BASIC STRUCTURE OF MACHINES

Size: px

Start display at page:

Download "BASIC STRUCTURE OF MACHINES"

Sherman Barton
6 years ago
Views:

1 BASIC STRUCTUR OF MACHINS Finite-state Automata: No output as such. Input = a a 2 a k a a 2 a 3 a j a k read-head control Control updates "internal memory" = state of the machine as it reads each input symbol a j only once. The read-head moves toright after reading a symbol. Push-down Automata: No output as such. Input = a a 2 a k a a 2 a 3 a j a k xternal memory (stack) c c 2 c n read-head control read-write head Control now also updates the "external memory". Turing machine: Output = a string, written on input-output tape. Input = a a 2 a k a a 2 a 3 a j a k read-write head control Aninput/output-tape symbol may be read/written more than once.

2 2.2 FSA FOR VN NUMBR OF ZROS IN BINARY STRINGS string: zero-count: even: Y NYYNNY (before reading symbol) Key observations for designing the algorithm: Initially, the zero-count is even (= zero); we only need to keep track of odd/even status of the zero-count. Reading does not change the odd/even status of the zero-count. Reading changes the odd/even status to the opposite. Final desired state is even. Suffices to look at each symbol only once, in the left to right order. Representing the algorithm as a state-diagram: M -div-2 : L -div-2 ={λ,,,,,,,, } State = odd/even status of zero-count Start-state indicated by ">"; final-state indicated by thick-circle Transitions indicate the input-symbol read and the state-update. ach bad string (x L div 2 )ends in a non-final state starting at the start-state; each good string (x L div 2 )ends in a final-state.

3 2.3 XRCIS. Show that there are 2 n strings of length n ( ) in L -div-2 by giving an - and onto mapping f (x) from {x L -div-2 : x =n} to{y {, }*: y =n }. 2. Give the state-diagram for an FSA M -div-3 which accepts the binary strings L -div-3 ={x {, }*: #(, x) isdivisible by 3}. (Hint: First, write down some of the strings in L -div-3.) 3. Show the state-diagram of an FSA which accepts the complement language L c -div Show the state-diagrams of all FSA s with input alphabet Σ ={, }, two states (both must be reachable from the start-state), and at least one final-state. Give an nglish descriptions of the set of strings accepted by each FSA.

4 2.4 BINARY STRINGS WITH VN #(, x): ACOMPUTABL PATTRN Pattern-description: Binary strings with even number of zeros. This is a mathematically good description; it states a precise condition (a property) that a string must satisfy. It does not say how to go about in verifying the required condition (property) for a given string x. Some possible methods: () Count #(, x), divide it by 2, and verify that the remainder is zero. (Also, check that each symbol in x is or.) (2) Find x, count #(, x), subtract it from x toget #(, x), etc. Computational-description: The finite-state automaton M -div-2. Itdescribes an algorithm to verify the required property. Where/how to start (set initial count = even, i.e., position yourself at the start-state and at the first input symbol). When to stop (at the end of input string). What to do at each step (what to read, where to find it, how to update the state). The description is finite: the finite alphabet, the finite number of states (and the transitions), the start-state, and the final-states. How does M -div-2 describe infinitely many strings in a finite way? Via the dynamics of FSA: successive applications of transitions. Longer strings give longer chains of transitions.

5 2.5 Static-part: FORMAL FINITION OF AN FSA M =(Q, q, F, Σ, δ), where Q =Afinite, non-empty set of states q =The start-state; q Q F =The set of final-state(s) Q Σ =Afinite, non-empty input alphabet δ(q i, a j )= q k,the next-state after reading the input symbol a j in state q i ; δ is called the transition function. Wesometimes write δ(q i, a j )=q k as the triplet (q i, a j, q k ). ynamic-part: For input x = a a 2 a k,apply the transitions successively as indicated below: a q a q 2 q 2 a 3 a k a k xample: M -div-2 : Q ={, }, q =, F ={}, Σ ={, }, Transitions = {(,, ), (,, ), (,, ), (,, )} Aninput string x = a a 2 a k is accepted by M if the state reached after reading/processing all of x is a final-state. Language defined by M: L(M) ={x Σ : x accepted by M}. L(M div 2 )={λ,,,,, } Question: What is the language if both states and were finalstates in M div 2?

