Melanie Siebenhofer, Jänner 2018

Asymptotische Grenzverteilung¶

alt text

Baum 1¶

\begin{align*} s_0 + s_1 + s_2 + s_3 &= 1\\ 0 \leq s_0 &\leq 1\\ 0 \leq s_1 &\leq 1\\ 0 \leq s_2 &\leq 1\\ 0 \leq s_3 &\leq 1\\ \ \\ 2\cdot s_0 + 2 \cdot s_1 + 2\cdot s_2 + 2 \cdot s_3 &\leq c\\ s_2 + s_3 + 1 &\geq s_0\\ s_1 + s_2 + 1 &\geq s_0\\ s_1 + s_2 + 1 &\geq s_3\\ s_0 + s_1 + 1 &\geq s_3\\ \end{align*}

asymptotisch: \begin{align*} s_0 + s_1 + s_2 + s_3 & = 1\\ 0 \leq s_0 & \leq 1\\ 0 \leq s_1 & \leq 1\\ 0 \leq s_2 & \leq 1\\ 0 \leq s_3 & \leq 1\\ \ \\ 2\cdot s_0 + 2 \cdot s_1 + 2\cdot s_2 + 2 \cdot s_3 \leq & c\\ s_0 \leq & s_2 + s_3 \\ s_0 \leq & s_1 + s_2 \\ s_3 \leq & s_1 + s_2 \\ s_3 \leq & s_0 + s_1 \\ \end{align*}

from sage.symbolic.relation import solve_ineq

forget()
var('s0,s1,s2,s3,c')

(s0, s1, s2, s3, c)

assume(0 <= s0)
assume(s0 <= 1)
assume(0 <= s1)
assume(s1 <= 1)
assume(0 <= s2)
assume(s2 <= 1)
assume(0 <= s3)
assume(s3 <= 1)

setze $s_0 = 1 - s_1 - s_2 - s_3$

s0 = 1 - s1 - s2 - s3

forme Ungleichungen, die $s_0$ enthalten, um

2*s0 + 2*s1 + 2*s2 + 2*s3 <= c

2 <= c

solve_ineq([s0 <= s2 + s3],[s3,s2,s1])

[[s3 == -1/2*s1 - s2 + 1/2], [-1/2*s1 - s2 + 1/2 < s3]]

solve_ineq([s0 <= s1 + s2],[s3,s2,s1])

[[s3 == -2*s1 - 2*s2 + 1], [-2*s1 - 2*s2 + 1 < s3]]

solve_ineq([s3 <= s0 + s1],[s3,s2,s1])

[[s3 == -1/2*s2 + 1/2], [s3 < -1/2*s2 + 1/2]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq 1 - s_1 - s_2 - s_3 \leq & 1\\ 0 \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & 1\\ 0 \leq s_3 \leq & 1\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \quad &\\ 1 - 2s_1 - 2 s_2 \leq s_3 \quad &\\ s_3 \leq & s_1 + s_2\\ s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Forme 3. Zeile um:

solve_ineq([0 <= 1 - s1 - s2 - s3],[s3,s2,s1])

[[s3 == -s1 - s2 + 1], [s3 < -s1 - s2 + 1]]

bool(1 - s1 - s2 - s3 <= 1)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & 1\\ 0 \leq s_3 \leq & 1\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \quad &\\ 1 - 2s_1 - 2 s_2 \leq s_3 \quad &\\ s_3 \leq & s_1 + s_2\\ s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

prüfe Grenzen für $s_3$

bool(0 <= s1 + s2)

True

bool(0 <= 1/2 - s2/2)

True

solve_ineq([0 <= 1 - s1 - s2],[s2,s1])

[[s2 == -s1 + 1], [s2 < -s1 + 1]]

solve_ineq([1/2 - s1/2 - s2 <= s1 + s2],[s2,s1])

[[s2 == -3/4*s1 + 1/4], [-3/4*s1 + 1/4 < s2]]

bool(1/2 - s1/2 - s2 <= 1/2 - s2/2)

True

bool(1/2 - s1/2 - s2 <= 1 - s1 - s2)

True

solve_ineq([1 - 2*s1 - 2*s2 <= s1 + s2],[s2,s1])

[[s2 == -s1 + 1/3], [-s1 + 1/3 < s2]]

solve_ineq([1 - 2*s1 - 2*s2 <= 1/2 - s2/2],[s2,s1])

[[s2 == -4/3*s1 + 1/3], [-4/3*s1 + 1/3 < s2]]

bool(1 - 2*s1 - 2*s2 <= 1 - s1 - s2)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{3} - \frac{4}{3}s_1 \leq s_2 \quad &&\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \quad &\\ 1 - 2s_1 - 2 s_2 \leq s_3 \quad &\\ s_3 \leq & s_1 + s_2\\ s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

prüfe Grenzen für $s_2$:

bool(0 <= 1 - s1)

True

bool(1/4 - 3*s1/4 <= 1 - s1)

True

bool(1/3 - s1 <= 1 - s1)

True

bool(1/3 - 4*s1/3 <= 1 - s1)

True

bool(1/3 - s1 >= 1/3 - 4/3*s1)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \quad &\\ 1 - 2s_1 - 2 s_2 \leq s_3 \quad &\\ s_3 \leq & s_1 + s_2\\ s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Grenzen für s3¶

obere Grenze¶

solve_ineq([1 - s1 - s2 <= s1 + s2],[s2,s1])

[[s2 == -s1 + 1/2], [-s1 + 1/2 < s2]]

solve_ineq([1 - s1 - s2 <= 1/2 - s2/2],[s2,s1])

[[s2 == -2*s1 + 1], [-2*s1 + 1 < s2]]

solve_ineq([s1 + s2 <= 1/2 - s2/2],[s2,s1])

[[s2 == -2/3*s1 + 1/3], [s2 < -2/3*s1 + 1/3]]

