Placing square on a grid - Part 3

On a 3 by 3 grid, the red, yellow, blue and green 2 by 2 squares are placed in some order. They have to be placed in their location (IE not moved somewhere else). There are 4!=24 4 ! = 24 different ways to place these squares.

Of these 24 different ways, how many distinct results would we get?

For example, placing 1432 and 4132 will both yield

Note by Chung Kevin
6 years ago

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16 16 .

I'll be referring to the 2 X 2 squares as numbers, and each of the 9 squares as blocks. First, look at the four corner blocks. They can only be covered by the number on it. So we can disregard those. We are left with a central cross. The center of this cross can be covered in 4 4 ways, since whichever number we choose first covers it. Now, we are left with the four other blocks. Of these, 2 2 will be covered by the first number we chose, so they don't matter. We are left with two blocks. Each of these blocks can be covered by two numbers and each choice is possible, so we have 422=16 4 * 2 * 2 = 16 .

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I'm getting a total of 20,20, although your reasoning seems sound. As I understood the question the first number forms the bottom "layer", the second number the second layer and so on, so that the last number/color is on top. For the sequences that end in 23,32,14,4123, 32, 14, 41 it doesn't matter what the order of the first two layers is, so this produces 44 duplicates. For the other 88 possible pairs of top layers it will matter how the first two layers are chosen, so there are no duplicates here. Thus I get 244=2024 - 4 = 20 sequences that produce distinct patterns. Where is the flaw in my logic?

Edit: O.k., I see now. 13241324 gives the same result as 12341234, 43214321 gives the same result as 42314231, 31423142 gives the same result as 34123412 and 21432143 gives the same result as 2413.2413. This gives us 44 more duplicates, leaving us with 1616 distinct outcomes, as Siddhartha and Brock have found.

Brian Charlesworth - 6 years ago

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Wow, that's a nice analysis! It's certainly simpler than I thought it would be.

Calvin Lin Staff - 6 years ago

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Nice approach. I did it in a similar manner too :)

Chung Kevin - 6 years ago

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I'm finding 16\boxed{16}, gentlemen.

Python 2.7:

from itertools import permutations
found = set()
class Grid():
    def __init__(self):
        self.grid = [['w']*3,['w']*3,['w']*3]
    def __str__(self):
        string = ''
        for row in self.grid:
            for color in row:
                string = string + color
            string = string + '\n'
        return string
    def paste(self,origin,color):
        for x in xrange(2):
            for y in xrange(2):
                self.grid[origin[0]+x][origin[1]+y] = color
position = {'r':(0,0),'g':(1,1),'b':(1,0),'y':(0,1)}
for colors in permutations('rgby'):
    grid = Grid()
    for color in colors:
print "Answer:", len(found)

Brock Brown - 6 years ago

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Nice program, Brock. I can't argue with its output; the answer is indeed 16.16. :)

Brian Charlesworth - 6 years ago

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Thanks! :D

Brock Brown - 6 years ago

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