Q1. Prove that (a) Every point has some knight-move neighbour at smaller knight-move distance from the origin if and only if (b) Every point has some knight-move neighbour at greater knight-move distance from the origin. --first of all (a) is false, a counterexample is (0,0). However, (a') is true, meaning (a) for every point other than (0,0). This is simply because for ANY connected graph, and ANY two unequal vertices A and B in it, there exists a point 1-nearer to A than B. (Here we use the knight-adjacency graph.) And you can find it by breadth-first search from A to find all points at distance 1, then all at 2, then 3, etc. Now I don't think you really were trying to get me to prove (b) is false. You wanted me to prove (b) true. It seems to me that (a') is obvious as above but (b) is not so obvious. However, in your paper draft there is a result that not only is (a') true, but indeed a 1-nearer neighbor exists in a certain angular wedge. Given that lemma, and perhaps some manual examination of cases at small distance (say distance<9) too, it seems to me that (b) then follows because the set of points "1-nearer", after taking union with all angular rotations and also union with all small-distance cases verified manually, is the entire universe.

Q2. Prove(!) that there is just a single shortest knight path from (0,0) to (2r,r) taking distance r steps.

--this path is a "geodesic." That is, the furthest EUCLIDEAN distance you can hop in r steps, is sqrt(5)*r, obviously, which happens iff all steps are exactly parallel, which proves uniqueness.

Q3. Denote e(x, y) the number of shortest knight paths from (0,0) to (x,y) . Prove that e(1999999999, 3) = e(1500000000, 1499999998) . --Let me chop out 7 digits because otherwise I have trouble counting them: "Prove that e(199, 3) = e(150, 148)." Well... think I'll skip that one for now. If you edit the draft to incorporate my last round of abuse, I'll ponder the new draft.

-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)