1. Theory of computing, part 3

2. 1Introduction 2Theoretical background Biochemistry/molecular biology 3Theoretical background computer science 4History of the field 5Splicing systems 6P systems 7Hairpins 8Detection techniques 9Micro technology introduction 10Microchips and fluidics 11Self assembly 12Regulatory networks 13Molecular motors 14DNA nanowires 15Protein computers 16DNA computing - summery 17Presentation of essay and discussion Course outline

3. Finite automata

4.  Introduction  Deterministic finite automata (DFA’s)  Non-deterministic finite automata (NFA’s)  NFA’s to DFA’s  Simplifying DFA’s  Regular expressions  finite automata Outline

5. Consider the control system for a one-way swinging door: There are two states: Open and Closed It has two inputs, person detected at position A and person detected at position B If the door is closed, it should open only if a person is detected at A but not B Door should close only if no one is detected AB Automatic one way door

6. OpenClosed A, no B No A or B A and B A, no B B, no A A and B B, no A No A or B Control schematic

7.  A finite automaton is usually represented like this as a directed graph  Two parts of a directed graph: The states (also called nodes or vertices) The edges with arrows which represent the allowed transitions  One state is usually picked out as the starting point  For so-called ‘accepting automata,’ some states are chosen to be final states Finite automaton

8.  The input data are represented by a string over some alphabet and it determines how the machine progresses from state to state.  Beginning in the start state, the characters of the input string cause the machine to change from one state to another.  Accepting automata give only yes or no answers, depending on whether they end up in a ‘final state.’ Strings which end in a final state are accepted by the automaton. Strings and automata

9. The labeled graph in the figure above represents a FA over the alphabet Σ = {a, b} with start state 0 and final state 3. Final states are denoted by a double circle. Example

10.  The previous graph was an example of a deterministic finite automaton – every node had two edges (a and b) coming out  A DFA over a finite alphabet Σ is a finite directed graph with the property that each node emits one labeled edge for each distinct element of Σ. Deterministic finite automata (DFA’s)

11.  A DFA accepts a string w in Σ* if there is a path from the start state to some final state such that w is the concatenation of the labels on the edges of the path.  Otherwise, the DFA rejects w.  The set of all strings accepted by a DFA M is called the language of M and is denoted by L(M) More formally

12.  Construct a DFA to recognize the regular languages represented by the regular expression (a + b)* over alphabet Σ = {a, b}.  This is the set {a, b}* of all strings over {a, b}. This can be recognised by Example: (a+b)*

13.  Find a DFA to recognize the language represented by the regular expression a(a + b)* over the alphabet Σ = {a, b}.  This is the set of all strings in Σ* which begin with a. One possible DFA is: Example: a(a+b)*

14.  Build a DFA to recognize the regular language represented by the regular expression (a + b)*abb over the alphabet Σ = {a, b}.  The language is the set of strings that begin with anything, but must end with the string abb.  Effectively, we’re looking for strings which have a particular pattern to them Example: pattern recognition

15. The diagram below shows a DFA to recognize this language. If in state 1: the last character was a If in state 2: the last two symbols were ab If in state 3: the last three were abb Solution: (a+b)*abb

16. State transition function

17. We can also represent a DFA by a state transition function, which we'll denote by T, where any state transition of the form is represented by: T(i,a) = j To describe a full DFA we need to know:  what states there are,  which are the start and final ones,  the set of transitions between them. State transition function

18.  The class of regular languages is exactly the same as the class of languages accepted by DFAs!  Kleene (1956)  For any regular language, we can find a DFA which recognizes it! Regular languages

19.  DFA’s are very often used for pattern matching, e.g. searching for words/structures in strings  This is used often in UNIX, particularly by the grep command, which searches for combinations of strings and wildcards (*, ?)  grep stands for Global (search for) Regular Expressions Parser  DFA’s are also used to design and check simple circuits, verifying protocols, etc.  They are of use whenever significant memory is not required Applications of DFA’s

20.  DFA’s are called deterministic because following any input string, we know exactly which state its in and the path it took to get there  For NFA’s, sometimes there is more than one direction we can go with the same input character  Non-determinism can occur, because following a particular string, one could be in many possible states, or taken different paths to end at the same state! Non-deterministic finite automata

21.  A non-deterministic finite automaton (NFA) over an alphabet Σ is a finite directed graph with each node having zero or more edges,  Each edge is labelled either with a letter from Σ or with .  Multiple edges may be emitted from the same node with the same label.  Some letters may not have an edge associated with them. Strings following such paths are not recognised. NFA’s

