Programming Challenge from Kattis: Watchdog
$begingroup$
So I've tried to solve the kattis-challenge called Watchdog. My code works just fine, but it is too slow on the last test. I'm wondering if anyone sees any big algorithmic performance-issues.
A short description of the task: First line from .in states number of test-cases.
First line in each test-case gives length of a square and number of hatches respectively. For each hatch you'll have one line giving x and y-point of hatch respectively. The mission is to find a point x,y where the length from x,y to any hatch is not greater than the length from x,y to any point outside of the square. If no such point, print poodle
. If several such point print point with lowest x, if several points with equal x, print point with lowest y. Hatches and these points cannot overlap.
I think one issue might be extensive usage of lists, but I'm also not sure how to avoid them. Initially I started reading one and one line from stdin, using the data. That way things looked way more readable, but it also was slower. And it's the speed that I'm having issues with.
Example in:
3
10 2
6 6
5 4
20 2
1 1
19 19
10 3
1 1
1 2
1 3
Example out:
3 6
poodle
2 2
2 2
My code:
import sys
from math import sqrt
import itertools
def findmaxdist(x1, y1, list):
'''Finds the distance from x1, y1 to the most
distant point in the list.'''
dist = 0
for x2, y2 in list:
newdist = (x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)
if newdist>dist:
dist=newdist
return sqrt(dist)
fulltext = sys.stdin.readlines()
text = [w.rstrip('n') for w in fulltext]
cases = int(text[0])
i = 0
j = 1
while i < cases:
side, hatches = text[j].split()
side = int(side)
hatches = int(hatches)
j += 1
hatch_list = list()
for k in range(hatches):
x, y = text[j].split()
hatch_list.append((int(x), int(y)))
j += 1
possibles = list()
for x,y in itertools.product(range(side), range(side)):
if (x, y) not in hatch_list:
dist = findmaxdist(x, y, hatch_list)
if x+dist<=side and x-dist>=0:
if y+dist<=side and y-dist>=0:
possibles.append((x, y))
if len(possibles)==0:
print('poodle')
elif len(possibles)==1:
print(str(possibles[0][0])+' '+str(possibles[0][1]))
elif len(possibles)>1:
smallx = min(possibles, key = lambda t: t[0])[0]
semifinal = list((tuple for tuple in possibles if tuple[0] == smallx))
if len(semifinal)==1:
x,y = semifinal[0]
print(str(x)+' '+str(y))
elif len(semifinal)>1:
smally = min(semifinal, key = lambda t: t[1])[1]
final = list((tuple for tuple in semifinal if tuple[1] == smally))
if len(final)==1:
x,y = final[0]
print(str(x)+' '+str(y))
else:
print('Error: Wrong input-format?')
i+=1
performance python-3.x programming-challenge
New contributor
$endgroup$
add a comment |
$begingroup$
So I've tried to solve the kattis-challenge called Watchdog. My code works just fine, but it is too slow on the last test. I'm wondering if anyone sees any big algorithmic performance-issues.
A short description of the task: First line from .in states number of test-cases.
First line in each test-case gives length of a square and number of hatches respectively. For each hatch you'll have one line giving x and y-point of hatch respectively. The mission is to find a point x,y where the length from x,y to any hatch is not greater than the length from x,y to any point outside of the square. If no such point, print poodle
. If several such point print point with lowest x, if several points with equal x, print point with lowest y. Hatches and these points cannot overlap.
I think one issue might be extensive usage of lists, but I'm also not sure how to avoid them. Initially I started reading one and one line from stdin, using the data. That way things looked way more readable, but it also was slower. And it's the speed that I'm having issues with.
