User:Dummy index/Bimetallic MOS

See Metallic MOS. The article is uncomfortable with the definition of the split operation, so I'll write it in my own way.

Golden case

You know a process cutting the square from golden rectangle. Imagine [math]\displaystyle{ L = φ }[/math] and [math]\displaystyle{ s = 1 }[/math] and divide L into

[math]\displaystyle{ \qquad L_1:s_1 = φ-1:1 = [0; 1, 1, 1, 1, ...] }[/math].

No, [math]\displaystyle{ L_1 < s_1 }[/math]. Let new [math]\displaystyle{ s := L_1 }[/math] and new [math]\displaystyle{ L = s_1 }[/math]. (Rotate the rectangle 90°)

OK, now [math]\displaystyle{ L = 1 }[/math] and [math]\displaystyle{ s = φ-1 }[/math] and [math]\displaystyle{ L:s = φ }[/math].

Loop.

In thinking MOS pattern, direction of division is altered when every rotating.

L
sL
LLs (2L 1s)
sLsLL (3L 2s)
LLsLLsLs (5L 3s)
sLsLLsLsLLsLL (8L 5s)

Silver case

First, Imagine [math]\displaystyle{ L = δ_s }[/math] and [math]\displaystyle{ s = 1 }[/math] and divide L into

[math]\displaystyle{ \qquad L_1:s_1 = δ_s-1:1 = [1; 2, 2, 2, 2, ...] }[/math].

OK, now [math]\displaystyle{ L_1 = δ_s-1 }[/math] and [math]\displaystyle{ s_1 = s = 1 }[/math]. Next, divide L₁ into

[math]\displaystyle{ \qquad L_2:s_2 = δ_s-2:1 = [0; 2, 2, 2, 2, ...] }[/math].

No, [math]\displaystyle{ L_2 < s_2 }[/math]. Let new [math]\displaystyle{ s := L_2 }[/math] and new [math]\displaystyle{ L = s_2 }[/math]. (Rotate the rectangle 90°)

OK, now [math]\displaystyle{ L = 1 }[/math] and [math]\displaystyle{ s = δ_s-2 }[/math] and [math]\displaystyle{ L:s = δ_s }[/math].

Loop.

In this case, we actually can choose from two entry points:

• Let L = period and divide L, or
• Let L₁ = period and divide L₁.

L (first entry point)
L s (second entry point)
sL L
ssL sL (from 1st: 2L 3s, from 2nd: 1L 2s)
LLLs LLs (from 1st: 5L 2s, from 2nd: 3L 1s)
LsLsLss LsLss (from 1st: 5L 7s, from 2nd: 3L 4s)