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=57502&t=57484 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
