NestedLoop joins

1
Nested-Loop joins

• one-and-a-half pass method, since one relation
  will be read just once.
• Tuple-Based Nested-loop Join Algorithm
• FOR each tuple s in S DO
• FOR each tuple r in R DO
• IF r and s join to make a tuple t THEN
• output t
• Improvement to Take Advantage of Disk I/O Model
• Instead of retrieving tuples of R, T(S) times,
  load memory with as many tuples of S as can fit,
  and match tuples of R against all Stuples in
  memory.

2
Block-based nested loops

• Assume B(S) B(R), and B(S) gt M
• Read M-1 blocks of S into main memory and compare
  to all of R, block by block
• 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

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Example

• B(R) 1000, B(S) 500, M 101
• Important Aside 101 buffer blocks is not as
  unrealistic as it sounds. There may be many
  queries at the same time, competing for
  mainmemory buffers.
• Outer loop iterates 5 times at 100 I/Os each
• At each iteration we read M-1 (i.e. 100) blocks
  of S and all of R (i.e. 1000) blocks.
• Total time 5(100 1000) 5500 I/Os
• Question What if we reversed the roles of R and
  S?
• We would iterate 10 times, and in each we would
  read 100500 blocks, for a total of 6000 I/Os.
• Compare with one-pass join, if it could be done!
• We would need 1500 disk I/Os if B(S) ? M-1

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Analysis of blocks nested loops

• Number of disk I/Os
• B(S)/(M-1)(M-1 B(R))
• or
• B(S) B(S)B(R)/(M-1)
• or approximately B(S)B(R)/M

5
Two-pass algorithms based on sorting

• This special case of multi-pass algorithms is
  sufficient for most of the relation sizes.
• Main idea for unary operations on R
• Suppose B(R) ? M (main memory size in blocks)
• First pass
• Read M blocks of R into MM
• Sort the content of MM
• Write the sorted result (sublist/run) into M
  blocks on disk.
• Second pass create final result

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Duplicate elimination ? using sorting

• In the second phase (merging) we dont sort but
  copy each tuple just once.
• We can do that because the identical tuples will
  appear at the same time, i.e. they will be all
  the first ones at the buffers (for the sorted
  sublists).
• As usual, if one buffer gets empty we refill it.

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Duplicate-Elimination using Sorting Example

• Assume M3, each block holds 2 records and
  relation R consists of the following 17 tuples
• 2, 5, 2, 1, 2, 2, 4, 5, 4, 3, 4, 2, 1, 5, 2, 1,
  3
• After the first pass the following sorted
  sub-lists are created
• 1, 2, 2, 2, 2, 5
• 2, 3, 4, 4, 4, 5
• 1, 1, 2, 3, 5
• In the second pass we dedicate a memory buffer to
  each sub-list.

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Example (Contd)
9
Example (Contd)
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Example (Contd)
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Analysis of ?(R)

• 2B(R) when creating sorted sublists
• B(R) to read each sublist in phase 2
• Total 3B(R)
• How large can R be?
• There can be no more than M sublists since we
  need one buffer for each one.
• So, B(R)/M M, (B(R)/M is the number of
  sublists) i.e. B(R) M2
• To compute ?(R) we need at least sqrt(B(R))
  blocks of MM.

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Sort-based ?, ?, -

• Exampe set union.
• Create sorted sublists of R and S
• Use input buffers for sorted sublists of R and S,
  one buffer per sublist
• Output each tuple once. We can do that since all
  the identical tuples appear at the same time.
• Analysis 3(B(R) B(S)) disk I/Os
• Condition B(R) B(S) M2
• Similar algorithms for sort based intersection
  and difference (bag or set versions).

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Join

• A problem for joins but not for the previous
  operators The number of joining tuples from the
  two relations can exceed what fits in memory.
• First, we can try to maximize the number of
  available buffers for putting the joining tuples.
• How, we can do this?
• By minimizing the number of sorted sublists
  (since we need a buffer for each one of them).

