HASHING METHODS. Hanan Samet

Transcription

1 hs0 HASHING METHODS Hanan Samet Compute Science Depatment and Cente fo Automation Reseach and Institute fo Advanced Compute Studies Univesity of Mayland College Pak, Mayland Copyight 1996 Hanan Samet These notes may not be epoduced by any means (mechanical o electonic o any othe) without the expess witten pemission of Hanan Samet

2 hs1 HASHING OVERVIEW Task: compae the value of a key with a set of key values in a table Conventional solutions: 1. use a compaison on key values (tee-based) 2. banching pocess govened by the digits compising the key value (tie-based) Altenative solution is to find a 1-1 mapping (i.e., function) fom set of possible key values to a memoy addess and use table lookup methods to etieve the ecod O (1) pocess Poblem: the set of possible key values is much lage than the numbe of available memoy addesses 1. developing the 1-1 function h is time-consuming as it equies puzzle-solving abilities esult is called a pefect hashing function 2. once h is found, addition of a single key value may ende the function meaningless need to develop it anew 3. can eplace h by a pogam, which may itself be timeconsuming to compute Result: usually abandon goal of finding 1-1 mapping and use a special method to esolve any ambiguity (i.e., when moe than one key value is mapped to the same addess temed a collision)

3 hs2 HASHING Def: to mess things up Hashing function h(k) is used to calculate addess whee to stat the seach fo the ecod with key value k Issues 1. what kind of a function is h(k)? easy and fast to compute minimize the numbe of collisions 2. what if h(k) does not yield the desied esult? how to handle collisions Assume table of size m and 0 h(k) < m Example hashing functions: 1. division techniques often use h(k) = k mod m choice of m is impotant a. m even bad as h(k) even when k even and odd when k odd b. m is a powe of the adix of alphanumeic set of chaacte values bad as only least significant chaactes matte with m= 3, ABCDEF, IJKDEF, and KLMDEF all hash to the same location c. usually choose m to be pime 2. multiplicative techniques entie key value is used examples: a. multiply fields and take modulo b. add o exclusive-o of fields

4 SEPARATE CHAINING Hash table of size m 3 z 2 1 b hs3 One chain (linked list) fo each of m hash values containing all elements that hash to that location (known as a collision list ) Hash chains ae known as buckets Hash table locations ae known as bucket addesses Fo n key values, aveage chain size is n/m One chain (linked list) fo each of m hash values Retieval 1. use sequential seach though chain 2. speed up unsuccessful seach by soting chain by key value 3. speed up successful seach by self-oganizing methods move key value to stat of chain each time it is accessed Ex: h(k) NAME k=key NEXT 0 JIM 49 Λ 1 JOHN 22 Λ 2 RAY 30 Λ 3 SUZY 3 Λ LUCY 41 Λ JANE 14 Λ 1. add JANE(14) 0 2. add LUCY(41) 6

5 IN-PLACE CHAINING When m is lage, many of the chains ae empty 4 g 3 z 2 1 b hs4 Use empty locations in table fo the chain Must be able to distinguish between fee and occupied locations Insetion algoithm: 1. if key value not pesent, then allocate a fee location 2. link location to chain which was unsuccessfully seached Ex: h(k) NAME k=key NEXT 0 JIM 49 Λ 1 JOHN 22 Λ 2 RAY 30 Λ 3 SUZY 3 Λ LUCY JANE JANE LUCY Λ Λ Λ 1. add JANE(14) 0 which collides with JIM(49) 0 2. add LUCY(41) 6 which collides with JANE(14) 0 which is stoed at 6 esult in coalescing of chains of JANE and LUCY making unsuccessful seach longe as seveal chains must be seached Can avoid coalescing by moving JANE just befoe adding LUCY

6 hs5 IN-PLACE CHAINING INSERTION ALGORITHM location pocedue CHAINING_WITH_COALESCING_INSERTION(k); begin value key k; intege i; global intege ; /* is the most ecently allocated location */ global hashtable table; i h(k); if OCCUPIED(table[i]) then begin while NOT(NULL(NEXT(table[i])) do begin if k=key(table[i]) then etun(i) else i NEXT(table[i]); end; if k=key(table[i]) then etun(i); while OCCUPIED(table[]) do -1; if 0 then etun(`overflow') else begin NEXT(table[i]) ; i ; end; end; MARK(table[i],`OCCUPIED'); KEY(table[i]) k; NEXT(table[i]) NIL; etun(i); end;

