數(shù)據(jù)結(jié)構(gòu)與算法分析課后題答案

1、 Chapter5: Hashingm5.1 (a) On the assumption that we add collisions to the end of the list (which is the easierway ifoa hash table is being built by hand), the separate chaining hash table that results is shownhere.c01234567.8w網(wǎng)a案d答9419996791989后h課(b)k0123456789.www4199-25-(c)m0123456789oc.w4199網(wǎng)a(d

2、) 1989 cannot be inserted into the table because hash (1989) = 6, and the alternative locations案25, 1, 7, and3 arealreadytaken. Thetableat this point is as follows:d答0后h14371課23456789k.43444199ww5.2 When rehashing, we choose a table size that is roughly twice as large and prime. In ourwcase,the appr

3、opriatenewtablesizeis 19, with hash function h(x) = x(mod 19).(a) Scanning down the separate chaining hash table, the new locations are 4371 in list 1,1323 in list 12, 6173 in list 17, 4344 in list 12, 4199 in list 0, 9679 in list 8, and 1989 in list13.(b) The new locations are 9679 in bucket 8, 437

4、1 in bucket 1, 1989 in bucket 13, 1323 inbucket 12, 6173 in bucket 17, 4344 in bucket 14 because both 12 and 13 are already occu-pied, and4199 in bucket0.-26-(c) The new locations are 9679 in bucket 8, 4371 in bucket 1, 1989 in bucket 13, 1323 inbucket 12, 6173 in bucket 17, 4344 in bucket 16 becaus

5、e both 12 and 13 are already occu-pied, and4199 in bucket0.m(d) The new locations are 9679 in bucket 8, 4371 in bucket 1, 1989 in bucket 13, 1323 inbucket 12, 6173 in bucket 17, 4344 in bucket 15 because 12 is already occupied, and 4199in bucket0.o5.4 We must be carefulnot to rehash too often. Let p

6、 be the threshold (fractionof table size)atwhich we rehash to a smaller table. Then if the new table has size N, it contains 2pN ele-ments. This table will require rehashing after either 2N 2pNBalancing these costs suggests that a good choice is p = 2/ 3. For instance, suppose wehave a table of size

7、 300. If we rehash at 200 elements, then the new table size is N = 150,insertions or pN deletions.candwecando either100 insertions or 100 deletions u.ntil anewrehashis required.If we know that insertions are more frequent than deletions, then we might choose p to besomewhat larger. If p is too close

8、 to 1.0, however, then a sequence of a small number ofdeletions followed by insertions can cause frequent rehashing. In the worst case, if p = 1.0,then alternatingdeletions andinsertions both requirerehashing.a5.5 (a) Since each table slot is eventually probed, if the table is not empty, the collisi

9、on can be網(wǎng)ww-27-O(M +N) space.Another method of implementing these operations is to use a search tree instead of a hashtable; a balanced tree is required because elements are inserted in the tree with too muchmorder. A splay tree might be particularly well suited for this type of a problem because i

10、tdoes well with sequential accesses. Comparing the differentways of solving the problem isagood programmingassignment.o5.8 The table size would be roughly 60,000 entries. Each entry holds 8 bytes, for a total of480,000 bytes.c(d) As discussed in part (b), the algorithm will fail to detect one in ten

11、 misspellings on aver-5.9 (a)This statement is true.(b) If a word hashes to a location with value 1, there is no guarantee that the word is in thedictionary. It is possible that it just hashes to the same value as some other word in the dic-.tionary. In our case, the table is approximately 10% full

12、(30,000 words in a table of(e) A 20document would have about 60 misspellings. This algorithm would beexpected to detect 54. A table three times as large would still t in about 100K bytes ander of errors to two. This is good enough for many applications,300,007), so there is a 10% chance that a word

13、that is not in the dictionary happens to hashout to alocation with value1.w(c)300,007 bits is 37,501 bytes on most machines.網(wǎng)age.a案reduce the expected numbd答especially since spelling detection is a very inexact science. Many misspelled words (espe-cially short ones) are still words. For instance, ty

14、ping them instead of then is a misspelling后that wont bedetehctedby anyalgorithm.5.10 To each hash tab課le slot, we can add an extra eld that well call WhereOnStack, and we cankeepan extrastack. Whenaninsertion is rst performedinto aslot, wepush the address(ornumber)of the slot onto the stack and set the WhereOnStack eld to point to the top of thestack. When we accessa hash table slot, we checkthat WhereOnStack points to avalid partof the stack and that t