ADBMS Tut 3 solution Detailed

Okay, I’ll answer the questions based on the concepts from the PDF content you provided earlier (Chapters 15 and 16 on Query Processing and Optimization). I’ll use MathJax for equations and variables, and focus on providing clear, exam-oriented explanations.

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 of Hash Join:

Hash join is a join algorithm that uses hashing to efficiently find matching tuples in two relations, r and s. It’s particularly well-suited for natural joins and equi-joins. The core idea is to partition both relations based on a hash function applied to the join attributes. This ensures that tuples that could potentially join will reside in corresponding partitions.

Hash Function Use:

A hash function, h, is crucial. It maps the join attributes (JoinAttrs) to a set of integers {0, 1, …, $n_h$}. This function should ideally have the properties of:

• Randomness: Distributes join attribute values evenly across the hash range.
• Uniformity: Minimizes the chance of many values hashing to the same bucket (which would lead to large partitions and reduced efficiency).

The hash function is used in two key ways:

1. Partitioning: Both relations, r and s, are partitioned into $n_h$ partitions. A tuple $t_r$ from r is placed in partition $r_i$ where i = h($t_r$[JoinAttrs]). Similarly, a tuple $t_s$ from s is placed in partition $s_i$ where i = h($t_s$[JoinAttrs]).
2. Probing (within each partition): After partitioning, for each i from 0 to $n_h$, an in-memory hash index is built on partition $s_i$ (the “build” input). Then, each tuple in the corresponding partition $r_i$ (the “probe” input) is used to probe this index, looking for matches based on the join attributes. A different hash function (than the partitioning hash function) is typically used for building this in-memory index.

Phases of Hash Join: The basic hash join algorithm involves these phases:

1. Partitioning Phase (or Build Phase):
   • Scan relation s (smaller relation, called build input) and partition its tuples into $s_0,s_1,...,s_{n_h}$ using the hash function h on the join attributes.
   • Scan relation r (larger relation, called probe input) and partition its tuples into $r_0,r_1,...,r_{n_h}$ using the same hash function h on the join attributes.
2. Probing Phase (Join Phase):
   • For each i from 0 to $n_h$:
     • Build an in-memory hash index on partition $s_i$.
     • Read partition $r_i$. For each tuple $t_r$ in $r_i$:
       • Probe the in-memory hash index for $s_i$ using $t_r$`[JoinAttrs].
       • For each matching tuple $t_s$ found in $s_i$, form the joined tuple $t_r\bowtie t_s$ and add it to the result.

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

Hash Join Preference:

Hash join is generally preferred over nested-loop join when:

• Join Type: The join is a natural join or an equi-join (join condition involves equality). Nested-loop join can handle any join condition, but hash join is optimized for equality.
• Relation Size: The relations involved are not extremely small. If one relation fits entirely in memory, a simple nested-loop join (with the smaller relation as the inner loop) can be very efficient. Hash join has overhead due to partitioning.
• No suitable index: If there is an suitable index in the inner relation of nested loop join, indexed nested loop join will be much more performant.

Conditions for Hash Join Efficiency:

• Sufficient Memory: A key factor is having enough memory. Ideally, the smaller of the two relations (after any selections are applied) should fit comfortably in memory, or at least a significant portion of it for hybrid hash-join.
• Good Hash Function: A hash function that distributes tuples evenly across partitions is crucial. Poor hash functions lead to skewed partitions and reduce performance.
• Comparable Relation Sizes: Although not strictly required, hash join performs best when the relations aren’t drastically different in size. If one relation is much larger, the partitioning overhead on the larger relation becomes significant.

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:

If the smaller relation (the “build” relation, typically s) does not fit entirely in memory, the basic hash join algorithm needs modification. The core problem is that the in-memory hash index built during the probing phase must fit in memory.

Grace Hash Join is another name for Basic Hash Join.

Recursive Partitioning:

This is the primary technique to handle the situation where the build relation doesn’t fit in memory.

1. Initial Partitioning: Both relations r and s are partitioned using a hash function h, as in the basic hash join. However, the number of partitions, $n_h$, is chosen to be larger than the number of available memory blocks, meaning the partition will not fit in memory.

