Python：如何使我的4着色检查器更具可读性

Question

I was trying to write a general program to check The 17x17 problem SOLVED! , 4-coloring of a17x17 grid with no monochromatic rectangles. Solution link: 17.txt .

This is what I wrote:

from itertools import product

def is_solution(myfile,m,n):
    """ m-lines, n-columns """
    grid = [c.strip() for c in line.split(',')] for line in open(myfile).readlines()]
    for x0,y0 in product(xrange(m),xrange(n)):
        start = grid[x0][y0]
        for x in xrange(x0+1,m):
            if grid[x][y0] == start:
                for y in xrange(y0+1,n):
                    if grid[x0][y] == start == grid[x][y]:
                            return False
    return True


print is_solution('17.txt',17,17)

Is there a more readable, concise or efficient way (in that order of priority) to write this? Maybe a different approach with different data structures... Since i am learning Python at the moment, any advice is very welcome.

Answer 1

1. You should cleanly separate the logic for in-/output from the verification logic (this basically applies to any code). This also includes that you only put definitions at the module's top level. The actual program usually goes into a conditional like in the code below which is only executed if the file is called directly from the command line (and not if it is only import ed by another file).
2. The dimensions of the grid can be derived from the input. No need for separate parameters here.
3. You should work with integers instead of strings (this is optional, but more clean, IMO)

My attempt (taking the file from STDIN, can be called like python script.py < 17.txt ):

import itertools

def has_monochromatic_rectangles(grid):
  # use range instead of xrange here (xrange is not in Python 3)
  points = list(itertools.product(range(len(grid)), range(len(grid[0]))))
  # check if for any rectangle, all 4 colors are equal
  # (this is more brute-force than necessary, but you placed simplicity
  # above efficiency. Also, for 17x17, it doesn't matter at all ;)
  return any(grid[x1][y1] == grid[x1][y2] == grid[x2][y1] == grid[x2][y2]
             for (x1,y1), (x2,y2) in itertools.product(points, points)
             if x1 != x2 and y1 != y2)

def has_max_colors(grid, most):
  # collect all grid values and uniquify them by creating a set
  return len(set(sum(grid, []))) <= most

if __name__ == '__main__':
  # read from STDIN (could easily be adapted to read from file, URL, ...)
  import sys
  grid = [map(int, line.split(',')) for line in sys.stdin]

  assert has_max_colors(grid, 4)
  assert not has_monochromatic_rectangles(grid)

Answer 2

import urllib
grid=urllib.urlopen("http://www.cs.umd.edu/~gasarch/BLOGPAPERS/17.txt")
grid=[map(int,row.split(",")) for row in grid]
print grid

def check_grid(grid):
    for i in range(17):
        for j in range(17):
            for i2 in range(i):
                for j2 in range(j):
                    colours=[grid[a][b] for a in (i,i2) for b in (j,j2)]
                    assert(len(set(colours))>1)

check_grid(grid)
grid[1][1]=2
check_grid(grid)

Answer 3

Obligatory one-liner*!

Assuming you've loaded the 17x17 data into a numpy array named A (see @robertking's answer to use urllib), you can do it with one line with numpy!

 print (array([len(set(A[i:i+k1,j:j+k2][zip(*[(0,0), (0,-1),(-1,0),(-1,-1)])])) for i in xrange(16) for j in xrange(16) for k1 in xrange(2,17) for k2 in xrange(2,17)])!=1).all()

* Don't actually do this in one-line. Here it is expanded out a bit for clarity:

corners = zip(*[(0,0), (0,-1),(-1,0),(-1,-1)])

for k1 in xrange(2,17):
    for k2 in xrange(2,17):
        for i in xrange(16):
            for j in xrange(16):

                # Pull out each sub-rectange
                sub = A[i:i+k1, j:j+k2]

                # Only use the corners
                sub = sub[corners]

                # Count the number of unique elements
                uniq = len(set(sub))

                # Check if all corners are the same
                if uniq == 1: 
                    print False
                    exit()
print True

Answer 4

Here is another way to think about it.

import urllib
grid=urllib.urlopen("http://www.cs.umd.edu/~gasarch/BLOGPAPERS/17.txt")
grid=[map(int,row.split(",")) for row in grid]

def check(grid):
    colour_positions=lambda c,row:set(i for i,colour in enumerate(row) if colour==c) #given a row and a colour, where in the row does that colour occur
    to_check=[[colour_positions(c,row) for row in grid] for c in range(1,5)] #for each row and each colour, get the horizontal positions.
    from itertools import combinations
    for i in to_check:
        for a,b in combinations(i,2):
            if len(a&b)>1: #for each colour, for each combination of rows, do we ever get more than 1 horizontal position in common (e.g. a rectangle)
                return False
    return True

print check(grid)
grid[1][1]=2
print check(grid)