session 4: atomic language. take turing machine look back to 1 st session’s work on hardware, risc...
DESCRIPTION
S1 ? ?TRANSCRIPT
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Session 4: Atomic Language.
Take Turing Machine
Look back to 1st session’s work on hardware, RISC and CISC register-based machines.
Also look forward to next week’s work on Software, computer programming languages.
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S1
?
S1
?
S2
S1
S2
Sn
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S1
?
S1
?
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ax
eax
rax
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0a
q
s1a2a 0b 1b 2b
m n
4 2 1 1 2 4
(s,m,n,q) (s*,m*,n*,q*)
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4 2 1 1 2 4
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0a
q
s1a2a 0b 1b 2b
m n
4 2 1 1 2 4
(s,m,n,q) (s’,m’,n’,q’)
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sm n
8 4 2 1 1 2 4 8
0 1 1 0 1 1 1 0
(s,m,n) = ( , , )
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s*m* n*
8 4 2 1 1 2 4 8
0 1 1 0 1 1 1 0
(s*,m*,n*) = ( , , )
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sm n
8 4 2 1 1 2 4 8
0 1 1 1 0 1 1 0
(s,m,n) = ( , , )
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s*m* n*
8 4 2 1 1 2 4 8
0 1 1 1 0 1 1 0
(s*,m*,n*) = ( , , )
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* ( / 2)* 2* / 2
s rem nm s mn n
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mov reg,0 Put 0 into register “reg”
inc reg Add 1 to the contents of register “reg”
decjmpreg reg,lab If register “reg” is zero, jump to the label “lab” else subtract 1 from “reg” and do the next instruction
jmp lab Jump straight to the label “lab”
hlt Halt
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inc reg Add 1 to the contents of register “reg”
decjmpreg reg,lab If register “reg” is zero, jump to the label “lab” else subtract 1 from “reg” and do the next instruction
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assume ecx is zeroinc ecxinc ecxinc ecx
assume ecx is zeroL1: …
…decjmpreg ecx,L1
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mov ecx,3mov eax,0mov ebx,0
L3: decjmpreg ecx,L4inc ebxjmp L3
L4: nop
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov eax,0inc eaxhlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov ecx,3 decjmpreg ecx,L1L1: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov ecx,3L2: decjmpreg ecx,L1 jmp L2L1: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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is S = q?
read line
inc line pointer
end ofTable?
end (error)
is “read” = s? inc line pointer
move left/right
write in cell
Current state
read write moveL / R
New state
write in cell end ofTable?
end (error)
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main PROC
mov ecx,N
mov eax,0
l1: mov ebx,0
decjmpz l2
inc ebx
decjmpz l2
inc eax
jmp l1
l2: hlt
main ENDP
ecx ebx eax
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m
s
n
z
ALU
sm n
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m
s
n
z
ALU z = 2m
* ( / 2)* 2* / 2
s rem nm s mn n
z = z + s
m* = z
s* = rem(n/2)
n = z
z = n/2
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mov ecx,5mov eax,0
L3: mov ebx,0decjmpreg ecx,L4inc ebxdecjmpreg ecx,L4inc eaxjmp L3
L4: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov ecx,5mov eax,0
L3: mov ebx,0decjmpreg ecx,L4inc ebxdecjmpreg ecx,L4inc eaxjmp L3
L4: hlt
eax
ebx
ecx
edx
5
0
0
4
1
3
1
eax
ebx
ecx
edx
0
2
1
1
2
eax
ebx
ecx
edx
0
0
1
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mov ecx,3mov eax,0
L3: decjmpreg ecx,L4inc eaxinc eaxjmp L3
L4: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov eax,3mov edx,0
L3: decjmpreg eax,L4inc edxinc edxjmp L3
L4: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov ecx,3L3: mov eax,0
decjmpreg ecx,L4inc eaxjmp L3
L4: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov reg,0inc regdecjmpz Ljmp L
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mov ecx,3mov ebx,0
L3: decjmpreg ecx,L4inc ebxinc ebxjmp L3
L4: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov ecx,3mov ebx,0
L3: decjmpreg ecx,L4inc ebxinc ebxjmp L3
L4: hlt
eax
ebx
ecx
edx
3
0
2
1
2
eax
ebx
ecx
edx
1
3
4
eax
ebx
ecx
edx
0
5
6
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……
L1: decjmpreg ecx,L2jmp L1
L2 …
eax
ebx
ecx
edx
3
2
eax
ebx
ecx
edx
1
eax
ebx
ecx
edx
0
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L2: decjmpreg eax,L1inc edxinc edxjmp L2
L1: decjmpreg ebx,L3inc edx
L3: decjmpreg edx,L4inc eaxjmp L3
L4: mov ebx,0decjmpreg ecx,L6inc ebxdecjmpreg ecx,L6inc edxjmp L4
L6: decjmpreg edx,L7inc ecxjmp L6
L7: hlt
eax holds m ebx holds s ecx holds n edx holds z
If s is 1 z = z + 1
z = 2m
m* = z
s* = rem(n/2) z = n/2 whole
n* = z
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rem whole
1 2 4 8 1 2 4 8
8 4 2 1 8 4 2 1
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8 4 2 1 8 4 2 1
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decjmpreg ebx,L3inc edx
L3:
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eax holds m ebx holds s ecx holds n edx holds z
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sm n
8 4 2 1 1 2 4 8
0 1 1 0 1 1 1 0
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sm n
8 4 2 1 1 2 4 8
0 1 1 1 0 0 0 11
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sm n
8 4 2 1 1 2 4 8
0 1 1 1 0 0 0 11
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; get current state
; read current symbol s. (Read ebx)
; get new state using value of s (in ebx)
; write new symbol. (Write to ebx)
; get the movement direction
; execute the movement code (left or right)
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; assume we start in state 0
; write the starting symbol into ebx
; write the code for move right or move left
; write the new symbol into ebx
; jump to the new state code
; repeat lines 2 to 4
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The movement code (left or right) depending on the state
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1
2
x
y
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Head move
right code
Head move
left code
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read the symbol sstore sif(s == 0), set s = 0 and jmp to move-right codeif(s == 1), set s = 0 and jmp to move-right code
use the stored s
if(s == 0), jump to the even state code
if(s == 1), jump to the odd state code
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mov eax,3inc eaxinc ebx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov ecx,2L1: decjmpreg ecx,L2
jmp L1L2: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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mov eax,0mov ecx,2
L1: decjmpreg ecx,L2inc eaxjmp L1
L2: hlt
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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read the symbol sstore sif(s == 0), set s = 0 and jmp to move-right codeif(s == 1), set s = 0 and jmp to move-right code
use the stored s
if(s == 0), jump to the even state code
if(s == 1), jump to the odd state code
Code for read-write
head moveleft
Code for read-write
head moveright
Current State S
Symbol read s
Symbol to write s*
Direction to move d
New state S*
Even 0 0 R Even
Even 1 0 R Odd
Even @ 0 N Halt
StateEven
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0 1 1 1
8 4 2 1 8 4 2 1
0 1 1 1
8 4 2 1 8 4 2 1
1 11 1 1 0
1 2 4 8 1 2 4 8
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eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
eax
ebx
ecx
edx
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Turing MachineTable
Tape t
R/W Head