On 26/05/2014 22:28, Fred Lunnon wrote:
Immmediate reaction: why doesn't either formula give the correct answer when N = 1 , and both counts should equal 2 ?
Immediate metareaction: Why would you expect them to in so special a case? Further remark: the following paths from (0,0) to (4,3) all have length 3, which is obviously best possible since the length must be odd and cannot be 1: 0,0 -NNE-> 1,2 -NNE-> 2,4 -SEE-> 4,3 0,0 -NNE-> 1,2 -SEE-> 3,1 -NNE-> 4,3 0,0 -SEE-> 2,-1 -NNE-> 3,1 -NNE-> 4,3 so either I have terribly misunderstood the problem statement or the answer cannot be 2. Furtherer remark: my formula for the RHS in the case N=1 is (2 choose 1;0;1) which does in fact equal 2. To summarize. You claim - The formulae should apply when N=1. - In this case, LHS = RHS = 2. - In this case, my formula gives different, hence wrong, results for both LHS and RHS to which I reply - If the formulae turned out wrong for very small N it would not surprise me (though actually I think they don't). - In this case, LHS > 2 since I have exhibited 3 distinct elements of the set being counted. - In this case, my formula for the RHS gives the answer you say is right. As usual, though, I think it eminently possible that I'm just confused and have misunderstood or misremembered the problem or done 2+2=3 somewhere or something equally daft. What am I getting wrong? -- g