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|sort merge join|
sort( a ), Sort( b ); x, y = 0
while !r.empty() && !s.empty():
if ( a[ x ] == b[ y ] )
output += a[ x ]
elseif ( a[ x ] < b[ y ] )
Optimized sort merge join algorithm
1) Combine join with the merge phase of sort
2) Sort R and S in M runs of size Mon disk.
3) Merge and join the tuples in one pass.
1) Cost: 3b( r ) + 3b( s )
2) Memory requirement: b( r ) + b( s ) <= m power 2
- because we merge them in one pass
- more efficient but more strict requirement.
1) Cost of the sort is : A( logA )+ ( B logB ) + (A + B)
2) And the scanning of the cost is, ( A + B ), could be ( A*B )
- With ‘35’, ‘100’ or ‘300’ buffer pages, both can be sorted in 2 passes;
- Total cost of the join is = 7500.