Answer:
Yes, there is a sequence, and at the end of the sequence the following arrangement will be in each container;
10-pint container will contain 2 pints of water
7-pint container will contain 7 pints of water
4-pint container will contain 2 pints of water
Explanation:
The 7- and 4-pint containers are full, while the 10-pint container is initially full.
The following steps are followed;
1. pour the content of the 7-pint container into the 10-pint container, and pour 3-pint of water from the 4-pint container to fill up the 10-pint container, leaving the following in the three containers
10-pint container will contain 10 pints of water
7-pint container will contain 0 pints of water
4-pint container will contain 1 pints of water
2. Next, pour the 1 pint of water from the 4-pint container into the 7-pint container, and in turn pour from the 10-pint container into the 4-pint container until it is full, leaving the following in the three containers;
10-pint container will contain 6 pints of water
7-pint container will contain 1 pints of water
4-pint container will contain 4 pints of water
3. next, pour all the content of the 4-pint container into the 7-pint container, and refill the 4-pint container by pouring into it from the 10-pint container, leaving the following in the 3 containers;
10-pint container will contain 2 pints of water
7-pint container will contain 5 pints of water
4-pint container will contain 4 pints of water
4, Finally, pour from the 4-pint container into the 7-pint container until it is full, leaving the following in the 3 containers;
10-pint container will contain 2 pints of water
7-pint container will contain 7 pints of water
4-pint container will contain 2 pints of water.
There you go, this sequence leaves 2-pint of water in the 4-pint container.
Answer:
Here’s a way to solve this:
(7/7,4/4,0/10) Æ (7/7, 0/4, 4/10) Æ(1/7, 0/4, 10/10) Æ (1/7, 4/4, 6/10), (5/7, 0/4, 6/10) Æ
(5/7, 4/4, 2/10) Æ (7/7, 2/4, 2/10), hence we end up with 2 pints in the 4‐pint container .
To model this as a graph problem, we represent each potential state of the three containers with an ordered pair (A,B), where A represents the amount of water in the 7‐pint container and B represents the amount of water in the 4‐pint container.
Hence the initial state would be(7,4). Note that we do not need an ordered triple as we know the total amount of water in all three containers (11 pints). Therefore, the total state is completely determined by water in the 7 and 4 pint containers.
Hence, we have 5 x 8 = 40 possible states for the system. We model these states as nodes in a directed graph and draw an edge from one node to another node if we can go from the configuration represented by one node to the configuration represented by another node through one pour.
For example, there would be an edge from (7,4) to (0,4) but no edge from (1,3) to (5,1).
Thus, if we can find a path from one node to another, then we can go from one state to another
through a series of pourings. The nodes are represented in the attached file (the edges are left out).
Explanation:
Since we want to find a path from (7,4) to one of the nodes contained in one of the rectangles. Our solution above is depicted in the attached file
Looking at the graph in attachment, we will start at node (7,4) and run DFS to see if we can reach one of the nodes boxed by a rectangle (i.e. has 2 as one of the coordinates). If we can, then there is a solution.
Otherwise, there is no way to have two liters in one of the containers through a series of pouring from
the original state of (7/7,4/4,0/10).
