Hofstadter Figure-Figure sequences: Difference between revisions

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Latest revision as of 18:53, 20 April 2024

These two sequences of positive integers are defined as:

{\begin{aligned}R(1)&=1\ ;\ S(1)=2\\R(n)&=R(n-1)+S(n-1),\quad n>1.\end{aligned}}

The sequence $S(n)$ is further defined as the sequence of positive integers not present in $R(n)$ .

Sequence $R$ starts:

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

Sequence $S$ starts:

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

Task

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).
No maximum value for n should be assumed.
Calculate and show that the first ten values of R are:
1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
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

Sloane's A005228 and A030124.
Wolfram MathWorld
Wikipedia: Hofstadter Figure-Figure sequences.

11l

Translation of: Python

V cR = [1]
V cS = [2]

F extend_RS()
   V x = :cR[:cR.len-1] + :cS[:cR.len-1]
   :cR [+]= (x)
   :cS [+]= :cS.last+1 .< x
   :cS [+]= (x + 1)

F ff_R(n)
   assert(n > 0)
   L n > :cR.len
      extend_RS()
   R :cR[n - 1]

F ff_S(n)
   assert(n > 0)
   L n > :cS.len
      extend_RS()
   R :cS[n - 1]

print((1..10).map(i -> ff_R(i)))

V arr = [0] * 1001
L(i) (40.<0).step(-1)
   arr[ff_R(i)]++
L(i) (960.<0).step(-1)
   arr[ff_S(i)]++

I all(arr[1..1000].map(a -> a == 1))
   print(‘All Integers 1..1000 found OK’)
E
   print(‘All Integers 1..1000 NOT found only once: ERROR’)

Output:

[1, 3, 7, 12, 18, 26, 35, 45, 56, 69]
All Integers 1..1000 found OK

ABC

PUT {[1]: 1} IN r.list
PUT {[1]: 2} IN s.list

HOW TO EXTEND R TO n:
    SHARE r.list, s.list
    WHILE n > #r.list:
        PUT r.list[#r.list] + s.list[#r.list] IN next.r
        FOR i IN {s.list[#s.list]+1 .. next.r-1}:
            PUT i IN s.list[#s.list+1]
        PUT next.r IN r.list[#r.list+1]
        PUT next.r + 1 IN s.list[#s.list+1]

HOW TO EXTEND S TO n:
    SHARE r.list, s.list
    WHILE n > #s.list: EXTEND R TO #r.list + 1

HOW TO RETURN ffr n:
    SHARE r.list
    IF n > #r.list: EXTEND R TO n
    RETURN r.list[n]

HOW TO RETURN ffs n:
    SHARE s.list
    IF n > #s.list: EXTEND S TO n
    RETURN s.list[n]

WRITE "R[1..10]:"
FOR i IN {1..10}: WRITE ffr i
WRITE /

PUT {} IN thousand
FOR i IN {1..40}: INSERT ffr i IN thousand
FOR i IN {1..960}: INSERT ffs i IN thousand
IF thousand = {1..1000}:
    WRITE "R[1..40] + S[1..960] = [1..1000]"/

Output:

R[1..10]: 1 3 7 12 18 26 35 45 56 69
R[1..40] + S[1..960] = [1..1000]

Ada

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)

APL

Works with: Dyalog APL

:Class HFF
    :Field Private Shared RBuf←,1   
    ∇r←ffr n 
        :Access Public Shared
        r←n⊃RBuf←(⊢,⊃∘⌽+≢⊃(⍳1+⌈/)~⊢)⍣(0⌈n-≢RBuf)⊢RBuf
    ∇
    ∇s←ffs n;S 
        :Access Public Shared
        :Repeat
            S←((⍳1+⌈/)~⊢)RBuf
            :If n≤≢S ⋄ :Leave ⋄ :EndIf
            S←ffr 1+≢RBuf
        :EndRepeat
        s←n⊃S
    ∇
    ∇Task;th 
        :Access Public Shared
        ⎕←'R(1 .. 10):', ffr¨⍳10
        :If (⍳1000) ∧.∊ ⊂th←(ffr¨⍳40) ∪ (ffs¨⍳960)
            ⎕←'1..1000 ∊ (ffr 1..40) ∪ (ffs 1..960)'
        :Else
            ⎕←'Missing values: ', (⍳1000)~th
        :EndIf
    ∇
:EndClass

Output:

      HFF.Task
R(1 .. 10): 1 3 7 12 18 26 35 45 56 69
1..1000 ∊ (ffr 1..40) ∪ (ffs 1..960)

AutoHotkey

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,...

