Hofstadter Q sequence: Difference between revisions
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Q(1000000) = 512066 |
Q(1000000) = 512066 |
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=={{header|GW-BASIC}}== |
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<lang gwbasic>10 DIM Q!(1000) |
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20 Q(1) = 1: Q(2) = 1 |
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30 FOR N = 3 TO 1000 |
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40 Q(N) = Q(N - Q(N - 1)) + Q(N - Q(N - 2)) |
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50 NEXT N |
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60 FOR N = 1 TO 10 |
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70 PRINT Q(N) |
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80 NEXT N |
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90 PRINT Q(1000)</lang> |
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=={{header|Haskell}}== |
=={{header|Haskell}}== |
||
Revision as of 20:32, 30 March 2021
You are encouraged to solve this task according to the task description, using any language you may know.
The Hofstadter Q sequence is defined as:
It is defined like the Fibonacci sequence, but whereas the next term in the Fibonacci sequence is the sum of the previous two terms, in the Q sequence the previous two terms tell you how far to go back in the Q sequence to find the two numbers to sum to make the next term of the sequence.
- Task
- Confirm and display that the first ten terms of the sequence are: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6
- Confirm and display that the 1000th term is: 502
- Optional extra credit
- Count and display how many times a member of the sequence is less than its preceding term for terms up to and including the 100,000th term.
- Ensure that the extra credit solution safely handles being initially asked for an nth term where n is large.
(This point is to ensure that caching and/or recursion limits, if it is a concern, is correctly handled).
11l
<lang 11l>V qseq = [0] * 100001 qseq[1] = 1 qseq[2] = 1
L(i) 3 .< qseq.len
qseq[i] = qseq[i - qseq[i-1]] + qseq[i - qseq[i-2]]
print(‘The first 10 terms are: ’qseq[1..10].map(q -> String(q)).join(‘, ’)) print(‘The 1000'th term is ’qseq[1000])
V less_than_preceding = 0 L(i) 2 .< qseq.len
I qseq[i] < qseq[i-1]
less_than_preceding++
print(‘Times a member of the sequence is less than its preceding term: ’less_than_preceding)</lang>
- Output:
The first 10 terms are: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 The 1000'th term is 502 Times a member of the sequence is less than its preceding term: 49798
360 Assembly
<lang 360asm>* Hofstrader q sequence for any n - 18/10/2015 HOFSTRAD CSECT
USING HOFSTRAD,R15 set base register
MVC Q,=F'1' q(1)=1
MVC Q+4,=F'1' q(2)=1
LA R4,1 i=1
LOOPI C R4,N do i=1 to n
BH ELOOPI
C R4,=F'3' if i>=3 then
BL NOTREC
LR R1,R4 i
SLA R1,2 i*4
L R2,Q-8(R1) q(i-1)
LR R1,R4 i
SR R1,R2 i-q(i-1)
SLA R1,2 *4
L R2,Q-4(R1) r2=q(i-q(i-1))
LR R1,R4 i
SLA R1,2 i*4
L R3,Q-12(R1) q(i-2)
LR R1,R4 i
SR R1,R3 i-q(i-2)
SLA R1,2 *4
L R3,Q-4(R1) r3=q(i-q(i-2))
AR R2,R3 r2=r2+r3
LR R1,R4 i
SLA R1,2 i*4
ST R2,Q-4(R1) q(i)=q(i-q(i-1))+q(i-q(i-2))
NOTREC C R4,=F'10' if i<=10
BNH PRT
C R4,N or i=n then
BNE NOPRT
PRT XDECO R4,XD edit i
MVC PG+2(4),XD+8 output i
LR R1,R4 i
SLA R1,2 i*4
L R2,Q-4(R1) q(i)
XDECO R2,XD edit q(i)
MVC PG+10(4),XD+8 output q(i)
XPRNT PG,80 print buffer
NOPRT LA R4,1(R4) i=i+1
B LOOPI
ELOOPI XR R15,R15 set return code
BR R14 return to caller
PG DC CL80'n=...., q=....' buffer XD DS CL12 temporary variable
LTORG insert literals for addressability
N DC F'1000' n=1000 Q DS 1000F array q(1000)
YREGS
END HOFSTRAD</lang>
- Output:
n= 1, q= 1 n= 2, q= 1 n= 3, q= 2 n= 4, q= 3 n= 5, q= 3 n= 6, q= 4 n= 7, q= 5 n= 8, q= 5 n= 9, q= 6 n= 10, q= 6 n=1000, q= 502
8080 Assembly
<lang 8080asm>puts: equ 9 ; CP/M call to print a string org 100h ;;; Generate the first 1000 members of the Q sequence lxi b,3 ; Start at 3rd element (1 and 2 already defined) genq: dcx b ; BC = N-1 call q mov e,m ; DE = Q(N-1) inx h mov d,m inx b ; BC = (N-1)+1 = N xchg ; HL = Q(N-1) call neg ; HL = -Q(N-1) dad b ; HL = N-Q(N-1) push b ; Keep N mov b,h ; BC = N-Q(N-1) mov c,l call q ; HL = *Q(N-Q(N-1)) mov e,m ; DE = Q(N-Q(N-1)) inx h mov d,m pop b ; Restore N push d ; push Q(N-Q(N-1)) dcx b ; BC = N-2 dcx b call q ; DE = Q(N-2) mov e,m inx h mov d,m inx b ; BC = (N-2)+2 = N inx b xchg ; HL = Q(N-2) call neg ; HL = -Q(N-2) dad b ; HL = N-Q(N-2) push b ; Keep N mov b,h ; BC = N-Q(N-2) mov c,l call q ; HL = *Q(N-Q(N-2)) mov a,m ; HL = Q(N-Q(N-2)) inx h mov h,m mov l,a pop b ; Restore N pop d ; pop Q(N-Q(N-1)) dad d ; HL = Q(N-Q(N-1))+Q(N-Q(N-2)) xchg ; DE = Q(N-Q(N-1))+Q(N-Q(N-2)) call q ; HL = *Q(N) mov m,e ; Store Q(N) inx h mov m,d inx b ; N = N+1 lxi h,-1001 dad b ; Are we there yet? jnc genq ;;; Print first 10 terms lxi d,m10 mvi c,puts call 5 lxi b,1 ; Start at term 1 mvi d,10 ; 10 terms p10: push b ; Save counters push d call prterm ; Print current term pop d ; Restore counters pop b inx b ; Next term dcr d ; Repeat 10 times jnz p10 ;;; Print 1000th term lxi d,m1000 mvi c,puts call 5 lxi b,1000 ; 1000th term ;;; Print Q(BC) prterm: call q ; Load term into HL mov a,m inx h mov h,m mov l,a lxi b,num ; Push pointer to end of number buffer push b lxi b,-10 ; Divisor dgt: lxi d,-1 ; Quotient divlp: inx d dad b jc divlp mvi a,'0'+10 add l ; Make ASCII digit pop h ; Get pointer dcx h mov m,a ; Store digit push h xchg ; HL = next quotient mov a,h ; More digits? ora l jnz dgt pop d ; Print string mvi c,puts jmp 5 ;;; Set HL = -HL neg: dcx h mov a,h cma mov h,a mov a,l cma mov l,a ret ;;; Set HL to memory location of Q(BC) q: push d ; Keep DE mov h,b ; HL = 2*(BC-1) mov l,c dcx h dad h lxi d,qq ; Add to start of sequence dad d pop d ret m10: db 'The first 10 terms are: $' m1000: db 13,10,'The 1000th term is: $' db '*****' ; Placeholder for number num: db ' $' qq: dw 1,1 ; Q sequence stored here, starting with 1, 1</lang>
- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502
8086 Assembly
<lang asm>puts: equ 9 ; MS-DOS syscall to print a string cpu 8086 org 100h section .text ;;; Generate first 1000 elements of Q sequence mov dx,3 ; DX = N mov di,Q+4 ; DI = place to store elements mov cx,998 ; Generate 998 more terms genq: mov si,dx ; SI = N sub si,[di-2] ; SI -= Q[N-1] mov bp,dx ; BP = N sub bp,[di-4] ; BP -= Q[N-2] dec si ; SI = 2*(SI-1) (0-indexed, 2 bytes/term) shl si,1 dec bp ; Same for BP shl bp,1 mov ax,[si+Q] ; Load Q[n-Q[n-1]] add ax,[bp+Q] ; Add Q[n-Q[n-2]] stosw ; Store as Q[n] inc dx ; Increment N loop genq ;;; Print first 10 elements mov ah,puts mov dx,m10 int 21h mov cx,10 mov bx,1 p10: call prterm inc bx loop p10 ;;; Print 1000th element mov ah,puts mov dx,m1000 int 21h mov bx,1000 ;;; Print the term in BX prterm: push bx ; Save term dec bx shl bx,1 mov ax,[bx+Q] ; Load term into AX mov bp,10 ; Divisor mov bx,num ; Number buffer pointer .dgt: xor dx,dx div bp ; Divide number by 10 dec bx add dl,'0' ; DX = remainder, add '0' mov [bx],dl ; Stored digit test ax,ax ; Done yet? jnz .dgt ; If not, find next digit mov dx,bx ; Print the number mov ah,puts int 21h pop bx ; Restore term ret section .data m10: db 'First 10 terms are: $' m1000: db 13,10,'1000th term is: $' db '*****' ; Number placeholder num: db ' $' Q: dw 1,1 </lang>
- Output:
First 10 terms are: 1 1 2 3 3 4 5 5 6 6 1000th term is: 502
Ada
<lang Ada>with Ada.Text_IO;
procedure Hofstadter_Q_Sequence is
type Callback is access procedure(N: Positive);
procedure Q(First, Last: Positive; Q_Proc: Callback) is -- calls Q_Proc(Q(First)); Q_Proc(Q(First+1)); ... Q_Proc(Q(Last)); -- precondition: Last > 2
Q_Store: array(1 .. Last) of Natural := (1 => 1, 2 => 1, others => 0);
-- "global" array to store the Q(I)
-- if Q_Store(I)=0, we compute Q(I) and update Q_Store(I)
-- else we already know Q(I) = Q_Store(I)
function Q(N: Positive) return Positive is
begin
if Q_Store(N) = 0 then
Q_Store(N) := Q(N - Q(N-1)) + Q(N-Q(N-2));
end if;
return Q_Store(N);
end Q;
begin
for I in First .. Last loop
Q_Proc(Q(I));
end loop;
end Q;
procedure Print(P: Positive) is
begin
Ada.Text_IO.Put(Positive'Image(P));
end Print;
Decrease_Counter: Natural := 0; Previous_Value: Positive := 1;
procedure Decrease_Count(P: Positive) is
begin
if P < Previous_Value then
Decrease_Counter := Decrease_Counter + 1;
end if;
Previous_Value := P;
end Decrease_Count;
begin
Q(1, 10, Print'Access); -- the first ten terms of the sequence are: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6 Ada.Text_IO.New_Line;
Q(1000, 1000, Print'Access); -- the 1000'th term is: 502 Ada.Text_IO.New_Line;
Q(2, 100_000, Decrease_Count'Access); Ada.Text_IO.Put_Line(Integer'Image(Decrease_Counter)); -- how many times a member of the sequence is less than its preceding term -- for terms up to and including the 100,000'th term
end Hofstadter_Q_Sequence;</lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 49798
ALGOL 68
Note: This specimen retains the original C coding style.
File: Hofstadter_Q_sequence.a68<lang algol68>#!/usr/local/bin/a68g --script #
INT n = 100000; main: (
INT flip;
[n]INT q;
q[1] := q[2] := 1;
FOR i FROM 3 TO n DO
q[i] := q[i - q[i - 1]] + q[i - q[i - 2]] OD;
FOR i TO 10 DO
printf(($g(0)$, q[i], $b(l,x)$, i = 10)) OD;
printf(($g(0)l$, q[1000]));
flip := 0;
FOR i TO n-1 DO
flip +:= ABS (q[i] > q[i + 1]) OD;
printf(($"flips: "g(0)l$, flip))
)</lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 flips: 49798
APL
<lang APL>∇ Q_sequence;seq;size
size←100000
seq←{⍵,+/⍵[(1+⍴⍵)-¯2↑⍵]}⍣(size-2)⊢1 1
⎕←'The first 10 terms are:', seq[⍳10] ⎕←'The 1000th term is:', seq[1000] ⎕←(+/ 2>/seq),'terms were preceded by a larger term.'
∇</lang>
- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502 49798 terms were preceded by a larger term.
