Hofstadter Figure-Figure sequences: Difference between revisions
rearranges in order of the language. |
Kotlin version |
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=={{header| |
=={{header|Kotlin}} == |
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Translated from Java. |
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<lang scala>package hofstadter |
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fun ffr(n: Int) = get(n, 0)[n - 1] |
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fun ffs(n: Int) = get(0, n)[n - 1] |
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internal fun get(rlistSize: Int, slistSize: Int): List<Int> { |
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val rlist = arrayListOf(1, 3, 7) |
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val slist = arrayListOf(2, 4, 5, 6) |
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val list = if (rlistSize > 0) rlist else slist |
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val targetSize = if (rlistSize > 0) rlistSize else slistSize |
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while (list.size() > targetSize) |
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list.remove(list.size() - 1) |
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while (list.size() < targetSize) { |
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val lastIndex = rlist.lastIndex |
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val lastr = rlist[lastIndex] |
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val r = lastr + slist[lastIndex] |
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rlist += r |
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var s = lastr + 1 |
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while (s < r && list.size() < targetSize) |
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slist += s++ |
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} |
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return list |
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} |
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fun main(args: Array<String>) { |
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print("R():") |
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1..10 forEach { print(" " + ffr(it)) } |
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println() |
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val first40R = 1..40 map { ffr(it) } |
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val first960S = 1..960 map { ffs(it) } |
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for (n in 1..1000) |
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if (n in first40R == n in first960S) |
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println("Integer $n either in both or neither set") |
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println("Done") |
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}</lang> |
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=={{header|Mathematica}} == |
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1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively. |
1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively. |
Revision as of 19:26, 10 November 2015
You are encouraged to solve this task according to the task description, using any language you may know.
These two sequences of positive integers are defined as:
- The sequence is further defined as the sequence of positive integers not present in .
Sequence R starts: 1, 3, 7, 12, 18, ...
Sequence S starts: 2, 4, 5, 6, 8, ...
Task:
- 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.
Ada
Specifying a package providing the functions FFR and FFS: <lang Ada>package Hofstadter_Figure_Figure is
function FFR(P: Positive) return Positive;
function FFS(P: Positive) return Positive;
end Hofstadter_Figure_Figure;</lang>
The implementation of the package internally uses functions which generate an array of Figures or Spaces: <lang Ada>package body Hofstadter_Figure_Figure is
type Positive_Array is array (Positive range <>) of Positive;
function FFR(P: Positive) return Positive_Array is Figures: Positive_Array(1 .. P+1); Space: Positive := 2; Space_Index: Positive := 2; begin Figures(1) := 1; for I in 2 .. P loop Figures(I) := Figures(I-1) + Space; Space := Space+1; while Space = Figures(Space_Index) loop Space := Space + 1; Space_Index := Space_Index + 1; end loop; end loop; return Figures(1 .. P); end FFR;
function FFR(P: Positive) return Positive is Figures: Positive_Array(1 .. P) := FFR(P); begin return Figures(P); end FFR;
function FFS(P: Positive) return Positive_Array is Spaces: Positive_Array(1 .. P); Figures: Positive_Array := FFR(P+1); J: Positive := 1; K: Positive := 1; begin for I in Spaces'Range loop while J = Figures(K) loop J := J + 1; K := K + 1; end loop; Spaces(I) := J; J := J + 1; end loop; return Spaces; end FFS;
function FFS(P: Positive) return Positive is Spaces: Positive_Array := FFS(P); begin return Spaces(P); end FFS;
end Hofstadter_Figure_Figure;</lang>
Finally, a test program for the package, solving the task at hand: <lang Ada>with Ada.Text_IO, Hofstadter_Figure_Figure;
procedure Test_HSS is
use Hofstadter_Figure_Figure;
A: array(1 .. 1000) of Boolean := (others => False); J: Positive;
begin
for I in 1 .. 10 loop Ada.Text_IO.Put(Integer'Image(FFR(I))); end loop; Ada.Text_IO.New_Line;
for I in 1 .. 40 loop J := FFR(I); if A(J) then raise Program_Error with Positive'Image(J) & " used twice"; end if; A(J) := True; end loop;
for I in 1 .. 960 loop J := FFS(I); if A(J) then raise Program_Error with Positive'Image(J) & " used twice"; end if; A(J) := True; end loop;
for I in A'Range loop if not A(I) then raise Program_Error with Positive'Image(I) & " unused"; end if; end loop; Ada.Text_IO.Put_Line("Test Passed: No overlap between FFR(I) and FFS(J)");
exception
when Program_Error => Ada.Text_IO.Put_Line("Test Failed"); raise;
end Test_HSS;</lang>
The output of the test program: <lang> 1 3 7 12 18 26 35 45 56 69 Test Passed: No overlap between FFR(I) and FFS(J)</lang>
AutoHotkey
<lang 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 }</lang> Examples:<lang AutoHotkey>Loop MsgBox, 262144, , % "R(" A_Index ") = " R(A_Index) "`nS(" A_Index ") = " S(A_Index)</lang>
Outputs:
R(1) = 1, 3, 7, 12, 18, 26, 35,... S(1) = 2, 4, 5, 6, 8, 9, 10,...
