CompSci 367/761 ASSIGNMENT

CompSci 367/761 ASSIGNMENT 2: TRAVELLING SALESPERSON
PROBLEM (TSP) : FINDING OPTIMAL TOUR
Due 27 August 11:59PM worth 5%.
1 Introduction
This assignment will be released in stages. This phase of the assignment will introduce
you to the problem you are to solve. The next phase will introduce you
to example road networks. These example road networks will provide you with
test cases for your code. The final phase will provide the wrapper prolog code
that will use your code to find the optimal tour for a TSP. Phase 2 will be available
on Monday 5 August and the final phase will be available on Wednesday
7 August. You can begin trying to understand the assignment now. However,
the asignment should become clearer as you learn prolog in the tutorial and the

代写CompSci 367/761作业、代做Java/Python编程设计作业、c/c++实验作业代写
lectures this week. Please use Piazza to ask questions about and to discuss the
assignments.
1. Declarative Programming
The goal of declarative programming is to separate the what from the how.
When you were "programming" the logic puzzle in assignment 1, you were
doing, to a large extent, declarative programming. A large part of your
program described what relationships must exist between the values of
different variables. You did not describe how to find those values.
In this assignment, you will be describing what it means for a list of cities
to be a tour for a given start city and a road network. You should NOT
be describing how that tour is found.
2. Context
In this assignment you need to write prolog code to generate solutions for
the asymmetric travelling salesman problem (see https://en.wikipedia.
org/wiki/Travelling_salesman_problem#Description). The goal is to
write a "prototype" solver for travelling salesman problems. Your assignment
is make the code as declarative and as obviously correct as possible.
Efficiency is of no concern here. Basically your code will "define" which it
means for a list of cities to be a tour of a road network starting from and
ending at a specified city.
3. Inputs and Outputs
You must write the code for:
solution(+Path, +RoadNetwork, -SolutionCost, -SolutionPath).
See "Notation of Predicate Descriptions" (https://www.swi-prolog.org/
pldoc/man?section=preddesc) for descriptions of the meaning of "+"
and "-" in front of the argument names.
The inputs are:
1
Path a list of cities, in reverse order of being visited (e.g., the last city in
the list is the first city in the tour
RoadNetwork a description of all the directed roads between cities and
their cost
The outputs are:
SolutionCost the cost of a tour of network;
SolutionPath the tour for which SolutionCost was computed.
2 Domain Description
1. Road Network
The road network is a directed asymmetric weighted graph, the graph is
not nessarily complete (i.e., there need not be edges between every two
nodes in the graph). The interpretation is that the nodes are cities, the
edges are one-way roads, and an edge weight is the distance between the
two edge nodes (i.e., cities) in that direction. For example, the "road"
between city a and city b might have cost 5 and the "road" between city
b and city a might have cost 2 (or there might not a "road" going from
city b to city a).
2. Tour Definition
Given a starting city and a road network, a tour is defined as follows. A
tour is defined to be a list of cities, such that:
every city in the network appears in the list at least once;
only the starting city can appear in the list more than once;
the starting city is both the first and the last element in the list;
the starting city appears in the list exactly twice.
the list contains exactly one more element than the number of distinct
cities in the road network.
For example, for a road network that only contains one city, there would
only be one tour with two elements, that city appearing twice, e.g., let the
city be a, then the tour is the list [a, a]. Note that since the graph need
not be complete, not all road networks have tours. In this case, solution/4
(the predicate solution with 4 arguments) should simply fail.
The cost of a tour is the sum of the edge costs between the successive
nodes in the tour list. The distance between any city and itself is zero.
Therefore the cost of the tour [a, a] is 0. If there were two cities, a and b,
and the distance from a to b is 4 and the distance from b to a is 3, and a
is the starting city then the only legal tour is the list [a, b, a], which has
a cost of 7. If there were no road (edge) from b to a then solution/4 will
fail.

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来源： https://www.cnblogs.com/comp9321/p/11425814.html