User:Dummy index/Bimetallic MOS

See Metallic MOS. The article is unelegant with the definition of the split operation, so l write it in my own way.

Golden case

You know a process cutting off a square from a 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 = [1; 1, 1, 1, 1, ...] }[/math].

Loop.

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

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 = [2; 2, 2, 2, 2, ...] }[/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 (left: 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)

Bimetallic case

First, Imagine [math]\displaystyle{ L = \sqrt{3}+1 }[/math] and [math]\displaystyle{ s = 1 }[/math] and divide L into

[math]\displaystyle{ \qquad L_1:s_1 = \sqrt{3}:1 = [1; 1, 2, 1, 2, ...] }[/math].

OK, now [math]\displaystyle{ L_1 = \sqrt{3} }[/math] and [math]\displaystyle{ s_1 = s = 1 }[/math]. Next, divide L₁ into

[math]\displaystyle{ \qquad L_2:s_2 = \sqrt{3}-1:1 = [0; 1, 2, 1, 2, ...] }[/math].

No, [math]\displaystyle{ L_2 < s_2 }[/math]. Let [math]\displaystyle{ s_3 := L_2 }[/math] and [math]\displaystyle{ L_3 := s_2 }[/math]. (Rotate the rectangle 90°)

OK, now [math]\displaystyle{ L_3 = 1 }[/math] and [math]\displaystyle{ s_3 = \sqrt{3}-1 }[/math] and [math]\displaystyle{ L_3:s_3 = (\sqrt{3}+1) / 2 = [1; 2, 1, 2, 1, ...] }[/math]. Next, divide L₃ into

[math]\displaystyle{ \qquad L_4:s_4 = (\sqrt{3}-1)/2:1 = [0; 2, 1, 2, 1, ...] }[/math].

No, [math]\displaystyle{ L_4 < s_4 }[/math]. Let new [math]\displaystyle{ s := L_4 }[/math] and new [math]\displaystyle{ L := s_4 }[/math]. (Rotate the rectangle 90°)

OK, now [math]\displaystyle{ L = \sqrt{3}-1 }[/math] and [math]\displaystyle{ s = 2-\sqrt{3} }[/math] and [math]\displaystyle{ L_3:s_3 = \sqrt{3}+1 = [2; 1, 2, 1, 2, ...] }[/math].

Loop.

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

• Let L = period and divide L, or
• Let L₁ = period and divide L₁, or
• Let L₃ = period and divide L₃.

L (first entry point)
L s (left: second entry point)
sL L (right: third entry point)
LLs Ls (from 1st: 3L 2s, from 2nd: 2L 1s)
LsLss Lss (from 1st: 3L 5s, from 2nd: 2L 3s, from 3rd: 1L 2s)
sLLsLLL sLLL (from 1st: 8L 3s, from 2nd: 5L 2s, from 3rd: 3L 1s)
LLsLsLLsLsLs LLsLsLs (from 2nd: 7L 5s, from 3rd: 4L 3s)