The Magic Encyclopedia ™

The Bent Hyperagonal
{note: investigative article}
(by Aale de Winkel)

As a generalisation of the bent diagonals found in the squares of Ben Franklin, the bent hyperagonal serves a simular purpose within a hypercube. Presented here a studies of these bent hyperagonals, for completeness and introduction of the made notational effort, the diagonal and bent diagonal are also depicted. Also use of the notation to depict other types of paths will be briefly shown.
Currently this article is showing the form of the paths, which might be positioned with an n-Point when so desired
so [0,0] m {<1,1>} thus depicting a squares regular main diagonal
Noticed a slight flaw in the depicted notational effort in that the path restarts the bended line-part in an adjecent position to the half-line ended, this flaw might be corrected by moving the path length in front of every n-Vector and mentioning the cell-move explicitely between both path-parts. I currently assume this cell-move implicitly in the notation, and use the more explicit form when this cell-move deviates from this implicit cell-move
ie: m/2 {<1,1>,<1,-1>} = {m/2<1,1>,<1,0>,m/2<1,-1>}

The Diagonal
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
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
bent hypertriagonal as a sample al the bent hyper triagonals with the cube
I do think the formulae catch all possibilities (needs verification)
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} ;
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
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)
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