# searching a list of lists as a two-dimensional array?

Discussion in 'Python' started by agent-s, Feb 12, 2007.

1. ### agent-sGuest

Basically I'm programming a board game and I have to use a list of
lists to represent the board (a list of 8 lists with 8 elements each).
I have to search the adjacent cells for existing pieces and I was
wondering how I would go about doing this efficiently. Thanks

agent-s, Feb 12, 2007

2. ### James StroudGuest

agent-s wrote:
> Basically I'm programming a board game and I have to use a list of
> lists to represent the board (a list of 8 lists with 8 elements each).
> I have to search the adjacent cells for existing pieces and I was
> wondering how I would go about doing this efficiently. Thanks
>

This isn't very clear. What do you mean by "I have to search the
adjacent cells for existing pieces"?

If piece is 1 and empty is 0 and piece is at ary[row][col]:

import operator
srch = [(i,j) for i in [-1,0,1] for j in [-1,0,1] if (i,j) != (0,0)]
is_adj = reduce(operator.or_, [ary[row+i][col+j] for (i,j) in srch]])

James

James Stroud, Feb 12, 2007

3. ### Samuel Karl PetersonGuest

James Stroud <> on Sun, 11 Feb 2007 16:53:16 -0800
didst step forth and proclaim thus:

> agent-s wrote:
> > Basically I'm programming a board game and I have to use a list of
> > lists to represent the board (a list of 8 lists with 8 elements each).
> > I have to search the adjacent cells for existing pieces and I was
> > wondering how I would go about doing this efficiently. Thanks
> >

>
> This isn't very clear. What do you mean by "I have to search the
> adjacent cells for existing pieces"?
>
> If piece is 1 and empty is 0 and piece is at ary[row][col]:
>
> import operator
> srch = [(i,j) for i in [-1,0,1] for j in [-1,0,1] if (i,j) != (0,0)]
> is_adj = reduce(operator.or_, [ary[row+i][col+j] for (i,j) in srch]])

Wow, maybe it's just me (I'm a pretty bad programmer) but this is
where list comprehensions begin to look unreadable to me. Here's a
C-like way to do it, (warning, untested in python):

for i in range(8):
for j in range(8):
for offset_i in range(-1,2):
for offset_j in range(-1, 2):
row = i + offset_i
col = j + offset_j
if (row < 0 or row > 7) or (col < 0 or col > 8) \
or ((row,col) == (i,j)):
continue
# else do something with board[row][col]

I realize this is gross and un-Pythonic and does the same thing the
above code does, but it's probably the way I'd choose to do it .
Then again, I've been negatively influenced by doing a game of life in
C a few months back.

--
Sam Peterson
skpeterson At nospam ucdavis.edu
"if programmers were paid to remove code instead of adding it,
software would be much better" -- unknown

Samuel Karl Peterson, Feb 12, 2007
4. ### Gabriel GenellinaGuest

En Mon, 12 Feb 2007 02:24:54 -0300, Samuel Karl Peterson
<> escribiÃ³:

> James Stroud <> on Sun, 11 Feb 2007 16:53:16 -0800
> didst step forth and proclaim thus:
>
>> agent-s wrote:
>> > Basically I'm programming a board game and I have to use a list of
>> > lists to represent the board (a list of 8 lists with 8 elements each).
>> > I have to search the adjacent cells for existing pieces and I was
>> > wondering how I would go about doing this efficiently. Thanks

> Wow, maybe it's just me (I'm a pretty bad programmer) but this is
> where list comprehensions begin to look unreadable to me. Here's a
> C-like way to do it, (warning, untested in python):

Just for fun, and to add one more way. This is a generator for adjacent
indexes, that can be used to search for occupied cells, for locating a
suitable next move, or whatever:

py> def adj(i, j):
.... for ni in (i-1, i, i+1):
.... if ni not in range(8): continue
.... for nj in (j-1, j, j+1):
.... if nj not in range(8): continue
.... if ni!=i or nj!=j:
.... yield ni,nj
....
py>
[(3, 2), (3, 3), (3, 4), (4, 2), (4, 4), (5, 2), (5, 3), (5, 4
[(6, 2), (6, 3), (6, 4), (7, 2), (7, 4)]
[(3, 6), (3, 7), (4, 6), (5, 6), (5, 7)]
[(0, 6), (1, 6), (1, 7)]