22.  If an edge is labelled with the empty string , then we can travel the edge without consuming an input letter. Effectively we could be in either state, and so the possible paths could branch.  If there are two edges with the same label, we can take either path.  NFA’s recognise a string if any one of its many possible states following it is a final state  Otherwise, it rejects it. Non-determinism

23. DFA for a*a : Why is the top an NFA while the bottom is a DFA? NFA for a*a : NFA’s versus DFA’s

24.  Draw two NFAs to recognize the language of the regular expression ab + a*a.  This NFA has a  edge, which allows us to travel to state 2 without consuming an input letter.  The upper path corresponds to ab and the lower one to a*a Example

25. This NFA also recognizes the same language. Perhaps it's easier to see this by considering the equality ab + a*a = ab + aa* An equivalent NFA

26.  Since there may be non-determinism, we'll let the values of this function be sets of states.  For example, if there are no edges from state k labelled with a, we'll write T(k, a) =   If there are three edges from state k, all labelled with a, going to states i, j and k, we'll write T(k, a) = {i, j, k} NFA transition functions

27.  All digital computers are deterministic; quantum computers may be another story!  The usual mechanism for deterministic computers is to try one particular path and to backtrack to the last decision point if that path proves poor.  Parallel computers make non-determinism almost realizable. We can let each process make a random choice at each branch point, thereby exploring many possible trees. Comments on non-determinism

28.  The class of regular languages is exactly the same as the class of languages accepted by NFAs!  Rabin and Scott (1959)  Just like for DFA’s!  Every NFA has an equivalent DFA which recognises the same language. Some facts

29.  We prove the equivalence of NFA’s and DFA’s by showing how, for any NFA, to construct a DFA which recognises the same language  Generally the DFA will have more possible states than the NFA. If the NFA has n states, then the DFA could have as many as 2 n states!  Example: NFA has three states {A}, {B}, {C} the DFA could have eight: {  }, {A}, {B}, {C}, {A, B}, {A, C}, {B, C}, {A, B, C}  These correspond to the possible states the NFA could be in after any string From NFA’s to DFA’s

30.  Begin in the NFA start state, which could be a multiple state if its connected to any by   Determine the set of possible NFA states you could be in after receiving each character. Each set is a new DFA state, and is connected to the start by that character.  Repeat for each new DFA state, exploring the possible results for each character until the system is closed  DFA final states are any that contain a NFA final state DFA construction

31. The start state is A, but following an a you could be in A or B; following a b you could only be in state A AB C a b a,b AA,B a b b A,C a b a NFA DFA Example (a+b)*ab

32.  Regular expressions represent the regular languages.  DFA’s recognize the regular languages.  NFA’s also recognize the regular languages. Summary

33.  So far, we’ve introduced two kinds of automata: deterministic and non-deterministic.  We’ve shown that we can find a DFA to recognise anything language that a given NFA recognises.  We’ve asserted that both DFA’s and NFA’s recognise the regular languages, which themselves are represented by regular expressions.  We prove this by construction, by showing how any regular expression can be made into a NFA and vice versa. Finite automata

34.  Given a regular expression, we can find an automata which recognises its language.  Start the algorithm with a machine that has a start state, a single final state, and an edge labelled with the given regular expression as follows: Regular expressions  finite automata

35. 1.If an edge is labelled with , then erase the edge. 2.Transform any diagram like into the diagram Four step algoritm

36. 3. Transform any diagram like into the diagram Four step algoritm

37. 4. Transform any diagram like into the diagram Four step algoritm

38. Construct a NFA for the regular expression, a* + ab Start with Apply rule 2 a* + ab ab a* Example a*+ab

39. ab   a   a a b Apply rule 4 to a* Apply rule 3 to ab Example a*+ab

40. 1Create a new start state s, and draw a new edge labelled with  from s to the original start state. 2Create a new final state f, and draw new edges labelled with  from all the original final states to f Finite automata  regular expressions

41. 3For each pair of states i and j that have more than one edge from i to j, replace all the edges from i to j by a single edge labelled with the regular expression formed by the sum of the labels on each of the edges from i to j. 4Construct a sequence of new machines by eliminating one state at a time until the only states remaining are s and the f. Finite automata  regular expressions