Example in:
3
10 2
6 6
5 4
20 2
1 1
19 19
10 3
1 1
1 2
1 3
Example out:
3 6
poodle
2 2
2 2
My code:
import sys
from math import sqrt
import itertools
def findmaxdist(x1, y1, list):
'''Finds the distance from x1, y1 to the most
distant point in the list.'''
dist = 0
for x2, y2 in list:
newdist = (x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)
if newdist>dist:
dist=newdist
return sqrt(dist)
fulltext = sys.stdin.readlines()
text = [w.rstrip('n') for w in fulltext]
cases = int(text[0])
i = 0
j = 1
while i < cases:
side, hatches = text[j].split()
side = int(side)
hatches = int(hatches)
j += 1
hatch_list = list()
for k in range(hatches):
x, y = text[j].split()
hatch_list.append((int(x), int(y)))
j += 1
possibles = list()
for x,y in itertools.product(range(side), range(side)):
if (x, y) not in hatch_list:
dist = findmaxdist(x, y, hatch_list)
if x+dist<=side and x-dist>=0:
if y+dist<=side and y-dist>=0:
possibles.append((x, y))
if len(possibles)==0:
print('poodle')
elif len(possibles)==1:
print(str(possibles[0][0])+' '+str(possibles[0][1]))
elif len(possibles)>1:
smallx = min(possibles, key = lambda t: t[0])[0]
semifinal = list((tuple for tuple in possibles if tuple[0] == smallx))
if len(semifinal)==1:
x,y = semifinal[0]
print(str(x)+' '+str(y))
elif len(semifinal)>1:
smally = min(semifinal, key = lambda t: t[1])[1]
final = list((tuple for tuple in semifinal if tuple[1] == smally))
if len(final)==1:
x,y = final[0]
print(str(x)+' '+str(y))
else:
print('Error: Wrong input-format?')
i+=1
performance python-3.x programming-challenge
New contributor
$endgroup$
$begingroup$
To specifically answer your question, you'd need to use a profiler to look at the code and see how long it takes. Given there is no sample input with your code, I'm unable to demonstrate how you would do this yourself. Could you add it to your question?
$endgroup$
– C. Harley
8 mins ago
add a comment |
$begingroup$
So I've tried to solve the kattis-challenge called Watchdog. My code works just fine, but it is too slow on the last test. I'm wondering if anyone sees any big algorithmic performance-issues.
A short description of the task: First line from .in states number of test-cases.
First line in each test-case gives length of a square and number of hatches respectively. For each hatch you'll have one line giving x and y-point of hatch respectively. The mission is to find a point x,y where the length from x,y to any hatch is not greater than the length from x,y to any point outside of the square. If no such point, print poodle
. If several such point print point with lowest x, if several points with equal x, print point with lowest y. Hatches and these points cannot overlap.
I think one issue might be extensive usage of lists, but I'm also not sure how to avoid them. Initially I started reading one and one line from stdin, using the data. That way things looked way more readable, but it also was slower. And it's the speed that I'm having issues with.
Example in:
3
10 2
6 6
5 4
20 2
1 1
19 19
10 3
1 1
1 2
1 3
Example out:
3 6
poodle
2 2
2 2
My code:
import sys
from math import sqrt
import itertools
def findmaxdist(x1, y1, list):
'''Finds the distance from x1, y1 to the most
distant point in the list.'''
dist = 0
for x2, y2 in list:
newdist = (x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)
if newdist>dist:
dist=newdist
return sqrt(dist)
fulltext = sys.stdin.readlines()
text = [w.rstrip('n') for w in fulltext]
cases = int(text[0])
i = 0
j = 1
while i < cases:
side, hatches = text[j].split()
side = int(side)
hatches = int(hatches)
j += 1
hatch_list = list()
for k in range(hatches):
x, y = text[j].split()
hatch_list.append((int(x), int(y)))
j += 1
possibles = list()
for x,y in itertools.product(range(side), range(side)):
if (x, y) not in hatch_list:
dist = findmaxdist(x, y, hatch_list)
if x+dist<=side and x-dist>=0:
if y+dist<=side and y-dist>=0:
possibles.append((x, y))
if len(possibles)==0:
print('poodle')
elif len(possibles)==1:
print(str(possibles[0][0])+' '+str(possibles[0][1]))
elif len(possibles)>1:
smallx = min(possibles, key = lambda t: t[0])[0]
semifinal = list((tuple for tuple in possibles if tuple[0] == smallx))
if len(semifinal)==1:
x,y = semifinal[0]
print(str(x)+' '+str(y))
elif len(semifinal)>1:
smally = min(semifinal, key = lambda t: t[1])[1]
final = list((tuple for tuple in semifinal if tuple[1] == smally))
if len(final)==1:
x,y = final[0]
print(str(x)+' '+str(y))
else:
print('Error: Wrong input-format?')
i+=1
performance python-3.x programming-challenge
New contributor
$endgroup$
So I've tried to solve the kattis-challenge called Watchdog. My code works just fine, but it is too slow on the last test. I'm wondering if anyone sees any big algorithmic performance-issues.