AWK

# Hofstadter Figure-Figure sequences
#
#    R(1) = 1; S(1) = 2;
#    R(n) = R(n-1) + S(n-1), n > 1
#    S(n) is the values not in R(n)

BEGIN {

    # start with the first two values of R and S to simplify finding S[n]:
    R[ 1 ] = 1;
    R[ 2 ] = 3;
    S[ 1 ] = 2;
    S[ 2 ] = 4;
    # maximum n we currently have of R and S
    rMax   = 2;
    sMax   = 2;

    # calculate and show the first 10 values of R:
    printf( "R[1..10]:" );
    for( n = 1; n < 11; n ++ )
    {
        printf( " %d", ffr( n ) );
    }
    printf( "\n" );
    # check that R[1..40] and S[1..960] contain the numbers 1..1000 once each
    # add the values of R[ 1..40 ] to the set V
    for( n = 1; n <= 40; n ++ )
    {
        V[ ffr( n ) ] ++;
    }
    # add the values of S[ 1..960 ] to the set V
    for( n = 1; n <= 960; n ++ )
    {
        V[ ffs( n ) ] ++;
    }
    # check all numbers are present and not duplicated
    ok = 1;
    for( n = 1; n <= 1000; n ++ )
    {
        if( ! ( n in V ) )
        {
            printf( "%d not present in R[1..40], S[1..960]\n", n );
            ok = 0;
        }
        else if( V[ n ] != 1 )
        {
            printf( "%d occurs %d times in R[1..40], S[1..960]\n", n, V[ n ] );
            ok = 0;
        }
    }
    if( ok )
    {
        printf( "R[1..40] and S[1..960] uniquely contain all 1..1000\n" );
    }

} # BEGIN

function ffr( n )
{
    # calculate R[n]
    if( ! ( n in R ) )
    {
        # we haven't calculated R[ n ] yet
        R[ n ]  = ffs( n - 1 );
        R[ n ] += ffr( n - 1 );
    }
return R[ n ];
} # ffr

function ffs( n )
{
    # calculate S[n]
    if( ! ( n in S ) )
    {
        # starting at the highest known R, calculate the next one and fill in the S values
        # continuing until we have enough S values
        do
        {
            R[ rMax + 1 ] = R[ rMax ] + S[ rMax ];
            for( sValue = R[ rMax ] + 1; sValue < R[ rMax + 1 ]; sValue ++ )
            {
                S[ sMax ++ ] = sValue;
            }
            rMax ++;
        }
        while( sMax < n );
    }
return S[ n ];
} # ffs

Output:

R[1..10]: 1 3 7 12 18 26 35 45 56 69
R[1..40] and S[1..960] uniquely contain all 1..1000

BBC BASIC

Works with: BBC BASIC for Windows

      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

#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#

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

C++

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

CLU

figfig = cluster is ffr, ffs
    rep = null
    ai = array[int]
    own R: ai := ai$[1]
    own S: ai := ai$[2]
    
    % Extend R and S until R(n) is known
    extend = proc (n: int)
        while n > ai$high(R) do
            next: int := ai$top(R) + S[ai$high(R)]
            ai$addh(R, next)
            while ai$top(S) < next-1 do
                ai$addh(S, ai$top(S)+1)
            end
            ai$addh(S, next+1)
        end
    end extend
    
    ffr = proc (n: int) returns (int)
        extend(n)
        return(R[n])
    end ffr
    
    ffs = proc (n: int) returns (int)
        while n > ai$high(S) do
            extend(ai$high(R) + 1)
        end
        return(S[n])
    end ffs
end figfig

start_up = proc ()
    ai = array[int]
    po: stream := stream$primary_output()
    
    % Print R[1..10]
    stream$puts(po, "R[1..10] =")
    for i: int in int$from_to(1,10) do
        stream$puts(po, " " || int$unparse(figfig$ffr(i)))
    end
    stream$putl(po, "")
    
    % Count the occurrences of 1..1000 in R[1..40] and S[1..960]
    occur: ai := ai$fill(1, 1000, 0)
    for i: int in int$from_to(1, 40) do
        occur[figfig$ffr(i)] := occur[figfig$ffr(i)] + 1
    end
    for i: int in int$from_to(1, 960) do
        occur[figfig$ffs(i)] := occur[figfig$ffs(i)] + 1
    end
    
    % See if they all occur exactly once
    begin
        for i: int in int$from_to(1, 1000) do
            if occur[i] ~= 1 then exit wrong(i) end
        end
        stream$putl(po, 
            "All numbers 1..1000 occur exactly once in R[1..40] U S[1..960].")
    end except when wrong(i: int):
        stream$putl(po, "Error: " ||
            int$unparse(i) || " occurs " || int$unparse(occur[i]) || " times.")
    end
end start_up

Output:

R[1..10] = 1 3 7 12 18 26 35 45 56 69
All numbers 1..1000 occur exactly once in R[1..40] U S[1..960].

