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fun_with_mov_turing_completeness.cpp
695 lines (640 loc) · 28.5 KB
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fun_with_mov_turing_completeness.cpp
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/*
fun_with_mov_turing_completeness.c - Just having fun with ideas taken from that paper:
http://www.cl.cam.ac.uk/~sd601/papers/mov.pdf
Copyright (C) 2013 Axel "0vercl0k" Souchet - http://www.twitter.com/0vercl0k
This program is free software: you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation, either version 3 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program. If not, see <http://www.gnu.org/licenses/>.
If you are interested in about the turing completeness, you may want
to read a relevant discusion on reddit with rolfr & PaxTeam here:
-> http://www.reddit.com/r/ReverseEngineering/comments/1lfm21/mov_is_turingcomplete_by_stephen_dolan_pdf/
I'm quite sure if Walter Bishop was a computer science guy, he would have loved to play
with that crazy machine.
So, Walter, please accept this gift.
______________ ______________
\__ ___/ | \_ _____/
| | / ~ \ __)_
| | \ Y / \
|____| \___|_ /_______ /
\/ \/
__________.___ _________ ___ ___ ________ __________
\______ \ |/ _____// | \\_____ \\______ \
| | _/ |\_____ \/ ~ \/ | \| ___/
| | \ |/ \ Y / | \ |
|______ /___/_______ /\___|_ /\_______ /____|
\/ \/ \/ \/
_____ _____ _________ ___ ___ .___ _______ ___________
/ \ / _ \ \_ ___ \ / | \| |\ \ \_ _____/
/ \ / \ / /_\ \/ \ \// ~ \ |/ | \ | __)_
/ Y \/ | \ \___\ Y / / | \| \
\____|__ /\____|__ /\______ /\___|_ /|___\____|__ /_______ /
\/ \/ \/ \/ \/ \/
Special thanks to Stephen Dolan for the discussions by emails.
Example output for the lazy ones:
D:\Codes\fun_with_mov_turing_completeness\Release>fun_with_mov_turing_completeness.exe
OK, mem allocated @ 00935FE8
S0=0x00935fe8, S1=0x00935fec
OK, here's the initial tape (0087F704)
0 1 1 1 0 1 1 0
N=0x00935ff0, @S=0x0087f700, @L=0x0087f6f8, @R=0x0087f6f4
Q0=0x0087f6cc, Q0_=0x0087f668, Q0__=0x0087f58c
Q1=0x0087f694, Q1_=0x0087f65c, Q1__=0x0087f57c
Q2=0x0087f68c, Q2_=0x0087f650, Q2__=0x0087f56c
Q3=0x0087f644, Q3_=0x0087f55c
Q4=0x0087f6d4, Q4_=0x0087f684, Q4__=0x0087f54c
c1=0x0087f6ac, c2=0x0087f6bc, c3=0x0087f69c, c4=0x0087f6e4
c5=0x0087f6dc, c6=0x0087f6b4, c7=0x0087f6a4, c8=0x0087f6c4
c9=0x0087f52c
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 [1] 1 1 0 1 1 0
[0x00935ff0] [0x0087f6ac] [0x0087f6e4] [0x0087f6dc] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6bc, L=0x0087f6ac, R=0x0087f69c, T=0x0087f668
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 [1] 1 0 1 1 0
[0x00935ff0] [0x0087f6ac] [0x0087f6e4] [0x0087f6dc] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f69c, L=0x0087f6bc, R=0x0087f6e4, T=0x0087f694
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 [1] 0 1 1 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f6dc] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6e4, L=0x0087f69c, R=0x0087f6dc, T=0x0087f694
