Pearson eText for Concepts of Programming Languages -- Instant Access (Pearson+)
12th Edition
ISBN: 9780135102268
Author: Robert Sebesta
Publisher: PEARSON+
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Expert Solution & Answer
Chapter 4, Problem 1RQ
Explanation of Solution
Syntax analyzers are based on grammars:
Syntax analyzer phase in compilation process is the step in which the “tokens” in the programs are identified and recognize whether they are specified in a valid way.
So, syntax analyzers are based on the syntax of the language. The syntax is often described formally by Context Free Grammars (CFG) or Backus-Naur Form (BNF).
Context Free Grammars (CFG):
There were four classes of grammars which describes the four classes of languages. Context Free Grammars was one of the four classes which described the syntax of whole
Backus-Naur Form (BNF):
BNF is a natural notation for describing syntax described by John Backus and Peter Naur. It is very similar to Context Free Grammars.
Thus, syntax is based on BNF (or context free grammar from) due to three main reasons as described below:
- Instead of using some informal notations, syntax description with context free grammars are precise and unambiguous. This is easy to understand for both human and software systems.
- The formal description of the syntax, CFG or BNF both are used as the direct basis of syntax analyzer.
- Context free grammars or BNF are easy to implement because of their modular nature. That is, they can be implemented in modules.
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By dragging statements from the left column to the right column below, give a proof by induction of the following statement:
an
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= 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao
=
: 0.
The correct proof will use 8 of the statements below.
Statements to choose from:
Note that a₁ =
9a0 + 8.
Now assume that P(n) is true for all n ≥ 0.
Your Proof: Put chosen statements in order in this
column and press the Submit Answers button.
Let P(n) be the predicate, "a = 9″ – 1".
απ = 90 − 1 =
Note that
Let P(n) be the predicate, "an
9" - 1 is a
solution to the recurrence relation an = 9an-1 +8
with ao = 0."
-
Now assume that P(k + 1) is true.
Thus P(k) is true for all k.
Thus P(k+1) is true.
Then ak+1
=
9ak +8, so P(k + 1) is true.
= 1 − 1 = 0, as required.
Then = 9k — 1.
ak
Now assume that P(k) is true for an arbitrary
integer k ≥ 1.
By the recurrence relation, we have
ak+1 =
ak+1 =
= 9ak + 8 = 9(9k − 1) + 8
This simplifies to
9k+19+8 = 9k+1 − 1
Then 9k+1 − 1 = 9(9*…
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Chapter 4 Solutions
Pearson eText for Concepts of Programming Languages -- Instant Access (Pearson+)
Ch. 4 - Prob. 1RQCh. 4 - Prob. 2RQCh. 4 - Prob. 3RQCh. 4 - Prob. 4RQCh. 4 - Prob. 5RQCh. 4 - Prob. 6RQCh. 4 - Prob. 7RQCh. 4 - Prob. 8RQCh. 4 - Prob. 9RQCh. 4 - Prob. 10RQ
Ch. 4 - Prob. 11RQCh. 4 - Prob. 12RQCh. 4 - Prob. 13RQCh. 4 - Prob. 14RQCh. 4 - Prob. 15RQCh. 4 - Prob. 16RQCh. 4 - Prob. 17RQCh. 4 - Prob. 18RQCh. 4 - Prob. 19RQCh. 4 - Prob. 20RQCh. 4 - Prob. 21RQCh. 4 - Prob. 22RQCh. 4 - Prob. 23RQCh. 4 - What was Knuths insight in developing the LR...Ch. 4 - Prob. 25RQCh. 4 - Prob. 26RQCh. 4 - Is left recursion a problem for LR parsers?Ch. 4 - Prob. 1PSCh. 4 - Prob. 2PSCh. 4 - Prob. 3PSCh. 4 - Prob. 4PSCh. 4 - Prob. 7PSCh. 4 - Prob. 8PS
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