--
Gabriel Genellina

Gabriel Genellina, Feb 12, 2007
5. ### Gabriel GenellinaGuest

En Mon, 12 Feb 2007 02:24:54 -0300, Samuel Karl Peterson
<> escribiÃ³:

> James Stroud <> on Sun, 11 Feb 2007 16:53:16 -0800
> didst step forth and proclaim thus:
>
>> agent-s wrote:
>> > Basically I'm programming a board game and I have to use a list of
>> > lists to represent the board (a list of 8 lists with 8 elements each).
>> > I have to search the adjacent cells for existing pieces and I was
>> > wondering how I would go about doing this efficiently. Thanks

> Wow, maybe it's just me (I'm a pretty bad programmer) but this is
> where list comprehensions begin to look unreadable to me. Here's a
> C-like way to do it, (warning, untested in python):

Just for fun, and to add one more way. This is a generator for adjacent
indexes, that can be used to search for occupied cells, for locating a
suitable next move, or whatever:

py> def adj(i, j):
.... for ni in (i-1, i, i+1):
.... if ni not in range(8): continue
.... for nj in (j-1, j, j+1):
.... if nj not in range(8): continue
.... if ni!=i or nj!=j:
.... yield ni,nj
....
py>
[(3, 2), (3, 3), (3, 4), (4, 2), (4, 4), (5, 2), (5, 3), (5, 4
[(6, 2), (6, 3), (6, 4), (7, 2), (7, 4)]
[(3, 6), (3, 7), (4, 6), (5, 6), (5, 7)]
[(0, 6), (1, 6), (1, 7)]

--
Gabriel Genellina

Gabriel Genellina, Feb 12, 2007
6. ### Paul RubinGuest

"agent-s" <> writes:
> Basically I'm programming a board game and I have to use a list of
> lists to represent the board (a list of 8 lists with 8 elements each).
> I have to search the adjacent cells for existing pieces and I was
> wondering how I would go about doing this efficiently. Thanks

You're getting a bunch of Pythonic suggestions which are easy to
understand though not so efficient. If you're after efficiency you
might look at a book like Welsh and Baczynskyj "Computer Chess II" for
some techniques (warning, this is a rather old book though) and
program in a closer-to-the-metal language. One common approach these
days is "bit boards". Basically you represent the board state as a
bunch of 64-bit words, one bit per square. So for checking occupancy,
you'd have a word having the bits set for occupied squares. If you
only want to check adjacent squares (king moves), you could have a
bunch of bit masks (one for each square) with bits set only for the
adjacent squares (similarly for bishop moves, rook moves, etc.) Then
to check adjacency you'd mask off the appropriate bits for that
square, then AND it against the occupancy word. Normally you'd have
separate occupancy words for your own pieces and your opponent's.

Paul Rubin, Feb 12, 2007
7. ### James StroudGuest

Paul Rubin wrote:
> "agent-s" <> writes:
>
>>Basically I'm programming a board game and I have to use a list of
>>lists to represent the board (a list of 8 lists with 8 elements each).
>>I have to search the adjacent cells for existing pieces and I was
>>wondering how I would go about doing this efficiently. Thanks

>
>
> You're getting a bunch of Pythonic suggestions which are easy to
> understand though not so efficient. If you're after efficiency you
> might look at a book like Welsh and Baczynskyj "Computer Chess II" for
> some techniques (warning, this is a rather old book though) and
> program in a closer-to-the-metal language. One common approach these
> days is "bit boards". Basically you represent the board state as a
> bunch of 64-bit words, one bit per square. So for checking occupancy,
> you'd have a word having the bits set for occupied squares. If you
> only want to check adjacent squares (king moves), you could have a
> bunch of bit masks (one for each square) with bits set only for the
> adjacent squares (similarly for bishop moves, rook moves, etc.) Then
> to check adjacency you'd mask off the appropriate bits for that
> square, then AND it against the occupancy word. Normally you'd have
> separate occupancy words for your own pieces and your opponent's.

In defense of the less efficient suggestions, he did say he had to use a
list of lists. But what you describe is pretty cool.