$\qquad \mathbf{\leq} \qquad \quad$	$\mathbf{1-s_1-s_2}$	$\mathbf{s_1 + s_2}$	$\mathbf{\frac{1}{2} - \frac{1}{2} s_2 \quad}$
$\mathbf{1 - s_1 - s_2}$	immer $\qquad \qquad$	${\frac{1}{2}-s_1\leq s_2}$	${1 - 2s_1 \leq s_2}$
$\mathbf{s_1 + s_2}$	${s_2 \leq \frac{1}{2} - s_1}$	immer $\qquad \qquad$	${s_2 \leq \frac{1}{3}-\frac{2}{3}s_1}$
$\mathbf{\frac{1}{2}-\frac{1}{2}s_2}$	${s_2 \leq 1 - 2s_1 }$	${\frac{1}{3}-\frac{2}{3}s_1 \leq s_2}$	immer $\qquad \qquad$

solve_ineq([1/2 - s1 >= 1 - 2*s1],[s2,s1])

[[s1 == (1/2)], [(1/2) < s1]]

solve_ineq([1/2 - s1 <= 1/3 - 2*s1/3],[s2,s1])

[[s1 == (1/2)], [(1/2) < s1]]

solve_ineq(1/3-2*s1/3<=1-2*s1)

[[s1 <= (1/2)]]

obere Grenze:

obere Grenze $\quad$	im Fall $\qquad \qquad \qquad \qquad \quad$
$\mathbf{1-s_1-s_2}$	${1-2s_1 \leq s_2}$ und $s_1 \leq \frac{1}{2}$
$\mathbf{1-s_1-s_2}$	${\frac{1}{2}-s_1 \leq s_2}$ und $\frac{1}{2} \leq s_1$
$\mathbf{s_1 + s_2}$	${s_2 \leq \frac{1}{3} - \frac{2}{3}s_1}$ und ${s_1 \leq \frac{1}{2}}$
$\mathbf{s_1 + s_2}$	${s_2 \leq \frac{1}{2} - s_1}$ und ${\frac{1}{2} \leq s_1}$
$\mathbf{\frac{1}{2} - \frac{1}{2}s_2}$	${\frac{1}{3}-\frac{2}{3}s_1 \leq s_2 \leq 1 - 2s_1}$ und $s_1 \leq \frac{1}{2}$

Fall 4 ergibt eine Nullmenge, da $0 \leq s_2$ und damit:

solve_ineq(0 <= 1/2 - s1)

[[s1 <= (1/2)]]

neue obere Grenzen:

obere Grenze $\quad$	im Fall $\qquad \qquad \qquad \qquad \quad$
$\mathbf{1-s_1-s_2}$	${1-2s_1 \leq s_2}$ und $s_1 \leq \frac{1}{2}$
$\mathbf{1-s_1-s_2}$	${\frac{1}{2}-s_1 \leq s_2}$ und $\frac{1}{2} \leq s_1$
$\mathbf{s_1 + s_2}$	${s_2 \leq \frac{1}{3} - \frac{2}{3}s_1}$ und ${s_1 \leq \frac{1}{2}}$
$\mathbf{\frac{1}{2} - \frac{1}{2}s_2}$	${\frac{1}{3}-\frac{2}{3}s_1 \leq s_2 \leq 1 - 2s_1}$ und $s_1 \leq \frac{1}{2}$

untere Grenze¶

solve_ineq([0 >= 1/2 - s1/2 - s2],[s2,s1])

[[s2 == -1/2*s1 + 1/2], [-1/2*s1 + 1/2 < s2]]

solve_ineq([0 >= 1 - 2*s1 - 2*s2],[s2,s1])

[[s2 == -s1 + 1/2], [-s1 + 1/2 < s2]]

solve_ineq([1/2 - s1/2 - s2 >= 1 - 2*s1 - 2*s2],[s2,s1])

[[s2 == -3/2*s1 + 1/2], [-3/2*s1 + 1/2 < s2]]

$\qquad \mathbf{\geq} \qquad \qquad$	$\mathbf{0}$	$\mathbf{\frac{1}{2} - \frac{1}{2}s_1 - s_2}$	$\mathbf{1 - 2s_1 - 2s_2}$
$\mathbf{0}$	immer $\qquad \qquad$	${\frac{1}{2}-\frac{1}{2}s_1 \leq s_2}$	${\frac{1}{2}-s_1\leq s_2}$
$\mathbf{\frac{1}{2} - \frac{1}{2}s_1 - s_2}$	${s_2 \leq \frac{1}{2}-\frac{1}{2}s_1}$	immer $\qquad \qquad$	${\frac{1}{2}-\frac{3}{2}s_1 \leq s_2}$
$\mathbf{1 - 2s_1 - 2s_2}$	${s_2 \leq \frac{1}{2} - s_1}$	${s_2 \leq \frac{1}{2}-\frac{3}{2}s_1}$	immer $\qquad \qquad$

bool(1/2 - s1/2 >= 1/2 - s1)

True

bool(1/2 - 3*s1/2 <= 1/2 - s1/2)

True

bool(1/2 - 3/2*s1 <= 1/2 -s1)

True

untere Grenze:

untere Grenze $\quad$	im Fall $\qquad \qquad \qquad \qquad \quad$
$\mathbf{0}$	${\frac{1}{2}-\frac{1}{2}s_1 \leq s_2}$
$\mathbf{\frac{1}{2} - \frac{1}{2}s_1 - s_2}$	${\frac{1}{2}-\frac{3}{2}s_1 \leq s_2 \leq \frac{1}{2}-\frac{1}{2}s_1}$
$\mathbf{1 - 2s_1 - 2s_2}$	${s_2 \leq \frac{1}{2}-\frac{3}{2}s_1}$

Fall 1¶

untere Grenze: $0$

obere Grenze: $1 - s_1 - s_2$ und $s_1 \leq \frac{1}{2}$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ 1-2\cdot s_1 \leq s_2 \quad & \textrm{(neu)}\\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \quad & \textrm{(neu)} \\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

prüfe Grenzen für $s_2$

bool(1-2*s1 <= 1 - s1)

True

bool(1/2 - s1/2 <= 1 - s1)

True

bool(0 <= 1/2 - 1/2*s1)

True

bool(1/4 - 3*s1/4 <= 1/2 - 1/2*s1)

True

bool(1/3 - s1 <= 1/2 - 1/2*s1)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \\ 1-2\cdot s_1 \leq s_2 \leq & 1 - s_1\\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \quad & \\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

solve_ineq(1-2*s1 >= 1/2 - s1/2)

