Hofstadter Q sequence
The Hofstadter Q sequence is defined as:
You are encouraged to solve this task according to the task description, using any language you may know.
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
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)- 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
* 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- 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
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- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502
8086 Assembly
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
- Output:
First 10 terms are: 1 1 2 3 3 4 5 5 6 6 1000th term is: 502
AArch64 Assembly
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program hofstader64.s */
/*******************************************/
/* Constantes */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
/*******************************************/
/* macros */
/*******************************************/
//.include "../../ficmacros64.inc" // for developper debugging
/*********************************/
/* Initialized data */
/*********************************/
.data
szMessDebutPgm: .asciz "Program 64 bits start. \n"
szCarriageReturn: .asciz "\n"
szMessFinOK: .asciz "Program normal end. \n"
szMessErreur: .asciz "Error !!!\n"
szMessHofs: .asciz "Hofstader numbers :\n"
szSpace: .asciz " "
.align 4
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
.align 4
/*********************************/
/* code section */
/*********************************/
.text
.global main
main:
ldr x0,qAdrszMessDebutPgm
bl affichageMess // start message
ldr x0,qAdrszMessHofs
bl affichageMess
mov x0,10 // maxi
mov x1,1 // display number
bl genererHofs
ldr x0,qAdrszCarriageReturn
bl affichageMess
mov x0,1000 // maxi
mov x1,0 // display last number
bl genererHofs
ldr x0,qAdrszCarriageReturn
bl affichageMess
ldr x0,qMax100000 // maxi
mov x1,2 // display counter
bl genererHofs
bl displayNumber // display return number
ldr x0,qAdrszCarriageReturn
bl affichageMess
ldr x0,qAdrszMessFinOK
bl affichageMess
b 100f
99:
ldr x0,qAdrszMessErreur // error
bl affichageMess
mov x0, #1 // return code error
b 100f
100:
mov x8,EXIT
svc #0 // system call
qAdrszMessDebutPgm: .quad szMessDebutPgm
qAdrszMessFinOK: .quad szMessFinOK
qAdrszMessErreur: .quad szMessErreur
qAdrsZoneConv: .quad sZoneConv
qAdrszMessHofs: .quad szMessHofs
qMax100000: .quad 100000
/***************************************************/
/* Generation Fibonacci numbers */
/***************************************************/
/* x0 contains limit number */
/* x1 display the last number */
genererHofs:
stp x1,lr,[sp,-16]!
stp x2,x3,[sp,-16]!
stp x4,x5,[sp,-16]!
stp x6,x7,[sp,-16]!
stp x8,x9,[sp,-16]!
stp x10,x11,[sp,-16]!
mov x2,x0
lsl x10,x0,3
sub sp,sp,x10 // reserve area on stack
mov x7,sp
mov x4,#0
mov x5,#0
1: // init area loop
str x4,[x7,x5,lsl #3]
add x5,x5,#1
cmp x5,x2
blt 1b
mov x4,#1
str x4,[x7] // store value 1 in first item
str x4,[x7,8] // store value 1 in second item
cmp x1,1 // display number ?
bne 2f
mov x0,1 // display L(1)
bl displayNumber
mov x0,1 // display L(2)
bl displayNumber
2:
mov x9,0
mov x4,#2
3:
sub x3,x4,#1 // L -1
ldr x11,[x7,x3,lsl #3] // load ancien result
sub x6,x4,x11
ldr x5,[x7,x6,lsl #3]
sub x3,x3,#1 // L - 2
ldr x8,[x7,x3,lsl #3] // load ancien result
sub x6,x4,x8
ldr x8,[x7,x6,lsl #3]
add x5,x5,x8
str x5,[x7,x4,lsl #3]
cmp x5,x11
cinc x9,x9,lt
cmp x1,1
bne 4f
mov x0,x5
bl displayNumber
4:
add x4,x4,#1 // increment counter
cmp x4,x2 // end compute ?
blt 3b
cmp x1,1
bge 100f
mov x0,x5
bl displayNumber
100:
mov x0,x9
add sp,sp,x10 // free reserved area
ldp x10,x11,[sp],16
ldp x8,x9,[sp],16
ldp x6,x7,[sp],16
ldp x4,x5,[sp],16
ldp x2,x3,[sp],16
ldp x1,lr,[sp],16
ret
/***************************************************/
/* display number */
/***************************************************/
/* x0 contains number */
displayNumber:
stp x1,lr,[sp,-16]!
stp x2,x3,[sp,-16]!
ldr x1,qAdrsZoneConv
bl conversion10
mov x2,#0
add x1,x1,x0
strb w2,[x1]
ldr x0,qAdrsZoneConv
bl affichageMess
ldr x0,qAdrszSpace
bl affichageMess
100:
ldp x2,x3,[sp],16
ldp x1,lr,[sp],16
ret
qAdrszCarriageReturn: .quad szCarriageReturn
qAdrszSpace: .quad szSpace
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeARM64.inc"- Output:
Program 64 bits start. Hofstader numbers : 1 1 2 3 3 4 5 5 6 6 502 49798 Program normal end.
