Q:

square puzzle solution

Question: given an integer number n, print the numbers from 1 up to n2 like this:
n = 4
result is:
01 02 03 04
12 13 14 05
11 16 15 06
10 09 08 07

How do you solve it (apart from the solution provided in the link below)?
http://www.programmersheaven.com/mb/CandCPP/81986/81986/problem-in-making-ap-c++-program/?S=B20000
I'm looking in another direction. So far, I'm trying to figure out if I could obtain the ordered list of positions I have to fill in.
Here's what Im looking into: is there a way to obtain the "fdisp" so as to solve the problem that way, instead of "walk" in the matrix?
matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
n = len(matrix)

# final disposition wrote by hand: how to get it for arbitrary n?
fdisp = [(0,0), (0,1), (0,2), (0,3), (1,3), (2,3), (3,3), (3,2),
         (3,1), (3,0), (2,0), (1,0), (1,1), (1,2), (2,2), (2,1)]

for val,i in enumerate(fdisp):
    matrix[i[0]][i[1]] = val + 1

def show_matrix(matrix, n):
    for i,l in enumerate(matrix):
        for j in range(n):
            print "%d\t" % matrix[i][j],
        print

show_matrix(matrix, n)

A:

Here's a different approach. It relies on spotting that the movements you make cycle between: right, down, left, up, right, .... Further, the number of times you move goes: 3 right, 3 down, 3 left, 2 up, 2 right, 1 down, 1 left. So without further ado, I will code this up in Python.
First, I will use some itertools and some numpy:
from itertools import chain, cycle, imap, izip, repeat
from numpy import array

The directions cycle between: right, down, left, up, right, ...:
directions = cycle(array(v) for v in ((0,1),(1,0),(0,-1),(-1,0)))

(I'm using numpy's arrays here so I can easily add directions together. Tuples don't add nicely.)
Next, the number of times I move counts down from n-1 to 1, repeating each number twice, and the first number three times:
countdown = chain((n-1,), *imap(repeat, range(n-1,0,-1), repeat(2)))

So now my sequence of directions can be created by repeating each successive direction by the paired number in countdown:
dirseq = chain(*imap(repeat, directions, countdown))

To get my sequence of indices, I can just sum this sequence, but (AFAIK) Python does not provide such a method, so let's quickly throw one together:
def sumseq(seq, start=0):
  v = start
  yield v
  for s in seq:
    v += s
    yield v

Now to generate the original array, I can do the following:
a = array(((0,)*n,)*n) # n-by-n array of zeroes
for i, v in enumerate(sumseq(dirseq, array((0,0)))):
  a[v[0], v[1]] = i+1
print a

Which, for n = 4, gives:
[[ 1  2  3  4]
 [12 13 14  5]
 [11 16 15  6]
 [10  9  8  7]]

and, for n = 5, gives:
[[ 1  2  3  4  5]
 [16 17 18 19  6]
 [15 24 25 20  7]
 [14 23 22 21  8]
 [13 12 11 10  9]]

This approach can be generalised to rectangular grids; I leave this as an exercise for the reader ;)

