UGC NET June-2019 CS Paper-2
October 6, 2023UGC NET June-2019 CS Paper-2
October 6, 2023UGC NET June-2019 CS Paper-2
| Question 24 |
Consider double hashing of the form
h(k,i)=(h 1 (k)+ih 2 (k)) mod m
Where h 1 (k)=k mod m
h 2 (k)=1+(k mod n)
Where n=m-1and m=701
for k=123456, what is the difference between first and second probes in terms of slots?
h(k,i)=(h 1 (k)+ih 2 (k)) mod m
Where h 1 (k)=k mod m
h 2 (k)=1+(k mod n)
Where n=m-1and m=701
for k=123456, what is the difference between first and second probes in terms of slots?
| 255 | |
| 256 | |
| 257 | |
| 258 |
Question 24 Explanation:
Given h(k,i)=h1(k)+ih2(k)
where h1(k)=k mod m
h2(k)=1+k mod n
where n=m-1 and m=701. For k=123456
h1(k)=123456 mod 701
h1(k)=80
h2(k)=1+(123456 mod 700)
h2(k)=1+256=257
1st probe: i=1
h(k,i)=h1(k)+ih2(k)
h(k,1)=h1(k)+ih2(k)
h(k,1)=80+257=337
h(k,1)=337
2nd probe: i=2
h(k,2)=80+2257
h(k,2)=80+514
h(k,2)=594
∴ Difference h(k,2)-h(k,1)
594-337
257
where h1(k)=k mod m
h2(k)=1+k mod n
where n=m-1 and m=701. For k=123456
h1(k)=123456 mod 701
h1(k)=80
h2(k)=1+(123456 mod 700)
h2(k)=1+256=257
1st probe: i=1
h(k,i)=h1(k)+ih2(k)
h(k,1)=h1(k)+ih2(k)
h(k,1)=80+257=337
h(k,1)=337
2nd probe: i=2
h(k,2)=80+2257
h(k,2)=80+514
h(k,2)=594
∴ Difference h(k,2)-h(k,1)
594-337
257
Correct Answer: C
Question 24 Explanation:
Given h(k,i)=h1(k)+ih2(k)
where h1(k)=k mod m
h2(k)=1+k mod n
where n=m-1 and m=701. For k=123456
h1(k)=123456 mod 701
h1(k)=80
h2(k)=1+(123456 mod 700)
h2(k)=1+256=257
1st probe: i=1
h(k,i)=h1(k)+ih2(k)
h(k,1)=h1(k)+ih2(k)
h(k,1)=80+257=337
h(k,1)=337
2nd probe: i=2
h(k,2)=80+2257
h(k,2)=80+514
h(k,2)=594
∴ Difference h(k,2)-h(k,1)
594-337
257
where h1(k)=k mod m
h2(k)=1+k mod n
where n=m-1 and m=701. For k=123456
h1(k)=123456 mod 701
h1(k)=80
h2(k)=1+(123456 mod 700)
h2(k)=1+256=257
1st probe: i=1
h(k,i)=h1(k)+ih2(k)
h(k,1)=h1(k)+ih2(k)
h(k,1)=80+257=337
h(k,1)=337
2nd probe: i=2
h(k,2)=80+2257
h(k,2)=80+514
h(k,2)=594
∴ Difference h(k,2)-h(k,1)
594-337
257
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