ARM Assembly
<lang>.text .global _start _start: ldr r6,=qs @ R6 = base register for Q array @@@ Write first 2 elements mov r0,#1 @ Q(1) and Q(2) are 1 strh r0,[r6,#4] strh r0,[r6,#8] @@@ Generate 100 thousand elements mov r1,#0x86A0 movt r1,#1 @ 0x186A0 = 100.000 mov r0,#3 @ Starting at element 3 1: sub r2,r0,#1 @ r2 = n-1 ldr r2,[r6,r2,lsl#2] @ r2 = Q[r2] sub r2,r0,r2 @ r2 = n-Q[r2] ldr r2,[r6,r2,lsl#2] @ r2 = Q[r2] sub r3,r0,#2 @ r3 = n-2 ldr r3,[r6,r3,lsl#2] @ r3 = Q[r3] sub r3,r0,r3 @ r3 = n-Q[r3] ldr r3,[r6,r3,lsl#2] @ r3 = Q[r3] add r2,r2,r3 @ r2 += r3 str r2,[r6,r0,lsl#2] @ Q[n] = r2 add r0,r0,#1 @ n++ cmp r0,r1 bls 1b @ If r0<=r1, generate next @@@ Print first 10 elements ldr r1,=f10m bl pstr mov r8,#1 @ Start at element 1 1: ldr r0,[r6,r8,lsl#2] @ Grab current element bl pnum @ Print it ldr r1,=space @ Print a space bl pstr add r8,r8,#1 cmp r8,#10 @ Keep going until 10 elements printed bls 1b ldr r1,=nl @ Print newline bl pstr @@@ Print 1000th element ldr r1,=f1000m bl pstr mov r8,#1000 @ Grab 1000th element ldr r0,[r6,r8,lsl#2] bl pnum ldr r1,=nl @ Print newline bl pstr @@@ Find how many times a member is less than its preceding term mov r0,#0 @ counter mov r1,#0x86A0 @ max element movt r1,#1 mov r2,#1 @ value of previous element mov r3,#2 @ number of current element 2: ldr r4,[r6,r3,lsl#2] @ get value of current element cmp r2,r4 @ if previous more than current addhi r0,r0,#1 @ then increment counter mov r2,r4 @ current el is now prevous el add r3,r3,#1 @ increment element index cmp r3,r1 @ are we there yet? bls 2b @ if not, keep going bl pnum @ otherwise, print the number ldr r1,=ltermm @ and the corresponding message bl pstr mov r0,#0 @ and then exit mov r7,#1 swi #0 @@@ Print a length-prefixed string (in r1) pstr: push {r7,lr} @ Save syscall and link registers mov r0,#1 @ 1 = stdout ldrb r2,[r1],#1 @ Get length and advance r1 mov r7,#4 @ Write swi #0 pop {r7,pc} @@@ Print unsigned number in r0 using Linux pnum: push {r7,lr} @ Save syscall and link registers ldr r7,=qs @ May as well use R7 as buffer pointer 1: mov r1,#10 @ Div-mod by 10 bl divmod add r1,r1,#'0 @ This makes an ASCII digit strb r1,[r7,#-1]! @ Store it in the buffer tst r0,r0 @ Are there more digits? bne 1b @ If so, calculate them mov r0,#1 @ 1 = stdout mov r1,r7 @ Start of number in R1 ldr r2,=qs @ Calculate length sub r2,r2,r1 mov r7,#4 @ 4 = write swi #0 pop {r7,pc} @@@ Division routine: r0=r0/r1, r1=r0%r1 divmod: mov r2,#0 @ R2 = counter 1: cmp r1,r0 @ Double R1 until R1>R0 lslls r1,r1,#1 addls r2,r2,#1 bls 1b mov r3,#0 2: lsl r3,r3,#1 subs r0,r0,r1 @ Trial subtraction addhs r3,r3,#1 @ If it worked, mark addlo r0,r0,r1 @ If it didn't, undo lsr r1,r1,#1 @ Halve R1 subs r2,r2,#1 @ Decrement counter bhs 2b @ Keep going until zero mov r1,r0 @ R1 = modulus mov r0,r3 @ R0 = quotient bx lr .data space: .ascii "\x1 " nl: .ascii "\x1\n" f10m: .ascii "\x18The first 10 terms are: " f1000m: .ascii "\x14The 1000th term is: " ltermm: .ascii "' terms were preceded by a larger term.\n" .bss .align 4 .space 8 @ Buffer for number output qs: .space 4 * 100001 @ One word per term</lang>
- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502 49798 terms were preceded by a larger term.
AutoHotkey
<lang AutoHotkey>SetBatchLines, -1 Q := HofsQSeq(100000)
Loop, 10 Out .= Q[A_Index] ", "
MsgBox, % "First ten:`t" Out "`n" . "1000th:`t`t" Q[1000] "`n" . "Flips:`t`t" Q.flips
HofsQSeq(n) { Q := {1: 1, 2: 1, "flips": 0} Loop, % n - 2 { i := A_Index + 2 , Q[i] := Q[i - Q[i - 1]] + Q[i - Q[A_Index]] if (Q[i] < Q[i - 1]) Q.flips++ } return Q }</lang>
- Output:
First ten: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 1000th: 502 Flips: 49798
AWK
<lang awk>#!/usr/bin/awk -f BEGIN {
N = 100000
print "Q-sequence(1..10) : " Qsequence(10)
Qsequence(N,Q)
print "1000th number of Q sequence : " Q[1000]
for (n=2; n<=N; n++) {
if (Q[n]<Q[n-1]) NN++
} print "number of Q(n)<Q(n+1) for n<=100000 : " NN
}
function Qsequence(N,Q) {
Q[1] = 1
Q[2] = 1
seq = "1 1"
for (n=3; n<=N; n++) {
Q[n] = Q[n-Q[n-1]]+Q[n-Q[n-2]]
seq = seq" "Q[n]
}
return seq
} </lang>
Q-sequence(1..10) : 1 1 2 3 3 4 5 5 6 6 1000th number of Q sequence : 502 number of Q(n)<Q(n+1) for n<=100000 : 49798
BASIC256
<lang BASIC256> limite = 100000 dim Q[limite+1] cont = 0 Q[1] = 1 Q[2] = 1 for i = 3 to limite Q[i] = Q[i-Q[i-1]] + Q[i-Q[i-2]] if Q[i] < Q[i-1] then cont += 1 next i
print "Primeros 10 términos: "; for i = 1 to 10 print Q[i] + " "; next i
print "Término número 1000: "; Q[1000] print "Términos menores que los anteriores: "; cont </lang>
- Output:
Igual que la entrada de FreeBASIC.
BBC BASIC
<lang bbcbasic> PRINT "First 10 terms of Q = " ;
FOR i% = 1 TO 10 : PRINT ;FNq(i%, c%) " "; : NEXT : PRINT
PRINT "1000th term = " ; FNq(1000, c%)
PRINT "100000th term = " ; FNq(100000, c%)
PRINT "Term is less than preceding term " ; c% " times"
END
DEF FNq(n%, RETURN c%)
LOCAL i%,q%()
IF n% < 3 THEN = 1 ELSE IF n% = 3 THEN = 2
DIM q%(n%)
q%(1) = 1 : q%(2) = 1 : q%(3) = 2
c% = 0
FOR i% = 3 TO n%
q%(i%) = q%(i% - q%(i%-1)) + q%(i% - q%(i%-2))
IF q%(i%) < q%(i%-1) THEN c% += 1
NEXT
= q%(n%)</lang>
- Output:
First 10 terms of Q = 1 1 2 3 3 4 5 5 6 6 1000th term = 502 100000th term = 48157 Term is less than preceding term 49798 times
Bracmat
<lang bracmat>( 0:?memocells & tbl$(memo,!memocells+1) { allocate array } & ( Q
=
. !arg:(1|2)&1
| !arg:>2
& ( !arg:>!memocells:?memocells { Array is too small. }
& tbl$(memo,!memocells+1) { Let array grow to needed size. }
| { Array is not too small. }
)
& ( !(!arg$memo):>0 { Set index to !arg. Return value at index if > 0 }
| Q$(!arg+-1*Q$(!arg+-1))+Q$(!arg+-1*Q$(!arg+-2))
: ?(!arg$?memo) { Set index to !arg. Store value just found. }
)
)
& 0:?i & whl
' (1+!i:~>10:?i&put$(str$(Q$!i " ")))
& put$\n & whl'(1+!i:~>1000:?i&Q$!i) & out$(Q$1000) & 0:?previous:?lessThan:?i & whl
' ( 1+!i:~>100000:?i
& Q$!i
: ( <!previous&1+!lessThan:?lessThan
| ?
)
: ?previous
)
& out$!lessThan );</lang> Output:
1 1 2 3 3 4 5 5 6 6 502 49798
C
<lang c>#include <stdio.h>
- include <stdlib.h>
- define N 100000
int main() { int i, flip, *q = (int*)malloc(sizeof(int) * N) - 1;
q[1] = q[2] = 1;
for (i = 3; i <= N; i++) q[i] = q[i - q[i - 1]] + q[i - q[i - 2]];
for (i = 1; i <= 10; i++) printf("%d%c", q[i], i == 10 ? '\n' : ' ');
printf("%d\n", q[1000]);
for (flip = 0, i = 1; i < N; i++) flip += q[i] > q[i + 1];
printf("flips: %d\n", flip); return 0; }</lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 flips: 49798
C#
<lang C sharp>using System; using System.Collections.Generic;
namespace HofstadterQSequence {
class Program
{
// Initialize the dictionary with the first two indices filled.
private static readonly Dictionary<int, int> QList = new Dictionary<int, int>
{
{1, 1},
{2, 1}
};
private static void Main()
{
int lessThanLast = 0;
/* Initialize our variable that holds the number of times
* a member of the sequence was less than its preceding term. */
for (int n = 1; n <= 100000; n++)
{
int q = Q(n); // Get Q(n).
if (n > 1 && QList[n - 1] > q) // If Q(n) is less than Q(n - 1),
lessThanLast++; // then add to the counter.
if (n > 10 && n != 1000) continue; /* If n is greater than 10 and not 1000,
* the rest of the code in the loop does not apply,
* and it will be skipped. */
if (!Confirm(n, q)) // Confirm Q(n) is correct.
throw new Exception(string.Format("Invalid result: Q({0}) != {1}", n, q));
Console.WriteLine("Q({0}) = {1}", n, q); // Write Q(n) to the console.
}
Console.WriteLine("Number of times a member of the sequence was less than its preceding term: {0}.",
lessThanLast);
}
private static bool Confirm(int n, int value)
{
if (n <= 10)
return new[] {1, 1, 2, 3, 3, 4, 5, 5, 6, 6}[n - 1] == value;
if (n == 1000)
return 502 == value;
throw new ArgumentException("Invalid index.", "n");
}
private static int Q(int n)
{
int q;
if (!QList.TryGetValue(n, out q)) // Try to get Q(n) from the dictionary.
{
q = Q(n - Q(n - 1)) + Q(n - Q(n - 2)); // If it's not available, then calculate it.
QList.Add(n, q); // Add it to the dictionary.
}
return q;
}
}
}</lang>
- Output:
Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 Number of times a member of the sequence was less than its preceding term: 49798.
C++
solution modeled after Perl solution
<lang Cpp>#include <iostream>
int main() {
const int size = 100000;
int hofstadters[size] = { 1, 1 };
for (int i = 3 ; i < size; i++)
hofstadters[ i - 1 ] = hofstadters[ i - 1 - hofstadters[ i - 1 - 1 ]] +
hofstadters[ i - 1 - hofstadters[ i - 2 - 1 ]];
std::cout << "The first 10 numbers are: ";
for (int i = 0; i < 10; i++)
std::cout << hofstadters[ i ] << ' ';
std::cout << std::endl << "The 1000'th term is " << hofstadters[ 999 ] << " !" << std::endl;
int less_than_preceding = 0;
for (int i = 0; i < size - 1; i++)
if (hofstadters[ i + 1 ] < hofstadters[ i ])
less_than_preceding++;
std::cout << "In array of size: " << size << ", "; std::cout << less_than_preceding << " times a number was preceded by a greater number!" << std::endl; return 0;
}</lang>
- Output:
The first 10 numbers are: 1 1 2 3 3 4 5 5 6 6 The 1000'th term is 502 ! In array of size: 100000, 49798 times a number was preceded by a greater number!
