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*}

In [1]:
from sage.symbolic.relation import solve_ineq
In [2]:
forget()
var('s0,s1,s2,s3,c')
Out[2]:
(s0, s1, s2, s3, c)
In [3]:
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 $s_0 = 1 - s_1 - s_2 - s_3$

In [4]:
s0 = 1 - s1 - s2 - s3

forme Ungleichungen, die $s_0$ enthalten, um

In [5]:
2*s0 + 2*s1 + 2*s2 + 2*s3 <= c
Out[5]:
2 <= c
In [6]:
solve_ineq([s0 <= s2 + s3],[s3,s2,s1])
Out[6]:
[[s3 == -1/2*s1 - s2 + 1/2], [-1/2*s1 - s2 + 1/2 < s3]]
In [7]:
solve_ineq([s0 <= s1 + s2],[s3,s2,s1])
Out[7]:
[[s3 == -2*s1 - 2*s2 + 1], [-2*s1 - 2*s2 + 1 < s3]]
In [8]:
solve_ineq([s3 <= s0 + s1],[s3,s2,s1])
Out[8]:
[[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:

In [9]:
solve_ineq([0 <= 1 - s1 - s2 - s3],[s3,s2,s1])
Out[9]:
[[s3 == -s1 - s2 + 1], [s3 < -s1 - s2 + 1]]
In [10]:
bool(1 - s1 - s2 - s3 <= 1)
Out[10]:
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 $s_3$

In [11]:
bool(0 <= s1 + s2)
Out[11]:
True
In [12]:
bool(0 <= 1/2 - s2/2)
Out[12]:
True
In [13]:
solve_ineq([0 <= 1 - s1 - s2],[s2,s1])
Out[13]:
[[s2 == -s1 + 1], [s2 < -s1 + 1]]
In [14]:
solve_ineq([1/2 - s1/2 - s2 <= s1 + s2],[s2,s1])
Out[14]:
[[s2 == -3/4*s1 + 1/4], [-3/4*s1 + 1/4 < s2]]
In [15]:
bool(1/2 - s1/2 - s2 <= 1/2 - s2/2)
Out[15]:
True
In [16]:
bool(1/2 - s1/2 - s2 <= 1 - s1 - s2)
Out[16]:
True
In [17]:
solve_ineq([1 - 2*s1 - 2*s2 <= s1 + s2],[s2,s1])
Out[17]:
[[s2 == -s1 + 1/3], [-s1 + 1/3 < s2]]
In [18]:
solve_ineq([1 - 2*s1 - 2*s2 <= 1/2 - s2/2],[s2,s1])
Out[18]:
[[s2 == -4/3*s1 + 1/3], [-4/3*s1 + 1/3 < s2]]
In [19]:
bool(1 - 2*s1 - 2*s2 <= 1 - s1 - s2)
Out[19]:
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 $s_2$:

In [20]:
bool(0 <= 1 - s1)
Out[20]:
True
In [21]:
bool(1/4 - 3*s1/4 <= 1 - s1)
Out[21]:
True
In [22]:
bool(1/3 - s1 <= 1 - s1)
Out[22]:
True
In [23]:
bool(1/3 - 4*s1/3 <= 1 - s1)
Out[23]:
True
In [24]:
bool(1/3 - s1 >= 1/3 - 4/3*s1)
Out[24]:
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

In [25]:
solve_ineq([1 - s1 - s2 <= s1 + s2],[s2,s1])
Out[25]:
[[s2 == -s1 + 1/2], [-s1 + 1/2 < s2]]
In [26]:
solve_ineq([1 - s1 - s2 <= 1/2 - s2/2],[s2,s1])
Out[26]:
[[s2 == -2*s1 + 1], [-2*s1 + 1 < s2]]
In [27]:
solve_ineq([s1 + s2 <= 1/2 - s2/2],[s2,s1])
Out[27]:
[[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 $\qquad \qquad$
In [28]:
solve_ineq([1/2 - s1 >= 1 - 2*s1],[s2,s1])
Out[28]:
[[s1 == (1/2)], [(1/2) < s1]]
In [29]:
solve_ineq([1/2 - s1 <= 1/3 - 2*s1/3],[s2,s1])
Out[29]:
[[s1 == (1/2)], [(1/2) < s1]]
In [30]:
solve_ineq(1/3-2*s1/3<=1-2*s1)
Out[30]:
[[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:

In [31]:
solve_ineq(0 <= 1/2 - s1)
Out[31]:
[[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

In [32]:
solve_ineq([0 >= 1/2 - s1/2 - s2],[s2,s1])
Out[32]:
[[s2 == -1/2*s1 + 1/2], [-1/2*s1 + 1/2 < s2]]
In [33]:
solve_ineq([0 >= 1 - 2*s1 - 2*s2],[s2,s1])
Out[33]:
[[s2 == -s1 + 1/2], [-s1 + 1/2 < s2]]
In [34]:
solve_ineq([1/2 - s1/2 - s2 >= 1 - 2*s1 - 2*s2],[s2,s1])
Out[34]:
[[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 $\qquad \qquad$
In [35]:
bool(1/2 - s1/2 >= 1/2 - s1)
Out[35]:
True
In [36]:
bool(1/2 - 3*s1/2 <= 1/2 - s1/2)
Out[36]:
True
In [37]:
bool(1/2 - 3/2*s1 <= 1/2 -s1)
Out[37]:
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 $s_2$

