## Description

CSci 2041: Advanced Programming Principles

You will solve two searching problems in this assignment. The first

is graph coloring, the second is the so-called “water jug” problem.

## Part 1: Graph Coloring

Write your solution to this problem in a file named

“graph_coloring.ml“ in a directory named “Hwk_07“.

### Getting Started

Graph coloring is a famous problem in mathematics and computer

science. To familiarize yourself with this problem, read the

beginning of the Wikipedia entry on this topic, found

here: https://en.wikipedia.org/wiki/Graph_coloring. You can stop

reading when you get to the part labeled Chromatic Polynomial.

In our approach to solving this problem we will represent nodes and

colors as integers, and edges are a pair of nodes. Since it is easy

to confuse the integers representing colors and nodes we will wrap

these in an OCaml type with a simple one letter constructor to

distinguish them.

Below are the type declarations that we will use:

“`

type node = N of int

type edge = node * node

type graph = node list * edge list

type color = C of int

type coloring = (node * color) list

“`

Nodes are “int“ values under the “N“ constructor. These are

distinguished from colors which are under the “C“ constructor. You

might wonder if it is worth the trouble to do this. In my original

solution to this problem I just used “int“ for nodes and colors and

my solution was returning wrong results. After changing the types of

“node“ and “color“ to those above OCaml immediately identified the

place where I was mistakenly using a node value as a

color. Additionally, they make graph literals and colorings easier to

read. Thus, these simple constructors are well worth the small added

effort.

As an example, below is a declaration of simple 4-node graph and its

3-coloring:

“`

let g1 = ( [N 1; N 2; N 3; N 4], [ (N 1,N 2); (N 1,N 3); (N 2,N 3); (N 3,N 4) ] )

let g1_coloring = [ (N 1,C 1); (N 2,C 2); (N 3,C 3); (N 4,C 2) ]

“`

This is the second example graph from the Wikipedia that has 12

different 3-colorings.

Below you are asked to write two functions that each check **if there is

a coloring of a given graph using 3 colors**. These functions are not

looking for the lowest number of colors that provide a valid coloring,

they are only checking if there is a solution that uses 3 colors.

### Graph coloring using options

Write a function named “color_option“ with the type “graph ->

coloring option“. It should use the “option“ type and its values

as the mechanism for controlling the search of the search space.

If you test your function with “g1“ above your function should

return a valid coloring. Here is a sample output:

“`

Some [(N 1,C 1); (N 2,C 2); (N 3,C 3); (N 4,C 2)]

“`

Note that your solution may be different

from the sample solution “g1_coloring“ given above since there are

many possible ways to assign colors to this graph.

The feedback tests, when complete, will only tests that your coloring

is a valid and complete one. They will not test for a specific

solution since there are many.

### Graph coloring using exceptions

Now, write a function named “color_exception“ with type “graph ->

unit“. This should raise an exception named “FoundColoring“

defined as follows:

“`

exception FoundColoring of coloring

“`

This solution will use exceptions to control the search process.

Here is a sample output:

“`

Exception: FoundColoring [(N 1,C 1); (N 2,C 2); (N 3,C 3); (N 4,C 2)].

“`

Similarly, the feedback tests will check for valid and complete

colorings, not for a specific solution.

### Explaining your solution

Write a comment near the top of your file that answers the following

questions. The answers will very likely be the same for both of your

solutions. But you can answer the questions by referring to only one

of your solutions if you like. Be clear which solution you are

discussing in your answers.

1. What is the search space that your solution explores? Specifically,

explain what decisions are made at each step in exploring this

space. In the case of the subset-sum problem, we made a decision

about including, or not including, an element from the original set in

the partial subset we were constructing. Explain the decisions being

made in your solution, with references to those parts of your

solution.

2. In exploring the potential coloring of a graph, your solution must

not continue searching along that path if two adjacent nodes are

colored with the same color. Explain how your solution avoids this

potential inefficiency.

Note that we did not have this concern in the subset-sum problem.

In choosing certain elements of the set, we could not tell if this

was a dead end because the last element in the set could result in

getting a sum of 0. In this problem, that is not the case. If we

color adjacent nodes with the same color early in the process then

there is no point in continuing.

## Part 2: The Water Jug Problem

Write your solution to this problem in a file named

“water_jug.ml“ in a directory named “Hwk_07“.

### Getting Started

The water jug problem is a simple one. We start with 2 jugs, one is a

4 gallon jug, the other is a 3 gallon jug. Initially both jugs are

empty. The goal is to get 2 gallons into the 4 gallon jug.

There are 8 permitted operations and a solution to this problem is a

list of operations that take us from the initial state in which both

jugs are empty to the goal state in which the 4 gallon jug has 2

gallons in it and the 3 gallon jug is empty.

Valid operations are enumerated by the following type:

“`

type operation = Fill4GallonJugFromTap

| Fill3GallonJugFromTap

| Empty4GallonJugOnGround

| Empty3GallonJugOnGround

| Fill4GallonJugFrom3GallonJug

| Fill3GallonJugFrom4GallonJug

| Empty4GallonJugInto3GallonJug

| Empty3GallonJugInto4GallonJug

“`

A solution to the problem has type “operation list“.

In this problems, some operations have pre-conditions. That is, even

though they may lead to a valid state, they are not allowed.