Common 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")

Output:

First of R: (1 3 7 12 18 26 35 45 56 69)
Ok

Cowgol

include "cowgol.coh";
include "strings.coh";
include "malloc.coh";

# An uint16 is big enough to deal with the figures from the task,
# but it is good practice to allow it to be easily redefined.
typedef N is uint16;

# There is no extensible vector type included in the standard library,
# so it is necessary to define one. 
record VecR is
    len: intptr;
    alloc: intptr;
    data: [N];
end record;
    
typedef Vec is [VecR];

sub NewVec(): (v: Vec) is
    v := Alloc(@bytesof VecR) as Vec;
    MemZero(v as [uint8], @bytesof VecR);
    v.alloc := 256;
    v.data := Alloc(@bytesof N * 256) as [N];
    MemZero(v.data as [uint8], @bytesof N * 256);
end sub;

sub VecGet(v: Vec, i: intptr): (r: N) is
    if i >= v.len then
        print("index error\n");
        ExitWithError();
    end if;
    r := [v.data + i * @bytesof N];
end sub;

sub VecSet(v: Vec, i: intptr, n: N) is
    if i >= v.alloc then
        var newsize := v.alloc;
        while i >= newsize loop
            newsize := newsize + 256;
        end loop;
        var newbytes := newsize * @bytesof N;
        var oldbytes := v.alloc * @bytesof N;
        var newdata := Alloc(newbytes) as [N];
        MemCopy(v.data as [uint8], oldbytes, newdata as [uint8]);
        MemZero(newdata as [uint8] + oldbytes, newbytes - oldbytes);
        Free(v.data as [uint8]);
        v.data := newdata;
        v.alloc := newsize;
    end if;
    [v.data + i * @bytesof N] := n;
    if i >= v.len then
        v.len := i+1;
    end if;
end sub;
    
sub Last(v: Vec): (r: N) is r := VecGet(v, v.len-1); end sub;
sub Append(v: Vec, n: N) is VecSet(v, v.len, n); end sub;

# We also need to define a flag array, to avoid taking up 1K of memory
# for a thousand bit flags.
sub GetFlag(bitarr: [uint8], n: intptr): (s: uint8) is
    s := ([bitarr + (n >> 3)] >> (n as uint8 & 7)) & 1;
end sub;
sub SetFlag(bitarr: [uint8], n: intptr) is
    var p := bitarr + (n >> 3);
    var f: uint8 := 1;
    [p] := [p] | (f << (n as uint8 & 7));
end sub;
    
# Define and initialize vectors holding the R and S sequences
var R := NewVec(); Append(R, 1);
var S := NewVec(); Append(S, 2);

# Extend the sequences until R(n) is known.
sub Extend(n: intptr) is
    while n > R.len loop
        var newR := Last(R) + VecGet(S, R.len-1);
        Append(R, newR);
        while Last(S) < newR - 1 loop
            Append(S, Last(S) + 1);
        end loop;
        Append(S, newR + 1);
    end loop;
end sub;

# Get R
sub ffr(n: intptr): (r: N) is
    Extend(n);
    r := VecGet(R, n-1);
end sub;

# Get S
sub ffs(n: intptr): (s: N) is
    while n > S.len loop
        Extend(R.len + 1);
    end loop;
    s := VecGet(S, n-1);
end sub;

# Print the first 10 values of R.
print("R(1 .. 10): ");
var n: intptr := 1;
while n <= 10 loop
    print_i32(ffr(n) as uint32);
    print_char(' ');
    n := n + 1;
end loop;
print_nl();


print("Checking that (1 .. 1000) are in R(1 .. 40) U S(1 .. 960)...\n");
# Reserve 1000 bits to use as flags, and set them all to zero
var flags: uint8[1000 / 8];
MemZero(&flags[0], @bytesof flags);

# Set the flags corresponding to FFR(1 .. 40) and FFS(1 .. 960)
n := 1;
while n <= 40 loop 
    SetFlag(&flags[0], (ffr(n)-1) as intptr); 
    n := n + 1; 
end loop;

n := 1;
while n <= 960 loop
    SetFlag(&flags[0], (ffs(n)-1) as intptr);
    n := n + 1;
end loop;

# Check all flags
var ok: uint8 := 1;
n := 1;
while n <= 1000 loop
    if GetFlag(&flags[0], (n-1) as intptr) == 0 then
        print_i32(n as uint32);
        print(" not found!\n");
        ok := 0;
    end if;
    n := n + 1;
end loop;

if ok != 0 then
    print("All numbers 1 .. 1000 found!\n");
end if;

Output:

R(1 .. 10): 1 3 7 12 18 26 35 45 56 69
Checking that (1 .. 1000) are in R(1 .. 40) U S(1 .. 960)...
All numbers 1 .. 1000 found!

