Question: Insert the keys 211, 376, 259, 127, 69, 84, 261, 273, 217 into the hash table of length m=13 using open addressing with the hash
Insert the keys 211, 376, 259, 127, 69, 84, 261, 273, 217 into the hash table of length m=13 using open addressing with the hash function h1(k)=k mod m. Use double hashing as the collision resolution strategy with h2(k)=R-(k mod R) where R is determined by following our in-class discussion; calculate the collision frequencies (CF) into the table!
R=11 since 11 is the greatest prime less than m. Hence, h2(k)=11-(k mod 11)
In general R is taken as the greatest prime number less than m.
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| 273 | 261 |
| 211 | 259 |
| 84 | 69 |
| 217 | 127 |
| 376 |
Find a key, X, in the last line of the table above (i.e., after all keys are in part a are placed in the hash table) that can be placed in the hash table after 9 collisions.
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