Clojure
The qs function, given the initial subsequence of Q of length n, produces the initial subsequence of length n+1. The subsequences are vectors for efficient indexing. qfirst iterates qs so the nth iteration is Q{1..n]. <lang clojure>(defn qs [q]
(let [n (count q)]
(condp = n
0 [1]
1 [1 1]
(conj q (+ (q (- n (q (- n 1))))
(q (- n (q (- n 2)))))))))
(defn qfirst [n] (-> (iterate qs []) (nth n)))
(println "first 10:" (qfirst 10)) (println "1000th:" (last (qfirst 1000))) (println "extra credit:" (->> (qfirst 100000) (partition 2 1) (filter #(apply > %)) count))</lang>
- Output:
<lang>first 10: [1 1 2 3 3 4 5 5 6 6] 1000th: 502 extra credit: 49798</lang>
CoffeeScript
<lang coffeescript>hofstadterQ = do ->
memo = [ 1 ,1, 1]
Q = (n) ->
result = memo[n]
if typeof result != 'number'
result = memo[n] = Q(n - Q(n - 1)) + Q(n - Q(n - 2))
result
- some results:
console.log 'Q(' + i + ') = ' + hofstadterQ(i) for i in [1..10] console.log 'Q(1000) = ' + hofstadterQ(1000)</lang>
- Output:
Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502
Common Lisp
<lang lisp>(defparameter *mm* (make-hash-table :test #'equal))
- generic memoization macro
(defmacro defun-memoize (f (&rest args) &body body)
(defmacro hash () `(gethash (cons ',f (list ,@args)) *mm*))
(let ((h (gensym)))
`(defun ,f (,@args)
(let ((,h (hash)))
(if ,h ,h (setf (hash) (progn ,@body)))))))
- def q
(defun-memoize q (n)
(if (<= n 2) 1
(+ (q (- n (q (- n 1))))
(q (- n (q (- n 2)))))))
- test
(format t "First of Q: ~a~%Q(1000): ~a~%Bumps up to 100000: ~a~%" (loop for i from 1 to 10 collect (q i)) (q 1000) (loop with c = 0 with last-q = (q 1) for i from 2 to 100000 do (let ((next-q (q i))) (if (< next-q last-q) (incf c)) (setf last-q next-q)) finally (return c)))</lang>
- Output:
First of Q: (1 1 2 3 3 4 5 5 6 6) Q(1000): 502 Bumps up to 100000: 49798
Although the above definition of q is more general, for this specific problem the following is faster:<lang lisp>(let ((cc (make-array 3 :element-type 'integer
:initial-element 1
:adjustable t
:fill-pointer 3)))
(defun q (n)
(when (>= n (length cc)) (loop for i from (length cc) below n do (q i)) (vector-push-extend (+ (aref cc (- n (aref cc (- n 1)))) (aref cc (- n (aref cc (- n 2))))) cc)) (aref cc n)))</lang>
Cowgol
<lang cowgol>include "cowgol.coh";
- Generate 1000 terms of the Q sequence
var Q: uint16[1001]; Q[1] := 1; Q[2] := 1;
var n: @indexof Q := 3; while n <= 1000 loop
Q[n] := Q[n-Q[n-1]] + Q[n-Q[n-2]]; n := n + 1;
end loop;
- Print first 10 terms
print("The first 10 terms are: "); n := 1; while n <= 10 loop
print_i16(Q[n]);
print_char(' ');
n := n + 1;
end loop; print_nl();
- Print 1000th term
print("The 1000th term is: "); print_i16(Q[1000]); print_nl();</lang>
- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502
D
<lang d>import std.stdio, std.algorithm, std.functional, std.range;
int Q(in int n) nothrow in {
assert(n > 0);
} body {
alias mQ = memoize!Q;
if (n == 1 || n == 2)
return 1;
else
return mQ(n - mQ(n - 1)) + mQ(n - mQ(n - 2));
}
void main() {
writeln("Q(n) for n = [1..10] is: ", iota(1, 11).map!Q);
writeln("Q(1000) = ", Q(1000));
writefln("Q(i) is less than Q(i-1) for i [2..100_000] %d times.",
iota(2, 100_001).count!(i => Q(i) < Q(i - 1)));
} </lang>
- Output:
Q(n) for n = [1..10] is: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] Q(1000) = 502 Q(i) is less than Q(i-1) for i [2..100_000] 49798 times.
Faster Version
Same output. <lang d>import std.stdio, std.algorithm, std.range, std.array;
uint Q(in int n) nothrow in {
assert(n > 0);
} body {
__gshared static Appender!(int[]) s = [0, 1, 1];
foreach (immutable i; s.data.length .. n + 1)
s ~= s.data[i - s.data[i - 1]] + s.data[i - s.data[i - 2]];
return s.data[n];
}
void main() {
writeln("Q(n) for n = [1..10] is: ", iota(1, 11).map!Q);
writeln("Q(1000) = ", Q(1000));
writefln("Q(i) is less than Q(i-1) for i [2..100_000] %d times.",
iota(2, 100_001).count!(i => Q(i) < Q(i - 1)));
} </lang>
Even Faster Version
This code is here to show that you don't have to use all fancy features of D. Straightforward simple code is often clearer, and faster.
<lang d> import std.stdio;
int[100_000] Q;
void main() { Q[0] = 1; Q[1] = 1;
for (int i = 2; i < 100_000; i++) { Q[i] = Q[i - Q[i - 1]] + Q[i - Q[i - 2]]; }
write("Q(1..10) : "); for (int i = 0; i < 10; i++) { write(" ", Q[i]); } writeln;
write("Q(1000) : "); writeln(Q[999]);
int lt = 0; for (int i = 1; i < 100_000; i++) { if( Q[i-1] > Q[i] ) lt++; }
writefln("Q(i) is less than Q(i-1) for i [2..100_000] %d times.", lt); } </lang>
Dart
Naive version using only recursion (Q(1000) fails due to browser script runtime restrictions) <lang dart>int Q(int n) => n>2 ? Q(n-Q(n-1))+Q(n-Q(n-2)) : 1;
main() {
for(int i=1;i<=10;i++) {
print("Q($i)=${Q(i)}");
}
print("Q(1000)=${Q(1000)}");
}</lang>
Version featuring caching. <lang dart>class Q {
Map<int,int> _table;
Q() {
_table=new Map<int,int>();
_table[1]=1;
_table[2]=1;
}
int q(int n) {
// if the cache is not filled until n-1, fill it starting with the lowest entries first
// this avoids doing a recursion from n to 2 (e.g. if you call q(1000000) first)
// this doesn't happen in the tasks calls since the cache is filled ascending
if(_table[n-1]==null) {
for(int i=_table.length;i<n;i++) {
q(i); }
}
if(_table[n]==null) {
_table[n]=q(n-q(n-1))+q(n-q(n-2));
}
return _table[n]; }
}
main() {
Q q=new Q();
for(int i=1;i<=10;i++) {
print("Q($i)=${q.q(i)}");
}
print("Q(1000)=${q.q(1000)}");
int count=0;
for(int i=2;i<=100000;i++) {
if(q.q(i)<q.q(i-1)) {
count++;
}
}
print("value is smaller than previous $count times");
}</lang>
- Output:
Q(1)=1 Q(2)=1 Q(3)=2 Q(4)=3 Q(5)=3 Q(6)=4 Q(7)=5 Q(8)=5 Q(9)=6 Q(10)=6 Q(1000)=502 value is smaller than previous 49798 times
If the maximum number is known, filling an array is probably the fastest solution. <lang dart>main() {
List<int> q=new List<int>(100001);
q[1]=q[2]=1;
int count=0;
for(int i=3;i<q.length;i++) {
q[i]=q[i-q[i-1]]+q[i-q[i-2]];
if(q[i]<q[i-1]) {
count++;
}
}
for(int i=1;i<=10;i++) {
print("Q($i)=${q[i]}");
}
print("Q(1000)=${q[1000]}");
print("value is smaller than previous $count times");
}</lang>
EchoLisp
<lang scheme> (define RECURSE_BUMP 500) ;; minimum of chrome:500 safari:1000 firefox:2000
- count flips
(define (flips N) (for/sum ((n (in-range 2 (1+ N)))) #:when (< (Q n) (Q (1- n))) 1))
(cache-size 120000) (define (Q n) ;; prevent browser stack overflow at low-cost (when (zero? (modulo n RECURSE_BUMP)) (for ((i (in-range 0 n RECURSE_BUMP ))) (Q i))) (+ (Q (- n (Q (1- n)))) (Q (- n (Q (- n 2)))))) (remember 'Q #(1 1 1)) ;; memoize and init
- first call
- check stack OK
(Q 100000) → 48157
(for ((i 11)) (write (Q i))) 1 1 1 2 3 3 4 5 5 6 6
(Q 1000) → 502 (flips 100000) → 49798 </lang>
Eiffel
<lang Eiffel> class APPLICATION
create make
feature
make -- Test output of the feature hofstadter_q_sequence. local count, i: INTEGER test: ARRAY [INTEGER] do io.put_string ("%NFirst ten numbers: %N") test := hofstadter_q_sequence (10) across test as ar loop io.put_string (ar.item.out + "%T") end test := hofstadter_q_sequence (100000) io.put_string ("1000th:%N") io.put_integer (test [1000]) io.put_string ("%NNumber of Flips:%N") from i := 2 until i > 100000 loop if test [i] < test [i - 1] then count := count + 1 end i := i + 1 end io.put_integer (count) end
hofstadter_q_sequence (lim: INTEGER): ARRAY [INTEGER] -- Hofstadter Q Sequence up to 'lim'. require lim_positive: lim > 0 local q: ARRAY [INTEGER] i: INTEGER do create Result.make_filled (1, 1, lim) Result [1] := 1 Result [2] := 1 from i := 3 until i > lim loop Result [i] := Result [i - Result [i - 1]] + Result [i - Result [i - 2]] i := i + 1 end end
end
</lang>
- Output:
First ten numbers: 1 1 2 3 3 4 5 5 6 6 1000th: 502 Number of Flips: 49798
Elixir
changed collection (Erlang array => Map) <lang elixir>defmodule Hofstadter do
defp flip(v2, v1) when v1 > v2, do: 1
defp flip(_v2, _v1), do: 0
defp list_terms(max, n, acc), do: Enum.map_join(n..max, ", ", &acc[&1])
defp hofstadter(n, n, acc, flips) do
IO.puts "The first ten terms are: #{list_terms(10, 1, acc)}"
IO.puts "The 1000'th term is #{acc[1000]}"
IO.puts "Number of flips: #{flips}"
end
defp hofstadter(max, n, acc, flips) do
qn1 = acc[n-1]
qn = acc[n - qn1] + acc[n - acc[n-2]]
hofstadter(max, n+1, Map.put(acc, n, qn), flips + flip(qn, qn1))
end
def main(max \\ 100_000) do
acc = %{1 => 1, 2 => 1}
hofstadter(max+1, 3, acc, 0)
end
end
Hofstadter.main</lang>
- Output:
The first ten terms are: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 The 1000'th term is 502 Number of flips: 49798
Erlang
<lang erlang>%% @author Jan Willem Luiten <jwl@secondmove.com> %% Hofstadter Q Sequence for Rosetta Code
-module(hofstadter). -export([main/0]). -define(MAX, 100000).
flip(V2, V1) when V1 > V2 -> 1; flip(_V2, _V1) -> 0.
list_terms(N, N, Acc) -> io:format("~w~n", [array:get(N, Acc)]); list_terms(Max, N, Acc) -> io:format("~w, ", [array:get(N, Acc)]), list_terms(Max, N+1, Acc).
hofstadter(N, N, Acc, Flips) -> io:format("The first ten terms are: "), list_terms(9, 0, Acc), io:format("The 1000'th term is ~w~n", [array:get(999, Acc)]), io:format("Number of flips: ~w~n", [Flips]); hofstadter(Max, N, Acc, Flips) -> Qn1 = array:get(N-1, Acc), Qn = array:get(N - Qn1, Acc) + array:get(N - array:get(N-2, Acc), Acc), hofstadter(Max, N+1, array:set(N, Qn, Acc), Flips + flip(Qn, Qn1)).
main() -> Tmp = array:set(0, 1, array:new(?MAX)), Acc = array:set(1, 1, Tmp), hofstadter(?MAX, 2, Acc, 0). </lang>
- Output:
The first ten terms are: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 The 1000'th term is 502 Number of flips: 49798
ERRE
- ERRE:
<lang ERRE> PROGRAM HOFSTADER_Q
! ! for rosettacode.org !
DIM Q%[10000]
PROCEDURE QSEQUENCE(Q,FLAG%->SEQ$) ! if FLAG% is true accumulate sequence in SEQ$ ! (attention to string var lenght=255) ! otherwise calculate values in Q%[] only
LOCAL N Q%[1]=1 Q%[2]=1 SEQ$="1 1" IF NOT FLAG% THEN Q=NUM END IF FOR N=3 TO Q DO Q%[N]=Q%[N-Q%[N-1]]+Q%[N-Q%[N-2]] IF FLAG% THEN SEQ$=SEQ$+STR$(Q%[N]) END IF END FOR
END PROCEDURE
BEGIN
NUM=10000
QSEQUENCE(10,TRUE->SEQ$)
PRINT("Q-sequence(1..10) : ";SEQ$)
QSEQUENCE(1000,FALSE->SEQ$)
PRINT("1000th number of Q sequence : ";Q%[1000])
FOR N=2 TO NUM DO
IF Q%[N]<Q%[N-1] THEN NN+=1 END IF
END FOR
PRINT("Number of Q(n)<Q(n+1) for n<=10000 : ";NN)
END PROGRAM </lang> Note: The extra credit was limited to 10000 because memory addressable range is limited to 64K. If you want to implement extra credit for 100,000 you must use external file for array Q%[].
