Shuigeng Zhou. May 18, 2016 School of Computer Science Fudan University

Transcription

1 Query Processing Shuigeng Zhou May 18, 2016 School of Comuter Science Fudan University

2 Overview Outline Measures of Query Cost Selection Oeration Sorting Join Oeration Other Oerations Evaluation of Exressions 16/5/25 School of C.S., Fudan Univ. 2

3 Basic Stes in Query Processing 1. Parsing and translation 2. Otimization 3. Evaluation 16/5/25 School of C.S., Fudan Univ. 3

4 Basic Stes in Query Processing (Cont.) Parsing and translation n translate the query into its internal form, which is then translated into relational algebra n Parser checks syntax, verifies relations Evaluation n The query-execution engine takes a queryevaluation lan, executes that lan, and returns the answers to the query 16/5/25 School of C.S., Fudan Univ. 4

5 Basic Stes in Query Processing: Otimization A relational algebra exression may have many equivalent exressions Each relational algebra oeration can be evaluated using one of several different algorithms Annotated exression secifying detailed evaluation strategy is called an evaluation-lan n can use an index on balance to find accounts with balance < 2500, n or can erform comlete relation scan and discard accounts with balance /5/25 School of C.S., Fudan Univ. 5

6 Basic Stes in Query Processing: Otimization Examle: select * from account where balance < 2500 can use an index on balance to find accounts with balance < 2500, or can erform comlete relation scan and discard accounts with balance /5/25 School of C.S., Fudan Univ. 6

7 Basic Stes: Otimization (Cont.) Query otimization: amongst all equivalent evaluation lans choose the one with lowest cost. n Cost is estimated using statistical information from the database catalog In this lecture we study n How to measure query costs n Algorithms for evaluating relational algebra oerations n How to combine algorithms for individual oerations in order to evaluate a comlete exression In next lecture n We study how to otimize queries, that is, how to find an evaluation lan with lowest estimated cost 16/5/25 School of C.S., Fudan Univ. 7

8 Measures of Query Cost

9 Measures of Query Cost Cost is generally measured as total elased time for answering query n disk accesses, CPU, or even network communication Tyically disk access is the redominant cost, and is also relatively easy to estimate. Measured by taking into account n Number of seeks --- average-seek-cost n Number of blocks read --- average-block-read-cost n Number of blocks written --- average-block-write-cost l Cost to write a block is greater than cost to read a block data is read back after being written to ensure that the write was successful 16/5/25 School of C.S., Fudan Univ. 9

10 Measures of Query Cost (Cont.) For simlicity we just use number of block transfers from disk as the cost measure n n We ignore the difference in cost between sequential and random I/O for simlicity We also ignore CPU costs for simlicity Costs deends on the size of the buffer in main memory n n Having more memory reduces need for disk access We often use worst case estimates, assuming only the minimum amount of memory needed for the oeration is available Real systems take CPU cost into account, differentiate between sequential and random I/O, and take buffer size into account We do not include cost to writing outut to disk in our cost formulae 16/5/25 School of C.S., Fudan Univ. 10

11 Selection σ (r) or select * from r where

12 Selection Oeration File scan search algorithms that locate and retrieve records that fulfill a selection condition. Algorithm A1 (linear search) n Cost estimate (number of disk blocks scanned) = b r n If selection is on a key attribute, average cost = (b r /2) l sto on finding record n Linear search can be alied regardless of l selection condition or l ordering of records in the file, or l availability of indices 16/5/25 School of C.S., Fudan Univ. 12

13 Selection Oeration (Cont.) A2 (binary search) n Alicable if selection is an equality comarison on the attribute on which file is ordered. n Assume that the blocks of a relation are stored contiguously n Cost estimate (number of disk blocks to be scanned): l log 2 (b r ) cost of locating the first tule by a binary search on the blocks l Plus number of blocks containing records that satisfy selection condition Will see how to estimate this cost in next lecture 16/5/25 School of C.S., Fudan Univ. 13

14 Selections Using Indices Index scan search algorithms that use an index n selection condition must be on search-key of index. A3 (rimary index on candidate key, equality) n Retrieve a single record that satisfies the corresonding equality condition n Cost = HT i + 1 (B + -tree) A4 (rimary index on nonkey, equality) : Retrieve multile records n Records will be on consecutive blocks n Cost = HT i + number of blocks containing retrieved records 16/5/25 School of C.S., Fudan Univ. 14

