Hash function h(a) = (3x + 1) mod 5
Sequence
1,2,4,1,2,4,4,1,2,4,4,1,7
Hash Function h(a) Substitution
h (1) = (3(1) + 1) mod 5 = 4 mod 5 = 4
h (2) = (3(2) + 1) mod 5 = 7 mod 5 = 2
h (4) = (3(4) + 1) mod 5 = 13 mod 5 = 3
h (7) = (3(7) + 1) mod 5 = 22 mod 5 = 2
Replace the sequence with the hash
values
4,2,3,4,2,3,3,4,2,3,3,4,2
Convert into binary format
2 = 010
3 = 011
4 = 100
100, 010, 011, 100, 010, 011, 011, 100, 010, 011,
011, 100, 010
Calculate r(a)
Check Number of trailing zeros [i.e. number of 0’s
after 1]
2,1,0,2,1,0,0,2,1,0,0,2,1
R= max(r(a))
R =2
Estimate 2R
22 = 4