D

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

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;
}

EasyLang

Translation of: C

global rs[] ss[] .
procdecl RS_append . .
func R n .
   while n > len rs[]
      RS_append
   .
   return rs[n]
.
func S n .
   while n > len ss[]
      RS_append
   .
   return ss[n]
.
proc RS_append . .
   n = len rs[]
   r = R n + S n
   s = S len ss[]
   rs[] &= r
   repeat
      s += 1
      until s = r
      ss[] &= s
   .
   ss[] &= r + 1
.
rs[] = [ 1 ]
ss[] = [ 2 ]
write "R(1 .. 10): "
for i to 10
   write R i & " "
.
print ""
len seen[] 1000
for i to 40
   seen[R i] = 1
.
for i to 960
   seen[S i] = 1
.
for i to 1000
   if seen[i] = 0
      print i & " not seen"
      return
   .
.
print "first 1000 ok"

Output:

R(1 .. 10): 1 3 7 12 18 26 35 45 56 69 
first 1000 ok

EchoLisp

(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

>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

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

FreeBASIC

Translation of: BBC BASIC

function ffr( n as integer ) as integer
    if n = 1 then return 1
    dim as integer i, j, r=1, s=1, v(1 to 2*n+1)
    v(1) = 1
    for i = 2 to n
        for j = s to 2*n
            if v(j) = 0 then exit for
        next j
        v(j) = 1
        s = j
        r += s
        if r <= 2*n then v(r) = 1
    next i
    return r
end function

function ffs( n as integer ) as integer
    if n = 1 then return 2
    dim as integer i, j, r=1, s=2, v(1 to 2*n+1)
    for i = 1 to n
        for j = s to 2*n
            if v(j) = 0 then exit for
        next j
        v(j) = 1
        s = j
        r += s
        if r <= 2*n then v(r) = 1
    next i
    return s
end function

dim as integer i
print " R"," S"
print
for i = 1 to 10
    print ffr(i), ffs(i)
next i

dim as boolean found(1 to 1000), failed
for i = 1 to 40
    found(ffr(i)) = true
next i
for i = 1 to 960
    found(ffs(i)) = true
next i
for i = 1 to 1000
    if found(i) = false then failed = true
next i

if failed then print "Oh no!" else print "All integers from 1 to 1000 accounted for"

Output:

R             S

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

All integers from 1 to 1000 accounted for

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")
}

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

import Data.List (delete, sort)

-- Functions by Reinhard Zumkeller
ffr :: Int -> Int
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 :: Int -> Int
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 :: IO ()
main = do
  print $ ffr <$> [1 .. 10]
  let i1000 = sort (fmap ffr [1 .. 40] ++ fmap 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 :: [Int]
r = scanl (+) 1 s

s :: [Int]
s = 2 : 4 : tail (complement (tail r))
  where
    complement = concat . interval
    interval x = zipWith (\x y -> [succ x .. pred y]) x (tail x)

main :: IO ()
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

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

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

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

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

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.

jq

Works with: jq

Works with gojq, the Go implementation of jq

In this entry, the functions `ffr` and `ffs` are defined as per the task requirements, but neither is used. Instead, for efficiency, a function for extending the two sequences is defined. This function is then used to create generators, which are then harnessed to accomplish the specific tasks.

def init: {r: [0, 1], s: [0, 2] };

# input: {r,s}
# output: {r,s,emit} where .emit is either null or the next R and where either .r or .s on output has been extended.
# .emit is provided in case an unbounded stream of R values is desired.
def extend_ff:
  (.r|length) as $rn
  | if .s[$rn - 1]
    then .emit = .r[$rn - 1] + .s[$rn - 1]
    | .r[$rn] = .emit
    | reduce range( [.r[$rn-1], .s[-1]] | max + 1; .r[$rn] ) as $i (.; .s += [$i] )
    else .emit = null
    | .s += [.r[$rn - 1] + 1]
    end;

def ffr($n):
  first(init | while(true; extend_ff) | select(.r[$n])).r[$n] ;

def ffs($n):
  first(init | while(true; extend_ff) | select(.s[$n])).s[$n] ;

def task1($n):
  "The first \($n) values of R are:",
  (init | until( .r | length > $n; extend_ff) | .r[1:]) ;

def task2:
  "The result of checking that the first 40 values of R and the first 960 of S together cover the interval [1,1000] is:",
  ( init | until( (.r|length) > 40 and (.s|length) > 960; extend_ff)
   | (.r[1:41] + .s[1:961] | sort) == [range(1;1001)] ) ;

task1(10), task2

Output:

The first 10 values of R are:
[1,3,7,12,18,26,35,45,56,69]
The result of checking that the first 40 values of R and the first 960 of S together cover the interval [1,1000] is:
true

Julia

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

Translated from Java.

