COMP9414 Artificial Intelligence - Assignment 1: Constraint Satisfaction Search
COMP9414 Artificial Intelligence
Assignment 1: Constraint Satisfaction Search
Objective
This assignment concerns developing optimal solutions to a scheduling problem inspired by the scenario of hosting a number of visitors to an organization such as a university department. Each visitor must have a number of meetings, however there are both constraints on availability (of rooms and hosts), and preferences of each visitor for the days and times of each meeting. Some of the constraints are "hard" constraints (cannot be violated in any solution), while the preferences are "soft" constraints (can be satisfied to more or less degree). Each soft constraint has a cost function giving the "penalty" for scheduling the meeting at a given time (lower costs are preferred). The aim is to schedule all the required meetings so that the sum total of all the penalties is minimized, and all the constraints are satisfied.
More technically, this assignment is an example of a constraint optimization problem, a problem that has constraints like a standard Constraint Satisfaction Problem (CSP), but also costs associated with each solution. For this assignment, we will use a greedy algorithm to find optimal solutions to visitor hosting problems that are specified as text strings. However, unlike the greedy search algorithm described in the lectures on search, this greedy algorithm has the property that it is guaranteed to find an optimal solution for any problem (if a solution exists).
The assignment will use the AIPython code of Poole & Mackworth. You are given code to translate visitor hosting problems specified as text strings into CSPs with a cost, and you are given code for several constraint solving algorithms -- based on domain splitting and arc consistency, and based on depth-first search. The assignment will be to implement some missing procedures and to analyse the performance of the constraint solving methods, both analytically and experimentally.
Submission Instructions
-
This is an individual assignment.
-
Submit this Jupyter notebook on the CSE give system (or vlab) using the following command: give cs9414 ass1 ass1.ipynb
-
This means that the Jupyter notebook file must be called
ass1.ipynb. -
Make sure you set up AIPython (as done below) so the code can be run on either CSE machines or a marker\'s own machine.
-
Do not submit any AIPython code. Hence do not change any AIPython code to make your code run.
-
Make sure your notebook runs cleanly (restart the kernel and run each cell to check).
-
Do not modify the existing code in this notebook except to answer the questions. Marks will be given as and where indicated.
-
If you want to submit additional code (e.g. for generating plots), add that at the end of the notebook.
-
Do not distribute any of this code on the Internet.
Late Penalties
Standard UNSW late penalties apply.
Plagiarism
Remember that ALL work submitted for this assignment must be your own work and no sharing or copying of code or answers is allowed. You may discuss the assignment with other students but must not collaborate on developing answers to the questions. You may use code from the Internet only with suitable attribution of the source. You may not use ChatGPT or any similar software to generate any part of your explanations, evaluations or code. Do not use public code repositories on sites such as github or file sharing sites such as Google Drive to save any part of your work -- make sure your code repository or cloud storage is private and do not share any links. This also applies after you have finished the course, as we do not want next year's students accessing your solution, and plagiarism penalties can still apply after the course has finished.
All submitted assignments will be run through plagiarism detection software to detect similarities to other submissions, including from past years. You should carefully read the UNSW policy on academic integrity and plagiarism (linked from the course web page), noting, in particular, that collusion (working together on an assignment, or sharing parts of assignment solutions) is a form of plagiarism.
- Finally, do not use any contract cheating "academies" or online
- "tutoring" services. This counts as serious misconduct with heavy
- penalties up to automatic failure of the course with 0 marks, and
- expulsion from the university for repeat offenders.
- ::
:: {.cell .markdown}
The Visitor Hosting Problem
A CSP for this assignment is a set of variables representing meetings, binary constraints on pairs of meetings, and unary constraints (hard or soft) on meetings. The domains are all working hours in one week, and meetings are all assumed to be 1 hour duration. Days are represented (in the input and output) as strings 'mon', 'tue', 'wed', 'thu' and 'fri', and times are represented as strings '9am', '10am', '11am', '12pm', '1pm', '2pm', '3pm' and '4pm'. The only possible values are a combination of a day and time, e.g. 'mon 9am'. Each meeting name is a string (with no spaces), and each constraint is hard or soft.
There are three types of constraint:
- Binary Constraints: These specify a hard requirement for the relationship between two meetings.
- Hard Domain Constraints: These specify hard requirements for the meetings themselves.
- Soft Domain Constraints: These constraints are non-critical requirements for a meeting that represent preferences.
Each soft constraint has a function defining the cost associated with violating the preference, that the constraint solver must minimize, while respecting all the hard constraints. The cost of a solution is simply the sum of the costs for the soft constraints that the solution violates (and is always a non-negative integer).
This is the list of possible constraints for a visitor hosting problem (comments below are for explanation and do not appear in the input specification; however, the code we supply should work with comments that take up a full line):
# binary constraints
constraint, ⟨m1⟩ before ⟨m2⟩ # m1 must start before m2 starts but m2 could start when m1 finishes
constraint, ⟨m1⟩ same-day ⟨m2⟩ # m1 and m2 must start on the same day
constraint, ⟨m1⟩ one-day-between ⟨m2⟩ # 1 whole day between m1 and m2
constraint, ⟨m1⟩ one-hour-between ⟨m2⟩ # 1 hour between the end of m1 and the start of m2
# hard domain constraints
domain, ⟨m⟩, ⟨day⟩, hard # must start on day
domain, ⟨m⟩, ⟨time⟩, hard # must start at time but on any day
domain, ⟨m⟩, ⟨day1⟩ ⟨time1⟩-⟨day2⟩ ⟨time2⟩, hard # day-time range for start time; includes day1, time1 and day2, time2
domain, ⟨m⟩, morning, hard # finishes at or before 12pm
domain, ⟨m⟩, afternoon, hard # starts on or after 12pm
# all of these are strictly before/after, not "on or before/after"
domain, ⟨m⟩, before ⟨day⟩, hard # must be on a previous day
domain, ⟨m⟩, before ⟨time⟩, hard # must start strictly before time but on any day
domain, ⟨m⟩, before ⟨day⟩ ⟨time⟩, hard # must start strictly before day, time
domain, ⟨m⟩, after ⟨day⟩, hard # must be on a following day
domain, ⟨m⟩, after ⟨time⟩, hard # must start after time but on any day
domain, ⟨m⟩, after ⟨day⟩ ⟨time⟩, hard # must start after day, time and could be on a following day
# soft domain constraints # cost for scheduling at day, time
domain, ⟨m⟩, early-week, soft # the number of days from mon to day (0 if day = mon)
domain, ⟨m⟩, late-week, soft # the number of days from day to fri (0 if day = fri)
domain, ⟨m⟩, early-morning, soft # the number of hours from 9am to time (0 if time = 9am)
domain, ⟨m⟩, midday, soft # the number of hours from 12pm to time (0 if time = 12pm)
domain, ⟨m⟩, late-afternoon, soft # the number of hours from time to 4pm (0 if time = 4pm)The input specification will consist of several "blocks", listing the meetings, binary constraints, hard unary constraints and soft unary constraints for the given problem. So a declaration of each meeting will be included before it is used in a constraint. A sample input specification is as follows. Comments are for explanation and do not have to be included in the input.
# two meetings with one binary constraint and the same domain constraints
meeting, m1
meeting, m2
# one binary constraint
constraint, m1 before m2
# domain constraints
domain, m1, mon, hard
domain, m2, mon, hard
domain, m1, early-morning, soft
domain, m2, early-morning, soft- :::
:: {.cell .markdown}
Preparation
1. Set up AIPython
You will need AIPython for this assignment. To find the aipython files, the aipython directory has to be added to the Python path.
Do this temporarily, as done here, so we can find AIPython and run your code (you will not submit any AIPython code).
- You can add either the full path (using
os.path.abspath), or as in the - code below, the relative path.
- ::
:: {.cell .code}
import sys
sys.path.append('aipython') # change to your directory
sys.path # check that aipython is now on the path- :::
:: {.cell .markdown}
2. Representation of Day Times
Input and output are day time strings such as \'mon 10am\' or a range of day time strings such as \'mon 10am-mon 4pm\'.
The CSP will represent these as integer hour numbers in the week, ranging from 0 to 39.
- The following code handles the conversion between day times and hour
- numbers.