15 Selections Using Indices (Cont.) A5 (equality on search-key of secondary index) n Retrieve a single record if the search-key is a candidate key l Cost = HT i + 1 n Retrieve multile records if search-key is not a candidate key l Cost = HT i + number of records retrieved Can be very exensive! l Each record may be on a different block One block access for each retrieved record 16/5/25 School of C.S., Fudan Univ. 15

16 Selections Involving Comarisons Can imlement selections of the form σ A V (r) or σ A V (r) by using n a linear file scan or binary search, n or by using indices in the following ways: A6 (rimary index, comarison) n Relation is sorted on A n For σ A V (r) use index to find first tule v and scan relation sequentially from there n For σ A V (r) just scan relation sequentially till first tule > v; do not use index 16/5/25 School of C.S., Fudan Univ. 16

17 Selections Involving Comarisons (Cont.) A7 (secondary index, comarison) n For σ A V (r) use index to find first index entry v and scan index sequentially from there, to find ointers to records. n For σ A V (r) just scan leaf ages of index finding ointers to records, till first entry > v n In either case, retrieve records that are ointed to l requires an I/O for each record l Linear file scan may be cheaer if many records are to be fetched! 16/5/25 School of C.S., Fudan Univ. 17

18 Imlementation of Comlex Selections Conjunction ( 合取 ): σ θ 1 θ 2... θ n(r) A8 (conjunctive selection using one index) n Select a condition of θ i and algorithms A1 through A7 that results in the least cost forσ θi (r). n Test other conditions on tule after fetching it into memory buffer. A9 (conjunctive selection using multile-key index) n Use aroriate comosite (multile-key) index if available. 16/5/25 School of C.S., Fudan Univ. 18

19 Imlementation of Comlex Selections (Cont.) A10 (conjunctive selection by intersection of identifiers) n Requires indices with record ointers. n Use corresonding index for each condition, and take intersection of all the obtained sets of record ointers. n Then fetch records from file n If some conditions do not have aroriate indices, aly test in memory. 16/5/25 School of C.S., Fudan Univ. 19

20 Imlementation of Comlex Selections (Cont.) Disjunction ( 析取 ):σ θ 1 θ 2... θ n (r). A11 (disjunctive selection by union of identifiers) n Alicable if all conditions have available indices. l Otherwise use linear scan. n Use corresonding index for each condition, and take union of all the obtained sets of record ointers. n Then fetch records from file 16/5/25 School of C.S., Fudan Univ. 20

21 Imlementation of Comlex Selections (Cont.) Negation: σ θ (r) n Use linear scan on file n If very few records satisfy θ, and an index is alicable to θ l Find satisfying records using index and fetch from file n Examle: σ (A 0) (r)= σ A=0 (r) 16/5/25 School of C.S., Fudan Univ. 21

22 Sorting

23 Why sorting? Sorting n For a SQL query with order by clause, query answers must be returned in order n For join oeration with two relations, if the relations are sorted by the join attributes, the join oeration can be evaluated very efficiently n Sorting for constructing ordered indices 16/5/25 School of C.S., Fudan Univ. 23

24 Sorting How to sort relations? n For relations that fit in memory, techniques like quick-sort can be used n We may build an index on the relation, and then use the index to read the relation in sorted order. May lead to one disk block access for each tule (for non-rimary indices) n For relations that don t fit in memory, external sort-merge is a good choice 16/5/25 School of C.S., Fudan Univ. 24

25 External Sort-Merge Let M denote memory size (in ages) 1. Create sorted runs. Let i be 0 initially. Reeatedly do the following till the end of the relation: (a) Read M blocks of relation into memory (b) Sort the in-memory blocks (c) Write sorted data to run R i ; increment i. Let the final value of i be N 2. Merge the runs (N-way merge). We assume (for now) that N < M. 1. Use N blocks of memory to buffer inut runs, and 1 block to buffer outut. Read the first block of each run into its buffer age 2. reeat 1. Select the first record (in sort order) among all buffer ages 2. Write the record to the outut buffer. If the outut buffer is full write it to disk. 3. Delete the record from its inut buffer age. If the buffer age becomes emty then read the next block of the run into the buffer. 3. until all inut buffer ages are emty: 16/5/25 School of C.S., Fudan Univ. 25

