S --> L . R L.pos = 0; R.pos = -1; S.val = L.val + R.val S --> L L.pos = 0; S.val = L.val; L --> L1 B L1.pos = L.pos + 1; B.pos = L.pos; L.val = L1.val + B.val; L --> B B.pos = L.pos; L.val = B.val; R --> R1 B R1.pos = R.pos - 1; B.pos = R.pos; L.val = L1.val + B.val; R --> B B.pos = R.pos; L.val = B.val; B --> 0 B.val = 0; B --> 1 B.val = 1*2B.pos;

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Productions Semantic Rules
S --> L . R L.pos = 0; R.pos = -1; S.val = L.val + R.val
S --> L L.pos = 0; S.val = L.val;
L --> L1 B L1.pos = L.pos + 1; B.pos = L.pos; L.val = L1.val + B.val;
L --> B B.pos = L.pos; L.val = B.val;
R --> R1 B R1.pos = R.pos - 1; B.pos = R.pos; L.val = L1.val + B.val;
R --> B B.pos = R.pos; L.val = B.val;
B --> 0 B.val = 0;
B --> 1 B.val = 1*2B.pos;

Create an annotated parse tree for the following input strings:

b) 010.1111

Expert Solution
Step 1

S --> L . R L.pos = 0; R.pos = -1; S.val = L.val + R.val
S --> L L.pos = 0; S.val = L.val;
L --> L1 B L1.pos = L.pos + 1; B.pos = L.pos; L.val = L1.val + B.val;
L --> B B.pos = L.pos; L.val = B.val;
R --> R1 B R1.pos = R.pos - 1; B.pos = R.pos; L.val = L1.val + B.val;
R --> B B.pos = R.pos; L.val = B.val;
B --> 0 B.val = 0;
B --> 1 B.val = 1*2B.pos;

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