F#
The function
<lang fsharp> // Populate an array with values of Hofstadter Q sequence. Nigel Galloway: August 26th., 2020 let fQ N=let g=Array.length N in N.[0]<-1; N.[1]<-1;(for g in 2..g-1 do N.[g]<-N.[g-N.[g-1]]+N.[g-N.[g-2]]) </lang>
The Tasks
<lang fsharp> let Q=Array.zeroCreate<int>10 in fQ Q; printfn "%A" Q let Q=Array.zeroCreate<int>1000 in fQ Q; printfn "%d" (Array.last Q) </lang>
- Output:
[|1; 1; 2; 3; 3; 4; 5; 5; 6; 6|] 502
Extra Credit
<lang fsharp> let Q=Array.zeroCreate<int>100000 in fQ Q; printfn "%d" (Q|>Seq.pairwise|>Seq.sumBy(fun(n,g)->if n>g then 1 else 0)) </lang>
- Output:
49798
- What is a large number?
<lang fsharp> let Q=Array.zeroCreate<int>2500000000 in fQ Q; printfn "%d" (Array.last Q) let Q=Array.zeroCreate<int>5000000000 in fQ Q; printfn "%d" (Array.last Q) </lang>
- Output:
121648520 (in 0m14.347s) 247777817 (in 0m37.757s)
Factor
We define a method next that takes a sequence of the first n Q values and appends the next one to it. Then we perform it 1000 times on { 1 1 } and show the first 10 and 999th (because the list is zero-indexed) elements.
<lang factor>( scratchpad ) : next ( seq -- newseq )
dup 2 tail* over length [ swap - ] curry map
[ dupd swap nth ] map 0 [ + ] reduce suffix ;
( scratchpad ) { 1 1 } 1000 [ next ] times dup 10 head . 999 swap nth . { 1 1 2 3 3 4 5 5 6 6 } 502</lang>
Fermat
<lang>Func Hq(n) = if n<2 then 1 else
Array qq[n+1]; qq[1] := 1; qq[2] := 1; for i = 3, n do qq[i]:=qq[i-qq[i-1]]+qq[i-qq[i-2]] od; Return(qq[n]); fi; .
for i=1 to 10 do !Hq(i);!' ' od; Hq(1000)</lang>
- Output:
1 1 2 3 3 4 5 5 6 6
502
Forth
<lang forth>100000 constant N
- q ( n -- addr ) cells here + ;
- qinit
1 0 q ! 1 1 q ! N 2 do i i 1- q @ - q @ i i 2 - q @ - q @ + i q ! loop ;
- flips
." flips: " 0 N 1 do i q @ i 1- q @ < if 1+ then loop . cr ;
- qprint ( n -- )
0 do i q @ . loop cr ;
qinit 10 qprint 999 q @ . cr flips bye</lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 flips: 49798
Fortran
The latter-day function COUNT(logical expression) could easily be replaced by a simple test-and-count in the DO-loop preparing the array. One hopes that the compiler produces sensible code rather than creating an auxiliary array of boolean results then counting the true values. Rather more clunky is the need to employ odd structure for the input loop so as to handle possible bad input (text, rather than a valid number, for example) and who knows, end-of-file might happen also.
<lang Fortran> Calculate the Hofstadter Q-sequence, using a big array rather than recursion.
INTEGER ENUFF
PARAMETER (ENUFF = 100000)
INTEGER Q(ENUFF) !Lots of memory these days.
Q(1) = 1 !Initial values as per the definition.
Q(2) = 1
Q(3:) = -123456789!This will surely cause trouble!
DO I = 3,ENUFF !For values beyond the second,
Q(I) = Q(I - Q(I - 1)) + Q(I - Q(I - 2)) !Reach back according to the last two values.
END DO
Cast forth results as per the specification.
WRITE (6,1) Q(1:10) !Should be 1 1 2 3 3 4 5 5 6 6...
1 FORMAT ("First ten values:",10I2) !Known to be one-digit numbers.
WRITE (6,*) "Q(1000) =",Q(1000) !Should be 502.
WRITE (6,3) ENUFF,COUNT(Q(2:ENUFF) < Q(1:ENUFF - 1)) !Please don't create a temporary array!
3 FORMAT ("Count of those elements 2:",I0,
1 " which are less than their predecessor: ",I0) !Should be 49798.
Curry favour by allowing enquiries.
10 WRITE (6,11) ENUFF
11 FORMAT ("Nominate an index (in 1:",I0,"): ",$) !Obviously, the $ says don't start a new line.
READ (5,*,END = 999, ERR = 999) I !Ask for a number, with precautions.
IF (I.GT.0 .AND. I.LE.ENUFF) THEN !A good number, but, within range?
WRITE (6,12) I,Q(I) !Yes. Reveal the requested value.
12 FORMAT ("Q(",I0,") = ",I0) !This should do.
GO TO 10 !And ask again.
END IF ! WHILE read(5,*) i & i > 0 & i < enuff DO write(6,*) "Q(",i,")=",Q(i);
Closedown.
999 WRITE (6,*) "Bye."
END
</lang>
Output:
First ten values: 1 1 2 3 3 4 5 5 6 6 Q(1000) = 502 Count of those elements 2:100000 which are less than their predecessor: 49798 Nominate an index (in 1:100000): 100000 Q(100000) = 48157 Nominate an index (in 1:100000): 0 Bye.
FreeBASIC
<lang freebasic> Const limite = 100000
Dim As Long Q(limite), i, cont = 0
Q(1) = 1 Q(2) = 1 For i = 3 To limite
Q(i) = Q(i-Q(i-1)) + Q(i-Q(i-2)) If Q(i) < Q(i-1) Then cont += 1
Next i
Print "Primeros 10 terminos: "; For i = 1 To 10
Print Q(i) &" ";
Next i Print
Print "Termino numero 1000: "; Q(1000)
Print "Terminos menores que los anteriores: " &cont End </lang>
- Output:
Primeros 10 terminos: 1 1 2 3 3 4 5 5 6 6 Termino numero 1000: 502 Terminos menores que los anteriores: 49798
Fōrmulæ
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Go
Sure there are ways that run faster or handle larger numbers; for the task though, maps and recursion work just fine. <lang go>package main
import "fmt"
var m map[int]int
func initMap() {
m = make(map[int]int) m[1] = 1 m[2] = 1
}
func q(n int) (r int) {
if r = m[n]; r == 0 {
r = q(n-q(n-1)) + q(n-q(n-2))
m[n] = r
}
return
}
func main() {
initMap()
// task
for n := 1; n <= 10; n++ {
showQ(n)
}
// task
showQ(1000)
// extra credit
count, p := 0, 1
for n := 2; n <= 1e5; n++ {
qn := q(n)
if qn < p {
count++
}
p = qn
}
fmt.Println("count:", count)
// extra credit
initMap()
showQ(1e6)
}
func showQ(n int) {
fmt.Printf("Q(%d) = %d\n", n, q(n))
}</lang>
- Output:
Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 count: 49798 Q(1000000) = 512066
GW-BASIC
<lang gwbasic>10 DIM Q!(1000) 20 Q(1) = 1: Q(2) = 1 30 FOR N = 3 TO 1000 40 Q(N) = Q(N - Q(N - 1)) + Q(N - Q(N - 2)) 50 NEXT N 60 FOR N = 1 TO 10 70 PRINT Q(N) 80 NEXT N 90 PRINT Q(1000)</lang>
Haskell
The basic task:
<lang Haskell>qSequence = tail qq where
qq = 0 : 1 : 1 : map g [3..] g n = qq !! (n - qq !! (n-1)) + qq !! (n - qq !! (n-2))
-- Output:
- Main> (take 10 qSequence, qSequence !! (1000-1))
([1,1,2,3,3,4,5,5,6,6],502) (0.00 secs, 525044 bytes)</lang>
Extra credit task:
<lang Haskell>import Data.Array
qSequence n = arr
where
arr = listArray (1,n) $ 1:1: map g [3..n]
g i = arr!(i - arr!(i-1)) +
arr!(i - arr!(i-2))
gradualth m k arr -- gradually precalculate m-th item
| m <= v = pre `seq` arr!m -- in steps of k where -- to prevent STACK OVERFLOW pre = foldl1 (\a b-> a `seq` arr!b) [u,u+k..m] (u,v) = bounds arr
qSeqTest m n = let arr = qSequence $ max m n in
( take 10 . elems $ arr -- 10 first items
, gradualth m 10000 $ arr -- m-th item
, length . filter (> 0) -- reversals in n items
. _S (zipWith (-)) tail . take n . elems $ arr )
_S f g x = f x (g x)</lang>
- Output:
<lang Haskell>Prelude Main> qSeqTest 1000 100000 -- reversals in 100,000 ([1,1,2,3,3,4,5,5,6,6],502,49798) (0.09 secs, 18879708 bytes)
Prelude Main> qSeqTest 1000000 100000 -- 1,000,000-th item ([1,1,2,3,3,4,5,5,6,6],512066,49798) (2.80 secs, 87559640 bytes)</lang>
Using a list (more or less) seemlessly backed up by a double resizing array: <lang haskell>q = qq (listArray (1,2) [1,1]) 1 where
qq ar n = (arr!n) : qq arr (n+1) where
l = snd (bounds ar)
step n =arr!(n - (fromIntegral (arr!(n - 1)))) +
arr!(n - (fromIntegral (arr!(n - 2))))
arr :: Array Int Integer
arr | n <= l = ar
| otherwise = listArray (1, l*2)$
([ar!i | i <- [1..l]] ++
[step i | i <- [l+1..l*2]])
main = do
putStr("first 10: "); print (take 10 q)
putStr("1000-th: "); print (q !! 999)
putStr("flips: ")
print $ length $ filter id $ take 100000 (zipWith (>) q (tail q))</lang>
- Output:
first 10: [1,1,2,3,3,4,5,5,6,6] 1000-th: 502 flips: 49798
List backed up by a list of arrays, with nominal constant lookup time. Somehow faster than the previous method. <lang haskell>import Data.Array import Data.Int (Int64)
q = qq [listArray (1,2) [1,1]] 1 where
qq a n = seek aa n : qq aa (1 + n) where
aa | n <= l = a
| otherwise = listArray (l+1,l*2) (take l $ drop 2 lst):a
where
l = snd (bounds $ head a)
lst = seek a (l-1):seek a l:(ext lst (l+1))
ext (q1:q2:qs) i = (g (i-q2) + g (i-q1)):ext (q2:qs) (1+i)
g = seek aa
seek (ar:ars) n
| n >= fst (bounds ar) = ar ! n
| otherwise = seek ars n
-- Only a perf test. Task can be done exactly the same as above main = print $ sum qqq
where
qqq :: [Int64]
qqq = map fromIntegral $ take 3000000 q</lang>
Icon and Unicon
<lang Icon>link printf
procedure main()
V := [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] every i := 1 to *V do
if Q(i) ~= V[i] then stop("Assertion failure for position ",i)
printf("Q(1 to %d) - verified.\n",*V)
q := Q(n := 1000) v := 502 printf("Q[%d]=%d - %s.\n",n,v,if q = v then "verified" else "failed")
invcount := 0 every i := 2 to (n := 100000) do
if Q(i) < Q(i-1) then {
printf("Q(%d)=%d < Q(%d)=%d\n",i,Q(i),i-1,Q(i-1))
invcount +:= 1
}
printf("There were %d inversions in Q up to %d\n",invcount,n) end
procedure Q(n) #: Hofstader Q sequence static S initial S := [1,1]
if q := S[n] then return q else {
q := Q(n - Q(n - 1)) + Q(n - Q(n - 2))
if *S = n - 1 then {
put(S,q)
return q
}
else
runerr(500,n)
}
end</lang>
printf.icn provides formatting
- Output:
Q(1 to 10) - verified. Q[1000]=502 - verified. Q(16)=9 < Q(15)=10 Q(25)=14 < Q(24)=16 Q(32)=17 < Q(31)=20 Q(36)=19 < Q(35)=21 ... Q(99996)=48252 < Q(99995)=50276 Q(99999)=48456 < Q(99998)=50901 Q(100000)=48157 < Q(99999)=48456 There were 49798 inversions in Q up to 100000
IS-BASIC
<lang IS-BASIC>100 PROGRAM "QSequen.bas" 110 LET LIMIT=1000 120 NUMERIC Q(1 TO LIMIT) 130 LET Q(1),Q(2)=1 140 FOR I=3 TO LIMIT 150 LET Q(I)=Q(I-Q(I-1))+Q(I-Q(I-2)) 160 NEXT 170 PRINT "First 10 terms:" 180 FOR I=1 TO 10 190 PRINT Q(I); 200 NEXT 210 PRINT :PRINT "Term 1000:";Q(1000)</lang>
J
Solution (bottom-up):<lang j> Qs=:0 1 1
Q=: verb define
n=. >./,y
while. n>:#Qs do.
