ADBMS Tut 3 Solution Short

1. Explain the process of Hash Join. How is the hash function used in the join process? What are the phases of hash join?

Process: Hash Join works by partitioning tuples of both relations based on a hash function applied to the join attributes. It then compares tuples only within corresponding partitions.

Hash Function Use: The hash function (h) maps join attribute values to a range of integers (0 to $n_h$). This determines which partition a tuple belongs to. A good hash function distributes tuples uniformly across partitions.

Phases:

1. Partitioning (Build) Phase: Both relations (r and s) are partitioned into $r_0,r_1,...,r_{n_h}$ and $s_0,s_1,...,s_{n_h}$ using the hash function h on the join attributes.
2. Probing (Join) Phase: For each partition i, an in-memory hash index is built on $s_i$ (build input). Then, tuples from $r_i$ (probe input) are used to probe this index, finding matching tuples.

2. When is Hash Join preferred over Nested Loop Join? Under what conditions is Hash Join more efficient?

Hash Join Preference:

• Equi-joins: Hash Join is highly preferred for natural joins and equi-joins (where the join condition is equality).
• Larger Relations: When relations are not very small. Nested loop join is preferred when one of the relation is very small.

Efficiency Conditions:

• When there is enough memory to hold at least the smaller relation’s hash partitions (with their index) in memory.
• When the join attributes have a relatively uniform distribution (to avoid skewed partitions).

3. What happens in a Hash Join when the smaller relation does not fit into memory? Explain how grace hash join and recursive partitioning work.

Smaller Relation Doesn’t Fit:

• Recursive Partitioning: If the smaller (build) relation doesn’t fit in memory, the hash join uses recursive partitioning. Both relations are partitioned into smaller sub-partitions repeatedly, using different hash functions in each pass, until each build-relation partition (and its hash index) fits in memory.

Grace Hash Join vs. Recursive Partitioning in simple way:

Grace hash join is just another name for the standard hash join algorithm using recursive partitioning.

4. Explain the working of Index Nested Loop Join. How does the presence of an index improve performance?

Indexed Nested-Loop Join:

• For each tuple in the outer relation, an index on the inner relation’s join attribute is used to find matching tuples directly.

Performance Improvement:

• Avoids Full Scans: The index avoids scanning the entire inner relation for each outer tuple.
• Faster Lookups: Index lookups (e.g., using a B+-tree) are significantly faster than linear scans, especially for large relations.

5. When should we use Index Nested Loop Join instead of Hash Join? Compare performance based on data size and indexing availability.

When to Use Indexed Nested-Loop Join:

• Small Outer Relation: When the outer relation is very small, and the inner has an index on the join attribute.
• Selective Joins: When the join condition is very selective (results in few matching tuples).
• Index is available on inner relation’s join attribute.

Performance Comparison:

Factor	Indexed Nested-Loop Join	Hash Join
Outer Relation Size	Prefers small	Less sensitive to size
Inner Relation Size	Less sensitive to size(because of Index)	Size impacts partitioning and memory needs
Index Availability	Requires index on inner join attribute	Index not required (but can improve partitioning)
Join Selectivity	Prefers highly selective joins	Less sensitive to selectivity
Memory	Requires less memory	Requires more memory.

6. How does an Indexed Nested Loop Join differ from a regular Nested Loop Join? Explain how index lookup reduces the number of comparisons.

Difference:

• Regular Nested-Loop: Scans the entire inner relation for each tuple in the outer relation.
• Indexed Nested-Loop: Uses an index on the inner relation to find only matching tuples for each outer tuple.

Reduced Comparisons:

• The index allows direct access to potentially matching tuples, avoiding comparisons with all tuples in the inner relation.

7. When does a Nested Loop Join perform better than a Hash Join? Discuss scenarios where Nested Loop Join is preferable.

Nested Loop Join Preferable When:

• One relation is very small (especially if it fits entirely in memory).
• Join condition is not an equality condition (e.g., range conditions, inequalities), where hash join cannot be used.
• When indexed nested loop join is applicable and selection condition on outer relation is very selective.

8. Why is Nested Loop Join considered inefficient for large datasets? How does it compare to Sort-Merge and Hash Join?

Inefficiency:

• Quadratic Complexity: In the worst case (no index, large relations), it has a time complexity of O($n_r\ast n_s$), where $n_r$ and $n_s$ are the number of tuples in the relations. This means the number of comparisons grows very rapidly with relation size.