[[s1 <= (1/3)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{3}\\ 1 - 2 \cdot s_1 \leq s_2 \leq & 1 - s_1\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ \frac{1}{2} - \frac{1}{2} \cdot s_1 \leq s_2 \leq & 1 - s_1\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

Fall 2¶

untere Grenze: $0$

obere Grenze: $1 - s_1 - s_2$ und $\frac{1}{2} \leq s_1$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{2} \leq s_1 \leq & 1 \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{2} - s_1 \leq s_2 \quad & \textrm{(neu)} \\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \quad & \textrm{(neu)} \\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

bool(1/2 - s1 <= 1 - s1)

True

bool(1/2 - s1/2 <= 1 - s1)

True

untere Grenze für $s_2$

bool(0 <= 1/2 - s1/2)

True

bool(1/2 - s1 <= 1/2 - s1/2)

True

bool(1/4 - 3*s1/4 <= 1/2 - s1/2)

True

bool(1/3 - s1 <= 1/2 - s1/2)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{2} \leq s_1 \leq & 1 \\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \leq & 1 - s_1\\ 0 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

Fall 3¶

untere Grenze: $0$

obere Grenze: $s_1 + s_2$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1 \quad \textrm{(neu)}\\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \quad & \textrm{(neu)} \\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ 0 \leq s_3 \leq & s_1 + s_2\\ \end{align*}

Dieser Fall ist eine Nullmenge, da für die Grenzen von $s_2$ gilt:

solve_ineq(1/2 - 1/2*s1 <= 1/3 - 2/3*s1)

[[s1 <= -1]]

Fall 4¶

untere Grenze: $0$

obere Grenze: $\frac{1}{2} - \frac{1}{2}s_2$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & 1 - 2 s_1 \quad \textrm(neu)\\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \quad & \textrm(neu) \\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ 0 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2 \\ \end{align*}

Prüfe Grenzen für $s_2$:

bool(1-2*s1 <= 1 - s1)

True

solve_ineq(0 <= 1 - 2*s1)

[[s1 <= (1/2)]]

solve_ineq(1/3-2/3*s1 <= 1 - 2*s1)

[[s1 <= (1/2)]]

solve_ineq(1/2 - s1/2 <= 1 - 2*s1)

[[s1 <= (1/3)]]

solve_ineq(1/4 - 3/4*s1 <= 1 - 2*s1)

[[s1 <= (3/5)]]

bool(1/3 <= 3/5)

True

solve_ineq(1/3 - s1 <= 1 - 2*s1)

[[s1 <= (2/3)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{3}\\ 0 \leq s_2 \leq & 1 - 2 s_1\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \quad &\\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \quad & \\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ 0 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2 \\ \end{align*}

Finde untere Grenze für $s_2$:

bool(0 <= 1/2 - s1/2)

True

bool(1/3 - 2*s1/3 <= 1/2 - s1/2)

True

bool(1/4 - 3*s1/4 <= 1/2 - s1/2)

True

bool(1/3 - s1 <= 1/2 - s1/2)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{3}\\ \frac{1}{2} - \frac{1}{2}s_1 \leq s_2 \leq & 1 - 2 s_1\\ 0 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2 \\ \end{align*}

Fall 5¶

untere Grenze: $\frac{1}{2} - \frac{1}{2}s_1 - s_2$

obere Grenze: $1 - s_1 - s_2 $ und $s_1 \leq \frac{1}{2}$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \quad \textrm{(neu)} \\ 1 - 2 s_1 \leq s_2 \quad & \textrm{(neu)}\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

prüfe Grenzen für $s_2$

bool(0 <= 1/2 - s1/2)

True

bool(1/4 - 3/4*s1 <= 1/2 - s1/2)

True

bool(1/3 - s1 <= 1/2 - s1/2)

True

bool(1/2 - 3/2*s1 <= 1 - s1)

True

bool(1/2 - 3/2*s1 <= 1/2 - 1/2*s1)

True

bool(1 - 2*s1 <= 1 - s1)

True

solve_ineq(1 - 2*s1 <= 1/2 - 1/2*s1)

[[s1 >= (1/3)]]

bool(1/2 - s1/2 <= 1 - s1)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ 0 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \quad & \\ 1 - 2 s_1 \leq s_2 \quad & \\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

finde untere Grenze für $s_2$

solve_ineq(0 <= 1 - 2*s1) #true

[[s1 <= (1/2)]]

solve_ineq(1/4 - 3/4*s1 <= 1 - 2*s1) #true

[[s1 <= (3/5)]]

bool(3/5 >= 1/2)

True

solve_ineq(1/3 - s1 <= 1 - 2*s1) #true

[[s1 <= (2/3)]]

bool(1/2 - 3/2*s1 <= 1 - 2*s1)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ 1 - 2s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

Fall 6¶

untere Grenze: $\frac{1}{2} - \frac{1}{2}s_1 - s_2$

obere Grenze: $1 - s_1 - s_2 $ und $\frac{1}{2} \leq s_1$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{2} \leq s_1 \leq & 1 \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{2} - s_1 \leq s_2 \quad & \textrm{(neu)}\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \quad \textrm{(neu)} \\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

bool(1/2 - s1/2 <= 1 - s1)

True

prüfe Grenzen für $s_2$

bool(0 <= 1/2 - 1/2*s1)

True

bool(1/4 - 3/4*s1 <= 1/2 - 1/2*s1)

True

bool(1/3 - s1 <= 1/2 - s1/2)

True

bool(1/2 - s1 <= 1/2 - s1/2)

True

bool(1/2 - 3/2*s1 <= 1/2 - 1/2*s1)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{2} \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{2} - s_1 \leq s_2 \quad & \\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \quad & \\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

Finde untere Grenze für $s_2$

solve_ineq(1/4 - 3/4 *s1 <= 0) #true

[[s1 >= (1/3)]]

solve_ineq(1/3 - s1 <= 0) #true

[[s1 >= (1/3)]]

solve_ineq(1/2 - s1 <= 0) #true

[[s1 >= (1/2)]]

solve_ineq(1/2 - 3/2*s1 <= 0) #true

[[s1 >= (1/3)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{2} \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

Fall 7¶

untere Grenze: $\frac{1}{2} - \frac{1}{2}s_1 - s_2$

obere Grenze: $s_1 + s_2$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \quad \textrm{(neu)} \\ s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1 \quad \textrm{(neu)}\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}

