LL(1) parsing table construction in Compiler Design - Time & Space Complexity
We want to understand how the time to build an LL(1) parsing table changes as the grammar size grows.
Specifically, how does the work increase when we add more grammar rules or symbols?
Analyze the time complexity of the following pseudocode for LL(1) parsing table construction.
for each nonterminal N:
for each terminal t in FIRST(N):
add production to table[N, t]
if EPSILON in FIRST(N):
for each terminal b in FOLLOW(N):
add production to table[N, b]
This code builds the LL(1) parsing table by filling entries based on FIRST and FOLLOW sets.
Look at the loops that repeat work:
- Primary operation: Iterating over all nonterminals and terminals in FIRST and FOLLOW sets.
- How many times: For each nonterminal, we loop over terminals in its FIRST set, and possibly over terminals in its FOLLOW set.
The work depends on the number of nonterminals and the size of their FIRST and FOLLOW sets.
| Input Size (n = number of nonterminals) | Approx. Operations |
|---|---|
| 10 | About 10 x (average FIRST + FOLLOW set size) operations |
| 100 | About 100 x (average FIRST + FOLLOW set size) operations |
| 1000 | About 1000 x (average FIRST + FOLLOW set size) operations |
Pattern observation: The operations grow roughly in direct proportion to the number of nonterminals and the size of their FIRST and FOLLOW sets.
Time Complexity: O(N x T)
This means the time to build the table grows linearly with the number of nonterminals (N) and the number of terminals (T) involved in FIRST and FOLLOW sets.
[X] Wrong: "The parsing table construction is constant time because it just fills a table once."
[OK] Correct: The table size and the number of entries depend on grammar size, so more rules and symbols mean more work to fill the table.
Understanding how parsing table construction scales helps you explain compiler design clearly and shows you can analyze algorithm costs in real systems.
"What if the grammar has many epsilon-productions? How would that affect the time complexity of building the LL(1) parsing table?"
Practice
LL(1) parsing table in compiler design?Solution
Step 1: Understand the role of LL(1) parsing table
The LL(1) parsing table helps the parser decide which production to use by looking at the current input symbol and the top of the parsing stack.Step 2: Eliminate incorrect options
Options A, B, and C describe unrelated tasks: optimizing the grammar by removing left recursion (A), storing code (B), and generating machine code (C), which are not the parsing table's purpose.Final Answer:
To decide which production rule to apply based on the current input symbol -> Option AQuick Check:
Parsing table guides production choice = D [OK]
- Confusing parsing table with code storage
- Thinking parsing table generates machine code
- Mixing parsing table with grammar optimization
Solution
Step 1: Identify what each cell in LL(1) parsing table holds
Each cell in the LL(1) parsing table contains exactly one production rule to apply when the parser sees a specific non-terminal and input terminal.Step 2: Exclude incorrect options
A set of terminals is incorrect because the cell does not hold a set of terminals. A list of all grammar symbols is wrong because it does not hold all grammar symbols. A sequence of input tokens is incorrect because input tokens are not stored in the table cell.Final Answer:
A single production rule -> Option BQuick Check:
Table cell = one production rule [OK]
- Thinking cell holds multiple terminals or symbols
- Confusing input tokens with table entries
- Assuming cell stores grammar symbols list
S -> A aA -> b | εand the FIRST and FOLLOW sets:
FIRST(A) = {b, ε}
FOLLOW(A) = {a}
What production will be placed in the parsing table cell for non-terminal
A and terminal a?Solution
Step 1: Use FIRST and FOLLOW sets to fill the table
Sinceais in FOLLOW(A) andεis in FIRST(A), the productionA -> εis placed in the cell forAanda.Step 2: Confirm no other production applies
The productionA -> bapplies only for inputb. There is no productionA -> a. So the correct entry isA -> ε.Final Answer:
A -> ε -> Option CQuick Check:
FOLLOW(A) contains 'a' and FIRST(A) has ε -> use A -> ε [OK]
- Ignoring ε in FIRST set
- Placing wrong production for terminal 'a'
- Confusing FOLLOW and FIRST sets
B and terminal c. What is the most likely cause?Solution
Step 1: Understand why multiple productions appear in one cell
Multiple productions in one cell indicate a conflict, meaning the grammar is not LL(1). This usually happens if the grammar is ambiguous or has left recursion.Step 2: Evaluate other options
You forgot to compute FOLLOW sets is incorrect because missing FOLLOW sets would cause incomplete table, not conflicts. You used the wrong terminals in the table rows is about table construction errors but does not explain conflicts. The input string is invalid is unrelated to table construction.Final Answer:
The grammar is not LL(1) because it is ambiguous or has left recursion -> Option AQuick Check:
Multiple entries = grammar not LL(1) [OK]
- Blaming input string instead of grammar
- Ignoring grammar ambiguity or left recursion
- Assuming FOLLOW sets cause conflicts
E -> T E'E' -> + T E' | εT -> int | ( E )Given FIRST and FOLLOW sets:
FIRST(E') = {+, ε}
FOLLOW(E') = {), $}
Which entries should be placed in the LL(1) parsing table for non-terminal
E' and terminal )?Solution
Step 1: Check if terminal ')' is in FIRST(E')
Since FIRST(E') = {+, ε}, ')' is not in FIRST(E'). But since ε is in FIRST(E'), we look at FOLLOW(E') which contains ')'.Step 2: Use ε-production for terminals in FOLLOW(E')
Because ')' is in FOLLOW(E') and ε is in FIRST(E'), the productionE' -> εis placed in the cell forE'and).Final Answer:
E' -> ε -> Option DQuick Check:
FOLLOW(E') has ')' and FIRST(E') has ε -> use E' -> ε [OK]
- Placing wrong production for terminal ')'
- Ignoring ε in FIRST set
- Confusing FOLLOW and FIRST sets
Ready to prove it?
Pass the test and this pattern moves unknown