For example, one cannot fill up the 4 gallon jug if it is already

full. We also cannot fill the 4 gallon jug using the contents of the

3 gallon jug if the total number of gallons in both is not at least 4

gallons.

Performing either of these actions may lead to a “valid state”, but

they are still not allowed. This is a situation we did not encounter

in the problems solved in class and thus you need to consider this

here.

A description and the pre-condition for each operation are listed below:

– “Fill4GallonJugFromTap“. This operation fills the 4 gallon jug from

the tap. It is only allowed when the 4 gallon holds less than 4

gallons.

– “Fill3GallonJugFromTap“. This operation fills the 3 gallon jug from

the tap. It is only allowed when the 3 gallon holds less than 3

gallons.

– “Empty4GallonJugOnGround“. This operations empties the 4 gallon jug

onto the ground. It is only allowed when the jug is not empty.

– “Empty3GallonJugOnGround“. This operations empties the 3 gallon jug

onto the ground. It is only allowed when the jug is not empty.

– “Fill4GallonJugFrom3GallonJug“. This operation pours some of the

contents of the 3 gallon jug into the 4 gallon jug, filling the 4

gallon jug. Some water may remain in the 3 gallon jug. This is

only allowed with the 3 gallon jug is not empty and the 4 gallon

jug would be full after the operation.

– “Fill3GallonJugFrom4GallonJug“. This operation pours some of the

contents of the 4 gallon jug into the 3 gallon jug, filling the 3

gallon jug. Some water may remain in the 4 gallon jug. This is

only allowed with the 4 gallon jug is not empty and the 3 gallon

jug would be full after the operation.

– “Empty4GallonJugInto3GallonJug“. This operation pours ALL the contents

of the 4 gallon jug into the 3 gallon jug. After this operation the 4

gallon jug will be empty and the 3 gallon jug will not be empty (but

it is possible that the 3 gallon jug is not full. This is only

allowed if 4 gallon jug is not empty and no water is wasted (meaning

that there is not more than 3 gallons between the two jugs before

the operation).

– “Empty3GallonJugInto4GallonJug“. This operation pours ALL the contents

of the 3 gallon jug into the 4 gallon jug. After this operation the 3

gallon jug will be empty and the 4 gallon jug will not be empty (but

it is possible that the 4 gallon jug is not full. This is only

allowed if 3 gallon jug is not empty and no water is wasted (meaning

that there is not more than 4 gallons between the two jugs before

the operation).

Note that some operations may result in the same state. But we still

consider them to be different operations.

## Finding a solution

Write a function named “play“ with the type “unit -> (operation *

string) list option“. No real input is needed since we know the

initial state is when both jugs are empty.

Your function will compute the list of operations needed to get to the

solution. With each operation we associate a string that describes the

situation after the operation is complete.

Below is an interaction in “utop“ of a solution. Note that there

are different possible solutions to this problem – this is just one of

them.

“`

utop # play () ;;

– : (operation * string) list option =

Some

[(Fill4GallonJugFromTap,

“The 4 gallon jug contains 4 gallons, The 3 gallon jug is empty.”);

(Fill3GallonJugFromTap,

“The 4 gallon jug contains 4 gallons, The 3 gallon jug contains 3 gallons.”);

(Empty4GallonJugOnGround,

“The 4 gallon jug is empty, The 3 gallon jug contains 3 gallons.”);

(Empty3GallonJugInto4GallonJug,

“The 4 gallon jug contains 3 gallons, The 3 gallon jug is empty.”);

(Fill3GallonJugFromTap,

“The 4 gallon jug contains 3 gallons, The 3 gallon jug contains 3 gallons.”);

(Fill4GallonJugFrom3GallonJug,

“The 4 gallon jug contains 4 gallons, The 3 gallon jug contains 2 gallons.”);

(Empty4GallonJugOnGround,

“The 4 gallon jug is empty, The 3 gallon jug contains 2 gallons.”);

(Empty3GallonJugInto4GallonJug,

“The 4 gallon jug contains 2 gallons, The 3 gallon jug is empty.”)]

─( 11:52:30 )─< command 45 >──────────────────────────────────────{ counter: 0 }─

utop #

“`

The “string“ description can be generated by using the following

function which you are free to copy into your file.

“`

let describe (four:int) (three:int) : string =

let describe’ jug amount =

“The ” ^ string_of_int jug ^ ” gallon jug ” ^

match amount with

| 0 -> ” is empty”

| 1 -> ” contains 1 gallon”

| x -> ” contains ” ^ string_of_int x ^ ” gallons”

in

describe’ 4 four ^ “, ” ^ describe’ 3 three ^ “.”

“`

It takes the number of gallons from the 4 gallon and and 3 gallon jug

as input and generates the string.

Feedback tests, when complete, will check for any valid sequence of

moves. They will not explain what is incorrect with your set of moves

since that is something you should figure out on your own.

### Explaining your solution

Write a comment near the top of your “water_jug.ml“ file explaining

how you represent the state of the problem. That is, what type do

you use to represent the amount of water in the jugs?

You should define a type named “state“ in your solution and use it

for this purpose. Place this definition near the comment with the

answer to this question.

## Turning in your work

Push these two files, “graph_coloring.ml“ and “water_jug.ml“ to

the “Hwk_07“ directory of your repository by the due date.

As in past homework assignments, part of your score will be based on

your code quality. (See “Writing transparent code” in Homework 4.)