15.3 Nested-Loop Joins By: Saloni Tamotia (215)

Introduction to Nested-Loop Joins  Used for relations of any side.  Not necessary that relation fits in main memory  Uses “One-and-a-half” pass method in which for each variation: One argument read just once. Other argument read repeatedly.  Two kinds: Tuple-Based Nested Loop Join Block-Based Nested Loop Join

ADVANTAGES OF NESTED-LOOP JOIN  Fits in the iterator framework.  Allows us to avoid storing intermediate relation on disk.

Tuple-Based Nested-Loop Join  Simplest variation of the nested-loop join  Loop ranges over individual tuples

Tuple-Based Nested-Loop Join  Algorithm to compute the Join R(X,Y) | | S(Y,Z) FOR each tuple s in S DO FOR each tuple r in R DO IF r and s join to make tuple t THEN output t  R and S are two Relations with r and s as tuples.  carelessness in buffering of blocks causes the use of T(R)T(S) disk I/O’s

IMPROVEMENT & MODIFICATION To decrease the cost  Method 1: Use algorithm for Index-Based joins We find tuple of R that matches given tuple of S We need not to read entire relation R  Method 2: Use algorithm for Block-Based joins Tuples of R & S are divided into blocks Uses enough memory to store blocks in order to reduce the number of disk I/O’s.

Block-Based Nested-Loop Join Algorithm  Access to arguments is organized by block. While reading tuples of inner relation we use less number of I/O’s disk.  Using enough space in main memory to store tuples of relation of the outer loop. Allows to join each tuple of the inner relation with as many tuples as possible.

Block-Based Nested-Loop Join Algorithm ALGORITHM: FOR each chunk of M - 1 blocks of S DO FOR each block b of R DO FOR each tuple t of b DO find the tuples of S in memory that join with t output the join of t with each of these tuples

Block-Based Nested-Loop Join Algorithm Assumptions: B(S) ≤ B(R) B(S) > M This means that the neither relation fits in the entire main memory.

Analysis of Nested-Loop Join Number of disk I/O’s: [B(S)/(M- 1 )]*(M-1 +B(R)) or B(S) + [B(S)B(R)/(M- 1 )] or approximately B(S)*B(R)/M

QUESTIONS