Hofstadter Figure-Figure sequences
You are encouraged to solve this task according to the task description, using any language you may know.
These two sequences of positive integers are defined as:
- The sequence is further defined as the sequence of positive integers not present in .
Sequence R starts: 1, 3, 7, 12, 18, ...
Sequence S starts: 2, 4, 5, 6, 8, ...
Task:
- 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>
Common Lisp
<lang lisp>;;; equally doable with a list (flet ((seq (i) (make-array 1 :element-type 'integer :initial-element i :fill-pointer 1 :adjustable t)))
(let ((rr (seq 1)) (ss (seq 2))) (labels ((extend-r ()
(let* ((l (1- (length rr))) (r (+ (aref rr l) (aref ss l))) (s (elt ss (1- (length ss))))) (vector-push-extend r rr) (loop while (<= s r) do (if (/= (incf s) r) (vector-push-extend s ss))))))
(defun seq-r (n)
(loop while (> n (length rr)) do (extend-r)) (elt rr (1- n)))
(defun seq-s (n)
(loop while (> n (length ss)) do (extend-r)) (elt ss (1- n))))))
(defun take (f n)
(loop for x from 1 to n collect (funcall f x)))
(format t "First of R: ~a~%" (take #'seq-r 10))
(mapl (lambda (l) (if (and (cdr l) (/= (1+ (car l)) (cadr l))) (error "not in sequence")))
(sort (append (take #'seq-r 40)
(take #'seq-s 960)) #'<)) (princ "Ok")</lang>output<lang>First of R: (1 3 7 12 18 26 35 45 56 69) Ok</lang>
C#
Creates an IEnumerable for R and S and uses those to complete the task <lang C#>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
D
<lang d>import std.stdio, std.array, std.range, std.algorithm;
int delegate(in int) nothrow ffr, ffs;
static this() {
auto r = [0, 1], s = [0, 2];
ffr = (in int n) {
while (r.length <= n) {
int nrk = r.length - 1;
int rNext = r[nrk] + s[nrk];
r ~= rNext;
foreach (sn; r[nrk] + 2 .. rNext)
s ~= sn;
s ~= rNext + 1;
}
return r[n];
};
ffs = (in int n) {
while (s.length <= n)
ffr(r.length);
return s[n];
};
}
void main() {
writeln(map!ffr(iota(1, 11))); auto t = chain(map!ffr(iota(1, 41)), map!ffs(iota(1, 961))); writeln(equal(sort(array(t)), iota(1, 1001)));
}</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 int[] r = [int.min, 1];
static int opCall(in int n) {
assert(n > 0);
if (n < r.length) {
return r[n];
} else {
int ffr_n_1 = ffr(n - 1);
int lastr = r[$ - 1];
// extend s up to, and one past, last r
ffs.s ~= array(iota(ffs.s[$ - 1] + 1, lastr));
if (ffs.s[$ - 1] < lastr)
ffs.s ~= lastr + 1;
// access s[n-1] temporarily extending s if necessary
size_t len_s = ffs.s.length;
int ffs_n_1 = len_s > n ? ffs.s[n - 1] :
(n - len_s) + ffs.s[$-1];
int ans = ffr_n_1 + ffs_n_1;
r ~= ans;
return ans;
}
}
}
struct ffs {
static int[] s = [int.min, 2];
static int opCall(in int n) {
assert(n > 0);
if (n < s.length) {
return s[n];
} else {
foreach (i; ffr.r.length .. n+2) {
ffr(i);
if (s.length > n)
return s[n];
}
assert(0, "Whoops!");
}
}
}
void main() {
writeln(map!ffr(iota(1, 11))); auto t = chain(map!ffr(iota(1, 41)), map!ffs(iota(1, 961))); writeln(equal(sort(array(t)), iota(1, 1001)));
}</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
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
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>
PicoLisp
<lang PicoLisp>(setq *RNext 2)
(de ffr (N)
(cache '(NIL) (pack (char (hash N)) N)
(if (= 1 N)
1
(+ (ffr (dec N)) (ffs (dec N))) ) ) )
(de ffs (N)
(cache '(NIL) (pack (char (hash N)) N)
(if (= 1 N)
2
(let S (inc (ffs (dec N)))
(when (= S (ffr *RNext))
(inc 'S)
(inc '*RNext) )
S ) ) ) )</lang>
Test: <lang PicoLisp>: (mapcar ffr (range 1 10)) -> (1 3 7 12 18 26 35 45 56 69)
- (=
(range 1 1000) (sort (conc (mapcar ffr (range 1 40)) (mapcar ffs (range 1 960)))) )
-> T</lang>
Perl 6
<lang perl6>my @ffr; my @ffs;
@ffr.plan: 0, 1, gather take @ffr[$_] + @ffs[$_] for 1..*; @ffs.plan: 0, 2, 4..6, gather take @ffr[$_] ^..^ @ffr[$_+1] for 3..*;
say @ffr[1..10];
say "Rawks!" if (1...1000) eqv sort @ffr[1..40], @ffs[1..960];</lang> Output:
1 3 7 12 18 26 35 45 56 69 Rawks!