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

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

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

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): int =
  assert n > 0
  while n > cr.len: extendRS()
  cr[n - 1]

proc ffs(n: int): 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

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

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";
}

Phix

Initialising such that length(S)>length(F) simplified things significantly.

with javascript_semantics
sequence F = {1,3,7},
         S = {2,4,5,6}
integer fmax = 3 -- (ie F[3], ==7, already in S)
 
forward function ffs(integer n)
 
function ffr(integer n)
    integer l = length(F)
    while n>l do
        F &= F[l]+ffs(l)
        l += 1
    end while
    return F[n]
end function
 
function ffs(integer n)
    while n>length(S) do
        fmax += 1
        if fmax>length(F) then {} = ffr(fmax) end if
        S &= tagset(lim:=F[fmax]-1,start:=F[fmax-1]+1)
        -- ie/eg if fmax was 3, then F[2..3] being {3,7} 
        --       ==> tagset(lim:=6,start:=4), ie {4,5,6}.
    end while
    return S[n]
end function
 
{} = ffr(10)    -- (or collect one by one)
printf(1,"The first ten values of R: %v\n",{F[1..10]})
{} = ffr(40)    -- (not actually needed)
{} = ffs(960)
if sort(F[1..40]&S[1..960])=tagset(1000) then
    puts(1,"test passed\n")
else
    puts(1,"some error!\n")
end if

Output:

The first ten values of R: {1,3,7,12,18,26,35,45,56,69}
test passed

PicoLisp

(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

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

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

:- 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

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

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

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

Quackery

from, index, and end are defined at Loops/Increment loop index within loop body#Quackery.

As with the Phix solution, initialising to the first few elements simplified things significantly.

  [ ' [ 1 3 7 ]
    ' [ 2 4 5 6 ] ]      is initialise (       --> r s   )

  [ over size 1 -
    over swap peek
    dip [ over -1 peek ]
    + swap dip join
    over -2 split nip do
    temp put
    1 + from
      [ temp share
        index = iff
          end done
        index join ]
    temp release ]       is extend     (   r s --> r s   )

  [ temp put
    [ over size
      temp share < while
      extend again ]
    over
    temp take 1 - peek ] is ffr        ( r s n --> r s n )

  [ temp put
    [ dup size
      temp share < while
      extend again ]
    dup
    temp take 1 - peek ] is ffs        ( r s n --> r s n )

  initialise
  say "R(1)..R(10): "
  10 times
    [ i^ 1+ ffr echo sp ]
  cr cr
  960 ffs drop
  960 split drop
  dip [ 40 split drop ]
  join sort
  [] 1000 times
    [ i^ 1+ join ]
  !=
  say "All integers from 1 to 1000"
  if say " not"
  say " found once and only once."

Output:

R(1)..R(10): 1 3 7 12 18 26 35 45 56 69 

All integers from 1 to 1000 found once and only once.

R

Global variables aren't idiomatic R, but this is otherwise an ideal task for the language. Comments aside, this is easily one of the shortest solutions on this page. This is mostly due to how R treats most things as a vector. For example, rValues starts out as the number 1, but repeatedly has new values appended to it without much ceremony.

rValues <- 1
sValues <- 2
ffr <- function(n)
{
  if(!is.na(rValues[n])) rValues[n] else (rValues[n] <<- ffr(n-1) + ffs(n-1))
}

#In theory, generating S requires computing ALL values not in R.
#That would be infinitely many values.
#However, to generate S(n) we only need to observe that its value cannot exceed R(n)+1.
ffs <- function(n)
{
  if(!is.na(sValues[n])) sValues[n] else (sValues[n] <<- setdiff(seq_len(1 + ffr(n)), rValues)[n])
}

#Task 1
invisible(ffr(10))
print(rValues)

#Task 2
#If we try to call ffs(960) directly, R will complain about the stack being too big.
#Calling ffs(500) first solves this problem.
invisible(ffs(500))
invisible(ffs(960))
#In R, "the first 40 values of ffr plus the first 960 values of ffs" can easily be misread.
#rValues[1:40]+sValues[1:960] is valid R code. It will duplicate the first 40 rValues 23
#times, append them to the original, and add that vector to the first 960 sValues.
This gives an output of length 960, which clearly cannot contain 1000 different values.
#Presumably, the task wants us to append rValues[1:40] and sValues[1:960].
print(table(c(rValues[1:40], sValues[1:960])))

Output:

> print(rValues)
 [1]  1  3  7 12 18 26 35 45 56 69
> print(table(c(rValues[1:40], sValues[1:960])))