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 [0] 1 1 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6dc, L=0x0087f6e4, R=0x0087f6b4, T=0x0087f694
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 [0] 1 1 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6dc, L=0x0087f6e4, R=0x0087f6b4, T=0x0087f65c
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 1 [1] 1 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6b4, L=0x0087f6dc, R=0x0087f6a4, T=0x0087f68c
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 1 1 [1] 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6dc] [0x0087f6c4] [0x0087f52c]
S=0x0087f6a4, L=0x0087f6b4, R=0x0087f6c4, T=0x0087f68c
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 1 1 1 [0]
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6dc] [0x0087f6b4] [0x0087f52c]
S=0x0087f6c4, L=0x0087f6a4, R=0x0087f52c, T=0x0087f68c
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 1 1 1 [0]
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6dc] [0x0087f6b4] [0x0087f6a4]
S=0x0087f6c4, L=0x0087f6a4, R=0x0087f52c, T=0x0087f650
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 1 1 [1] 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6dc] [0x0087f6b4] [0x0087f52c]
S=0x0087f6a4, L=0x0087f6b4, R=0x0087f6c4, T=0x0087f644
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 1 [1] 0 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6dc] [0x0087f6c4] [0x0087f52c]
S=0x0087f6b4, L=0x0087f6dc, R=0x0087f6a4, T=0x0087f6d4
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 1 [1] 1 0 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6dc, L=0x0087f6e4, R=0x0087f6b4, T=0x0087f6d4
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 1 [1] 1 1 0 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f69c] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6e4, L=0x0087f69c, R=0x0087f6dc, T=0x0087f6d4
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 1 [1] 1 1 1 0 0
[0x00935ff0] [0x0087f6ac] [0x0087f6bc] [0x0087f6dc] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f69c, L=0x0087f6bc, R=0x0087f6e4, T=0x0087f6d4
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
0 [1] 1 1 1 1 0 0
[0x00935ff0] [0x0087f6ac] [0x0087f6e4] [0x0087f6dc] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6bc, L=0x0087f6ac, R=0x0087f69c, T=0x0087f6d4
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
[0] 1 1 1 1 1 0 0
[0x00935ff0] [0x0087f69c] [0x0087f6e4] [0x0087f6dc] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6ac, L=0x00935ff0, R=0x0087f6bc, T=0x0087f6d4
------------------------------------
-----------DEBUG--------------------
OK, done, inspecting the tape now..
[0] 1 1 1 1 1 0 0
[0x0087f6bc] [0x0087f69c] [0x0087f6e4] [0x0087f6dc] [0x0087f6b4] [0x0087f6a4] [0x0087f6c4] [0x0087f52c]
S=0x0087f6ac, L=0x00935ff0, R=0x0087f6bc, T=0x0087f684
------------------------------------
OK, machine done, inspecting the tape now..
0 [1] 1 1 1 1 0 0
*/
#include <stdio.h>
#include <windows.h>
void turing_machine_simple()
{
// This is a simple way to add two number via a turing machine