2. Recursive Steps: If any partition $s_i$ is still too large to fit in memory (along with its hash index), recursive partitioning is applied:

   • A new hash function, $h^{\prime }$, is used.
   • $s_i$ is repartitioned into smaller partitions $s_{i1},s_{i2},...$.
   • The corresponding partition $r_i$ is also repartitioned using the same new hash function $h^{\prime }$.
   • This process continues recursively until all partitions of the build relation (and their hash indices) fit in memory.

3. Join the partitions After partitioning is complete, the algorithm proceeds with join of $r_i$ and $s_i$.

Key Points about Recursive Partitioning:

• Multiple Passes: Partitioning may require multiple passes over the data. Each pass reads and writes the entire (remaining) relation.
• Different Hash Functions: A different hash function is used in each level of recursion to avoid repeating the same partitioning pattern.
• Memory Management: The number of partitions ($n_h$) in each pass is limited by the number of available buffer blocks in memory.

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

Working of Indexed Nested-Loop Join:

Indexed nested-loop join is a variation of the nested-loop join that leverages an index on the inner relation’s join attribute(s) to speed up the join process.

4. Outer Loop: Iterate through each tuple $t_r$ in the outer relation r.
5. Inner Loop (Index Lookup): For each $t_r$, use the index on the inner relation s to directly find tuples $t_s$ in s that satisfy the join condition with $t_r$. This avoids a full scan of s for every tuple in r.
6. Join: For each matching tuple $t_s$ found via the index, form the joined tuple $t_r\bowtie t_s$ and add it to the result.

Performance Improvement with an Index:

The presence of an index drastically improves performance because:

• Avoids Full Scans: Instead of scanning the entire inner relation s for each tuple in r, the index allows direct access to only the relevant tuples in s.
• Reduced I/O: This significantly reduces the number of disk I/Os, especially if the index is a clustering index (where data records are stored in the same order as the index). Even with a non-clustering index, the number of I/Os is typically much lower than a full scan.
• Faster Lookups: Index lookups (e.g., using a B+-tree) are typically very fast, with logarithmic time complexity (O(log n)) in the number of entries in the index, as opposed to linear time complexity (O(n)) for a full scan.

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 (INLJ) over Hash Join (HJ):

• Index Availability: The most critical factor is the presence of a suitable index on the inner relation’s join attribute(s). If no such index exists, INLJ is not possible.
• Join Selectivity: INLJ is particularly advantageous when the join is highly selective, meaning that for each tuple in the outer relation, only a small number of tuples in the inner relation are expected to match. This is often the case with foreign key joins.
• Outer Relation Size: INLJ can be very efficient if the outer relation is relatively small. Even if the inner relation is large, the index allows for quick lookups.
• Non-Equi Joins: INLJ can be applied in certain cases of non-equi joins (range queries) if suitable index is available.

Performance Comparison:

Feature	Indexed Nested-Loop Join	Hash Join
Index Required?	Yes, on the inner relation’s join attribute(s).	No, but benefits from partitioning.
Data Size	Performs well if the outer relation is small or the join is highly selective.	Generally better for larger relations, especially if the smaller relation fits (mostly) in memory.
Join Type	Works for any join condition, but most efficient with equality and highly selective joins.	Primarily for equi-joins and natural joins. Can be adapted for some others.
I/O Cost	Low I/O if the index lookups are efficient (few matching tuples per outer tuple).	I/O cost depends on partitioning and whether recursive partitioning is needed.
CPU Cost	Can have lower CPU cost if few index lookups are needed.	Higher CPU cost for hashing, but often less than the I/O savings.
Memory Usage	Relatively low memory usage (primarily for the index).	Requires more memory, especially for the build phase (hash table).

In Summary:

• If a suitable index exists and the join is selective or the outer relation is small, INLJ is often the best choice.
• If no suitable index exists, or if the relations are large and the join is not highly selective, Hash Join is generally preferred (assuming sufficient 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 between Indexed Nested-Loop Join (INLJ) and Regular Nested-Loop Join (NLJ):

• Regular Nested-Loop Join (NLJ):

  • For each tuple $t_r$ in the outer relation r, it scans the entire inner relation s to find matching tuples.
  • This is a brute-force approach, comparing $t_r$ with every tuple in s.

