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 wraparound 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 hyperagonal has within a hypercube, below only a generic formulae is given, a full listing of bent hypertriagonals is provided as a sample 

bent hyperragonal  a line through the rdimensional subhypercube bending halfway it length in perpendicular direction  
m/2 {<v_{i}; i = 0..r1>,<w_{i}; i = 0..r1>} both nVectors 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> = ab cos (angle(a,b)) ==> angle(a,b) = acos(<a,b> / ab) working out the inner product <a,b> and the respective lengths of the hyperagonal 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 nvector 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 hypernagonally opposed I therefore conclude that one needs only consider all bent hyperragonals starting from the (0)position and its hyperragonally opposed (m1)position to be complete, so the various hyperragonals [0_{i} i = 0..r] m/2 {<1_{i} i = 0..r>,"other"} and [m1_{i} i = 0..r] m/2 {<1_{i} 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 (2^{n1}1) corners can be reached by a bent hypernagonal so 4 (2^{n1}1) bent hypernagonals are there in n dimensional hypercube 