7 LAMPSON S IN-PLACE CHAINING hs6 Avoid exta space fo NEXT field by not stoing entie key value with ecod k = m q(k) + h(k), q(k) = k/m, h(k) = k mod m Stoe q(k) in table instead of k Can compute k given m, q(k), and h(k), Ex: 0 k < 2 32 q(k) h(k) Since only compae q(k), all elements in same collision list must have the same value of h(k) and thus no coalescing is allowed Data stuctue: 1. cicula collision lists 2. flag FIRST denoting if fist element on collision list 3. pointe NEXT to next element in cicula list with same h(k) value Ex: h(k) NAME k=key FIRST q(k) NEXT 0 JIM 49 T JOHN 22 T RAY 30 T SUZY 3 T JANE 14 F JANE LUCY F T add JANE(14) 0 2. add LUCY(41) 6 but 6 contains JANE if at least one element of the hash chain stating at 6 exists, then it must be stoed thee must move JANE as it does not belong in 6 Nice compomise between use of a key value as an index to a table, which is impossible due to lage numbe of possible key values, and stoing the entie key value as in a conventional hashing method 4 g 3 z 2 1 b

8 OPEN ADDRESSING hs7 Like chaining but NEXT link field is open o unspecified Pobe sequence: set of locations compising collision list of a key Goal: cycle though all locations with little o no duplication Linea pobing: h(k), h(k)+1, h(k)+2,, m 1, 0, 1, h(k) 1 Insetion Algoithm: 1. calculate hash addess i 2. if TABLE(i ) is empty then inset and exit; else i i+1 mod m and epeat step 2 until exhausting TABLE 6 z 5 4 g 3 z 2 1 b Ex: h(k) NAME k=key 0 JIM 49 1 JOHN 22 2 RAY 30 3 SUZY 3 4 JANE LUCY 41 DELETED N N Y N N N N 1. adding JANE(14) 0 yields a collision; cyclic pobe sequence causes its insetion in 4 2. adding LUCY(41) 6 3. delete RAY(30) 2 poblem: if look up JANE then don t find he since a collision exists at location 0, and pobe sequence finds location 2 unoccupied solution: add DELETED flag to each enty to halt the seach duing insetion but not duing lookup

9 hs8 PILEUP PHENOMENON Next key to be inseted goes into one of the vacant locations Not all vacant locations ae equally pobable Ex: inset k into location 3 if 0 h(k) 3 inset k into location 6 if h(k)=6 3 is fou times as likely as 6 Coalescing in open addessing with linea pobing can lead to big gowth (e.g., inseting into location 9 makes the list of 8 gow by 4) Diffeent fom in-place chaining whee coalescing causes a list to gow by only one element Pileup phenomenon aises wheneve consecutive values ae likely to occu Ovecome by a numbe of techniques: 1. use additive constant instead of 1 should be elatively pime to m so can cycle though table 2. use a pseudo andom numbe geneato to ceate successive offsets fom h(k) (andom pobing) make sue cycle though table unifom hashing = when all possible configuations of empty and occupied locations ae equally likely = model of hashing fo compaing vaious methods

10 hs9 ANALYSIS OF PERFORMANCE n= numbe of key values in table m= maximum size of table α = n/m = load facto Expected numbe of pobes fo successful seach: α linea pobing andom pobing (unde unifom hashing model) sepaate chaining (1 α/2)/(1-α) (ln (1-α))/α 1+α/

