On Mon, Nov 8, 2010 at 10:58 PM, Greg <g...@kinostudios.com> wrote: > On Nov 8, 2010, at 7:49 PM, Greg wrote: >> So I'm unclear on what 3 (mod 3) means... > > I may have answered my own question, let me know: > > 6 = 3 (mod 3) > > That means that *both* sides are modulo 3, in which case 0 = 0. > > Whereas, > > (a + c) = 3 != 2 (mod 3) > > Makes sense because: > > 0 != 2
Yes; x and y are equal mod 3 if (= (rem x 3) (rem y 3)). > If so, then now the only thing is I'm not sure how you saw that the wheels > satisfied that equation in the first place... As I said, one wheel makes a and b increase by one mod 3 and the other makes b and c increase. Mod is linear -- if a mod 3 cycles by 1, (a + b) mod 3 goes up by 1. So each wheel increases b by one, and increases exactly one of a and c, so (a + c) mod 3 increases by 1. Since both were zero and every move increases both by one (modulo 3), they stay equal. Or you can visualize a 3x3x3 cube where the three numbers are x, y, z coordinates. You have a starting position in one corner and one move goes diagonally parallel to one face and the other goes diagonally parallel to the other. Those moves in various combinations give you a tilted plane inside the cube that amounts to 9 of its 27 cells. (2 2 1) happens to be in one of the other two planes of 9 cells at that angle. Or you can resort to linear algebra and note right away that you have a vector space over Z_3 and two vectors, (1 1 0) and (0 1 1), which will span a planar subspace. Their cross product (1 -1 1) should be perpendicular. Projecting (2 2 1) onto that vector is a simple matter of a dot product, which comes out to 1, showing that (2 2 1) does not lie in the span of those vectors. Since the initial state (3 3 3) = (0 0 0) (mod 3) is the origin and is in that plane, the target (2 2 1) isn't reachable from there using any of those moves in any combination. (Note that (1 -1 1) dot (a b c) = (a + c - b) which goes right back to the invariant of (a + c) mod 3 = b mod 3; that invariant is really "the dot product with (1 -1 1) must be 0" in disguise.) Or you can resort to group theory... -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
