Creating the shortest Turing-complete interpreter [closed]

Answer 1

Assemble the code below using A86 to get a 150 byte BF interpreter!

    add dh,10h
    push dx
    add dh,10h
    push dx
    mov bl,80h
    lea dx,[bx+2]
    add bl,byte ptr [bx]
    mov byte ptr [bx+1],al
    mov ah,3dh
    int 21h
    pop ds,es
    jc l14
    mov bx,ax
    mov ah,3fh  
    mov cx,di
    xor dx,dx
    int 21h
    jc l14
    mov bx,ax
    xor ax,ax
    rep stosw
    xor di,di
    xor si,si
    inc ch
l1:
    cmp si,bx
    jae l14
    lodsb
    mov bp,8
    push l1
l2: 
    dec bp
    js ret
    cmp al,[bp+l15]
    jne l2
l3:
    mov cl,[bp+8+l15]
    jmp cx
l4:
    inc di  
    ret
l5:
    dec di
    ret
l6:
    es inc byte ptr [di]
    ret
l7:
    es dec byte ptr [di]
    ret
l8:
    mov ah,2
    es mov dl,[di]
    int 21h
    ret
l9:
    mov ah,8
    int 21h
    es mov [di],al
    ret
l10:
    es cmp byte ptr [di],dh
    jne ret
l11:    
    cmp si,bx
    jae l14
    lodsb
    cmp al,']'
    jne l11
    ret
l12:
    es cmp byte ptr [di],dh
    je ret
l13:    
    dec si
    jz l14
    cmp byte ptr [si-1],'['
    jne l13
l14:
    ret
l15:
    db '>'
    db '<'
    db '+'
    db '-'
    db '.'
    db ','
    db '['
    db ']'
    db offset l4-100h
    db offset l5-100h
    db offset l6-100h
    db offset l7-100h
    db offset l8-100h
    db offset l9-100h
    db offset l10-100h
    db offset l12-100h

Specify a filename on the command line, no need for double quotes, as the BF source.

Answer 2

Not mine, but take a look at the 210 bit binary lambda calculus self-interpreter here.

Answer 3

Python program that interprets a language I just created:

from random import randint
z = [randint(0,1), None]  # example: input data is 0 or 1
x = '_b_ed_a_i____d_ai'   # this program will set p = data[1] = not data[0]

# input x[i], data z[k]
# jumper j
# p is +1 or -1 each step
# commands:
#    a   set data = 0 or 1
#    b   j += 0 or +-9 depending on data
#    d   move data left or right
#    e   perform jump left or right
#    j   set jumper to 0
#    i   end program
# output: p or some data[x], both possible

g = globals()
p = i = 1
a = b = d = j = e = k = 0
while i:
    h = x[i]
    g[h] = p = -p
    i += 1 + j*e
    if a: z[k] = i % 2
    if z[k]: j += 9*b
    k += d
    g[h] = 0        
#    print(a, b, d, e, i, j, k, h, z)

print('Input:', z, 'Output:', p > 0)

Optimized version:

g=globals()
p=i=1
a=b=d=j=e=k=0
while i:
 h=x[i]
 g[h]=p=-p
 i+=1+j*e
 if a:z[k]=i%2
 if z[k]:j+=9*b
 k+=d
 g[h]=0

114 bytes

Update: I want to add that the point of my program is not to create any practical language, and not even to somehow have the 'best' (=='shortest') interpreter, but rather to demonstrate interesting programming tricks. For example, I am relying on direct access to global variables through globals(), so that I never even test for j command, saving precious bytes :)

Answer 4

Building on my previous code-golf entry, here is a (slightly generalized for IO) OISC emulator in

Fortran77

Obsfucated and without the loading scaffold (655 characters):

  subroutine o(n,m)
  integer m(n)              
  l=1;                      
  do while (l.ne.0)
     i=m(l)
     j=m(l+1)
     k=m(l+2)
     mi=mg(n,m,i)
     mj=mg(n,m,j)
     if(j.eq.(n+2)) then
        write(6,*)mj-mi
     else
        m(l+1)=mj-mi
     endif
     if (m(l+1).lt.0) then
        l=mg(n,m,k)
     else
        l=l+3
     endif
  end do
  return
  end
  function mg(n,m,i)
  integer m(n)
  if (i.eq.n+2) then
     read(5,*)mg
  elseif (i.eq.n+1) then
     mg=0
  else
     mg=m(i)
  endif
  return
  end

