Table 1.

The iterative algorithm of group sparse CCA.

1. Initialize u0 and v0 by traditional CCA decomposition, ∥u0∥2 = 1, ∥v0∥2 = 1.

2. Solve Uj, Vj using the following iterations until it convergence:

(a) $\mathrm{Fix}\mathit{v}={\mathit{v}}^{\mathit{j}-1},{\mathit{u}}^{\mathit{j}}\leftarrow \underset{\mathit{u},\mathit{d}}{\arg \min }{\Vert \mathit{K}-{\mathit{duv}}^{\mathit{t}}\Vert }_{\mathit{F}}^2+\Psi \left(\mathit{u}\right)s.t.{\Vert \mathit{u}\Vert }_2^2\le 1$

(b) $\mathrm{Fix}\mathit{u}={\mathit{u}}^{\mathit{j}},{\mathit{v}}^{\mathit{j}}\leftarrow \underset{\mathit{v},\mathit{d}}{\arg \min }{\Vert \mathit{K}-{\mathit{duv}}^{\mathit{t}}\Vert }_{\mathit{F}}^2+\Phi \left(\mathit{v}\right)s.t.{\Vert \mathit{v}\Vert }_2^2\le 1$

(c) dj ← tr (Kvj (uj)t) or tr (Kuj (vj)t)

3. Update the remaining matrix K ← K – tr(Kvut)uvt; go to Step (1) to obtain the next pair of loading vectors (u, v).