// Comes from http://turingmaschine.klickagent.ch/einband/?&lang=en#2_+_3
// Each transition is written this way: (TriggeringSymbol, NewSymbol, Direction)
// +--+(0,0,R) +--+(1,1,R)
// | | | |
// | v | v
// +--+ (1,1,R)+--+ (0,1,R)
// |00+--------->|01|---------+
// +--+ +--+ |
// |
// v
// +--+ (1,1,L) +--+-------+
// | | |02| |(1,1,R)
// | | +--+<------+
// | v |
// +--+ (1,0,L) +--+ (0,0,L)|
// |04|<---------+03|<---------+
// ++-+ +--+
// |
// |(0,0,R)
// v
// +--+
// |05|
// +--+
///// SETUP THE MACHINE
unsigned int *mem = (unsigned int*)malloc(3 * sizeof(unsigned int));
printf("OK, mem allocated @ %p\n", mem);
unsigned int S0 = (unsigned int)mem;
unsigned int S1 = (unsigned int)(mem + 1);
unsigned int N = (unsigned int)(mem + 2);
unsigned int DirR = 1, DirL = 0;
printf("S0=%#.8x, S1=%#.8x\n", S0, S1);
//// Building the transition table now
unsigned int Q0[2], Q0_[2], Q0__[2];
unsigned int Q1[2], Q1_[2], Q1__[2];
unsigned int Q2[2], Q2_[2], Q2__[2];
unsigned int Q3[2], Q3_[2];
unsigned int Q4[2], Q4_[2], Q4__[2];
/// Q0
// This one represent the transition (0, 0, R) from Q0
// Q0[0] -> 0 -> 0 -> 1 -> Q0
unsigned int Q0_first_trans[2];
Q0[0] = (unsigned int)Q0_first_trans;
Q0_first_trans[0] = S0;
unsigned int Q0_first_trans_[2];
Q0_first_trans[1] = (unsigned int)Q0_first_trans_;
Q0_first_trans_[0] = S0;
unsigned int Q0_first_trans__[2];
Q0_first_trans_[1] = (unsigned int)Q0_first_trans__;
Q0_first_trans__[0] = DirR;
unsigned int Q0_first_trans___;
Q0_first_trans__[1] = (unsigned int)&Q0_first_trans___;
Q0_first_trans___ = (unsigned int)Q0;
// This one represent the transition (1, 1, R) from Q0
// Q0[1] -> 1 -> 1 -> 1 -> Q1
unsigned int Q0_second_trans[2];
Q0[1] = (unsigned int)Q0_;
Q0_[0] = (unsigned int)Q0_second_trans;
Q0_second_trans[0] = S1;
unsigned int Q0_second_trans_[2];
Q0_second_trans[1] = (unsigned int)Q0_second_trans_;
Q0_second_trans_[0] = S1;
unsigned int Q0_second_trans__[2];
Q0_second_trans_[1] = (unsigned int)Q0_second_trans__;
Q0_second_trans__[0] = DirR;
unsigned int Q0_second_trans___;
Q0_second_trans__[1] = (unsigned int)&Q0_second_trans___;
Q0_second_trans___ = (unsigned int)Q1;
// end
Q0_[1] = (unsigned int)Q0__;
Q0__[0] = N;
/// Q1
// This one represent the transition (1, 1, R) from Q1
// Q1[0] -> 1 -> 1 -> R -> Q1
unsigned int Q1_first_trans[2];
Q1[0] = (unsigned int)Q1_first_trans;
Q1_first_trans[0] = S1;
unsigned int Q1_first_trans_[2];
Q1_first_trans[1] = (unsigned int)Q1_first_trans_;
Q1_first_trans_[0] = S1;
unsigned int Q1_first_trans__[2];
Q1_first_trans_[1] = (unsigned int)Q1_first_trans__;
Q1_first_trans__[0] = DirR;
unsigned int Q1_first_trans___;
Q1_first_trans__[1] = (unsigned int)&Q1_first_trans___;
Q1_first_trans___ = (unsigned int)Q1;
// This one represent the transition (0, 1, R) from Q1
// Q1[1] -> 0 -> 1 -> 1 -> Q2
unsigned int Q1_second_trans[2];
Q1[1] = (unsigned int)Q1_;
Q1_[0] = (unsigned int)Q1_second_trans;
Q1_second_trans[0] = S0;
unsigned int Q1_second_trans_[2];
Q1_second_trans[1] = (unsigned int)Q1_second_trans_;