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Simple sort-based join

• For R(X,Y) S(Y,Z) with M buffers of memory
• Sort R on Y, sort S on Y
• Merge phase
• Use 2 input buffers 1 for R, 1 for S.
• Pick tuple t with smallest Y value in the buffer
  for R (or for S)
• If t doesnt match with the first tuple in the
  buffer for S, then just remove t.
• Otherwise, read all the tuples from R with the
  same Y value as t and put them in the M-2 part of
  the memory.
• When the input buffer for R is exhausted fill it
  again and again.
• Then, read the tuples of S that match. For each
  one we produce the join of it with all the tuples
  of R in the M-2 part of the memory.

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Example of sort join

• B(R) 1000, B(S) 500, M 101
• To sort R, we need 4B(R) I/Os, same for S.
• Total disk I/Os 4(B(R) B(S))
• Doing the join in the merge phase
• Total disk I/Os B(R) B(S)
• Total disk I/Os 5(B(R) B(S)) 7500
• Memory Requirement To be able to do the sort,
  should have B(R) M2 and B(S) M2
• Recall for nested-loop join, we needed 5500 disk
  I/Os, but the memory requirement was quadratic
  (it is linear, here), i.e., nested-loop join is
  not good for joining relations that are much
  larger than MM.

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Potential problem ...
S(Y, Z) --------- a z1 a z2 ...
a zm
R(X , Y) ----------- x1 a x2 a
xn a
What if n1 gt M-1 and m1 gt M-1?

• If the tuples from R (or S) with the same value y
  of Y do not fit in M-1 buffers, then we use all
  M-1 buffers to do a nested-loop join on the
  tuples with Y-value y from both relations.
• Observe that we can smoothly continue with the
  nested loop join when we see that the R tuples
  with Y-value y do not fit in M-1 buffers.

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Can We Improve on Sort Join?

• Do we really need the fully sorted files?

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A more efficient sort-based join

• Suppose we are not worried about many common Y
  values
• Create Y-sorted sublists of R and S
• Bring first block of each sublist into a buffer
  (assuming we have at most M sublists)
• Find smallest Y-value from heads of buffers. Join
  with other tuples in heads of buffers, use other
  possible buffers, if there are many tuples with
  the same Y values.
• Disk I/O 3(B(R) B(S))
• Requirement B(R) B(S) M2

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Example of more efficient sort-join

• B(R) 1000, B(S) 500, M 101
• Total of 15 sorted sublists
• If too many tuples join on a value y, use the
  remaining 86 MM buffers for a one pass join on y
• Total cost 3(1000 500) 4500 disk I/Os
• M2 10201 gt B(R) B(S), so the requirement is
  satisfied

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Summary of sort-based algorithms
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Two-pass algorithms based on hashing

• Main idea Let B(R) gt M
• instead of sorted sublists, create partitions,
  based on hashing
• Second pass to create result from partitions

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Creating partitions

• Here partitions are created based on all
  attributes of the relation except for grouping
  and join, where the partitions are based on the
  grouping and join-attributes respectively.
• Why bucketize? Tuples with matching values end
  up in the same bucket.
• Initialize M-1 buckets using M-1 empty buffers
• FOR each block b of relation R DO
• read block b into the M-th buffer
• IF the buffer for bucket h(t) has no room for t
  THEN
• copy the buffer to disk
• initialize a new empty block in that buffer
• ENDIF
• copy t to the buffer for bucket h(t)
• ENDFOR
• FOR each bucket DO
• IF the buffer for this bucket is not empty THEN
• write the buffer to disk

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Simple Example
Hash even / odd
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Hash-based duplicate elimination

• Pass 1 create partitions by hashing on all
  attributes
• Pass 2 for each partition, use the one-pass
  method for duplicate elimination
• Cost 3B(R) disk I/Os
• Requirement B(R) M(M-1)
• (B(R)/M is the approximate size of one bucket)
• i.e. the req. is approximately B(R) M2

25
Hash-based grouping and aggregation

• Pass 1 create partitions by hashing on grouping
  attributes
• Pass 2 for each partition, use one-pass method.
• Cost 3B(R), Requirement B(R) M2
• If B(R) gt M2
• Read blocks of partition one by one
• Create one slot in memory for each group-value
• Requirement
• where L is the list of grouping attributes

