partage/ift7025-tp1
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity

Merge branch 'master' of git.francoispelletier.org:francois/ift7025-tp1

Browse source
This commit is contained in:
François Pelletier 2019-04-10 23:58:28 -04:00
commit 3c87d54f68
3 changed files with 16 additions and 4 deletions
Show stats Download patch file Download diff file Expand all files Collapse all files

6
README.md
View file

@ -1,2 +1,6 @@
# ift7025-tp1
# IFT-7025 Hiver 2019 - Travail pratique #1
IFT-7025 Hiver 2019
Travail pratique #1
remis par François Pelletier 908144032

4
search/search.py
View file

@ -10,6 +10,10 @@ al.ca# search.py
# (denero@cs.berkeley.edu) and Dan Klein (klein@cs.berkeley.edu).
# Student side autograding was added by Brad Miller, Nick Hay, and
# Pieter Abbeel (pabbeel@cs.berkeley.edu).
#
# IFT-7025 Hier 2019
# François Pelletier
# 908144032
"""

10
search/searchAgents.py
View file

@ -10,7 +10,10 @@
# (denero@cs.berkeley.edu) and Dan Klein (klein@cs.berkeley.edu).
# Student side autograding was added by Brad Miller, Nick Hay, and
# Pieter Abbeel (pabbeel@cs.berkeley.edu).
#
# IFT-7025 Hier 2019
# François Pelletier
# 908144032
"""
This file contains all of the agents that can be selected to control Pacman. To
@ -307,7 +310,7 @@ class CornersProblem(search.SearchProblem):
"*** YOUR CODE HERE ***"
currentPosition, currentCornerSet = state
# print "Validation du but a ", currentPosition, "Coins restants: ", currentCornerSet
# On atteint le but s'il reste un seul coin a atteindre, et que c'est l'etat actuel
# On atteint le but s'il restait un seul coin a atteindre, et que c'est l'etat actuel
# print "But atteint"
return (len(currentCornerSet)==1 and currentPosition in currentCornerSet)
@ -393,7 +396,8 @@ def cornersHeuristic(state, problem):
# Liste des distances de Manhattan vers les sommets restants
mh=[]
for i in range(nb_corners):
mh.append(util.manhattanDistance(currentPosition,currentCornerList[i]))
current_distance = util.manhattanDistance(currentPosition,currentCornerList[i])
mh.append(current_distance)
# On retourne la plus grande distance, qui est le meilleur scenario possible pour terminer
return max(mh)