Q1_second_trans_[0] = S1;
unsigned int Q1_second_trans__[2];
Q1_second_trans_[1] = (unsigned int)Q1_second_trans__;
Q1_second_trans__[0] = DirR;
unsigned int Q1_second_trans___;
Q1_second_trans__[1] = (unsigned int)&Q1_second_trans___;
Q1_second_trans___ = (unsigned int)Q2;
// end
Q1_[1] = (unsigned int)Q1__;
Q1__[0] = N;
/// Q2
// This one represent the transition (1, 1, R) from Q2
// Q2[0] -> 1 -> 1 -> R -> Q2
unsigned int Q2_first_trans[2];
Q2[0] = (unsigned int)Q2_first_trans;
Q2_first_trans[0] = S1;
unsigned int Q2_first_trans_[2];
Q2_first_trans[1] = (unsigned int)Q2_first_trans_;
Q2_first_trans_[0] = S1;
unsigned int Q2_first_trans__[2];
Q2_first_trans_[1] = (unsigned int)Q2_first_trans__;
Q2_first_trans__[0] = DirR;
unsigned int Q2_first_trans___;
Q2_first_trans__[1] = (unsigned int)&Q2_first_trans___;
Q2_first_trans___ = (unsigned int)Q2;
// This one represent the transition (0, 0, L) from Q2
// Q2[1] -> 0 -> 0 -> L -> Q3
unsigned int Q2_second_trans[2];
Q2[1] = (unsigned int)Q2_;
Q2_[0] = (unsigned int)Q2_second_trans;
Q2_second_trans[0] = S0;
unsigned int Q2_second_trans_[2];
Q2_second_trans[1] = (unsigned int)Q2_second_trans_;
Q2_second_trans_[0] = S0;
unsigned int Q2_second_trans__[2];
Q2_second_trans_[1] = (unsigned int)Q2_second_trans__;
Q2_second_trans__[0] = DirL;
unsigned int Q2_second_trans___;
Q2_second_trans__[1] = (unsigned int)&Q2_second_trans___;
Q2_second_trans___ = (unsigned int)Q3;
// end
Q2_[1] = (unsigned int)Q2__;
Q2__[0] = N;
/// Q3
// This one represent the transition (1, 0, L) from Q3
// Q3[0] -> 1 -> 0 -> L -> Q4
unsigned int Q3_first_trans[2];
Q3[0] = (unsigned int)Q3_first_trans;
Q3_first_trans[0] = S1;
unsigned int Q3_first_trans_[2];
Q3_first_trans[1] = (unsigned int)Q3_first_trans_;
Q3_first_trans_[0] = S0;
unsigned int Q3_first_trans__[2];
Q3_first_trans_[1] = (unsigned int)Q3_first_trans__;
Q3_first_trans__[0] = DirL;
unsigned int Q3_first_trans___;
Q3_first_trans__[1] = (unsigned int)&Q3_first_trans___;
Q3_first_trans___ = (unsigned int)Q4;
// end
Q3[1] = (unsigned int)Q3_;
Q3_[0] = N;
/// Q4
// This one represent the transition (1, 1, R) from Q4
// Q4[0] -> 1 -> 1 -> L -> Q4
unsigned int Q4_first_trans[2];
Q4[0] = (unsigned int)Q4_first_trans;
Q4_first_trans[0] = S1;
unsigned int Q4_first_trans_[2];
Q4_first_trans[1] = (unsigned int)Q4_first_trans_;
Q4_first_trans_[0] = S1;
unsigned int Q4_first_trans__[2];
Q4_first_trans_[1] = (unsigned int)Q4_first_trans__;
Q4_first_trans__[0] = DirL;
unsigned int Q4_first_trans___;
Q4_first_trans__[1] = (unsigned int)&Q4_first_trans___;
Q4_first_trans___ = (unsigned int)Q4;
// This one represent the transition (0, 0, R) from Q4
// Q4[1] -> 0 -> 0 -> R -> N
unsigned int Q4_second_trans[2];
Q4[1] = (unsigned int)Q4_;
Q4_[0] = (unsigned int)Q4_second_trans;
Q4_second_trans[0] = S0;
unsigned int Q4_second_trans_[2];
Q4_second_trans[1] = (unsigned int)Q4_second_trans_;
Q4_second_trans_[0] = S0;
unsigned int Q4_second_trans__[2];
Q4_second_trans_[1] = (unsigned int)Q4_second_trans__;