26 External Sort-Merge INPUT 1... INPUT 2... Disk Make runs INPUT M M Main memory buffers INPUT 1 Disk... INPUT 2... OUTPUT INPUT N Merge Disk M Main memory buffers Disk 16/5/25 School of C.S., Fudan Univ. 26

27 External Sort-Merge (Cont.) If N M, several merge asses are required. n In each ass, contiguous grous of M - 1 runs are merged. n A ass reduces the number of runs by a factor of M -1. l E.g. If M=11, and there are 90 runs, one ass reduces the number of runs to 9, each 10 times the size of the initial runs n Reeated asses are erformed till all runs have been merged into one. 16/5/25 School of C.S., Fudan Univ. 27

28 Examle: External Sorting Using Sort-Merge Sort on the first column! M=3 16/5/25 School of C.S., Fudan Univ. 28

29 External Merge Sort (Cont.) Cost analysis: n Total number of merge asses required: log M 1 (b r /M). n Disk accesses for initial run creation as well as in each ass is 2b r l for final ass, we don t count write cost we ignore final write cost for all oerations since the outut of an oeration may be sent to the arent oeration without being written to disk Thus total number of disk accesses for external sorting: b r ( 2 log M 1 (b r / M) + 1)=b r ( 2 log M 1 (b r / M) )-b r + 2b r 16/5/25 School of C.S., Fudan Univ. 29

30 Join n Nested-loo join n Block nested-loo join n Indexed nested-loo join n Merge-join n Hash-join

31 Join Oeration Several different algorithms to imlement joins n Nested-loo join n Block nested-loo join n Indexed nested-loo join n Merge-join n Hash-join Choice based on cost estimate Examles use the following information n Number of records of customer: 10,000 deositor: 5000 n Number of blocks of customer: 400 deositor: /5/25 School of C.S., Fudan Univ. 31

32 Tule Nested-Loo Join To comute the theta join r θ s for each tule t r in r do begin for each tule t s in s do begin test air (t r,t s ) to see if they satisfy the join condition θ if they do, add t r t s to the result. end end r is called the outer relation and s the inner relation Requires no indices and can be used with any kind of join condition. Exensive since it examines every air of tules in the two relations. 16/5/25 School of C.S., Fudan Univ. 32

33 Tule nested-loo Join (Cont.) In the worst case, the estimated cost is n r b s + b r disk accesses. If the best case, reduces cost to b r + b s. Assuming worst case memory availability cost estimate is n = 2,000,100 disk accesses with deositor as outer relation, and n = 1,000,400 disk accesses with customer as the outer relation. If smaller relation (deositor) fits entirely in memory, the cost estimate will be 500 disk accesses. Block nested-loos algorithm (next slide) is referable. 16/5/25 School of C.S., Fudan Univ. 33

34 Block Nested-Loo Join Variant of nested-loo join in which every block of inner relation is aired with every block of outer relation. for each block B r of r do begin for each block B s of s do begin for each tule t r in B r do begin for each tule t s in B s do begin Check if (t r,t s ) satisfy the join condition if they do, add t r t s to the result. end end end end 16/5/25 School of C.S., Fudan Univ. 34

35 Block Nested-Loo Join (Cont.) r join s Hash table for block of r (M-2 blocks)... Join Result Inut buffer for s Outut buffer 16/5/25 School of C.S., Fudan Univ. 35

36 Block Nested-Loo Join (Cont.) Worst case estimate: b r b s + b r block accesses. Best case: b r + b s block accesses. Imrovements : n In block nested-loo, use (M 2) disk blocks as blocking unit for outer relations, where M = memory size in blocks; use remaining two blocks to buffer inner relation and outut l Cost = b r / (M-2) b s + b r n If equi-join attribute forms a key of inner relation, sto inner loo on first match n Scan inner relation forward and backward alternately, to make use of the blocks remaining in buffer. 16/5/25 School of C.S., Fudan Univ. 36

37 Indexed Nested-Loo Join Index lookus can relace file scans if n join is an equi-join or natural join and n an index is available on the inner relation s join attribute For each tule t r in the outer relation r, use the index to look u tules in s Cost of the join: b r + n r c n Where c is the cost of traversing index and fetching all matching s tules for one tule of r If indices are available on join attributes of both r and s, use the relation with fewer tules as the outer relation. 16/5/25 School of C.S., Fudan Univ. 37

