Wednesday, November 8, 2017

Transcription

1 Wednesday, November 8, 207 Topics for today Grammars and Languages (hapter 7) Finite State Machines Semantic actions ode generation - Overview Finite State Machines (see 7.2) If a language is regular (Type 3) we can answer the question Is string X in the language? by means of an appropriate Finite State Machine (FSM). For every regular language there is an FSM that can recognize strings in the language. (The reverse is also true - for every FSM, we can represent the language it recognizes by means of a regular expression). Visually an FSM might be a graph with the states as nodes and transitions as arcs. One of the states is identified as the start state. The machine is input driven, the combination of current state and input determines what the next state is 2. Only strings that are in the language will cause the FSM to end up in one of the "final" states. In our graphs, the start state is labeled with -, final states are labeled with +. In Warford s diagrams he has an arrow pointing to the start state and the final states have a double circle. FSM Example : recognizer for label letter or letter colon As a transition table we might have something like the following where the entries tell us the next state given the current state (row) and input (column). This is the case for a deterministic FSM. In a non-deterministic FSM there may be multiple transitions from a particular state for a given input. 2 Algorithms exist that will minimize the number of states in an FSM. This is useful if the FSM is to be implemented in hardware. omp 62 Notes Page of 9 November 8, 207

2 Input urrent Letter Digit olon Other State 2 Error error error error If we start in state then any string that takes us to state 3 is a label. If input is not a label then the FSM will crash in some error state. The following general algorithm will determine if the input is in the language defined by a transition table. current_state = Start while ( current_state!= error && more_input ) { get(input) current_state = transition_table[current_state][typeof(input)] } if (ismember(finish_states, current_state) output( Yes ) else output( No ) FSM Example 2: recognizer for train On a certain railroad, all trains begin with one or more locomotives followed by zero or more wagons followed by a caboose. Using W for wagons, L for locomotives, and for cabooses, the following regular expression defines valid trains. The following FSM recognizes valid trains. LL*W* L 3 W - L 2 W + 4 omp 62 Notes Page 2 of 9 November 8, 207

3 As a table Input urrent State L W Other 2 error error error error 3 error 3 4 error Semantic actions Semantic actions are actions associated with particular transitions from one state to another. They are useful for performing such operations as forming input characters into a string or creating a variable containing the value of an input number. Example We can attach semantic actions to our label checker that build up the string as it is read characterby-character then, when we read the colon, enters the string into the symbol table. Here are the actions Action When done Details A On reading the first character String S = character B On reading a subsequent non-colon character Append character to S On reading a colon Enter S into symbol table Here is the modified FSM with transitions labeled input/action Letter/B or /B - letter /A colon/ omp 62 Notes Page 3 of 9 November 8, 207

4 FSM Example 3: recognizer for real numbers Assume that in a real number we must have at least one before the decimal point or at least one after the decimal point. A number can begin with an optional sign character. Thus, Valid Invalid Regular expression: [sign] ( *.* *. * ) [x] = none or one x Here is a FSM that recognizes real numbers defined in this way. 2 dec. pt - sign dec. pt dec. pt 4 We can attach semantic actions to our real number parser to create a variable containing the value of the number. There will be an action performed before input is read and another action performed after input has finished and actions associated with state-to-state transitions. The actions will use global variables. Here are the actions: omp 62 Notes Page 4 of 9 November 8, 207

5 Action When done Details Initially dec_places = 0, sign =, N = 0 A On reading a decimal point dec_places = 0 B On reading a sign character if (input = '-') sign = - On reading a N = N * 0 + input; dec_places++ At the end N *= sign, N = N / (0 dec_places ) So our augmented FSM with transitions now depicting input/action is 2 / / / / dec. pt/a - sign/b dec. pt/a dec. pt/a 4 / Here is a trace of what happens when input is the 7-character string Input Sign N Dec_places <initially> <end> omp 62 Notes Page 5 of 9 November 8, 207

6 ode generation (see section 7.4) A compiler, having verified that the input string is legal in the particular programming language, will generate appropriate assembly language. Warford s example in section 7.4 is not very general; we describe and solve a more general problem: Translate an arithmetic expression into Pep/9 assembly language Like a compiler, we can produce the assembly code in two phases: () Analysis: verify input and build an appropriate data structure. (2) Synthesis: generate code from the data structure Example - simple assignment statements such as X = A + ( B + ) * ( D + E ) Analysis: A binary tree representing the expression is constructed using semantic actions as the expression is read symbol by symbol. If input were the expression above we get this tree that correctly represents the priorities of the arithmetic operations. = X + A * + + B D E Synthesis: We traverse the tree in a systematic way to generate assembly code. For example, from the tree above, the sequence of Pep/9 assembly language begins subsp 2,i ldwa B,d stwa 0,s subsp 2,i ldwa,d etc We will lead up to the Analysis and Synthesis algorithms involved by first looking at three simpler ones. omp 62 Notes Page 6 of 9 November 8, 207

7 Algorithm. Evaluation of a postfix expression Algorithm 2. onversion of an infix expression to postfix form Algorithm 3. Evaluation of an infix expression (algorithms and 2 combined). Then we can look at the Analysis algorithm Algorithm 4: Building a tree from an infix expression And finally at the Synthesis algorithm Algorithm 5: Generating assembly code We know that the language of arithmetic expressions is not Type 3 so a simple finite state machine will not be sufficient to process it. Our algorithm will use stacks. Algorithm will use a stack of operands (numbers) Algorithm 2 will use a stack of operators Algorithm 3, because it combines Algorithms and 2 will use two stacks one of operands and one of operators. We will keep our algorithm as simple as possible by limiting our identifiers to single letter and constants to single s. Reading We are looking at an alternative to Warford s section 7.4 omp 62 Notes Page 7 of 9 November 8, 207

8 Review Questions. For each of the following regular expressions, draw a FSM that recognizes the same language. (a) a (b c)* d (b) a (bc)* d 2. Derive a regular expression for the language recognized by the following FSM A D E G - A, + F B I H 3. onsider the real number recognizer. For each of the following invalid inputs identify as appropriate the state where the FSM crashes or the non-final state it is in when the input ends. (a) (b) 624 (c) (d) The notes show the first few instructions generated from the tree representing X = A + ( B + ) * ( D + E ) What are likely to be the final three instructions? 5. Suppose we remove the parentheses from our example assignment resulting in X = A + B + * D + E Draw a tree that represents this revised statement omp 62 Notes Page 8 of 9 November 8, 207

9 Review Answers. (a) A D - + (b) B A D - + B 2. (AD) (B) E* F (G (HI) )* 3. (a) 5 (b) 2 (c) 3 (d) 5 4. ldwa 0,s stwa X,d addsp 2,i 5. = X + A + B + * E D omp 62 Notes Page 9 of 9 November 8, 207