42. As each state is eliminated, a new machine is constructed from the previous machine as follows:  Let old(i,j) denote the label on edge  i,j  of the current machine. If no edge exists, label it .  Assume that we wish to eliminate state k. For each pair of edges  i,k  (incoming edge) and  k,j  (outgoing edge) we create a new edge label new(i, j) Eliminating states

43.  The label of this new edge is given by: new(i,j) = old(i,j) + old(i, k) old(k, k)* old(k,j)  All other edges, not involving state k, remain the same: new(i, j) = old(i, j) After eliminating all states except s and f, we wind up with a two-state machine with the single edge  s, f  labelled with the desired regular expression new(s, f) Eliminate state k

44. Initial DFA Steps 1 and 2 Add start and final states Example

45. Eliminate state 2 (No path to f) Eliminate state 0 Eliminate state 1 Final regular expression Example

46.  Sometimes our constructions lead to more complicated automata than we need, having more states than are really necessary  Next, we look for ways of making DFA’s with a minimum number of states  Myhill-Nerode theorem: ‘Every regular expression has a unique* minimum state DFA’ * up to a simple renaming of the states Finding simpler automata

47. Two steps to minimizing DFA: 1Discover which, if any, pairs of states are indistinguishable. Two states, s and t, are equivalent if for all possible strings w, T(s,w) and T(t,w) are both either final or non-final. 2 Combine all equivalent states into a single state, modifying the transition functions appropriately. Finding minimum state DFA

48. States 1 and 2 are indistinguishable! Starting in either, b* is rejected and anything with a in it is accepted. a a b b a b 1 2 a,b a b Consider the DFA

49. 1.Remove all inaccessible states, where no path exists to them from start. 2.Construct a grid of pairs of states. 3.Begin by marking those pairs which are clearly distinguishable, where one is final and the other non-final. 4.Next eliminate all pairs, which on the same input, lead to a distinguishable pair of states. Repeat until you have considered all pairs. 5.The remaining pairs are indistinguishable. Part 1, finding indistinguishable pairs

50. 1.Construct a new DFA where any pairs of indistinguishable states form a single state in the new DFA. 2.The start state will be the state containing the original start state. 3.The final states will be those which contain original final states. 4.The transitions will be the full set of transitions from the original states (these should all be consistent.) Part 2, construct minimum DFA

51. a a a,b a a b bb 0 3 2 1 4 What are the distinguishable pairs of states? Clearly, {0, 4} {1, 4} {2, 4} {3, 4} are all distinguishable because 4 is final but none of the others are. b Example

52.  We eliminate these as possible indistinguishable pairs.  Next consider {0, 1}. With input a, this becomes {3, 4} which is distinguishable, so {0, 1} is as well.  Similarly, we can show {0, 2} and {0, 3} are also distinguishable, leading to the modified grid… 12341234 ? ? ? ? x x 0 1 2 3 12341234 x x ? x ? ? x x 0 1 2 3 Grid of pairs of state

53. We are left with {1, 2}given a{4, 4} given b {2, 1} {2, 3} given a{4, 4} given b {1, 2} {1, 3} given a{4, 4} given b {2, 2} These do not lead to pairs we know to be distinguishable, and are therefore indistinguishable! Remaining pairs

54.  States 1, 2, and 3 are all indistinguishable, thus the minimal DFA will have three states: {0} {1, 2, 3} {4}  Since originally T(0, a) = 3 and T(0, b) = 1, the new transitions are T(0, a) = T(0, b) = {1,2,3}  Similarly, T({1,2,3}, a) = 4 and T({1,2,3}, b) = {1,2,3}  Finally, as before, T(4, a) = 4 and T(4, b) = 4 Construct minimal DFA

55. The resulting DFA is much simpler: aa,b b 01, 2, 34 This recognises regular expressions of the form, (a + b) b* a (a + b)* This is the simplest DFA which will recognise this language! Resulting minimal DFA

56.  We now have many equivalent ways of representing regular languages: DFA’s, NFA’s, regular expressions and regular grammars.  We can also now simply(?!) move between these various representations.  We’ll see next lecture that the automata representation leads to a simple way of recognising some languages which are not regular.  We’ll also begin to consider more powerful language types and correspondingly more powerful computing models! conclusions

57. M = (Q, Σ, δ, q 0, F) Q= statesa finite set Σ= alphabeta finite set δ= transition functiona total function in Q  Σ  Q q 0 = initial/starting stateq 0  Q F= final statesF  Q Formal definition