Qs=: Qs,+/(-_2{.Qs){Qs
end.
y{Qs
)</lang>
Solution (top-down):<lang j> Q=: 1:`(+&$:/@:- $:@-& 1 2)@.(>&2)"0 M.</lang>
Example:<lang j> Q 1+i.10 1 1 2 3 3 4 5 5 6 6
Q 1000
502
+/2>/\ Q 1+i.100000
49798</lang>
Note: The bottom-up solution uses iteration and doesn't risk failure due to recursion limits or cache overflows. The top-down solution uses recursion, and likely hews closer to the spirit of the task. While this latter uses memoization/caching, at some point it will still hit a recursion limit (depends on the environment; in mine, it barfs at N=4402). We use the bottom up version for the extra credit part of this task (the expression which compares adjacent numbers and gave us the result 49798).
It happens to be that the bottom-up version is written in the "explicit" style of code and the top-down version is written in the "tacit" (aka "point-free") style. This is incidental and it's possible to write bottom-up tacitly and/or top-down explicitly.
The top-down version may be interesting as an example of algebraic factorization of code: taking advantage of some unique function composition operations in J, it manages to only mention $: (aka recursion aka "Q") twice.
Java
This example also counts the number of times each n is used as an argument up to 100000 and reports the one that was used the most. <lang java5>import java.util.HashMap; import java.util.Map;
public class HofQ { private static Map<Integer, Integer> q = new HashMap<Integer, Integer>(){{ put(1, 1); put(2, 1); }};
private static int[] nUses = new int[100001];//not part of the task
public static int Q(int n){ nUses[n]++;//not part of the task if(q.containsKey(n)){ return q.get(n); } int ans = Q(n - Q(n - 1)) + Q(n - Q(n - 2)); q.put(n, ans); return ans; }
public static void main(String[] args){ for(int i = 1; i <= 10; i++){ System.out.println("Q(" + i + ") = " + Q(i)); } int last = 6;//value for Q(10) int count = 0; for(int i = 11; i <= 100000; i++){ int curr = Q(i); if(curr < last) count++; last = curr; if(i == 1000) System.out.println("Q(1000) = " + curr); } System.out.println("Q(i) is less than Q(i-1) for i <= 100000 " + count + " times");
//Optional stuff below here int maxUses = 0, maxN = 0; for(int i = 1; i<nUses.length;i++){ if(nUses[i] > maxUses){ maxUses = nUses[i]; maxN = i; } } System.out.println("Q(" + maxN + ") was called the most with " + maxUses + " calls"); } }</lang>
- Output:
Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 Q(i) is less than Q(i-1) for i <= 100000 49798 times Q(44710) was called the most with 19 calls
JavaScript
ES5
Based on memoization example from 'JavaScript: The Good Parts'. <lang JavaScript>var hofstadterQ = function() {
var memo = [1,1,1];
var Q = function (n) {
var result = memo[n];
if (typeof result !== 'number') {
result = Q(n - Q(n-1)) + Q(n - Q(n-2));
memo[n] = result;
}
return result;
};
return Q;
}();
for (var i = 1; i <=10; i += 1) {
console.log('Q('+ i +') = ' + hofstadterQ(i));
}
console.log('Q(1000) = ' + hofstadterQ(1000)); </lang>
- Output:
Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502
ES6
Memoising with the accumulator of a fold <lang JavaScript>(() => {
'use strict';
// hofQSeq :: Int -> [Int]
const hofQSeq = x =>
x > 2 ? tail(foldl((Q, n) =>
n < 3 ? Q : Q.concat(
Q[n - Q[n - 1]] + Q[n - Q[n - 2]]
), [0, 1, 1],
range(1, x))) : (x > 0 ? take(x, [1, 1]) : undefined);
// GENERIC FUNCTIONS -------------------------------------------
// foldl :: (b -> a -> b) -> b -> [a] -> b const foldl = (f, a, xs) => xs.reduce(f, a),
// range :: Int -> Int -> [Int]
range = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i),
// tail :: [a] -> [a]
tail = xs => xs.length ? xs.slice(1) : undefined,
// last :: [a] -> a
last = xs => xs.length ? xs.slice(-1)[0] : undefined,
// Int -> [a] -> [a]
take = (n, xs) => xs.slice(0, n);
// TEST --------------------------------------------------------
return {
firstTen: hofQSeq(10),
thousandth: last(hofQSeq(1000)),
'Q<Q-1UpTo10E5': hofQSeq(100000)
.reduce((a, x, i, xs) => x < xs[i - 1] ? a + 1 : a, 0)
};
})();</lang>
- Output:
<lang JavaScript>{"firstTen":[1, 1, 2, 3, 3, 4, 5, 5, 6, 6],
"thousandth":502, "Q<Q-1UpTo10E5":49798}</lang>
jq
For the tasks related to evaluating Q(n) directy, a recursive implementation is used, firstly because the task requirements refer to "recursion limits", and secondly to demonstrate one way to handle a cache in a functional language. To count the number of inversions, a non-recursive approach is used as it is faster and scales linearly.
For simplicity, we also define Q(0) = 1, so that the defining formula also holds for n == 2, and so that we can cache Q(n) at the n-th position of an array with index origin 0. <lang jq>
- For n>=2, Q(n) = Q(n - Q(n-1)) + Q(n - Q(n-2))
def Q:
def Q(n):
n as $n
| (if . == null then [1,1,1] else . end) as $q
| if $q[$n] != null then $q
else
$q | Q($n-1) as $q1
| $q1 | Q($n-2) as $q2
| $q2 | Q($n - $q2[$n - 1]) as $q3 # Q(n - Q(n-1))
| $q3 | Q($n - $q3[$n - 2]) as $q4 # Q(n - Q(n-2))
| ($q4[$n - $q4[$n-1]] + $q4[$n - $q4[$n -2]]) as $ans
| $q4 | setpath( [$n]; $ans)
end ;
. as $n | null | Q($n) | .[$n];
- count the number of times Q(i) > Q(i+1) for 0 < i < n
def flips(n):
(reduce range(3; n) as $n
([1,1,1]; . + [ .[$n - .[$n-1]] + .[$n - .[$n - 2 ]] ] )) as $q
| reduce range(0; n) as $i
(0; . + (if $q[$i] > $q[$i + 1] then 1 else 0 end)) ;
- The three tasks:
((range(0;11), 1000) | "Q(\(.)) = \( . | Q)"),
(100000 | "flips(\(.)) = \(flips(.))")</lang>
Transcript
<lang bash> $ uname -a Darwin Mac-mini 13.3.0 Darwin Kernel Version 13.3.0: Tue Jun 3 21:27:35 PDT 2014; root:xnu-2422.110.17~1/RELEASE_X86_64 x86_64 $ time jq -r -n -f hofstadter.jq Q(0) = 1 Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 flips(100000) = 49798
real 0m0.562s user 0m0.541s sys 0m0.011s</lang>
Julia
The following implementation accepts an argument that is a single integer, an array of integers, or a range: <lang julia>function hofstQseq(n, typerst::Type=Int)
nmax = maximum(n)
r = Vector{typerst}(nmax)
r[1] = 1
if nmax ≥ 2 r[2] = 1 end
for i in 3:nmax
r[i] = r[i - r[i - 1]] + r[i - r[i - 2]]
end
return r[n]
end
println("First ten elements of sequence: ", join(hofstQseq(1:10), ", ")) println("1000-th element: ", hofstQseq(1000)) </lang>
- Output:
First ten elements of sequence: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 1000-th element: 502
And we can also count the number of times a value is less than its predecessor by, for example: <lang julia>seq = hofstQseq(1:100_000) cnt = count(diff(seq) .< 0) println("$cnt elements are less than the preceding one.")</lang>
- Output:
49798 elements are less than the preceding one.
Since the implementation is non-recursive, there is no issue with recursion limits.
Kotlin
<lang scala>// version 1.1.4
fun main(args: Array<String>) {
val q = IntArray(100_001)
q[1] = 1
q[2] = 1
for (n in 3..100_000) q[n] = q[n - q[n - 1]] + q[n - q[n - 2]]
print("The first 10 terms are : ")
for (i in 1..10) print("${q[i]} ")
println("\n\nThe 1000th term is : ${q[1000]}")
val flips = (2..100_000).count { q[it] < q[it - 1] }
println("\nThe number of flips for the first 100,000 terms is : $flips")
}</lang>
- Output:
The first 10 terms are : 1 1 2 3 3 4 5 5 6 6 The 1000th term is : 502 The number of flips for the first 100,000 terms is : 49798
MAD
<lang MAD> NORMAL MODE IS INTEGER
VECTOR VALUES FMT = $2HQ(,I4,3H) =,I4*$
DIMENSION Q(1000)
Q(1) = 1
Q(2) = 1
THROUGH FILL, FOR N=3, 1, N.G.1000
FILL Q(N) = Q(N-Q(N-1)) + Q(N-Q(N-2))
THROUGH SHOW, FOR N=1, 1, N.G.10
SHOW PRINT FORMAT FMT, N, Q(N)
PRINT FORMAT FMT, 1000, Q(1000)
END OF PROGRAM</lang>
- Output:
Q( 1) = 1 Q( 2) = 1 Q( 3) = 2 Q( 4) = 3 Q( 5) = 3 Q( 6) = 4 Q( 7) = 5 Q( 8) = 5 Q( 9) = 6 Q( 10) = 6 Q(1000) = 502
Maple
We use automatic memoisation ("option remember") in the following. The use of "option system" assures that memoised values can be garbage collected. <lang Maple>Q := proc( n )
option remember, system;
if n = 1 or n = 2 then
1
else
thisproc( n - thisproc( n - 1 ) ) + thisproc( n - thisproc( n - 2 ) )
end if
end proc:</lang> From this we get: <lang Maple>> seq( Q( i ), i = 1 .. 10 );
1, 1, 2, 3, 3, 4, 5, 5, 6, 6
> Q( 1000 );
502</lang>
To determine the number of "flips", we proceed as follows. <lang Maple>> flips := 0: > for i from 2 to 100000 do > if L[ i ] < L[ i - 1 ] then > flips := 1 + flips > end if > end do: > flips;
49798</lang>
Alternatively, we can build the sequence in an array. <lang Maple>Qflips := proc( n )
local a := Array( 1 .. n );
a[ 1 ] := 1;
a[ 2 ] := 1;
for local i from 3 to n do
a[ i ] := a[ i - a[ i - 1 ] ] + a[ i - a[ i - 2 ] ]
end do;
local flips := 0;
for i from 2 to n do
if a[ i ] < a[ i - 1 ] then
flips := 1 + flips
end if
end do;
flips
end proc:</lang> This gives the same result. <lang Maple>> Qflips( 10^5 );
49798</lang>
Mathematica / Wolfram Language
<lang Mathematica>Hofstadter[1] = Hofstadter[2] = 1; Hofstadter[n_Integer?Positive] := Hofstadter[n] = Block[{$RecursionLimit = Infinity},
Hofstadter[n - Hofstadter[n - 1]] + Hofstadter[n - Hofstadter[n - 2]]
]</lang>
- Output:
<lang Mathematica>Hofstadter /@ Range[10] {1,1,2,3,3,4,5,5,6,6} Hofstadter[1000] 502 Count[Differences[Hofstadter /@ Range[100000]], _?Negative] 49798</lang>
MATLAB / Octave
This solution pre-allocates memory and is an iterative solution, so caching or recursion limits do not apply. <lang MATLAB>function Q = Qsequence(N)
%% zeros are used to pre-allocate memory, this is not strictly necessary but can significantly improve performance for large N Q = [1,1,zeros(1,N-2)]; for n=3:N Q(n) = Q(n-Q(n-1))+Q(n-Q(n-2)); end;
end; </lang> Confirm and display that the first ten terms of the sequence are: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6
>> Qsequence(10) ans = 1 1 2 3 3 4 5 5 6 6
Confirm and display that the 1000'th term is: 502
>> Q=Qsequence(1000); Q(end) ans = 502
Count and display how many times a member of the sequence is less than its preceding term for terms up to and including the 100,000'th term.