• Indexed Nested-Loop Join (INLJ):

  • For each tuple $t_r$ in the outer relation r, it uses an index on the join attribute(s) of the inner relation s to directly find matching tuples.
  • It does not scan the entire inner relation.

Reduction in Comparisons:

The index lookup in INLJ drastically reduces the number of comparisons because:

7. Targeted Search: The index acts as a guide, leading directly to the tuples in s that potentially match $t_r$, based on the join attribute value.
8. No Exhaustive Scan: Instead of comparing $t_r$ with every tuple in s (which has $n_s$ tuples), the index lookup typically examines a much smaller number of tuples. The exact number depends on the index type and the distribution of data, but it’s often logarithmic (e.g., O(log $n_s$)) for a B+-tree index.
9. Index Structure Efficiency: If index is a B+ tree, search will be logrithmic to the number of entries, if the index is a hash, lookup will take constant time in average case.

Example:

Suppose we have:

• instructor (outer relation) with 1000 tuples.

• teaches (inner relation) with 10,000 tuples.

• Join condition: instructor.ID = teaches.ID

• NLJ: For each of the 1000 instructor tuples, we scan all 10,000 teaches tuples. This results in 1000 * 10,000 = 10,000,000 comparisons.

• INLJ (with an index on teaches.ID): For each instructor tuple, we use the index to find matching teaches tuples. If, on average, an instructor teaches 4 courses, the index lookup might only need to examine a few index entries (e.g., 3-4 levels of a B+-tree) and then retrieve the 4 matching tuples. This results in significantly fewer comparisons (and I/Os).

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

Nested-Loop Join (NLJ) Outperforming Hash Join (HJ):

Although Hash Join is often superior for larger relations, there are scenarios where Nested-Loop Join (or its variants) can perform better:

10. Very Small Relations: If both relations are very small and fit entirely in memory, the overhead of hashing in Hash Join (partitioning, building hash tables) can outweigh its benefits. A simple nested-loop join in memory can be faster.

11. Highly Skewed Data (with INLJ): If the join attribute values are extremely skewed, and you have an indexed nested-loop join, INLJ can outperform Hash Join. Skewed data can lead to very large hash buckets in Hash Join, degrading its performance.

12. Non-Equi Joins (NLJ Only): Hash Join is primarily designed for equi-joins. While variations exist for other join types, nested-loop join is a more general-purpose algorithm that can handle any join condition (e.g., <, >, <=, >=, <>).

13. Low Memory Conditions: Hash Join has higher memory requirements than nested-loop join. If memory is extremely limited, nested-loop join (especially the block nested-loop variant) might be the only feasible option, even if it’s slower.

14. Cartesian Product: If no join condition exists between two tables, Nested loop join can be used to get cartesian product.

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

Inefficiency of NLJ for Large Datasets:

Nested-Loop Join is inefficient for large datasets primarily because of its quadratic time complexity.

• Brute-Force Nature: For every tuple in the outer relation, NLJ scans the entire inner relation. This means the number of comparisons (and potentially disk I/Os) grows proportionally to the product of the relation sizes ($n_r\ast n_s$).
• I/O Bottleneck: If the relations don’t fit in memory, NLJ results in a massive number of disk I/Os. Each scan of the inner relation for a tuple in the outer relation can potentially require reading the entire inner relation from disk.

Comparison with Sort-Merge and Hash Join:

