Blag

Still near Linz, at the RISC/SCIEnce Summer School. Different courses, interesting people, no time to blog. ;)

We were yesterday in Linz for the sightseeing and the conference dinner. ‚T was nice.

And by the way, I have a new puzzle for you. It was told to me by Padraig Ó Catháin. You have two rectangles with one being 3 times the area of other and other being 3 times the perimeter of the one. Find their sides in integral numbers. More difficult question: how many solutions are there?

The afternoon lecture by Alexander Konovalov was a huge fun. Aside from all the algebraical goodness in GAP, he demonstrated something really amazing even to generic audience. It is actually easy to represent Rubik’s Cube as a group.

Let’s just unfold the cube and number the faces.

                     +--------------+
                     |              |
                     |  1    2    3 |
                     |              |
                     |  4  top    5 |
                     |              |
                     |  6    7    8 |
                     |              |
      +--------------+--------------+--------------+--------------+
      |              |              |              |              |
      |  9   10   11 | 17   18   19 | 25   26   27 | 33   34   35 |
      |              |              |              |              |
      | 12  left  13 | 20 front  21 | 28 right  29 | 36  rear  37 |
      |              |              |              |              |
      | 14   15   16 | 22   23   24 | 30   31   32 | 38   39   40 |
      |              |              |              |              |
      +--------------+--------------+--------------+--------------+
                     |              |
                     | 41   42   43 |
                     |              |
                     | 44 bottom 45 |
                     |              |
                     | 46   47   48 |
                     |              |
                     +--------------+

Then you can represent the cube as

gap> cube := Group(
> ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19),
> ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35),
> (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11),
> (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24),
> (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27),
> (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40) );

GAP can tell you, how many different positions the puzzle has:

gap> Size( cube );
43252003274489856000

And that’s pretty damn much. But the best is: you can actually produce a solution from an arbitrary position just in three lines of GAP code. It looks like this: (mehr …)