- ::
:: {.cell .code execution_count="2"}
# -*- coding: utf-8 -*-
""" day_time string format is a day plus time, e.g. Mon 10am, Tue 4pm, or just Tue or 4pm
if only day or time, returns day number or hour number only
day_time strings are converted to and from integer hours in the week from 0 to 39
"""
class Day_Time():
num_hours_in_day = 8
num_days_in_week = 5
def __init__(self):
self.day_names = ['mon','tue','wed','thu','fri']
self.time_names = ['9am','10am','11am','12pm','1pm','2pm','3pm','4pm']
def string_to_week_hour_number(self, day_time_str):
""" convert a single day_time into an integer hour in the week """
value = None
value_type = None
day_time_list = day_time_str.split()
if len(day_time_list) == 1:
str1 = day_time_list[0].strip()
if str1 in self.time_names: # this is a time
value = self.time_names.index(str1)
value_type = 'hour_number'
else:
value = self.day_names.index(str1) # this is a day
value_type = 'day_number'
# if not day or time, throw an exception
else:
value = self.day_names.index(day_time_list[0].strip())*self.num_hours_in_day \
+ self.time_names.index(day_time_list[1].strip())
value_type = 'week_hour_number'
return (value_type, value)
def string_to_number_set(self, day_time_list_str):
""" convert a list of day-times or ranges 'Mon 9am, Tue 9am-Tue 4pm' into a list of integer hours in the week
e.g. 'mon 9am-1pm, mon 4pm' -> [0,1,2,3,4,7]
"""
number_set = set()
type1 = None
for str1 in day_time_list_str.lower().split(','):
if str1.find('-') > 0:
# day time range
type1, v1 = self.string_to_week_hour_number(str1.split('-')[0].strip())
type2, v2 = self.string_to_week_hour_number(str1.split('-')[1].strip())
if type1 != type2: return None # error, types in range spec are different
number_set.update({n for n in range(v1, v2+1)})
else:
# single day time
type2, value2 = self.string_to_week_hour_number(str1)
if type1 != None and type1 != type2: return None # error: type in list is inconsistent
type1 = type2
number_set.update({value2})
return (type1, number_set)
# function for morning constraint: starts 9am, 10am or 11am
def is_morning(self, week_hour_number):
h, d = self.hour_day_split(week_hour_number)
return h in [0, 1, 2]
# function for afternoon constraint: starts 12pm,..., 5pm
def is_afternoon(self, week_hour_number):
h, d = self.hour_day_split(week_hour_number)
return h in [3, 4, 5, 6, 7, 8]
# convert integer hour in week to day time string
def week_hour_number_to_day_time(self, week_hour_number):
hour = self.day_hour_number(week_hour_number)
day = self.day_number(week_hour_number)
return self.day_names[day]+' '+self.time_names[hour]
# convert integer hour in week to integer day and integer time in day
def hour_day_split(self, week_hour_number):
return (self.day_hour_number(week_hour_number), self.day_number(week_hour_number))
# convert integer hour in week to integer day in week
def day_number(self, week_hour_number):
return int(week_hour_number / self.num_hours_in_day)
# convert integer hour in week to integer time in day
def day_hour_number(self, week_hour_number):
return week_hour_number % self.num_hours_in_day
def __repr__(self):
day_hour_number = self.week_hour_number % self.num_hours_in_day
day_number = int(self.week_hour_number / self.num_hours_in_day)
return self.day_names[day_number]+' '+self.time_names[day_hour_number]- :::
:: {.cell .markdown}
3. Constraint Satisfaction Problems with Costs
Since AI Python does not provide the CSP class with an explicit cost, we
implement our own class that extends CSP.
- We also store the cost functions explicitly in the CSP.
- ::
:: {.cell .code execution_count="3"}
from cspProblem import CSP
# implement nodes as CSP problems as nodes with cost functions
class CSP_with_Cost(CSP):
""" cost_functions maps a CSP var, here a meeting name, to a list of functions for the constraints that apply """
def __init__(self, domains, constraints, cost_functions):
self.domains = domains
self.variables = self.domains.keys()
super().__init__("title of csp", self.variables, constraints)
self.cost_functions = cost_functions
self.cost = self.calculate_cost()
# specific to the visitor hosting csp
def calculate_cost(self):
""" this is really a function f = path cost + heuristic to be used by the constraint solver """
cost = 0
# TODO: write cost function
return cost
def __repr__(self):
""" string representation of an arc"""
return "CSP_with_Cost("+str(list(self.domains.keys()))+':'+str(self.cost)+")"- :::
:: {.cell .markdown} This formulates a solver for a CSP with cost as a search problem, using domain splitting with arc consistency to define the successors of a node.
