# Cycle Law

Difficulty: Advanced   Keywords: Ko, Rules, Theory

rec.games.go, post

1999-02-28

### Proposition:

In a situational cycle the difference (D) of passes equals the negative difference of removed stones.

### Conventions:

Let M_b and M_w be the numbers of black and white moves during the cycle.

Let B_b and B_w be the numbers of black and white plays during the cycle.

Let P_b and P_w be the numbers of black and white passes during the cycle.

Let S_b and S_w be the numbers of black and white stones on the board before the cycle.

Let T_b and T_w be the numbers of black and white stones on the board after the cycle.

Let R_b and R_w be the numbers of black and white stones removed during the cycle.

Let  D := P_b - P_w.

### Proof:

 bb((A)) The cycle is situational => M_b = M_w .
 bb((B)) It is a cycle => S_c = T_c" for "c in { b, w }.
 bb((C)) By definition of move, play, pass => M_c = B_c + P_c" for "c in { b, w }.
 bb((D)) From definition of D,  bb((A)),bb((C)) => B_b + P_b = B_w + P_w <=> P_b - P_w = B_w - B_b <=> D = B_w - B_b .
 bb((E)) By definition of move and of removals => S_c + B_c = T_c + R_c" for "c in { b, w }.
 bb((F)) From bb((B)),bb((E)) => B_c = R_c" for "c in { b, w } .
 bb((G)) From bb((D)),bb((F)) => D = R_w - R_b .
 cc(QED) .

### Corollary 1:

 P_b = P_w <=> R_b = R_w .
Note:
In particular,  P_b = P_w = 0 => R_b = R_w .

### Corollary 2:

If the cycle is positional but not situational, then D is adjusted by +-1.

### Corollary 3:

For an equivalence proof of area and territory scoring the following are requirements:
a) M_b = M_w.
b) The difference of pass stones equals D.

Cycle Law last edited by PJTraill on January 23, 2019 - 15:20