I am trying to solve apparatus problem, paraphrased below.
There is an apparatus with n on/off switches with a one-to-one correspondence to n lights (but with an unknown permutation). We have several different photographs of the apparatus showing different configurations of the switches and lights. Are these photos sufficient to fully describe the device?
Input
First line contains two integers: n (the number of switches, 1 ≤ n ≤ 1000) and m (the number of photographs, 0 ≤ m ≤ 1000).
Each subsequent pair of lines describes the switches and the lights as a string of
1
(on) and0
(off).Output
Write the number of possible different wirings, modulo 1000003.
I have a solution but it takes longer than 2 seconds which is the time limit. I've tried to optimize my code for speed but can't get it within the 2 second limit.
import sys
import math
for line in sys.stdin:
line = line.strip("\n").split(" ")
numSwitches = int(line[0])
numPics = int(line[1])
wiring = {}
inValid = False
for i in range(numPics):
if (inValid):
break
x = sys.stdin.readline().strip("\n")
f_x = sys.stdin.readline().strip("\n")
x_ones = 0
f_x_ones = 0
digit = 0
for i in range(numSwitches):
if f_x[i]=='1':
digit += 2**(numSwitches-i-1)
f_x_ones += 1
for switch in range(numSwitches):
if (x[switch]=='1'):
x_ones += 1
if not (switch in wiring.keys()):
wiring[switch] = digit
else:
wiring[switch] &= digit
if x_ones != f_x_ones:
inValid = True
break
if not inValid:
for i in wiring.values():
if i==0:
inValid = True
break
for possibilities in set(wiring.values()):
frequency = wiring.values().count(possibilities)
if frequency>1:
oneBits = 0
while (possibilities>0):
oneBits += (possibilities%2==1)
possibilities /= 2
if oneBits < frequency:
inValid = True
break
if not inValid:
print math.factorial(numSwitches-numPics)%1000003
else:
print 0
I'm looking for suggestions of ways I should have approached the problem or input on how I can optimize my current code.
Note: Consider the following test case:
3 2
110
011
011
011
My code finds that is invalid in the following manner. First, upon encountering the first photograph (110, 011). The wiring dictionary gets assigned the following keys and values:
wiring[0] = 011
wiring[1] = 011
This means that the first and second switch can light up either the second or third lights. Upon encountering the second photograph (011, 011). wiring is updated as follows:
wiring[1] = 011 & 011 = 011
wiring[2] = 011
Now observe that the state of wiring indicates that all three switches can light up either the second and third lights. This is an inconsistency since 3 switches have to light up three lights, here we have three switches lighting up 2 lights.
1 Answer 1
Disclaimer: I won't try and provide a different algorithm to improve the time complexity of the approach. Instead I’ll rather focus on introducing python constructs that will cleanup the code and may help gain speed a bit.
Reading data from standard input
I had trouble running your script from both IDLE and windows command prompt. Neither of them were good at handling sys.stdin
as a file. But we can improve things by using the builtin input
(raw_input
in Python 2) since we just want to read lines one by one.
Converting back and forth between integers and their binary representation
One of the way to improve your computation of oneBite
would be to use the divmod
builtin. But it is even more easier when using bin
and count
ing the number of '1'
in the returned string.
Same when computing digit
: the int
builtin accept a second argument which is the base in which the number is represented.
Use functions
You could make the code easier to read and understand by using functions: one to parse two lines (one photo) and one to iterate over the whole set of photos and interpret the results.
Functions will also allow you to import
your file into an interactive session and test things more easily. If you want to have code at the top-level to be run when invoking your script from the command line, it is recommended to put it under an if __name__ == "__main__"
clause.
collections
The last part of your code count the frequency for a certain possibility in a bizarre way. Time for you to learn about collections.Counter
.
PEP8
- Use snake_case instead of camelCase for your variable names.
- Use spaces around most of your operators (
frequency > 1
,possibilities%2 == 1
,possibilities > 0
...) - Remove parenthesis around your tests
EAFP
While
if not (switch in wiring.keys()):
wiring[switch] = digit
else:
wiring[switch] &= digit
is correct, I would first write it
if switch in wiring:
wiring[switch] &= digit
else:
wiring[switch] = digit
for readability (and less computation) but then change it to
try:
wiring[switch] &= digit
except KeyError:
wiring[switch] = digit
It is not necessarily faster in this case (it is when failures are much less than success) but I find it clearer.
Putting it all together
In Python 2 use raw_input
, xrange
and itervalues
instead of input
, range
and values
.
from math import factorial
from collections import Counter
def parse_photo(wiring):
switches = input()
lights = input()
if switches.count('1') != lights.count(1):
return False # invalid data
# Convert binary value to integer
lights_value = int(lights, 2)
for switch, switch_value in enumerate(switches):
if switch_value == "1":
try:
wiring[switch] &= lights_value
except KeyError:
wiring[switch] = lights_value
return True
def wiring_possibilities():
wiring = {}
num_switches, num_photos = map(int, input().split())
for _ in range(num_photos):
if not parse_photo(wiring):
return 0
frequencies = Counter(wiring.values())
if 0 in frequencies:
# Same as if 0 in wiring.values()
return 0
for possibility, frequency in frequencies.items():
switched_on_bits = bin(possibility).count('1')
if switched_on_bits < frequency:
return 0
return factorial(num_switches - num_photos) % 1000003
if __name__ == "__main__":
print(wiring_possibilities())
-
\$\begingroup\$ @Ragnar As suggested since my first comment, a trying-everything approach is certainly not what this problem is expecting. Even in case that you would be able to notably improve the performance and meet the target time (very unlikely scenario if the problem was properly created; although well... after reading a so unclear description I don't trust too much in the creators), it would be somehow against what is expected and wouldn't help you to learn how to face this kind of situations (you will fail the next time because of the same reason). \$\endgroup\$varocarbas– varocarbas2015年11月26日 08:29:36 +00:00Commented Nov 26, 2015 at 8:29
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\$\begingroup\$ Mathias, my message was addressed to the OP as a continuation of my comments above. You let very clear your position in the disclaimer you are mentioning (which I read and understood before writing my comment :)). \$\endgroup\$varocarbas– varocarbas2015年11月26日 08:37:10 +00:00Commented Nov 26, 2015 at 8:37
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\$\begingroup\$ @varocarbas OK. Wanted to be extra clear, just in case. \$\endgroup\$301_Moved_Permanently– 301_Moved_Permanently2015年11月26日 08:48:49 +00:00Commented Nov 26, 2015 at 8:48
Explore related questions
See similar questions with these tags.
110 011
also means that light 3 is turned off by switch 1, so you can infer an input001 100
to your logic. In this way (you should better test) you can probably cut off your inconsistency test. \$\endgroup\$