11 hs10 QUADRATIC PROBING Altenative to linea pobing Avoids pimay clusteing h i = (h(k)+i 2) mod m Locations in the pobe sequence can be computed with no multiplication 1. h i is location of element i 2. h i+1 = (h i + d i mod m ) whee d 0 = 1 and d i+1 = d i +2 Theoem: if m is pime, then quadatic pobing will seach though at least 50% of the table befoe seeing a paticula location again Ex: m=7 h 0 = 0, h 1 = 1, h 2 = 4 h 3 = 9 mod 7 = 2 and h 4 = 16 mod 7 =2 Poof: 1. let pobes i, j pobe the same location (assume i j ) 2. i 2 mod m = j 2 mod m 3. (i 2 j 2) mod m = (i+j) (i j) mod m = 0 mod m 4. but i,j ae both < m implying (i j ) c m 5. theefoe, i+j = c m and i o j must be at least m/2 since pobe sequence stats with i =1, and ecycling of values won t occu until at least 50% of table has been seached Sequence diffes fom one obtained fom the pseudo andom numbe geneato as the pseudo andom numbe geneato guaantees that evey location will be on a pobe sequence

12 hs11 DOUBLE HASHING Use an additional hash function g(k) to geneate a constant incement fo the pobe sequence Pobe sequence fo key value k: p 0 = h(k) p 1 = (h(k) + g(k)) mod m p 2 = (h(k) + 2 g(k)) mod m p 3 = (h(k) + 3 g(k)) mod m p i = (h(k) + i g(k)) mod m h(k) and g(k) should be independent g(k) geneates values between 1 and m 1 Two diffeent key values will have the same value fo h and g with pobability O (1/m 2) instead of O (1/m) Key value k is stoed at any one of the locations along its pobe sequence Key values stoed along the pobe sequence of k ae not necessaily pat of k s pobe sequence Ex: key values s and t can both hash to location u key value s : u = (h (s) + c g (s)) mod m key value t : u = (h (t ) + d g (t )) mod m

13 SELF-ORGANIZING DOUBLE HASHING 3 z 2 1 b hs12 Collision lists ae long as each location is fequently on the collision lists of many diffeent key values Develop techniques fo eaanging the elements on the collision lists so that subsequent seaches ae shote Assume ecods ae etieved many times once inseted into the table hence it pays to eaange the collision lists Assume tying to inset key value k and pobe locations p 0, p 1, p i, p t, whee p i = (h(k)+i g(k)) mod m befoe finding location p t empty p i = (h(k)+i g(k)) mod m Each p i (0 i <t ) is also pat of a hash chain consisting of locations: (p i +j g(key(p i ))) mod m fo abitay j Assume c i = g(key(p i )), and p i = h(key(p i )) mod m: p 0 p 1 p 2 p 3 p 4 p t p 0 + c 0 1 p 1 + c 1 2 p 2 + c 2 4 p 3 + c 3 7 p 4 + c 4 11 p 0 +2c 0 3 p 1 +2c 1 5 p 2 +2c 2 8 p 3 +2c 3 12 p 0 +3c 0 6 p 1 +3c 1 9 p 2 +3c 2 13 p 0 +4c 0 10 p 1 +4c 1 14 p 0 +5c 0 15 Actually, p i = (h(key(p i )) +d i g(key(p i ))) mod m Bent algoithm: ty to inset key value k in one of p i and move the contents of p i to an empty location along its pobe sequence (column) so as to minimize the effective incemental seach cost Ode fo testing candidate locations fo moving (along diagonals)

14 EXAMPLE OF BRENT ALGORITHM Attempt to inset RUDY 9 g 8 b 7 6 v 5 g 4 z 3 v 2 1 b hs13 p 0 =TIM p 1 =ALAN p 2 =JAY p 3 =KATY p 4 =? RUDY RUDY RUDY RUDY JOAN RUTH JAY KIM RON ALAN ALEX RITA BOB TIM KATY Fist fee location in RUDY s pobe sequence is p 4 fo an incease in cost of 4 Examine locations in diagonal ode fo fist fee location Altenatively, find fist fee location in each hash chain and check if oveall seach cost is inceased by a movement of it Movement must esult in an incease <4 in the seach cost Moving TIM inceases the seach cost by 4 Moving ALAN inceases its seach cost by 2, while inceasing that of RUDY by 1 fo a total of 3 Moving JAY inceases its seach cost by 1, while inceasing that of RUDY by 2 fo a total of 3 Moving KATY inceases its seach cost by 4, while inceasing that of RUDY by 3 fo a total of 7 Move JAY as its incease (3) was the least and was encounteed fist Need 2.5 pobes on the aveage fo successful seach Aveage numbe of pobes fo unsuccessful seach is not educed (as high as (m+1)/2 when the table is full)