Bestimme obere Grenze für $s_2$:

bool(1/3 - 2/3*s1 <= 1/2 - 1/2*s1)

True

bool(1/3 - 2/3*s1 <= 1 - s1)

True

Prüfe Grenzen für $s_2$:

solve_ineq(0 <= 1/3 - 2/3*s1) #true

[[s1 <= (1/2)]]

bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)

True

bool(1/3 - s1 <= 1/3 - 2/3 *s1)

True

solve_ineq(1/2 - 3/2*s1 <= 1/3 - 2/3 *s1)

[[s1 >= (1/5)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{5} \leq s_1 \leq & \frac{1}{2}\\ 0 \leq s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1 \\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \quad & \\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}

Bestimme untere Grenze für $s_2$:

solve_ineq(0 >= 1/4 - 3/4*s1)

[[s1 >= (1/3)]]

solve_ineq(0 >= 1/3 - s1)

[[s1 >= (1/3)]]

solve_ineq(0 >= 1/2 - 3/2*s1)

[[s1 >= (1/3)]]

solve_ineq(1/2 - 3/2*s1 >= 0)

[[s1 <= (1/3)]]

solve_ineq(1/2 - 3/2*s1 >= 1/4 - 3/4*s1)

[[s1 <= (1/3)]]

solve_ineq(1/2 - 3/2*s1 >= 1/3 - s1)

[[s1 <= (1/3)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ 0 \leq s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1 \quad\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{5} \leq s_1 \leq & \frac{1}{3}\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1 \quad\\ \frac{1}{2} - \frac{1}{2}s_1 - s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}

Fall 8¶

untere Grenze: $\frac{1}{2} - \frac{1}{2}s_1 - s_2$

obere Grenze: $\frac{1}{2} - \frac{1}{2}s_2$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & 1 - 2 s_1 \quad \textrm{(neu)}\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1 \quad \textrm{(neu)} \\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Prüfe Grenzen für $s_2$:

solve_ineq(1/2 - 1/2*s1 <= 1 - 2*s1)

[[s1 <= (1/3)]]

bool(0 <= 1/2 - 1/2*s1)

True

solve_ineq(0 <= 1 - 2*s1) #true

[[s1 <= (1/2)]]

solve_ineq(1/4 - 3/4*s1 <= 1 - 2*s1) #true (1/2 <= 3/5)

[[s1 <= (3/5)]]

bool(1/4 - 3/4*s1 <= 1/2 - 1/2*s1)

True

solve_ineq(1/3 - s1 <= 1- 2*s1) #true (1/2 <= 2/3)

[[s1 <= (2/3)]]

bool(1/3 - s1 <= 1/2 - 1/2*s1)

True

solve_ineq(1/3 - 2/3*s1 <= 1 - 2*s1) #true

[[s1 <= (1/2)]]

bool(1/3 - 2/3*s1 <= 1/2 - 1/2*s1)

True

bool(1/2 - 3/2*s1 <= 1 - 2*s1)

True

bool(1/2 - 3/2*s1 <= 1/2 - 1/2*s1)

True

Fall 1 (obere Grenze $1 - 2s_1$): \begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ 0 \leq s_2 \leq & 1 - 2 s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \quad & \\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \quad & \\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Bestimme untere Grenze für $s_2$:

solve_ineq(0 <= 1/3 - 2/3*s1) #true

[[s1 <= (1/2)]]

bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)

True

bool(1/3 - s1 <= 1/3 - 2/3*s1)

True

solve_ineq(1/2 - 3/2*s1 <= 1/3 - 2/3*s1) #true

[[s1 >= (1/5)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{3} \leq s_1 \leq & \frac{1}{2}\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & 1 - 2 s_1\\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Fall 2 (obere Grenze $\frac{1}{2} - \frac{1}{2} s_1$): \begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{3}\\ 0 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \quad & \\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \quad & \\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Bestimme untere Grenze für $s_2$:

solve_ineq(0 <= 1/3 - 2/3*s1) #true

[[s1 <= (1/2)]]

bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)

True

bool(1/3 - s1 <= 1/3 - 2/3*s1)

True

solve_ineq(0 <= 1/2 - 3/2*s1) #true

[[s1 <= (1/3)]]

solve_ineq(1/4 - 3/4*s1 <= 1/2 - 3/2*s1) #true

[[s1 <= (1/3)]]

solve_ineq(1/3 - s1 <= 1/2 - 3/2*s1) #true

[[s1 <= (1/3)]]

solve_ineq(1/3 - 2/3*s1 <= 1/2 - 3/2*s1)

[[s1 <= (1/5)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{5}\\ \frac{1}{2} - \frac{3}{2} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1\\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{5} \leq s_1 \leq & \frac{1}{3}\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{1}{2} s_1\\ \frac{1}{2} - \frac{1}{2} s_1 - s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Fall 9¶

untere Grenze: $1 - 2s_1 - 2s_2$

obere Grenze: $1 - s_1 - s_2$ und $s_1 \leq \frac{1}{2}$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ 1 - 2 s_1 \leq s_2 \quad & \textrm{(neu)} \\ s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1 \quad \textrm{(neu)}\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

Dieser Fall ist eine Nullmenge, da für die Grenzen von $s_2$ gilt:

solve_ineq(1 - 2*s1 <= 1/2 - 3/2*s1)

[[s1 >= 1]]

Fall 10¶

untere Grenze: $1 - 2s_1 - 2s_2$

obere Grenze: $1 - s_1 - s_2$ und $\frac{1}{2} \leq s_1$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{2} \leq s_1 \leq & 1 \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{2} - s_1 \leq s_2 \quad & \textrm{(neu)}\\ s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1 \quad \textrm{(neu)}\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & 1 - s_1 - s_2\\ \end{align*}

Dieser Fall ist eine Nullmenge, da für die Grenzen von $s_2$ gilt:

solve_ineq(1/2 - s1 <= 1/2 - 3/2*s1) #false

[[s1 <= 0]]

Fall 11¶

untere Grenze: $1 - 2s_1 - 2s_2$

obere Grenze: $s_1 + s_2$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1 \quad \textrm{(neu)}\\ s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1 \quad \textrm{(neu)}\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}

