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Consider the following context-free grammar:

S -> SS + | SS * | a

1. Explain how to use the grammar generates the string aa + a *

2. Construct a test for this string parse tree

3. What language is generated by the grammar? Prove

1. S -> S S * -> S S + S * -> a S + S * -> aa + S * -> aa + a *

2.

3. The operation is as a number, L = {expression supports the addition and multiplication representation suffix}

What language is generated below each grammar? Every answer your proof

1. S -> 0 S 1 | 0 1

2. S -> + SS | - SS | a

3. S -> S (S) S | ε

4. S -> a S b S | b S a S | ε

5. S -> a | S + S | SS | S * | (S)

1. L = {0 1 | N> = 1}

2. L = {prefix expressions of support for the addition and subtraction of expressions}

3. L = {matching parentheses and nested parentheses arbitrary string arrangement, including ε}

4. L = {a and b are the same number of symbol string consisting include ε}

5. ?

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2.2 Exercise

2.2.1

Answer

2.2.2

Answer

N N

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Previous question in which grammar is ambiguous

1. No

2. No

3. Have

4. Have

5. Have

Build unambiguous context for each language grammar below. Prove your grammar is correct.

1. Arithmetic expression represented by the suffix method

2.2.3

Answer

2.2.4

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2. Comma-separated list of identifiers left associative (identifier id said, the same below)

3. List of identifiers separated by commas right combination

4. There integer identifier, four binary operator +, -, *, / arithmetic expression composed

5. ! Increase Monocular Monocular + and the operator of a problem - consisting of arithmetic expressions

1. E -> EE op | num

2. list -> list, id | id

3. list -> id, list | id

4. expr -> expr + term | expr - term | term

term -> term * factor | term / factor | factor

factor -> id | num | (expr)

5. Note: The highest priority subtraction of monocular

expr -> expr + term | expr - term | term

term -> term * unary | term / unary | unary

unary -> + factor | - factor

factor -> id | num | (expr)

1. Proof: The following grammar generating all the values of the binary string can be divisible by 3. (Hint: parse tree node tree

using mathematical induction)

num -> 11 | 1001 | num 0 | num num

2. Are the above grammar can generate all binary strings divisible by 3?

1. Proof

Comply with the leftmost bit binary string grammar must be composed of any number of 11,1001 and 0 is not zero sequence

The decimal sequence is:

sum

Sigma = (2 + 2 ) * 2 + sigma (2 + 2 ) * 2

Sigma = 3 2 * + sigma 2 * 9

Is clearly divisible by 3

2. Is not. 10101 binary string, a value of 21 is divisible by 3, but can not deduce from the grammar.

Note: There are more general law permits it?

Build a context-free grammar is the Roman numeral

Note: The grammar does not consider Roman numerals on the underlined, it can only produce a number less than 4k

Answer

2.2.5

Answer

N1 0 N

m3 0 m

NN

mm

2.2.6

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Rules Reference Wikipedia: Roman numerals

According to the rules described in the wiki can be found digit representation can be divided into four categories:

I, II, III | IV | V, VI, V II, V III | IX

Ie production:

digit -> smallDigit | IV | V smallDigit | IX

smallDigit -> I | II | III | ε

Other digital representation similar.

You can also find: Roman numerals and Arabic numerals have relatively simple digital correspondence.

For example:

XII - X, II - 10 + 2 - 12

CXCIX - C, XC, IX - 100 + 90 + 9 - 199

MDCCCLXXX - M, DCCC, LXXX - 1000 + 800 + 80 - 1880

According to these two laws deduced production:

romanNum -> thousand hundred ten digit

thousand -> M | MM | MMM | ε

hundred -> smallHundred | CD | D smallHundred | CM

smallHundred -> C | CC | CCC | ε

ten -> smallTen | XL | L smallTen | XC

smallTen -> X | XX | XXX | ε

digit -> smallDigit | IV | V smallDigit | IX

smallDigit -> I | II | III | ε

Answer

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