The Diagonal | ||
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main diagonal | line in square a full square donwards | |
m {<1,1>} | ||
sub diagonal | line in square a full square upwards | |
m {<1,-1>} | The Bent Diagonal | |
\/ type bent diagonal | line in square half a square donwards and half a square upwards | |
m/2 {<1,1>,<1,-1>} | ||
/\ type bent diagonal | line in square half a square upwards and half a square downwards | |
m/2 {<1,-1>,<1,1>} | ||
> type bent diagonal | line in square half a square donwards and half a square downwards to the left | |
m/2 {<1,1>,<-1,1>} | ||
< type bent diagonal | line in square half a square downwards to the left and half a square downwards | |
m/2 {<-1,1>,<1,1>} | ||
notice that within a full "summing square" due to wrap-around the "/\" will be equivalent to "\/" as are the squares mentioning all sums to ">" and "<" type paths |
The Bent Hyperagonal | |
note that due to the many degrees of freedom a bent hyper-agonal has within a hypercube, below only a generic formulae is given, a full listing of bent hypertriagonals is provided as a sample |
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bent hyper-r-agonal | a line through the r-dimensional subhypercube bending halfway it length in perpendicular direction | |
m/2 {<vi; i = 0..r-1>,<wi; i = 0..r-1>} both n-Vectors elements -1 and 1 only |
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bent hypertriagonal |
as a sample al the bent hyper triagonals with the cube I do think the formulae catch all possibilities (needs verification) |
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m/2 {<1,1,1>,<1,1,-1>} ;
m/2 {<1,1,1>,<1,-1,1>} ;
m/2 {<1,1,1>,<-1,1,1>} ; {a,b} -> {a,-b} ; {a,b} -> {-a,b} ; {a,b} -> {-a,-b} ; |
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note: a formula from vector algebra states the relation between vectors and inrmediate angle as; <a,b> = |a||b| cos (angle(a,b)) ==> angle(a,b) = acos(<a,b> / |a||b|) working out the inner product <a,b> and the respective lengths of the hyper-agonal vectors gives angle(main,sub) = acos(1 - 2p / n) where 'main' denotes the main hyperagonal and 'sub' the relavant subdiagonal with p -1's in the paths n-vector which shows that the parts of a bent hyperagonals are not orthogonal to another |
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notice: looking from one corner at all the possible paths described above one reaches every other corner save the hyper-n-agonally opposed I therefore conclude that one needs only consider all bent hyper-r-agonals starting from the (0)-position and its hyper-r-agonally opposed (m-1)-position to be complete, so the various hyper-r-agonals [0i i = 0..r] m/2 {<1i i = 0..r>,"other"} and [m-1i i = 0..r] m/2 {<-1i i = 0..r>,"other"} where other stand for all the possible other directions (which are of course the same set of vectors) |
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Counting the number of possibilities from the 2 corners 2 (2n-1-1) corners can be reached by a bent hyper-n-agonal so 4 (2n-1-1) bent hyper-n-agonals are there in n dimensional hypercube |