BBC BASIC
<lang bbcbasic> PRINT "First 10 values of R:"
FOR i% = 1 TO 10 : PRINT ;FNffr(i%) " "; : NEXT : PRINT PRINT "First 10 values of S:" FOR i% = 1 TO 10 : PRINT ;FNffs(i%) " "; : NEXT : PRINT PRINT "Checking for first 1000 integers:" r% = 1 : s% = 1 ffr% = FNffr(r%) ffs% = FNffs(s%) FOR wanted% = 1 TO 1000 CASE TRUE OF WHEN wanted% = ffr% : r% += 1 : ffr% = FNffr(r%) WHEN wanted% = ffs% : s% += 1 : ffs% = FNffs(s%) OTHERWISE: EXIT FOR ENDCASE NEXT IF r% = 41 AND s% = 961 PRINT "Test passed" ELSE PRINT "Test failed" END DEF FNffr(N%) LOCAL I%, J%, R%, S%, V% DIM V% LOCAL 2*N%+1 V%?1 = 1 IF N% = 1 THEN = 1 R% = 1 S% = 2 FOR I% = 2 TO N% FOR J% = S% TO 2*N% IF V%?J% = 0 EXIT FOR NEXT V%?J% = 1 S% = J% R% += S% IF R% <= 2*N% V%?R% = 1 NEXT I% = R% DEF FNffs(N%) LOCAL I%, J%, R%, S%, V% DIM V% LOCAL 2*N%+1 V%?1 = 1 IF N% = 1 THEN = 2 R% = 1 S% = 2 FOR I% = 1 TO N% FOR J% = S% TO 2*N% IF V%?J% = 0 EXIT FOR NEXT V%?J% = 1 S% = J% R% += S% IF R% <= 2*N% V%?R% = 1 NEXT I% = S%</lang>
First 10 values of R: 1 3 7 12 18 26 35 45 56 69 First 10 values of S: 2 4 5 6 8 9 10 11 13 14 Checking for first 1000 integers: Test passed
C
<lang 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; }</lang>
C#
Creates an IEnumerable for R and S and uses those to complete the task <lang Csharp>using System; using System.Collections.Generic; using System.Linq;
namespace HofstadterFigureFigure { class HofstadterFigureFigure { readonly List<int> _r = new List<int>() {1}; readonly List<int> _s = new List<int>();
public IEnumerable<int> R() { int iR = 0; while (true) { if (iR >= _r.Count) { Advance(); } yield return _r[iR++]; } }
public IEnumerable<int> S() { int iS = 0; while (true) { if (iS >= _s.Count) { Advance(); } yield return _s[iS++]; } }
private void Advance() { int rCount = _r.Count; int oldR = _r[rCount - 1]; int sVal;
// Take care of first two cases specially since S won't be larger than R at that point switch (rCount) { case 1: sVal = 2; break; case 2: sVal = 4; break; default: sVal = _s[rCount - 1]; break; } _r.Add(_r[rCount - 1] + sVal); int newR = _r[rCount]; for (int iS = oldR + 1; iS < newR; iS++) { _s.Add(iS); } } }
class Program { static void Main() { var hff = new HofstadterFigureFigure(); var rs = hff.R(); var arr = rs.Take(40).ToList();
foreach(var v in arr.Take(10)) { Console.WriteLine("{0}", v); }
var hs = new HashSet<int>(arr); hs.UnionWith(hff.S().Take(960)); Console.WriteLine(hs.Count == 1000 ? "Verified" : "Oops! Something's wrong!"); } } } </lang> Output:
1 3 7 12 18 26 35 45 56 69 Verified
Common Lisp
<lang lisp>;;; equally doable with a list (flet ((seq (i) (make-array 1 :element-type 'integer :initial-element i :fill-pointer 1 :adjustable t)))
(let ((rr (seq 1)) (ss (seq 2))) (labels ((extend-r ()
(let* ((l (1- (length rr))) (r (+ (aref rr l) (aref ss l))) (s (elt ss (1- (length ss))))) (vector-push-extend r rr) (loop while (<= s r) do (if (/= (incf s) r) (vector-push-extend s ss))))))
(defun seq-r (n)
(loop while (> n (length rr)) do (extend-r)) (elt rr (1- n)))
(defun seq-s (n)
(loop while (> n (length ss)) do (extend-r)) (elt ss (1- n))))))
(defun take (f n)
(loop for x from 1 to n collect (funcall f x)))
(format t "First of R: ~a~%" (take #'seq-r 10))
(mapl (lambda (l) (if (and (cdr l) (/= (1+ (car l)) (cadr l))) (error "not in sequence")))
(sort (append (take #'seq-r 40)
(take #'seq-s 960)) #'<)) (princ "Ok")</lang>output<lang>First of R: (1 3 7 12 18 26 35 45 56 69) Ok</lang>
D
<lang d>int delegate(in int) nothrow ffr, ffs;
nothrow static this() {
auto r = [0, 1], s = [0, 2];
ffr = (in int n) nothrow { while (r.length <= n) { immutable int nrk = r.length - 1; immutable int rNext = r[nrk] + s[nrk]; r ~= rNext; foreach (immutable sn; r[nrk] + 2 .. rNext) s ~= sn; s ~= rNext + 1; } return r[n]; };
ffs = (in int n) nothrow { while (s.length <= n) ffr(r.length); return s[n]; };
}
void main() {
import std.stdio, std.array, std.range, std.algorithm;
iota(1, 11).map!ffr.writeln; auto t = iota(1, 41).map!ffr.chain(iota(1, 961).map!ffs); t.array.sort().equal(iota(1, 1001)).writeln;
}</lang>
- Output:
[1, 3, 7, 12, 18, 26, 35, 45, 56, 69] true
Alternative version
(Same output) <lang d>import std.stdio, std.array, std.range, std.algorithm;
struct ffr {
static r = [int.min, 1];
static int opCall(in int n) nothrow { assert(n > 0); if (n < r.length) { return r[n]; } else { immutable int ffr_n_1 = ffr(n - 1); immutable int lastr = r[$ - 1]; // Extend s up to, and one past, last r. ffs.s ~= iota(ffs.s[$ - 1] + 1, lastr).array; if (ffs.s[$ - 1] < lastr) ffs.s ~= lastr + 1; // Access s[n - 1] temporarily extending s if necessary. immutable size_t len_s = ffs.s.length; immutable int ffs_n_1 = (len_s > n) ? ffs.s[n - 1] : (n - len_s) + ffs.s[$ - 1]; immutable int ans = ffr_n_1 + ffs_n_1; r ~= ans; return ans; } }
}
struct ffs {
static s = [int.min, 2];
static int opCall(in int n) nothrow { assert(n > 0); if (n < s.length) { return s[n]; } else { foreach (immutable i; ffr.r.length .. n + 2) { ffr(i); if (s.length > n) return s[n]; } assert(false, "Whoops!"); } }
}
void main() {
iota(1, 11).map!ffr.writeln; auto t = iota(1, 41).map!ffr.chain(iota(1, 961).map!ffs); t.array.sort().equal(iota(1, 1001)).writeln;
}</lang>
EchoLisp
<lang scheme> (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)) </lang>
- Output:
<lang scheme> (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
</lang>
Euler Math Toolbox
<lang Euler Math Toolbox> >function RSstep (r,s) ... $ n=cols(r); $ r=r|(r[n]+s[n]); $ s=s|(max(s[n]+1,r[n]+1):r[n+1]-1); $ return {r,s}; $ endfunction >function RS (n) ... $ if n==1 then return {[1],[2]}; endif; $ if n==2 then return {[1,3],[2]}; endif; $ r=[1,3]; s=[2,4]; $ loop 3 to n; {r,s}=RSstep(r,s); end; $ return {r,s}; $ endfunction >{r,s}=RS(10); >r
[ 1 3 7 12 18 26 35 45 56 69 ]
>{r,s}=RS(50); >all(sort(r[1:40]|s[1:960])==(1:1000))
1
</lang>
Factor
We keep lists S and R, and increment them when necessary. <lang factor>SYMBOL: S V{ 2 } S set SYMBOL: R V{ 1 } R set
- next ( s r -- news newr )
2dup [ last ] bi@ + suffix dup [
[ dup last 1 + dup ] dip member? [ 1 + ] when suffix
] dip ;
- inc-SR ( n -- )
dup 0 <= [ drop ] [ [ S get R get ] dip [ next ] times R set S set ] if ;
- ffs ( n -- S(n) )
dup S get length - inc-SR 1 - S get nth ;
- ffr ( n -- R(n) )
dup R get length - inc-SR 1 - R get nth ;</lang>
<lang factor>( scratchpad ) 10 iota [ 1 + ffr ] map . { 1 3 7 12 18 26 35 45 56 69 } ( scratchpad ) 40 iota [ 1 + ffr ] map 960 iota [ 1 + ffs ] map append 1000 iota 1 v+n set= . t</lang>
Go
<lang go>package main
import "fmt"
var ffr, ffs func(int) int
// The point of the init function is to encapsulate r and s. If you are // not concerned about that or do not want that, r and s can be variables at // package level and ffr and ffs can be ordinary functions at package level. func init() {
// task 1, 2 r := []int{0, 1} s := []int{0, 2}
ffr = func(n int) int { for len(r) <= n { nrk := len(r) - 1 // last n for which r(n) is known rNxt := r[nrk] + s[nrk] // next value of r: r(nrk+1) r = append(r, rNxt) // extend sequence r by one element for sn := r[nrk] + 2; sn < rNxt; sn++ { s = append(s, sn) // extend sequence s up to rNext } s = append(s, rNxt+1) // extend sequence s one past rNext } return r[n] }
ffs = func(n int) int { for len(s) <= n { ffr(len(r)) } return s[n] }
}
func main() {
// task 3 for n := 1; n <= 10; n++ { fmt.Printf("r(%d): %d\n", n, ffr(n)) } // task 4 var found [1001]int for n := 1; n <= 40; n++ { found[ffr(n)]++ } for n := 1; n <= 960; n++ { found[ffs(n)]++ } for i := 1; i <= 1000; i++ { if found[i] != 1 { fmt.Println("task 4: FAIL") return } } fmt.Println("task 4: PASS")
}</lang> Output:
r(1): 1 r(2): 3 r(3): 7 r(4): 12 r(5): 18 r(6): 26 r(7): 35 r(8): 45 r(9): 56 r(10): 69 task 4: PASS
The following defines two mutually recursive generators without caching results. Each generator will end up dragging a tree of closures behind it, but due to the odd nature of the two series' growth pattern, it's still a heck of a lot faster than the above method when producing either series in sequence. <lang go>package main import "fmt"
type xint int64 func R() (func() (xint)) { r, s := xint(0), func() (xint) (nil) return func() (xint) { switch { case r < 1: r = 1 case r < 3: r = 3 default: if s == nil { s = S() s() } r += s() } if r < 0 { panic("r overflow") } return r } }
func S() (func() (xint)) { s, r1, r := xint(0), xint(0), func() (xint) (nil) return func() (xint) { if s < 2 { s = 2 } else { if r == nil { r = R() r() r1 = r() } s++ if s > r1 { r1 = r() } if s == r1 { s++ } } if s < 0 { panic("s overflow") } return s } }
func main() { r, sum := R(), xint(0) for i := 0; i < 10000000; i++ { sum += r() } fmt.Println(sum) }</lang>
Haskell
<lang haskell>import Data.List (delete, sort)
-- Functions by Reinhard Zumkeller ffr n = rl !! (n - 1) where
rl = 1 : fig 1 [2 ..] fig n (x : xs) = n' : fig n' (delete n' xs) where n' = n + x
ffs n = rl !! n where
rl = 2 : figDiff 1 [2 ..] figDiff n (x : xs) = x : figDiff n' (delete n' xs) where n' = n + x
main = do
print $ map ffr [1 .. 10] let i1000 = sort (map ffr [1 .. 40] ++ map ffs [1 .. 960]) print (i1000 == [1 .. 1000])</lang>
Output:
[1,3,7,12,18,26,35,45,56,69] True
Defining R and S literally: <lang haskell>import Data.List (sort)
r = scanl (+) 1 s s = 2:4:tail (compliment (tail r)) where compliment = concat.interval interval x = zipWith (\x y -> [x+1..y-1]) x (tail x)
