Conversion from booleans to signs $\{-1, +1\}$: true ↦ 1, false ↦ -1.
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Total Rademacher absolute expectation: $\sum_{y \in \{-1,+1\}^n} \left| \sum_j y_j \right|$, i.e., the (unnormalized) expected absolute value of a $\pm 1$ random walk of length $n$.
Instances For
The image of boolToSign lies in $\{-1, +1\}$.
Mapping a boolean vector pointwise through boolToSign yields a sign vector.
Averaging principle: for any function $f : \iota \to \mathbb{Z}$ on a finite nonempty type there exists $x \in \iota$ whose value, multiplied by $|\iota|$, exceeds the total sum $\sum_y f(y)$.
Greedy row selection: given a matrix $a$ and a vector $y$, choose for each row $i$ a sign $x_i \in \{-1, +1\}$ to align with the row-sum, so that $\sum_{i,j} a_{ij} x_i y_j = \sum_i \left|\sum_j a_{ij} y_j\right|$.
The sign flip operation is its own inverse.
Compatibility: multiplying a sign $a \in \{-1, +1\}$ by boolToSign b is the same
as boolToSign applied to the XOR-flipped bit.
Sign-symmetry of the Rademacher-row sum: for any sign row $a$, $\sum_y |\sum_j a_j y_j|$ over $y \in \{-1,+1\}^n$ equals $\sum_y |\sum_j y_j|$, since the sign flip is a bijection on $\{-1,+1\}^n$.
Total sum identity used in the unbalancing-lights proof: for a sign matrix $a$, $\sum_y \sum_i |\sum_j a_{ij} y_j| = n \cdot \mathbb{E}_{\mathrm{Rad}}(n)$, where the right-hand side is the (unnormalized) Rademacher walk expectation.
Unbalancing lights (Theorem 2.5.1, integer form). For any sign matrix $a \in \{-1, +1\}^{n \times n}$, there exist sign vectors $x, y \in \{-1, +1\}^n$ such that $2^n \cdot \sum_{i,j} a_{ij} x_i y_j \geq n \cdot \mathbb{E}_{\mathrm{Rad}}(n)$. This gives the asymptotic bound $\sum_{i,j} a_{ij} x_i y_j \geq (\sqrt{2/\pi} + o(1)) n^{3/2}$.
Improved unbalancing lights (Theorem 2.5.2). For each $k \geq 2$ there is a constant $c > 0$ such that for every $n \geq 1$ and every $\pm 1$ coloring of the $k$-element subsets of $\mathrm{Fin}\,k \times \mathrm{Fin}\,n$ that assigns $+1$ to all cross edges, some sub-hypergraph $S$ has discrepancy at least $c \cdot n^k$ in absolute value.