Comparison:

• Sort-Merge Join: Generally more efficient than nested-loop for large datasets if relations need to be sorted. Complexity is O(n log n) for sorting + O(n) for merging, where n is number of tuples.
• Hash Join: Often the most efficient for large datasets with equi-joins, especially if partitioning can be done in one pass. Complexity is close to O(n) if memory is sufficient.

9. Hash Join Numerical Problems. Consider two relations R and S: Relation R has 1,000 pages. Relation S has 500 pages. The system has 50 buffer pages. Calculate the I/O cost of performing a Grace Hash Join assuming uniform distribution. R has 1000 pages, S has 500 pages, M = 50.

1. Partitioning Phase:

   • Number of partitions: We need to partition S such that each partition fits within M-1 buffers (leaving 1 for input). So, $n_h$ > 500 / (50-1) ≈ 10.2, so lets take $n_h$ = 11.

   • I/O cost to read and write both relations: 2 * (1000 + 500) = 3000 I/Os.

2. Join (Probing) Phase: Read each partition of R and S once: 1000 + 500 = 1500 I/Os.

3. Total I/Os: 3000+1500 = 4500 I/Os

10. Consider a Hash Join between two tables: Table A has 1,000,000 tuples, each 200 bytes. Table B has 500,000 tuples, each 250 bytes. The block size is 8 KB. Estimate the number of disk I/Os required for the join.

4. Calculate Relation Sizes in Blocks:

   • $b_A$ = $\lceil$(1,000,000 * 200) / 8192$\rceil$ = 24415 blocks
   • $b_B$ = $\lceil$(500,000 * 250) / 8192$\rceil$ = 15259 blocks

5. Assume memory size M is greater than $\sqrt{b_B}$ (so recursive partitioning is not required), then Number of I/Os = 3($b_r$+$b_s$) = 3*(24415+15259) = 119022 I/Os.

11. Index Nested Loop Join Numerical Problems. Consider a table with the following statistics. Outer relation RRR has 10,000 tuples. Inner relation SSS has 1,000,000 tuples. The index lookup cost is 2 I/Os per tuple. Compute the total cost of Index Nested Loop Join.

• Outer Relation Scan: We’ll assume we have to scan RRR. Let’s say it takes ‘x’ I/Os to scan RRR.
• Index Lookups: For each of the 10,000 tuples in RRR, we perform an index lookup costing 2 I/Os. Total index lookup cost: 10,000 * 2 = 20,000 I/Os.
• Total Cost: x + 20,000 I/Os

12. A database contains two tables: Table A has 10,000 pages. Table B has 2,000 pages. The database system has 20 buffer pages. Compute the I/O cost for Block Nested Loop Join. Compare with the cost of Simple Nested Loop Join Index Nested Loop Join with Clustered vs. Unclustered Index

• Block Nested Loop Join Cost

  • Outer Relation A, Inner B:
    • $\lceil 10000/18\rceil$ * 2000 + 10000 = 1,121,112 I/Os
  • Outer Relation B, Inner A:
    • $\lceil 2000/18\rceil$ * 10000 + 2000 = 1,113,112 I/Os.

• Simple Nested Loop Join.

  • Using smaller relation B as outer:
    • 2000 + (nB * 10000) I/Os, where nB is number of tuples in relation B.

• Indexed Nested loop join:

  • With clustered index:
    • 2000 + (nB * 1)
  • With unclustered index:
    • 2000 + (nB * 5)

13. Suppose an Index Nested Loop Join is being performed on the following dataset:

• Outer Relation RRR: 100,000 tuples
• Inner Relation SSS: 500,000 tuples
• SSS is indexed using:
  • Clustered index: Each lookup requires 1 I/O.
  • Unclustered index: Each lookup requires 5 I/Os.

Question:

• Compute the I/O cost for both clustered and unclustered index-based joins.
• Which approach is more efficient?

Calculations:

6. Clustered Index:

   • Outer relation scan (assume cost = x I/Os to scan RRR).
   • Index lookups: 100,000 tuples * 1 I/O/lookup = 100,000 I/Os
   • Total Cost: x + 100,000 I/Os

7. Unclustered Index:

   • Outer relation scan (assume cost = x I/Os to scan RRR).
   • Index lookups: 100,000 tuples * 5 I/Os/lookup = 500,000 I/Os
   • Total Cost: x + 500,000 I/Os

Conclusion: The clustered index-based join is significantly more efficient because each lookup requires fewer I/Os. The unclustered index requires many more I/Os due to the potential need to fetch data blocks for each matching tuple.