   1    2    3    4    5    6    7    8    9   10   11   12   13   14   15   16   17   18   19 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  20   21   22   23   24   25   26   27   28   29   30   31   32   33   34   35   36   37   38 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  39   40   41   42   43   44   45   46   47   48   49   50   51   52   53   54   55   56   57 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  58   59   60   61   62   63   64   65   66   67   68   69   70   71   72   73   74   75   76 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  77   78   79   80   81   82   83   84   85   86   87   88   89   90   91   92   93   94   95 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
  96   97   98   99  100  101  102  103  104  105  106  107  108  109  110  111  112  113  114 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 115  116  117  118  119  120  121  122  123  124  125  126  127  128  129  130  131  132  133 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 134  135  136  137  138  139  140  141  142  143  144  145  146  147  148  149  150  151  152 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 153  154  155  156  157  158  159  160  161  162  163  164  165  166  167  168  169  170  171 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 172  173  174  175  176  177  178  179  180  181  182  183  184  185  186  187  188  189  190 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 191  192  193  194  195  196  197  198  199  200  201  202  203  204  205  206  207  208  209 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 210  211  212  213  214  215  216  217  218  219  220  221  222  223  224  225  226  227  228 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 229  230  231  232  233  234  235  236  237  238  239  240  241  242  243  244  245  246  247 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 248  249  250  251  252  253  254  255  256  257  258  259  260  261  262  263  264  265  266 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 267  268  269  270  271  272  273  274  275  276  277  278  279  280  281  282  283  284  285 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 286  287  288  289  290  291  292  293  294  295  296  297  298  299  300  301  302  303  304 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 305  306  307  308  309  310  311  312  313  314  315  316  317  318  319  320  321  322  323 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 324  325  326  327  328  329  330  331  332  333  334  335  336  337  338  339  340  341  342 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 343  344  345  346  347  348  349  350  351  352  353  354  355  356  357  358  359  360  361 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 362  363  364  365  366  367  368  369  370  371  372  373  374  375  376  377  378  379  380 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 381  382  383  384  385  386  387  388  389  390  391  392  393  394  395  396  397  398  399 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 400  401  402  403  404  405  406  407  408  409  410  411  412  413  414  415  416  417  418 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 419  420  421  422  423  424  425  426  427  428  429  430  431  432  433  434  435  436  437 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 438  439  440  441  442  443  444  445  446  447  448  449  450  451  452  453  454  455  456 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 457  458  459  460  461  462  463  464  465  466  467  468  469  470  471  472  473  474  475 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 476  477  478  479  480  481  482  483  484  485  486  487  488  489  490  491  492  493  494 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 495  496  497  498  499  500  501  502  503  504  505  506  507  508  509  510  511  512  513 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 514  515  516  517  518  519  520  521  522  523  524  525  526  527  528  529  530  531  532 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 533  534  535  536  537  538  539  540  541  542  543  544  545  546  547  548  549  550  551 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 552  553  554  555  556  557  558  559  560  561  562  563  564  565  566  567  568  569  570 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 571  572  573  574  575  576  577  578  579  580  581  582  583  584  585  586  587  588  589 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 590  591  592  593  594  595  596  597  598  599  600  601  602  603  604  605  606  607  608 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 609  610  611  612  613  614  615  616  617  618  619  620  621  622  623  624  625  626  627 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 628  629  630  631  632  633  634  635  636  637  638  639  640  641  642  643  644  645  646 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 647  648  649  650  651  652  653  654  655  656  657  658  659  660  661  662  663  664  665 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 666  667  668  669  670  671  672  673  674  675  676  677  678  679  680  681  682  683  684 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 685  686  687  688  689  690  691  692  693  694  695  696  697  698  699  700  701  702  703 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 704  705  706  707  708  709  710  711  712  713  714  715  716  717  718  719  720  721  722 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 723  724  725  726  727  728  729  730  731  732  733  734  735  736  737  738  739  740  741 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 742  743  744  745  746  747  748  749  750  751  752  753  754  755  756  757  758  759  760 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 761  762  763  764  765  766  767  768  769  770  771  772  773  774  775  776  777  778  779 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 780  781  782  783  784  785  786  787  788  789  790  791  792  793  794  795  796  797  798 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 799  800  801  802  803  804  805  806  807  808  809  810  811  812  813  814  815  816  817 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 818  819  820  821  822  823  824  825  826  827  828  829  830  831  832  833  834  835  836 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 837  838  839  840  841  842  843  844  845  846  847  848  849  850  851  852  853  854  855 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 856  857  858  859  860  861  862  863  864  865  866  867  868  869  870  871  872  873  874 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 875  876  877  878  879  880  881  882  883  884  885  886  887  888  889  890  891  892  893 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 894  895  896  897  898  899  900  901  902  903  904  905  906  907  908  909  910  911  912 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 913  914  915  916  917  918  919  920  921  922  923  924  925  926  927  928  929  930  931 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 932  933  934  935  936  937  938  939  940  941  942  943  944  945  946  947  948  949  950 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 951  952  953  954  955  956  957  958  959  960  961  962  963  964  965  966  967  968  969 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 970  971  972  973  974  975  976  977  978  979  980  981  982  983  984  985  986  987  988 
   1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1    1 
 989  990  991  992  993  994  995  996  997  998  999 1000 
   1    1    1    1    1    1    1    1    1    1    1    1

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/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

Raku

(formerly Perl 6)

Works with: Rakudo version 2018.03

my %r = 1 => 1;
my %s = 1 => 2;

sub ffr ($n) { %r{$n} //= ffr($n - 1) + ffs($n - 1) }
sub ffs ($n) { %s{$n} //= (grep none(map &ffr, 1..$n), max(%s.values)+1..*)[0] }

my @ffr = map &ffr, 1..*;
my @ffs = map &ffs, 1..*;

say @ffr[^10];
say "Rawks!" if 1...1000 eqv sort |@ffr[^40], |@ffs[^960];

Output:

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

REXX

version 1

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.