- ::
:: {.cell .code execution_count="4"}
from cspProblem import Constraint
from cspConsistency import Con_solver, select, partition_domain
from searchProblem import Arc, Search_problem
from operator import lt, gt
# rewrites rather than extends Search_with_AC_from_CSP
class Search_with_AC_from_Cost_CSP(Search_problem):
""" A search problem with domain splitting and arc consistency """
def __init__(self, csp):
self.cons = Con_solver(csp) # copy of the CSP with access to arc consistency algorithms
self.domains = self.cons.make_arc_consistent(csp.domains)
self.constraints = csp.constraints
self.cost_functions = csp.cost_functions
csp.domains = self.domains # after arc consistency
self.csp = csp
def is_goal(self, node):
""" node is a goal if all domains have exactly 1 element """
return all(len(node.domains[var]) == 1 for var in node.domains)
def start_node(self):
return self.csp
def neighbors(self, node):
""" returns the neighbouring nodes of the CSP_with_Cost node from domain splitting """
neighs = []
var = select(x for x in node.domains if len(node.domains[x]) > 1) # chosen at random
if var:
dom1, dom2 = partition_domain(node.domains[var])
self.display(2,"Splitting", var, "into", dom1, "and", dom2)
to_do = self.cons.new_to_do(var, None)
for dom in [dom1, dom2]:
newdoms = node.domains | {var: dom} # overwrite domain of var with dom
cons_doms = self.cons.make_arc_consistent(newdoms, to_do)
if all(len(cons_doms[v]) > 0 for v in cons_doms):
# all domains are non-empty
# make new CSP_with_Cost node to continue the search
csp_node = CSP_with_Cost(cons_doms, self.constraints, self.cost_functions)
neighs.append(Arc(node, csp_node))
else:
self.display(2,"...",var,"in",dom,"has no solution")
return neighs
def heuristic(self, n):
return n.cost- :::
:: {.cell .markdown}
4. Visitor Hosting Constraint Satisfaction Problems
The following code sets up a CSP problem from a given specification.
- Hard (unary) domain constraints are applied to reduce the domains of the
- variables before the constraint solver runs.
- ::
:: {.cell .code execution_count="5"}
# domain specific CSP builder for week schedule
class CSP_builder():
# list of text lines without comments and empty lines
type1, default_domain = Day_Time().string_to_number_set('mon 9am-fri 4pm') # should be 0,...,39
# hard unary constraints: domain is a list of values
def apply_hard_morning(self, domain):
domain_orig = domain.copy()
for val in domain_orig:
if not Day_Time().is_morning(val):
if val in domain: domain.remove(val)
return domain
def apply_hard_afternoon(self, domain):
domain_orig = domain.copy()
for val in domain_orig:
if not Day_Time().is_afternoon(val):
if val in domain: domain.remove(val)
return domain
# param is a single value
def apply_hard_before(self, param_type, param, domain):
domain_orig = domain.copy()
param_val = param.pop()
for val in domain_orig:
h, d = Day_Time().hour_day_split(val)
if param_type == 'hour_number' and not h < param_val:
if val in domain: domain.remove(val)
if param_type == 'day_number' and not d < param_val:
if val in domain: domain.remove(val)
if param_type == 'week_hour_number' and not val < param_val:
if val in domain: domain.remove(val)
return domain
def apply_hard_after(self, param_type, param, domain):
domain_orig = domain.copy()
param_val = param.pop()
for val in domain_orig:
h, d = Day_Time().hour_day_split(val)
if param_type == 'hour_number' and not h > param_val:
if val in domain: domain.remove(val)
if param_type == 'day_number' and not d > param_val:
if val in domain: domain.remove(val)
if param_type == 'week_hour_number' and not val > param_val:
if val in domain: domain.remove(val)
return domain
def apply_hard_same_day(self, param_type, param, domain):
domain_orig = domain.copy()
for val in domain_orig:
h, d = Day_Time().hour_day_split(val)
if param_type == 'hour_number' and h not in param:
if val in domain: domain.remove(val)
if param_type == 'day_number' and d not in param:
if val in domain: domain.remove(val)
if param_type == 'week_hour_number' and val not in param:
if val in domain: domain.remove(val)
return domain
def apply_hard_same_as(self, param_type, param, domain):
domain_orig = domain.copy()
for val in domain_orig:
h, d = Day_Time().hour_day_split(val)
if param_type == 'hour_number' and h not in param:
if val in domain: domain.remove(val)
if param_type == 'day_number' and d not in param:
if val in domain: domain.remove(val)
if param_type == 'week_hour_number' and val not in param:
if val in domain: domain.remove(val)
return domain
# soft unary constraints: return cost to break constraint
def early_morning_soft(self, day, hour):
