# Quartered Go

I had an idea a while ago for a very simple variant.

1. An even board is used, like 12x12, 14x14, 16x16, 18x18 or 20x20.

2. Moves may only be made in the quarter(s) of the board with the least stones.

Consider this position and sequence, for instance.

At the start of the position, there’s one stone in each corner, so any quarter can be played in.

(1) adds a stone to the lower left. It can no longer be played in until the three other quarters also have two stones, so White can’t respond to the approach.

(2) adds a stone to the top right. Black can’t continue the joseki yet.

(3) encloses the lower right.

(4) can only be played in the top left, since all other quarters have two stones.

(5) can again be played anywhere, as there are now two stones in each quarter. Black continues the joseki in the top right.

(6) pincers in the lower left. Black can’t yet do anything about this and must choose between enhancing his shimari in the lower right into a three-stone formation and shoulderhitting White’s in the top left. White will then have to play in the quarter that remains.

I think that Black’s advantage is larger than in standard Go, since he has the 4-quarter and 2-quarter choices in each cycle whereas White has the 3-quarter and 1-quarter ones.

So, I have a second suggestion that just came to me: Thue-Morse Quartered Go!

8 Likes

This sounds like a cool idea! My first thought is, as long as no stones get captured, black can freely place a stone every fourth move. This might give black some advantage in the opening, but as soon as stones get captured, the roles may reverse. Really interesting

Let’s try it sometime!

4 Likes

Is Thue-Morse quartered Go fairer than regular quartered Go?

Consider a metric I call “quarterl fairness”. This is the sum of the number of quarters in which Black can play at each move in the game divided by the number of quarters in which White can play one of his moves.

In the regular game, Black’s pattern (without captures) is 4, 2, 4, 2… which sums to 3.
White’s is 3, 1, 3, 1… which sums to 2.

Thus the quarterly fairness in an infinite sum is 2 / 3 = 0.666…, and the longer the captureless game the more QF will converge on that value.

So, what about Thue-Morse?

According to the On-Line Encyclopedia of Integer Sequences, this is the pattern for the first 64 moves, if we consider 0 as Black and 1 as White.

Move No. Pattern
1–8 BWWB WBBW
9–16 WBBW BWWB
17–24 WBBW WWBB
25–32 WWBW BBWW
33–40 WBBW BWWB
41–48 BWWB WBBW
49–56 BWWB WBBW
57–64 WBBW BBWB

We can analyse the quarter choice delta for each four-move block (BΔ, WΔ).

BWWB / WBBW = 5, 5
BBWW = 7, 3
WWBB = 3, 7
BBWB = 8, 2
WWBW - 2, 8

Move No. Pattern
1–8 5, 5 – 5, 5
9–16 5, 5 – 5, 5
17–24 5, 5 – 3, 7
25–32 2, 8 – 7, 3
33–40 5, 5 – 5, 5
41–48 5, 5 – 5, 5
49–56 5, 5 – 5, 5
57–64 5, 5 – 8, 2

QF = (8 + 7 + (12 x 15) + 3 + 2) / (8 + 7 + (12 x 5) + 3 + 2)
which is 185 / 185, which is of 1.

So, there is essentially no quarterly unfairness in Thue-Morse quartered Go.

[In] a book [by[ Mizokami Tomochika … he argued that by splitting the board in half and counting the stones in each half you could (indeed should) choose your move on the basis of which side had a preponderance of stones in that area.

Sounds like an interesting challenge, if he take the meaning as a suggestion to always play on the half with least stones.