main = do putStr "R: "; print (take 10 r) putStr "S: "; print (take 10 s) putStr "test 1000: "; print ([1..1000] == sort ((take 40 r) ++ (take 960 s)))</lang> output:
R: [1,3,7,12,18,26,35,45,56,69] S: [2,4,5,6,8,9,10,11,13,14] test 1000: True
Icon and Unicon
<lang Icon>link printf,ximage
procedure main()
printf("Hofstader ff sequences R(n:= 1 to %d)\n",N := 10) every printf("R(%d)=%d\n",n := 1 to N,ffr(n))
L := list(N := 1000,0) zero := dup := oob := 0 every n := 1 to (RN := 40) do if not L[ffr(n)] +:= 1 then # count R occurrence oob +:= 1 # count out of bounds
every n := 1 to (N-RN) do if not L[ffs(n)] +:= 1 then # count S occurrence oob +:= 1 # count out of bounds every zero +:= (!L = 0) # count zeros / misses every dup +:= (!L > 1) # count > 1's / duplicates printf("Results of R(1 to %d) and S(1 to %d) coverage is ",RN,(N-RN)) if oob+zero+dup=0 then printf("complete.\n") else printf("flawed\noob=%i,zero=%i,dup=%i\nL:\n%s\nR:\n%s\nS:\n%s\n", oob,zero,dup,ximage(L),ximage(ffr(ffr)),ximage(ffs(ffs)))
end
procedure ffr(n) static R,S initial {
R := [1] S := ffs(ffs) # get access to S in ffs } if n === ffr then return R # secret handshake to avoid globals :) if integer(n) > 0 then return R[n] | put(R,ffr(n-1) + ffs(n-1))[n]
end
procedure ffs(n) static R,S initial {
S := [2] R := ffr(ffr) # get access to R in ffr } if n === ffs then return S # secret handshake to avoid globals :) if integer(n) > 0 then { if S[n] then return S[n] else { t := S[*S] until *S = n do if (t +:= 1) = !R then next # could be optimized with more code else return put(S,t)[*S] # extend S } }
end</lang>
printf.icn provides formatting ximage.icn allows formatting entire structures
Output:
Hofstader ff sequences R(n:= 1 to 10) R(1)=1 R(2)=3 R(3)=7 R(4)=12 R(5)=18 R(6)=26 R(7)=35 R(8)=45 R(9)=56 R(10)=69 Results of R(1 to 40) and S(1 to 960) coverage is complete.
J
<lang j>R=: 1 1 3 S=: 0 2 4 FF=: 3 :0
while. +./y>:R,&#S do. R=: R,({:R)+(<:#R){S S=: (i.<:+/_2{.R)-.R end. R;S
) ffr=: { 0 {:: FF@(>./@,) ffs=: { 1 {:: FF@(0,>./@,)</lang>
Required examples:
<lang j> ffr 1+i.10 1 3 7 12 18 26 35 45 56 69
(1+i.1000) -: /:~ (ffr 1+i.40), ffs 1+i.960
1</lang>
Java
Code:
<lang java>import java.util.*;
class Hofstadter {
private static List<Integer> getSequence(int rlistSize, int slistSize) { List<Integer> rlist = new ArrayList<Integer>(); List<Integer> slist = new ArrayList<Integer>(); Collections.addAll(rlist, 1, 3, 7); Collections.addAll(slist, 2, 4, 5, 6); List<Integer> list = (rlistSize > 0) ? rlist : slist; int targetSize = (rlistSize > 0) ? rlistSize : slistSize; while (list.size() > targetSize) list.remove(list.size() - 1); while (list.size() < targetSize) { int lastIndex = rlist.size() - 1; int lastr = rlist.get(lastIndex).intValue(); int r = lastr + slist.get(lastIndex).intValue(); rlist.add(Integer.valueOf(r)); for (int s = lastr + 1; (s < r) && (list.size() < targetSize); s++) slist.add(Integer.valueOf(s)); } return list; } public static int ffr(int n) { return getSequence(n, 0).get(n - 1).intValue(); } public static int ffs(int n) { return getSequence(0, n).get(n - 1).intValue(); } public static void main(String[] args) { System.out.print("R():"); for (int n = 1; n <= 10; n++) System.out.print(" " + ffr(n)); System.out.println(); Set<Integer> first40R = new HashSet<Integer>(); for (int n = 1; n <= 40; n++) first40R.add(Integer.valueOf(ffr(n))); Set<Integer> first960S = new HashSet<Integer>(); for (int n = 1; n <= 960; n++) first960S.add(Integer.valueOf(ffs(n))); for (int i = 1; i <= 1000; i++) { Integer n = Integer.valueOf(i); if (first40R.contains(n) == first960S.contains(n)) System.out.println("Integer " + i + " either in both or neither set"); } System.out.println("Done"); }
}</lang>
Output:
R(): 1 3 7 12 18 26 35 45 56 69 Done
JavaScript
Translated from Ruby. <lang JavaScript>var R = [null, 1]; var S = [null, 2];
var extend_sequences = function (n) { var current = Math.max(R[R.length-1],S[S.length-1]); var i; while (R.length <= n || S.length <= n) { i = Math.min(R.length, S.length) - 1; current += 1; if (current === R[i] + S[i]) { R.push(current); } else { S.push(current); } } }
var ffr = function(n) { extend_sequences(n); return R[n]; };
var ffs = function(n) { extend_sequences(n); return S[n]; };
for (var i = 1; i <=10; i += 1) {
console.log('R('+ i +') = ' + ffr(i));
}
var int_array = [];
for (var i = 1; i <= 40; i += 1) { int_array.push(ffr(i)); } for (var i = 1; i <= 960; i += 1) { int_array.push(ffs(i)); }
int_array.sort(function(a,b){return a-b;});
for (var i = 1; i <= 1000; i += 1) { if (int_array[i-1] !== i) { throw "Something's wrong!" } else { console.log("1000 integer check ok."); } }</lang> Output:
R(1) = 1 R(2) = 3 R(3) = 7 R(4) = 12 R(5) = 18 R(6) = 26 R(7) = 35 R(8) = 45 R(9) = 56 R(10) = 69 1000 integer check ok.
