Farid Khan
Nov 9, 2020, 9:58:59 AM11/9/20
to Discussion forum for Computational Thinking, lasikared...@gmail.com
no of bins for least no of comparsions, k = 20/2 = 10
Factors of reduction = n-1/(n/k -1) = 20 - 1 / (20/10 -1) = 19
On Monday, 9 November 2020 at 18:17:45 UTC+5:30 lasikared...@gmail.com wrote:
How is the answer 19?
B Lasika Reddy
Nov 9, 2020, 10:02:45 AM11/9/20
to cs1001-...@nptel.iitm.ac.in, fari...@gmail.com
Thank you so much!
Farid Khan
Nov 9, 2020, 12:38:37 PM11/9/20
to B Lasika Reddy, cs1001-...@nptel.iitm.ac.in
Anytime!
Computational Thinking Support 2
Nov 10, 2020, 12:34:27 AM11/10/20
to Discussion forum for Computational Thinking, Farid Khan, cs1001-...@nptel.iitm.ac.in, lasikared...@gmail.com
Hi,
Whatever Farid has explained is correct.
Regards,
Deepak
IITM Online Degree Team
Nandni Gupta
Nov 10, 2020, 9:05:37 AM11/10/20
to Discussion forum for Computational Thinking, ct-su...@onlinedegree.iitm.ac.in, Farid Khan, cs1001-...@nptel.iitm.ac.in, lasikared...@gmail.com
but sir it is mentioned that we need least number of comparisons which could be 0 i.e., we can have 20 bins. Why do we have to solve for 1 comparison and solve by taking 10 bins although it is given that least umber of comparisons are required which could be 0.
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