James

James Stroud, Feb 12, 2007
8. ### Guest

James Stroud:
> import operator
> srch = [(i,j) for i in [-1,0,1] for j in [-1,0,1] if (i,j) != (0,0)]
> is_adj = reduce(operator.or_, [ary[row+i][col+j] for (i,j) in srch]])

Or maybe (untested), Python V.2.5:

srch = [(i,j) for i in [-1,0,1] for j in [-1,0,1] if (i,j) != (0,0)]
is_adj = any(ary[row+i][col+j] for (i,j) in srch)

Or:

is_adj = any(ary[row+i][col+j] for i in [-1,0,1] for j in [-1,0,1] if
(i,j) != (0,0))

Bye,
bearophile

, Feb 12, 2007
9. ### John MachinGuest

On Feb 12, 4:24 pm, Samuel Karl Peterson
<> wrote:
> C-like way to do it, (warning, untested in python):
>
> for i in range(8):
> for j in range(8):
> for offset_i in range(-1,2):
> for offset_j in range(-1, 2):
> row = i + offset_i
> col = j + offset_j
> if (row < 0 or row > 7) or (col < 0 or col > 8) \
> or ((row,col) == (i,j)):
> continue
> # else do something with board[row][col]
>
> I realize this is gross and un-Pythonic and does the same thing

It's also just a little bit inefficient. Never mind the algorithm,
we'll talk about that later. Let's just improve the code a little:

side_range = range(8)
delta_range = range(-1, 2)
for i in side_range:
for offset_i in delta_range:
row = i + offset_i
if not (0 <= row <= 7): continue
for j in side_range:
for offset_j in delta_range:
if not offset_i and not offset_j: continue
col = j + offset_j
if not(0 <= col <= 7): continue
# *now* it's safe to do something
# with (row, col)

Such hoisting of code out of inner loops is done for you by a C
compiler -- Python assumes you know what you are doing

Now for the algorithm: all of that testing to see if you are about to
sail off the end of the world is a bit ugly and slow. You can use bit-
bashing, as Paul suggested, even though it's on Steven D'Aprano's list
of 6 deadly sins

Another approach that has been used is to have a 9 x 9 storage area;
the squares off the edge contain a value that is neither "empty" nor
any value that represents a piece. So with careful coding, they
automatically fail tests like "can I capture the piece on this
square" [no, it's not a piece] and "can I move here" [no, it's not
empty]. If the game is chess, you need 10 x 10 to cater for those
pesky knights.

Cheers,
John

John Machin, Feb 12, 2007
10. ### Steven D'ApranoGuest

On Sun, 11 Feb 2007 23:20:11 -0800, John Machin wrote:

> Now for the algorithm: all of that testing to see if you are about to
> sail off the end of the world is a bit ugly and slow. You can use bit-
> bashing, as Paul suggested, even though it's on Steven D'Aprano's list
> of 6 deadly sins

Heh. Being a smart-alec is number 7

Seriously, this is Python. Are you *sure* bit-bashing is going to be
faster than the alternative? If this was C, or assembly, you'd almost
certainly be right. But Python is heavily object-oriented, and bit
manipulations are just methods, with all the overhead that entails.

>>> import timeit
>>> timeit.Timer("3 | 2", "").repeat()

[0.33678007125854492, 0.33447504043579102, 0.33331012725830078]
>>> timeit.Timer("3 < 2", "").repeat()

[0.30328893661499023, 0.29070115089416504, 0.28839397430419922]

The problem with bit-bashing, masking etc. is that except for the most
simple cases it is quite obfuscated. If you aren't going to gain a serious
performance boost, why bother?

--
Steven D'Aprano

Steven D'Aprano, Feb 12, 2007
11. ### Neil CeruttiGuest

On 2007-02-12, James Stroud <> wrote:
> Paul Rubin wrote:
>> "agent-s" <> writes:
>>>Basically I'm programming a board game and I have to use a
>>>list of lists to represent the board (a list of 8 lists with 8
>>>elements each). I have to search the adjacent cells for
>>>existing pieces and I was wondering how I would go about doing
>>>this efficiently. Thanks

>>
>> You're getting a bunch of Pythonic suggestions which are easy
>> to understand though not so efficient. If you're after
>> efficiency you might look at a book like Welsh and Baczynskyj
>> "Computer Chess II" for some techniques (warning, this is a
>> rather old book though) and program in a closer-to-the-metal
>> language. One common approach these days is "bit boards".
>> Basically you represent the board state as a bunch of 64-bit
>> words, one bit per square. So for checking occupancy, you'd
>> have a word having the bits set for occupied squares. If you
>> only want to check adjacent squares (king moves), you could
>> have a bunch of bit masks (one for each square) with bits set
>> only for the adjacent squares (similarly for bishop moves,
>> rook moves, etc.) Then to check adjacency you'd mask off the
>> appropriate bits for that square, then AND it against the
>> occupancy word. Normally you'd have separate occupancy words
>> for your own pieces and your opponent's.