return hour
def late_afternoon_soft(self, day, hour):
return 7 - hour
def midday_soft(self, day, hour):
# midday is 12pm
return abs(hour - 3)
def early_week_soft(self, day, hour):
return day
def late_week_soft(self, day, hour):
return 4 - day
def no_cost(self, day ,hour):
return 0
# hard binary constraints
# one day gap
def one_day_between(self, week_hour1, week_hour2):
h1, d1 = Day_Time().hour_day_split(week_hour1)
h2, d2 = Day_Time().hour_day_split(week_hour2)
return abs(d1 - d2) > 1
def one_hour_between(self, week_hour1, week_hour2):
h1, d1 = Day_Time().hour_day_split(week_hour1)
h2, d2 = Day_Time().hour_day_split(week_hour2)
return abs(h1 - h2) > 1 or d1 != d2
def same_day(self, week_hour1, week_hour2):
h1, d1 = Day_Time().hour_day_split(week_hour1)
h2, d2 = Day_Time().hour_day_split(week_hour2)
return d1 == d2
# domain is a list of values
def apply_hard_constraint(self, domain, spec):
if spec == 'morning':
return self.apply_hard_morning(domain)
elif spec == 'afternoon':
return self.apply_hard_afternoon(domain)
else:
dt = spec.strip()
if dt.find('before') == 0:
param_type,param = Day_Time().string_to_number_set(dt[len('before'):].strip())
if len(param) != 1:
return None # before, after should be single value
return self.apply_hard_before(param_type,param,domain)
elif dt.find('after') == 0:
param_type,param = Day_Time().string_to_number_set(dt[len('after'):].strip())
if len(param) != 1:
return None # before, after should be single value
return self.apply_hard_after(param_type,param,domain)
else:
# if not a keyword, must be day-time
param_type,param = Day_Time().string_to_number_set(dt)
return self.apply_hard_same_as(param_type,param,domain)
def get_cost_function(self, spec):
func_dict = {'early-morning':self.early_morning_soft, 'late-afternoon':self.late_afternoon_soft,
'midday':self.midday_soft, 'early-week':self.early_week_soft, 'late-week':self.late_week_soft, 'no-cost':self.no_cost}
return [func_dict[spec]]
def get_binary_constraint(self, spec):
tokens = spec.strip().split(' ')
if len(tokens) < 3: return None # error in spec
# meeting1 relation meeting2
fun_dict = {'before':lt, 'after':gt, 'one-day-between':self.one_day_between, 'one-hour-between':self.one_hour_between, 'same-day':self.same_day}
return Constraint((tokens[0].strip(), tokens[2].strip()), fun_dict[tokens[1].strip()])
def get_CSP_with_Cost(self, input_lines):
domains = dict()
constraints = []
cost_functions = dict()
# process each input line of the specification
for input_line in input_lines:
func_spec = None
input_line_tokens = input_line.strip().split(',')
if len(input_line_tokens) % 2 != 0:
return None # must have even number of tokens
if len(input_line_tokens) < 2:
return None
if input_line_tokens[0].strip() == 'meeting':
key = input_line_tokens[1].strip()
domains[key] = self.default_domain.copy() # meeting name and domain
# get zero cost function for this meeting as default, may add real cost later
cost_functions[key] = self.get_cost_function('no-cost')
elif input_line_tokens[0].strip() == 'domain':
key = input_line_tokens[1].strip()
for token1 in input_line_tokens[1:]:
if token1.strip() == 'hard':
# by now, fun_spec string should be set, because spec string comes before hard, soft
domains[key] = self.apply_hard_constraint(domains[key], func_spec)
elif token1.strip() == 'soft':
cost_functions[key] += self.get_cost_function(func_spec)
else:
func_spec = token1.strip()
elif input_line_tokens[0].strip() == 'constraint': # binary constraint
if len(input_line_tokens) < 2: return None # error in spec
constraints.append(self.get_binary_constraint(input_line_tokens[1].strip()))
else:
return None
return CSP_with_Cost(domains, constraints, cost_functions)
def create_CSP_from_spec(spec: str):
input_lines = list()
spec = spec.split('\n')
# strip comments
for input_line in spec:
input_line = input_line.split('#')
if len(input_line[0]) > 0:
input_lines.append(input_line[0])
print(input_line[0])
# construct initial CSP problem
csp = CSP_builder()
csp_problem = csp.get_CSP_with_Cost(input_lines)
return csp_problem- :::
:: {.cell .markdown}
5. Greedy Search Constraint Solver using Domain Splitting and Arc Consistency
Create a GreedySearcher to search over the CSP.
- The cost function for CSP nodes is used as the heuristic, but is
- actually a direct estimate of the total path cost function f used in
- A* Search.
- ::
:: {.cell .code execution_count="6"}
from searchGeneric import AStarSearcher
class GreedySearcher(AStarSearcher):
""" returns a searcher for a problem.
Paths can be found by repeatedly calling search().
