【编译原理】龙书 第二章课后题答案
Exercises for Section 2.2
2.2.1
Consider the context-free grammar:
S -> S S + | S S * | a
- Show how the string
aa+a*can be generated by this grammar. - Construct a parse tree for this string.
- What language does this grammar generate? Justify your answer.
Answer
S->SS * ->SS + S * -> aS+ S * -> a a +S* -> a a + a *
- L = {Postfix expression consisting of digits, plus and multiple signs}
2.2.2
What language is generated by the following grammars? In each case justify your answer.
- S -> 0 S 1 | 0 1
- S -> + S S | - S S | a
- S -> S ( S ) S | ε
- S -> a S b S | b S a S | ε
- S -> a | S + S | S S | S * | ( S )
Answer
- L = {0n1n | n>=1}
- L = {Prefix expression consisting of plus and minus signs}
- L = {Matched brackets of arbitrary arrangement and nesting, includes ε}
- L = {String has the same amount of a and b, includes ε}
- L = {Regular expressions used to describe regular languages} refer to wiki
2.2.3
Which of the grammars in Exercise 2.2.2 are ambiguous?
Answer
-
No
-
No
-
Yes
-
Yes
-
Yes
2.2.4
Construct unambiguous context-free grammars for each of
the following languages. In each case show that your grammar is correct.
- Arithmetic expressions in postfix notation.
- Left-associative lists of identifiers separated by commas.
- Right-associative lists of identifiers separated by commas.
- Arithmetic expressions of integers and identifiers with the four binary operators +, -, *, /.
- Add unary plus and minus to the arithmetic operators of 4.
Answer
1. E -> E E 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. expr -> expr + term | expr - term | term
term -> term * unary | term / unary | unary
unary -> + factor | - factor | factor
factor - > id | num | (expr)
2.2.5
-
Show that all binary strings generated by the following grammar have values divisible by 3. Hint. Use induction on the number of nodes in a parse tree.
num -> 11 | 1001 | num 0 | num num
-
Does the grammar generate all binary strings with values divisible by 3?
Answer
-
Proof
Any string derived from the grammar can be considered to be a sequence consisting of 11 and 1001, where each sequence element is possibly suffixed with a 0.
Let
nbe the set of positions where11is placed.11is said to be at positioniif the first1in11is at positioni, whereistarts at 0 and
grows from least significant to most significant bit.Let
mbe the equivalent set for1001.The sum of any string produced by the grammar is:
sum
= Σn (21 + 20) * 2 n + Σm (23 + 20) * 2m
= Σn 3 * 2 n + Σm 9 * 2m
This is clearly divisible by 3.
-
No. Consider the string “10101”, which is divisible by 3, but cannot be
derived from the grammar.Readers seeking a more formal proof can read about it below:
Proof:
Every number divisible by 3 can be written in the form
3k. We will considerk > 0(though it would be valid to considerkto be an arbitrary integer).Note that every part of num(11, 1001 and 0) is divisible by 3, if the grammar could generate all the numbers divisible by 3, we can get a production for binary k from num’s production:
3k = num -> 11 | 1001 | num 0 | num num k = num/3 -> 01 | 0011 | k 0 | k k k -> 01 | 0011 | k 0 | k kIt is obvious that any value of
kthat has more than 2 consecutive bits set to 1 can never be produced. This can be confirmed by the example given in the beginning:10101 is 3*7, hence, k = 7 = 111 in binary. Because 111 has more than 2
consecutive 1’s in binary, the grammar will never produce 21.
2.2.6
Construct a context-free grammar for roman numerals.
Note: we just consider a subset of roman numerals which is less than 4k.
Answer
wikipedia: Roman_numerals
-
via wikipedia, we can categorize the single roman numerals into 4 groups:
I, II, III | I V | V, V I, V II, V III | I Xthen get the production:
digit -> smallDigit | I V | V smallDigit | I X smallDigit -> I | II | III | ε -
and we can find a simple way to map roman to arabic numerals. For example:
- XII => X, II => 10 + 2 => 12
- CXCIX => C, XC, IX => 100 + 90 + 9 => 199
- MDCCCLXXX => M, DCCC, LXXX => 1000 + 800 + 80 => 1880
-
via the upper two rules, we can derive the production:
romanNum -> thousand hundred ten digit
thousand -> M | MM | MMM | ε
hundred -> smallHundred | C D | D smallHundred | C M
smallHundred -> C | CC | CCC | ε
ten -> smallTen | X L | L smallTen | X C
smallTen -> X | XX | XXX | ε
digit -> smallDigit | I V | V smallDigit | I X
smallDigit -> I | II | III | ε
Exercises for Section 2.2
2.2.1
Consider the context-free grammar:
S -> S S + | S S * | a
- Show how the string
aa+a*can be generated by this grammar. - Construct a parse tree for this string.
