Gregory Guthrie wrote:

Sorry for a simple question- but I don't understand how to parse this

use of a list comprehension.

The "or" clauses are odd to me.

It also seems like it is being overly clever (?) in using a lc

expression as a for loop to drive the recursion.

You are spot on there. It is a list comprehension, but the resulting list

is just thrown away, so using a list comprehension is a complete waste of

time serving only to confuse the issue. Presumably it saved the author a

character or two.

[ exp for var in seq ]

when the result isn't used can be rewritten as:

for var in seq:

exp

and:

exp1 or exp2

when the result is thrown away is just:

if not exp1:

exp2

So:

[ m in [(i-j)%9*(i/9^j/9)*(i/27^j/27|i%9/3^j%9/3)

or board[j] for j in range(81) ]

or solve(board[:i]+m+board[i+1:]) for m in'%d'%5**18 ]

is equivalent to:

inner = [(i-j)%9*(i/9^j/9)*(i/27^j/27|i%9/3^j%9/3) or board[j] for j in

range(81) ]

for m in '3814697265625':

if m not in inner:

solve(board[:i]+m+board[i+1:])

(That inner list comprehension doesn't depend on m, so it doesn't need to

be reevaluated each time round the loop except, again, to save a few

characters.)

The '%d'%5**18 looks to be a silly way to iterate through all possible

digits for m even though it does some of them twice while saving one

character over writing range(1,10).

The strange expression I called 'inner' is a list containing the string

value board[j] if j is in the same row, column or block as i, or an integer

for any other cells. So 'm not in inner' is true only if the value for m is

not already used in that row column or block and is therefore a possible

candidate for that location in the board.