"""
def add_to_frontier(self, path):
""" add path to the frontier with the appropriate cost """
# value = path.cost + self.problem.heuristic(path.end()) -- A* definition
value = path.end().cost
self.frontier.add(path, value) - :::
:: {.cell .markdown} Run the GreedySearcher on the CSP derived from the sample input.
- ::
:: {.cell .code execution_count="7"}
# Sample problem specification
sample_spec = """
# two meetings with one binary constraint and the same domain constraints
meeting, m1
meeting, m2
# one binary constraint
constraint, m1 before m2
# domain constraints
domain, m1, mon, hard
domain, m2, mon, hard
domain, m1, early-morning, soft
domain, m2, early-morning, soft
"""- :::
:: {.cell .code}
# display details (0 turns off)
Con_solver.max_display_level = 0
Search_with_AC_from_Cost_CSP.max_display_level = 2
GreedySearcher.max_display_level = 0
def test_csp_solver(searcher):
final_path = searcher.search()
if final_path == None:
print('No solution')
else:
domains = final_path.end().domains
result_str = ''
for name, domain in domains.items():
for n in domain:
result_str += '\n'+str(name)+': '+Day_Time().week_hour_number_to_day_time(n)
print(result_str[1:]+'\ncost: '+str(final_path.end().cost))
csp_problem = create_CSP_from_spec(sample_spec)
solver = GreedySearcher(Search_with_AC_from_Cost_CSP(csp_problem))
test_csp_solver(solver)- :::
:: {.cell .markdown}
6. Depth-First Search Constraint Solver
The Depth-First Constraint Solver in AIPython by default uses a random ordering of the variables in the CSP.
We need to modify this code to make it compatible with the arc consistency solver.
- Run the solver by calling
dfs_solve1(first solution) or dfs_solve_all(all solutions).- ::
:: {.cell .code execution_count="9"}
num_expanded = 0
display = False
def dfs_solver(constraints, domains, context, var_order):
""" generator for all solutions to csp
context is an assignment of values to some of the variables
var_order is a list of the variables in csp that are not in context
"""
global num_expanded, display
to_eval = {c for c in constraints if c.can_evaluate(context)}
if all(c.holds(context) for c in to_eval):
if var_order == []:
print("Nodes expanded to reach solution:", num_expanded)
yield context
else:
rem_cons = [c for c in constraints if c not in to_eval]
var = var_order[0]
for val in domains[var]:
if display:
print("Setting", var, "to", val)
num_expanded += 1
yield from dfs_solver(rem_cons, domains, context|{var:val}, var_order[1:])
def dfs_solve_all(csp, var_order=None):
""" depth-first CSP solver to return a list of all solutions to csp """
global num_expanded
num_expanded = 0
if var_order == None: # use an arbitrary variable order
var_order = list(csp.domains)
return list(dfs_solver(csp.constraints, csp.domains, {}, var_order))
def dfs_solve1(csp, var_order=None):
""" depth-first CSP solver """
global num_expanded
num_expanded = 0
if var_order == None: # use an arbitrary variable order
var_order = list(csp.domains)
for sol in dfs_solver(csp.constraints, csp.domains, {}, var_order):
return sol # return first one- :::
:: {.cell .markdown} Run the Depth-First Solver on the sample problem.
- ::
:: {.cell .code}
def test_dfs_solver(csp_problem):
solution = dfs_solve1(csp_problem)
if solution == None:
print('No solution')
else:
result_str = ''
for name in solution.keys():
result_str += '\n'+str(name)+': '+Day_Time().week_hour_number_to_day_time(solution[name])
print(result_str[1:])
# call the Depth-First Search solver
csp_problem = create_CSP_from_spec(sample_spec)
test_dfs_solver(csp_problem) # set display to True to see number of nodes expanded- :::
:: {.cell .markdown}
7. Depth-First Search Constraint Solver using Forward Checking with MRV Heuristic
The Depth-First Constraint Solver in AIPython by default uses a random ordering of the variables in the CSP.
We redefine the dfs_solver methods to implement the MRV (Minimum
Remaining Values) heuristic using forward checking.
- Because the AIPython code is designed to manipulate domain sets, we also
- need to redefine
can_evaluateto handle partial assignments. - ::
:: {.cell .code execution_count="11"}
num_expanded = 0
display = False
def can_evaluate(c, assignment):
""" assignment is a variable:value dictionary
returns True if the constraint can be evaluated given assignment
"""
return assignment != {} and all(v in assignment.keys() and type(assignment[v]) != list for v in c.scope)
def mrv_dfs_solver(constraints, domains, context, var_order):
""" generator for all solutions to csp.
context is an assignment of values to some of the variables.
var_order is a list of the variables in csp that are not in context.
