Turing Machine

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Turing Machine

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

|dw:1444363880290:dw|
so basically turing came up with this idea for a computer
Now the theorem is that, any intuitive thing you can think of doing, can be formulated into steps, or an algorithm for this "computer"

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

That shouldnt really say readstate, its not reading state, think about it like the machine knows the state its in, or some person or thing is there to change the state, so the machine is in a current state, no based on what it reads or doesnt read in the strip, you can make it change state or do other stuff
i can give u some example or you can try a fun excercise
Sure sounds fun
okay let me give u some example first, so u can play this game right
Lets say our strip is loaded with 0s everywhere to begin with |dw:1444364282679:dw|
The main thing to get from that example is how these steps are being used
this program is not technically "complete" because we left some cases out like (s2,1,....) (s3,1,....) the cases where its in this state and reads a 1 in the strip, which we know wont happen since we decided this stips starting condition was all 0
Ok I think I'm following so far.
okay now the turing theory is that, anything you can possibly think of, can be done with just what u have already
a fun problem to throw you into this mess is
suppose I give you a strip, that has a "number" somewhere in the stip, lets say there are n, 1s, side by side, and everything else is a 0, what is one way, you can write an algorithm that will always find this number
wait before that start with this,
|dw:1444364919462:dw|
since it's an infinite strip I guess I have to zig zag back and forth until I find it
yeah exactly
try to do that left most strip problem, i want to make sure u get how to write the steps
the halting state is S0 state, that marks the end of all computation
|dw:1444365057431:dw|
|dw:1444365079498:dw|
after some program, u want to leave the strip as it is, and get that pointer/ reader to the left most 1 in the strip
ill just get u started i think u need to see an example atleast
No what's that word say "will en--" got cut off
|dw:1444365168395:dw|
oh my bad okay
okay but do u get this example, u dont need to think about htis really, i want to make sure u understand rules
that first line is enough to cover everything that happens when it reads s1,1
this only works for the scneario that your reader is already somewhere in a list of 1s
Yeah I was working on it I am not clear on the rules yet but I see you put it, I'll pretend I didn't see it: (S1,1,S1,1,L) (S1,0,S0,?,?)
you have to make sure that direction is R on the 2nd step
and leave the 0 as 0 , so the strip is unchanged
Won't that make it off by one? like it'll be |dw:1444365416912:dw|
Since we moved right
the 2nd line of code basically means u were |dw:1444365428897:dw|
the 2nd line* is triggered once you are the 0 so u are fine
we should be in a call right now xD why are even wasting out time typing all this lol
Ok so let me explain this if I have it correctly: (s1,1,s1,1,L) This means this is what you do when s1 is 'called' by the third yeah true
okay so basically that line you wrote (s1,1,s1,1,L) is saying if you in state s1, and read 1 in that box, stay s1, leave box 1, and go left 1 space
as soon as it goes to the left spot, the reader will read again, and see what state u are in, and continue the process
\[(S_{2n-1},1,S_{2n},1,L)\]\[(S_{2n-1},0,S_0,0,R)\]\[(S_{2n},1,S_{2n+1},1,R)\]\[(S_{2n},0,S_0,0,L)\]
can you put 5 heads and get it done at one go avoiding backtracking ?
only 1 reader fort now :)
you can convert that state machine to turing program easily
** |dw:1444367192800:dw|
basically, the statemachine resets everytime it sees a 0
can u write out your program in that 5 step rule
|dw:1444367429426:dw|
your states correspond to above, right ?
okay i can kind of see what ua re doing
yeah, like if u are in a state, and u see a 1 or 2 u can change state or loop back to same stape and so on
those kind of drawings could be useful actually in just come up with some idea to do, before writing al gorithm
