combinatorial topology?
401893
First, topology is 'rubber sheet geometry' where things like angles and distances aren't important, but the number of holes is (since you can change angles and distances by stretching the rubber, but you can't change whether there is a hole.\n\nCombinatorics is the mathematics of counting things.\n\nAn example of combinatorial topology is given by Euler's equation. If you imagine paving a sphere with countries, and count the number of countries (F, for faces), the number of boundaries (E, for edges) and the number of places where three or more countries come together (V, for vertices), then it will always happen that F+V=E+2.\n\nIf, instead, you do this with a donut shape (a torus), you will find that F+V=E. In fact, each different surface has a slightly different formula that will hold no matter how it is paved with countries. This is a first example in combinatorial topology. You count various things (like countries) and find something that doesn't depend on specific paving, but rather on the type of surface it is.
683015
It actually becomes trivial at other dimensions.\n\nConsider 1 dimension, or basically a line. This space can be colored in 2 colors. One just alternates the two colors on the line.\n\nConsider 3 dimensions, or our normal space. Imagine a stack of cat food cans as high as you want. Next imagine that each can has a very long chop stick glued to it, and the stick and can are the same color. If the can is sitting flat on the ground, the stick will be straight up in the air - or perpendicular to the lids.\n\nStack the cans so that the sticks all ring around the cans, so that every can's stick touches every other can. Because each can/stick pair is one color, and each stick touches every other stick, in 3-space, you may need a unique color for each entity.\n\nBack to 2-dimensions for a moment.\n\nColoring a planar graph with 5 colors is fairly trival proof in graph theory.\n\n3 colors is never sufficient. Consider a graph where three nodes are all touching a central fourth node. You'll definitely need 4 colors for that.\n\nHowever, in order to determine whether any graph is colorable in 3 colors or not is not easy. In fact it's NP-Complete. Believe me I know. I struggled over that as a homework problem in complexity theory in graduate about 20 years ago. I think I still remember how to prove it.
358650
poly-more than one\nmono-only one.
240497
Here is some information about tiling regular polygons (tiling is also called tessellation):\nhttp://mathforum.org/sum95/suzanne/whattess.html
397805
This problem is certainly tractable, you just need to\nfind a consistent/methodical way of counting the\nrectangles -- without missing any and without counting\nany of them multiple times.\n\nThe way I chose is to define each rectangle as being\ndefined by the square in the upper left corner and the\nsquare in the lower right corner.  Consider the following\nsimplified 2x3 case (where each square is labelled by\nits X and Y coordinates:\n\n0,0     0,1   0,2\n1,0     1,1   1,2\n\n\nThis 2X3 example has the following rectangles starting\nat 0,0.  Remember, each rectangle is defined by the\nsquare in its upper left and the square in the lower right:\n0,0   0,0\n0,0   0,1\n0,0   0,2\n0,0   1,0\n0,0   1,1\n0,0   1,2\n\nThen it also has the set anchored in the upper left by 0,1:\n0,1   0,1\n0,1   0,2\n0,1   1,1\n0,1   1,2\n\nThen it has the set anchored in the upper left by 0,2:\n0,2   0,2\n0,2   1,2\n\nThen those anchored by 1,0:\n1,0   1,0\n1,0   1,1\n1,0   1,2\n\nThose anchored by 1,1:\n1,1   1,1\n1,1   1,2\n\nAnd finally those anchored by 1,2:\n1,2   1,2\n\n\nRewriting the matrix another way, showing how many\nrectangles are anchored by each location in the matrix:\n\n2x3    2x2   2,1\n1x3    1x2   1x1\n\n\nand then doing the math:\n\n6      4     2\n3      2     1\n\nfor a total of 6+4+2+3+2+1 = 18\n\n\nDoing the same analysys for a 4x5 gives:\n\n4x5   4x4   4x3   4x2  4x1\n3x5   3x4   3x3   3x2  3x1\n2x5   2x4   2x3   2x2  2x1\n1x5   1x4   1x3   1x2  1x1\n\nor rewriting after doing the multiplication:\n\n20    16    12    8    4\n15    12     9    6    3\n10     8     6    4    2\n 5     4     3    2    1\n\nAdding up the columns gives:\n\n50    40    30   20   10\n\nFor a total of 50+40+30+20+10 = 150 total rectangles
703879
This sounds like plane geometry where you have an n sided polygon and want to know how many diagonals you can draw.  From each vertex, you can draw a diagonal to every other vertex except to yourself and the two next to you (those two line make sides of the polygon not diagonals).  So for n sides (which means n vertices), you can draw n-3 diagonals at each vertex.  This gives (n-3)*n.  But this just counts the starting end of each line. Whenever you draw one of these lines, it goes to another vertex.  If we drew (n-3)*n lines, we would traces over each diagonal twice, once from each end so there are only half as many diagonals as that.\n\nThe answer then is: (n-3)*n/2
1092713
triskaidecagon\n\nor a 13-sided polygon, haha.\n\nCheck the source for polygon names from 1-20 sides  :)
903397
You might try asking this question under "Mathematics."
100955
Cube, pyramid, sphere, cone, cylinder.  There are others.
231650
No.  With six nodes each with degree three, you can have both the nonplanar K3,3 as well as the planar graph with an outer triangle ABC (draw A centered above a horizontal BC) and an inner triangle DEF (as a shrunken ABC inside ABC) with added edges AD, BE, CF.
243807
Combinatorics (as well as most mathmatics) play a very practical role in computer science when it comes to programming and algorithm design.  For example if I was to create a program that designed LAN networks.  I would need to know the number of computers and the number of connections each computer had.  (for example each computer has 2 unique nic cards).  Then I would use combinatorics to count the number of unique connections that exist in the LAN and from that I would know if there is a feasible design.\n\nhope that makes sense.
1093966
Organizing the different legs of a network
967530
May be a Sphere/Circle of defined polygons or Trees.... with infinite edges and finite branching points....
922060
My suspicion upon seeing this question is that it can be solved using mathematical induction.  After all, couldn't we simply 'build' such a (maximally connected) graph starting with n=2 vertices and 'grow' it from there?\n\nRather than take on the proof myself -- mostly because I am unsure of the definition of 'components' -- I leave it as an exercise for you.  However, it bears noting that the formula shown is the commonly accepted "sum of the first n integers" formula used in almost every text on induction.\n\nGood luck!
731311
I use GrafEq for my graphs, you could give it a try at http://www.peda.com/grafeq/
608619
By considering a change of base point, it is clear that π\n1\n(X, {x\n0\n} depends only on the\npath component of x\n0\n. Since π\n1\n(X × Y, {(x\n0\n, y\n0\n)} = π\n1\n(X, {x\n0\n}) × π\n1\n(Y, {y\n0\n}), we can\nassume the space is path connected and omit the base point.\nSimply connected: an arcwise-connected space with trivial fundamental group. Ho-\nmotopy equivalent spaces have the same fundamental group. Thus, contractible\nspaces, which are homotopy equivalent to a point, are simply connected. Actually,\nany loop S\n1\n→ X which is homotopically trivial extends to a map S\n\n\nSorry, copy pasted it from my e-book.
128133
A hexakaidecagon (hexadecagon).\n\nFor the rest of the polygon questions you'll have, try the link below: