Introduction to Syntax/Exercise 1: Difference between revisions

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== The Problem ==
#REDIRECT [[ist:Introduction to Syntax/Exercise 1]]
 
Consider the following grammar:
 
  bexpr  -> bexpr or bexpr | bterm
  bterm  -> bterm and bterm | bfactor
  bfactor -> not bfactor | ( bexpr ) | true | false
 
# Identify the terminal and non-terminal symbols of the grammar.
# Show that the grammar is ambiguous by deriving two different trees for the same input sequence.
# Write a non-ambiguous grammar for the same language.
# Build the tree corresponding to the analysis of the following input sequence: '''not ( true or false and true )'''
 
== Solution ==
 
1. The terminal symbols are all the symbols not defined by a rule: '''<tt>or</tt>''', '''<tt>and</tt>''', '''<tt>not</tt>''', '''<tt>(</tt>''', '''<tt>)</tt>''', '''<tt>true</tt>''', '''<tt>false</tt>'''.
 
2. The following trees are possible for '''true and false and true''' (an analogous analysis could be done for '''or'''). Terminals are in blue and non-terminals in black.
 
<graph>
graph nfa {
node [shape=plaintext]
  0 [id=0,label="bexpr"];
  1 [id=1,label="bterm"];
  2 [id=2,label="bterm"];
  3 [id=3,label="bterm"];
  4 [id=4,label="bfactor"];
  5 [id=5,label="true", fontcolor=blue];
  6 [id=6,label="and", fontcolor=blue];
  7 [id=7,label="bterm"];
  8 [id=8,label="bfactor"];
  9 [id=9,label="false", fontcolor=blue];
  10 [id=10,label="and", fontcolor=blue];
  11 [id=11,label="bterm"];
  12 [id=12,label="bfactor"];
  13 [id=13,label="true", fontcolor=blue];
 
  20 [id=20,label="bexpr"];
  21 [id=21,label="bterm"];
  22 [id=22,label="bterm"];
  23 [id=23,label="bfactor"];
  24 [id=24,label="true", fontcolor=blue];
  25 [id=25,label="and", fontcolor=blue];
  26 [id=26,label="bterm"];
  27 [id=27,label="bterm"];
  28 [id=28,label="bfactor"];
  29 [id=29,label="false", fontcolor=blue];
  30 [id=30,label="and", fontcolor=blue];
  31 [id=31,label="bterm"];
  32 [id=32,label="bfactor"];
  33 [id=33,label="true", fontcolor=blue];
 
  { rank = same; 5; 6; 9; 10; 13; 24; 25; 29; 30; 33 }
  { rank = same; 4; 8; 12; 23; 28; 32 }
 
  0 -- 1
  1-- 2
  1-- 10
  1 -- 11
  2 -- 3 -- 4 -- 5
  2 -- 6
  2 -- 7 -- 8 -- 9
  11 -- 12 -- 13
 
  20 -- 21
  21-- 22
  21-- 25
  21-- 26
  22 -- 23 -- 24
  26 -- 27 -- 28 -- 29
  26 -- 30
  26 -- 31 -- 32 -- 33
}
</graph>
 
 
3. The following grammar selects left associativity for the binary operators ('''and''' and '''or'''):
 
bexpr  -> bexpr or bterm | bterm
bterm  -> bterm and bfactor | bfactor
bfactor -> not bfactor | ( bexpr ) | true | false
 
4. The following analysis considers the grammar defined in 3.
 
<graph>
graph nfa {
node [shape=plaintext]
  0 [id=0,label="bexpr"];
  1 [id=1,label="bterm"];
  2 [id=2,label="bfactor"];
  3 [id=3,label="not",fontcolor=blue];
  4 [id=4,label="bfactor"];
  5 [id=5,label="(", fontcolor=blue];
  6 [id=6,label="bexpr"];
  7 [id=7,label="bexpr"];
  8 [id=8,label="bterm"];
  9 [id=9,label="bfactor"];
  10 [id=10,label="true", fontcolor=blue];
  11 [id=11,label="or", fontcolor=blue];
  12 [id=12,label="bterm"];
  13 [id=13,label="bterm"];
  14 [id=14,label="bfactor"];
  15 [id=15,label="false", fontcolor=blue];
  16 [id=16,label="and", fontcolor=blue];
  17 [id=17,label="bfactor"];
  18 [id=18,label="true", fontcolor=blue];
  19 [id=19,label=")", fontcolor=blue];
 
  { rank = same; 3; 5; 10; 11; 15; 16; 18; 19 }
  { rank = same; 9; 14; 17 }
 
  0 -- 1 -- 2
  2 -- 3
  2 -- 4
  4 -- 5
  4 -- 6
 
  6 -- 7 -- 8 -- 9 -- 10
  6 -- 11
  6 -- 12
  12 -- 13 -- 14 -- 15
  12 -- 16
  12 -- 17 -- 18
 
  4 -- 19
}
</graph>
 
[[category:Compilers]]
[[category:Teaching]]

Latest revision as of 22:15, 5 December 2018