Action!
PROC Main()
DEFINE MAX="1000"
INT ARRAY q(MAX+1)
INT i
q(1)=1 q(2)=1
FOR i=3 TO MAX
DO
q(i)=q(i-q(i-1))+q(i-q(i-2))
OD
FOR i=1 TO 10
DO
PrintF("%I: %I%E",i,q(i))
OD
PrintF("%I: %I%E",MAX,q(MAX))
RETURN- Output:
Screenshot from Atari 8-bit computer
1: 1 2: 1 3: 2 4: 3 5: 3 6: 4 7: 5 8: 5 9: 6 10: 6 1000: 502
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;
- Output:
1 1 2 3 3 4 5 5 6 6 502 49798
ALGOL 68
File: Hofstadter_Q_sequence.a68
BEGIN
[100000]INT q;
INT flips := 0;
q[1] := q[2] := 1;
FOR i FROM 3 TO UPB q DO
q[i] := q[i - q[i - 1]] + q[i - q[i - 2]];
IF q[i] < q[i - 1] THEN flips +:= 1 FI
OD;
FOR i TO 10 DO
print((whole(q[i],0), IF i = 10 THEN newline ELSE space FI)) OD;
print((whole(q[1000],0), newline));
print(("flips: ", whole(flips,0), newline))
END- Output:
1 1 2 3 3 4 5 5 6 6 502 flips: 49798
ALGOL-M
begin
integer array Q[1:1000];
integer n;
Q[1] := Q[2] := 1;
for n := 3 step 1 until 1000 do
Q[n] := Q[n-Q[n-1]] + Q[n-Q[n-2]];
write("The first 10 terms are:");
write("");
for n := 1 step 1 until 10 do writeon(Q[n]);
write("The 1000th term is:", Q[1000]);
end- Output:
The first 10 terms are:
1 1 2 3 3 4 5 5 6 6
The 1000th term is: 502
ALGOL W
begin % find elements of the Hofstader Q sequence Q(1) = Q(2) = 1 %
% Q(n) = Q( n - Q( n - 1 ) ) + Q( n - Q( n - 2 ) ) for n > 2 %
integer MAX_Q;
max_Q := 100000;
begin
integer array Q ( 1 :: MAX_Q );
integer array xQ ( 1 :: 10 );
integer ltCount;
logical valuesOk;
% expected values of the first 10 elements %
xQ( 1 ) := xQ( 2 ) := 1;
xQ( 3 ) := 2; xQ( 4 ) := xQ( 5 ) := 3; xQ( 6 ) := 4;
xQ( 7 ) := xQ( 8 ) := 5; xQ( 9 ) := xQ( 10 ) := 6;
% calculate the sequence and count how often Q( n ) < Q( n - 1 ) %
ltCount := 0;
Q( 1 ) := Q( 2 ) := 1;
for n := 3 until MAX_Q do begin
Q( n ) := Q( n - Q( n - 1 ) ) + Q( n - Q( n - 2 ) );
if Q( n ) < Q( n - 1 ) then ltCount := ltCount + 1
end for_n ;
valuesOk := true;
write( "The first 10 terms of the Hofstader Q sequence:" );
for i := 1 until 10 do begin
writeon( i_w := 1, s_w := 0, " ", Q( i ) );
if Q( i ) not = xQ( i ) then begin
writeon( i_w := 1, s_w := 0, "-EXPECTED-", xQ( i ) );
valuesOk := false
end if_Q_i_ne_xQ_i
end for_i ;
write( i_w := 1, s_w := 0, "The 1000th term is: ", Q( 1000 ) );
if Q( 1000 ) not = 502 then begin
writeon( "-EXPECTED-502" );
valuesOk := false
end if_Q_100_ne_502 ;
if valuesOk then write( " (Computed values are as expected)" )
else write( "Values NOT as expected" );
write( i_w := 1, s_w := 0, "Q(n) < Q(n-1) ", ltCount," times for n up to ", MAX_Q )
end
end.- Output:
The first 10 terms of the Hofstader Q sequence: 1 1 2 3 3 4 5 5 6 6
The 1000th term is: 502
(Computed values are as expected)
Q(n) < Q(n-1) 49798 times for n up to 100000
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.'
∇
- 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
.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
- 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.
Arturo
q: new [1 1]
n: 2
while [n<1001][
'q ++ (get q n-q\[n-1]) + get q n-q\[n-2]
n: n+1
]
print ["First ten items:" first.n: 10 q]
print ["1000th item:" q\[999]]
- Output:
First ten items: [1 1 2 3 3 4 5 5 6 6] 1000th item: 502
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
}
- Output:
First ten: 1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 1000th: 502 Flips: 49798
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
}
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
BASIC
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- Output:
Igual que la entrada de FreeBASIC.
