代做CSE 240 Homework 08, Fall 2024代写C/C++程序

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CSE 240 Homework 08, Fall 2024 (50 points)

Due Saturday, December 7th, 2024 at 11:59PM, plus one day grace period

Introduction

The aim of this assignment is to make sure that you understand and are familiar with the concepts covered in the lectures on Prolog.  By the end of the assignment, you should have

•     strong concept of logic / declarative paradigm;

•     strong concept of facts, rules, and questions in Prolog.

•     Compared and contrasted the programming C++, Scheme, and Prolog.

•     Prolog list operations and manipulations

•     Prolog recursive rules with multiple nested calls

Reading: Text Chapter 5 and lecture slides. Also read Text Appendix B.4 Prolog Tutorial

You are expected to do most of the assignment outside the class meetings. Should you need assistance, or have questions about the assignment, please contact the instructor or the TA during their office hours.

You are encouraged to ask and answer questions on the course discussion board and in the course Discord server.  (However, do not share your answers in the course discussion board.)

Practice Exercises (no submission required)

1.      Answer the  multiple-choice questions in text section 5.9 that have been covered in the lectures.

2.      Complete the questions 2 through 5 in text section 5.9.

3.      Lookup the Unix command table in Appendix B.1 and read the Prolog tutorial in B.4.

4.      Log onto general.asu.edu and execute the Unix commands and Prolog exercises.

5.      Follow the Prolog tutorial in Text Appendix B.4 or the tutorials given in the course Web site, start GNU Prolog and try following simple programs (build-in functions). Don't forget the period at the end of the statement.


|  ?- write(hello) .                  % hello is a constant */

|  ?- write(Hello) .                % Hello is avariable. Its address will be displayed*/

|  ?- write('helloworld') .      % a string is printed */

| ?- read(Y), write('The variable entered is '), write(Y), nl. /* nl prints a newline. Type a period and an enter at the end of the input */ |  ?- X is 2+2.

|  ?- Y is 5*8.

|  ?- Y is 2**10.

|  ?- length([a,b, x, y, 2, 45, z], L).

|  ?- append([a,b, c, d], [4, 6, 8], LL).

|  ?- append(X,Y,[a,b,c]). Then, type ";" to obtain all possible answers. 

|  ?-     X is [1 | [2 | [3 | []]]], write(X). Explain the output.

6.      Use trace and notrace before and after your questions in Question 5. For example

| ?- trace.

| ?-      X is [1 | [2 | [3 | []]]], write(X). Explain the output.

| ?- notrace.

Programming Assignment (50 points)

1.    Consider the following diagram of a familytree:

 

Implement the family relationships in a Prolog factbase:

Note: The section in the highlighted box above has been completed for you.



/* Factbase for a family tree. It consists of facts and rules. */ /* Facts */ male(abe). male(rob). male(jim). female(joy). female(ana). father_of(abe, ana). /* abe is the father of ana */ father_of(abe, rob). /* abe is the father of rob */ father_of(abe, jim). /* abe is the father of jim */ mother_of(joy, rob). /* joy is the mother of rob */ mother_of(joy, jim). /* joy is the mother of jim */ mother_of(joy, ana). /* joy is the mother of ana */ /* Complete the facts given in the diagram above */ /* Rules */




Open the given file 1FamilyTreeSolution.pl using a text editor such as pico or vim on a Unix operating system. You may write the program on your own computer using any editor such as notepad and upload the program to the general server to compile and execute.

Compile the program using the command:

$ gplc 1FamilyTreeSolution.pl

This should create an executable that you can then run using the following command:

$ ./1FamilyTreeSolution

This will enter you into the GNU Prolog programming environment where your program has already been executed.

Ask questions by typing, e.g.:

|?- father_of(mark, beth).

|?- mother_of(beth, tom).

To exit from GNU Prolog, type the end-of-file character at the main Prolog prompt ^d (Ctrl-d).

|?- ^d

You can find a complete set of GNU Prolog commands at:

http://www.gprolog.org/manual/gprolog.html



For all the following questions, please label the question number in a comment. For example, if Question 2.1 asks you to define a rule named is_male (X) that returns "yes" (or "true") if X is the father of a member of the family, then your code should look like:

/* Question 1.1 */

is_male(X) :-

male(X);

father_of(X, _).

This is a reading check. If you see this, congratulations. You may use this in your solution. Now, you can start to add your code into the program.

Complete the program by adding facts for the remaining members on the familytree. The section inside of the highlighted box has already been completed for you for clarification. Please pay close attention when adding the remaining family members. spelling counts and all letters should be lowercase.

