in Algorithms
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2 votes
2 votes
While inserting keys 12,44,13,88,23,94,11,39,20,16 and 5 in a 11 item hash table using the hash function  $h(i) =  (2i+5) \mod 11$, total number of collisions that occur is _________ (On collision no insertion takes place)
in Algorithms
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1 Answer

4 votes
4 votes
Hashvalues will be

12 - 29 mod 11 = 7
44 - 93 mod 11 = 5
13 - 31 mod 11 = 9
88 - 181 mod 11 = 5 - collission
23 - 51 mod 11 = 7 - collision
94 - 193 mod 11 = 6
11 - 27 mod 11 = 5 - collision
39 - 83 mod 11 = 6 - collision
20 - 45 mod 11 = 1
16 - 37 mod 11 = 4
5 - 15 mod 11 = 4 - collision

So, 5 collisions.
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4 Comments

No need to follow !! just count the collisions .
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m asking when coliisio happed ex: 45 is collide then where to push 45 in hash table,
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@Anirudh yes, that must have been mentioned in question - otherwise its ambiguous. Added now.
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Answer:

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