Two Curves
Two curves can be placed back-to-back, as if starting from a line segment traversed in both directions.
<--------- back-to-back
---------> lines
For example to N=16,
11-----10-----9,7-----6------5 k=4 3
| | |
13-----12 8 4------3 2
| |
14 2 1
| |
15-----16 0------1 <- Y=0
1------0 16----15 <- Y=0
| |
2 14 -1
| |
3------4 8 12-----13 -2
| | |
5------6-----7,9----10-----11 -3
The boundary and area of this shape are
Btwo[k] = / 2 if k=0
| 8*2^h - 8 if k even >= 2
\ 12*2^h - 8 if k odd
= 2, 4, 8, 16, 24, 40, 56, 88, 120, 184, 248, 376, 504, 760, 1016, 1528, 2040, ...
Atwo[k] = / 0 if k=0
| (7/2)*2^k - 7*2^h + 4 if k even >= 2
\ (7/2)*2^k - 10*2^h + 4 if k odd
= 0, 1, 4, 12, 32, 76, 172, 372, 788, 1636, 3364, 6852, 13892, 28036, 56452, 113412, 227588
The straight and zigzag parts are the two middle sides of the right and convex hull shapes shown above. So the boundary
Btwo[k] = 4*S[h] + 4*Z[h-1] for k even >= 2
= 4*(2^h) + 4*(2*2^(h-1) - 2)
= 8*2^h - 8
Btwo[k] = 4*S[h] + 4*Z[h] for k odd
= 4*(2^h) + 4*(2*2^h - 2)
= 12*2^h - 8
The area can be calculated from the enclosing square S[h]+Z[h-1] from which subtract the four zigzag triangles at the corners.
Atwo[k] = 4*(S[h]+Z[h-1])^2 + 4*Z[h-1]/2*(Z[h-1]/2 + 1)/2
for k even >= 2
Atwo[k] = 4*(S[h]+Z[h])^2 + 4*Z[h]/2 *(Z[h]/2 + 1)/2
for k odd
X positive Xpos[8i+0] = e[8i+0] + a[8i+0] + 1 = 2*d[8i-7] + d[8i-3] + 1 Xpos[8i+1] = d[8i+1] Xpos[8i+2] = c[8i+2] = d[8i+1] Xpos[8i+3] = b[8i+3] = d[8i+1] Xpos[8i+4] = a[8i+4] = d[8i+1] Xpos[8i+5] = h[8i+5] = d[8i+1] Xpos[8i+6] = g[8i+6] = d[8i+1] Xpos[8i+7] = f[8i+7] = d[8i+1] 1,1,1,1,1,1,1,1, 7, 17,17,17,17,17,17,17, 103, 273,273
X axis X[4i+0] = 2*a[4i+0] + e[4i+0] = 2*d[4i-3] + 2*d[4i-7] X[4i+1] = d[4i+1] + h[4i+1] = d[4i+1] + d[4i-3] X[4i+2] = c[4i+2] + g[4i+2] = d[4i+1] + d[4i-3] X[4i+3] = b[4i+3] + f[4i+3] = d[4i+1] + d[4i-3] 2,2,2,2, 4, 6,6,6, 12, 22,22,22, 44, 86,86,86, 172, 342,342,342, 684