1.1 Define (add into the factbase) a rule called is_male(X) that returns “yes” (or true”) if X is a male or the father of a member of the family.

Note: the system will return a "yes", if it finds a "true" answer and there exist no more true answers. The system will return "true ?" if it finds a "true" answer and there are still possibly further matches. In this case, if you press enter, it will return "yes" and stop. If you type " ;", it will continue to search for further answers.

1.2       Define (add into the factbase) a rule called is_female(X) that returns "yes" (or "true") if X is a female or the mother of a member of the family.

1.3  Define  a  rule  called  grandmother_of(X,  Z) that  returns  yes”  (or  true”)  if X is  a  grandmother of Z. Define another rule called grandfather_of(X, Z) that returns yes” (or “true”) if X is a grandfather of Z. (you can use 1.5 parent_of(X, Y) rule)

1.4     Define a rule called sibling_of(X, Y) that returns yes” (or true”) if X is a sibling of Y. Note: a family member cannot be their own sibling. (you can use 1.5 parent_of(X, Y)

rule)

1.5    Define a rule called parent_of(X, Y) that returns yes” (or true”) if X is aparent of Y.

1.6 Define a rule called descendent_of(X, Y) that returns yes” (or true”) if X is a descendent of Y. A descendant, in the context of a family tree, refers to any individual directly related through a series of parent-child relationships, e.g. if joy is the parent of jim, and jim is the parent of kim, then kim is considered a descendant of both joy and jim. Note: you will need to use recursion as well as the parent_of rule.  (you can use 1.5 parent_of(X, Y) rule)


2.    Consider the following factbase for the games planetDrop and mechaTech. [12 points]

planetDrop

mechaTech

genre: action

genre: rpg

focus:

focus:

gameplay

story

-    mechanics

-    characters

-    framerate

-    plot

graphics

world

-    rendering

-    design

-    meshes

-    culture

In 2GamesSolution.pl, complete the following:

Create facts for the genres: genre(X, Y) where X is a game and Y is a genre.

Create facts for the focuses: focus(X, Y) where X is a game and Y is a focus.

Create facts for the elements: element(X, Y) where X is a focus and Y is an element.

2.1    Create a rule game(X, Y) where X is a game and Y are the genre and the focuses of that game. For example, game(planetDrop, X) should return (use  ; to browse through the results):

X = action ? ;

X = gameplay ? ;

X = graphics

2.2    Create a rule creation(X, Y) where X is a game and Y are the elements of that game. For example, creation(planetDrop, X) should return:

X = mechanics ? ;

X = framerate ? ;

X = rendering ? ;

X = meshes



3.    In 3QuicksortSolution.pl, you will re-implement the Quicksort given in textbook and lecture slides. In the given example, the first (left-most) element of the given list is selected as the pivot. In this question, you must choose the second element of the list as the pivot. Hint: You can represent the input list into pairs: [First|[Pivot|Tail]]. You must write comments to indicate the size-n problem, stopping condition and its return value, size m-problems, and construction of the size-n problem from size-m problems.

Test case:

| ?- qsort2([8, 3, 4, 12, 25, 4, 6, 1, 9, 22, 6], Sorted).

It returns:

Sorted = [1,3,4,4,6,6,8,9,12,22,25]

4.    A premium pizza is comprised of exactly 40 ounces of toppings. The available toppings are listed below with their corresponding weight (in ounces). There can be multiple entries of each topping, as long as the total weight is equal to 40 ounces.

Topping

Weight (ounces)

Pepperoni

Sausage

Bacon

Onion

Mushroom

4

10

6

5

7

For example, a pizza can contain 1 topping ofpepperoni, 2 toppings ofsausage, 1 topping of bacon, and 2 toppings of onion : 1*4 + 2*10 + 1*6 + 2*5 = 40 (ounces)

A pizza cannot contain 7 toppings of bacon :  7 * 6 = 42 > 40

A pizza cannot contain only 3 toppings ofsausage : 3*10 = 30 < 40

In 4PizzaSolution.pl, define a rule pizza(P, S, B, O, M) to find out how many of each topping can be contained on a pizza, where arguments P, S, B, O, and M represent the weights (in ounces) of the Pepperoni, Sausage, Bacon, Onion, and Mushroom toppings, respectively.

Your pizza rule will be used to find the outputs to questions similar to below: Test Case:

| ?- pizza(1, S, 1, O, M).

It returns: (use  ; to browse through the results)


M = 0 M = 0 M = 0 M = 0 O = 6 O = 4 O = 2 O = 0 S = 0            S = 1 S = 2 S = 3




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