Homework Answers
In the hash function ha(x) = ((a*x)%p)%s . Since a=2, therefore a*x will be the set of even number.
Now let's test the output ha(x) for x= 0,1,2,3,4,5,6 . Since p = 7 and s=4.
ha(0) = ((2*0)%7)%4 = 0 % 4 = 0
ha(1) = ((2*1)%7)%4 = 2 % 4 = 2
ha(2) = ((2*2)%7)%4 = 4% 4 = 0
ha(3) = ((2*3)%7)%4 = 6 % 4 = 2
ha(4) = ((2*4)%7)%4 = 1 % 4 = 1
ha(5) = ((2*5)%7)%4 = 3 % 4 = 3
ha(6) = ((2*6)%7)%4 = 5 % 4 = 1
So we can see that out of 4 possible output of ha(x) which are 0,1,2,3, output 3 occurs only once for x = 5 while every other output occurs twice.
Thus from above distribution of x, we get that output 3 is less probable.
Please comment for any clarification.
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