"""
global num_expanded, display
if display:
print("Context", context)
to_eval = {c for c in constraints if can_evaluate(c, context)}
if all(c.holds(context) for c in to_eval):
if var_order == []:
print("Nodes expanded to reach solution:", num_expanded)
yield context
else:
rem_cons = [c for c in constraints if c not in to_eval] # constraints involving unset variables
var = var_order[0]
rem_vars = var_order[1:]
for val in domains[var]:
if display:
print("Setting", var, "to", val)
num_expanded += 1
rem_context = context|{var:val}
# apply forward checking on remaining variables
if len(var_order) > 1:
rem_vars_original = list((v, list(domains[v].copy())) for v in rem_vars)
if display:
print("Original domains:", rem_vars_original)
# constraints that can't already be evaluated in rem_cons
rem_cons_ff = [c for c in constraints if c in rem_cons and not can_evaluate(c, rem_context)]
for rem_var in rem_vars:
# constraints that can be evaluated by adding a value of rem_var to rem_context
any_value = list(domains[rem_var])[0]
rem_to_eval = {c for c in rem_cons_ff if can_evaluate(c, rem_context|{rem_var: any_value})}
# new domain for rem_var are the values for which all newly evaluable constraints hold
rem_vals = domains[rem_var].copy()
for rem_val in domains[rem_var]:
# no constraint with rem_var in the existing context can be violated
for c in rem_to_eval:
if not c.holds(rem_context|{rem_var: rem_val}):
if rem_val in rem_vals:
rem_vals.remove(rem_val)
domains[rem_var] = rem_vals
# order remaining variables by MRV
rem_vars.sort(key=lambda v: len(domains[v]))
if display:
print("After forward checking:", list((v, domains[v]) for v in rem_vars))
if rem_vars == [] or all(len(domains[rem_var]) > 0 for rem_var in rem_vars):
yield from mrv_dfs_solver(rem_cons, domains, context|{var:val}, rem_vars)
# restore original domains if changed through forward checking
if len(var_order) > 1:
if display:
print("Restoring original domain", rem_vars_original)
for (v, domain) in rem_vars_original:
domains[v] = domain
if display:
print("Nodes expanded so far:", num_expanded)
def mrv_dfs_solve_all(csp, var_order=None):
""" depth-first CSP solver to return a list of all solutions to csp """
global num_expanded
num_expanded = 0
if var_order == None: # order variables by MRV
var_order = list(csp.domains)
var_order.sort(key=lambda var: len(csp.domains[var]))
return list(mrv_dfs_solver(csp.constraints, csp.domains, {}, var_order))
def mrv_dfs_solve1(csp, var_order=None):
""" depth-first CSP solver """
global num_expanded
num_expanded = 0
if var_order == None: # order variables by MRV
var_order = list(csp.domains)
var_order.sort(key=lambda var: len(csp.domains[var]))
for sol in mrv_dfs_solver(csp.constraints, csp.domains, {}, var_order):
return sol # return first one- :::
:: {.cell .markdown} Run this solver on the sample problem.
- ::
:: {.cell .code}
def test_mrv_dfs_solver(csp_problem):
solution = mrv_dfs_solve1(csp_problem)
if solution == None:
print('No solution')
else:
result_str = ''
for name in solution.keys():
result_str += '\n'+str(name)+': '+Day_Time().week_hour_number_to_day_time(solution[name])
print(result_str[1:])
# call the Depth-First MRV Search solver
csp_problem = create_CSP_from_spec(sample_spec)
test_mrv_dfs_solver(csp_problem) # set display to True to see number of nodes expanded- :::
:: {.cell .markdown}
Assignment
- :::
:: {.cell .markdown} Name:
- zID:
- ::
:: {.cell .markdown}
Question 1 (4 marks)
Consider the search spaces for the CSP solvers -- domain splitting with arc consistency and the DFS solver (without forward checking).
- Describe the search spaces informally in terms of start state, successor functions and goal state(s) (1 mark)
- What is the branching factor and maximum depth to find any solution for the two algorithms (ignoring costs)? (1 mark)
- What is the worst case time and space complexity of the two search algorithms (give the general form)? (1 mark)
-
- Give an example of a problem that is easier for the domain
- splitting with arc consistency solver than it is for the DFS solver,
- and explain why (1 mark)
- ::
::: {.cell .markdown} Answers for Question 1
- Write the answers here.
