Re: [Tutor] Creating Sudoku
A good place to look at : http://www.norvig.com/sudoku.html
On Mon, Apr 7, 2008 at 6:53 PM, Luke Paireepinart <[EMAIL PROTECTED]>
wrote:
> W W wrote:
> > On 4/7/08, Luke Paireepinart <[EMAIL PROTECTED]> wrote:
> >
> >> W W wrote:
> >> What are you talking about? I don't understand what you mean by
> "ignores
> >> whitespace between dictionary elements."
> >>
> >>
> >>> foo = {'1a': 1, '1b':2, '1c':3,
> >>> '2a': 0, '2b': 9, '2c': 6}
> >>>
> >
> > Exactly that. If you were to write:
> >
> > foo = {'1a': 1, '1b':2, '1c':3}
> > foo['2a'] = 0
> >
> > You would get a nifty error.
> >
> You mean that the dictionary _definition_ ignores whitespace between
> elements?
> That's quite different than the dictionary itself ignoring whitespace.
> That implies that
> foo['1b'] is the same element as foo['1 b'], hence the source of my
> confusion.
>
> That's not a feature of dictionaries, but of the comma.
> You can easily do the following:
> x = [1, 2,
> 3, 4]
> if you want.
> Same with tuples and various other things.
> Python just realizes that if it doesn't see a closing brace/bracket, but
> sees a comma, that more will probably be coming on the next line.
> You can do the same thing with backslash, if your statement does not end
> in a comma: for example,
> x = 1 + 1 + \
> 2 + 3 + 5 \
> + 8 + 13
>
> Also, did you test the code that "generates an error?"
> It works fine for me.
> >>> foo = {'1a': 'b'}
> >>> foo['2b'] = 0
> >>> print foo['2b']
> 0
> >>>
>
> Hope that helps,
> -Luke
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Re: [Tutor] crashed with Runtime Error: NZEC (non-zero exit code)
i'm thinking the same way Eric do. On Wed, Feb 27, 2008 at 11:18 PM, Eric Brunson <[EMAIL PROTECTED]> wrote: > Alan Gauld wrote: > > "bob gailer" <[EMAIL PROTECTED]> <[EMAIL PROTECTED]> wrote > > > > i don't really understand that. you are trying to say that the > tests > from the online judge are not "exactly good" ? > > > Yes that's exactly what I'm saying. At least one test case has a > non-decimal character. > > > Which actually makes it a very good test since thats > exactly the kind of thing you should be testing for :-) > A test suite that only checks valid data is a bad test. > > > > > That's fine if writing a test suite for production software, but from the > problem description: "There is a string containing only decimal digit > characters." > > Testing for valid input seems to be outside the scope of the problem > definition. :-) > > > > ___ > Tutor maillist - Tutor@python.org > http://mail.python.org/mailman/listinfo/tutor > > ___ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
[Tutor] crashed with Runtime Error: NZEC (non-zero exit code)
Hello,
I wrote a code for a problem and submitted to an online judge, a program
that runs my program against a set of predefined input data.
I wrote it for exercise my python writing. In don't want to discuss the
algorithm here!
I dare to ask for help because the answer was : crashed with Runtime Error:
NZEC (non-zero exit code). I heard that is a general error for most
interpreters and elsewhere that NZEC error code has to do with exceptions.
So, i'm seeking for a possible problem with the coding, not with the
algorithm.
Here, the code. (l also attached it in a file)
Using characters of the string, the program should construct the maximum
number which divides by fifteen without remainder.
