Question

Do 17.。

identify the counterfeit colll alu a. What is the number of final outcomes (the number U this problem in b. Find a lower bound on the number of comparisons required to solvt the worst case. c. Devise an algorithm that meets this lower bound (draw its decision tree) is to 17. One of four coins is counterfeit and is either too heavy or too light. The probl identify the counterfeit coin but not to determine whether it is heavy or light a. What is the number of final outcomes (the number of leaves in the decision b. Find a lower bound on the number of comparisons required to solve th tree)? the worst case. C. 18. One of four coins is counterfeit and is either too heavy or too light. The pr identify the counterfeit coin and determine whether it is heavy or light. oblem is to a. What is the number of final outcomes (the number of leaves in the decision tree)? b. Find a lower bound on the number of comparisons required to solve this problem in the worst case. c. Prove that no algorithm exists that can meet this lower bound. (Hint: The first com parison can be made with either two coins or four coins. Consider each case.) 19. Devise an algorithm to solve the problem of Exercise 18 using three comparisons in the

Answer 1

Assume the our coins be namely 1, 2, 3 and 4.

We can group the coins in two different ways, [(12, 34)] or [(1, 2) and (3, 4)]. Let us consider the combination (12, 34),

Note: 12 it means that for first weighing the coins 1 and 2 are placed together, similarly for 34 coins 3 and 4 are placed together.

Corresponding decision tree is given below:

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Weighing 1:

The outcome can be (12) < (34) i.e. we go on to left subtree or (12) > (34) i.e. we go on to right subtree.

The left subtree is possible in two ways,

A) Either 1 or 2 can be lighter (represented as L in tree) OR

B) Either 3 or 4 can be heavier (represented as H in tree).

Weighing 2:

Further on the left subtree, as second trial, we weigh (1, 2) or (3, 4). Let us consider (3, 4) as the analogy for (1, 2) is similar. The outcome of second trail can be three ways

A) (3) < (4) yielding 4 as defective heavier coin, OR

B) (3) > (4) yielding 3 as defective heavier coin OR

C) (3) = (4), yielding ambiguity. Here we need one more weighing to check a genuine coin against 1 or 2.

Weighing 3:

In tree we took (3, 2) where 3 is confirmed as genuine. We can get (3) > (2) in which 2 is lighter, or (3) = (2) in which 1 is lighter. Note that it impossible to get (3) < (2), it contradicts our assumption leaned to left side.

Similarly we can analyze the right subtree. We need two more weighings on right subtree as well.

Hence we need 3 weighings to find the counterfeit coin. Note that middle branch terminated after first weighing as the coins weight was equal which was contrary to questions as it says we definetly have one counterfeit coins and hence marked as impossible outcome. We got 11 leaves(boxed in decision tree) including impossible cases.

Answers:(a) 8 leaves (excluding impossible cases)

(b) 3

(c) Refer above explanation for algorithim and figure for decision tree.

Given N coins, if only one coin is counterfeit(which may be lighter or heavier). We need a decision tree with atleast (2N) leaves correspond to the outputs.