Given a set of points in the plane, we define the convex layers of Q inductively. The first convex layer of Q consists of those points in Q that are vertices of CHQ). For i > 1, define Qi to consist of the points of Q with all points in convex layers 1,2,...,i-1 removed. Then, the ith convex layer of Q is CHQ:) if Qi 70 and is undefined otherwise. Give a O(n)-time algorithm to find the convex layers of a set of n points. Justify the correctness of your algorithm and its running time. Do NOT use Jarvin March algorithm, just adapt Graham Scan.

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Step-by-step explanation:

Package wrapping technique can he used to compute convex null of a set Q which was given by Jarvis march  

The running time of the algorithm runs in 0(M), where h is the number of vertices of CH (0). The total running time for K convex la, and the I. layer contains li points is.  

Kindly check the attached image below to see the step by step explanation to the question above .

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.