Pandiagonal Stacking S_{m} = Stack(S_{p},S_{q},R,P,C) construction method for compound order panmagic squares (orders: m = p * q) |
||
---|---|---|
S_{m,k,l} = q * {S_{p,k/q,l/q} P_{i} + R_{i} + C_{i}(?) + i(hc)}
_{=[perm(q)]} + i(lc)_{=[perm(q)]} S_{q,k%q,l%q} = i(hc)_{=[perm(q)]} + i(lc)_{=[perm(q)]} |
||
the above formula seems to depict the entire stacking process, seen the various factors seem to depend on the stacked squares high component latin square forced me to split up this stackey into it's components, qualitatively lined up by the components digits changing permutation The role of the compensator is yet to be determined. (currently I haven't found an order 15 square yet with the regular number range) |
||
Stacker S_{p} factor: (p!)^{2} (preliminairy) |
The order p square which stacks the order q square | |
The order p square which is needed to stack panmagic square to create panmagic square needs to be of "skew magic" quality (ie: every (broken) 2-agonal sums to the same sum) curently I hold these as row/column permutations of the "natural squares" (N_{i,j} = i * p + j ; i,j = 0 .. p-1 suggesting the factor of (p!)^{2} |
||
Stackey S_{q} factor: (see order q) |
The order q square which stacked by the order p square | |
The panmagic square to be stacked adheres of course to the order q theory, currently there is only a definite number for prime orders q as well as for some low non-prime orders The Stacking method seems to rely heavely on the highest latin component square of the Stackey, as it distributes the numbers of the "pandiagoliser" and the "regulator" |
||
Pandiagoliser P factor: (undetermined) |
A panmagic square summing 0 | |
The crux of the stacking process is the pandiagoliser, the pandiagolising square is placed on the stackers celpositions and multiplied by the stackers number. The pandiagolising square can be seen as a digit changing permutation changing permutation of the stackey's highest component latin square, for even orders q the pandiagoliser can consist of mere +/-1 with a total sum of 0 (easiest to stack). For odd order q mere +/-1 won't do, the other values can be compensated for quite easily by the compensator (NOTE: presently this is an assumption due to the absence of a regular order 15 sample) |
||
Regulator R factor: 1 (preliminairy) |
A panmagic square lining up the order q squares to the range of the order m square | |
currently I hold R_{i} = (p - 1) * i + if (P_{i} < 0 ; p ; 0) i = 0 .. p-1 | ||
The Regulator lines up the numbers thus to avoid doubly appearing numbers, trick is also to have them exactly line up, in order to obtain a regular panmagic square of order m The Regulator can be seen as a digit changing permutation to be applied to the Stackeys high LS, values seem to depend on the used pandiagolisor so currently I hold the factor as 1. |
||
Compensator C factor: 1 (preliminairy) |
A function compensating pandiagoliser steps | |
C_{i} = S_{p,k,l} * if (P_{i} < 0 ; 1 ; -1) k,l = 0 .. p-1; i = 0 .. p - 1 |
||
This function merely reduces the steps taken by the pandiagoliser back to the steps of the lowest value. (The above reduces steps to 1, it might well be possible to take bigger steps. The main concern is to avoid double numbers in the determinator) (NOTE: currently this is under scruteny since I havn't found an actual order 15 square yet) |
||
Determinator | Expression to determine whether the values gives order m square in the regular range | |
D_{i} = S_{p,k,l} * (P_{i} + R_{i} + C_{i}(?)) + i k,l = 0 .. p-1; i = 0 .. p - 1 |
||
in order for the order m square to have the full number range the above expression must yield the full range of numbers 0 .. p m - 1 |