38 Examle of Nested-Loo Join Costs Comute deositor customer. Let customer have a rimary B + -tree index on the join attribute customer-name, which contains 20 entries in each index node. Since customer has 10,000 tules (400 blocks), the height of the tree is 4 = [log_20 ^10,000]+1, and one more access to find the actual data deositor has 5000 tules -> 100 blocks Cost of block nested loos join n 400* = 40,100 disk accesses Cost of indexed nested loos join n * 5 = 25,100 disk accesses. 16/5/25 School of C.S., Fudan Univ. 38

39 Merge-Join 1. Sort both relations on their join attribute (if not already sorted on the join attributes). 2. Merge the sorted relations to join them 1. Join ste is similar to the merge stage of the sort-merge algorithm. 2. Main difference is handling of dulicate values in join attribute every air with similar value on join attribute must be matched 3. Detailed algorithm in book 16/5/25 School of C.S., Fudan Univ. 39

40 Merge-Join (Cont.) Can be used only for equi-joins and natural joins Each block needs to be read only once (assuming all tules for any given value of the join attributes fit in memory) Thus number of block accesses for merge-join is b r + b s + the cost of sorting if relations are unsorted. hybrid merge-join (combining indices with merge-join): If one relation is sorted, and the other has a secondary B + -tree index on the join attribute n n n Merge the sorted relation with the leaf entries of the B + -tree, the result file contains tules from the sorted file and the addresses from the unsorted file Sort the result file on the addresses of the unsorted relation s tules Scan the unsorted relation in hysical address order and merge with revious result, to relace addresses by the actual tules 16/5/25 School of C.S., Fudan Univ. 40

41 Hash-Join Alicable for equi-joins and natural joins. A hash function h is used to artition tules of both relations h mas JoinAttrs values to {0, 1,..., n}. n r 0, r 1,..., r n denote artitions of r tules n s 0, s 1..., s n denotes artitions of s tules Note: in the book, r i is denoted as H ri,, s i is denoted as H si and n is denoted as n h 16/5/25 School of C.S., Fudan Univ. 41

42 Hash-Join (Cont.) 16/5/25 School of C.S., Fudan Univ. 42

43 Hash-Join (Cont.) r tules in r i need only to be comared with s tules in s i. n an r tule and an s tule that satisfy the join condition will have the same value for the join attributes. n If that value is hashed to some value i, the r tule has to be in r i and the s tule in s i. 16/5/25 School of C.S., Fudan Univ. 43

44 Hash-Join Algorithm The hash-join of r and s is comuted as follows: 1. Partition the relation s using hashing function h. When artitioning a relation, one block of memory is reserved as the outut buffer for each artition 2. Partition r similarly 3. For each i: (a) Load s i into memory and build an in-memory hash index on it using the join attribute. This hash index uses a different hash function than the earlier one h. (b) Read the tules in r i from the disk one by one. For each tule t r locate each matching tule t s in s i using the in-memory hash index. Outut the concatenation of their attributes Relation s is called the build inut and r is called the robe inut 16/5/25 School of C.S., Fudan Univ. 44

45 Hash-Join Algorithm Partition both relations using hash fn h: r tules in artition i will only match s tules in artition i Original Relation... INPUT hash function h OUTPUT 1 n 2 Partitions 1 2 n Disk M main memory buffers Disk Read in a artition of r, hash it using h2 (<> h!). Scan matching artition of s, search for matches Partitions of r & s hash fn h2 Disk Hash table for artition si (k < M-1 ages) h2 Inut buffer For ri Outut buffer M main memory buffers Join Result Disk 16/5/25 School of C.S., Fudan Univ. 45

46 Hash-Join algorithm (Cont.) The value n (# artitions) and the hash function h is chosen such that each s i should fit in memory (say M blocks) n Tyically n is chosen as b s /M * f where f is a fudge factor, tyically around 1.2 n The robe relation artitions r i need not fit in memory Recursive artitioning required if number of artitions n is greater than number of ages M of memory. n instead of artitioning n ways, use M 1 artitions for s n Further artition the M 1 artitions using a different hash function n Use same artitioning method on r 16/5/25 School of C.S., Fudan Univ. 46

