# Attrition method approach move map

Charles This is not such a novel idea, maybe; but one of those things that is worth getting out in the open (reifying) at some juncture.

I want simply to propose looking at diagrams like (for example, in relation with the farmer's hat)

. 0 . 1 3 2

where the numbers label the number of plays made internally to capture at particular points.

So this says this much:

One method for White to capture these stones by the attrition method involves

**Pass 1** Play at *c*, *d*, *b*

**Pass 2** Play at *c*, *d*

**Pass 3** Play at *c*.

That much reconstructs from the map. * Attrition* means that each pass puts Black in

*atari*assuming all outside liberties filled. Black then captures.

We know that for success with the method the first of the plays in Pass 1 and Pass 2 must be nakade: namely *c* in the diagam. This isn't true for Pass 3: at this point the snapback stage is reached, and White could equally play at *d*. That gives another valid map:

. 0 . 1 2 3 .

We also have no reason to make Pass 1 involve *b* and *d* rather than (say) *a* or *b*. That gives

. 2 . 1 3 0.

Altogether there are a dozen maps in all that are valid.

I thought of this first in relation to the question of ** where** the approach moves are. The answer seems to be: 2.5 at

*c*and 1.1666... at each of

*a*,

*b*,

*c*.

For a total of 6. The proverb here (Four is Five and Five is Eight and Six is Twelve) counts 5 approach moves, because the assumption is that we begin with a white stone at *c* (or else Black at *c* lives, naturally).

anonymous no, first a and c is 2 moves, then b is sente(black has to capture) and therefore not counted as a move, since after it's still whites turn. then C is one move, and b is sente. then two more moves to finish black off. together 5