CSE341 – Programming Languages Homework #4 Solved

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Part 1. In the graph below you see the possible flights between some of the cities in Turkey. Write the predicate “route(X,Y) – a route between X and Y exists” that returns true of if there is a route between any given two cities.

Your program should have all the facts and predicates/rules. See the following:

% knowledge base

flight(istanbul,antalya).  % the fact that Istanbul and Antalya has a flight.

% rules  … route(X,Y) :- flight(X,Y). % a predicate indicating there exist a route between                                       % X and Y if there is flight between X and Y.

A single query to complete your program should check if there is a direct route between two given cities. Alternatively, it can list all the connected cities for a given city. See the following:

?- route(edirne,X).

X = erzincan ;

X = edremit ;

Make sure that your predicate implementation handles cycles properly avoiding infinite loops.

Part 2. Continuing with the previous problem, you are asked to write a program that checks if a route exists between two cities and if so, provides the shortest route.

In the first step, you are to expand the knowledge by adding distances for the direct flights. E.g.,

% knowledge base

flight(istanbul, antalya).  % the fact that Istanbul and Antalya has a flight. distance(istanbul, antalya, 481).  % flight distance – calculated using

% https://www.distancecalculator.net

% complete all the flights and distances …

A query to complete your program should check if there is a direct route between two given cities and the shortest distance between them. See the following example:

?- sroute(edremit,erzincan,X).

X = 1044 ;

 

Part 3. You are given the following database about classes, classrooms and student enrollment.

Classes     Enrollment
Class Time Room   Student Class
102 10 z23   a 102
108 12 z11   a 108
341 14 z06   b 102
455 16 207   c 108
452 17 207   d 341
        e 455

 

Write the predicates “when(X,Y) – time of the course X is Y”, “where(X,Y) – place of the course X is Y”, and “enroll(X,Y) – student X is enrolled in course Y”. For example:

% facts.. when(102,10).

3.1. Define/write a predicate “schedule(S,P,T)” that associates a student to a place and time of class. See the example query and its result.

?- schedule(a,P,T).

P = 102

T = 10 ;

 

P = 108

T = 12 ;

3.2. Define/write another predicate “usage(P,T)” that gives the usage times of a classroom. See the example query and its result.

?- usage(207,T).

T = 455 ;

T = 456 ;

3.3. Define/write another predicate “conflict(X,Y)” that gives true if X and Y conflicts due to classroom or time.

3.4. Define/write another predicate “meet(X,Y)” that gives true if student X and student Y are present in the same classroom at the same time.

Part 4. Write the following predicates operating on sets.

4.1. Define a Prolog predicate “element(E,S)” that returns true if E is in S.

4.2. Define a Prolog predicate “union(S1,S2,S3)” that returns true if S3 is the union of S1 and S2.

4.3. Define a Prolog predicate “intersect(S1,S2,S3)” that returns true if S3 is the intersection of of S1 and S2.

4.3. Define a Prolog predicate “equivalent(S1,S2)” that returns true if S1 and S2 are equivalent sets.

 

Part 5. Given a list of integes, find a correct way of inserting arithmetic (operators) such that the result is a correct equation. Example: With the list of numbers [5,3,5,7,49] we can form the equations (53+5*7) = 11.  Please pay attention to the arithmetic operator precedence. 

 

 

 

 

 

 

 

 

Part 6.  In this part you will solve a puzzle, Essentially, each row and column of a   rectangular bitmap is annotated with the respective lengths of   its distinct strings of occupied cells. The person who solves the puzzle must complete the bitmap given only these lengths.

Problem statement:          Solution:

|_|_|_|_|_|_|_|_| 3          |_|X|X|X|_|_|_|_| 3

|_|_|_|_|_|_|_|_| 2 1       |X|X|_|X|_|_|_|_| 2 1

|_|_|_|_|_|_|_|_| 3 2       |_|X|X|X|_|_|X|X| 3 2

|_|_|_|_|_|_|_|_| 2 2       |_|_|X|X|_|_|X|X| 2 2

|_|_|_|_|_|_|_|_| 6          |_|_|X|X|X|X|X|X| 6

|_|_|_|_|_|_|_|_| 1 5       |X|_|X|X|X|X|X|_| 1 5

|_|_|_|_|_|_|_|_| 6          |X|X|X|X|X|X|_|_| 6

|_|_|_|_|_|_|_|_| 1          |_|_|_|_|X|_|_|_| 1

|_|_|_|_|_|_|_|_| 2          |_|_|_|X|X|_|_|_| 2

  • 3 1 7 5 3 4 3 1 3 1 7 5 3 4 3
  • 1 5 1 2 1 5 1

For       the       example      above,       the       problem      can       be       stated       as       the     two              lists

[[3],[2,1],[3,2],[2,2],[6],[1,5],[6],[1],[2]] and   [[1,2],[3,1],[1,5],[7,1],[5],[3],[4],[3]] which give the   “solid” lengths of the rows and columns, top-to-bottom and    left-to-right, respectively.

 

Test Cases:  Top-to-bottom or Left-to-right 

  1. ([[3], [2,1], [3,2], [2,2], [6], [1,5], [6], [1], [2]],

[[1,2], [3,1], [1,5], [7,1], [5], [3], [4], [3]])

  1. ( [[3,1], [2,4,1], [1,3,3], [2,4], [3,3,1,3], [3,2,2,1,3], [2,2,2,2,2], [2,1,1,2,1,1], [1,2,1,4], [1,1,2,2], [2,2,8],  [2,2,2,4], [1,2,2,1,1,1],

[3,3,5,1], [1,1,3,1,1,2],  [2,3,1,3,3], [1,3,2,8], [4,3,8], [1,4,2,5], [1,4,2,2],

[4,2,5], [5,3,5], [4,1,1], [4,2], [3,3]],

[[2,3], [3,1,3], [3,2,1,2], [2,4,4], [3,4,2,4,5], [2,5,2,4,6], [1,4,3,4,6,1], [4,3,3,6,2], [4,2,3,6,3], [1,2,4,2,1], [2,2,6],  [1,1,6], [2,1,4,2], [4,2,6], [1,1,1,1,4], [2,4,7], [3,5,6], [3,2,4,2], [2,2,2], [6,3]])

 

  1. ( [[5], [2,3,2], [2,5,1], [2,8], [2,5,11], [1,1,2,1,6], [1,2,1,3],[2,1,1], [2,6,2], [15,4], [10,8], [2,1,4,3,6], [17], [17], [18], [1,14], [1,1,14], [5,9], [8], [7]], [[5], [3,2], [2,1,2], [1,1,1], [1,1,1], [1,3], [2,2], [1,3,3], [1,3,3,1], [1,7,2], [1,9,1], [1,10], [1,10], [1,3,5], [1,8],

[2,1,6],[3,1,7], [4,1,7], [6,1,8], [6,10], [7,10], [1,4,11], [1,2,11], [2,12], [3,13]] )

 

  • HW4-sk9rpy.zip