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 <lang Julia> 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 </lang>
Main <lang Julia> 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.") </lang>
- 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. <lang scala>package hofstadter
fun ffr(n: Int) = get(n, 0)[n - 1]
fun ffs(n: Int) = get(0, n)[n - 1]
internal fun get(rlistSize: Int, slistSize: Int): List<Int> {
val rlist = arrayListOf(1, 3, 7) val slist = arrayListOf(2, 4, 5, 6) val list = if (rlistSize > 0) rlist else slist val targetSize = if (rlistSize > 0) rlistSize else slistSize
while (list.size() > targetSize) list.remove(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) } for (n in 1..1000) if (n in first40R == n in first960S) println("Integer $n either in both or neither set") println("Done")
}</lang>
Mathematica
1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively.
The instructions call for two functions. Because S[n] is generated while computing R[n], one would normally avoid redundancy by combining R and S into a single function that returns both sequences.
2. No maximum value for n should be assumed.
<lang Mathematica>
ffr[j_] := Module[{R = {1}, S = 2, k = 1}, Do[While[Position[R, S] != {}, S++]; k = k + S; S++; R = Append[R, k], {n, 1, j - 1}]; R]
ffs[j_] := Differences[ffr[j + 1]]
</lang>
3. Calculate and show that the first ten values of R are: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
<lang Mathematica>
ffr[10]
(* out *) {1, 3, 7, 12, 18, 26, 35, 45, 56, 69}
</lang>
4. Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once.
<lang Mathematica>
t = Sort[Join[ffr[40], ffs[960]]];
t == Range[1000]
(* out *) True
</lang>
MATLAB / Octave
1. Create two functions named ffr and ffs that when given n return R(n) or S(n) respectively. 2. No maximum value for n should be assumed.
<lang MATLAB> function [R,S] = ffr_ffs(N)
t = [1,0]; T = 1; n = 1; %while T<=1000, while n<=N, R = find(t,n); S = find(~t,n); T = R(n)+S(n);
% pre-allocate memory, this improves performance
if T > length(t), t = [t,zeros(size(t))]; end;
t(T) = 1; n = n + 1; end; if nargout>0, r = max(R); s = max(S); else printf('Sequence R:\n'); disp(R); printf('Sequence S:\n'); disp(S); end; end; </lang>
3. Calculate and show that the first ten values of R are: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69
>>ffr_ffs(10) Sequence R: 1 3 7 12 18 26 35 45 56 69 Sequence S: 2 4 5 6 8 9 10 11 13 14
4. This is self-evident from the function definition, but also because R and S are complementary in t and ~t. However, one can also Calculate and show that the first 40 values of ffr plus the first 960 values of ffs include all the integers from 1 to 1000 exactly once. Modify the function above in such a way that, instead of r and s, R and S are returned, and run
[R1,S1] = ffr_ffs(40); [R2,S2] = ffr_ffs(960); all(sort([R1,S2])==1:1000) ans = 1
Nim
<lang 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 =
assert n > 0 while n > cr.len: extendRS() cr[n - 1]
proc ffs(n): int =
assert n > 0 while n > cs.len: extendRS() cs[n - 1]
for i in 1..10: stdout.write ffr i," " echo ""
var bin: array[1..1000, int] for i in 1..40: inc bin[ffr i] for i in 1..960: inc bin[ffs i] var all = true for x in bin:
if x != 1: all = false break
if all: echo "All Integers 1..1000 found OK" else: echo "All Integers 1..1000 NOT found only once: ERROR"</lang> Output:
/home/deen/git/nim-unsorted/hofstadter 1 3 7 12 18 26 35 45 56 69 All Integers 1..1000 found 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.
<lang perl>#!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"; } </lang>
Perl 6
<lang perl6>my @ffr; my @ffs;
@ffr.plan: 0, 1, gather take @ffr[$_] + @ffs[$_] for 1..*; @ffs.plan: 0, 2, 4..6, gather take @ffr[$_] ^..^ @ffr[$_+1] for 3..*;
say @ffr[1..10];
say "Rawks!" if (1...1000) eqv sort @ffr[1..40], @ffs[1..960];</lang> Output:
1 3 7 12 18 26 35 45 56 69 Rawks!