This REXX version is over 15,000% faster than REXX version 2.

/*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

/* 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

Ring

# Project : Hofstadter Figure-Figure sequences

hofr = list(20)
hofr[1] = 1
hofs = []
add(hofs,2)
for n = 1 to 10
      hofr[n+1] = hofr[n] + hofs[n]
      if n = 1
         add(hofs,4)
      else
         for p = hofr[n] + 1 to hofr[n+1] - 1
               if p != hofs[n]
                  add(hofs,p)
               ok
         next
      ok
next
see "First 10 values of R:" + nl
showarray(hofr)
see "First 10 values of S:" + nl
showarray(hofs)

func showarray(vect)
         svect = ""
        for n = 1 to 10
              svect = svect + vect[n] + " "
        next
        svect = left(svect, len(svect) - 1)
        see svect + nl

Output:

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

RPL

Works with: Halcyon Calc version 4.2.8

RPL code

Comment

 ≪ { 1 3 } 'R' STO { 2 } 'S' STO 
≫ 'INITFF' STO

≪ 
   S DUP SIZE GET 
   DO 1 + UNTIL R OVER POS NOT END 
   S OVER + 'S' STO 
   R DUP SIZE GET + R SWAP + 'R' STO 
≫ 'NXTFF' STO

≪ 
   WHILE R SIZE OVER < REPEAT NXTFF END
   R SWAP GET
≫ 'FFR' STO

≪ 
   WHILE S SIZE OVER < REPEAT NXTFF END
   S SWAP GET
≫ 'FFS' STO

≪ INITFF
   40 FFR DROP R 
   960 FFS DROP S + 
   1 SF 1 1000 FOR j 
      IF DUP j POS NOT THEN 1 CF END NEXT DROP 
   1 FS? "Passed" "Failed" IFTE
≫ 'TASK4' STO

INITFF ( -- ) 
Initialize R(1..2) and S(1)

NXTFF ( -- ) 
n = last stored item of S()
n += 1 until n not in R()
append n to S()
append (n + last item of R()) to R()


FFR ( n -- R(n) ) 
if R(n) not stored, develop R()
Get R(n)


FFS ( n -- S(n) ) 
if S(n) not stored, develop S()
Get S(n)


TASK4 ( -- "Result" ) 
Get R(40) and put R(1..40) in stack
Get S(960), append S(1..960) to R(1..40)
set flag ; for j=1 to 1000
  if j not in the merged list then clear flag
Flag is still set iff all 1..1000 were in list once

Input:

10 FFR DROP R
TASK4

Output:

2: { 1 3 7 12 18 26 35 45 56 69 }
1: "Passed"

Ruby

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

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

Rust

use std::collections::HashMap;

struct Hffs {
    sequence_r: HashMap<usize, usize>,
    sequence_s: HashMap<usize, usize>,
}

impl Hffs {
    fn new() -> Hffs {
        Hffs {
            sequence_r: HashMap::new(),
            sequence_s: HashMap::new(),
        }
    }
    fn ffr(&mut self, n: usize) -> usize {
        // first try the cache
        let new_r = if let Some(result) = self.sequence_r.get(&n) {
            *result
        } else if n == 0 {
            1
        } else {
            // call recursively
            self.ffr(n - 1) + self.ffs(n - 1)
        };

        // insert into the cache and return value
        *self.sequence_r.entry(n).or_insert(new_r)
    }

    fn ffs(&mut self, n: usize) -> usize {
        // first try the cache
        let new_s = if let Some(result) = self.sequence_s.get(&n) {
            *result
        } else if n == 0 {
            2
        } else {
            let lower = self.ffs(n - 1) + 1_usize;
            let upper = self.ffr(n) + 1_usize;
            let mut min_s: usize = 0;
            // find next available S
            for i in lower..=upper {
                if !self.sequence_r.values().any(|&val| val == i) {
                    min_s = i;
                    break;
                }
            }
            min_s
        };

        // insert into the cache and return value
        *self.sequence_s.entry(n).or_insert(new_s)
    }
}

impl Default for Hffs {
    fn default() -> Self {
        Self::new()
    }
}
fn main() {
    let mut hof = Hffs::new();

    for i in 0..10 {
        println!("H:{} -> R: {}, S: {}", i, hof.ffr(i), hof.ffs(i));
    }

    let r40 = (0..40).map(|i| hof.ffr(i)).collect::<Vec<_>>();
    let mut s960 = (0..960).map(|i| hof.ffs(i)).collect::<Vec<_>>();

    s960.extend(&r40);
    s960.sort_unstable();
    let f1000 = (1_usize..=1000).collect::<Vec<_>>();

    assert_eq!(f1000, s960, "Does NOT match");
}

Output:

