Mathematics of Scoring


With some simple algebra, one can reason about how different scoring systems affect the results. This treatment looks at the mathematics of the scoring phase. It is understood that different rule sets will also affect the outcome.

See Territory and Area Scoring for a non-mathematical treatment of much of the same material and further discussion of the practical implications. See Logical proof of the equivalence of territory and area scoring for a similar treatment of a more limited case.


At the end of the game, we have these quantities:

   Ws, Bs :: stones on the board for White and Black
   Wt, Bt :: territory on board for White and Black
     Nt   :: non-scoring territory
     S    :: number of points on a board edge
   Wp, Bp :: White and Black prisoners
   Wz, Bz :: number of passes by White and Black
     H    :: number of handicap stones played by Black save first
     K    :: komi
     P    :: 1 if Black passed last, 0 if White passed last

All of these values are positive or zero. Different rule sets may assign different values to these quantities even for the same final board position. For example, points in seki may be considered Nt or Wt, Bt depending on the rule set.

All of the board is accounted for:

   (Wt + Ws) + (Bt + Bs) + Nt = S*S

The parity of the board depends on the size, and the number of non-scoring points:

   (Wt + Ws) + (Bt + Bs) = S*S - Nt

The difference between two terms on the left above has the same parity:

   if (S*S - Nt) is odd:  then (Wt + Ws) - (Bt + Bs) is odd
           "     is even:                "           is even

This is because if a+b is odd, then a-b is odd, and similarly if a+b is even, then a-b is even.

In area scoring (see below), Nt is 0 (except when there are odd points in seki), and so possible scores can only be odd, and thus differ by multiples of two.

Number of Moves

The number of moves made by each player is equal to the number of stones on the board (adjusted for handicap), the number of prisoners, and the number of passes.

   Wm, Bm :: moves made by White and Black
   Wm = Ws + Wp + Wz
   Bm = Bs + Bp + Bz - H

Since the game has strict alternation, Black starting, if the games ends with White's pass, players have made the same number of moves. If the game ends on Black's pass, Black has made one more move than White:

   Bm - P = Wm
   Bm - P - Wm = 0

The Score

The mathematical score is defined as a single value:

   Positive values mean a win for Black.
   Negative values mean a win for White.
   Zero is a jigo.

Territory (Japanese) scoring, scores territory less prisoners for each player:

   Sj = (Bt - Bp) - (Wt - Wp) - K

Area (Chinese) scoring, scores territory plus stones for each player:

   Sc = (Bt + Bs) - (Wt + Ws) - K

Note these definitions are the "common" ones, and don't correspond to any particular rule sets. They correspond to how people generally count the score. It will be shown that AGA rules define a score that differs from both, but is easily computed via either counting method.

For any given game, these two versions of the score will differ by the sum of three values related to the handicap, 'excess' passes, and who played last:

   Sc - Sj = ((Bt + Bs) - (Wt + Ws) - K) - ((Bt - Bp) - (Wt - Wp) - K)
           = (Bt + Bs) - (Wt + Ws) - K - (Bt - Bp) + (Wt - Wp) + K
           = (Bt + Bs) - (Bt - Bp) - (Wt + Ws) + (Wt - Wp)
           = (Bs + Bp) - (Ws + Wp)
           = (Bs + Bp + Bz - H - Bz + H) - (Ws + Wp + Wz - Wz)
           = (Bm - Bz + H) - (Wm - Wz)
           = (Bm - Wm) - Bz + H + Wz
           = (Bm - P - Wm + P) - Bz + H + Wz
           = (Bm - P - Wm) + P + H + (Wz - Bz)
           = P + H + (Wz - Bz)

How the Scores Might Differ

Looking at how these two scores for a game differ, it is important to keep a neutral point of view. When the two scores differ, we just compute by how much they differ, and then show which one favors White vs. which one favors Black.

Even Games

In an even game, where the only passes were the final two, ending with White, Then all three of these quantities is zero and the two scores are the same:

       H = 0
       Wz = Bz = 1
       P = 0
       Sc = Sj
   W <------+-------> B

If the passes ended with Black, the scores differ by one point, with territory scoring favoring White, and area scoring favoring Black:

       H = 0
       Wz = Bz = 1
       P = 1
       Sc = Sj + 1
   W <------+---+-------> B
           Sj   Sc

Handicap Games

In an n+1 stone handicap game where the only passes end the game, the scores differ by n, with territory scoring favoring White, and area scoring favoring Black:

       H = n
       Wz = Bz = 1
       P = 0
       Sc = Sj + n
   W <------+---+ ... +---+-------> B
           Sj             Sc

This implies to me that a difference in one handicap will be bigger in communities that use area scoring, vs. those that use territory scoring. Though note, if the area scorers add compensation (as the AGA rules do), then the rank difference will be the same as territory scorers. I have never seen this pointed out anywhere and wonder if it is known.

Extra Passes

In an even game White passing last, the scores differ by the number of passes one side has in excess of the other. Territory scoring favors the excessive passer, and area scoring the other player.

   if                  if
       H = 0               H = 0
       Wz = z              Wz = z + wx
       Bz = z + bx         Bz = z
       P = 0               P = 0
   then                then
       Sc = Sj - bx        Sc = Sj + wx
   W <------+---+ ... +---+-------> B
           Sc             Sj
   W <------+---+ ... +---+-------> B
           Sj             Sc

AGA Scoring

The AGA defines the score in such a way that it comes out the same whether one counts via territory or area. The rules stipulate that White must pass last, that pass stones are used (increases the prisoners by passes), and that, if counting via area, handicap compensation is used:

   P = 0
   Saj = (Bt - Bp - Bz) - (Wt - Wp - Wz) - K
       = (Bt - Bp) - Bz - (Wt - Wp) + Wz - K
       = (Bt - Bp) - (Wt - Wp) - K + (Wz - Bz)
       = Sj + (Wz - Bz)
   Sac = (Bt + Bs) - (Wt + Ws) - K - H
         Sc - H

These can be shown to be equal:

       Sac = Saj
    Sc - H = Sj + (Wz - Bz)
        Sc = H + Sj + (Wz - Bz)
   Sc - Sj = H + (Wz - Bz)
   Sc - Sj = P + H + (Wz - Bz)
       -- which is true from above

Mathematics of Scoring last edited by on August 15, 2013 - 08:33
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