Q4_second_trans__[0] = DirR;
unsigned int Q4_second_trans___;
Q4_second_trans__[1] = (unsigned int)&Q4_second_trans___;
Q4_second_trans___ = N;
// end
Q4_[1] = (unsigned int)Q4__;
Q4__[0] = N;
// Now we have to build the tape -- it's supposed to be infinite, but no for our simple PoC
unsigned int cell1[2], cell2[2], cell3[2], cell4[2], cell5[2], cell6[2], cell7[2], cell8[2], cell9[2];
unsigned int tape[8] = {
(unsigned int)cell1, (unsigned int)cell2,
(unsigned int)cell3, (unsigned int)cell4,
(unsigned int)cell5, (unsigned int)cell6,
(unsigned int)cell7, (unsigned int)cell8
};
// At the begining we have
// S = cell2
// L = cell1
// R = cell3 (cell3->cell4->cell5->cell6->cell7->cell8)
// 0 | 1 | 1 | 1 | 0 | 1 | 1 | 0
// 3 + 2
//
// At the end the tape should look:
// .. | 1 | 1 | 1 | 1 | 1 | 0
// L
cell1[0] = S0;
cell1[1] = N;
// S
cell2[0] = S1;
cell2[1] = 0xdeadbeef;
// R
cell3[0] = S1;
cell3[1] = (unsigned int)cell4;
cell4[0] = S1;
cell4[1] = (unsigned int)cell5;
cell5[0] = S0;
cell5[1] = (unsigned int)cell6;
cell6[0] = S1;
cell6[1] = (unsigned int)cell7;
cell7[0] = S1;
cell7[1] = (unsigned int)cell8;
cell8[0] = S0;
cell8[1] = (unsigned int)cell9;
printf("OK, here's the initial tape (%p)\n", tape);
for(unsigned int i = 0; i < sizeof(tape) / sizeof(tape[0]); ++i)
printf("%d ", ((((unsigned int*)tape[i])[0] & 4) == 0 ? 0 : 1));
printf("\n");
// We use a register T to hold the current transition to be tested, and a
// register S to hold a cell containing the current symbol.
unsigned int T, S;
// The register L holds the list representing the part of the tape to the left of the
// current position, and the register R holds the list representing the part to the right.
unsigned int R, L;
printf("N=%#.8x, @S=%#.8x, @L=%#.8x, @R=%#.8x\n", N, &S, &L, &R);
printf("Q0=%#.8x, Q0_=%#.8x, Q0__=%#.8x\n", Q0, Q0_, Q0__);
printf("Q1=%#.8x, Q1_=%#.8x, Q1__=%#.8x\n", Q1, Q1_, Q1__);
printf("Q2=%#.8x, Q2_=%#.8x, Q2__=%#.8x\n", Q2, Q2_, Q2__);
printf("Q3=%#.8x, Q3_=%#.8x\n", Q3, Q3_);
printf("Q4=%#.8x, Q4_=%#.8x, Q4__=%#.8x\n", Q4, Q4_, Q4__);
printf("c1=%#.8x, c2=%#.8x, c3=%#.8x, c4=%#.8x\n", cell1, cell2, cell3, cell4);
printf("c5=%#.8x, c6=%#.8x, c7=%#.8x, c8=%#.8x\n", cell5, cell6, cell7, cell8);
printf("c9=%#.8x\n", cell9);
// At program startup, T holds the address Q0
// Q0 is the program startup state
T = (unsigned int)Q0;
// S holds the address T1
S = (unsigned int)cell2;
// The register L holds the address N, and R holds the address T2
L = (unsigned int)cell1;
R = (unsigned int)cell3;
///// LAUNCH THE MACHINE
__try
{
here_we_go:
__asm
{
pushad
; First, we check whether the current transition should fire, by
; comparing S and the symbol on the current transition T.
mov eax, T
mov eax, [eax] ;; get transition
mov eax, [eax] ;; get trigger symbol, eax got a real symbol now
mov ebx, S
mov ebx, [ebx] ;; get current symbol, ebx got the current symbol
mov dword ptr [ebx], 0 ;; we write @ ebx 0
mov dword ptr [eax], 1 ;; we write @ eax 1
mov ecx, [ebx] ;; ecx is 1 if the transition matches, and 0 otherwise.