H:0 -> R: 1, S: 2
H:1 -> R: 3, S: 4
H:2 -> R: 7, S: 5
H:3 -> R: 12, S: 6
H:4 -> R: 18, S: 8
H:5 -> R: 26, S: 9
H:6 -> R: 35, S: 10
H:7 -> R: 45, S: 11
H:8 -> R: 56, S: 13
H:9 -> R: 69, S: 14

Scala

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

SETL

program figure_figure;
    init R := [1], S := [2];

    print("R(1..10):", [ffr(n) : n in [1..10]]);
    print("R(1..40) + S(1..960) = {1..1000}:",
        {ffr(n) : n in {1..40}} + {ffs(n) : n in {1..960}} = {1..1000});

    proc ffr(n);
        loop while n > #R do
            nextR := R(#R) + S(#R);
            S +:= [S(#S)+1 .. nextR-1];
            R with:= nextR;
            S with:= nextR + 1;
        end loop;
        return R(n);
    end proc;

    proc ffs(n);
        loop while n > #S do
            ffr(#R + 1);
        end loop;
        return S(n);
    end proc;
end program;

Output:

R(1..10): [1 3 7 12 18 26 35 45 56 69]
R(1..40) + S(1..960) = {1..1000}: #T

Sidef

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

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

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

VBA

Private Function ffr(n As Long) As Long
    Dim R As New Collection
    Dim S As New Collection
    R.Add 1
    S.Add 2
    'return R(n)
    For i = 2 To n
        R.Add R(i - 1) + S(i - 1)
        For j = S(S.Count) + 1 To R(i) - 1
            S.Add j
        Next j
        For j = R(i) + 1 To R(i) + S(i - 1)
            S.Add j
        Next j
    Next i
    ffr = R(n)
    Set R = Nothing
    Set S = Nothing
End Function
Private Function ffs(n As Long) As Long
    Dim R As New Collection
    Dim S As New Collection
    R.Add 1
    S.Add 2
    'return S(n)
    For i = 2 To n
        R.Add R(i - 1) + S(i - 1)
        For j = S(S.Count) + 1 To R(i) - 1
            S.Add j
        Next j
        For j = R(i) + 1 To R(i) + S(i - 1)
            S.Add j
        Next j
        If S.Count >= n Then Exit For
    Next i
    ffs = S(n)
    Set R = Nothing
    Set S = Nothing
End Function
Public Sub main()
    Dim i As Long
    Debug.Print "The first ten values of R are:"
    For i = 1 To 10
        Debug.Print ffr(i);
    Next i
    Debug.Print
    Dim x As New Collection
    For i = 1 To 1000
        x.Add i, CStr(i)
    Next i
    For i = 1 To 40
        x.Remove CStr(ffr(i))
    Next i
    For i = 1 To 960
        x.Remove CStr(ffs(i))
    Next i
    Debug.Print "The first 40 values of ffr plus the first 960 values of ffs "
    Debug.Print "include all the integers from 1 to 1000 exactly once is "; Format(x.Count = 0)
End Sub

Output:

The first ten values of R are:
 1  3  7  12  18  26  35  45  56  69 
The first 40 values of ffr plus the first 960 values of ffs 
include all the integers from 1 to 1000 exactly once is True

VBScript

'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!!!

Wren

Translation of: Go

var r = [0, 1]
var s = [0, 2]

var ffr = Fn.new { |n|
    while (r.count <= n) {
        var nrk = r.count - 1         // last n for which r[n] is known
        var rNxt = r[nrk] + s[nrk]    // r[nrk+1]
        r.add(rNxt)                   // extend r by one element
        for (sn in r[nrk]+2...rNxt) {
            s.add(sn)                 // extend sequence s up to rNxt
        }
        s.add(rNxt + 1)               // extend sequence s one past rNxt
    }
    return r[n]
}

var ffs = Fn.new { |n|
    while (s.count <= n) ffr.call(r.count)
    return s[n]
}

System.print("The first 10 values of R are:")
for (i in 1..10) System.write(" %(ffr.call(i))")
System.print()
var present = List.filled(1001, false)
for (i in 1..40)  present[ffr.call(i)] = true
for (i in 1..960) present[ffs.call(i)] = true
var allPresent = present.skip(1).all { |i| i == true }
System.print("\nThe first 40 values of ffr plus the first 960 values of ffs")
System.print("includes all integers from 1 to 1000 exactly once is %(allPresent).")

Output:

The first 10 values of R are:
 1 3 7 12 18 26 35 45 56 69

The first 40 values of ffr plus the first 960 values of ffs
includes all integers from 1 to 1000 exactly once is true.

zkl

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].pump(List))  // [1..n].pump(List)-->(1,2,3...)
   .println("<-- should be True");

Output:

True<-- should be True

Latest revision as of 18:53, 20 April 2024

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MATLAB / Octave

Constraint Handling Rules

Alternative

Using cyclic iterators

version 1

Version 2 from PL/I

Using cyclic iterators