PicoLisp
<lang 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 ) ) ) )</lang>
Test: <lang PicoLisp>: (mapcar ffr (range 1 10)) -> (1 3 7 12 18 26 35 45 56 69)
- (=
(range 1 1000) (sort (conc (mapcar ffr (range 1 40)) (mapcar ffs (range 1 960)))) )
-> T</lang>
PL/I
<lang pli>ffr: procedure (n) returns (fixed binary(31));
declare n fixed binary (31); declare v(2*n+1) bit(1); declare (i, j) fixed binary (31); declare (r, s) fixed binary (31);
v = '0'b; v(1) = '1'b;
if n = 1 then return (1);
r = 1; do i = 2 to n; do j = 2 to 2*n; if v(j) = '0'b then leave; end; v(j) = '1'b; s = j; r = r + s; if r <= 2*n then v(r) = '1'b; end; return (r);
end ffr;</lang> Output:
Please type a value for n: 1 3 7 12 18 26 35 45 56 69 83 98 114 131 150 170 191 213 236 260 285 312 340 369 399 430 462 495 529 565 602 640 679 719 760 802 845 889 935 982
<lang pli>ffs: procedure (n) returns (fixed binary (31));
declare n fixed binary (31); declare v(2*n+1) bit(1); declare (i, j) fixed binary (31); declare (r, s) fixed binary (31);
v = '0'b; v(1) = '1'b;
if n = 1 then return (2);
r = 1; do i = 1 to n; do j = 2 to 2*n; if v(j) = '0'b then leave; end; v(j) = '1'b; s = j; r = r + s; if r <= 2*n then v(r) = '1'b; end; return (s);
end ffs;</lang> Output of first 960 values:
Please type a value for n: 2 4 5 6 8 9 10 11 13 14 15 16 17 19 20 21 22 23 24 25 27 28 29 30 31 32 33 34 36 37 ... 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000
Verification using the above procedures: <lang pli>
Dcl t(1000) Bit(1) Init((1000)(1)'0'b); put skip list ('Verification that the first 40 FFR numbers and the first'); put skip list ('960 FFS numbers result in the integers 1 to 1000 only.'); do i = 1 to 40; j = ffr(i); if t(j) then put skip list ('error, duplicate value at ' || i); else t(j) = '1'b; end; do i = 1 to 960; j = ffs(i); if t(j) then put skip list ('error, duplicate value at ' || i); else t(j) = '1'b; end; if all(t = '1'b) then put skip list ('passed test');
</lang> Output:
Verification that the first 40 FFR numbers and the first 960 FFS numbers result in the integers 1 to 1000 only. passed test
Prolog
Constraint Handling Rules
CHR is a programming language created by Professor Thom Frühwirth.
Works with SWI-Prolog and module chr written by Tom Schrijvers and Jan Wielemaker
<lang Prolog>:- use_module(library(chr)).
- - chr_constraint ffr/2, ffs/2, hofstadter/1,hofstadter/2.
- - chr_option(debug, off).
- - chr_option(optimize, full).
% to remove duplicates ffr(N, R1) \ ffr(N, R2) <=> R1 = R2 | true. ffs(N, R1) \ ffs(N, R2) <=> R1 = R2 | true.
% compute ffr ffr(N, R), ffr(N1, R1), ffs(N1,S1) ==>
N > 1, N1 is N - 1 |
R is R1 + S1.
% compute ffs ffs(N, S), ffs(N1,S1) ==>
N > 1, N1 is N - 1 |
V is S1 + 1, ( find_chr_constraint(ffr(_, V)) -> S is V+1; S = V).
% init hofstadter(N) ==> ffr(1,1), ffs(1,2). % loop hofstadter(N), ffr(N1, _R), ffs(N1, _S) ==> N1 < N, N2 is N1 +1 | ffr(N2,_), ffs(N2,_).
</lang> Output for first task :
?- hofstadter(10), bagof(ffr(X,Y), find_chr_constraint(ffr(X,Y)), L). ffr(10,69) ffr(9,56) ffr(8,45) ffr(7,35) ffr(6,26) ffr(5,18) ffr(4,12) ffr(3,7) ffr(2,3) ffr(1,1) ffs(10,14) ffs(9,13) ffs(8,11) ffs(7,10) ffs(6,9) ffs(5,8) ffs(4,6) ffs(3,5) ffs(2,4) ffs(1,2) hofstadter(10) L = [ffr(10,69),ffr(9,56),ffr(8,45),ffr(7,35),ffr(6,26),ffr(5,18),ffr(4,12),ffr(3,7),ffr(2,3),ffr(1,1)].
Code for the second task <lang Prolog>hofstadter :- hofstadter(960), % fetch the values of ffr bagof(Y, X^find_chr_constraint(ffs(X,Y)), L1), % fetch the values of ffs bagof(Y, X^(find_chr_constraint(ffr(X,Y)), X < 41), L2), % concatenate then append(L1, L2, L3), % sort removing duplicates sort(L3, L4), % check the correctness of the list ( (L4 = [1|_], last(L4, 1000), length(L4, 1000)) -> writeln(ok); writeln(ko)), % to remove all pending constraints fail. </lang> Output for second task
?- hofstadter. ok false.