>
> In defense of the less efficient suggestions, he did say he had
> to use a list of lists. But what you describe is pretty cool.

Precomputing and storing the adjecent indexes for each square
would be a possible hybrid solution. In effect every square would
contain pointers to all its neighbors.

--
Neil Cerutti

Neil Cerutti, Feb 12, 2007
12. ### Hendrik van RooyenGuest

"John Machin" <> wrote:

> Now for the algorithm: all of that testing to see if you are about to
> sail off the end of the world is a bit ugly and slow. You can use bit-
> bashing, as Paul suggested, even though it's on Steven D'Aprano's list
> of 6 deadly sins

Thou shallt not bit - bash

What are the other five?

- Hendrik

Hendrik van Rooyen, Feb 13, 2007
13. ### John MachinGuest

On Feb 13, 4:57 pm, "Hendrik van Rooyen" <> wrote:
> "John Machin" <> wrote:
>
> > Now for the algorithm: all of that testing to see if you are about to
> > sail off the end of the world is a bit ugly and slow. You can use bit-
> > bashing, as Paul suggested, even though it's on Steven D'Aprano's list
> > of 6 deadly sins

>
> Thou shallt not bit - bash
>
> What are the other five?

"""
.... regexes have their place, together with pointer arithmetic, bit
manipulations, reverse polish notation and goto. The problem is when
people use them inappropriately ...
"""

The sixth is never mentioned. Aliter: I can't count. Believe whichever
hypothesis you like

John Machin, Feb 13, 2007
14. ### Hendrik van RooyenGuest

"John Machin" <> wrote:

> On Feb 13, 4:57 pm, "Hendrik van Rooyen" <> wrote:
> > "John Machin" <> wrote:
> >
> > > Now for the algorithm: all of that testing to see if you are about to
> > > sail off the end of the world is a bit ugly and slow. You can use bit-
> > > bashing, as Paul suggested, even though it's on Steven D'Aprano's list
> > > of 6 deadly sins

> >
> > Thou shallt not bit - bash
> >
> > What are the other five?

>
> """
> ... regexes have their place, together with pointer arithmetic, bit
> manipulations, reverse polish notation and goto. The problem is when
> people use them inappropriately ...
> """
>

I find this unsatisfactory - I was hoping for a five commandment
style of thing, setting out proper prohibitions against evil stuff.

It seems that I am a closet fundamentalist.

A sixth, btw is:

Thou shallt not use globals.

- Hendrik

Hendrik van Rooyen, Feb 13, 2007
15. ### agent-sGuest

Thanks for the help guys but I'm a newbie at this and from what I
understand from the code, it looks like you guys are using a two
dimensional array. I am not using a two dimensional array. Basically,
it looks like this:

[
[' ',' ',' ',' ',' ',' ',' ',' '],
[' ',' ',' ',' ',' ',' ',' ',' '],
[' ',' ',' ',' ',' ',' ',' ',' '],
[' ',' ',' ',' ',' ',' ',' ',' '],
[' ',' ',' ',' ',' ',' ',' ',' '],
[' ',' ',' ',' ',' ',' ',' ',' '],
[' ',' ',' ',' ',' ',' ',' ',' '],
[' ',' ',' ',' ',' ',' ',' ',' ']
]

above: a list of lists NOT a two-dimensional array

agent-s, Feb 14, 2007
16. ### agent-sGuest

OK I just realized that a list of lists can be accessed in the same
way a 2d array would be accessed. Thanks anyways guys.