Bestimme obere Grenze für $s_2$:

bool( 1/3 - 2/3*s1 <= 1 - s1)

True

solve_ineq(1/3 - 2/3*s1 <= 1/2 - 3/2*s1)

[[s1 <= (1/5)]]

Prüfe Grenzen für $s_2$:

solve_ineq(0 <= 1/3 - 2/3*s1) #true

[[s1 <= (1/2)]]

bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)

True

bool(1/3 - s1 <= 1/3 - 2/3*s1)

True

solve_ineq(0 <= 1/2 - 3/2*s1)

[[s1 <= (1/3)]]

solve_ineq(1/4 - 3/4*s1 <= 1/2 - 3/2*s1)

[[s1 <= (1/3)]]

solve_ineq(1/3 - s1 <= 1/2 - 3/2*s1)

[[s1 <= (1/3)]]

Bestimme untere Grenze für $s_2$:

solve_ineq(0 <= 1/3 - s1) #true

[[s1 <= (1/3)]]

solve_ineq(1/4 - 3/4*s1 <= 1/3 - s1) #true

[[s1 <= (1/3)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ \frac{1}{5} \leq s_1 \leq & \frac{1}{3}\\ \frac{1}{3} - s_1 \leq s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{5}\\ \frac{1}{3} - s_1 \leq s_2 \leq & \frac{1}{3} - \frac{2}{3} s_1\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & s_1 + s_2\\ \end{align*}

Fall 12¶

untere Grenze: $1 - 2s_1 - 2s_2$

obere Grenze: $\frac{1}{2}-\frac{1}{2}s_2$

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{2} \quad \textrm{(neu)}\\ 0 \leq s_2 \leq & 1 - s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & 1 - 2 s_1 \quad \textrm{(neu)}\\ s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1 \quad \textrm{(neu)}\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Bestimme obere Grenze für $s_2$:

bool(1/2 - 3/2*s1 <= 1 - s1)

True

bool(1/2 - 3/2*s1 <= 1 - 2*s1)

True

Prüfe Grenzen für $s_2$:

solve_ineq(0 <= 1/2 - 3/2*s1)

[[s1 <= (1/3)]]

solve_ineq(1/4 - 3/4*s1 <= 1/2 - 3/2*s1)

[[s1 <= (1/3)]]

solve_ineq(1/3 - s1 <= 1/2 - 3/2*s1)

[[s1 <= (1/3)]]

solve_ineq(1/3 - 2/3*s1 <= 1/2 - 3/2*s1)

[[s1 <= (1/5)]]

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{5} \\ 0 \leq s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1\\ \frac{1}{4} - \frac{3}{4} s_1 \leq s_2 \quad &\\ \frac{1}{3} - s_1 \leq s_2 \quad &\\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \quad & \\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Bestimme die untere Grenze für $s_2$:

solve_ineq(0 <= 1/3 - 2/3*s1) #true

[[s1 <= (1/2)]]

bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)

True

bool(1/3 - s1 <= 1/3 - 2/3*s1)

True

\begin{align*} 2 \leq c \quad &\\ s_0 = & 1 - s_1 - s_2 - s_3\\ 0 \leq s_1 \leq & \frac{1}{5} \\ \frac{1}{3} - \frac{2}{3} s_1 \leq s_2 \leq & \frac{1}{2} - \frac{3}{2} s_1\\ 1 - 2s_1 - 2 s_2 \leq s_3 \leq & \frac{1}{2} - \frac{1}{2} s_2\\ \end{align*}

Baum 3¶

(Bereiche für Baum 2 wegen Symmetrie analog)¶

\begin{align*} s_0 + s_1 + s_2 + s_3 & = 1\\ 0 \leq s_0 & \leq 1\\ 0 \leq s_1 & \leq 1\\ 0 \leq s_2 & \leq 1\\ 0 \leq s_3 & \leq 1\\ \ \\ 3 \cdot s_0 + 3 \cdot s_1 + 2 \cdot s_2 + s_3 & \leq c \\ s_3 & \geq s_0 + s_1 + 1 \\ s_2 & \geq s_0 \\ \end{align*}

forget()
var('s0,s1,s2,s3,c')

(s0, s1, s2, s3, c)

assume(0 <= s0)
assume(0 <= s1)
assume(0 <= s2)
assume(0 <= s3)
assume(s0 <= 1)
assume(s1 <= 1)
assume(s2 <= 1)
assume(s3 <= 1)

Asymptotisch:

\begin{align*} s_0 + s_1 + s_2 + s_3 & = 1\\ 0 \leq s_0 \leq &1\\ 0 \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & 1\\ 0 \leq s_3 \leq & 1\\ \ \\ 3 \cdot s_0 + 3 \cdot s_1 + 2 \cdot s_2 + s_3 & \leq c \\ s_0 + s_1 \leq s_3 \quad & \\ s_0 \leq s_2 \quad &\\ \end{align*}

Setze $s_1 = 1 - s_0 - s_2 - s_3$

s1 = 1 - s0 - s2 - s3

Forme Ungleichungen, die $s_2$ enthalten, um:

solve_ineq([3*s0+3*s1+2*s2+s3 <= c],[c,s3,s2,s0])

[[c == -s2 - 2*s3 + 3], [-s2 - 2*s3 + 3 < c]]

bool(s1 <= 1)

True

solve_ineq([0 <= s1],[s3,s2,s0,c])

[[s3 == -s0 - s2 + 1], [s3 < -s0 - s2 + 1]]

solve_ineq([s0 + s1 <= s3],[s3,s2,s0,c])

[[s3 == -1/2*s2 + 1/2], [-1/2*s2 + 1/2 < s3]]

\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 3 - s_2 - 2 \cdot s_3 \leq c \quad & \\ 0 \leq s_0 \leq &1 \\ s_0 \leq s_2 \leq& 1 \\ 0 \leq s_3 \leq &1 \\ \frac{1}{2} - \frac{1}{2} \cdot s_2 \leq s_3 \quad & \\ s_3 \leq & 1 - s_0 - s_2 \end{align*}

Prüfe neue Grenzen:

bool(1 - s0 - s2 <= 1)

True

bool(0 <= 1/2 - 1/2*s2)