- ::
:: {.cell .markdown}
Question 2 (5 marks)
Define the cost function for a visitor hosting CSP (i.e. a node in the search space for domain splitting and arc consistency) as the total cost of the soft constraints violated for all of the variables, assuming that each variable is assigned the best possible value from its domain, where the "best" value for a variable v is the one that has the lowest total cost to violate the soft constraints for that variable v -- note that there can be more than one soft constraint applying to v so we add up the costs of violating each of those constraints to define the best value for v.
- Implement the cost function in the indicated cell and place a copy of the code below (3 marks)
- What is its computational complexity (give a general form)? (1 mark)
-
- Show that the cost function f never decreases along a path, and
- explain why this means the search algorithm is optimal (1 mark)
- ::
::: {.cell .code}
# Code for Question 2
# Place a copy of your code here and run it in the relevant cell- :::
:: {.cell .markdown} Answers for Question 2
- Write the other answers here.
- ::
:: {.cell .markdown}
Question 3 (4 marks)
Perform an empirical evaluation of the domain splitting CSP solver using the cost function defined as above compared to using no cost function (i.e. the zero cost function, as originally defined in the above cell). Use the average number of nodes expanded as a metric to compare the two algorithms.
- Write a function
generate_problem(n)that takes an integernand generates a problem specification withnmeetings and a random set of hard and soft constraints in the correct format for the constraint solvers (2 marks)
Run the CSP solver (with and without the cost function) over a number of
problems of size n for a range of values of n.
- Plot the performance of the two constraint solving algorithms on the
above metric against
n(1 mark) -
- Quantify the performance gain (if any) achieved by the use of this
- cost function (1 mark)
- ::
::: {.cell .code}
# Code for Question 3
# Place your code here- :::
:: {.cell .markdown} Answers for Question 3
- Write the other answers here.
- ::
:: {.cell .markdown}
Question 4 (5 marks)
Compare the Depth-First Search (DFS) solver to the Depth-First Search solver using forward checking with Minimum Remaining Values heuristic (DFS-MRV). For this question, ignore the costs associated with the CSP problems.
- What is the worst case time and space complexity of each algorithm? (1 mark)
- What are the properties of the search algorithms (completeness, optimality)? (1 mark)
- Give an example of a problem that is easier for the DFS-MRV solver than it is for the DFS solver, and explain why (1 mark)
- Empirically compare the quality of the first solution found by DFS and DFS-MRV compared to the optimal solution (1 marks)
- Empirically compare DFS-MRV with DFS in terms of the number of nodes expanded (1 mark)
- For the empirical evaluations, run the two algorithms on a variety of
- problems of size
nfor varyingn. Note that the domain splitting CSP - solver with costs should always find an optimal solution.
- ::
:: {.cell .markdown} Answers for Question 4
- If you want to submit additional code, put this at the end of the
- notebook. Here just give the answers (including plots or tables).
- ::
:: {.cell .markdown}
Question 5 (4 marks)
The DFS solver chooses variables in random order, and systematically explores all values for those variables in no particular order.
Incorporate costs into the DFS constraint solver. Similar to the cost function for the domain splitting solver, for a given variable v, the cost of assigning the value val to v is the total cost of violating all the soft constraints associated with v for the value val. The minimum cost for v is the lowest cost from amongst the values in the domain of v. The DFS solver should choose the variable v with lowest minimum cost, and explore its values in order of cost from lowest to highest.
- Implement this behaviour by modifiying the code in
dfs_solverand place a copy of the code below (2 marks) - Empirically compare the performance of DFS with and without these heuristics (2 marks)
- For the empirical evaluations, again run the two algorithms on a variety
- of problems of size
nfor varyingn. - ::
:: {.cell .code}
# Code for Question 5
# Place a copy of your code here and run it in the relevant cell- :::
:: {.cell .markdown} Answers for Question 5
- Write the other answers here.
- ::
:: {.cell .markdown}
Question 6 (3 marks)
The CSP solver with domain splitting splits a CSP variable domain into exactly two partitions. Poole & Mackworth claim that in practice, this is as good as splitting into a larger number of partitions. In this question, empirically evaluate this claim for the visitor hosting CSP.
- Write a new
partition_domainfunction that partitions a domain into a list ofkpartitions, wherekis a parameter to the function (1 mark) -
- Modify the CSP solver to use the list of
kpartitions and evaluate - the performance of the solver using the above metric for a range of
- values of
k(2 marks) - ::
- Modify the CSP solver to use the list of
::: {.cell .code}
# Code for Question 6
# Place a copy of your code here and run it in the relevant cell- :::
:: {.cell .markdown} Answers for Question 6
- Write the other answers here.
- ::
