I like your methods. Since we're networking people, I think the best answer is, run the STP on it first, and the answer is:
ZERO triangles! That wasn't really what I had in mind, but it's a good one! Thanks for the insight. :-) Priscilla Marc Thach Xuan Ky wrote: > > Hi Priscilla, > At the end of the slideshow you ask for other methods, well > I've got one > and it's really easy. Before I start you should note that my > emoticons > have broken down so you may need to insert your own. > Unfortunately my first attempt to implement the method that I'm > about to > describe was error-prone and gave the answer as 31 triangles. > Now, the > shape is five-way symmetrical (which indicated that 31 was > probably not > correct), it's a five-point star with the pointy nodes joined > together > by extra links. We'll call the five pointy bits distribution > nodes, and > the five intersections in the middle we'll call core nodes. > The area > outside the shape is the access area. Now any given triangle > can have > either 3 distribution, 2 distribution / 1 core, 1 distribution > / 2 core, > or 3 core (except that the core isn't meshed so this is zero). > We will > abbreviate these types as 3D, 2D/1C, 1D/2C, and 3C because we > like > jargon. Inspection also shows that the 3D types can be > subdivided into > long triangles and fat triangles (3LD and 3FD) 2D/1C types can > also be > subdivided, into adjacent D's and non-adjacent D's (2AD/1C and > 2ND/1C). > With me so far? Good because we now subdivide the 2AD/IC into > three > subtypes: straight down, hanging left and hanging right > (2AD/1Cbis, > 2AD/1C(L) 2AD/1C(R)). Anyway all told we now have eight > categories of > triangle, we can count each category (please don't count the > 3Cs during > your leisure time). > So by breaking the problem down this way, it is easier to count > and thus > much quicker to implement. In fact we now just have to count > from one to > five several times. Of course if we employed a project manager > the > probleem could be shared between seven triangle-counters > working in > parallel. This could bring the end-date in by a full ten > percent. > Disclaimer: Note that if working in a quality-assured > environment you > will need eight triangle-counters. The 3C type cannot be > assumed to > have no triangles. Time-savings shown are for example only and > cannot > be guaranteed. > Just to close, there is a further refinement of the technique. > Because > the shape is five-way symmetrical, you in fact only have to > count to > one, what could be more straightforward than that? This has > the added > benefit of enabling the project to be broken up into even > smaller and > more manageable tasks. > One more thing, perhaps it's a trick question. All nodes may > run STP so > all loops are removed, hence the correct answer could be zero. > BTW if you were wondering about the access area, it's not > actually > relevant. > rgds > Marc TXK > > Priscilla Oppenheimer wrote: > > > > I added a Topology Troubleshooting Puzzle to my Web site. > It's not > > Cisco-specific. Well, to be honest, it's not even networking > specific! ;-) > > But it does make you think and wonder how you could be so > blind, if you're > > like me when I first did it. Be sure to actually try it > before going on to > > the solution. OK, is that enough filler? The URL is here: > > > > http://www.troubleshootingnetworks.com/triangles/index.htm > > > > Offline, let me know what you think (if you have my address, > which I can't > > publish due to commercial unsolicited e-mail.) > > > > Priscilla > > Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=57521&t=57484 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