True

solve_ineq([1/2 - 1/2*s2 <= 1 - s0 - s2],[s2,s0])

[[s2 == -2*s0 + 1], [s2 < -2*s0 + 1]]

\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 3 - s_2 - 2 \cdot s_3 \leq c \quad & \\ 0 \leq s_0 \leq &1 \\ s_0 \leq s_2 \leq& 1 \\ s_2 \leq & 1 - 2 \cdot s_0 \\ \frac{1}{2} - \frac{1}{2} \cdot s_2 \leq s_3 \leq &1- s_0 - s_2 \\ \end{align*}

bool(1 - 2*s0 <= 1)

True

solve_ineq(s0 <= 1 - 2*s0)

[[s0 <= (1/3)]]

\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 3 - s_2 - 2 \cdot s_3 \leq c \quad & \\ 0 \leq s_0 \leq & \frac{1}{3} \\ s_0 \leq s_2 \leq& 1 - 2 \cdot s_0 \\ \frac{1}{2} - \frac{1}{2} \cdot s_2 \leq s_3 \leq &1- s_0 - s_2 \\ \end{align*}

Forme Ungleichung um, um Grenze für $s_3$ zu erhalten:

solve_ineq([3 - s2 - 2*s3 <= c], [s3, s2, c])

[[s3 == -1/2*c - 1/2*s2 + 3/2], [-1/2*c - 1/2*s2 + 3/2 < s3]]

\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 0 \leq s_0 \leq & \frac{1}{3} \\ s_0 \leq s_2 \leq& 1 - 2 \cdot s_0 \\ \frac{1}{2} - \frac{1}{2} \cdot s_2 \leq s_3 \leq &1- s_0 - s_2 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_2 \leq s_3 \quad &\\ \end{align*}

Prüfe Grenzen und daraus erhaltene neue Grenzen:

solve_ineq([s0 <= c - 2*s0 - 1],[s0,c])

[[s0 == 1/3*c - 1/3], [s0 < 1/3*c - 1/3]]

solve_ineq([0 <= 1/3*c - 1/3])

[[c == 1], [1 < c]]

\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 1 \leq c \quad & \\ 0 \leq s_0 \leq & \frac{1}{3} \\ s_0 \leq & \frac{1}{3}c - \frac{1}{3}\\ s_0 \leq s_2 \leq& 1 - 2 \cdot s_0 \\ s_2 \leq & c - 2s_0 -1 \\ \frac{1}{2} - \frac{1}{2} \cdot s_2 \leq s_3 \leq &1- s_0 - s_2 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_2 \leq s_3 \quad &\\ \end{align*}

Bestimme Grenzen:

solve_ineq([1/2 - 1/2*s2 >= 3/2 - 1/2*c - 1/2*s2],[s2,s0,c])

[[c == 2], [2 < c]]

solve_ineq([1-2*s0 <= c - 2*s0 - 1],[s0,c])

[[c == 2], [2 < c]]

solve_ineq([1/3 <= 1/3*c - 1/3])

[[c == 2], [2 < c]]

\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 1 \leq c \leq & 2\\ 0 \leq s_0 \leq & \frac{1}{3}c - \frac{1}{3} \\ s_0 \leq s_2 \leq& c - 2s_0 -1 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_2 \leq s_3 \leq &1- s_0 - s_2 \\ \end{align*}

\begin{align*} s_1 = & 1 - s_0 - s_2 - s_3 \\ 2 \leq c \quad & \\ 0 \leq s_0 \leq & \frac{1}{3} \\ s_0 \leq s_2 \leq& 1 - 2 \cdot s_0 \\ \frac{1}{2} - \frac{1}{2} \cdot s_2 \leq s_3 \leq &1- s_0 - s_2 \\ \end{align*}

Baum 4¶

(Bereiche für Baum 5 wegen Symmetrie analog)¶

\begin{align*} s_0 + s_1 + s_2 + s_3 & = 1 \quad &\\ 0 \leq s_0 \leq & 1\\ 0 \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & 1\\ 0 \leq s_3 \leq & 1\\ \ \\ s_0 + 3 \cdot s_1 + 3 \cdot s_2 + 2 \cdot s_3 \leq c \quad & \\ s_0 \geq & s_1 + s_2 + 1 \\ s_3 \geq & s_1 \\ s_0 \geq & s_3 \\ \end{align*}

Asymptotisch: \begin{align*} s_0 + s_1 + s_2 + s_3 & = 1 \quad &\\ 0 \leq s_0 \leq & 1\\ 0 \leq s_1 \leq & 1\\ 0 \leq s_2 \leq & 1\\ 0 \leq s_3 \leq & 1\\ \ \\ s_0 + 3 \cdot s_1 + 3 \cdot s_2 + 2 \cdot s_3 \leq c \quad & \\ s_1 + s_2 \leq s_0 \quad &\\ s_1 \leq s_3 \quad & \\ s_3 \leq s_0 \quad &\\ \end{align*}

setze $s_2 = 1 - s_0 - s_1 - s_3$

forget()
var('s0,s1,s2,s3,c')

(s0, s1, s2, s3, c)

assume(0 <= s0)
assume(0 <= s1)
assume(0 <= s2)
assume(0 <= s3)
assume(s0 <= 1)
assume(s1 <= 1)
assume(s2 <= 1)
assume(s3 <= 1)

s2 = 1 - s0 - s1 - s3

Forme Ungleichungen, die $s_2$ enthalten, um:

s0 + 3*s1 + 3*s2 +2*s3 <= c

-2*s0 - s3 + 3 <= c

solve_ineq([s1 + s2 <= s0], [s0,s3,s1])

[[s0 == -1/2*s3 + 1/2], [-1/2*s3 + 1/2 < s0]]

bool(s2 <= 1)

True

solve_ineq([0 <= s2],[s0,s3,s1])

[[s0 == -s1 - s3 + 1], [s0 < -s1 - s3 + 1]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 3 - 2 \cdot s_0 - s_3 \leq c \\ 0 \leq s_1 \leq & 1 \\ s_1 \leq s_3 \leq & 1 \\ s_3 \leq s_0 \leq & 1 \\ s_0 \leq & 1 - s_1 - s_3 \\ \frac{1}{2} - \frac{1}{2} \cdot s_3 \leq s_0 \quad &\\ \end{align*}

bool(1 - s1 - s3 <= 1)

