Challenge - Keypad puzzle

[pinkman]

I still haven't finished a clean implementation in HPPL, but I think I found an elegant solution in python.
Elegant but brut force (find all numbers and divide them by 11) as I requested.

A few preliminary comments about the methods:
What I call a 'move' is the path followed by your finger on the keypad between 2 keys.
Suppose you start from the '5' key (coordinates 2x2 on the grid) : you have 8 possible moves : 4 (left, right, up, down) x 2 (clockwise and counter-clockwise). Starting from other keys reduces the number of authorized moves, but the principle is the same.
A move from '5' key is -1 ('4' key), +1 ('6' key), -3 ('2' key) or +3 ('8' key).
To find the number formed by the square, you make 3 moves after having used the initial key, which makes 4 digits.
After a horizontal move (+/-1) you have to make a vertical move (+/-3), and then an horizontal move in the opposite direction of the previous horizontal one, then a vertical move in the opposite direction of the previous vertical one.

Mathematically there are two properties to follow a consecutive square on the keypad :
1- If we call m our first move (being -1, +1, -3 or +3), then the third move is -m.
2- If we call m' our second move (being -1, +1, -3, +3), then you'll have ton constraint |m|≠|m'| (in readable language: +/-1 is followed by +/-3, +/-3 is followed by +/-1).

For each key, the moves are m, m', -m.

PHP Code:

result = []
movr = [[3], [-3,3], [-3]]      # authorized moves on each row from bottom to top (first to third)
movc = [[1], [-1,1], [-1]]      # authorized moves on each column (left to right)
a = 0
for i in movr:                  # step accross each row...
    for j in movc:              # and accross each column
        a += 1                  # name (number) of the key
        for k in i:             # start moving along the row
            for l in j:         # and along the column
                result.append(a*1000 + (a + k)*100 + (a + k + l)*10 + (a + k + l - k))
        for k in j:             # then start moving along the column
            for l in i:         # and along the row
                result.append(a*1000 + (a + k)*100 + (a + k + l)*10 + (a + k + l - k))
for i in result:
    print(i, " modulo (11) -> ", i %11) 


And the result :

Code:


1452  modulo (11) ->  0
1254  modulo (11) ->  0
2541  modulo (11) ->  0
2563  modulo (11) ->  0
2145  modulo (11) ->  0
2365  modulo (11) ->  0
3652  modulo (11) ->  0
3256  modulo (11) ->  0
4125  modulo (11) ->  0
4785  modulo (11) ->  0
4521  modulo (11) ->  0
4587  modulo (11) ->  0
5214  modulo (11) ->  0
5236  modulo (11) ->  0
5874  modulo (11) ->  0
5896  modulo (11) ->  0
5412  modulo (11) ->  0
5478  modulo (11) ->  0
5632  modulo (11) ->  0
5698  modulo (11) ->  0
6325  modulo (11) ->  0
6985  modulo (11) ->  0
6523  modulo (11) ->  0
6589  modulo (11) ->  0
7458  modulo (11) ->  0
7854  modulo (11) ->  0
8547  modulo (11) ->  0
8569  modulo (11) ->  0
8745  modulo (11) ->  0
8965  modulo (11) ->  0
9658  modulo (11) ->  0
9856  modulo (11) ->  0