- What language does this grammar generate? Justify your answer.
Answer
S->SS * ->SS + S * -> aS+ S * -> a a +S* -> a a + a *- L = {Postfix expression consisting of digits, plus and multiple signs}
2.2.2
What language is generated by the following grammars? In each case justify your answer.
- S -> 0 S 1 | 0 1
- S -> + S S | - S S | a
- S -> S ( S ) S | ε
- S -> a S b S | b S a S | ε
- S -> a | S + S | S S | S * | ( S )
Answer
- L = {0n1n | n>=1}
- L = {Prefix expression consisting of plus and minus signs}
- L = {Matched brackets of arbitrary arrangement and nesting, includes ε}
- L = {String has the same amount of a and b, includes ε}
- L = {Regular expressions used to describe regular languages} refer to wiki
2.2.3
Which of the grammars in Exercise 2.2.2 are ambiguous?
Answer
-
No
-
No
-
Yes
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-
Yes
-
Yes
2.2.4
Construct unambiguous context-free grammars for each of
the following languages. In each case show that your grammar is correct.
- Arithmetic expressions in postfix notation.
- Left-associative lists of identifiers separated by commas.
- Right-associative lists of identifiers separated by commas.
- Arithmetic expressions of integers and identifiers with the four binary operators +, -, *, /.
- Add unary plus and minus to the arithmetic operators of 4.
Answer
1. E -> E E 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. expr -> expr + term | expr - term | term
term -> term * unary | term / unary | unary
unary -> + factor | - factor | factor
factor - > id | num | (expr)
2.2.5
-
Show that all binary strings generated by the following grammar have values divisible by 3. Hint. Use induction on the number of nodes in a parse tree.
num -> 11 | 1001 | num 0 | num num
-
Does the grammar generate all binary strings with values divisible by 3?
Answer
-
Proof
Any string derived from the grammar can be considered to be a sequence consisting of 11 and 1001, where each sequence element is possibly suffixed with a 0.
Let
nbe the set of positions where11is placed.11is said to be at positioniif the first1in11is at positioni, whereistarts at 0 and
grows from least significant to most significant bit.Let
mbe the equivalent set for1001.The sum of any string produced by the grammar is:
sum
= Σn (21 + 20) * 2 n + Σm (23 + 20) * 2m
= Σn 3 * 2 n + Σm 9 * 2m
This is clearly divisible by 3.
-
No. Consider the string “10101”, which is divisible by 3, but cannot be
derived from the grammar.Readers seeking a more formal proof can read about it below:
Proof:
Every number divisible by 3 can be written in the form
3k. We will considerk > 0(though it would be valid to considerkto be an arbitrary integer).Note that every part of num(11, 1001 and 0) is divisible by 3, if the grammar could generate all the numbers divisible by 3, we can get a production for binary k from num’s production:
3k = num -> 11 | 1001 | num 0 | num num k = num/3 -> 01 | 0011 | k 0 | k k k -> 01 | 0011 | k 0 | k kIt is obvious that any value of
kthat has more than 2 consecutive bits set to 1 can never be produced. This can be confirmed by the example given in the beginning:10101 is 3*7, hence, k = 7 = 111 in binary. Because 111 has more than 2
consecutive 1’s in binary, the grammar will never produce 21.
2.2.6
Construct a context-free grammar for roman numerals.
Note: we just consider a subset of roman numerals which is less than 4k.
Answer
wikipedia: Roman_numerals
-
via wikipedia, we can categorize the single roman numerals into 4 groups:
I, II, III | I V | V, V I, V II, V III | I Xthen get the production:
digit -> smallDigit | I V | V smallDigit | I X smallDigit -> I | II | III | ε -
and we can find a simple way to map roman to arabic numerals. For example:
- XII => X, II => 10 + 2 => 12
- CXCIX => C, XC, IX => 100 + 90 + 9 => 199
- MDCCCLXXX => M, DCCC, LXXX => 1000 + 800 + 80 => 1880
-
via the upper two rules, we can derive the production:
romanNum -> thousand hundred ten digit
thousand -> M | MM | MMM | ε
hundred -> smallHundred | C D | D smallHundred | C M
smallHundred -> C | CC | CCC | ε
ten -> smallTen | X L | L smallTen | X C
smallTen -> X | XX | XXX | ε
digit -> smallDigit | I V | V smallDigit | I X
smallDigit -> I | II | III | ε
—# 2.3 Exercises for Section 2.3
2.3.1
Construct a syntax-directed translation scheme that translates arithmetic
expressions from infix notation into prefix notation in which an operator
appears before its operands; e.g. , -xy is the prefix notation for x - y. Give
annotated parse trees for the inputs 9-5+2 and 9-5*2.
