prime order pan diagonal hypercubes | ||||||||
---|---|---|---|---|---|---|---|---|
basic ingredients of pandiagonal hypercubes | ||||||||
Latin hypercube generating formula LH(a_{j}) factor: F = (^{(m-3)/2}_{n-1}) |
Latin hypercubes obtained by formula | |||||||
LH(a_{j}): LH[_{j}i] = (_{j=0}∑^{n}a_{j} _{j}i) % m ;
j ε [0,..,n-1]; i ε [0,..,m-1]; a_{j} < a_{j+1}; a_{0} = 1; a_{j} = 2 .. (m - 1)/2 |
||||||||
The latin hypercubes obtained by the above formula are in normalized position due to the condition a_{j} < a_{j+1} (can't be equal because that spoils pandiagonality). a_{0} = 1 because because of digit changing, thus parameters define the LH's structure the range of a_{j} avoids pan-flip variants introduces by parameter range (m+1)/2 .. m-1. |
||||||||
(^{(m - 3)/2}_{n-1}) | ||||||||
m \ n | 2 | 3 | 4 | 5 | 6 | 7 | ||
5 7 11 13 17 19 |
1 2 4 5 7 8 |
0 1 6 10 21 28 |
0 0 4 10 35 56 |
0 0 1 5 35 70 |
0 0 0 1 21 56 |
0 0 0 0 7 28 |
||
pandiagonal hypercubes | ||||||||
basic pandiagonal hypercubes factor: G = F (^{F 2(n-1) - 1} _{n-1}) (n-1)! |
the basic pandiagonal hypercubes | |||||||
H(a_{j,i}) = _{i=0}∑^{n-1} m^{n-i-1} LH(a_{j,i}) | ||||||||
In order to retain the normalized position the highest component (denoted with i=0) need to remain in normalized position, the other component are added in posible panflip and transpositional variants, leaving the x-axis as is there are 2^{n-1} panflips, resulting in 2^{n-1} F posibilities of which 1 is already chosen, and n-1 LH's need be randomly selected, this explains the listed factor. The factor of (n-1)! is due to reordering of the lower components |
||||||||
F (^{F 2(n-1) - 1} _{n-1}) (n-1)! (note: values ?? too large for excell) |
||||||||
m \ n | 2 | 3 | 4 | 5 | 6 | 7 | ||
5 7 11 13 17 19 |
1 6 28 45 91 120 |
0 6 3.036 14.820 142.926 341.880 |
0 0 107.880 4.744.740 ?? ?? |
0 0 32.760 180.300.120 ?? ?? |
0 0 0 20.389.320 ?? ?? |
0 0 0 0 ?? ?? |
||
pandiagonal hypercubes factor: m!^{n}G/(2m)^{n}= G((m-1)!/2)^{n} |
the pandiagonal hypercubes | |||||||
{pandiagonal monagonal [!]} | ||||||||
H(a_{j,i}) = _{i=0}∑^{n-1} m^{n-i-1} LH(a_{j,i})_{=[perm(i)]} | ||||||||
Applying independently digits changers to the various components generate all(?) possible pandiagonal hypercubes, which introduces a factor (m!)^{n} (if not mistaken). This however also introduces all panvariants which gives a deviding factor of (2m)^{n} with: 2^{n} (reflection) ; m^{n} (panrelocation) |
||||||||
pandiagonal magic hypercubes factors (see text): G_{P}m! G_{N}(m-1)! (G - G_{P} + G_{N})*(?) The various qualities determine possible dividing factors |
the pandiagonal magic hypercubes | |||||||
{pandiagonal magic} | ||||||||
The hypercubes n-agonals are given by
[_{0}0,_{k}0,_{l}(m-1)]
<_{0}1,_{k}1,_{l}-1> if the parameters (a_{j}) if thus that: (a_{0} _{0}1 + ∑a_{k} _{k}1 - ∑a_{l} _{l}1) % m = 0 the digit (∑a_{l} (m-1)) % m needs to be changed to (m-1)/2 since that digit is singular on the given n-agonal, the change to (m-1)/2 the n-agonal sums to the magic constant since the sum then also is m(m-1)/2 |
||||||||