>> sum(diff(Qsequence(100000))<0) ans = 49798
MiniScript
<lang MiniScript>cache = {1:1, 2:1}
Q = function(n)
if not cache.hasIndex(n) then
q = Q(n - Q(n-1)) + Q(n - Q(n-2))
cache[n] = q
end if
return cache[n]
end function
for i in range(1,10)
print "Q(" + i + ") = " + Q(i)
end for print "Q(1000) = " + Q(1000)</lang>
- Output:
Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502
Nim
<lang nim>var q = @[1, 1] for n in 2 ..< 100_000: q.add q[n-q[n-1]] + q[n-q[n-2]]
echo q[0..9] assert q[0..9] == @[1, 1, 2, 3, 3, 4, 5, 5, 6, 6]
echo q[999] assert q[999] == 502
var lessCount = 0 for n in 1 ..< 100_000:
if q[n] < q[n-1]: inc lessCount
echo lessCount</lang>
- Output:
@[1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 502 49798
Oberon-2
Works with oo2c version 2 <lang oberon2> MODULE Hofstadter; IMPORT
Out;
VAR
i,count,q,prev: LONGINT;
founds: ARRAY 100001 OF LONGINT;
PROCEDURE Q(n: LONGINT): LONGINT;
BEGIN
IF founds[n] = 0 THEN
CASE n OF
1 .. 2:
founds[n] := 1
ELSE founds[n] := Q(n - Q(n - 1)) + Q(n - Q(n - 2))
END
END;
RETURN founds[n]
END Q;
BEGIN
(* first ten numbers in the sequence *)
FOR i := 1 TO 10 DO
Out.String("At ");Out.LongInt(i,0);Out.String(":> ");Out.LongInt(Q(i),4);Out.Ln
END;
Out.String("1000th value: ");Out.LongInt(Q(1000),4);Out.Ln;
prev := 1;
FOR i := 2 TO 100000 DO
q := Q(i);
IF q < prev THEN INC(count) END;
prev := q
END;
Out.String("terms less than the previous: ");Out.LongInt(count,4);Out.Ln
END Hofstadter. </lang> Output:
At 1:> 1 At 2:> 1 At 3:> 2 At 4:> 3 At 5:> 3 At 6:> 4 At 7:> 5 At 8:> 5 At 9:> 6 At 10:> 6 1000th value: 502 terms less than the previous: 49798
Oforth
<lang Oforth>: QSeqTask | q i |
ListBuffer newSize(100000) dup add(1) dup add(1) ->q
0 3 100000 for: i [
q add(q at(i q at(i 1-) -) q at(i q at(i 2 -) -) +)
q at(i) q at(i 1-) < ifTrue: [ 1+ ]
]
q left(10) println q at(1000) println println ; </lang>
- Output:
[1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 502 49798
PARI/GP
Straightforward, unoptimized version; about 1 ms. <lang parigp>Q=vector(1000);Q[1]=Q[2]=1;for(n=3,#Q,Q[n]=Q[n-Q[n-1]]+Q[n-Q[n-2]]); Q1=vecextract(Q,"1..10"); print("First 10 terms: "Q1,if(Q1==[1, 1, 2, 3, 3, 4, 5, 5, 6, 6]," (as expected)"," (in error)")); print("1000-th term: "Q[1000],if(Q[1000]==502," (as expected)"," (in error)"));</lang>
- Output:
First 10 terms: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] (as expected) 1000-th term: 502 (as expected)
Pascal
<lang pascal>Program HofstadterQSequence (output);
const
limit = 100000;
var
q: array [1..limit] of longint; i, flips: longint;
begin
q[1] := 1;
q[2] := 1;
for i := 3 to limit do
q[i] := q[i - q[i - 1]] + q[i - q[i - 2]];
for i := 1 to 10 do
write(q[i], ' ');
writeln;
writeln(q[1000]);
flips := 0;
for i := 1 to limit - 1 do
if q[i] > q[i+1] then
inc(flips);
writeln('Flips: ', flips);
end.</lang>
- Output:
:> ./HofstadterQSequence 1 1 2 3 3 4 5 5 6 6 502 Flips: 49798
Perl
<lang Perl>my @Q = (0,1,1); push @Q, $Q[-$Q[-1]] + $Q[-$Q[-2]] for 1..100_000; say "First 10 terms: [@Q[1..10]]"; say "Term 1000: $Q[1000]"; say "Terms less than preceding in first 100k: ",scalar(grep { $Q[$_] < $Q[$_-1] } 2..100000);</lang>
- Output:
First 10 terms: [1 1 2 3 3 4 5 5 6 6] Term 1000: 502 Terms less than preceding in first 100k: 49798
A more verbose and less idiomatic solution: <lang Perl>#!/usr/bin/perl use warnings; use strict;
my @hofstadters = ( 1 , 1 ); while ( @hofstadters < 100000 ) {
my $nextn = @hofstadters + 1;
- array index counting starts at 0 , so we have to subtract 1 from the numbers!
push @hofstadters , $hofstadters [ $nextn - 1 - $hofstadters[ $nextn - 1 - 1 ] ]
+ $hofstadters[ $nextn - 1 - $hofstadters[ $nextn - 2 - 1 ]];
} for my $i ( 0..9 ) {
print "$hofstadters[ $i ]\n";
} print "The 1000'th term is $hofstadters[ 999 ]!\n"; my $less_than_preceding = 0; for my $i ( 0..99998 ) {
$less_than_preceding++ if $hofstadters[ $i + 1 ] < $hofstadters[ $i ];
} print "Up to and including the 100000'th term, $less_than_preceding terms are less " .
"than their preceding terms!\n";
</lang>
- Output:
1 1 2 3 3 4 5 5 6 6 The 1000'th term is 502! Up to and including the 100000'th term, 49798 terms are less than their preceding terms!
This different solution uses tie to make the Q sequence look like a regular array, and only fills the cache on demand. Some pre-allocation is done which provides a minor speed increase for the extra credit. I could have chosen to do recursion instead of iteration, as perl has no limit on how deeply one may recurse, but did not see the benefit of doing so.
<lang Perl>#!perl use strict; use warnings; package Hofstadter; sub TIEARRAY {
bless [undef, 1, 1], shift;
} sub FETCH {
my ($self, $n) = @_;
die if $n < 1;
if( $n > $#$self ) {
my $start = $#$self + 1;
$#$self = $n; # pre-allocate for efficiency
for my $nn ( $start .. $n ) {
my ($a, $b) = (1, 2);
$_ = $self->[ $nn - $_ ] for $a, $b;
$_ = $self->[ $nn - $_ ] for $a, $b;
$self->[$nn] = $a + $b;
}
}
$self->[$n];
}
package main;
tie my (@q), "Hofstadter";
print "@q[1..10]\n"; print $q[1000], "\n";
my $count = 0; for my $n ( 2 .. 100_000 ) {
$count++ if $q[$n] < $q[$n - 1];
} print "Extra credit: $count\n"; </lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 Extra credit: 49798
Phix
Just to be flash, I also calculated the 100 millionth term - the only limiting factor here would be the length of Q (on 32 bit, theoretical max sequence length is 402,653,177). <lang Phix>sequence Q = {1,1}
function q(integer n)
integer l = length(Q)
while n>l do
l += 1
Q &= Q[l-Q[l-1]]+Q[l-Q[l-2]]
end while
return Q[n]
end function
{} = q(10) -- (or collect one by one) printf(1,"First ten terms: %s\n",{sprint(Q[1..10])}) printf(1,"1000th: %d\n",q(1000)) printf(1,"100,000th: %d\n",q(100_000)) integer n = 0 for i=2 to 100_000 do
n += Q[i]<Q[i-1]
end for printf(1,"Flips up to 100,000: %d\n",{n}) atom t0 = time() printf(1,"100,000,000th: %d (%3.2fs)\n",{q(100_000_000),time()-t0})</lang>
- Output:
First ten terms: {1,1,2,3,3,4,5,5,6,6}
1000th: 502
100,000th: 48157
Flips up to 100,000: 49798
100,000,000th: 50166508 (7.53s)
PicoLisp
<lang PicoLisp>(de q (N)
(cache '(NIL) N
(if (>= 2 N)
1
(+
(q (- N (q (dec N))))
(q (- N (q (- N 2)))) ) ) ) )</lang>
Test: <lang PicoLisp>: (mapcar q (range 1 10)) -> (1 1 2 3 3 4 5 5 6 6)
- (q 1000)
-> 502
- (let L (mapcar q (range 1 100000))
(cnt < (cdr L) L) )
-> 49798</lang>
PL/I
<lang PL/I> /* Hofstrader Q sequence for any "n". */
H: procedure options (main); /* 28 January 2012 */
declare n fixed binary(31);
put ('How many values do you want? :');
get (n);
begin;
declare Q(n) fixed binary (31); declare i fixed binary (31);
Q(1), Q(2) = 1;
do i = 1 upthru n;
if i >= 3 then Q(i) = ( Q(i - Q(i-1)) + Q(i - Q(i-2)) );
if i <= 20 then put skip list ('n=' || trim(i), Q(i));
end;
put skip list ('n=' || trim(i), Q(i));
end; end H; </lang>
- Output:
How many values do you want? : n=1 1 n=2 1 n=3 2 n=4 3 n=5 3 n=6 4 n=7 5 n=8 5 n=9 6 n=10 6 n=11 6 n=12 8 n=13 8 n=14 8 n=15 10 n=16 9 n=17 10 n=18 11 n=19 11 n=20 12 n=1000 502
- Output:
for n=100,000
n=100000 48157
Bonus to produce the count of unordered values: <lang>
declare tally fixed binary (31) initial (0);
do i = 1 to n-1;
if Q(i) > Q(i+1) then tally = tally + 1;
end;
put skip data (tally);
</lang>
- Output:
n=100000 48157 TALLY= 49798;
PureBasic
<lang PureBasic>If Not OpenConsole("Hofstadter Q sequence")
End 1
EndIf
- N = 100000
Define i.i, flip.i = 0 Dim q.i(#N) q(1) = 1 q(2) = 1 For i = 3 To #N
q(i) = q(i - q(i - 1)) + q(i - q(i - 2))
Next For i = 1 To #N - 1
flip + Bool(q(i) > q(i + 1))
Next
Print(~"First ten:\t") For i = 1 To 10 : Print(LSet(Str(q(i)), 3)) : Next PrintN(~"\n1000th:\t\t" + Str(q(1000))) PrintN(~"Flips:\t\t" + Str(flip)) Input() End</lang>
- Output:
First ten: 1 1 2 3 3 4 5 5 6 6 1000th: 502 Flips: 49798
Python
<lang python>def q(n):
if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1")
try:
return q.seq[n]
except IndexError:
ans = q(n - q(n - 1)) + q(n - q(n - 2))
q.seq.append(ans)
return ans
q.seq = [None, 1, 1]
if __name__ == '__main__':
first10 = [q(i) for i in range(1,11)]
assert first10 == [1, 1, 2, 3, 3, 4, 5, 5, 6, 6], "Q() value error(s)"
print("Q(n) for n = [1..10] is:", ', '.join(str(i) for i in first10))
assert q(1000) == 502, "Q(1000) value error"
print("Q(1000) =", q(1000))</lang>
- Extra credit
If you try and initially compute larger values of n then you tend to hit the Python recursion limit.
The function q1 gets around this by calling function q to extend the Q series in increments below the recursion limit.
The following code is to be concatenated to the code above: <lang python>from sys import getrecursionlimit
def q1(n):
if n < 1 or type(n) != int: raise ValueError("n must be an int >= 1")
try:
return q.seq[n]
except IndexError:
len_q, rlimit = len(q.seq), getrecursionlimit()
if (n - len_q) > (rlimit // 5):
for i in range(len_q, n, rlimit // 5):
q(i)
ans = q(n - q(n - 1)) + q(n - q(n - 2))
q.seq.append(ans)
return ans
if __name__ == '__main__':
tmp = q1(100000)
print("Q(i+1) < Q(i) for i [1..100000] is true %i times." %
sum(k1 < k0 for k0, k1 in zip(q.seq[1:], q.seq[2:])))</lang>
- Combined output:
Q(n) for n = [1..10] is: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6 Q(1000) = 502 Q(i+1) < Q(i) for i [1..10000] is true 49798 times.
Alternative
<lang python>def q(n):
l = len(q.seq)
while l <= n:
q.seq.append(q.seq[l - q.seq[l - 1]] + q.seq[l - q.seq[l - 2]])
l += 1
return q.seq[n]
q.seq = [None, 1, 1]
print("Q(n) for n = [1..10] is:", [q(i) for i in range(1, 11)]) print("Q(1000) =", q(1000)) q(100000) print("Q(i+1) < Q(i) for i [1..100000] is true %i times." %
sum([q.seq[i] > q.seq[i + 1] for i in range(1, 100000)]))</lang>
R
<lang rsplus> cache <- vector("integer", 0) cache[1] <- 1 cache[2] <- 1
Q <- function(n) {
if (is.na(cache[n])) {
value <- Q(n-Q(n-1)) + Q(n-Q(n-2))
cache[n] <<- value
}
cache[n]
}
for (i in 1:1e5) {
Q(i)
}
for (i in 1:10) {
cat(Q(i)," ",sep = "")
} cat("\n") cat(Q(1000),"\n")
count <- 0 for (i in 2:1e5) {
if (Q(i) < Q(i-1)) count <- count + 1
} cat(count,"terms is less than its preceding term\n") </lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 49798 terms is less than its preceding term
Racket
<lang racket>
- lang racket
(define t (make-hash)) (hash-set! t 0 0) (hash-set! t 1 1) (hash-set! t 2 1)
(define (Q n)
(hash-ref! t n (λ() (+ (Q (- n (Q (- n 1))))
(Q (- n (Q (- n 2))))))))
(for/list ([i (in-range 1 11)]) (Q i)) (Q 1000)
- extra credit
(for/sum ([i 100000]) (if (< (Q (add1 i)) (Q i)) 1 0)) </lang>
- Output:
'(1 1 2 3 3 4 5 5 6 6) 502 49798
Raku
(formerly Perl 6)
OO solution
Similar concept as the perl5 solution, except that the cache is only filled on demand.
<lang perl6>class Hofstadter {
has @!c = 1,1;
method AT-POS ($me: Int $i) {
@!c.push($me[@!c.elems-$me[@!c.elems-1]] +
$me[@!c.elems-$me[@!c.elems-2]]) until @!c[$i]:exists;
return @!c[$i]; }
}
- Testing:
my Hofstadter $Q .= new();
say "first ten: $Q[^10]"; say "1000th: $Q[999]";
my $count = 0; $count++ if $Q[$_ +1 ] < $Q[$_] for ^99_999; say "In the first 100_000 terms, $count terms are less than their preceding terms";</lang>
- Output:
first ten: 1 1 2 3 3 4 5 5 6 6 1000th: 502 In the first 100_000 terms, 49798 terms are less than their preceding terms
Idiomatic solution
With a lazily generated array, we automatically get caching. <lang perl6>my @Q = 1, 1, -> $a, $b {
(state $n = 1)++; @Q[$n - $a] + @Q[$n - $b]
} ... *;
- Testing:
say "first ten: ", @Q[^10]; say "1000th: ", @Q[999]; say "In the first 100_000 terms, ",
[+](@Q[1..100000] Z< @Q[0..99999]), " terms are less than their preceding terms";</lang>
(Same output.)
REXX
non-recursive
The REXX language doesn't allow expressions for stemmed array indices, so a temporary variable must be used. <lang rexx>/*REXX program generates the Hofstadter Q sequence for any specified N. */ parse arg a b c d . /*obtain optional arguments from the CL*/ if a== | a=="," then a= 10 /*Not specified? Then use the default.*/ if b== | b=="," then b= -1000 /* " " " " " " */ if c== | c=="," then c= -100000 /* " " " " " " */ if d== | d=="," then d= -1000000 /* " " " " " " */ @.= 1; ac= abs(c) /* [↑] negative #'s don't show values.*/ call HofstadterQ a; say call HofstadterQ b; say 'HofstadterQ ' commas(abs(b))th(b) " term is: " commas(result) call HofstadterQ c; say downs= 0; do j=2 for ac-1; jm= j - 1
downs= downs + (@.j<@.jm)
end /*j*/
say commas(downs) ' HofstatdterQ terms are less then the previous term,' ,
' HofstatdterQ('commas(ac) || th(ac)") term is: " commas(@.ac)
call HofstadterQ d; ad= abs(d); say say 'The ' commas(ad) || th(ad) ' HofstatdterQ term is: ' commas(@.ad) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ HofstadterQ: procedure expose @.; parse arg x 1 ox /*get number to generate through.*/
/* [↑] OX is the same as X. */
x= abs(x); w= length( commas(x) ) /*use absolute value; get length.*/
do j=1 for x /* [↓] use short─circuit IF test*/
if j>2 then if @.j==1 then do; jm1= j - 1; jm2= j - 2
one= j - @.jm1; two= j - @.jm2
@.j= @.one + @.two
end
if ox>0 then say 'HofstadterQ('right(j, w)"): " right(@.j,max(w,length(@.j)))
end /*j*/
return @.x /*return the │X│th term to caller*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg _; do ?=length(_)-3 to 1 by -3; _=insert(',', _, ?); end; return _ th: procedure; #=abs(arg(1)); return word('th st nd rd',1+#//10*(#//100%10\==1)*(#//10<4))</lang>
- output when using the internal default inputs:
HofstadterQ( 1): 1 HofstadterQ( 2): 1 HofstadterQ( 3): 2 HofstadterQ( 4): 3 HofstadterQ( 5): 3 HofstadterQ( 6): 4 HofstadterQ( 7): 5 HofstadterQ( 8): 5 HofstadterQ( 9): 6 HofstadterQ(10): 6 HofstadterQ 1,000th term is: 502 49,798 HofstatdterQ terms are less then the previous term, HofstatdterQ(100,000th) term is: 48,157 The 1,000,000th HofstatdterQ term is: 512,066
non-recursive, simpler
This REXX example is identical to the first version except that it uses a function to retrieve array elements which may have index expressions. <lang rexx>/*REXX program generates the Hofstadter Q sequence for any specified N. */ parse arg a b c d . /*obtain optional arguments from the CL*/ if a== | a=="," then a= 10 /*Not specified? Then use the default.*/ if b== | b=="," then b= -1000 /* " " " " " " */ if c== | c=="," then c= -100000 /* " " " " " " */ if d== | d=="," then d= -1000000 /* " " " " " " */ @.= 1; ac= abs(c) /* [↑] negative #'s don't show values.*/ call HofstadterQ a; say call HofstadterQ b; say 'HofstadterQ ' commas(abs(b))th(b) " term is: " commas(result) call HofstadterQ c; say downs= 0; do j=2 for ac-1; jm= j - 1
downs= downs + (@.j<@.jm)
end /*j*/
say commas(downs) ' HofstatdterQ terms are less then the previous term,' ,
' HofstatdterQ('commas(ac) || th(ac)") term is: " commas(@.ac)
call HofstadterQ d; ad= abs(d); say say 'The ' commas(ad) || th(ad) ' HofstatdterQ term is: ' commas(@.ad) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ HofstadterQ: procedure expose @.; parse arg x 1 ox /*get number to generate through.*/
/* [↑] OX is the same as X. */
x= abs(x); w= length( commas(x) ) /*use absolute value; get length.*/
do j=1 for x
if j>2 then if @.j==1 then @.j= @(j - @(j-1)) + @(j - @(j-2))
if ox>0 then say 'HofstadterQ('right(j, w)"): " right(@.j,max(w,length(@.j)))
end /*j*/
return @.x /*return the │X│th term to caller*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ @: parse arg ?; return @.? /*return value of @.? to invoker.*/ th: procedure; #=abs(arg(1)); return word('th st nd rd',1+#//10*(#//100%10\==1)*(#//10<4)) commas: parse arg _; do ?=length(_)-3 to 1 by -3; _=insert(',', _, ?); end; return _</lang>
- output is identical to the 1st REXX version.
Because of the additional subroutine (function) invokes, this REXX version is about half as fast as the 1st REXX version.
recursive
<lang rexx>/*REXX program generates the Hofstadter Q sequence for any specified N. */ parse arg a b c d . /*obtain optional arguments from the CL*/ if a== | a=="," then a= 10 /*Not specified? Then use the default.*/ if b== | b=="," then b= -1000 /* " " " " " " */ if c== | c=="," then c= -100000 /* " " " " " " */ if d== | d=="," then d= -1000000 /* " " " " " " */ @.= 0; @.1= 1; @.2= 1; ac= abs(c) /* [↑] negative #'s don't show values.*/ call HofstadterQ a; say call HofstadterQ b; say 'HofstadterQ ' commas(abs(b))th(b) " term is: " commas(result) call HofstadterQ c; say downs= 0; do j=2 for ac-1; jm= j - 1
downs= downs + (@.j<@.jm)
end /*j*/
say commas(downs) ' HofstatdterQ terms are less then the previous term,' ,
' HofstatdterQ('commas(ac) || th(ac)") term is: " commas(@.ac)
call HofstadterQ d; ad= abs(d); say say 'The ' commas(ad) || th(ad) ' HofstatdterQ term is: ' commas(@.ad) exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ HofstadterQ: procedure expose @.; parse arg x 1 ox /*get number to generate through.*/
/* [↑] OX is the same as X. */
x= abs(x); w= length( commas(x) ) /*use absolute value; get length.*/
do j=1 for x
if @.j==0 then @.j= QR(j) /*Not defined? Then define it.*/
if ox>0 then say 'HofstadterQ('right(j, w)"): " right(@.j,max(w,length(@.j)))
end /*j*/
return @.x /*return the │X│th term to caller*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ QR: procedure expose @.; parse arg n /*this QR function is recursive.*/
if @.n==0 then @.n= QR(n-QR(n-1)) + QR(n-QR(n-2)) /*Not defined? Then define it.*/ return @.n /*return the value to the invoker*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ th: procedure; #=abs(arg(1)); return word('th st nd rd',1+#//10*(#//100%10\==1)*(#//10<4)) commas: parse arg _; do ?=length(_)-3 to 1 by -3; _=insert(',', _, ?); end; return _</lang>
- output is identical to the 1st REXX version.
The recursive version is almost ten times slower than the (1st) non-recursive version.
Ring
<lang ring> n = 20 aList = list(n) aList[1] = 1 aList[2] = 1 for i = 1 to n
if i >= 3 aList[i] = ( aList[i - aList[i-1]] + aList[i - aList[i-2]] ) ok if i <= 20 see "n = " + string(i) + " : "+ aList[i] + nl ok
next </lang>
Ruby
<lang ruby>@cache = [] def Q(n)
if @cache[n].nil? case n when 1, 2 then @cache[n] = 1 else @cache[n] = Q(n - Q(n-1)) + Q(n - Q(n-2)) end end @cache[n]
end
puts "first 10 numbers in the sequence: #{(1..10).map {|n| Q(n)}}" puts "1000'th term: #{Q(1000)}"
prev = Q(1) count = 0 2.upto(100_000) do |n|
q = Q(n) count += 1 if q < prev prev = q
end puts "number of times in the first 100,000 terms where Q(i)<Q(i-1): #{count}"</lang>
- Output:
first 10 numbers in the sequence: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 1000'th term: 502 number of times in the first 100,000 terms where Q(i)<Q(i-1): 49798
Run BASIC
<lang Runbasic>input "How many values do you want? :";n dim Q(n) Q(1) = 1 Q(2) = 1 for i = 1 to n
if i >= 3 then Q(i) = ( Q(i - Q(i-1)) + Q(i - Q(i-2)) )
if i <= 20 then print "n=";using("####",i);" ";using("###",Q(i))
next i if i > 20 then print "n=";using("####",i);using("####",Q(i)) end </lang>
- Output:
How many values do you want? :?1000 n= 1 1 n= 2 1 n= 3 2 n= 4 3 n= 5 3 n= 6 4 n= 7 5 n= 8 5 n= 9 6 n= 10 6 n= 11 6 n= 12 8 n= 13 8 n= 14 8 n= 15 10 n= 16 9 n= 17 10 n= 18 11 n= 19 11 n= 20 12 n=1000 502
Rust
Rust doesn't allow static Vec's (but there's lazy_static crate), thus memoization storage is allocated in main.
<lang rust>fn hofq(q: &mut Vec<u32>, x : u32) -> u32 {
let cur_len=q.len()-1;
let i=x as usize;
if i>cur_len {
// extend storage
q.reserve(i+1);
for j in (cur_len+1)..(i+1) {
let qj=(q[j-q[j-1] as usize]+q[j-q[j-2] as usize]) as u32;
q.push(qj);
}
}
q[i]
}
fn main() {
let mut q_memo: Vec<u32>=vec![0,1,1];
let mut q=|i| {hofq(&mut q_memo, i)};
for i in 1..11 {
println!("Q({})={}", i, q(i));
}
println!("Q(1000)={}", q(1000));
let q100001=q(100_000); // precompute all
println!("Q(100000)={}", q100000);
let nless=(1..100_000).fold(0,|s,i|{if q(i+1)<q(i) {s+1} else {s}});
println!("Term is less than preceding term {} times", nless);
} </lang>
- Output:
Q(1)=1 Q(2)=1 Q(3)=2 Q(4)=3 Q(5)=3 Q(6)=4 Q(7)=5 Q(8)=5 Q(9)=6 Q(10)=6 Q(1000)=502 Q(100001)=53471 Term is less than preceding term 49798 times
Scala
Naive but elegant version using only recursion doesn't work because runtime is excessive increasing ... <lang scala>object HofstadterQseq extends App {
val Q: Int => Int = n => {
if (n <= 2) 1
else Q(n-Q(n-1))+Q(n-Q(n-2))
}
(1 to 10).map(i=>(i,Q(i))).foreach(t=>println("Q("+t._1+") = "+t._2))
println("Q("+1000+") = "+Q(1000))
}</lang>
Unfortunately the function Q isn't tail recursiv,
therefore the compiler can't optimize it.
Thus we are forced to use a caching featured version.
<lang scala>object HofstadterQseq extends App {
val HofQ = scala.collection.mutable.Map((1->1),(2->1))
val Q: Int => Int = n => {
if (n < 1) 0
else {
val res = HofQ.keys.filter(_==n).toList match {
case Nil => {val v = Q(n-Q(n-1))+Q(n-Q(n-2)); HofQ += (n->v); v}
case xs => HofQ(n)
}
res
}
}
(1 to 10).map(i=>(i,Q(i))).foreach(t=>println("Q("+t._1+") = "+t._2))
println("Q("+1000+") = "+Q(1000))
println((3 to 100000).filter(i=>Q(i)<Q(i-1)).size)
}</lang>
- Output:
Q(1) = 1 Q(2) = 1 Q(3) = 2 Q(4) = 3 Q(5) = 3 Q(6) = 4 Q(7) = 5 Q(8) = 5 Q(9) = 6 Q(10) = 6 Q(1000) = 502 49798
Scheme
I wish there were a portable way to define-syntax,
or to resize arrays, or to do formated output--anything to make the code
less silly looking while still run under more than one interpreter.
