In the game of domineering, exactly as defined in Winning Ways, is it possible to build any number of the form $\frac{a}{2^b}, a\in\mathbb{Z}, b\in\mathbb{N}$? Is there an pragmatic/algorithmic way of instancing such numbers?
I was able to find 0, $\pm\frac{1}{2}$,$\pm\frac{1}{4}$,$\pm\frac{3}{4}$, in several different ways. When trying to extend previos boards by following a pattern I came up with I always end up in one of the previous calculated cases again. When I mix in squares in arbitrary positions, I end up getting non-numbers. In case it is not possible to build all such numbers, what dyadics can one build?
I am using CGSuite to calculate the examples I build. In case it helps: When trying to get to $\frac{7}{8}$, for example, I tried the following cases:
##.|##.|...|.## = 3/4
.##|.##|...|##.|...|.## = 3/4
.##|.##|...|##.|##.|...|.## = 3/4
####.|####.|##...|##.##|...##|.#### = 3/4
All of them.
In the 2015 Theoretical Computer Science paper "New results for Domineering from combinatorial game theory endgame databases" (arXiv link) by Uiterwijk and Barton, a construction is provided to produce any dyadic rational in a (connected!) board of Domineering.
They cite the 1996 paper New values in Domineering by Yonghoan Kim, which showed a construction for connected boards of value $2^{-n}$.