; After this sequence, the register M (ecx) is 1 if the transition matches,
; and 0 otherwise. Next, we update S: if the transition matches, we
; use the transition’s new symbol, otherwise we leave it unchanged
mov eax, T
mov eax, [eax] ;; get transition
mov eax, [eax + 4] ;; skip trigger symbol
mov eax, [eax] ;; get new symbol
mov ebx, S
mov ebx, [ebx] ;; get current symbol
; another trick to select between eax & ebx based on the previous comparaison
mov edx, N
mov [edx], ebx
mov [edx + 4], eax
; we will load in edx, either ebx (current symbol) if ecx = 0 (current symbol != triggering symbol), or
; eax (new symbol) if ecx = 1 (current symbol == triggering symbol)
mov edx, [edx + ecx * 4]
mov ebx, S
mov [ebx], edx ;; write the new symbol
; This updates S if the transition matches. Next, if the transition
; matches, we need to advance the tape in the appropriate direction.
; We do this in two stages. First, we push the cell S to one of the
; tape stacks, and then we pop a new S from the other tape stack.
; If the transition does not match, we push and pop S from the same
; tape stack, which has no effect. To determine whether the transition
; moves left or right, we use the following sequence:
mov eax, T
mov eax, [eax] ;; get transition
mov eax, [eax + 4] ;; skip the trigger symbol
mov eax, [eax + 4] ;; skip the new symbol
mov eax, [eax] ;; load direction
; After this, the register D holds the direction of tape movement:
; 0 for left, and 1 for right. If we are to move left, then the cell S
; must be added to the tape stack R, and vice versa. Adding the cell
; to a tape stack is done by first writing the tape stack’s current top
; to [S+1], and then modifying the tape stack register to point at S.
mov ebx, N ;; select new value for [S+1]
mov edx, R
mov [ebx], edx
mov edx, L
mov [ebx + 4], edx
mov edi, [ebx + eax * 4]
mov edx, S
mov [edx + 4], edi
mov edx, L
mov [ebx], edx ;; select new value for L
mov edx, S
mov [ebx + 4], edx
mov edx, [ebx + eax * 4]
mov L, edx
mov edx, S
mov [ebx], edx ;; select new value for R
mov edx, R
mov [ebx + 4], edx
mov edx, [ebx + eax * 4]
mov R, edx
; We must ensure that no movement of the tape happens if the
; transition does not match (that is, if M = 0). To this end, we flip the
; value of D if the transition does not match, so that we pop the cell
; we just pushed.
mov ebx, N
mov dword ptr [ebx], 1 ;; set X = not D
mov dword ptr [ebx + 4], 0
mov edi, [ebx + eax * 4]
mov [ebx], edi ;; select between D and X
mov [ebx + 4], eax
mov eax, [ebx + ecx * 4]
; Next, we pop a cell from a direction indicated by D: if D = 0,
; we pop a cell from L, and if D = 1 we pop one from R.
mov ebx, N
mov edi, L
mov [ebx], edi ;; select new value of S
mov edi, R
mov [ebx + 4], edi
mov edx, [ebx + eax * 4]
mov S, edx
mov edx, [edx + 4] ;; get new start of L or R
mov [ebx], edx ;; select new value of L
; WARNING: I believe this is a typo in the reference paper ; instead of 'mov edi, R', it is 'mov edi, L' (confirmed with the author)
mov edi, L
mov [ebx + 4], edi
mov edi, [ebx + eax * 4]
mov L, edi
mov edi, R
mov [ebx], edi ;; select new value for R
mov [ebx + 4], edx
mov edi, [ebx + eax * 4]
mov R, edi
; So, if the current transition matches, this code writes a symbol to
; the tape and advances in the appropriate direction. If the transition
; doesn’t match, this code has no effect.
; All that remains is to find the next transition to consider. If the
; current transition matches, then we should look at the next state’s
; list of transitions. Otherwise, we continue to the next transition in
; the current state.
mov eax, T
mov eax, [eax + 4] ;; get next transition of this state
mov ebx, T
mov ebx, [ebx] ;; get current transition
mov ebx, [ebx + 4] ;; skip trigger symbol
mov ebx, [ebx + 4] ;; skip new symbol
mov ebx, [ebx + 4] ;; skip direction
mov ebx, [ebx] ;; load transition list of the next state
mov edx, N
mov [edx], eax
mov [edx + 4], ebx
mov edi, [edx + ecx * 4]
mov T, edi
; This finds the next transition our Turing machine should consider.
