On 1/3/2014 7:16 PM, Wiktor wrote:
Hi,
it's my first post on this newsgroup so welcome everyone. :)
I'm still learning Python (v3.3), and today I had idea to design (my first)
recursive function, that generates (filled out) board to 'Towers' Puzzle:
http://www.chiark.greenend.org.uk/~sgtatham/puzzles/js/towers.html
(so I could in future write algorithm to solve it ;-))
I'm pretty proud of myself - that it works, and that took me only 4 hours
to debug. ;-)
But on Project Euler site sometimes I'm also proud, that I solved some
problem in 30-line script, and then on forum there's one lined solution...
So maybe You might look at this script, and tell me if this can be more
pythonic. It's nothing urgent. I can wait - it works after all. ;-)
Idea is that function "generate()" 'finds' one number at a time (well,
besides first row), then checks if there are no repetitions in column
(because in row there cannot be by design - I pop out numbers from shuffled
list [1, 2, 3, ..., size] for every row.)
If no repetition - calls the same function to find next number, and so on.
If there is repetition at some point - recursion jumps back, and try
different number on previous position.
import random
def check(towers, x=None):
if x:
c = x % len(towers) # check only column with
column = [] # value added on pos. x
for i in range(len(towers)):
column.append(towers[i][c])
column = [x for x in column if x != 0]
# print(column) # debugging leftovers ;-)
return len(column) == len(set(column))
else:
for c in range(len(towers)): # 'x' not provided,
column = [] # so check all columns
for i in range(len(towers)):
column.append(towers[i][c])
column = [x for x in column if x != 0]
# print(column)
if len(column) != len(set(column)):
return False
return True
def generate(size=4, towers=None, row=None, x=0):
if not towers: # executed only once.
row = [a for a in range(1, size+1)] # Then I'll pass towers list
random.shuffle(row) # at every recursion
towers = []
# not so pretty way to generate
for i in range(size): # matrix filled with 0's
towers.append([]) # I tried: towers = [[0]*size]*size
for j in range(size): # but this doesn't work. ;-)
towers[i].append(0) # I don't know how to do this with
# list comprehension (one inside
[0]*size] is fine for one row
towers = [[0]*size] for i in range(size)]
should do what you want for a 2-d array instead of the above.
I cannot look at the rest right now.
--
Terry Jan Reedy
--
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