like for example if i were to use your drawings to make this left most finding 1 problem it would look like --->
|dw:1444367903673:dw|
To identify the pattern 01111 w/o backtracking : ``` (START,0,s1,0,R) (s1, 0, s1, 0, R) (s1, 1, s2, 0, R) (s2, 0, s1, 0, R) (s2, 1, s3, 0, R) (s3, 0, s1, 0, R) (s3, 1, s4, 0, R) (s4, 0, s1, 0, R) (s4, 1, END, 0, R) ```
because the turing machine has state knowledge, it never has to backtrack
okay so can you tell me which problem you are trying to solve ganeshie
to identify the pattern 01111 on a sequential line using turing machine
(S1,0,S1,0,R) (S1,1,S2,0,R) (S2,0,S3,1,L) (S2,1,S0,1,L) (S3,0,S3,0,L) (S3,1,S4,0,L) (S4,0,S1,0,R) (S4,1,S0,1,R)
but remember this is an infnite strip ganeshie, you dont knwo if ur 1s are on the left or right
How does that matter if the turing maching is keeping track of the state that is in
s1, s2, s3, s4, s5 correspond to specific states that give us the information about what state we are in finding the pattern, right ?
are you currently just going only right ganeshie
|dw:1444368331824:dw|
Yes, only forward...
you have to make sure you find your number even if its in infinite time, just gotta make sure ur program is complete, right now you program has a flaw that if the number is to the left then you would be lost
the idea is that this is an infinite strip and u are somewhere lost on this strip, and u have to find a the 1s
I am assuming we have reset thingy and the tape has a well defined start point
no reset thing xD
there is the halting state which just stops, and move the reader to the left or right before it does so, thats about it
so the tape starts somewhere in the middle and the pattern we want to find "can" be to the either left or right is it ? so there is no "start point" for the tape ?
nope
i mean your reader is somewhere we dont know where it is, we just know its somewhere
jut some infinite line
Okay the previous statemaching works just fine, I just need to add what happens when it encounters "1", one sec...
If the current value is 0, it can simply move forward, no need to backtrack, agree ? |dw:1444368962345:dw|
|dw:1444369006730:dw|
Because, we're trying to find the pattern "01111", and, if the current value is 0, we know for sure what the next state will be
you gotta find a way to do that zigzag thign in the above pic
we have to back and forth
we have to back and forth only if the current value is 1 when we start
if u start on a 1, u are pretty much done, you will just go back until you go to the left most 1 and u found the start of your number sequence
the whhole problem is 0 0000000001111111000000000 you dont know if ur in the 0s on the left or the right
...0000001111110000......
if you start at 1, you don't know what the next state will be
Start at S5 heh (S5,0,S6,0,L) (S6,0,S7,0,L) (S7,0,S1,1,L) (S6,1,S0,1,R) (S7,1,S0,1,R) (S1,0,S1,0,R) (S1,1,S2,0,R) (S2,0,S3,1,L) (S2,1,S0,1,L) (S3,0,S3,0,L) (S3,1,S4,0,L) (S4,0,S1,0,R) (S4,1,S0,1,R)
`01111` it could be the first, second, third or fourth 1
however if the head reads 0 as its starting value, you always know that the next state is s1 it moves forward and looks for 1 next, if it doesn't find 1 there, it goes back to the state s1
oh i see i think u are misunderstading what a state is
enlighten me then
a state has nothing to do with where your reader is or how your strip is right now or anything
Start at S5 heh (S5,0,S6,0,L) (S6,0,S7,0,L) (S7,0,S1,1,R) (S6,1,S0,1,R) (S7,1,S0,1,R) (S1,0,S1,0,R) (S1,1,S2,0,R) (S2,0,S3,1,L) (S2,1,S0,1,L) (S3,0,S3,0,L) (S3,1,S4,0,L) (S4,0,S1,1,R) (S4,1,S0,1,R) |dw:1444369525101:dw|
its just this like variable you are using to keep track of what command u want to do right now
are you defining own terminology here ? because thats not how i remember using it..
consider the state diagram of a simpler circuit, mod10 counter or something, how do you define a state in this ?
http://prntscr.com/8pcu8g
|dw:1444369816762:dw|
don't take me seriosuly though, i could be wrong... its been so many years...
|dw:1444369924025:dw|
|dw:1444369944055:dw|
https://ms.mcmaster.ca/~speisseg/blog/?page_id=2731
|dw:1444370662565:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question