4.3 Area Rules III and Territory Rules I

4.3 Area Rules III and Territory Rules I

Prisoners do not count in area rules III, so the prisoners given for passes in territory rules I to ensure an equal number of moves would be meaningless. Under area rules III it does not matter whether the players make different numbers of moves in the final stage, as area rules III allow, or the same number of moves, as territory rules I require. The players can play area rules III up to the first pass, then apply territory rules I for the rest of the game by giving prisoners for passes, and the result will be in complete agreement. Therefore, you can transform territory rules I into area rules III simply by replacing the rule of scoring.
For proof that the results agree, look at Table 1.

Table 1

On an N x N go board:

Black

White

At the preliminary end

Number of moves

Number of stones on board

Number of prisoners

M1 - L1

M2 - L2

After the preliminary end

Number of stones played

Number of stones passed

Final number of stones on board

Number of prisoners captured

L1 + T1 - Q1

L2 + T2 - Q2

after the preliminary end

Territory

Total number of prisoners

M1 + T1 - Q1 + P1

M2 + T2 - Q2 + P2

By the rule of equal numbers of moves: T1 + P1 = T2 + P2

Territory rules I

Black's score = S1 - M1 - T1 + Q1 - P1
White's score = S2 - M2 - T2 + Q2 - P2
Difference D = (S1 - S2) + (Q1 - Q2) - (M1 - M2)

Area rules III

Black's score = Q1 + S1 - (M1 - M2)/2
White's score = Q2 + S2 + (M1 - M2)/2
Difference D = (S1 - S2) + (Q1 - Q2) - (M1 - M2)

On an N x N board, suppose that prior to the preliminary end of the game (the first two consecutive passes in territory rules I, equivalent to the first pass in area rules III) Black has played M1 moves and White has played M2 moves. We will continue to use 1 to designate Black and 2 to designate White. Let Li (i = 1, 2) be the number of stones the board. Then Black has lost M1 - L1 prisoners and White has lost M2 - L2. Let Ti be the number of stones played after the preliminary end, and Pi be the number passed as prisoners. Let Qi be the number of stones on the board at the end of the game, and let Si be the amount of territory surrounded.
The number of prisoners captured after the preliminary end is Li + Ti - Qi. The total number of prisoners is therefore Mi + Ti - Qi + Pi.

Under area rules III, the scores are:

Black Q1 + S1 - (M1 - M2)/2
White Q2 + S2 + (M1 - M2)/2

The final term (M1 - M2)/2 is the half point added and subtracted when Black makes the last competitive move, which occurs when M1 - M2 = 1. The difference D(area III) between Black's score and White's is:

D(area III) = black - white = (S1 - S2) + (Q1 - Q2) - (M1 - M2)

Under territory rules I, the scores (territory minus prisoners) are:

Black S1 - M1 - T1 + Q1 - P1
White S2 - M2 - T2 + Q2 - P2
D(ter. I) = (S1 - S2) + (Q1 - Q2) - (M1 - M2) - (T1 + P1 - T2 - P2)

By the rule of equal number of moves after the preliminary end of the game,

T1 + P1 = T2 + P2
T1 + P1 - T2 - P2 = 0
D(ter. I) = (S1 - S2) + (Q1 - Q2) - (M1 - M2)
D(ter. I) = D(area III)

So we have proved that area rules III and territory rules I are in complete agreement, even though one of them counts stones plus territory while the other counts territory minus prisoners. This comes from the addition of the last competitive move condition to area rules III.
Another consequence of this proof is that since area rules II do not have the half-point adjustment, when Black makes the last competitive move, area rules II are one point different from area rules III and territory rules I. Note that this does not depend on the size N of the N x N go board. There is a misconception that the one-point difference arises only when N is an odd number, but we have seen that this is not true.
Table 2 gives the numerical values for the Go-Miyamoto game.

Table 2

N = 9 Go Seigen vs. Miyamoto Naoki

	Prisoners	Territory
M1 = 40, L1 = 31, T1 = 2, P1 = 1, Q1 = 33, M1 +	T1 - Q1 + P1 = 10,	S1 = 10
M2 = 39, L2 = 33, T2 = 2, P2 = 1, Q2 = 27, M2 +	T2 - Q2 + P2 = 15,	S2 = 11

D(Terri. I) = (10 - 10) - (11 - 15) = 4
D(Area III) = (33 + 10) - (27 + 11) - 1 = 4