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# Lab 07

Here is a pdf version of the ppt I covered during the lab. And the codes that I (skimmed through) / (demonstrated live).

AMtoEL.cpp (The code shown for converting Adjacency Matrix to Edge List)

MockPE3Solution.cpp (The code for the Mock PE3 which is same as CS2040C PE(Practical Exam) S1 (Semester 1) Q1 (Question 1))

## Challenge Questions -

Q) Water jug problem. Given 3 jugs of size a, b and c liters respectively and an infinite supply of water from a tap. You can tranfer water from one jug to another, but note that you cannot measure things, meaning if you are pouring from jug 1 to jug 2, then you just pour until either jug 1 becomes empty or till the time jug 2 gets full, whichever happens first. Also when you fill a jug using the tap, you fill it to the full. You are also allowed to empty a jug by throwing its content into a infinite sized dustbin. Given all these conditions, determine the minimum no. of steps required to obtain exactly d liters of water in jug 1 which has capacity of a liters. Filling from a tap is a single step, tranferring from one jug to another is 1 step and also emptying a jug is 1 step.
Assume all the integers a, b, c and d lie between [1, 50]. Also if it is not possible to obtain exactly d liters of water in jug 1 of capacity a liters, then print -1.
A better descrption of the problem can be seen from here

Example -
a = 3, b = 5, c = 8, d = 2
Answer = 4 Explanation - Fill b. Then tranfer from b to a. Then empty a. Then again tranfer from b to a.

Hint - This question should have a pure mathematical approach (probably, although I don’t know the method for 3 jugs). The approach I am expecting is based on graph theory. Note, you would need to be know about bfs and dfs before trying this question.

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Solution The solution is basically model a graph having (a + 1) * (b + 1) * (c + 1) nodes, where each node looks like -
(x, y, z), i.e there is x liters of water in the first jug of capacity a, there is y litres of water in second jug of capacity b and z litres of water in third jug of capacity c.
Now add directed edges from (x1, y1, z1) and (x2, y2, z2) if you can move from the state (x1, y1, z1) to state (x2, y2, z2) in one step.

Now the question is reduced to finding the shortest path from (0, 0, 0) to (d, 0, 0). This can be done using a BFS / Djikstra. Note that if no path exists then the answer is -1.

The time complexity of BFS approach would be O(V + E), where V is the number of vertices and E is the number of edges. In this question V = (a + 1) * (b + 1) * (c + 1) and each node roughly has 12 outgoing edges (can be seen by considering all possible transfers/fills/empty kind of varations which cost 1 step).

So O(V + E) = O(V + 12*V) = O(13 * V) = O(13 * 50 * 50 * 50)