On 4/4/14, Warren D Smith <warren.wds@gmail.com> wrote:
WDS: I don't agree that your lemma 2 is proven. (Same objection I gave before, I don't see that it was impacted in the slightest.) MInd you, I believe this general approach should work, just a lot more cases will be needed.
WDS's latest postings discussed f(x, y) , the formula function. Lemma 2 is concerned exclusively with d(x, y) , the distance function. Regardless, if the proof of either lemma is incorrect or incomplete, it would be helpful to know at what precise point my logical sequence breaks down or lapses into obscurity. If there is a gap, what case is still missing? As with programmming, it is all too easy when constructing a proof to assume obvious some step which is either nontrivial, or may even be faulty. --------------------- Which reminds me that when the holes in my previous version surfaced, I became aware that I had subconsciously been unhappy about it from the start --- it somehow seemed too easy to be true. I have often had a similar experience previously, yet rarely succeed in articulating it sufficiently to incorporate it into the activity. There should be a word for this situation --- wishful thinking at work, perhaps. But it is intriguing to reflect that the big parallel processor in the background may well have a better grasp of formal logical processes than the feeble I/O unit which is all we can directly access. It was said of the pre-war world chess champion Alexander Alekhine that he seemed incapable of explaining why his own best moves were sound. So maybe this sort of experience is common in many fields. WFL On 4/3/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It is a pleasure to report the defeat of the horrid spectre of topology, by dint of abandoning attempts to reason about attaching an extra step at the far end [x-2, y-1] in favour of considering (shorter) paths from neighbours [2, 1] etc. of the origin instead.
Then straightforwardly the distance function d(x, y) satisfies the same recursion as the explicit expression f(x, y) , whence they are equal.
A corrected and slightly expanded screed is posted at https://www.dropbox.com/s/nzmzjswtctju23f/knights_path.txt Reports of attempts to blow further holes in it would be welcome, even if unsuccessful. Particularly if unsuccessful!
Fred Lunnon