agent-s, Feb 14, 2007
17. ### Gerard FlanaganGuest

On Feb 12, 1:27 am, "agent-s" <> wrote:
> Basically I'm programming a board game and I have to use a list of
> lists to represent the board (a list of 8 lists with 8 elements each).
> I have to search the adjacent cells for existing pieces and I was
> wondering how I would go about doing this efficiently. Thanks

def iter_grid(cols):
i=0
while True:
yield divmod(i,cols)
i += 1

gen = iter_grid(3)
for i in range(9):
x, y = gen.next()
if x == 1 and y ==1:
continue
else:
yield row - x + 1, col - y + 1

def make_grid_dict(rows, cols):
return dict( zip(iter_grid(cols), ['#'] * (rows*cols)) )

def make_adjacent_list(row, col, row_length, col_length):
ret = []
for x, y in iter_adjacents(row, col):
if x > -1 and y > -1 and x < row_length and y < col_length:
ret.append((x,y))
return ret

grid = make_grid_dict(8, 8)
for x,y in sorted(adjacents.keys()):
print '(%s, %s) - %s ' % (x, y, adjacents[x,y]

(0, 0) - [(1, 1), (1, 0), (0, 1)]
(0, 1) - [(1, 2), (1, 1), (1, 0), (0, 2), (0, 0)]
(0, 2) - [(1, 3), (1, 2), (1, 1), (0, 3), (0, 1)]
(0, 3) - [(1, 4), (1, 3), (1, 2), (0, 4), (0, 2)]
(0, 4) - [(1, 5), (1, 4), (1, 3), (0, 5), (0, 3)]
(0, 5) - [(1, 6), (1, 5), (1, 4), (0, 6), (0, 4)]
(0, 6) - [(1, 7), (1, 6), (1, 5), (0, 7), (0, 5)]
(0, 7) - [(1, 7), (1, 6), (0, 6)]
(1, 0) - [(2, 1), (2, 0), (1, 1), (0, 1), (0, 0)]
(1, 1) - [(2, 2), (2, 1), (2, 0), (1, 2), (1, 0), (0, 2), (0, 1), (0,
0)]
(1, 2) - [(2, 3), (2, 2), (2, 1), (1, 3), (1, 1), (0, 3), (0, 2), (0,
1)]
(1, 3) - [(2, 4), (2, 3), (2, 2), (1, 4), (1, 2), (0, 4), (0, 3), (0,
2)]
(1, 4) - [(2, 5), (2, 4), (2, 3), (1, 5), (1, 3), (0, 5), (0, 4), (0,
3)]
(1, 5) - [(2, 6), (2, 5), (2, 4), (1, 6), (1, 4), (0, 6), (0, 5), (0,
4)]
(1, 6) - [(2, 7), (2, 6), (2, 5), (1, 7), (1, 5), (0, 7), (0, 6), (0,
5)]
(1, 7) - [(2, 7), (2, 6), (1, 6), (0, 7), (0, 6)]
(2, 0) - [(3, 1), (3, 0), (2, 1), (1, 1), (1, 0)]
(2, 1) - [(3, 2), (3, 1), (3, 0), (2, 2), (2, 0), (1, 2), (1, 1), (1,
0)]
(2, 2) - [(3, 3), (3, 2), (3, 1), (2, 3), (2, 1), (1, 3), (1, 2), (1,
1)]
(2, 3) - [(3, 4), (3, 3), (3, 2), (2, 4), (2, 2), (1, 4), (1, 3), (1,
2)]
(2, 4) - [(3, 5), (3, 4), (3, 3), (2, 5), (2, 3), (1, 5), (1, 4), (1,
3)]
(2, 5) - [(3, 6), (3, 5), (3, 4), (2, 6), (2, 4), (1, 6), (1, 5), (1,
4)]
(2, 6) - [(3, 7), (3, 6), (3, 5), (2, 7), (2, 5), (1, 7), (1, 6), (1,
5)]
(2, 7) - [(3, 7), (3, 6), (2, 6), (1, 7), (1, 6)]
(3, 0) - [(4, 1), (4, 0), (3, 1), (2, 1), (2, 0)]
(3, 1) - [(4, 2), (4, 1), (4, 0), (3, 2), (3, 0), (2, 2), (2, 1), (2,
0)]
(3, 2) - [(4, 3), (4, 2), (4, 1), (3, 3), (3, 1), (2, 3), (2, 2), (2,
1)]
(3, 3) - [(4, 4), (4, 3), (4, 2), (3, 4), (3, 2), (2, 4), (2, 3), (2,
2)]
(3, 4) - [(4, 5), (4, 4), (4, 3), (3, 5), (3, 3), (2, 5), (2, 4), (2,
3)]
(3, 5) - [(4, 6), (4, 5), (4, 4), (3, 6), (3, 4), (2, 6), (2, 5), (2,
4)]
(3, 6) - [(4, 7), (4, 6), (4, 5), (3, 7), (3, 5), (2, 7), (2, 6), (2,
5)]
(3, 7) - [(4, 7), (4, 6), (3, 6), (2, 7), (2, 6)]
(4, 0) - [(5, 1), (5, 0), (4, 1), (3, 1), (3, 0)]
(4, 1) - [(5, 2), (5, 1), (5, 0), (4, 2), (4, 0), (3, 2), (3, 1), (3,
0)]
(4, 2) - [(5, 3), (5, 2), (5, 1), (4, 3), (4, 1), (3, 3), (3, 2), (3,
1)]
(4, 3) - [(5, 4), (5, 3), (5, 2), (4, 4), (4, 2), (3, 4), (3, 3), (3,
2)]
(4, 4) - [(5, 5), (5, 4), (5, 3), (4, 5), (4, 3), (3, 5), (3, 4), (3,
3)]
(4, 5) - [(5, 6), (5, 5), (5, 4), (4, 6), (4, 4), (3, 6), (3, 5), (3,
4)]
(4, 6) - [(5, 7), (5, 6), (5, 5), (4, 7), (4, 5), (3, 7), (3, 6), (3,
5)]
(4, 7) - [(5, 7), (5, 6), (4, 6), (3, 7), (3, 6)]
(5, 0) - [(6, 1), (6, 0), (5, 1), (4, 1), (4, 0)]
(5, 1) - [(6, 2), (6, 1), (6, 0), (5, 2), (5, 0), (4, 2), (4, 1), (4,
0)]
(5, 2) - [(6, 3), (6, 2), (6, 1), (5, 3), (5, 1), (4, 3), (4, 2), (4,
1)]
(5, 3) - [(6, 4), (6, 3), (6, 2), (5, 4), (5, 2), (4, 4), (4, 3), (4,
2)]
(5, 4) - [(6, 5), (6, 4), (6, 3), (5, 5), (5, 3), (4, 5), (4, 4), (4,
3)]
(5, 5) - [(6, 6), (6, 5), (6, 4), (5, 6), (5, 4), (4, 6), (4, 5), (4,
4)]
(5, 6) - [(6, 7), (6, 6), (6, 5), (5, 7), (5, 5), (4, 7), (4, 6), (4,
5)]
(5, 7) - [(6, 7), (6, 6), (5, 6), (4, 7), (4, 6)]
(6, 0) - [(7, 1), (7, 0), (6, 1), (5, 1), (5, 0)]
(6, 1) - [(7, 2), (7, 1), (7, 0), (6, 2), (6, 0), (5, 2), (5, 1), (5,
0)]
(6, 2) - [(7, 3), (7, 2), (7, 1), (6, 3), (6, 1), (5, 3), (5, 2), (5,
1)]
(6, 3) - [(7, 4), (7, 3), (7, 2), (6, 4), (6, 2), (5, 4), (5, 3), (5,
2)]
(6, 4) - [(7, 5), (7, 4), (7, 3), (6, 5), (6, 3), (5, 5), (5, 4), (5,
3)]
(6, 5) - [(7, 6), (7, 5), (7, 4), (6, 6), (6, 4), (5, 6), (5, 5), (5,
4)]
(6, 6) - [(7, 7), (7, 6), (7, 5), (6, 7), (6, 5), (5, 7), (5, 6), (5,
5)]
(6, 7) - [(7, 7), (7, 6), (6, 6), (5, 7), (5, 6)]
(7, 0) - [(7, 1), (6, 1), (6, 0)]
(7, 1) - [(7, 2), (7, 0), (6, 2), (6, 1), (6, 0)]
(7, 2) - [(7, 3), (7, 1), (6, 3), (6, 2), (6, 1)]
(7, 3) - [(7, 4), (7, 2), (6, 4), (6, 3), (6, 2)]
(7, 4) - [(7, 5), (7, 3), (6, 5), (6, 4), (6, 3)]
(7, 5) - [(7, 6), (7, 4), (6, 6), (6, 5), (6, 4)]
(7, 6) - [(7, 7), (7, 5), (6, 7), (6, 6), (6, 5)]
(7, 7) - [(7, 6), (6, 7), (6, 6)]

Gerard Flanagan, Feb 14, 2007