Answer
productions:
expr -> expr + term
| expr - term
| term
term -> term * factor
| term / factor
| factor
factor -> digit | (expr)
translation schemes:
expr -> {print("+")} expr + term
| {print("-")} expr - term
| term
term -> {print("*")} term * factor
| {print("/")} term / factor
| factor
factor -> digit {print(digit)}
| (expr)
2.3.2
Construct a syntax-directed translation scheme that translates arithmetic
expressions from postfix notation into infix notation. Give annotated parse
trees for the inputs 95-2* and 952*-.
Answer
productions:
expr -> expr expr +
| expr expr -
| expr expr *
| expr expr /
| digit
translation schemes:
expr -> expr {print("+")} expr +
| expr {print("-")} expr -
| {print("(")} expr {print(")*(")} expr {print(")")} *
| {print("(")} expr {print(")/(")} expr {print(")")} /
| digit {print(digit)}
Another reference answer
E -> {print("(")} E {print(op)} E {print(")"}} op | digit {print(digit)}
2.3.3
Construct a syntax-directed translation scheme that translates integers into
roman numerals.
Answer
assistant function:
repeat(sign, times) // repeat('a',2) = 'aa'
translation schemes:
num -> thousand hundred ten digit
{ num.roman = thousand.roman || hundred.roman || ten.roman || digit.roman;
print(num.roman)}
thousand -> low {thousand.roman = repeat('M', low.v)}
hundred -> low {hundred.roman = repeat('C', low.v)}
| 4 {hundred.roman = 'CD'}
| high {hundred.roman = 'D' || repeat('X', high.v - 5)}
| 9 {hundred.roman = 'CM'}
ten -> low {ten.roman = repeat('X', low.v)}
| 4 {ten.roman = 'XL'}
| high {ten.roman = 'L' || repeat('X', high.v - 5)}
| 9 {ten.roman = 'XC'}
digit -> low {digit.roman = repeat('I', low.v)}
| 4 {digit.roman = 'IV'}
| high {digit.roman = 'V' || repeat('I', high.v - 5)}
| 9 {digit.roman = 'IX'}
low -> 0 {low.v = 0}
| 1 {low.v = 1}
| 2 {low.v = 2}
| 3 {low.v = 3}
high -> 5 {high.v = 5}
| 6 {high.v = 6}
| 7 {high.v = 7}
| 8 {high.v = 8}
2.3.4
Construct a syntax-directed translation scheme that trans lates roman numerals into integers.
Answer
productions:
romanNum -> thousand hundred ten digit
thousand -> M | MM | MMM | ε
hundred -> smallHundred | C D | D smallHundred | C M
smallHundred -> C | CC | CCC | ε
ten -> smallTen | X L | L smallTen | X C
smallTen -> X | XX | XXX | ε
digit -> smallDigit | I V | V smallDigit | I X
smallDigit -> I | II | III | ε
translation schemes:
romanNum -> thousand hundred ten digit {romanNum.v = thousand.v || hundred.v || ten.v || digit.v; print(romanNun.v)}
thousand -> M {thousand.v = 1}
| MM {thousand.v = 2}
| MMM {thousand.v = 3}
| ε {thousand.v = 0}
hundred -> smallHundred {hundred.v = smallHundred.v}
| C D {hundred.v = smallHundred.v}
| D smallHundred {hundred.v = 5 + smallHundred.v}
| C M {hundred.v = 9}
smallHundred -> C {smallHundred.v = 1}
| CC {smallHundred.v = 2}
| CCC {smallHundred.v = 3}
| ε {hundred.v = 0}
ten -> smallTen {ten.v = smallTen.v}
| X L {ten.v = 4}
| L smallTen {ten.v = 5 + smallTen.v}
| X C {ten.v = 9}
smallTen -> X {smallTen.v = 1}
| XX {smallTen.v = 2}
| XXX {smallTen.v = 3}
| ε {smallTen.v = 0}
digit -> smallDigit {digit.v = smallDigit.v}
| I V {digit.v = 4}
| V smallDigit {digit.v = 5 + smallDigit.v}
| I X {digit.v = 9}
smallDigit -> I {smallDigit.v = 1}
| II {smallDigit.v = 2}
| III {smallDigit.v = 3}
| ε {smallDigit.v = 0}
2.3.5
Construct a syntax-directed translation scheme that translates postfix
arithmetic expressions into equivalent prefix arithmetic expressions.
Answer
production:
expr -> expr expr op | digit
translation scheme:
expr -> {print(op)} expr expr op | digit {print(digit)}