text stringlengths 0 8.13M |
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paper [17]. In fact, most modern informal descriptions are just rephrasings of |
Turing’s own words. This is true also for what follows here. |
The machine consists of a memory, a read-and-write head and a processing |
unit. |
Thememoryisatapewhichisdividedintodistinctsquares,alsocalledcells. |
It is infinite to the left and to the right6. The memory tape is used for giving |
inputtothe machine,forstoringintermediatedataduringcomputationandfor |
writing output. |
The read-and-write head can move along the tape. It can read symbols |
written on the tape (this is called scanning) and it can write symbols on the |
tape. |
Thesymbolscanbeanysymbols,buttheymustcomefromafinitealphabet |
Γ= S ,S ,...,S . |
0 1 n |
{ } |
The machinehasafinite setofelementaryoperationsthatitcanperformat |
each step in the computation. These are |
move one step to the right |
• |
move one step to the left |
• |
write a symbol |
• |
erase a symbol |
• |
halt |
• |
This can symbolically be written as a set of operations |
O= moveright,moveleft,write(S ),erase,halt |
i |
{ } |
Note that reading a symbol need not be considered to be an operation. In |
fact, the machine always reads the symbol written on the scanned square. In |
some formulations, the operation erase is replaced by writing a special symbol |
called a blank, i.e. by the operation write(blank). |
The halt operation can be implemented in different ways. |
The machine is controlled by a set of instructions. This is the program. |
In order to distinguish the instructions, the machine is considered to be in a |
set of different machine states. The states are numbered or given symbolic |
names from a set Q= q ,q ,...,q . Each instruction consists of four symbols |
0 1 n |
{ } |
(present state, scanned symbol on tape, operation, new state) or (q ,S,op,q ) |
i j |
where q Q,S Σ,op O,q Q. |
i j |
∈ ∈ ∈ ∈ |
The program is executed by a control unit. Execution starts in a special |
initial machine state q scanning the leftmost symbol on the tape. At the |
0 |
6Forpracticalpurposes,ifonewouldliketowriteacomputerprogramemulatingaTuring |
machine,itmightbeeasiertoconsideraone-waytapewithastartsquaretoleftandinfinite |
totheright. |
21 |
beginningofthecomputation,allbutafinite numberoftapesquaresareblank. |
This is true throughout the computation. At each step of the execution, the |
control unit checks through the list of instructions to find an instruction that |
matches the present state of the machine and the scanned symbol. Each cycle |
of the execution therefore consists of the following actions: |
get present state q |
present |
• |
get scanned symbol α |
scanned |
• |
find matching instruction (q ,α ,op,q ) |
present scanned new |
• |
execute the instruction op |
• |
change to the new state q as given in the matching instruction |
new |
• |
Some comments |
Infinite memoryinthe formofaninfinite tape is ofcourseimpossible inreality. |
But this is not a problem. At any stage of the computation, only a finite set of |
squares is needed. Should the machine ever run out of tape, a finite amount of |
newtapecanalwaysbeaddedtotherightortotheleft. Thenumberofsquares |
on the tape thus need not be actually infinite, only potentially infinite. |
As regards the symbols, the simplest choice is 0,1 where ’0’ serves the |
{ } |
purpose of a blank. Numbers are coded as strings of ’1’ separated by ’0’. The |
number 0 itself must be coded as ’1’ in order to distinguish it from a blank, |
and consequently 1 is coded as ”11” and so on. If one wants to use a more |
efficientbinarycoding,onecanuseanalphabetconsistingof’0’,’1’andablank |
separator ’#’. |
In some formulations of Turing machines, the operations of writing and |
movingarecombinedintoasingleoperation. Inthatcaseaninstructionconsists |
of five symbols q S S q M, where M denotes a move. |
i k l j |
Furthermore,a specialhalt instruction is not needed. The machine stops or |
halts when the control unit cannot find any matching instruction. In practice, |
though, it is convenient to include an explicit halt instruction. In fact, when |
discussing decision problems in terms of Turing machines, it is natural to have |
two halting states, for example named by yes and no. |
The names of the states are arbitrary, they can be named in any way that |
serves the purpose of clarity. |
In the next section, a formal definition of a Turing machine is given. It |
does not entirely conform to the informal description given above. The reader |
unfamiliar with Turing machines might benefit from comparing the details. |
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