with comments, loading scaffold and so on (2435 characters):

      program c
      parameter (n=1024)        ! The size to use for memeory
      integer m(n)              ! represent the memory
c     Load a program into memory
      i=1
 1    read(5,*,end=2)l
c     write (6,*) "Load ",l," into location ",i
      m(i)=l
      i=i+1
      goto 1
c     Run the computer
 2    call o(n,m)
      stop
      end

      subroutine o(n,m)
c     Simulate a simple computer that supports only a single
c     instruction. Memory is a fixed size array of integers.
c
c     The supported instruction is subtract-branch-negative which we
c     give the assembly syntax
c
c     sbn i j k
c
c     and acts by subtracting the value in memeory location i from that
c     in location j and storing the result in j. If the result is
c     negative, the PC is set to k. Because there is only one opcode, it
c     is not represented, so a program consists simply of a series of
c     triplet (i,j,k), and the PC is generally incremented by 3. The
c     program counter starts at 1
c
c     Povisions for IO and halting are by way of special memory
c     location:
c
c     * PC=0 means halt
c     * writes (opcode argument j) to memory location n+1 send
c       output, reads from n+1 evaluate as 0
c     * reads (opcode argument i, j, k) from memory location n+2 fetch
c       input, writes to n+2 are forbidden
c     * All others reads and writes outside of memeory are forbidden

c     n                         ! the size of the computers memory 
      integer m(n)              ! an array representing the computers memory
      l=1;                      ! program counter
      do while (l.ne.0)
         i=m(l)
         j=m(l+1)
         k=m(l+2)
c        write (6,*) "Executing from PC=",l,": (",i,", ",j,", ", k,")"
c        handle the write special case for output
         mi=mg(n,m,i)
         mj=mg(n,m,j)
         if(j.eq.(n+1)) then
            write(6,*)mj-mi
         else
c           write (6,*) "Setting location ",j," to ",mj-mi
            m(l+1)=mj-mi
         endif
         if (m(l+1).lt.0) then
            l=mg(n,m,k)
         else
            l=l+3
         endif
      end do
      return
      end

c     Handle the various read special cases
      function mg(n,m,i)
      integer m(n)
      if (i.eq.n+2) then
         read(5,*)mg
      elseif (i.eq.n+1) then
         mg=0
      else
         mg=m(i)
      endif
c     write (6,*) "Read ",mg," from location ",i
      return
      end

Sample program:

13
1025
0

14 
1025
0

14
15
0

0
0
0

-1
1
0

resulting in output:

$ cat trivial.oisc | ./a.out 
           1
          -1

which is as expected.

It really is exceedingly straightforward code. The point here is not really how tightly I can code it, but how simple a language you need for Turing completeness.

Answer 5

#include "/dev/tty"

Answer 6

URM interpreter in CoffeeScript, 143 bytes (167 with new lines).

This version of URM has consists of an unbounded amount of registers, with zero, successor and jump operators. It is well known that this is turing-complete.

The URM program is written in the array c (commands), and the inputs are in the array r (registers). After the calculation the output is at r[0] and this value is displayed.

The interpreter, with sample program and input that computes 32+13 (and indeed outputs 45):

# Addition program, similar to http://planetmath.org/examplesofunlimitedregistermachines
c = [['j', 1, 2, 4], ['s', 0], ['s', 2], ['j', 1, 1, 0]]

# Input: 32, 13, thus the desired output is: 45
r = [32, 13]

k = 0
while k < c.length
    t = c[k]
    k += 1
    if t[0] == 'z'
        r[t[1]] = 0
    if t[0] == 's'
        if !r[t[1]]?
            r[t[1]] = 0
        r[t[1]] += 1
    if t[0] == 'j' && r[t[1]] == r[t[2]]
        k = t[3]
alert r[0]

Minified version:

k=0
while k<c.length
 t=c[k]
 k+=1
 if t[0]=='z'
  r[t[1]]=0
 if t[0]=='s'
  if !r[t[1]]?
   r[t[1]]=0
  r[t[1]]+=1
 if t[0]=='j'&&r[t[1]]==r[t[2]]
  k=t[3]
alert r[0]

What I really like in this code is that is really straightforward and easy to understand.