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Hash-based set union

• Pass 1 create partitions R1,,RM-1 of R, and
  S1,,SM-1 of S (with the same hash function)
• Pass 2 for each pair Ri, Si compute Ri ? Si
  using the one-pass method.
• Cost 3(B(R) B(S))
• Requirement?
• min(B(R),B(S)) M2
• Similar algorithms for intersection and
  difference (set and bag versions)

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Partition hash-join

• Pass 1 create partitions R1, ..,RM-1 of R, and
  S1, ..,SM-1 of S, based on the join attributes
  (the same hash function for both R and S)
• Pass 2 for each pair Ri, Si compute Ri ?? Si
  using the one-pass method.
• Cost 3(B(R) B(S))
• Requirement min(B(R),B(S)) M2

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Example

• B(R) 1000 blocks
• B(S) 500 blocks
• Memory available 101 blocks
• R ?? S on common attribute C
• Use 100 buckets
• Read R
• Hash
• Write buckets

• Same for S

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• Read one R bucket
• Build memory hash table
• Read corresponding S bucket block by block.

S
R
...
R
...
Memory

• Cost
• Bucketize
• Read write R
• Read write S
• Join
• Read R
• Read S
• Total cost 31000500 4500

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Saving some disk I/Os (I)

• If we have more memory than we need to hold one
  block per bucket, then we can use several buffers
  for each bucket, and write them out as a group
  saving in seek time and rotational latency.
• Also, we can read the buckets in group in the
  second pass and saving in seek time and
  rotational latency.
• Well, these techniques dont save disk I/Os, but
  make them faster.
• What about saving some I/Os?

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Saving some disk I/Os (II)

• Suppose that to join R with S we decide to create
  k buckets where k is much smaller than M.
• When we hash S we can keep m of the k buckets in
  memory, while keeping only one block for each of
  the other k-m buckets.
• We can do so provided
• m(B(S)/k) (k-m) ? M
• B(S)/k is the approximate size of a bucket of S.
• Now, when we read the tuples of R, to hash them
  into buckets, we keep in memory
• The m buckets of S that were never written out to
  disk, and
• One block for each of the k-m buckets of R whose
  corresponding buckets of S were written to disk.

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Saving some disk I/Os (III)

• If a tuple t of R hashes to one of the first m
  buckets, then we immediately join it with all the
  tuples of the corresponding S-bucket.
• If a tuple t of R hashes to a bucket whose
  corresponding S-bucket is on disk,
  then t is sent to the main memory buffer for that
  bucket, and eventually migrates to disk, as for a
  two pass, hash-based join.
• In the second pass, we join the corresponding
  buckets of R and S as usual (but only m-k).
• The savings in I/Os is equal to two for every
  block of the S-buckets that remain in
  memory, and their corresponding
  R-buckets. Since m/k of the buckets are in memory
  we save 2(m/k)(B(S)B(R)) .

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Saving some disk I/Os (IV) How can we choose m
and k?

• All but k-m of the memory buffers can be used to
  hold tuples of S, and the more of these tuples,
  the fewer the disk I/Os.
• Thus, we want to minimize k, the number of
  buckets.
• We do so by making each bucket about as big as
  can fit in memory, i.e. the buckets are of
  (approximately) M size, and therefore kB(S)/M.
• If that is the case, then there is room for one
  bucket in memory, i.e. m1.
• We have to make the bucket actually M-k blocks,
  but we are talking here approximately, when kltltM.
• So, we have that the savings in I/Os are
• 2(M/B(S))(B(R) B(S))
• And, the total cost is
• (3 - 2(M/B(S)))(B(R) B(S))

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Summary of hash-based methods


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Sort vs. Hash based algorithms

• Hash-based algorithms have a size requirement
  that depends only on the smaller of the two
  arguments rather than on the sum of the argument
  sizes, as for sort-based algorithms.
• Sort-based algorithms allow us to produce the
  result in sorted order and take advantage of that
  sort later. The result can be used in another
  sort-based algorithm later.
• Hash-based algorithms depend on the buckets being
  of nearly equal size. Well, what about a join
  with a very few values for the join attribute