47 Handling of Overflows Hash-table overflow occurs in artition s i if s i does not fit in memory. Reasons could be n n Many tules in s with same value for join attributes Bad hash function Overflow resolution can be done in build hase n n Partition s i is further artitioned using different hash function. Partition r i must be similarly artitioned. Overflow avoidance erforms artitioning carefully to avoid overflows during build hase n E.g. artition build relation into many artitions, then combine them Both aroaches fail with large numbers of dulicates n Fallback otion: use block nested loos join on overflowed artitions 16/5/25 School of C.S., Fudan Univ. 47

48 Cost of Hash-Join If recursive artitioning is not required: cost of hash join is 2(b r + b s ) +(b r + b s ) + 4n If recursive artitioning required, number of asses required for artitioning s is log M 1 (b s ) 1. the number of asses for artitioning of r is also the same as for s. Therefore it is best to choose the smaller relation as the build relation. Total cost estimate is: 2(b r + b s ) log M 1 (b s ) 1 + b r + b s If the entire build inut can be ket in main memory, n can be set to 0 and the algorithm does not artition the relations into temorary files. Cost estimate goes down to b r + b s. 16/5/25 School of C.S., Fudan Univ. 48

49 Examle of Cost of Hash-Join customer deositor Assume that memory size is 22 blocks b deositor = 100 and b customer = 400. deositor is to be used as build inut. Partition it into five artitions, each of size 20 blocks. This artitioning can be done in one ass. Similarly, artition customer into five artitions,each of size 80. This is also done in one ass. Therefore total cost: 3( ) = 1500 block transfers n ignores cost of writing artially filled blocks 16/5/25 School of C.S., Fudan Univ. 49

50 Hybrid Hash Join Useful when memory size are relatively large, and the build inut is bigger than memory. Main feature of hybrid hash join: Kee the first artition of the build relation in memory. E.g. With memory size of 25 blocks, deositor can be artitioned into five artitions, each of size 20 blocks. Division of memory: n n The first artition occuies 20 blocks of memory 1 block is used for inut, and 1 block each for buffering the other 4 artitions. customer is similarly artitioned into five artitions each of size 80; the first is used right away for robing, instead of being written out and read back. Cost of 3( ) = 1300 block transfers for hybrid hash join, instead of 1500 with lain hash-join. Hybrid hash-join most useful if M >> bs 16/5/25 School of C.S., Fudan Univ. 50

51 Comlex Joins Join with a conjunctive condition: n r θ1 θ 2... θ n s Either use nested loos/block nested loos, or n Comute the result of one of the simler joins r θi s l final result comrises those tules in the intermediate result that satisfy the remaining conditions θ 1... θ i 1 θ i θ n Join with a disjunctive condition n r θ1 θ2... θns Either use nested loos/block nested loos, or n Comute as the union of the records in individual joins r θ i s: (r θ1s) (r θ2 s)... (r θn s) 16/5/25 School of C.S., Fudan Univ. 51

52 Comlex Joins Join involving three relations: loan deositor customer Strategy 1. Comute deositor customer; use result to comute loan (deositor customer) Strategy 2. Comuter loan deositor first, and then join the result with customer. Strategy 3. Perform the air of joins at once. Build and index on loan for loan-number, and on customer for customer-name. n n For each tule t in deositor, look u the corresonding tules in customer and the corresonding tules in loan. Each tule of deositor is examined exactly once. Strategy 3 combines two oerations into one secial-urose oeration that is more efficient than imlementing two joins of two relations. 16/5/25 School of C.S., Fudan Univ. 52

53 Other Oerations

54 Other Oerations Dulicate elimination can be imlemented via hashing or sorting. n On sorting dulicates will come adjacent to each other, and all but one coy of dulicates can be deleted. Otimization: dulicates can be deleted during run generation as well as at intermediate merge stes in external sort-merge. n Hashing is similar dulicates will come into the same bucket. Projection is imlemented by erforming rojection on each tule followed by dulicate elimination. 16/5/25 School of C.S., Fudan Univ. 54

55 Other Oerations: Aggregation Aggregation can be imlemented in a manner similar to dulicate elimination. n Sorting or hashing can be used to bring tules in the same grou together, and then the aggregate functions can be alied on each grou. n Otimization: combine tules in the same grou during run generation and intermediate merges, by comuting artial aggregate values l For count, min, max, sum: kee aggregate values on tules found so far in the grou. l For avg, kee sum and count, and divide sum by count at the end 16/5/25 School of C.S., Fudan Univ. 55

