Did I just trick you into making the vector version? :)

I would also suggest the use of PixelConstrained...

Given that all lines are straight it should really make no difference.

Yes, I suppose one can:

t1p = SparseArray[{{0, 0, 1}, {0, 0, 0}, {1, 0, 0}}];
t1m = SparseArray[1 - t1p, Automatic, 0];
t2p = Reverse[t1p];
t2m = Reverse[t1m];
t3p = SparseArray[{{0, 0, 0}, {0, 0, 0}, {1, 0, 1}}];
t3m = SparseArray[1 - t3p, Automatic, 0];
t4p = Reverse[t3p];
t4m = Reverse[t3m];

With[{m = 20, n = 53}, 
     ArrayPlot[ArrayFlatten[Table[Switch[Mod[j, 2, 1],
                                         1, Switch[Mod[k, 18, 1],
                                                   x_ /; x < 9, If[Mod[k, 2, 1] == 1,
                                                                   t1m, t1p],
                                                   9, t3m,
                                                   x_ /; x < 18, If[Mod[k, 2, 1] == 1,
                                                                    t2m, t2p],
                                                   18, t4p],
                                         2, Switch[Mod[k, 18, 1],
                                                   x_ /; x < 9, If[Mod[k, 2, 1] == 1,
                                                                   t1p, t1m],
                                                   9, t3p,
                                                   x_ /; x < 18, If[Mod[k, 2, 1] == 1,
                                                                    t2p, t2m],
                                                   18, t4m]], {j, m}, {k, n}]], 
                ColorFunction -> "M10DefaultDensityGradient", 
                Mesh -> {Range[0, 3 m, 3], Range[0, 3 n, 3]}, Frame -> None]]