; If T has the value N, then this means there are no more transitions to consider:
; either we got to the end of a state’s list of transitions with no matches,
; or we just transitioned to a state that has no outgoing transitions.
; Either way, the machine should halt in this case.
; First, we check whether this is the case by setting the register H to 1 if T is N
mov eax, T
mov eax, [eax]
mov ebx, N
mov dword ptr [ebx], 0
mov ecx, T
mov dword ptr [ecx], 1
mov edi, [ebx]
mov [ecx], eax
; If H is 1, we must halt the machine. We do so by reading from
; the invalid memory address 0:
; WARNING: there is an error in the paper ; the code and the comments
; are doing exactly the contrary. (got a confirmation from the author)
; That is why I did:
; mov dword ptr [edx], edx
; mov dword ptr [edx + 4], 0
; Instead of (as stated in the paper):
; mov dword ptr [edx], 0
; mov dword ptr [edx + 4], edx
mov edx, N
mov dword ptr [edx], edx
mov dword ptr [edx + 4], 0
mov eax, [edx + edi * 4]
mov eax, [eax]
popad
}
printf("-----------DEBUG--------------------\n");
printf("OK, done, inspecting the tape now..\n");
for(unsigned int i = 0; i < sizeof(tape) / sizeof(tape[0]); ++i)
{
unsigned int value = (((unsigned int*)tape[i])[0]) == S0 ? 0 : 1;
if(S == tape[i])
printf("[%d] ", value);
else
printf("%d ", value);
}
printf("\n");
for(unsigned int i = 0; i < sizeof(tape) / sizeof(tape[0]); ++i)
printf("[%#.8x] ", ((unsigned int*)tape[i])[1]);
printf("\n");
printf("S=%#.8x, L=%#.8x, R=%#.8x, T=%#.8x\n", S, L, R, T);
printf("------------------------------------\n");
goto here_we_go;
}
__except(EXCEPTION_EXECUTE_HANDLER)
{}
printf("OK, machine done, inspecting the tape now..\n");
for(unsigned int i = 0; i < sizeof(tape) / sizeof(tape[0]); ++i)
{
unsigned int value = (((unsigned int*)tape[i])[0]) == S0 ? 0 : 1;
if(S == tape[i])
printf("[%d] ", value);
else
printf("%d ", value);
}
printf("\n");
free(mem);
}
int main()
{
turing_machine_simple();
return 0;
}
/*
If like me, you didn't really understand how the tape was working, here is
an email Stephen sent me, it explains perfectly:
"""
Here's an example. The tape contains cells c0, c1, c2, c3 and c4. The
current cell is c2. I'll write c2 -> c3 to mean that c2's next pointer
(that is, [c2 + 1]) points to c3. Currently, we have:
S = c2
L = c1
c1 -> c0
R = c3
c3 -> c4
Suppose we're moving right. We want to end up with c2 -> c1 -> c0, L =
c2, S = c3, R = c4. First, we need to hook up the c2 -> c1 link, by
doing "mov [S+1], c1" (where c1 is the old value of L). Next, we need
to update L (by making it point to c2, the old value of S) and update
R (by leaving it unchanged). This completes the PUSH phase, and we end
up with:
S = c2
L = c2
c2 -> c1 -> c0
R = c3
c3 -> c4
We've pushed the S cell onto the left tape, but haven't popped a new
cell from the right yet. So, we set S to be the old value of R (c3),
and set X to be the new value of R by "mov X, [S+1]". Then, we update
R (by setting it to c3), and update L (by leaving it unchanged). We're
left with:
S = c3
L = c2
c2 -> c1 -> c0
R = c4
"""
*/