On Jun 11, 12:56 pm, Virgil <virg...@nowhere.com> wrote: > In article > <2614a3b6-17bb-48b3-aa1d-37fe32a7a...@o18g2000yqi.googlegroups.com>, > > > > > > WM <mueck...@rz.fh-augsburg.de> wrote: > > On 11 Jun., 13:00, William Hughes <wpihug...@hotmail.com> wrote: > > > On Jun 11, 6:21 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 11 Jun., 00:12, William Hughes <wpihug...@hotmail.com> wrote: > > > > > > - There is more that one path through "that node" > > > > > > - There are countably many nodes below "that node" > > > > We are discussing WM's statement: > > > > Let us choose one of them and map it one that node. > > > For the others we will find other nodes, below that one chosen. > > > > WM has agreed > > > > - There is more that one path through "that node" > > > > - There are countably many nodes below "that node" > > > > We are not at > > > > > > - Assuming that "For the others we > > > > > will find other nodes, below that one chosen" > > > > > is equivalent to assuming that there are only > > > > > countably many paths through "that node" > > > > > No. It is not assuming this but *proving* this. > > > > Nope, "assuming" is not "proving". You give a putative > > > proof that starts out by assuming what you want > > > to prove. This is circular. > > > Not at all. I show that there are no nodes left over, after a > > countable number of infinite paths have been constructed. > > Since infinitely many paths use each node, and you have only eliminated > one per node, you would then need to show that infinity minus one is > zero for your argument to hold. > > > My only > > assumption is that in a tree, the nodes of which are completely > > covered by paths, no further path can be constructed. > > But that is not the case for a MAXIMAL infinite binary tree. > For example, WM's own construction of a tree in which no path has more > than finitely many 1 branchings and infinitely many 0 branchings > By his own argument, correct for once, there is a path through every > node of even a maximal infinite binary tree. > > But such a tree is far from maximal as there are more paths missing than > present. > > > > > My proof rests upon the fact that after the set of all finite paths of > > the form 0.1, 0.11, 0.111, ... has been constructed, there is no > > chance to construct the path 0.111... in addition. > > Nevertheless, if a maximal tree that path must be present, so that WM's > construction does not produce the desired maximal infinite binary tree. > > > > > Regards, WM > > > > > Because paths cannot be distinguished without nodes. > > > > Proves nothing. We know that a countable > > > number of elements can distinguish an uncountable > > > number of subsets. A countable number of nodes can > > > distinguish an uncountable number of paths. > > > > - William Hughes > > -- > Virgil- Hide quoted text - > > - Show quoted text -
P-versus-NP pageIt has been archived in June 2004, and it proves that P=NP. ... In October 2005, Moustapha Diaby also constructed a linear programming formulation of the ... www.win.tue.nl/~gwoegi/P-versus-NP.htm
[PDF] The P versus NP ProblemFile Format: PDF/Adobe Acrobat - View as HTML search problem for every NP problem has a polynomial-time algorithm. For .... generators have been constructed assuming the hardness of integer factor- ... www.claymath.org/millennium/P_vs_NP/Official_Problem_Description.pdf