BBC BASIC
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%)
- 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
Gambas
Public Const limite As Integer = 100000
Public Q[limite + 1] As Long
Public Sub Main()
Dim i As Long, cont As Long = 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
Print "Primeros 10 terminos: ";
For i = 1 To 10
Print Q[i] & " ";
Next
Print "\nTermino numero 1000: "; Q[1000]
Print "Terminos menores que los anteriores: " & cont
End
- Output:
Same as FreeBASIC entry.
OxygenBasic
uses console
int limite = 100000
dim long Q[100000]
long 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 cr
printl "Termino numero 1000: " Q(1000)
printl "Terminos menores que los anteriores: " cont
printl cr "Enter ..."
waitkey
- Output:
Similar to FreeBASIC entry.
QBasic
CONST limite = 10000
DIM Q(limite)
Q(1) = 1
Q(2) = 1
cont = 0
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 = cont + 1
NEXT i
PRINT "First 10 terms: ";
FOR i = 1 TO 10
PRINT Q(i); " ";
NEXT i
PRINT
PRINT "Term 1000: "; Q(1000)
PRINT "Terms less than preceding in first 100k: "; cont
QB64
Const limite = 100000
Dim As Long Q(limite)
Q(1) = 1
Q(2) = 1
cont = 0
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 = cont + 1
Next i
Print "First 10 terms:";
For i = 1 To 10
Print Q(i);
Next i
Print
Print "Term 1000: "; Q(1000)
Print "Terms less than preceding in first 100k:"; cont
- Output:
Same as FreeBASIC entry.
True BASIC
LET limite = 100000
DIM q(0)
MAT REDIM q(limite)
LET q(1) = 1
LET q(2) = 1
LET count = 0
FOR i = 3 TO limite
LET q(i) = q(i-q(i-1))+q(i-q(i-2))
IF q(i) < q(i-1) THEN
LET count = count + 1
END IF
NEXT i
PRINT "First 10 terms: ";
FOR i = 1 TO 10
PRINT q(i);
NEXT i
PRINT
PRINT "Term 1000: "; q(1000)
PRINT "Terms less than preceding in first 100k: "; count
END
- Output:
Same as FreeBASIC entry.
XBasic
PROGRAM "Hofstadter Q sequence"
VERSION "0.0000"
DECLARE FUNCTION Entry ()
FUNCTION Entry ()
limite = 1e5
DIM q[limite]
q[1] = 1
q[2] = 1
count = 0
FOR i = 3 TO limite
q[i] = q[i-q[i-1]] + q[i-q[i-2]]
IF q[i] < q[i-1] THEN
INC count
END IF
NEXT i
PRINT "First 10 terms: ";
FOR i = 1 TO 10
PRINT q[i];
NEXT i
PRINT "\nTerm 1000: "; q[1000]
PRINT "Terms less than preceding in first 100k: "; count
END FUNCTION
END PROGRAM
- Output:
Same as FreeBASIC entry.
Yabasic
limite = 1e5
dim q(limite)
q(1) = 1
q(2) = 1
count = 0
for i = 3 to limite
q(i) = q(i-q(i-1)) + q(i-q(i-2))
if q(i) < q(i-1) count = count + 1
next i
print "First 10 terms: ";
for i = 1 to 10
print q(i), " ";
next i
print "\nTerm 1000: ", q(1000)
print "Terms less than preceding in first 100k: ", count
end
- Output:
Same as FreeBASIC entry.
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
);Output:
1 1 2 3 3 4 5 5 6 6 502 49798
BCPL
get "libhdr"
let start() be
$( let Q = vec 1000
Q!1 := 1
Q!2 := 1
for n = 3 to 1000 do
Q!n := Q!(n-Q!(n-1)) + Q!(n-Q!(n-2))
writes("The first 10 terms are:")
for n = 1 to 10 do writef(" %N", Q!n)
writef("*NThe 1000th term is: %N*N", Q!1000)
$)- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502
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;
}
- Output:
1 1 2 3 3 4 5 5 6 6 502 flips: 49798
C#
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;
}
}
}- 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
#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;
}
- 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].
(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))
- Output:
first 10: [1 1 2 3 3 4 5 5 6 6]
1000th: 502
extra credit: 49798
CLU
q_seq = proc (n: int) returns (sequence[int])
q: array[int] := array[int]$[1,1]
for i: int in int$from_to(3,n) do
array[int]$addh(q, q[i-q[i-1]] + q[i-q[i-2]])
end
return(sequence[int]$a2s(q))
end q_seq
start_up = proc ()
po: stream := stream$primary_output()
q: sequence[int] := q_seq(100000)
stream$puts(po, "First 10 terms:")
for i: int in int$from_to(1,10) do
stream$puts(po, " " || int$unparse(q[i]))
end
stream$puts(po, "\n1000th term: " || int$unparse(q[1000]))
flips: int := 0
for i: int in int$from_to(2, sequence[int]$size(q)) do
if q[i-1]>q[i] then flips := flips + 1 end
end
stream$putl(po, "\nflips: " || int$unparse(flips))
end start_up- Output:
First 10 terms: 1 1 2 3 3 4 5 5 6 6 1000th term: 502 flips: 49798
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)
- 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
COBOL
IDENTIFICATION DIVISION.
