Spencer's bound from the partial colouring step and the residual colouring

Assume both of the following.

1. (Partial colouring step.) For every $0/1$ matrix, every nonempty active set $T$ of columns and all parameters $0<\theta\le 1/2$, $\nu\ge2$ obeying $n\cdot 2e^{-\nu^{2}/2}(\tfrac34\nu^{2}+2)\le\tfrac{2}{3}\theta^{2}|T|$, there is a partial colouring of $T$ leaving at most a $\theta$ fraction of $T$ uncoloured whose row sums are at most $\nu\sqrt{|T|}$ in absolute value.
2. (Residual colouring.) Every set $T$ of columns carries a $\pm1$ colouring whose row sums are at most $\sqrt{2|T|\log(4n)}$ in absolute value.

Then every $n\times n$ matrix with entries in $\{0,1\}$ admits signs $\varepsilon_j\in\{\pm1\}$ with

which is Spencer's theorem. What has to be supplied is the scheduling: the fraction colored at each application, the parameter $\nu$ used there, the point at which the recursion is stopped in favour of the residual colouring, and the verification that the resulting geometric series of costs stays below $6$. The constant $6$ is the one from the original paper, and it is what makes the schedule delicate: a random colouring alone gives $\Theta(\sqrt{n\log n})$, and the recursion has to pay for every round.

Formalization Note The two inputs appear as explicit hypotheses so that this statement is exactly the combinatorial bookkeeping of the argument and nothing else; they are the separately stated partial colouring and residual colouring problems. Colourings are encoded as real-valued vectors, with $0$ marking an uncoloured coordinate in the partial colourings.