carlosayam
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 Dec1 comment How can I find the lenght of the third side of any triangle Sorry, apart from moving I don't see another option. If you can move to another position, please ensure you get reasonable measurements; this is, they need to be reasonably different from the ones you have (or put other way, you have to move at least 40 or 50 yards from your original point if your measurements are only precise +- 1 yards, I believe). Nov30 answered How can I find the lenght of the third side of any triangle Nov29 revised Halting probability of random tree-generating algorithm i stand corrected Nov29 comment Halting probability of random tree-generating algorithm @goos, so your understanding of the problem is that although it halts on a given node, the machine could still work on any other node that is still "open". Like a deep-first algorithm, right? Nov29 comment Halting probability of random tree-generating algorithm @goos, yes I understood it differently. Not sure of any edits by the OP now :( Nov28 comment Halting probability of random tree-generating algorithm @Vortico, although the algorithm seems to produce a tree, it is actually a list as it never recurses on previous nodes to generate new ones there - as I understood it. But even in that case, you could apply the very same reasoning to the following: either it creates a node in any place on the already generated tree with probability 1/2 or it stops with probability 1/2. The answer would be the same. Nov28 comment Halting probability of random tree-generating algorithm @Leo163, your first claim seems wrong. Please see my answer on how to approach this recursively. Nov28 comment Halting probability of random tree-generating algorithm @Vortico, if you prefer a recursive approach, I just added that. Nov28 revised Halting probability of random tree-generating algorithm recursive approach Nov28 comment Halting probability of random tree-generating algorithm @Vortico, I rephrased all for equal probability of stopping or not stopping. Nov28 revised Halting probability of random tree-generating algorithm modified to prob = 1/2 Nov28 comment Halting probability of random tree-generating algorithm @Vortico, the essential part of the argument is considering any past history and the current event as independent events; hence one can multiply their probabilities. Nov28 comment Halting probability of random tree-generating algorithm Probability of halting in step n is $(1/2)^{n-1} (1/2)$ = $(1/2)^n$ for n=1,..; which becomes $\sum_1^\infty{(1/2)^{k}}$ = 1. Even easier :) Nov28 revised Halting probability of random tree-generating algorithm added 228 characters in body Nov28 comment Halting probability of random tree-generating algorithm Hum, it does not, I understood that 3 new nodes and no node were all equiprobable, not as you say. But the outcome is the same, $(1/2) \frac{1}{1-(1/2)}$ = 1. Nov28 comment Halting probability of random tree-generating algorithm I have expanded the answer to your request, if I understood the problem correctly. Hope it helps. Nov28 revised Halting probability of random tree-generating algorithm detailed explanation Nov28 revised Halting probability of random tree-generating algorithm added 36 characters in body Nov28 answered Halting probability of random tree-generating algorithm Nov23 comment $x,y,z$ are positive real numbers and $x+y+z=1$ $\implies$ $\bigg(1+\dfrac 1x\bigg)\bigg(1+\dfrac 1y \bigg)\bigg(1+\dfrac 1z \bigg)\ge 64$? Hum, Jensen's inequality seems to give an smaller bound ... sorry.