PROGRAM-ID. Q-SEQ.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 SEQ.
02 Q PIC 9(3) OCCURS 1000 TIMES.
02 Q-TMP1 PIC 9(3).
02 Q-TMP2 PIC 9(3).
02 N PIC 9(4).
01 DISPLAYING.
02 ITEM PIC Z(3).
02 IX PIC Z(4).
PROCEDURE DIVISION.
MAIN-PROGRAM.
PERFORM GENERATE-SEQUENCE.
PERFORM SHOW-ITEM
VARYING N FROM 1 BY 1
UNTIL N IS GREATER THAN 10.
SET N TO 1000.
PERFORM SHOW-ITEM.
STOP RUN.
GENERATE-SEQUENCE.
SET Q(1) TO 1.
SET Q(2) TO 1.
PERFORM GENERATE-ITEM
VARYING N FROM 3 BY 1
UNTIL N IS GREATER THAN 1000.
GENERATE-ITEM.
COMPUTE Q-TMP1 = N - Q(N - 1).
COMPUTE Q-TMP2 = N - Q(N - 2).
COMPUTE Q(N) = Q(Q-TMP1) + Q(Q-TMP2).
SHOW-ITEM.
MOVE N TO IX.
MOVE Q(N) TO ITEM.
DISPLAY 'Q(' IX ') = ' ITEM.
- 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
(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)))
- 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:
(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)))
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();- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502
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)));
}
- 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.
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)));
}
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.
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);
}
Dart
Naive version using only recursion (Q(1000) fails due to browser script runtime restrictions)
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)=${Q(1000)}");
}
Version featuring caching.
class Q {
Map<int,int> _table;
Q() {
_table=new Map<int,int>();
_table[1]=1;
_table[2]=1;
}
int q(int n) {
// if the cache is not filled until n-1, fill it starting with the lowest entries first
// this avoids doing a recursion from n to 2 (e.g. if you call q(1000000) first)
// this doesn't happen in the tasks calls since the cache is filled ascending
if(_table[n-1]==null) {
for(int i=_table.length;i<n;i++) {
q(i);
}
}
if(_table[n]==null) {
_table[n]=q(n-q(n-1))+q(n-q(n-2));
}
return _table[n];
}
}
main() {
Q q=new Q();
for(int i=1;i<=10;i++) {
print("Q($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 $count times");
}
- Output:
Q(1)=1 Q(2)=1 Q(3)=2 Q(4)=3 Q(5)=3 Q(6)=4 Q(7)=5 Q(8)=5 Q(9)=6 Q(10)=6 Q(1000)=502 value is smaller than previous 49798 times
If the maximum number is known, filling an array is probably the fastest solution.
main() {
List<int> q=new List<int>(100001);
q[1]=q[2]=1;
int count=0;
for(int i=3;i<q.length;i++) {
q[i]=q[i-q[i-1]]+q[i-q[i-2]];
if(q[i]<q[i-1]) {
count++;
}
}
for(int i=1;i<=10;i++) {
print("Q($i)=${q[i]}");
}
print("Q(1000)=${q[1000]}");
print("value is smaller than previous $count times");
}
Draco
proc nonrec make_Q([*] word q) void:
word n;
q[1] := 1;
q[2] := 1;
for n from 3 upto dim(q,1)-1 do
q[n] := q[n-q[n-1]] + q[n-q[n-2]]
od
corp
proc nonrec main() void:
word MAX = 1000;
word i;
[MAX+1] word q;
make_Q(q);
write("The first 10 terms are:");
for i from 1 upto 10 do write(" ", q[i]) od;
writeln();
writeln("The 1000th term is: ", q[1000])
corp- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502
EasyLang
proc hofstadter limit . q[] .
q[] = [ 1 1 ]
for n = 3 to limit
q[] &= q[n - q[n - 1]] + q[n - q[n - 2]]
.
.
proc count . q[] cnt .
for i = 2 to len q[]
if q[i] < q[i - 1]
cnt += 1
.
.
.
hofstadter 100000 hofq[]
for i = 1 to 10
write hofq[i] & " "
.
print ""
print hofq[1000]
count hofq[] cnt
print cnt
EchoLisp
(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
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
- 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)
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
- 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
%% @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).
- 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:
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 PROGRAMNote: 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%[].
