Evo ti link koji moze da pojasni stvar:
http://www.rubiks.com/cube_online.html

Resenje je ogroman broj!!! Ni sam ne verujem da je tako ali .....
nemoze da se izracuna pomocu olovke i papira.......
Zato kazem, ako neko zna da izracuna pomocu racunara nekako, zato sto me zanima da li je tacno resenje koje imam....

I dalje mi nije jasno kakve to veze ima.. znam rubikova kocka sta je i znam da ima online resenja...

Cube Mathematical Permutations

The centre pieces of each face are always in the same relationship to each other.
Therefore, the number of ways the other pieces can be arranged in relation
to them (ie the possible arrangements of the cube) is:
Total possible arrangements = (possible arrangements of Corner Pieces) x (possible arrangements of Edge Pieces).
Possible Arrangements of Corner Pieces
There are 8 Corner Pieces. Therefore, the possible different arrangements of them is 8! (ie 8x7x6x5x4x3x2x1) = 40320.
Each Corner Piece has three different orientations, so this must then be
multiplied by 3 to the power 8 (ie 3x3x3x3x3x3x3x3) which equals 6561.
However, with the actual cube, once the second from last Corner Piece is
placed, the last piece can have only one automatic orientation so this
should be divided by 3 (effectively 3 to the power 7) which equals 2187.
Thus, total possible arrangements of Corner Pieces = 40320 x 2187 = 88,179,840.
Possible Arrangements of Edge Pieces
There are 12 Edge Pieces. Therefore, the possible different arrangements of them is 12! (ie 12x11x10x9x8x7x6x5x4x3x2x1) = 479,001,600. However, with the actual Cube (and unlike Corner Pieces) it is impossible to exchange just two Edge Pieces, so once the third from last is placed, the remaining two can have only one possible arrangement, so this total must be divided by 2, which equals 239,500,800.
Each Edge Piece has two different orientations, so this must then be multiplied by 2 to the power 12 (ie 2x2x2x2x2x2x2x2x2x2x2x2) which equals 6561. However, with the actual cube, once the third from last Edge Piece is placed, although the last two pieces will be in fixed positions, one can be reoriented but the last will always have a fixed orientation in relation to it. So this must be divided by 2 (effectively 2 to the power 11) which equals 2048.
Thus, total possible arrangements of Edge Pieces = 239,500,800 x 2048 = 490,497,638,400.
So - Total Possible Arrangements of Rubik's Cube = (possible arrangements of Corner Pieces) x (possible arrangements of Edge Pieces) = 88,179,840 x 490,497,638,400 = 43,252,003,274,489,856,000. Roughly speaking, 4.3 times 10 to the power 19.

Ima li nešto novo...dajte ljudi malo problemčića, ne spavajte...

AJ nek neko resi sa crncima

Radi lakseg razumevanja B-belac,C-crnac i V-crnac veslac.
1.predju BiC vrati se B
2.predju ViC vrati se V
3.predju BiB vrate se BiC
4.predju ViB vrate se BiC
5.predju BiB vrati se V
6.predju ViC vrati se V(ili B)
7.predju ViC(ili BiC)
The End.
Nadam se da nisma nigde pogresio.Aj' proverite.

Hm... ako se ne varam, ovo je ok.
Nisam 100% siguran jer me mrzi da gledam da li u bilo kom trenutku ima vise C nego B, ali ovako mi se za sada cini ok.

Dobri je, ne trebaju ti one zagrade na kraju...

Evo dopune Leggy-ijevog problema, mada je i pola resenja sadrzano u njemu, za ovo sto postaviti.

Imate kanap koji gori tacno 60 minuta (brzina gorenja kanapa nije uniformna, menja se po nekom nepoznatom zakonu).

Kako cete izmeriti 45 minuta sa dva takva kanapa?

P.S. Nadam se da nisam ponovio vec postavljen problem.

jedan zapalis sa krajeva a jedan i sa krajeva i sa sredine....