Odd Order Construction

(consecutively concentric squares)

Constraints

The bones 4(n-1) border numbers form 2(n-1) ± pairs. The + numbers are in the range:

((n-2)²+1)/2 to (n²-1)/2

The sum of each bones row, column, and main diagonal is 0.
Therefore, the border constraints are:

the sum of row 1, (row n), is 0
the sum of column 1, (column n), is 0
the other rows and columns, and the main diagonals have matching ±numbers at each end

Algorithn

The square is divided into the following regions:

center cell
NW,SE main diagonal
NE,SW main diagonal
W,E center row
N,S center column
NNW sector, SSW sector
NNE sector, SSE sector
WNW sector, ENE sector
WSW sector, ESE sector

Define rrel = row - (n+1)/2, crel = column - (n+1)/2.
Then, the basic formula used here to compute a cell number is 2x² + 2y + c, where, for each region:

x is rrel, or crel
y is crel, or rrel, or 0
c is a constant

The algorithm is:

assign x: rrel
assign y: crel
determine the region of the square
reset y to 0 in NW, SE, NE, SW
swap x and y in WNW, ENE, WSW, ESE, W, E
assign c:
- -1 in NNW, SSW, NE, SW
- 0 in WNW, ENE, W, center, E
- 1 in NNE, SSE, NW, SE
- 2 in WSW, ESE, N, S
compute 2x² + 2y + c
change sign in NNE, ESE, SSW, WNW, SE, SW, N, W
optionally, add (n²+1)/2

Summary:

    region          x              y             c          sign
   --------    -----------    -----------     -------    ----------
    center         rrel           crel           0           0

     NW            rrel            0             1           +
     SE            rrel            0             1           -

     NNW           rrel           crel          -1           +
     SSW           rrel           crel          -1           -

     N             rrel           crel           2           -
     S             rrel           crel           2           +

     NNE           rrel           crel           1           -
     SSE           rrel           crel           1           +

     NE            rrel            0            -1           +
     SW            rrel            0            -1           -

     ENE           crel           rrel           0           +
     WNW           crel           rrel           0           -

     E             crel           rrel           0           +
     W             crel           rrel           0           -

     ESE           crel           rrel           2           -
     WSW           crel           rrel           2           +

Correctness

Is the algorithm correct for all odd order squares?

It is sufficient to check the border, since the test applies recursively through the nested squares down to order 3.

Let B be the bones, m = (n+1)/2, and X = 2(1-m)² = 2(n-m)².

Rows

In row 1, rrel = 1-m, crel ranges from 1-m to n-m.

B(1,1) =   X + 1
B(1,m) = -(X + 2)
B(1,n) =   X - 1
The sum of these is X + 1 - X - 2 + X - 1 = X - 2.

NNW numbers are   X + 2crel - 1.
NNE numbers are -(X + 2crel + 1).
Summing NNW,NNE number pairs from left to right, we have:
B(1,2) + B(1,m+1)   = X + 2(2-m) - 1 - X - 2(1) - 1    = -(n+1)
	…
B(1,m-1) + B(1,n-1) = X + 2(-1) - 1 - X - 2(n-1-m) - 1 = -(n+1)
The sum of these (n-3)/2 pairs is -(n+1)(n-3)/2 = -(X - 2).

Row 1 total is (X - 2) - (X - 2) = 0.
Rows 2 to n-1 have ± the same numbers at each end.
Row n contains ± the same numbers as row 1.

Columns

In column 1, rrel ranges from 1-m to n-m, crel = 1-m.

B(1,1) =   X + 1
B(m,1) =  -X
B(n,1) = -(X - 1)
The sum of these is X + 1 - X - X + 1 = -(X - 2).

WNW numbers are -(X + 2rrel).
WSW numbers are   X + 2rrel + 2.
Summing the WNW,WSW number pairs from top to bottom, we have:
B(2,1) + B(m+1,1)   = -X - 2(2-m) + X + 2(1) + 2        = n+1
	…
B(m-1,1) + B(n-1,1) = -X - 2(-1) + X + 2(n - 1 - m) + 2 = n+1
The sum of these (n-3)/2 pairs is (n+1)(n-3)/2 = X - 2.

Column 1 total is -(X - 2) + (X - 2) = 0.
Columns 2 to n-1 have ± the same numbers at each end.
Column n contains ± the same numbers as column 1.

Border Numbers

Are the border numbers correct, that is, in the right range with no duplicates?

There are 4(n-1) border numbers or 2(n-1) ± pairs.
There is 1 pair in each of the main diagonals,
1 in the center row, and 1 in the center column.
The remaining 2(n-1) - 4 pairs are in the (NNW,SSW), (NNE,SSE), (WNW,ENE),
and (WSW,ESE) sectors.

The placing of the 2(n-1) positive numbers is:

----- n-3 numbers alternating NNW, ENE
NNW: B(1,2)   = X + 2(2-m) - 1   = X - n + 2
ENE: B(2,n)   = X + 2(2-m)       = X - n + 3
	…
NNW: B(1,m-1) = X + 2(-1) - 1    = X - 3
ENE: B(m-1,n) = X + 2(-1)        = X - 2


NE:  B(1,n)                      = X - 1
E:   B(m,n)                      = X
NW:  B(1,1)                      = X + 1
S:   B(n,m)                      = X + 2

----- n-3 numbers alternating SSE, WSW
SSE: B(n,m+1) = X + 2(1) + 1     = X + 3
WSW: B(m+1,1) = X + 2(1) + 2     = X + 4
	…
SSE: B(n,n-1) = X + 2(n-1-m) + 1 = X + n - 2
WSW: B(n-1,1) = X + 2(n-1-m) + 2 = X + n - 1

The next smaller odd order square is order k = n-2.
Its biggest border number is 2(1-(k+1)/2)² + k - 1 = X - n + 1.
The smallest border number of the order n is X - n + 2.
That is, the smallest border number of the order n is 1 greater
than the biggest border number of the order n-2.