Delphi
type TIntArray = array of integer;
procedure FillHofstadterArray(var HA: TIntArray);
{Fill array with Hofstader numbers}
{Preset array size to the number of terms you want}
var I: integer;
begin
{Starting condition}
HA[1]:=1; HA[2]:=1;
{Fill array up to last item}
for I:=3 to High(HA) do HA[I]:=HA[I-HA[I-1]]+HA[I-HA[I-2]];
end;
procedure ShowHofstadterNumbers(Memo: TMemo);
{Fill array with a }
var I, LessCount: integer;
var QArray: TIntArray;
begin
{Select the number of items we want}
SetLength(QArray,100000);
{Fill array}
FillHofstadterArray(QArray);
{Display first 10}
for I:=1 to 10 do Memo.Lines.Add(Format('%4d: %4d',[I,QArray[I]]));
Memo.Lines.Add(Format('%4d: %4d',[1000,QArray[1000]]));
{Count number the number of times Q(n)<Q(n-1)}
LessCount:=0;
for I:=1 to High(QArray) do
if QArray[I]>QArray[I-1] then Inc(LessCount);
Memo.Lines.Add('Count of Q(n)<Q(n-1) = '+IntToStr(LessCount));
end;
- Output:
1: 1 2: 1 3: 2 4: 3 5: 3 6: 4 7: 5 8: 5 9: 6 10: 6 1000: 502 Count of Q(n)<Q(n-1) = 49997
F#
The function
// 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]])
The Tasks
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)
- Output:
[|1; 1; 2; 3; 3; 4; 5; 5; 6; 6|] 502
Extra Credit
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))
- Output:
49798
- What is a large number?
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)
- 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.
( 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
Fermat
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)
- Output:
1 1 2 3 3 4 5 5 6 6
502
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
- 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.
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
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
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
- 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
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. 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 storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website.
In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.
Solution
The following function calculate the given number of terms of the Hofstadter Q sequence:
Case 1 First 10 terms
Case 2 Confirm and display that the 1000th term is 502
Case 3 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.
Go
Sure there are ways that run faster or handle larger numbers; for the task though, maps and recursion work just fine.
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))
}
- 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
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)
Haskell
The basic task:
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)
Extra credit task:
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)
- Output:
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)
Using a list (more or less) seemlessly backed up by a double resizing array:
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))
- 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.
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
Icon and Unicon
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
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)J
Solution (bottom-up):
Qs=:0 1 1
Q=: verb define
n=. >./,y
while. n>:#Qs do.
Qs=: Qs,+/(-_2{.Qs){Qs
end.
y{Qs
)
Solution (top-down):
Q=: 1:`(+&$:/@:- $:@-& 1 2)@.(>&2)"0 M.
Example:
Q 1+i.10
1 1 2 3 3 4 5 5 6 6
Q 1000
502
+/2>/\ Q 1+i.100000
49798
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.
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");
}
}
- 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'.
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));
- 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
(() => {
'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)
};
})();
- Output:
{"firstTen":[1, 1, 2, 3, 3, 4, 5, 5, 6, 6],
"thousandth":502,
"Q<Q-1UpTo10E5":49798}
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.
# 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(.))")Transcript
$ 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
Julia
The following implementation accepts an argument that is a single integer, an array of integers, or a range:
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))
- 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:
seq = hofstQseq(1:100_000)
cnt = count(diff(seq) .< 0)
println("$cnt elements are less than the preceding one.")
- Output:
49798 elements are less than the preceding one.
Since the implementation is non-recursive, there is no issue with recursion limits.
Kotlin
// version 1.1.4
fun main(args: Array<String>) {
val q = IntArray(100_001)
q[1] = 1
q[2] = 1
for (n in 3..100_000) q[n] = q[n - q[n - 1]] + q[n - q[n - 2]]
print("The first 10 terms are : ")
for (i in 1..10) print("${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")
}
- 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
Using Memoization
fun Q(n: Int): List<Int> {
val mem = mutableMapOf<Int, Int>().also {
it[1] = 1
it[2] = 1
}
q(n, mem)
return mem.values.toList()
}
private fun q(n: Int, mem: MutableMap<Int, Int>): Int {
if (!mem.containsKey(n)) {
mem[n] =
q(n - q(n - 1, mem), mem) + q(n - q(n - 2, mem), mem)
}
return mem[n]!!
}
fun main() {
val n = 1000
Q(n).also { qList ->
println("Q[1..10] = ${qList.take(10)}")
println("Q($n) = ${qList[1000 - 1]}") // 502
}
}
- Output:
Q[1..10] = [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] Q(1000) = 502
Lua
Here, the whole sequence up to the 100,000th term is generated for the first task, so this is where we risk hitting the recursion limit. As it happens, we do not. The function is called using 'pcall' so that any error would be caught. By increasing the argument on line 19 from 1e5 to 1e8, we can cause LuaJIT to run out of memory, but that is not necessary for this task.
function hofstadter (limit)
local Q = {1, 1}
for n = 3, limit do
Q[n] = Q[n - Q[n - 1]] + Q[n - Q[n - 2]]
end
return Q
end
function countDescents (t)
local count = 0
for i = 2, #t do
if t[i] < t[i - 1] then
count = count + 1
end
end
return count
end
local noError, hofSeq = pcall(hofstadter, 1e5)
if noError == false then
print("The sequence could not be calculated up to the specified limit.")
os.exit()
end
for i = 1, 10 do
io.write(hofSeq[i] .. " ")
end
print("\n" .. hofSeq[1000])
print(countDescents(hofSeq))
- Output:
1 1 2 3 3 4 5 5 6 6 502 49798
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- 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.