A short description of the task: First line from .in states number of test-cases.
First line in each test-case gives length of a square and number of hatches respectively. For each hatch you'll have one line giving x and y-point of hatch respectively. The mission is to find a point x,y where the length from x,y to any hatch is not greater than the length from x,y to any point outside of the square. If no such point, print poodle
. If several such point print point with lowest x, if several points with equal x, print point with lowest y. Hatches and these points cannot overlap.
I think one issue might be extensive usage of lists, but I'm also not sure how to avoid them. Initially I started reading one and one line from stdin, using the data. That way things looked way more readable, but it also was slower. And it's the speed that I'm having issues with.
Example in:
3
10 2
6 6
5 4
20 2
1 1
19 19
10 3
1 1
1 2
1 3
Example out:
3 6
poodle
2 2
2 2
My code:
import sys
from math import sqrt
import itertools
def findmaxdist(x1, y1, list):
'''Finds the distance from x1, y1 to the most
distant point in the list.'''
dist = 0
for x2, y2 in list:
newdist = (x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)
if newdist>dist:
dist=newdist
return sqrt(dist)
fulltext = sys.stdin.readlines()
text = [w.rstrip('n') for w in fulltext]
cases = int(text[0])
i = 0
j = 1
while i < cases:
side, hatches = text[j].split()
side = int(side)
hatches = int(hatches)
j += 1
hatch_list = list()
for k in range(hatches):
x, y = text[j].split()
hatch_list.append((int(x), int(y)))
j += 1
possibles = list()
for x,y in itertools.product(range(side), range(side)):
if (x, y) not in hatch_list:
dist = findmaxdist(x, y, hatch_list)
if x+dist<=side and x-dist>=0:
if y+dist<=side and y-dist>=0:
possibles.append((x, y))
if len(possibles)==0:
print('poodle')
elif len(possibles)==1:
print(str(possibles[0][0])+' '+str(possibles[0][1]))
elif len(possibles)>1:
smallx = min(possibles, key = lambda t: t[0])[0]
semifinal = list((tuple for tuple in possibles if tuple[0] == smallx))
if len(semifinal)==1:
x,y = semifinal[0]
print(str(x)+' '+str(y))
elif len(semifinal)>1:
smally = min(semifinal, key = lambda t: t[1])[1]
final = list((tuple for tuple in semifinal if tuple[1] == smally))
if len(final)==1:
x,y = final[0]
print(str(x)+' '+str(y))
else:
print('Error: Wrong input-format?')
i+=1
performance python-3.x programming-challenge
performance python-3.x programming-challenge
New contributor
New contributor
New contributor
asked 36 mins ago
RoyMRoyM
1
1
New contributor
New contributor
$begingroup$
To specifically answer your question, you'd need to use a profiler to look at the code and see how long it takes. Given there is no sample input with your code, I'm unable to demonstrate how you would do this yourself. Could you add it to your question?
$endgroup$
– C. Harley
8 mins ago
add a comment |
$begingroup$
To specifically answer your question, you'd need to use a profiler to look at the code and see how long it takes. Given there is no sample input with your code, I'm unable to demonstrate how you would do this yourself. Could you add it to your question?
$endgroup$
– C. Harley
8 mins ago
$begingroup$
To specifically answer your question, you'd need to use a profiler to look at the code and see how long it takes. Given there is no sample input with your code, I'm unable to demonstrate how you would do this yourself. Could you add it to your question?
$endgroup$
– C. Harley
8 mins ago
$begingroup$
To specifically answer your question, you'd need to use a profiler to look at the code and see how long it takes. Given there is no sample input with your code, I'm unable to demonstrate how you would do this yourself. Could you add it to your question?
$endgroup$
– C. Harley
8 mins ago
add a comment |
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RoyM is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
To specifically answer your question, you'd need to use a profiler to look at the code and see how long it takes. Given there is no sample input with your code, I'm unable to demonstrate how you would do this yourself. Could you add it to your question?
$endgroup$
– C. Harley
8 mins ago