Python
<lang python>def ffr(n):
if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1") try: return ffr.r[n] except IndexError: r, s = ffr.r, ffs.s ffr_n_1 = ffr(n-1) lastr = r[-1] # extend s up to, and one past, last r s += list(range(s[-1] + 1, lastr)) if s[-1] < lastr: s += [lastr + 1] # access s[n-1] temporarily extending s if necessary len_s = len(s) ffs_n_1 = s[n-1] if len_s > n else (n - len_s) + s[-1] ans = ffr_n_1 + ffs_n_1 r.append(ans) return ans
ffr.r = [None, 1]
def ffs(n):
if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1") try: return ffs.s[n] except IndexError: r, s = ffr.r, ffs.s for i in range(len(r), n+2): ffr(i) if len(s) > n: return s[n] raise Exception("Whoops!")
ffs.s = [None, 2]
if __name__ == '__main__':
first10 = [ffr(i) for i in range(1,11)] assert first10 == [1, 3, 7, 12, 18, 26, 35, 45, 56, 69], "ffr() value error(s)" print("ffr(n) for n = [1..10] is", first10) # bin = [None] + [0]*1000 for i in range(40, 0, -1): bin[ffr(i)] += 1 for i in range(960, 0, -1): bin[ffs(i)] += 1 if all(b == 1 for b in bin[1:1000]): print("All Integers 1..1000 found OK") else: print("All Integers 1..1000 NOT found only once: ERROR")</lang>
- Output
ffr(n) for n = [1..10] is [1, 3, 7, 12, 18, 26, 35, 45, 56, 69] All Integers 1..1000 found OK
Alternative
<lang python>cR = [1] cS = [2]
def extend_RS(): x = cR[len(cR) - 1] + cS[len(cR) - 1] cR.append(x) cS += range(cS[-1] + 1, x) cS.append(x + 1)
def ff_R(n): assert(n > 0) while n > len(cR): extend_RS() return cR[n - 1]
def ff_S(n): assert(n > 0) while n > len(cS): extend_RS() return cS[n - 1]
- 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")</lang>output<lang>[1, 3, 7, 12, 18, 26, 35, 45, 56, 69] Ok</lang>
Using cyclic iterators
Defining R and S as mutually recursive generators. Follows directly from the definition of the R and S sequences. <lang python>from itertools import islice
def R(): n = 1 yield n for s in S(): n += s yield n;
def S(): yield 2 yield 4 u = 5 for r in R(): if r <= u: continue; for x in range(u, r): yield x u = r + 1
def lst(s, n): return list(islice(s(), n))
print "R:", lst(R, 10) print "S:", lst(S, 10) print sorted(lst(R, 40) + lst(S, 960)) == list(range(1,1001))
- perf test case
- print sum(lst(R, 10000000))</lang>
- Output:
R: [1, 3, 7, 12, 18, 26, 35, 45, 56, 69] S: [2, 4, 5, 6, 8, 9, 10, 11, 13, 14] True
Racket
We store the values of r and s in hash-tables. The first values are added by hand. The procedure extend-r-s! adds more values.
<lang Racket>#lang racket/base
(define r-cache (make-hash '((1 . 1) (2 . 3) (3 . 7)))) (define s-cache (make-hash '((1 . 2) (2 . 4) (3 . 5) (4 . 6))))
(define (extend-r-s!)
(define r-count (hash-count r-cache)) (define s-count (hash-count s-cache)) (define last-r (ffr r-count)) (define new-r (+ (ffr r-count) (ffs r-count))) (hash-set! r-cache (add1 r-count) new-r) (define offset (- s-count last-r)) (for ([val (in-range (add1 last-r) new-r)]) (hash-set! s-cache (+ val offset) val)))</lang>
The functions ffr and ffs simply retrieve the value from the hash table if it exist, or call extend-r-s until they are long enought.
<lang Racket>(define (ffr n)
(hash-ref r-cache n (lambda () (extend-r-s!) (ffr n))))
(define (ffs n)
(hash-ref s-cache n (lambda () (extend-r-s!) (ffs n))))</lang>
Tests: <lang Racket>(displayln (map ffr (list 1 2 3 4 5 6 7 8 9 10))) (displayln (map ffs (list 1 2 3 4 5 6 7 8 9 10)))
(displayln "Checking for first 1000 integers:") (displayln (if (equal? (sort (append (for/list ([i (in-range 1 41)])
(ffr i)) (for/list ([i (in-range 1 961)]) (ffs i))) <) (for/list ([i (in-range 1 1001)]) i)) "Test passed" "Test failed"))</lang>
Sample Output:
(1 3 7 12 18 26 35 45 56 69) (2 4 5 6 8 9 10 11 13 14) Checking for first 1000 integers: Test passed
REXX
version 1
This REXX example makes use of sparse arrays.