56 Other Oerations: Set Oerations Set oerations (, and ): can either use variant of mergejoin after sorting, or variant of hash-join. E.g., Set oerations using hashing: 1. Partition both relations using the same hash function, thereby creating, r 1,.., r n and s 1, s 2.., s n 2. Process each artition i as follows. Using a different hashing function, build an in-memory hash index on r i after it is brought into memory. 3. r s: Add tules in s i to the hash index if they are not already in it. At end of s i, add the tules in the hash index to the result. r s: outut tules in s i to the result if they are already there in the hash index. r s: for each tule in s i, if it is there in the hash index, delete it from the index. At end of s i, add remaining tules in the hash index to the result. 16/5/25 School of C.S., Fudan Univ. 56

57 Other Oerations: Outer Join Outer join can be comuted either as n n A join followed by addition of null-added non-articiating tules. by modifying the join algorithms. Modifying merge join to comute r s n In r s, non articiating tules are those in r Π R (r s) n Modify merge-join to comute r s: During merging, for every tule t r from r that do not match any tule in s, outut t r added with nulls. n Right outer-join and full outer-join can be comuted similarly. Modifying hash join to comute r s n n If r is robe relation, outut non-matching r tules added with nulls If r is build relation, when robing kee track of which r tules matched s tules. At end of s i outut non-matched r tules added with nulls 16/5/25 School of C.S., Fudan Univ. 57

58 Evaluation of Exressions

59 Evaluation of Exressions So far: we have seen algorithms for individual oerations Alternatives for evaluating an entire exression tree n Materialization: generate results of an exression whose inuts are relations or are already comuted, materialize (store) it on disk. n Pielining: ass on tules to arent oerations even as an oeration is being executed We study above alternatives in more detail 16/5/25 School of C.S., Fudan Univ. 59

60 Materialization Materialized evaluation: evaluate one oeration at a time, starting at the lowest-level. E.g., in figure below, comute and store σ ( account balance<2500 ) then comute and store the revious result join with customer, and finally comute the rojections on customer-name. 16/5/25 School of C.S., Fudan Univ. 60

61 Materialization (Cont.) Materialized evaluation is always alicable Cost of writing results to disk and reading them back can be quite high n Our cost formulas for oerations ignore cost of writing results to disk, so l Overall cost = Sum of costs of individual oerations + cost of writing intermediate results to disk Double buffering: use two outut buffers for each oeration, when one is full write it to disk while the other is getting filled n Allows overla of disk writes with comutation and reduces execution time 16/5/25 School of C.S., Fudan Univ. 61

62 Pielining Pielined evaluation : evaluate several oerations simultaneously, assing the results of one oeration on to the next. E.g., in revious exression tree, don t store result of σ balance<2500 ( account ) n instead, ass tules directly to the join.. Similarly, don t store result of join, ass tules directly to rojection. Much cheaer than materialization. Pielining may not always be ossible e.g., sort, hash-join. Pielines can be executed in two ways: demand driven and roducer driven 16/5/25 School of C.S., Fudan Univ. 62

63 Pielining (Cont.) In demand driven or lazy evaluation n n n n system reeatedly requests next tule from to level oeration Each oeration requests next tule from children oerations as required, in order to outut its next tule Between calls, oeration has to maintain state so it knows what to return next Each oeration is imlemented as an iterator imlementing the following oerations l oen() l E.g. file scan: initialize file scan, store ointer to beginning of file as state E.g.merge join: sort relations and store ointers to beginning of sorted relations as state next() E.g. for file scan: Outut next tule, and advance and store file ointer E.g. for merge join: continue with merge from earlier state till next outut tule is found. Save ointers as iterator state. l close() 16/5/25 School of C.S., Fudan Univ. 63

64 Pielining (Cont.) In roducer-driven or eager ielining n Oerators roduce tules eagerly and ass them u to their arents l Buffer maintained between oerators, child uts tules in buffer, arent removes tules from buffer l if buffer is full, child waits till there is sace in the buffer, and then generates more tules n System schedules oerations that have sace in outut buffer and can rocess more inut tules 16/5/25 School of C.S., Fudan Univ. 64

65 Assignments Practice exercises: 13.3, 13.6 Exercises: 13.11, /5/25 School of C.S., Fudan Univ. 65