The parametervector that combines with a n-agonal pathfinder into a multiple of m gives an n-agonal with a single digit, changing that digit into (m-1)/2 the latin hypercube n-agonal sums to m (m-1) / 2 and thus to its magic sum this pathfinder corresponds with a corner [c_{i}] c_{1} = m-1 if Pf_{i} = -1; c_{1} = 0 if Pf_{i} = 1 the digit to change to (m-1)/2 thus is (_{i=0}∑^{n}a_{i}c_{i}) % m the permutation =[(m-1)/2|d,p{{0..m-1}\(m-1)/2] changes the digit 'd' to (m-1)/2 while the remainder p[{0..m-1}\(m-1)/2] applies to all but the digit d of course This only applies to parametervectors that combine multiple m with one of the n-agonal pathfinders the other parmetervectors no restrictions are neccessairy future analysis might reveal a factor which is sligthly less then the (m!)^{n} above though the panrelocation factor can't be used,currenty I hold the figure therefore at for order 7: (^{4 * 6!}_{3}) = 3,977,165,760 |
||||||||
Of the F latin hypercubes F_{P} have all n-agonals showing all digits and the part F_{N} satisfy the 0%m affliction and have single digit n-agonals which gives: G_{P} = F_{P}(^{FP 2(n-1) - 1} _{n-1}) (n-1)! {pan n-agonal pan diagonal monagonal} hypercubes while: G_{N} = F_{N}(^{FN 2(n-1) - 1} _{n-1}) (n-1)! {n-agonal pan diagonal monagonal} hypercubes when each component are associated with careful chosen digit changers G - G_{P} - G_{N} {n-agonal pan diagonal monagonal} hypercubes have both types of components. (calculating the total of these is thus more difficult) (F_{P} and F_{N} are yet to be determined) |
||||||||
F_{P} factors. Unknown counting argument (numbers manually obtained) | ||||||||
m \ n | 2 | 3 | 4 |
Cube numbers where obtained in the {perfect} hypercube article. The tesseract numbers I'll have to correct for the non-pantriagonal tesseracts also in future upload the non-panquadragonal probably will get investigated | ||||
5 7 11 13 17 19 |
1 2 4 5 7 8 |
0 0 3 6 15 21 |
0 0 0? 0? 2? 5? |
SAMPLE: the 6 basic order 7 pandiagonal cubes | |||
---|---|---|---|
{pandiagonal monagonal} | |||
The 4 possible LH's | |||
LH(1,2,3) | LH(1,2,4) | LH(1,5,3) | LH(1,5,4) |
<LH(1,2,3),LH(1,2,4),LH(1,5,3)> <LH(1,2,3),LH(1,5,3),LH(1,2,4)> |
<LH(1,2,3),LH(1,2,4),LH(1,5,4)> <LH(1,2,3),LH(1,5,4),LH(1,2,4)> |
<LH(1,2,3),LH(1,5,3),LH(1,5,4)> <LH(1,2,3),LH(1,5,4),LH(1,5,3)> |
note that '5' = '2' panflipped and '4' = '3' panflipped and '6' would be '1' panflipped. |
The upload onto the database of the above mentioned show clearly the correlations LH(1,2,4) = LH(1,2,3)^{~2} LH(1,5,3) = LH(1,2,3)^{~4} and LH(1,5,4) = LH(1,2,3)^{~6} Thus showing that all order 7 cubes are based on the single latin cube LH(1,2,3) |
|||
{pandiagonal magic} | |||
The 4 possible LH's (and digit chaging permutations) | |||
LH(1,2,3)_{=[3|4,P[0,1,2,4,5,6]]} | LH(1,2,4)_{=[3|0,P[0,1,2,4,5,6]]} | LH(1,5,3)_{=[3|6,P[0,1,2,4,5,6]]} | LH(1,5,4)_{=[3|2,P[0,1,2,4,5,6]]} |
LH(1,2,3) has single digit along the <1,1,-1> direction (1+2-3)%7 = 0, the corner position is [6,6,0] thus the digit to change to 3 is (1*6+2*6+3*0)%7 = 18%7 = 4 LH(1,2,4) has single digit along the <1,1,1> direction (1+2+3)%7 = 0, the corner position is [0,0,0] thus the digit to change to 3 is (1*0+2*0+4*0)%7 = 0 LH(1,5,3) has single digit along the <-1,1,1> direction (-1+5+3)%7 = 7%7 = 0, the corner position is [6,0,0] thus the digit to change to 3 is (1*6+5*0+3*0)%7 = 6 LH(1,5,4) has single digit along the <1,-1,1> direction (1-5+4)%7 = 0, the corner position is [0,6,0] thus the digit to change to 3 is (1*0+5*6+4*0)%7 = 30%7 = 2 |