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:From this we get:
> seq( Q( i ), i = 1 .. 10 );
1, 1, 2, 3, 3, 4, 5, 5, 6, 6
> Q( 1000 );
502To determine the number of "flips", we proceed as follows.
> flips := 0:
> for i from 2 to 100000 do
> if L[ i ] < L[ i - 1 ] then
> flips := 1 + flips
> end if
> end do:
> flips;
49798Alternatively, we can build the sequence in an array.
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:This gives the same result.
> Qflips( 10^5 );
49798Mathematica / Wolfram Language
Hofstadter[1] = Hofstadter[2] = 1;
Hofstadter[n_Integer?Positive] := Hofstadter[n] = Block[{$RecursionLimit = Infinity},
Hofstadter[n - Hofstadter[n - 1]] + Hofstadter[n - Hofstadter[n - 2]]
]
- Output:
Hofstadter /@ Range[10]
{1,1,2,3,3,4,5,5,6,6}
Hofstadter[1000]
502
Count[Differences[Hofstadter /@ Range[100000]], _?Negative]
49798
MATLAB / Octave
This solution pre-allocates memory and is an iterative solution, so caching or recursion limits do not apply.
function Q = Qsequence(N)
%% zeros are used to pre-allocate memory, this is not strictly necessary but can significantly improve performance for large N
Q = [1,1,zeros(1,N-2)];
for n=3:N
Q(n) = Q(n-Q(n-1))+Q(n-Q(n-2));
end;
end;
Confirm and display that the first ten terms of the sequence are: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6
>> Qsequence(10) ans = 1 1 2 3 3 4 5 5 6 6
Confirm and display that the 1000'th term is: 502
>> Q=Qsequence(1000); Q(end) ans = 502
Count and display how many times a member of the sequence is less than its preceding term for terms up to and including the 100,000'th term.
>> sum(diff(Qsequence(100000))<0) ans = 49798
Maxima
/* Function that return the terms of the Hofstadter Q sequence */
hofstadter(n):=block(
if member(n,[1,2]) then L[n]:1 else L[n]:L[n-L[n-1]]+L[n-L[n-2]],
L[n])$
/* Test cases */
/* First ten terms */
makelist(hofstadter(i),i,1,10);
/* 1000th term */
last(makelist(hofstadter(i),i,1,1000));
- Output:
[1,1,2,3,3,4,5,5,6,6] 502
Modula-2
MODULE QSequence;
FROM InOut IMPORT WriteString, WriteCard, WriteLn;
VAR n: CARDINAL;
Q: ARRAY [1..1000] OF CARDINAL;
BEGIN
Q[1] := 1;
Q[2] := 1;
FOR n := 3 TO 1000 DO
Q[n] := Q[n-Q[n-1]] + Q[n-Q[n-2]];
END;
WriteString("The first 10 terms are:");
FOR n := 1 TO 10 DO
WriteCard(Q[n],2);
END;
WriteLn();
WriteString("The 1000th term is:");
WriteCard(Q[1000],4);
WriteLn();
END QSequence.
- Output:
The first 10 terms are: 1 1 2 3 3 4 5 5 6 6 The 1000th term is: 502
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)
- 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
Miranda
main :: [sys_message]
main = [Stdout (lay (map showq ([1..10] ++ [1000])))]
where showq n = "q!" ++ show n ++ " = " ++ show (q!n)
q :: [num]
q = 0 : 1 : 1 : map f [3..] where f n = q!(n - q!(n-1)) + q!(n - q!(n-2))- 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
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
- Output:
@[1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 502 49798
Oberon-2
Works with oo2c version 2
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.
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
OCaml
(* valid results for n in 0..119628 *)
let seq_hofstadter_q n =
let a = Bigarray.(Array1.create int16_unsigned c_layout n) in
let () =
for i = 0 to pred n do
a.{i} <- if i < 2 then 1 else a.{i - a.{pred i}} + a.{i - a.{i - 2}}
done
in
Seq.init n (Bigarray.Array1.get a)
let () =
let count_backflip (a, c) b = b, if b < a then succ c else c
and hq = seq_hofstadter_q 100_000 in
let () = Seq.(hq |> take 10 |> iter (Printf.printf " %u")) in
let () = Seq.(hq |> drop 999 |> take 1 |> iter (Printf.printf "\n%u\n")) in
hq |> Seq.fold_left count_backflip (0, 0) |> snd |> Printf.printf "%u\n"
- Output:
1 1 2 3 3 4 5 5 6 6 502 49798
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 ;- Output:
[1, 1, 2, 3, 3, 4, 5, 5, 6, 6] 502 49798
PARI/GP
Straightforward, unoptimized version; about 1 ms.