Feature	Nested-Loop Join (NLJ)	Sort-Merge Join	Hash Join
Time Complexity	O($n_r\ast n_s$) (quadratic)	O($n_r$ log $n_r$ + $n_s$ log $n_s$ + $n_r$ + $n_s$) (n log n, due to sorting)	O($n_r$ + $n_s$) (linear, ideally, but can degrade with poor hashing or recursive partitioning)
Best Case	Both relations very small and fit in memory.	Relations are already sorted on the join attribute(s).	Smaller relation fits entirely in memory; good hash function.
Worst Case	Large relations; no indexes; no sorting.	Large relations; unsorted; requires full sort.	Skewed data; many hash collisions; recursive partitioning needed.
Memory Usage	Low (can be as little as 3 blocks).	Moderate (depends on sorting algorithm and buffer size).	Higher (needs space for hash table/partitions).
Join Type	Any join condition.	Primarily equi-joins and natural joins. Can be adapted for some others.	Primarily equi-joins and natural joins.
Index Use	Basic NLJ doesn’t use indexes. Indexed NLJ is a separate, much more efficient variant.	Can benefit if relations are already sorted due to an index, avoids sorting cost.	Doesn’t directly use indexes in the same way as INLJ, but partitioning is crucial.
Suitable for	small dataset, non-equi joins	Sorted large dataset, equi-joins	Large dataset, equi-joins

In summary:

• NLJ: Simple but very inefficient for large, unsorted relations without indexes. The quadratic complexity is the killer.
• Sort-Merge: More efficient than basic NLJ for larger relations, especially if already sorted. The sorting cost is the main factor.
• Hash Join: Often the most efficient for large relations and equi-joins, provided there’s enough memory and a good hash function. Partitioning and potential recursion add overhead.

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. Here, Grace Hash Join is another name for Basic Hash Join. Given:

• $b_r$ (blocks in R) = 1000
• $b_s$ (blocks in S) = 500
• M (memory buffer pages) = 50

Since S is the smaller relation, it will be the build relation. The optimal number of partitions, $n_h$ to avoid recursive partitioning will be = \lceil$$b_s/M$\rceil$ = $\lceil$500/50$\rceil$ = 10.

But, to split relation into 10 partitions, only 11 buffer blocks will be required (10 for output partitions + 1 for input).

We assume the grace hash join uses hybrid hash join technique, so that it does not need to write out and read in first partition of build input relation.

Cost Calculation:

15. Partitioning Phase: In hash join we do not need to write first partition to disk, number of blocks in first partition will be 500/10 = 50 blocks.

    • Read both R and S: 1000 + 500 = 1500 I/Os.
    • Write 9 partitions of R and S to disk: (1000-100)+(500-50) = 900+450 = 1350 I/Os.

16. Probing Phase:

    • Read 9 partitions of R and S, and full relation of R and S, : 1000 + 500 = 1500 I/Os.

Total I/O Cost: 1500 + 1350+ 1500 = 4350 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.

Given:

• $n_A$ = 1,000,000 tuples
• $l_A$ = 200 bytes/tuple
• $n_B$ = 500,000 tuples
• $l_B$ = 250 bytes/tuple
• Block size = 8 KB = 8192 bytes

Calculations:

17. Blocking Factors:

    • $f_A$ = $\lfloor$8192 / 200$\rfloor$ = 40 tuples/block
    • $f_B$ = $\lfloor$8192 / 250$\rfloor$ = 32 tuples/block

18. Number of Blocks:

    • $b_A$ = $\lceil$1,000,000 / 40$\rceil$ = 25,000 blocks
    • $b_B$ = $\lceil$500,000 / 32$\rceil$ = 15,625 blocks

19. Hash Join Cost: We’ll assume enough memory is available to avoid recursive partitioning, and that the hash function distributes tuples uniformly. We also use basic formula of Hash join = 3($b_r$+$b_s$).

    • Total I/O= 3x(25000+15625)=121,875 I/Os.

Total Estimated I/O Cost: 121,875 I/Os (This is an estimate, as it depends on memory size, hash function quality, and potential overheads like partially filled blocks).

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.

Given:

• $n_{RRR}$ = 10,000 tuples
• $n_{SSS}$ = 1,000,000 tuples
• Index lookup cost = 2 I/Os per tuple

Cost Calculation:

20. Outer Relation Scan: We assume a simple scan of the outer relation RRR. The cost depends on how RRR is stored, so it is not given, we’ll call it $Cos{t}_{scan(RRR)}$. In the best case, if RRR is clustered, this cost could be relatively small.
21. Index lookups: For each tuple in RRR, we perform an index lookup on SSS.
    • Cost per outer tuple: 2 I/Os
    • Total index lookup cost: $n_{RRR}$ * 2 = 10,000 * 2 = 20,000 I/Os

Total Cost:

$Cos{t}_{scan(RRR)}$ + 20,000 I/Os

The total cost is dominated by the index lookup cost. The cost of scanning RRR is likely to be small relative to the 20,000 I/Os for the index lookups, especially if RRR is clustered. The precise total cost will depend on the blocking factor of RRR.

