rules

CHR is a programming language created by Professor Thom Frühwirth.
Works with SWI-Prolog and module chr written by Tom Schrijvers and Jan Wielemaker <lang Prolog>:- 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,_).

</lang> 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 <lang Prolog>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. </lang> Output for second task 

 ?- hofstadter.
ok
false.

Python

<lang python>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")</lang>
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

<lang python>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]

  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)]

  1. the fact that we got here without a key error

print("Ok")</lang>output<lang>[1, 3, 7, 12, 18, 26, 35, 45, 56, 69] Ok</lang>


Using cyclic iterators

Translation of: Haskell

Defining R and S as mutually recursive generators. Follows directly from the definition of the R and S sequences. <lang python>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))

  1. perf test case
  2. print sum(lst(R, 10000000))</lang>
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

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>#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)))</lang>

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.

<lang Racket>(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))))</lang>

Tests: <lang Racket>(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"))</lang>

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

version 1

This REXX example makes use of sparse arrays.
Almost half of the code was for verification of the first thousand numbers in the Figure-Figure sequences. <lang rexx>/*REXX pgm to calculate & verify the Hofstadter Figure-Figure sequences.*/ parse arg x highV . /*obtain any C.L. specifications.*/ if x== then x=10; if highV== then highV=1000 /*use the defaults?*/ low=1 /*use unity as the starting point*/ if x<0 then low=abs(x) /*only show a single │X│ value.*/ r.=0; r.1=1; rr.=r.; rr.1=1 /*initialize the R and RR arrays.*/ s.=0; s.1=2; ss.=s.; ss.2=1 /* " " S " SS " .*/ errs=0 

                 do i=low  to abs(x)  /*show first X values of  R & S  */
                 say right('R('i") =",20) right(ffr(i),7),  /*show nice*/
                     right('S('i") =",20) right(ffs(i),7)   /*  R & S  */
                 end   /*i*/

if x<1 then exit /*stick a fork in it, we're done.*/ /*═══════════════════════════════════════verify 1st 1k: unique & present*/ both.=0 /*initialize the BOTH array. */ 

                                      /*build list of 1st  40 R values.*/
 do m=1 for  40;     r=ffr(m)         /*calculate 1st   40  R   values.*/
 both.r=1                             /*build the  BOTH  array.        */
 end   /*m*/

    do n=1 for  960;    s=ffs(n)      /*calculate 1st  960  S   values.*/
    if both.s  then call sayErr 'duplicate number in R and S lists:' s
    both.s=1                          /*add to the  BOTH  array.       */
    end   /*n*/
                                      /*verify presence and uniqueness.*/
       do v=1  for highV              /*verify all  1 ≤ # ≤ 1k present.*/
       if \both.v  then call sayErr 'missing R │ S:'  v
       end   /*v*/

say @v='verification'; @i=" [inclusive]." /*shortcuts to shorten prog width*/ if errs==0 then say @v 'completed for all numbers from 1 ──►' highV @i 

           else say @v 'failed with' errs "errors."

exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────FFR subroutine──────────────────────*/ ffr: procedure expose r. s. rr. ss.; parse arg n if r.n\==0 then return r.n /*Defined? Then return the value.*/ _=ffr(n-1)+ffs(n-1) /*calculate the FFR value. */ r.n=_; rr._=1 /*assign the value to R and RR.*/ return _ /*return the value to the invoker*/ /*──────────────────────────────────FFS subroutine──────────────────────*/ ffs: procedure expose r. s. rr. ss.; parse arg n if s.n\==0 then return s.n /*Defined? Then return the value.*/ 

          do k=1  for n  while s.n==0 /*search for not null  R │ S num.*/
          if s.k\==0  then  if ffr(k)\==0  then iterate /*short circuit*/
          km=k-1;    _=s.km+1         /*the next  SS  number, possibly.*/
          _=_+rr._                    /*maybe adjust for the  FRR  num.*/
          s.k=_;     ss._=1           /*define couple of  FFS  numbers.*/
          end    /*k*/

return s.n /*return the value to the invoker*/ /*──────────────────────────────────SAYERR subroutine───────────────────*/ sayErr: errs=errs+1; say; say '***error***!'; say; say arg(1); say; return</lang> output when using the defaults 

              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

<lang rexx>/* 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</lang>

Verification that the first 40 FFR numbers snd 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
TSO interpreted: 139.699246 seconds elapsed
TSO compiled:      9.749457 seconds elapsed

Ruby

Translation of: Tcl

<lang ruby>$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</lang>

outputs 

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

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

Scala

Translation of: Go

<lang Scala>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())

}</lang> 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

Tcl

Library: Tcllib (Package: struct::set)

<lang tcl>package require Tcl 8.5 package require struct::set

  1. 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 } 

   }

}

  1. Accessor procedures

proc ffr {n} { 

   buildSeq $n
   lindex $::R $n

} proc ffs {n} { 

   buildSeq $n
   lindex $::S $n

}

  1. 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}}]"</lang> 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
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