<lang lisp>(define qc '#(0 1 1))
(define filled 3)
(define len 3)
- chicken scheme
- vector-resize!
- gambit
- vector-append
(define (extend-qc)
(let* ((new-len (* 2 len))
(new-qc (make-vector new-len)))
(let copy ((n 0))
(if (< n len)
(begin (vector-set! new-qc n (vector-ref qc n)) (copy (+ 1 n)))))
(set! len new-len) (set! qc new-qc)))
(define (q n)
(let loop ()
(if (>= filled len) (extend-qc))
(if (>= n filled)
(begin
(vector-set! qc filled (+ (q (- filled (q (- filled 1)))) (q (- filled (q (- filled 2)))))) (set! filled (+ 1 filled)) (loop))
(vector-ref qc n))))
(display "Q(1 .. 10): ") (let loop ((i 1))
;; (print) behave differently regarding newline across compilers (display (q i)) (display " ") (if (< i 10) (loop (+ 1 i)) (newline)))
(display "Q(1000): ") (display (q 1000)) (newline)
(display "bumps up to 100000: ") (display
(let loop ((s 0) (i 1))
(if (>= i 100000) s
(loop (+ s (if (> (q i) (q (+ 1 i))) 1 0)) (+ 1 i)))))
(newline)</lang>
- Output:
Q(1 .. 10): 1 1 2 3 3 4 5 5 6 6 Q(1000): 502 bumps up to 100000: 49798
Seed7
<lang seed7>$ include "seed7_05.s7i";
const type: intHash is hash [integer] integer;
var intHash: qHash is intHash.value;
const func integer: q (in integer: n) is func
result
var integer: q is 1;
begin
if n in qHash then
q := qHash[n];
else
if n > 2 then
q := q(n - q(pred(n))) + q(n - q(n - 2));
end if;
qHash @:= [n] q;
end if;
end func;
const proc: main is func
local
var integer: n is 0;
var integer: less_than_preceding is 0;
begin
writeln("q(n) for n = 1 .. 10:");
for n range 1 to 10 do
write(q(n) <& " ");
end for;
writeln;
writeln("q(1000)=" <& q(1000));
for n range 2 to 100000 do
if q(n) < q(pred(n)) then
incr(less_than_preceding);
end if;
end for;
writeln("q(n) < q(n-1) for n = 2 .. 100000: " <& less_than_preceding);
end func;</lang>
- Output:
q(n) for n = 1 .. 10: 1 1 2 3 3 4 5 5 6 6 q(1000)=502 q(n) < q(n-1) for n = 2 .. 100000: 49798
Sidef
Using a memoized function: <lang ruby>func Q(n) is cached {
n <= 2 ? 1
: Q(n - Q(n-1))+Q(n-Q(n-2))
} say "First 10 terms: #{ {|n| Q(n) }.map(1..10) }" say "Term 1000: #{Q(1000)}" say "Terms less than preceding in first 100k: #{2..100000->count{|i|Q(i)<Q(i-1)}}"</lang>
Using an array: <lang ruby>var Q = [0, 1, 1] 100_000.times {
Q << (Q[-Q[-1]] + Q[-Q[-2]])
} say "First 10 terms: #{Q.ft(1, 10)}" say "Term 1000: #{Q[1000]}" say "Terms less than preceding in first 100k: #{2..100000->count{|i|Q[i]<Q[i-1]}}"</lang>
- Output:
First 10 terms: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] Term 1000: 502 Terms less than preceding in first 100k: 49798
Swift
<lang swift>let n = 100000
var q = Array(repeating: 0, count: n) q[0] = 1 q[1] = 1
for i in 2..<n {
q[i] = q[i - q[i - 1]] + q[i - q[i - 2]]
}
print("First 10 elements of the sequence: \(q[0..<10])") print("1000th element of the sequence: \(q[999])")
var count = 0 for i in 1..<n {
if q[i] < q[i - 1] {
count += 1
}
} print("Number of times a member of the sequence is less than the preceding term for terms up to and including the 100,000th term: \(count)")</lang>
- Output:
First 10 elements of the sequence: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 1000th element of the sequence: 502 Number of times a member of the sequence is less than the preceding term for terms up to and including the 100,000th term: 49798
Tailspin
<lang tailspin> templates q
def outputFrom: $(1); def until: $(2); @: [1,1]; 1..$until -> # when <$@::length~..> do ..|@: $@($ - $@($ - 1)) + $@($ - $@($ - 2)); $ -> # when <$outputFrom..> do $@($) !
end q
[1,10] -> q -> '$; ' -> !OUT::write ' ' -> !OUT::write
[1000,1000] -> q -> '$; ' -> !OUT::write
templates countDownSteps
@: 0; def qs: $; 2..$qs::length -> # $@ ! when <?($qs($) <..~$qs($-1)>)> do @: $@ + 1;
end countDownSteps
[[1, 100000] -> q] -> countDownSteps -> 'Less than previous $; times' -> !OUT::write </lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 Less than previous 49798 times
Tcl
<lang tcl>package require Tcl 8.5
- Index 0 is not used, but putting it in makes the code a bit shorter
set tcl::mathfunc::Qcache {Q:-> 1 1} proc tcl::mathfunc::Q {n} {
variable Qcache
if {$n >= [llength $Qcache]} {
lappend Qcache [expr {Q($n - Q($n-1)) + Q($n - Q($n-2))}]
} return [lindex $Qcache $n]
}
- Demonstration code
for {set i 1} {$i <= 10} {incr i} {
puts "Q($i) == [expr {Q($i)}]"
}
- This runs very close to recursion limit...
puts "Q(1000) == [expr Q(1000)]"
- This code is OK, because the calculations are done step by step
set q [expr Q(1)] for {set i 2} {$i <= 100000} {incr i} {
incr count [expr {$q > [set q [expr {Q($i)}]]}]
} puts "Q(i)<Q(i-1) for i \[2..100000\] is true $count times"</lang>
- Output:
Q(1) == 1 Q(2) == 1 Q(3) == 2 Q(4) == 3 Q(5) == 3 Q(6) == 4 Q(7) == 5 Q(8) == 5 Q(9) == 6 Q(10) == 6 Q(1000) == 502 Q(i)<Q(i-1) for i [2..100000] is true 49798 times
uBasic/4tH
uBasic/4tH simply lacks the memory to make it through to the 1000th term. 256 is the best it can do. <lang>Print "First 10 terms of Q = " ; For i = 1 To 10 : Print FUNC(_q(i));" "; : Next : Print Print "256th term = ";FUNC(_q(256))
End
_q Param(1)
Local(2)
If a@ < 3 Then Return (1) If a@ = 3 Then Return (2)
@(0) = 1 : @(1) = 1 : @(2) = 2 c@ = 0
For b@ = 3 To a@-1 @(b@) = @(b@ - @(b@-1)) + @(b@ - @(b@-2)) If @(b@) < @(b@-1) Then c@ = c@ + 1 Next
Return (@(a@-1))</lang>
- Output:
First 10 terms of Q = 1 1 2 3 3 4 5 5 6 6 256th term = 123 0 OK, 0:320
VBA
<lang vb>Public Q(100000) As Long Public Sub HofstadterQ()
Dim n As Long, smaller As Long
Q(1) = 1
Q(2) = 1
For n = 3 To 100000
Q(n) = Q(n - Q(n - 1)) + Q(n - Q(n - 2))
If Q(n) < Q(n - 1) Then smaller = smaller + 1
Next n
Debug.Print "First ten terms:"
For i = 1 To 10
Debug.Print Q(i);
Next i
Debug.print
Debug.Print "The 1000th term is:"; Q(1000)
Debug.Print "Number of times smaller:"; smaller
End Sub</lang>
- Output:
First ten terms: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502 Number of times smaller: 49798
VBScript
<lang vb> Sub q_sequence(n) Dim Q() ReDim Q(n) Q(1)=1 : Q(2)=1 : Q(3)=2 less_precede = 0 For i = 4 To n Q(i)=Q(i-Q(i-1))+Q(i-Q(i-2)) If Q(i) < Q(i-1) Then less_precede = less_precede + 1 End If Next WScript.StdOut.Write "First 10 terms of the sequence: " For j = 1 To 10 If j < 10 Then WScript.StdOut.Write Q(j) & ", " Else WScript.StdOut.Write "and " & Q(j) End If Next WScript.StdOut.WriteLine WScript.StdOut.Write "1000th term of the sequence: " & Q(1000) WScript.StdOut.WriteLine WScript.StdOut.Write "Number of times the member of the sequence is less than its preceding term: " &_ less_precede End Sub
q_sequence(100000) </lang>
- Output:
First 10 terms of the sequence: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6 1000th term of the sequence: 502 Number of times the member of the sequence is less than its preceding term: 49798
Visual FoxPro
<lang vfp> LOCAL p As Integer, i As Integer CLEAR p = 0 ? "Hofstadter Q Sequence" ? "First 10 terms:" FOR i = 1 TO 10 ?? Q(i, @p) ENDFOR ? "1000th term:", Q(1000, @p) ? "100000th term:", q(100000, @p) ? "Number of terms less than the preceding term:", p
FUNCTION Q(n As Integer, k As Integer) As Integer LOCAL i As Integer LOCAL ARRAY aq[n] aq[1] = 1 IF n > 1
aq[2] = 1
ENDIF k = 0 FOR i = 3 TO n
aq[i] = aq[i - aq[i-1]] + aq[i-aq[i-2]] IF aq(i) < aq(i-1) k = k + 1 ENDIF
ENDFOR RETURN aq[n] ENDFUNC </lang>
- Output:
Hofstadter Q Sequence First 10 terms: 1 1 2 3 3 4 5 5 6 6 1000th term: 502 100000th term: 48157 Number of terms less than the preceding term: 49798
Wren
<lang ecmascript>var N = 1e5 var q = List.filled(N + 1, 0) q[1] = 1 q[2] = 1 for (n in 3..N) q[n] = q[n - q[n-1]] + q[n - q[n-2]]
System.print("The first ten terms of the Hofstadter Q sequence are:") System.print(q[1..10]) System.print("\nThe thousandth term is %(q[1000]).") var flips = 0 for (n in 2..N) {
if (q[n] < q[n-1]) flips = flips + 1
} System.print("\nThere are %(flips) flips in the first %(N) terms.")</lang>
- Output:
The first ten terms of the Hofstadter Q sequence are: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] The thousandth term is 502. There are 49798 flips in the first 100000 terms.
XPL0
<lang XPL0>code ChOut=8, CrLf=9, IntOut=11; int N, C, Q(100_001); [Q(1):= 1; Q(2):= 1; C:= 0; for N:= 3 to 100_000 do
[Q(N):= Q(N-Q(N-1)) + Q(N-Q(N-2));
if Q(N) < Q(N-1) then C:= C+1;
];
for N:= 1 to 10 do
[IntOut(0, Q(N)); ChOut(0, ^ )];
CrLf(0); IntOut(0, Q(1000)); CrLf(0); IntOut(0, C); CrLf(0); ]</lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 49798
zkl
<lang zkl>const n = 0d100_000; q:=(n+1).pump(List.createLong(n+1).write); // (0,1,2,...,n) base 1 q[1] = q[2] = 1;
foreach i in ([3..n]) { q[i] = q[i - q[i - 1]] + q[i - q[i - 2]] }
q[1,10].concat(" ").println(); println(q[1000]);
flip := 0; foreach i in (n){ flip += (q[i] > q[i + 1]) } println("flips: ",flip);</lang>
- Output:
1 1 2 3 3 4 5 5 6 6 502 flips: 49798
ZX Spectrum Basic
Extra credit 100000 is not implemented because of memory limitations. <lang zxbasic>10 PRINT "First 10 terms of Q = " 20 FOR i=1 TO 10: GO SUB 1000: PRINT s;" ";: NEXT i: PRINT 30 LET i=1000 40 PRINT "1000th term = ";: GO SUB 1000: PRINT s 50 PRINT "Term is less than preceding term ";c;" times" 100 STOP 1000 REM Qsequence subroutine 1010 IF i<3 THEN LET s=1: RETURN 1020 IF i=3 THEN LET s=2: RETURN 1030 DIM q(i) 1040 LET q(1)=1: LET q(2)=1: LET q(3)=2 1050 LET c=0 1060 FOR j=3 TO i 1070 LET q(j)=q(j-q(j-1))+q(j-q(j-2)) 1080 IF q(j)<q(j-1) THEN LET c=c+1 1090 NEXT j 1100 LET s=q(i) 1110 RETURN</lang>
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