6 2.6 A variant of M -div-2. SOM OTHR FSA FOR L -div-2 State is a duplicate of in the sense that δ(, a) =δ(, a) for a =,. does not participate in accepting any input string since it is not reachable from the start-state. (Changing δ(, a) for any of a =,does not alter the language accepted either.) To make reachable from the start-state without changing the language, we can change one or more transitions going to (from or ) togoto, making sure that this does not make itself unreachable. This can be done in three ways: Both and correspond to "odd number of s". There are countably many different FSAs for L -div-2. M -div-2 is the only one with the smallest number of states among all of them.

7 2.7 XRCIS. Consider duplicating the state in M -div-2 and making sure that the new state (and all other states) is reachable from the startstate. Carefully determine whether the new state should be a final-state since the original state is a final-state. (We should not have two start-states, however.) Show the state-diagram of your new FSA(s). 2. Modify the original M -div-2 so that λ is not accepted but all other strings in L -div-2 are accepted. (Hint: Its start-state should be similar to the state in M -div-2,but it should not be a final-state.)

8 2.8 FSA FOR BINARY STRINGS CONTAINING "" Key Steps: L has- ={,,,,,,,,,,, } Write down systematically some of the strings in the language. etermine what is to be remembered as you process each symbol in an input string in order to decide if the part processed so far is "good" or "bad". The things remembered must not grow arbitrarily large as more and more input symbols are processed. etermine the initial and the final value(s) of the thing being remembered. The final value(s) for good strings must be different from those for the bad strings. etermine how this memory (= the control memory) is updated as each input symbol is processed. For L has- : The memory has three possible values: No part of "" is seen, the first in a possible "" is seen, and the second in an "" is seen. We can represent them as "λ", "", and "". Initial value = λ; final value =. The update rules for the memory are the transitions below. M has- : λ, Question: How tomodify M has- so that it accepts the complement language L c has-?

9 2.9 XRCIS. Let f (n) =the number of strings of length n which are in L has- and g(n) =the number of strings of length n which are not in L has-. Then show that g(n) =g(n ) + g(n 2) for n 2, and f (n) =2 n g(n). 2. elete the transition from state to the state λ (labeled by the input symbol ) from each of M has- and M c has- above, and call the new automatons M and M 2. Then verify that the following is false: L(M 2 )=L(M ) c. 3. escribe in nglish the language L(M ). Then, modify M to make it completely defined, without changing its language and show the complement FSA for the modified M.

10 2. Complete efinition: COMPLTLY FIN FSA For eash state q Q and each input sysmbol a Σ,give the transition δ (q, a) for the next-state. xample. For L ={(rw) n : n }, the completely defined FSA is r w r w dead/error state w, r We often avoid writing dead/error states and transitions to/from them to simplify the state-diagram: r w

11 2. Past(q): PAST- AN FUTUR-BAS STATSCRIPTIONS The property of inputs x = a a 2 a n that brings the FSA to the state q, i.e., when do we arrive at q. Future(q): The property of inputs x = a a 2 a n that takes the FSM from q to a final state, i.e., what remains to do at q to reach afinite state. M -div-2 : L -div-2 ={λ,,,,,,, } State Past-based description Future-based description Have seen even many s Need to see even many s Have seen odd many s Need to see odd many s M has- : q λ q, q L has- ={,,,, } State Past-based description Future-based description q λ Have not seen any part of "" Need to see "" q Have only seen "", or the Need to see an immediate last two symbols seen are "" or a future "" q Have seen "" Any thing is fine Question: What is wrong with Past(q )=Have seen? Future-based state descriptions are often more useful.

12 2.2 XRCIS. Show anfsa for L bin even ={x: x {, }* represents a binary even number}. Then, give the past and future based descriptions of its states. 2. Give the past and future descriptions of the states of the FSA below. 3. Give the future based state descriptions for each of the three alternative FSA (with 3 states in each) shown earlier for M -div-2.

13 2.3 Problem: APROBLM INVOLVING COUNTING IS SOLV WITHOUT COUNTING Consider the language {x: x is a binary string with equal number of "" and ""}. xamples of Good Strings of Length 4: (y) (y) (n) (n) (n) (n) (y) (y) (y) (y) (n) (n) (n) (n) (y) (y) quivalent Property: The binary string x begins and ends with the same symbol. An FSA: A B C C

Part I: Definitions and Properties

Turing Machines Part I: Definitions and Properties Finite State Automata Deterministic Automata (DFSA) M = {Q, Σ, δ, q 0, F} -- Σ = Symbols -- Q = States -- q 0 = Initial State -- F = Accepting States