solveStepBruteForce

Method of continuousTimeMdss.

Solve the algebraic mode problem of a hybrid mdss model at a fixed state and input.

Source: src/simulator/@continuousTimeMdss/solveStepBruteForce.m

What is solved

At fixed state $x_0$ and input $u$ , find the state derivatives $\dot x$ , algebraic signals $y$ and binaries $z$ that satisfy

$F = \Phi_{\mathrm{eq}}\, m(\dot x, x_0, u, y, z) = 0, \qquad G = \Phi_{\mathrm{ineq}}\, m(\dot x, x_0, u, y, z) \le 0 ,$

where $\Phi = \Phi_c + \Phi_{+1} - \Phi_{-1}$ is the parameter-matrix split.

The monomial vector $m$

$m\in\mathbb R^{R}$ has one entry per monomial (one per column of the structure matrix). Each entry is a product of one factor per signal:

$m_k \;=\; \prod_{n} \varphi\!\big(t_{nk}, f_{nk}, c_{nk};\, v_n\big),$

so the residual $F=\Phi_{\mathrm{eq}}m$ is just a weighted sum of those monomials. The solver never forms $m$ from scratch — it starts at $m=\mathbf 1$ and, each time one signal’s value $v_n$ becomes known, multiplies that signal’s factor into every column at once (the $\odot$ is elementwise over the $R$ columns):

$m \;\leftarrow\; m \;\odot\; \underbrace{\big(1 + t_n f_n + t_n(v-1) - f_n v\big)}_{\text{true/false part}}\;\odot\;\underbrace{\big(1 - |c_n| + c_n v\big)}_{\text{continuous part}} .$

What one factor evaluates to

For a single column, the entries $t_n, f_n, c_n$ say how signal $n$ enters that monomial, and the factor collapses to:

structure entry in that column	meaning	factor value
$t_n = 1$ (true)	signal appears as $v$	$v$
$f_n = 1$ (false)	signal appears as its complement	$1 - v$
$t_n = f_n = c_n = 0$	signal absent from this monomial	$1$ (column unchanged)
$c_n \neq 0$ (continuous)	scaled/normalized entry	$1 - \lvert c_n\rvert + c_n v$

So substituting a value only touches the columns where that signal is present; absent columns are multiplied by $1$ and stay as they are.

Example. Take two states $x_1, x_2$ and a monomial $x_1 x_2$ (both appear as true). Its column starts at $1$ . Folding in $x_1 = 0.5$ multiplies it by $0.5$ ; folding in $x_2 = 4$ multiplies it by $4$ ; the column now equals $0.5\cdot 4 = 2 = x_1 x_2$ . A different column that does not contain $x_1$ is multiplied by $1$ when $x_1$ is substituted and is left untouched. Once every signal has been folded in, each $m_k$ holds the numeric value of its monomial.

Algorithm

The known $x_0$ and $u$ are folded into $m$ first. The routine then walks the solving order. Each row is one of three cases.

Explicit continuous unknown ( $\dot x_n$ or $y_n$ ): its equation $e$ is affine in that single unknown, $F_e = A\,v + B$ , so

$v = -\frac{B}{A},\quad A=\Phi_e\!\left[(t_n-f_n+c_n)\odot m\right],\quad B=\Phi_e\!\left[\big(1-|c_n+t_nf_n-t_n|\big)\odot m\right],$

and $v$ is folded into $m$ .

Explicit binary $z_n$ : evaluate its equation value at the previous value $z_n^0$ and flip on a positive value,

$F_0 = \Phi_e\, m\big|_{z_n=z_n^0},\qquad z_n = \begin{cases} z_n^0, & \operatorname{sign}(F_0)\le 0,\\[2pt] \lnot\, z_n^0, & \text{otherwise.}\end{cases}$

Coupled subproblem: delegated to solveSubProblemVariableZ, which enumerates the block’s binaries while solving its continuous unknowns.

After the whole order is processed, $F=\Phi_{\mathrm{eq}}m$ and $G=\Phi_{\mathrm{ineq}}m$ are returned.

Inputs / outputs

[xp, y, z, F, G] = solveStepBruteForce(sys, x0, u, solvingOrder, yguess, xpguess, zguess) — returns the state-derivative row xp ( $\dot x$ ), algebraic row y, binary row z, and the equality/inequality function-value rows F, G.