# Convex Iteration

(Difference between revisions)
 Revision as of 20:26, 5 February 2008 (edit)← Previous diff Revision as of 02:52, 29 June 2008 (edit) (undo)Next diff → Line 20: Line 20: such a vector is presumed existent in the feasible set.
Convex iteration works with a nonnegative variable; such a vector is presumed existent in the feasible set.
Convex iteration works with a nonnegative variable; absolute value $|x|$ is therefore needed. absolute value $|x|$ is therefore needed. - We propose that problem (1) can be equivalently written + We propose that nonconvex problem (1) can be equivalently written - + as a convex iteration: - $\begin{array}{cl}\text{minimize}_{x_{}\in_{_{}}\mathbb{R}^{^n},~y_{}\in_{_{}}\mathbb{R}^{^n}}&\langle_{}|x|_{}\,,\,y^{}\rangle\\ + - \mbox{subject to}&A_{}x=b\\ + - &0\preceq y\preceq\mathbf{1}\\ + - &y^T\mathbf{1}=n-_{}k + - \end{array}$ + - + - which moves the cardinality constraint to the objective. + - To express this nonconvex problem as a convex iteration, we separate it into halves: + for $\varepsilon$ a relatively small positive constant, for $\varepsilon$ a relatively small positive constant, Line 44: Line 36: \end{array}~~~~~~~~~~~~~~~~~~~(3)[/itex] \end{array}~~~~~~~~~~~~~~~~~~~(3)[/itex] - to find direction vector $y$. + to find ''direction vector'' $y$. + The cardinality constraint has been moved to the objective as a linear regularization. The term $\langle t\,,_{_{}}\varepsilon^{}\mathbf{1}\rangle$ in (2) is necessary to determine absolute value The term $\langle t\,,_{_{}}\varepsilon^{}\mathbf{1}\rangle$ in (2) is necessary to determine absolute value $|x|_{\!}=_{}t^{\star_{}}$ because vector $y$ can take zero values in its entries; $|x|_{\!}=_{}t^{\star_{}}$ because vector $y$ can take zero values in its entries;

## Revision as of 02:52, 29 June 2008

Convex iteration is method for constraining rank or cardinality in an otherwise convex optimization problem.
A rank or cardinality constraint is replaced by a weighted linear regularization term added to the objective.
Then two convex problems are iterated until convergence where, ideally, solution to the original problem is found.

## constraining cardinality of signed variable

(Excert from Semidefinite Programming). Consider a feasibility problem equivalent to the classical problem from linear algebra $LaTeX: A_{}x_{\!}=_{\!}b$ ,  but with an upper bound $LaTeX: k$ on cardinality $LaTeX: \|x\|_0$ :  for vector $LaTeX: b\!\in\!\mathcal{R}(A)$

$LaTeX: \begin{array}{rl}\text{find}&_{}x\in_{}\reals^n\\ \mbox{subject to}&A_{}x=b\\ &\|x\|_0\leq_{}k \end{array}~~~~~~~~~~(1)$

where $LaTeX: \|x\|_{0\!}\leq_{_{}\!}k$ means vector $LaTeX: x$ has at most $LaTeX: k$ nonzero entries; such a vector is presumed existent in the feasible set.
Convex iteration works with a nonnegative variable; absolute value $LaTeX: |x|$ is therefore needed. We propose that nonconvex problem (1) can be equivalently written as a convex iteration: for $LaTeX: \varepsilon$ a relatively small positive constant,

$LaTeX: \begin{array}{rl}\text{minimize}_{x_{}\in_{_{}}\mathbb{R}^{^n},~t_{}\in_{_{}}\mathbb{R}^{^n}}&\langle t_{}\,,\,y+\varepsilon^{}\mathbf{1}\rangle\\ \mbox{subject to}&A_{}x=b\\ &-t\preceq x\preceq_{_{}}t \end{array}~~~~~~~~~~~~~~(2)$

is iterated with

$LaTeX: \begin{array}{cl}\text{minimize}_{y_{}\in_{_{}}\mathbb{R}^{^n}}&\langle t^\star,\,y+\varepsilon^{}\mathbf{1}\rangle\\ \mbox{subject to}&0\preceq y\preceq\mathbf{1}\\ &y^T\mathbf{1}=n-_{}k \end{array}~~~~~~~~~~~~~~~~~~~(3)$

to find direction vector $LaTeX: y$. The cardinality constraint has been moved to the objective as a linear regularization. The term $LaTeX: \langle t\,,_{_{}}\varepsilon^{}\mathbf{1}\rangle$ in (2) is necessary to determine absolute value $LaTeX: |x|_{\!}=_{}t^{\star_{}}$ because vector $LaTeX: y$ can take zero values in its entries; a good estimate of $LaTeX: |x|$ is required for (3).

By initializing direction vector $LaTeX: y$ to $LaTeX: (1\!-_{}\!\varepsilon)\mathbf{1}_{}$, the first iteration of problem (2) is a 1-norm problem; i.e.,

$LaTeX: \begin{array}{ccc} \begin{array}{rl}\text{minimize}_{x_{}\in_{_{}}\mathbb{R}^{^n},~t_{}\in_{_{}}\mathbb{R}^{^n}}&\langle t\,,_{}\mathbf{1}\rangle\\ \mbox{subject to}&A_{}x=b\\ &-t\preceq x\preceq_{_{}}t \end{array} &\equiv&~ \begin{array}{cl}\text{minimize}_{x_{}\in_{_{}}\mathbb{R}^{^n}}&\|x\|_1\\ \mbox{subject to}&A_{}x=b \end{array} \end{array}~~~~~~~~~~(4)$

Subsequent iterations of problem (2) engaging cardinality term $LaTeX: \langle t_{}\,,\,y\rangle$ can be interpreted as corrections to this 1-norm problem leading to a 0-norm solution.
Iteration (2) (3) always converges to a locally optimal solution by virtue of a monotonically nonincreasing objective sequence.
There can be no proof of global optimality, defined by an optimal objective equal to 0. Local solutions are therefore detected by nonzero optimal objective.
Heuristics for reinitializing direction vector $LaTeX: y$ can lead to a globally optimal solution.