2 votes 2 votes Programming in C hashing + – KISHALAY DAS asked Nov 3, 2016 KISHALAY DAS 783 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Prabhanjan_1 commented Nov 3, 2016 reply Follow Share 0.2 ?? 0 votes 0 votes Aboveallplayer commented Nov 3, 2016 reply Follow Share 0.4? 1 votes 1 votes Kantikumar commented Nov 9, 2016 reply Follow Share $\frac{5}{11}$ ? 0 votes 0 votes vaishali jhalani commented Nov 9, 2016 reply Follow Share ans is 1/5. 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes after hashing elements map like this: 0: 1:20 2: 3: 4:16-5 5:44-88-11 6:94-39 7:12-23-1 8: 9: 13(edit) 10: so 5 position are vacant where any number can be placed without collision so probability :5/11 = 0.45 is it correct answer? kirti singh answered Nov 3, 2016 edited Nov 3, 2016 by kirti singh kirti singh comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments KISHALAY DAS commented Nov 3, 2016 reply Follow Share I too think this is the ans...thogh they given 1/5=0.2 0 votes 0 votes kirti singh commented Nov 3, 2016 reply Follow Share but here favorable cases are 5.. at any of the 5 position,, number can be placed without any collision out of 11 places... 0 votes 0 votes KISHALAY DAS commented Nov 3, 2016 reply Follow Share yes thats my point too..i guess its a wrong ans given 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes according to the has function 0th slot=----- 1st slot=20 2nd=-------- 3rd=-------- 4th=16 5th=44-88-11 6th=94-39 7th=12-23 8th------- 9th=13 10th--- so 5 vacant slots probability=5/11=0.45 Aboveallplayer answered Nov 3, 2016 Aboveallplayer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Answer : 5/11 Of 11 indexes only 2,3,8,9,10 are free after hashing. So prob for free collision = 5/11 Gate Mission 1 answered Nov 9, 2016 Gate Mission 1 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 1. (12*2 + 5)%11 = 7 2. (44*2 + 5)%11 = 5 3. (13*2 + 5)%11 = 9 4. (88*2 + 5)%11 = 5 5. (23*2 + 5)%11 = 7 like 94 -> 6 , 11 -> 5 , 39->6 , 20 -> 1, 16 -> 4 , 5->4; Free slots are : 0,2,3,8,10 .: Probability that x will not collide is : 5/11. sarveswara rao v answered Nov 10, 2016 sarveswara rao v comment Share Follow See all 0 reply Please log in or register to add a comment.