Hashing : Quadratic Probing

Question

Keys $9,19,29,39,49,59,69$ are inserted into a hash Table of size $10$ $(0-9)$ using the hash function $H = k mod 10$ and Quadratic Probing is used for collision resolution. What is the index into which 59 will be inserted ?

$a). 3$

$b). 6$

$c). 8$

$d). 5$

Comments

it seems to be 4... how the answer is 6 please explain

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Solution given in book :

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Veteran - your Book is wrong

( 

After the "see above H" line, 

4+1² mode 10 = 5 mode 10= 5

Next- 5+2²mode 10 = 5+4 mode 10= 9mode10, = 9., in your book it is taken as 0., which is a mistake.)

The answer is not 6. It is 8.

59 is at the index 8.

Answer 1

Given Element={9,19,29,39,49,59,69}

9%10=9

19%10=9=>9+1^2=10=>10%10=0

29%10=9==>9+1^2=10==>10%10=0==>9+2^2=13==>13%10=3

39=>9+3^2=18=>18%10=8

49=>9+4^2=25%10=5

59=>9+5^2=34%10=4

69=>9+6^2=45%10=5==>9+7^2=58=>58%10=8=>9+8^2=73%10=3=>9+9^2=90%10=0

Here no where we have location for 69...though for 59 it is 4 

Comments

Position of 29 = 29%10=9

*I=1, i​​​​​​​²= 1, 9+1=10%10=0.

*i=2, i²= 4, 0+ 4 = 4.

So, 4 th index is for 29 only.

"The answer is 8. 59 is at the index 8."

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Position of 29 = 29%10=9

*I=1, i²= 1, 9+1=10%10=0.

*i=2, i²= 4, 0+ 4 = 4.

So, 4 th index is for 29 only.

"The answer is 8. 59 is at the index 8."

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For quadratic probing what c1 and c2 we assumed nothing given in question'

For quadratic probing

H'(k,i) = (h(k) + c1*i + c2* (i^2))

how it is taken H'(k,i) = (h(k) + i^2)

Answer 2

8 is index to store 59

Answer 3

Caption