Cycle Law

RobertJasiek

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``.