True

Überprüfe neue Grenzen und daraus erhaltene neue Grenzen:

solve_ineq([s3 <= 1 - s1 - s3],[s3,s1])

[[s3 == -1/2*s1 + 1/2], [s3 < -1/2*s1 + 1/2]]

solve_ineq(s1 <= 1/2 - 1/2*s1)

[[s1 <= (1/3)]]

solve_ineq([1/2 - 1/2*s3 <= 1 - s1 - s3 ], [s3,s1])

[[s3 == -2*s1 + 1], [s3 < -2*s1 + 1]]

solve_ineq([s1 <= 1 - 2*s1])

[[s1 == (1/3)], [s1 < (1/3)]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 3 - 2 \cdot s_0 - s_3 \leq c \\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq s_3 \leq & 1 - 2s_1\\ s_3 \leq & \frac{1}{2} - \frac{1}{2}s_1 \\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \frac{1}{2} - \frac{1}{2} \cdot s_3 \leq s_0 \quad &\\ \end{align*}

solve_ineq( 1/2 - 1/2*s3 >= s3) # --> Fall 1 und Fall 2

[[s3 <= (1/3)]]

solve_ineq(1/2 -1/2*s1 <= 1 - 2*s1) #true

[[s1 <= (1/3)]]

solve_ineq(1/3 <= 1/2 - 1/2*s1) #true

[[s1 <= (1/3)]]

Fall 1¶

$(s_3 \leq \frac{1}{3})$ \begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 3 - 2 \cdot s_0 - s_3 \leq c \\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq s_3 \leq & \frac{1}{3}\\ \frac{1}{2} - \frac{1}{2} \cdot s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Forme Ungleichung um, um Grenze für $s_0$ zu erhalten:

solve_ineq([3 - 2*s0 - s3 <= c], [s0, s3, s1, c])

[[s0 == -1/2*c - 1/2*s3 + 3/2], [-1/2*c - 1/2*s3 + 3/2 < s0]]

Prüfe Grenzen:

solve_ineq([3/2 - 1/2*c - 1/2*s3 <= 1 - s1 - s3],[s3,s1,c])

[[s3 == c - 2*s1 - 1], [s3 < c - 2*s1 - 1]]

solve_ineq([s1 <= c - 2*s1 - 1],[s1,c])

[[s1 == 1/3*c - 1/3], [s1 < 1/3*c - 1/3]]

solve_ineq(0 <= 1/3*c - 1/3)

[[c >= 1]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 1 \leq c \quad &\\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{3}c - \frac{1}{3}\\ s_1 \leq s_3 \leq & \frac{1}{3}\\ s_3 \leq & c - 2s_1 - 1\\ \frac{1}{2} - \frac{1}{2} \cdot s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \quad & \end{align*}

Bestimme untere Grenze für $s_0$:

solve_ineq([1/2 - 1/2*s3 >= 3/2 - 1/2*c - 1/2*s3],[s3,s1,c])

[[c == 2], [2 < c]]

Bestimme obere Grenze für $s_3$:

solve_ineq([1/3 <= c - 2*s1 - 1],[s1,c]) # --> Fall 1.1 und Fall 1.2

[[s1 == 1/2*c - 2/3], [s1 < 1/2*c - 2/3]]

Fall 1.1¶

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 1 \leq c \quad &\\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{3}c - \frac{1}{3}\\ s_1 \leq & \frac{1}{2}c - \frac{2}{3} \quad \textrm{(neu)}\\ s_1 \leq s_3 \leq & \frac{1}{3}\\ \frac{1}{2} - \frac{1}{2} \cdot s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \quad & \end{align*}

prüfe neue Grenze:

solve_ineq(0 <= 1/2*c - 2/3)

[[c >= (4/3)]]

Bestimme obere Grenze für $s_1$:

solve_ineq(1/3 <= 1/3*c - 1/3)

[[c >= 2]]

solve_ineq(1/3 <= 1/2*c - 2/3)

[[c >= 2]]

solve_ineq(1/3*c - 1/3 <= 1/2*c - 2/3)

[[c >= 2]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \leq & 2\\ 0 \leq s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ s_1 \leq s_3 \leq & \frac{1}{3}\\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3\leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 2 \leq c \quad &\\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq s_3 \leq & \frac{1}{3}\\ \frac{1}{2} - \frac{1}{2} \cdot s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Fall 1.2¶

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 1 \leq c \quad &\\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{3}c - \frac{1}{3}\\ \frac{1}{2}c - \frac{2}{3} \leq s_1 \quad & \textrm{(neu)}\\ s_1 \leq s_3 \leq & c - 2s_1 - 1\\ \frac{1}{2} - \frac{1}{2} \cdot s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \quad & \end{align*}

Prüfe neue Grenze:

solve_ineq(1/2*c - 2/3 <= 1/3)

[[c <= 2]]

solve_ineq(1/2*c - 2/3 <= 1/3*c - 1/3)

[[c <= 2]]

Bestimme untere Grenze für $s_1$:

solve_ineq(1/2*c - 2/3 >= 0)

[[c >= (4/3)]]

Bestimme obere Grenze für $s_1$:

solve_ineq(1/3*c - 1/3 <= 1/3)

[[c <= 2]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \leq & 2\\ \frac{1}{2}c - \frac{2}{3} \leq s_1 \leq & \frac{1}{3}c - \frac{1}{3} \\ s_1 \leq s_3 \leq & c - 2s_1 - 1\\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 1 \leq c \leq & \frac{4}{3}\\ 0 \leq s_1 \leq & \frac{1}{3}c - \frac{1}{3} \\ s_1 \leq s_3 \leq & c - 2s_1 - 1\\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Fall 2¶

$(\frac{1}{3} \leq s_3)$

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 3 - 2 \cdot s_0 - s_3 \leq c \\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ \frac{1}{3} \leq s_3 \quad & \\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

da $s_1 \leq \frac{1}{3}$ gilt: \begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 3 - 2 \cdot s_0 - s_3 \leq c \\ 0 \leq s_1 \leq & \frac{1}{3} \\ \frac{1}{3} \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Forme Ungleichung um, um Grenze für $s_0$ zu erhalten:

solve_ineq([3 - 2*s0 - s3 <= c], [s0, s3, s1, c])

