Programming Challenge from Kattis: Watchdog The Next CEO of Stack Overflow

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Programming Challenge from Kattis: Watchdog



The Next CEO of Stack Overflow










0












$begingroup$


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








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    0












    $begingroup$


    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








    share







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    RoyM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      0












      0








      0





      $begingroup$


      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








      share







      New contributor




      RoyM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      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





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      Check out our Code of Conduct.










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      Check out our Code of Conduct.








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      asked 4 mins ago









      RoyMRoyM

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      RoyM is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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