Over a third of the program was for verification of the first thousand numbers in the Hofstadter Figure-Figure sequences. <lang rexx>/*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 S arrays.*/ errs=0; $.=0
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*/
if x<1 then exit
do m=1 for bot; r=ffr(m); $.r=1 end /*m*/ /* [↑] calculate the 1st 40 R values.*/
do n=1 for top-bot; s=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.*/
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 done.*/ /*────────────────────────────────────────────────────────────────────────────*/ ffr: procedure expose r. s. rr.; parse arg n /*obtain the number from the arg.*/
if r.n\==0 then return r.n /*Defined? Then return the value.*/ _=ffr(n-1) + ffs(n-1) /*calculate the FFR & FFS values.*/ r.n=_; rr._=1; return _ /*assign value to R & RR; return.*/
/*────────────────────────────────────────────────────────────────────────────*/ ffs: procedure expose r. s. rr.; parse arg n /*search for ¬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 call ffr k /*define R.k via FFR subroutine*/ km=k-1; _=s.km+1 /*the next S number, possibly.*/ _=_+rr._; s.k=_ /*define an element of S array.*/ end /*k*/ return s.n /*return S.n value to the invoker*/
/*────────────────────────────────────────────────────────────────────────────*/
ser: errs=errs+1; say '***error***!' arg(1); return</lang>
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
<lang rexx>/* REXX **************************************************************
- 21.11.2012 Walter Pachl transcribed from PL/I
- /
Call time 'R' Say 'Verification that the first 40 FFR numbers and the first' Say '960 FFS numbers result in the integers 1 to 1000 only.' t.=0 num.= do i = 1 to 40 j = ffr(i) if t.j then Say 'error, duplicate value at ' || i else t.j = 1 num.i=j end nn=0 Say time('E') 'seconds elapsed' Do i=1 To 3 ol= Do j=1 To 15 nn=nn+1 ol=ol right(num.nn,3) End Say ol End do i = 1 to 960 j = ffs(i) if t.j then Say 'error, duplicate value at ' || i else t.j = 1 end Do i=1 To 1000 if t.i=0 Then Say i 'was not set' End If i>1000 Then Say 'passed test' Say time('E') 'seconds elapsed' Exit
ffr: procedure Expose v. Parse Arg n v.= 0 v.1 = 1 if n = 1 then return 1 r = 1 do i = 2 to n do j = 2 to 2*n if v.j = 0 then leave end v.j = 1 s = j r = r + s if r <= 2*n then v.r = 1 end return r
ffs: procedure Expose v. Parse Arg n v.= 0 v.1 = 1 if n = 1 then return 2 r = 1 do i = 1 to n do j = 2 to 2*n if v.j = 0 then leave end v.j = 1 s = j r = r + s if r <= 2*n then v.r = 1 end return s</lang>
Verification that the first 40 FFR numbers snd the first 960 FFS numbers result in the integers 1 to 1000 only. 0.011000 seconds elapsed 1 3 7 12 18 26 35 45 56 69 83 98 114 131 150 170 191 213 236 260 285 312 340 369 399 430 462 495 529 565 602 640 679 719 760 802 845 889 935 982 passed test Windows (ooRexx) 33.183000 seconds elapsed TSO interpreted: 139.699246 seconds elapsed TSO compiled: 9.749457 seconds elapsed
Ruby
<lang ruby>$r = [nil, 1] $s = [nil, 2]
def buildSeq(n)
current = [ $r[-1], $s[-1] ].max while $r.length <= n || $s.length <= n idx = [ $r.length, $s.length ].min - 1 current += 1 if current == $r[idx] + $s[idx] $r << current else $s << current end end
end
def ffr(n)
buildSeq(n) $r[n]
end
def ffs(n)
buildSeq(n) $s[n]
end
require 'set' require 'test/unit'
class TestHofstadterFigureFigure < Test::Unit::TestCase
def test_first_ten_R_values r10 = 1.upto(10).map {|n| ffr(n)} assert_equal(r10, [1, 3, 7, 12, 18, 26, 35, 45, 56, 69]) end
def test_40_R_and_960_S_are_1_to_1000 rs_values = Set.new rs_values.merge( 1.upto(40).collect {|n| ffr(n)} ) rs_values.merge( 1.upto(960).collect {|n| ffs(n)} ) assert_equal(rs_values, Set.new( 1..1000 )) end
end</lang>
outputs
Loaded suite hofstadter.figurefigure Started .. Finished in 0.511000 seconds. 2 tests, 2 assertions, 0 failures, 0 errors, 0 skips
Using cyclic iterators
<lang ruby>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 </lang>
- Output:
[1, 3, 7, 12, 18, 26, 35, 45, 56, 69] [2, 4, 5, 6, 8, 9, 10, 11, 13, 14] true
Scala
<lang Scala>object HofstadterFigFigSeq extends App {
import scala.collection.mutable.ListBuffer
val r = ListBuffer(0, 1) val s = ListBuffer(0, 2)
def ffr(n: Int): Int = { val ffri: Int => Unit = i => { val nrk = r.size - 1 val rNext = r(nrk)+s(nrk) r += rNext (r(nrk)+2 to rNext-1).foreach{s += _} s += rNext+1 }
(r.size to n).foreach(ffri(_)) r(n) }
def ffs(n:Int): Int = { while (s.size <= n) ffr(r.size) s(n) }
(1 to 10).map(i=>(i,ffr(i))).foreach(t=>println("r("+t._1+"): "+t._2)) println((1 to 1000).toList.filterNot(((1 to 40).map(ffr(_))++(1 to 960).map(ffs(_))).contains)==List())
}</lang> Output:
r(1): 1 r(2): 3 r(3): 7 r(4): 12 r(5): 18 r(6): 26 r(7): 35 r(8): 45 r(9): 56 r(10): 69 true
Tcl
<lang tcl>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}}]"</lang> Output:
R(1) = 1 R(2) = 3 R(3) = 7 R(4) = 12 R(5) = 18 R(6) = 26 R(7) = 35 R(8) = 45 R(9) = 56 R(10) = 69 Considering {1..1000} vs {R(i)|i∈[1,40]}∪{S(i)|i∈[1,960]} set sizes: 1000 vs 1000 set equality: yes
VBScript
<lang vb> '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 </lang>
- Output:
First 10 Values of R: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69 Test for the first 1000 integers: Passed!!!
zkl
<lang 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) )</lang>
- Output:
(0).pump(10,List,genRS).apply("get",1).println(); L(1,3,7,12,18,26,35,45,56,69)
<lang zkl>genRS(True); // reset sink:=(0).pump(40,List, 'wrap(ns){ T(Void.Write,Void.Write,genRS()[1,*]) }); sink= (0).pump(960-40,sink,'wrap(ns){ T(Void.Write,genRS()[2]) }); (sink.sort()==[1..1000].walk()).println("<-- should be True");</lang>
- Output:
True<-- should be True