example
*Input:*
1
02041
*Output:*
4200
[code]
import sys
def write(number): # assamble the number
# assure the 5 divisibility
if number[0] != 0:
number[0] -= 1
last_d = 0
else:
number[5] -= 1
last_d = 5
zero = True
for i in reversed(range(1,10)):
for j in range(digits[i]):
sys.stdout.write(i)
zero = False
# leading zeroes should be omitted
if zero == False:
for i in range(digits[0]):
sys.stdout.write(0)
print last_d
def remove_item(d,item):
if type(item) != list:
d[item] -= 1
else:
d[item[0]] -= 1
d[item[1]] -= 1
def case5(d,item):
# return True if i try to remove the only five that assure the
divisibility with 5
return item == 5 and d[0] == 0 and d[item] == 1
def exist(d,item):
found = False
if type(item) != list:
if d[item] != 0 and case5(d,item) == False:
found = True
else:
p1 = d[item[0]] != 0 and case5(d,d[item[0]]) == False
d[item[0]] -= 1
p2 = d[item[1]] != 0 and case5(d,d[item[1]]) == False
d[item[0]] += 1
if p1 == True and p2 == True:
found = True
return found
def remove(d, mult):
found = False
for item in mult:
if exist(d,item) == True:
remove_item(d,item)
found = True
break
if found == False:
d = []
return d
def remove_all(digits,rest):
# if the number is the form 3k+1, i try to remove one 3k+1 digit or two
3k+2 digits
# if the number is the form 3k+2, i try to remove one 3k+2 digit or two
3k+2 digits
one = [1,4,7,[2,2],[2,5],[2,8],[5,5],[5,8],[8,8]]
two = [2,5,8,[1,1],[1,4],[1,7],[4,4],[4,7],[7,7]]
if rest == 1:
d = remove(digits, one)
else:
d = remove(digits, two)
return d
def resolve(digits):
suma = 0
for i in range(10):
suma += i*digits[i]
rest = suma % 3
if rest != 0: # if it's not divisible with 3 i try to remove some digits
digits = remove_all(digits,rest)
return digits
t = int(raw_input())
for tt in range(t):
line = raw_input()
digits = [0]*10
# counts the frequence of digits in the string
for i in range(len(line)):
digits[int(line[i])] += 1
# if it's not divisible with 5
if digits[0] == 0 and digits[5] == 0:
print("impossible")
else:
number = resolve(digits)
if(number == []):
print("impossible")
else:
write(number)
[/code]
Any alternatives for the personal rambling(if it's a wrong way to do it, of
course):
L = [1,2,[3,4],[5,5]]
for item in L:
if type(item) == list:
print L[item[0]], L[item[1]]
else:
print L[item]
Thanks,
Andrei
import sys
def write(number): # assamble the number
# assure the 5 divisibility
if number[0] != 0:
number[0] -= 1
last_d = 0
else:
number[5] -= 1
last_d = 5
zero = True
for i in reversed(range(1,10)):
for j in range(digits[i]):
sys.stdout.write(i)
zero = False
# leading zeroes should be omitted
if zero == False:
for i in range(digits[0]):
sys.stdout.write(0)
print last_d
def remove_item(d,item):
if type(item) != list:
d[item] -= 1
else:
d[item[0]] -= 1
d[item[1]] -= 1
def case5(d,item):
# return True if i try to remove the only five that assure the divisibility
with 5
return item == 5 and d[0] == 0 and d[item] == 1
def exist(d,item):
found = False
if type(item) != list:
if d[item] != 0 and case5(d,item) == False:
found = True
else:
p1 = d[item[0]] != 0 and case5(d,d[item[0]]) == False
d[item[0]] -= 1
p2 = d[item[1]] != 0 and case5(d,d[item[1]]) == False
d[item[0]] += 1
if p1 == True and p2 == True:
found = True
return found
def remove(d, mult):
found = False
for item in mult:
if exist(d,item) == True:
remove_item(d,item)
found = True
break
if found == False:
d = []
return
[Tutor] read from standard input
Hello,
I want to read from the standard input numbers until i reach a certain value
or to the end of the "file".
What is the simplest, straightforward, pythonic way to do it?
a sketch of how i tried to do it:
[code]
while 1 < 2:
x = raw_input()
if type(x) != int or x == 11:
break
else:
print x
[/code]
but don't work. and i'm interest in a general way to read until it is
nothing to read.
to ilustrate that in C:
[code]
int x;
while( scanf("%d",&x) == 1 && x != 11)
printf("%d\n", x);
[/code]
Thanks!
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[Tutor] tokenizing a simple string with split()
I want to split a string like "C:\My\Doc\;D:\backup\" with two separators: \
and ;
I found that \ is handled with *raw string* notation r"". But the problem i
encountered is with split() function.
In the 2.5 reference is said that "The sep argument of the split() function
may consist of multiple characters". But I cannot figured out why this
simple example not working:
s = "spam;egg mail"
s.split("; ")
output: ['spam;egg mail']
instead of ['spam', 'egg', 'mail']
any suggestion is welcome,
andrei
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