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

Given:

• $b_A$ = 10,000 pages
• $b_B$ = 2,000 pages
• M (memory buffer size) = 20 pages

Cost Calculations:

22. Block Nested-Loop Join (BNLJ): We’ll assume A is the outer relation and B is the inner relation. We’ll also assume that we use M-2 pages for the outer relation and 1 for input of inner relation, and 1 page for output buffer.

    • Number of outer relation “chunks”: $\lceil$10,000 / (20 - 2)$\rceil$ = $\lceil$10000/18$\rceil$ = 556
    • For each chunk of A, we read all of B: 556 * 2000 = 1,112,000 I/Os
    • Cost to read the outer relation A: 10,000
    • Total BNLJ Cost: 1,112,000 + 10000= 1,122,000 I/Os

23. Simple Nested-Loop Join (NLJ):

    • Cost: $n_A\ast b_B+b_A$ I/Os (assuming A is the outer relation). We don’t have $n_A$, the number of tuples in A, so we cannot directly compute this. However, we know it will be much higher than BNLJ, as every tuple in A requires a full scan of B.

24. Indexed Nested-Loop Join (INLJ):

    • Clustered Index on B: If the index on the join attribute(s) of B is clustered, then retrieving all tuples matching a given outer tuple is very efficient. Let c be the average number of tuples in B, matching a tuple in A.
      • Cost: $b_A+n_A\ast (cos{t}_{index\_lookup}+1)$ We add 1 to read the actual data page(assuming data is also stored with index).
    • Unclustered Index on B: If the index on B is unclustered, each matching tuple in B might require a separate I/O.
      • Cost: $b_A+n_A\ast (cos{t}_{index\_lookup}+c)$, where c is the cost of matching tuples, for each lookup.
    • We do not have enough information here ($n_A$, type of index and value of c, $cos{t}_{index\_lookup}$), to estimate the accurate value.

Comparison:

• Without tuple counts and index details, a precise comparison is difficult. However, we can make these general statements:
  • NLJ is by far the worst. It will almost certainly have millions more I/Os than any other method.
  • BNLJ is better than NLJ but still very expensive for large relations.
  • INLJ with a clustered index is likely the best, if the join is selective (few matches per outer tuple).
  • INLJ with an unclustered index can be worse than BNLJ if the join is not selective (many matches per outer tuple), as each match could require a separate I/O.
  • If there are no suitable indexes available, Hash join may be better.

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?

Given:

• $n_{RRR}$ = 100,000
• $n_{SSS}$ = 500,000
• Clustered index lookup cost = 1 I/O
• Unclustered index lookup cost = 5 I/Os

Cost Calculations:

25. Clustered Index:

    • Cost: $b_{RRR}+n_{RRR}\ast (cos{t}_{index\_lookup})$. Since we do not have value of $b_{RRR}$ we will compute the cost of index lookups.
    • Index Lookup Cost: 100,000 * 1 = 100,000 I/Os

26. Unclustered Index:

    • Cost: $b_{RRR}+n_{RRR}\ast (cos{t}_{index\_lookup})$. Since we do not have value of $b_{RRR}$ we will compute the cost of index lookups.
    • Index Lookup Cost: 100,000 * 5 = 500,000 I/Os

Comparison:

• Clustered Index: Total Cost (ignoring RRR scan) ≈ 100,000 I/Os
• Unclustered Index: Total Cost (ignoring RRR scan) ≈ 500,000 I/Os

Conclusion:

The clustered index approach is significantly more efficient in this case. The unclustered index requires five times as many I/Os for the index lookups alone, because each lookup is assumed to potentially access data from the relation, which can be on a different page than the index. This difference highlights the importance of clustered indexes for join operations, when available.