PL/I
<lang PL/I> ffr: procedure (n) returns (fixed binary(31));
declare n fixed binary (31); declare v(2*n+1) bit(1); declare (i, j) fixed binary (31); declare (r, s) fixed binary (31);
v = '0'b; v(1) = '1'b;
if n = 1 then return (1);
r = 1;
do i = 2 to n;
do j = 2 to 2*n;
if v(j) = '0'b then leave;
end;
v(j) = '1'b;
s = j;
r = r + s;
if r <= 2*n then v(r) = '1'b;
end;
return (r);
end ffr; </lang> Output:
Please type a value for n:
1 3 7 12 18 26 35 45 56 69 83 98 114 131 150
170 191 213 236 260 285 312 340 369 399 430 462 495 529 565
602 640 679 719 760 802 845 889 935 982
<lang> ffs: procedure (n) returns (fixed binary (31));
declare n fixed binary (31); declare v(2*n+1) bit(1); declare (i, j) fixed binary (31); declare (r, s) fixed binary (31);
v = '0'b; v(1) = '1'b;
if n = 1 then return (2);
r = 1;
do i = 1 to n;
do j = 2 to 2*n;
if v(j) = '0'b then leave;
end;
v(j) = '1'b;
s = j;
r = r + s;
if r <= 2*n then v(r) = '1'b;
end;
return (s);
end ffs; </lang> Output of first 960 values:
Please type a value for n:
2 4 5 6 8 9 10 11 13 14 15 16 17 19 20
21 22 23 24 25 27 28 29 30 31 32 33 34 36 37
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
103 104 105 106 107 108 109 110 111 112 113 115 116 117 118
119 120 121 122 123 124 125 126 127 128 129 130 132 133 134
135 136 137 138 139 140 141 142 143 144 145 146 147 148 149
151 152 153 154 155 156 157 158 159 160 161 162 163 164 165
166 167 168 169 171 172 173 174 175 176 177 178 179 180 181
182 183 184 185 186 187 188 189 190 192 193 194 195 196 197
198 199 200 201 202 203 204 205 206 207 208 209 210 211 212
214 215 216 217 218 219 220 221 222 223 224 225 226 227 228
229 230 231 232 233 234 235 237 238 239 240 241 242 243 244
245 246 247 248 249 250 251 252 253 254 255 256 257 258 259
261 262 263 264 265 266 267 268 269 270 271 272 273 274 275
276 277 278 279 280 281 282 283 284 286 287 288 289 290 291
292 293 294 295 296 297 298 299 300 301 302 303 304 305 306
307 308 309 310 311 313 314 315 316 317 318 319 320 321 322
323 324 325 326 327 328 329 330 331 332 333 334 335 336 337
338 339 341 342 343 344 345 346 347 348 349 350 351 352 353
354 355 356 357 358 359 360 361 362 363 364 365 366 367 368
370 371 372 373 374 375 376 377 378 379 380 381 382 383 384
385 386 387 388 389 390 391 392 393 394 395 396 397 398 400
401 402 403 404 405 406 407 408 409 410 411 412 413 414 415
416 417 418 419 420 421 422 423 424 425 426 427 428 429 431
432 433 434 435 436 437 438 439 440 441 442 443 444 445 446
447 448 449 450 451 452 453 454 455 456 457 458 459 460 461
463 464 465 466 467 468 469 470 471 472 473 474 475 476 477
478 479 480 481 482 483 484 485 486 487 488 489 490 491 492
493 494 496 497 498 499 500 501 502 503 504 505 506 507 508
509 510 511 512 513 514 515 516 517 518 519 520 521 522 523
524 525 526 527 528 530 531 532 533 534 535 536 537 538 539
540 541 542 543 544 545 546 547 548 549 550 551 552 553 554
555 556 557 558 559 560 561 562 563 564 566 567 568 569 570
571 572 573 574 575 576 577 578 579 580 581 582 583 584 585
586 587 588 589 590 591 592 593 594 595 596 597 598 599 600
601 603 604 605 606 607 608 609 610 611 612 613 614 615 616
617 618 619 620 621 622 623 624 625 626 627 628 629 630 631
632 633 634 635 636 637 638 639 641 642 643 644 645 646 647
648 649 650 651 652 653 654 655 656 657 658 659 660 661 662
663 664 665 666 667 668 669 670 671 672 673 674 675 676 677
678 680 681 682 683 684 685 686 687 688 689 690 691 692 693
694 695 696 697 698 699 700 701 702 703 704 705 706 707 708
709 710 711 712 713 714 715 716 717 718 720 721 722 723 724
725 726 727 728 729 730 731 732 733 734 735 736 737 738 739
740 741 742 743 744 745 746 747 748 749 750 751 752 753 754
755 756 757 758 759 761 762 763 764 765 766 767 768 769 770
771 772 773 774 775 776 777 778 779 780 781 782 783 784 785
786 787 788 789 790 791 792 793 794 795 796 797 798 799 800
801 803 804 805 806 807 808 809 810 811 812 813 814 815 816
817 818 819 820 821 822 823 824 825 826 827 828 829 830 831
832 833 834 835 836 837 838 839 840 841 842 843 844 846 847
848 849 850 851 852 853 854 855 856 857 858 859 860 861 862
863 864 865 866 867 868 869 870 871 872 873 874 875 876 877
878 879 880 881 882 883 884 885 886 887 888 890 891 892 893
894 895 896 897 898 899 900 901 902 903 904 905 906 907 908
909 910 911 912 913 914 915 916 917 918 919 920 921 922 923
924 925 926 927 928 929 930 931 932 933 934 936 937 938 939
940 941 942 943 944 945 946 947 948 949 950 951 952 953 954
955 956 957 958 959 960 961 962 963 964 965 966 967 968 969
970 971 972 973 974 975 976 977 978 979 980 981 983 984 985
986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000
Verification using the above procedures: <lang>
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>
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
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