15 hs14 GONNET-MUNRO ALGORITHM Moe geneal than the Bent algoithm Bent algoithm only attempts to move ecods on the pobe sequence of the ecod being inseted p 0 =TIM p 1 =ALAN p 2 =JAY p 3 =KATY p 4 =? JOAN RUTH KIM RON ALEX RITA BOB Gonnet-Muno attempts to move in seveal stages instead of just one stage 1. RUDY to TIM, TIM to JOAN, and JOAN to? 2. RUDY to ALAN, ALAN to RUTH, and RUTH to? Need emaining hash chains JOAN RITA KIM RUTH RON ALEX BOB JAY KIM KATY TIM ALAN JAY Can visualize seach fo best movement as a binay tee Best movement is the closest empty node to the oot

16 EXAMPLE OF GONNET-MUNRO ALGORITHM Attempt to inset RUDY RUDY, TIM 5 v 4 g 3 z 2 1 b hs15 denote puning due to epetition ALAN JOAN JAY RUTH RON KATY KIM RITA KIM JAY ALEX TIM JAY ALAN KATY * BOB Right son of a is next element in pobe sequence of KEY(a) Left son of a is next element in pobe sequence of a and a s fathe Seach geneates the tee level by level and chooses the fist empty node as the final taget fo a sequence of elocation steps RUDY can be elocated to any position in the leftmost pat of the tee Apply the elocation step as many times as needed to get to desied empty node Optimal solution moves RUDY to TIM, TIM to JOAN, and JOAN to its empty ight son 1. incease in total seach cost is 2 2. bette than 3 obtained by Bent algoithm Bent algoithm only applies one step and thus must find the empty node in just one iteation 1. move RUDY to JAY; JAY to empty ight son of JAY 2. incease in total seach cost is 3

17 SHORTCOMING OF GONNET-MUNRO ALGORITHM 1 b hs16 Only moves ecods in fowad diection along thei hash chains Sometimes can educe the cost by moving backwad along the chain Ex: suppose ALAN is not in the fist position along the hash chain stating at h(alan) mod m and that BOB immediately pecedes ALAN along the hash chain, although h(alan) h(bob) BOB p 1 =ALAN RUTH BOB BOB JAY ALAN RUDY RUTH Optimal solution moves RUDY to ALAN, ALAN to BOB, and BOB to its empty son 1. incease in total seach cost is 1 2. bette than 2 obtained by Gonnet-Muno algoithm Requies a tenay tee 1. need an additional link fom a to the pevious element in the pobe sequence of KEY(a) e.g., ALAN to BOB 2. seach pocess intepets pevious links as indicating a decease in cost 3. if incoming link to a is a pevious element link, then left son of a is the pio element in pobe sequence of a and its fathe 4. seach pocess is moe complex and empty node at closest level to the oot no longe epesents the cheapest elocation sequence 5. optimal solution may equie exhaustive seach TIM 3 z 2 JOAN

18 hs17 SUMMARY Advantages 1. sepaate chaining is supeio with espect to the numbe of pobes but need moe space 2. open addessing with linea pobing esults in moe accesses but this is compensated by its simplicity 3. compaes favoably with othe seach methods as the seach time is bounded as the numbe of ecods inceases (povided the table does not become too full) Disadvantages 1. size of hash table is usually fixed have to woy about ehashing sepaate chaining with oveflow buckets is good use linea hashing o spial hashing which just split one bucket and ehash its contents instead of ehashing the entie table 2. afte an unsuccessful seach we only know that the ecod is not pesent we don t know about the pesence o absence of othe ecods with simila key values such as the immediate pedecesso o successo contast with B-tees and othe methods based on binay seach which maintain the natual ode of the key values and pemit pocessing along this ode ode-peseving hashing methods such as those used to deal with multiattibute data (and spatial data) ae an exception 3. deletion may be cumbesome (e.g., open addessing) 4. only efficient on the aveage contast with B-tee methods which have guaanteed uppe bounds on seach time, etc. need faith in pobability theoy!