Hello, Erop, you wrote: E> Hello, Brice Tribbiani, you wrote: BT>> Orientation is precisely encoded by three , that is we lose 8 variants, all remaining our bits. Perhaps it is possible and is cheaper, but I do not know as. E> Well abruptly to prove, what it is impossible E> Actually there is an obvious decision with two bits of an overhead projector (interesting, the term "waybills" would be clear?) Instead of three What? BT>> in a square 33 I quickly counted 8 projections of invariants (one black corner, a corner and adjacent , the black side, here to you and 9). E> for the coding of 9 numbers it is enough to consider a prime number of black pixels E> But here look, from obvious, there are 4 angular pixels, 4 "side" and 1 center E> Number angular - 5 variants, number side - 5 variants, and central - 2 variants. E> it already 50 variants. It, of course, is less 2 ^ (9-3), but, it is possible to note that in both 4-kah pixels, it is possible to apply the same coding, as on a square 22 that already gives 72 variants... Yes, I there , it is necessary not to subtract 8, and to divide on 8 Then yes, orientation not the most favourable. BT>> on idea, you take the formula which was deduced by you with rg45, and you compare to my of the first answer. BT>> only something you for N=2 is not true, I do not know as for the big. E> we deduced the formula of number of the pictures which are playing back at turn on 90. For N == 2 it gave 2 variants. E> like true - all white and all black... Aha, understood. Yes, then your formula works, but to compare to it it is not necessary