[[s0 == -1/2*c - 1/2*s3 + 3/2], [-1/2*c - 1/2*s3 + 3/2 < s0]]

Prüfe die neue(n) Grenze(n):

solve_ineq([3/2 - 1/2*c - 1/2*s3 <= 1 - s1 - s3], [s3, s1, c])

[[s3 == c - 2*s1 - 1], [s3 < c - 2*s1 - 1]]

solve_ineq([1/3 <= c - 2*s1 - 1], [s1, c])

[[s1 == 1/2*c - 2/3], [s1 < 1/2*c - 2/3]]

solve_ineq(0 <= 1/2*c - 2/3)

[[c >= (4/3)]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \quad & \\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ \frac{1}{3} \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq & c - 2s_1 - 1 \\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \quad &\\ \end{align*}

Bestimme untere Grenze für $s_0$:

solve_ineq([s3 >= 3/2 - 1/2*c - 1/2*s3], [s3,s1,c]) # --> Fall 2.1 und Fall 2.2

[[s3 == -1/3*c + 1], [-1/3*c + 1 < s3]]

Fall 2.1¶

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \quad & \\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ \frac{1}{3} \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq & c - 2s_1 - 1 \\ 1 - \frac{1}{3}c \leq s_3 \quad & \textrm{(neu)}\\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Prüfe neue Grenze(n):

solve_ineq([1 - 1/3*c <= 1/2 - 1/2*s1],[s1,c])

[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]

solve_ineq([1 - 1/3*c <= c - 2*s1 - 1], [s1,c])

[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]

solve_ineq(0 <= 2/3*c - 1)

[[c >= (3/2)]]

bool(3/2 >= 4/3)

True

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{3}{2} \leq c \quad & \\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ s_1 \leq & \frac{2}{3}c - 1 \\ \frac{1}{3} \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq & c - 2s_1 - 1 \\ 1 - \frac{1}{3}c \leq s_3 \quad & \\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Grenzen für $s_3$:

solve_ineq(1/3 >= 1 - 1/3*c)

[[c >= 2]]

solve_ineq([1/2 - 1/2*s1 <= c - 2*s1 - 1], [s1, c]) #true

[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]

obere Grenze für $s_1$:

solve_ineq(1/3 <= 1/2*c - 2/3)

[[c >= 2]]

solve_ineq(1/3 <= 2/3*c - 1)

[[c >= 2]]

solve_ineq(2/3*c -1 <= 1/2*c - 2/3)

[[c <= 2]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{3}{2} \leq c \leq & 2\\ 0 \leq s_1 \leq & \frac{2}{3}c - 1 \\ 1 - \frac{1}{3}c \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ 2 \leq c \quad & \\ 0 \leq s_1 \leq & \frac{1}{3} \\ \frac{1}{3} \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Fall 2.2¶

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \quad & \\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ \frac{1}{3} \leq s_3 \leq &\frac{1}{2} - \frac{1}{2}s_1\\ s_3 \leq & c - 2s_1 - 1 \\ s_3 \leq & 1 - \frac{1}{3}c \quad \textrm{(neu)}\\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Prüfe neue Grenze:

solve_ineq(1/3 <= 1 - 1/3*c)

[[c <= 2]]

Bestimme Grenze für $s_3$:

solve_ineq([c - 2*s1 - 1 <= 1 - 1/3*c], [s1, c])

[[s1 == 2/3*c - 1], [2/3*c - 1 < s1]]

solve_ineq([c - 2*s1 - 1 <= 1/2 - 1/2*s1], [s1, c])

[[s1 == 2/3*c - 1], [2/3*c - 1 < s1]]

solve_ineq([1 - 1/3*c <= c - 2*s1 - 1], [s1, c])

[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]

solve_ineq([1 - 1/3*c <= 1/2 - 1/2*s1], [s1,c]) #--> Fall 2.2.1 und Fall 2.2.2

[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]

Fall 2.2.1 \begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \leq & 2\\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ \frac{2}{3}c - 1 \leq s_1 \quad & \textrm{(neu)}\\ \frac{1}{3} \leq s_3 \leq & c - 2s_1 - 1 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Prüfe neue Grenze:

solve_ineq(2/3*c - 1 <= 1/3) #true

[[c <= 2]]

solve_ineq(2/3*c - 1 <= 1/2*c - 2/3) #true

[[c <= 2]]

Grenzen für $s_1$:

solve_ineq(2/3*c - 1 >= 0)

[[c >= (3/2)]]

bool(3/2 > 4/3)

True

solve_ineq(1/2*c - 2/3 <= 1/3) #true

[[c <= 2]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \leq & \frac{3}{2}\\ 0 \leq s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ \frac{1}{3} \leq s_3 \leq & c - 2s_1 - 1 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{3}{2} \leq c \leq & 2\\ \frac{2}{3}c - 1 \leq s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ \frac{1}{3} \leq s_3 \leq & c - 2s_1 - 1 \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Fall 2.2.2 \begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{4}{3} \leq c \leq & 2 \\ 0 \leq s_1 \leq & \frac{1}{3} \\ s_1 \leq & \frac{1}{2}c - \frac{2}{3} \\ s_1 \leq & \frac{2}{3}c - 1 \quad \textrm{(neu)} \\ \frac{1}{3} \leq s_3 \leq & 1 - \frac{1}{3}c \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}

Prüfe neue Grenze:

solve_ineq(0 <= 2/3*c - 1)

[[c >= (3/2)]]

Bestimme Grenzen für $s_1$:

solve_ineq(2/3*c - 1 <= 1/3)

[[c <= 2]]

solve_ineq(2/3*c - 1 <= 1/2*c - 2/3)

[[c <= 2]]

\begin{align*} s_2 = 1 - s_0 - s_1 - s_3 \\ \frac{3}{2} \leq c \leq & 2 \\ 0 \leq s_1 \leq & \frac{2}{3}c - 1 \\ \frac{1}{3} \leq s_3 \leq & 1 - \frac{1}{3}c \\ \frac{3}{2} - \frac{1}{2}c - \frac{1}{2}s_3 \leq s_0 \leq & 1 - s_1 - s_3 \\ \end{align*}