Ken, Holy crap. Thank you for this wonderful message.
My apologies for taking a while to respond, but your email has somewhat disarmed me. I'm very impressed with your knowledge of math, and I felt at once eager to read and understand your email and was taken aback by it, as my math skills are a bit rusty unfortunately (a fact I'm not proud of). > 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. Thanks, that does indeed make it clearer, but I'm still impressed that you saw this. I guess you've had more experience with this sort of reasoning than I, but still, very well done sir! > Or you can visualize a 3x3x3 cube [...] 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. Indeed, a nice way of visualizing the situation, but one that is far from my way of thinking. I guess I really should probably dust off my copy of Princeton's Companion to Mathematics... not enough time unfortunately. :-\ > 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. I see that is true as well... Nicely done sir! :-D > Or you can resort to group theory... I am a bit grateful (though still curios) that you chose not to expand on this one, as my knowledge of mathematics doesn't extend to group theory, and I barely escaped your other examples with my ego intact! :-p Cheers! - Greg On Nov 8, 2010, at 8:20 PM, Ken Wesson wrote: > 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 -- 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