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)"));- Output:
First 10 terms: [1, 1, 2, 3, 3, 4, 5, 5, 6, 6] (as expected) 1000-th term: 502 (as expected)
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.
- Output:
:> ./HofstadterQSequence 1 1 2 3 3 4 5 5 6 6 502 Flips: 49798
PascalABC.NET
##
var q := |1, 1|.ToList;
for var n := 2 to 100_000 do
q.add(q[n - q[n - 1]] + q[n - q[n - 2]]);
q.take(10).println;
assert(q.Take(10).SequenceEqual(|1, 1, 2, 3, 3, 4, 5, 5, 6, 6|));
q[999].println;
assert(q[999] = 502);
var lessCount := 0;
for var n := 1 to 100_000 do
if q[n] < q[n - 1] then
lessCount += 1;
lessCount.Println;
- Output:
1 1 2 3 3 4 5 5 6 6 502 49798
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);
- 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:
#!/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, $less_than_preceding terms are less " .
"than their preceding terms!\n";
- Output:
1 1 2 3 3 4 5 5 6 6 The 1000'th term is 502! Up to and including the 100000'th term, 49798 terms are less than their preceding terms!
This different solution uses tie to make the Q sequence look like a regular array, and only fills the cache on demand. Some pre-allocation is done which provides a minor speed increase for the extra credit. I could have chosen to do recursion instead of iteration, as perl has no limit on how deeply one may recurse, but did not see the benefit of doing so.
#!perl
use strict;
use warnings;
package Hofstadter;
sub TIEARRAY {
bless [undef, 1, 1], shift;
}
sub FETCH {
my ($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";
- Output:
1 1 2 3 3 4 5 5 6 6 502 Extra credit: 49798
Phix
Just to be flash, I also (on the desktop only) calculated the 100 millionth term - the only limiting factor here is the length of Q (theoretically 402,653,177 on 32 bit).
with javascript_semantics 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: %v\n",{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}) if platform()!=JS then atom t0 = time() printf(1,"100,000,000th: %,d (%3.2fs)\n",{q(100_000_000),time()-t0}) end if
- Output:
First ten terms: {1,1,2,3,3,4,5,5,6,6}
1000th: 502
100,000th: 48,157
Flips up to 100,000: 49,798
100,000,000th: 50,166,508 (8.52s)
The last line shows fine under pwa/p2js, but would take about 20s.
Picat
go =>
println([q(I) : I in 1..10]),
println(q1000=q(1000)),
Q = {q(I) : I in 1..100_000},
println(flips=sum({1 : I in 2..100_000, Q[I-1] > Q[I]})),
nl.
table
q(1) = 1.
q(2) = 1.
q(N) = q(N-q(N-1)) + q(N-q(N-2)).- Output:
[1,1,2,3,3,4,5,5,6,6] q1000 = 502 flips = 49798
PicoLisp
(de q (N)
(cache '(NIL) N
(if (>= 2 N)
1
(+
(q (- N (q (dec N))))
(q (- N (q (- N 2)))) ) ) ) )Test:
: (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) )
-> 49798PL/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;- 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:
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);
- Output:
n=100000 48157 TALLY= 49798;
PL/M
100H:
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
PRINT$NUMBER: PROCEDURE (N);
DECLARE S (7) BYTE INITIAL ('..... $');
DECLARE (N, P) ADDRESS, C BASED P BYTE;
P = .S(5);
DIGIT:
P = P - 1;
C = N MOD 10 + '0';
N = N / 10;
IF N > 0 THEN GO TO DIGIT;
CALL PRINT(P);
END PRINT$NUMBER;
DECLARE Q (1001) ADDRESS;
DECLARE N ADDRESS;
Q(1)=1;
Q(2)=1;
DO N=3 TO LAST(Q);
Q(N) = Q(N-Q(N-1)) + Q(N-Q(N-2));
END;
CALL PRINT(.'THE FIRST 10 TERMS ARE: $');
DO N=1 TO 10;
CALL PRINT$NUMBER(Q(N));
END;
CALL PRINT(.(13,10,'THE 1000TH TERM IS: $'));
CALL PRINT$NUMBER(Q(1000));
CALL EXIT;
EOF- Output:
THE FIRST 10 TERMS ARE: 1 1 2 3 3 4 5 5 6 6 THE 1000TH TERM IS: 502
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
- Output:
First ten: 1 1 2 3 3 4 5 5 6 6 1000th: 502 Flips: 49798
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))
- 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:
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:])))
- 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
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)]))
Quackery
[ 2dup swap size dup negate swap within
not if
[ drop size 1+ number$
$ "Term " swap join
$ " of the Q sequence is not defined."
join message put bail ]
peek ] is qpeek ( [ n --> x )
[ dup dup -1 qpeek negate qpeek
dip [ dup dup -2 qpeek negate qpeek ]
+ join ] is next-q ( [ --> [ )
[ dup size 2 < iff
[ drop 0 ] done
0 swap behead swap
witheach
[ tuck > if [ dip 1+ ] ]
drop ] is drops ( [ --> n )
0 backup
[ ' [ 1 1 ]
998 times next-q
dup
-1 split swap 10 split drop
witheach [ echo sp ]
say "... "
0 peek echo cr
99000 times next-q
drops echo
say " decreasing terms" ]
bailed if
[ message take cr echo$ cr ]- Output:
1 1 2 3 3 4 5 5 6 6 ... 502 49798 decreasing terms
R
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")
- Output:
1 1 2 3 3 4 5 5 6 6 502 49798 terms is less than its preceding term
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))
- 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.
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";
- 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.
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";
(Same output.)
REXX
non-recursive
The REXX language doesn't allow expressions for stemmed array indices, so a temporary variable must be used.
/*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))
- 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.
/*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 _
- 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
/*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 _
- output is identical to the 1st REXX version.
The recursive version is almost ten times slower than the (1st) non-recursive version.
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
nextRPL
| RPL code | Comment |
|---|---|
≪
{ 1 1 } 3
WHILE DUP 4 PICK ≤ REPEAT
DUP2 2 - GETI ROT ROT GET → n q2 q1
≪ DUP n q1 - GET
OVER n q2 - GET + +
n 1 + SWAP
≫
END DROP
≫ 'HOFST' STO
|
HOFST ( m -- { Q(1)..Q(m) } )
initialize stack with Q1, Q2 and loop index n
loop
store n, Q(n-2) and Q(n-1)
get Q(n-Q(n-1))
add Q(n-Q(n-2)) and add result to list
put back n+1 in stack
|
- Input:
10 HOFST 1000 HOSFT DUP SIZE GET
- Output:
2: { 1 1 2 3 3 4 5 5 6 6 }
1: 502
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}"
- 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
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- 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.
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);
}
- 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 ...
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))
}
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.
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)
}
- 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.
(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)
- Output:
Q(1 .. 10): 1 1 2 3 3 4 5 5 6 6 Q(1000): 502 bumps up to 100000: 49798
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;- 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
SETL
program hofstadter_q;
q := [1,1];
loop for n in [3..100000] do
q(n) := q(n-q(n-1)) + q(n-q(n-2));
end loop;
print("First 10 terms: " + q(1..10));
print("1000th term: " + q(1000));
print("q(x) < q(x-1): " + #[x : x in [2..#q] | q(x) < q(x-1)]);
end program;- Output:
First 10 terms: [1 1 2 3 3 4 5 5 6 6] 1000th term: 502 q(x) < q(x-1): 49798
Sidef
Using a memoized function:
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)}}"
Using an array:
var Q = [0, 1, 1]
100_000.times {
Q << (Q[-Q[-1]] + Q[-Q[-2]])
}
say "First 10 terms: #{Q.slice(1).first(10)}"
say "Term 1000: #{Q[1000]}"
say "Terms less than preceding in first 100k: #{2..100000->count{|i|Q[i]<Q[i-1]}}"
- 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
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)")
- 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
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::writev0.5
q templates
outputFrom is $(1);
until is $(2);
@ set [1,1];
1..$until -> # !
when <|$@::length~..> do
..|@ set $@($ - $@($ - 1)) + $@($ - $@($ - 2));
$ -> # !
when <|$outputFrom..> do
$@($) !
end q
[1,10] -> q -> '$; ' !
'
' !
[1000,1000] -> q -> '$;
' !
countDownSteps templates
@ set 0;
qs is $;
2..$qs::length -> !#
$@ !
when <|?($qs($) matches <|..~$qs($-1)>)> do @ set $@ + 1;
end countDownSteps
[[1, 100000] -> q] -> countDownSteps -> 'Less than previous $; times' !- Output:
1 1 2 3 3 4 5 5 6 6 502 Less than previous 49798 times
Tcl
package require Tcl 8.5
# Index 0 is not used, but putting it in makes the code a bit shorter
set tcl::mathfunc::Qcache {Q:-> 1 1}
proc tcl::mathfunc::Q {n} {
variable Qcache
if {$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"- 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.
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))- Output:
First 10 terms of Q = 1 1 2 3 3 4 5 5 6 6 256th term = 123 0 OK, 0:320
VBA
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- 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
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)- 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
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- 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
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.")- 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
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);
]- Output:
1 1 2 3 3 4 5 5 6 6 502 49798
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);- 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.
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