In [38]:
bool(1-2*s1 <= 1 - s1)
Out[38]:
True
In [39]:
bool(1/2 - s1/2 <= 1 - s1)
Out[39]:
True
In [40]:
bool(0 <= 1/2 - 1/2*s1)
Out[40]:
True
In [41]:
bool(1/4 - 3*s1/4 <= 1/2 - 1/2*s1)
Out[41]:
True
In [42]:
bool(1/3 - s1 <= 1/2 - 1/2*s1)
Out[42]:
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*}

In [43]:
solve_ineq(1-2*s1 >= 1/2 - s1/2)
Out[43]:
[[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*}

In [44]:
bool(1/2 - s1 <= 1 - s1)
Out[44]:
True
In [45]:
bool(1/2 - s1/2 <= 1 - s1)
Out[45]:
True

untere Grenze für $s_2$

In [46]:
bool(0 <= 1/2 - s1/2)
Out[46]:
True
In [47]:
bool(1/2 - s1 <= 1/2 - s1/2)
Out[47]:
True
In [48]:
bool(1/4 - 3*s1/4 <= 1/2 - s1/2)
Out[48]:
True
In [49]:
bool(1/3 - s1 <= 1/2 - s1/2)
Out[49]:
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:

In [50]:
solve_ineq(1/2 - 1/2*s1 <= 1/3 - 2/3*s1)
Out[50]:
[[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 $s_2$:

In [51]:
bool(1-2*s1 <= 1 - s1)
Out[51]:
True
In [52]:
solve_ineq(0 <= 1 - 2*s1)
Out[52]:
[[s1 <= (1/2)]]
In [53]:
solve_ineq(1/3-2/3*s1 <= 1 - 2*s1)
Out[53]:
[[s1 <= (1/2)]]
In [54]:
solve_ineq(1/2 - s1/2 <= 1 - 2*s1)
Out[54]:
[[s1 <= (1/3)]]
In [55]:
solve_ineq(1/4 - 3/4*s1 <= 1 - 2*s1)
Out[55]:
[[s1 <= (3/5)]]
In [56]:
bool(1/3 <= 3/5)
Out[56]:
True
In [57]:
solve_ineq(1/3 - s1 <= 1 - 2*s1)
Out[57]:
[[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 $s_2$:

In [58]:
bool(0 <= 1/2 - s1/2)
Out[58]:
True
In [59]:
bool(1/3 - 2*s1/3 <= 1/2 - s1/2)
Out[59]:
True
In [60]:
bool(1/4 - 3*s1/4 <= 1/2 - s1/2)
Out[60]:
True
In [61]:
bool(1/3 - s1 <= 1/2 - s1/2)
Out[61]:
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 $s_2$

In [62]:
bool(0 <= 1/2 - s1/2)
Out[62]:
True
In [63]:
bool(1/4 - 3/4*s1 <= 1/2 - s1/2)
Out[63]:
True
In [64]:
bool(1/3 - s1 <= 1/2 - s1/2)
Out[64]:
True
In [65]:
bool(1/2 - 3/2*s1 <= 1 - s1)
Out[65]:
True
In [66]:
bool(1/2 - 3/2*s1 <= 1/2 - 1/2*s1)
Out[66]:
True
In [67]:
bool(1 - 2*s1 <= 1 - s1)
Out[67]:
True
In [68]:
solve_ineq(1 - 2*s1 <= 1/2 - 1/2*s1)
Out[68]:
[[s1 >= (1/3)]]
In [69]:
bool(1/2 - s1/2 <= 1 - s1)
Out[69]:
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 $s_2$

In [70]:
solve_ineq(0 <= 1 - 2*s1) #true
Out[70]:
[[s1 <= (1/2)]]
In [71]:
solve_ineq(1/4 - 3/4*s1 <= 1 - 2*s1) #true
Out[71]:
[[s1 <= (3/5)]]
In [72]:
bool(3/5 >= 1/2)
Out[72]:
True
In [73]:
solve_ineq(1/3 - s1 <= 1 - 2*s1) #true
Out[73]:
[[s1 <= (2/3)]]
In [74]:
bool(1/2 - 3/2*s1 <= 1 - 2*s1)
Out[74]:
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*}

In [75]:
bool(1/2 - s1/2 <= 1 - s1)
Out[75]:
True

prüfe Grenzen für $s_2$

In [76]:
bool(0 <= 1/2 - 1/2*s1)
Out[76]:
True
In [77]:
bool(1/4 - 3/4*s1 <= 1/2 - 1/2*s1)
Out[77]:
True
In [78]:
bool(1/3 - s1 <= 1/2 - s1/2)
Out[78]:
True
In [79]:
bool(1/2 - s1 <= 1/2 - s1/2)
Out[79]:
True
In [80]:
bool(1/2 - 3/2*s1 <= 1/2 - 1/2*s1)
Out[80]:
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 $s_2$

In [81]:
solve_ineq(1/4 - 3/4 *s1 <= 0) #true
Out[81]:
[[s1 >= (1/3)]]
In [82]:
solve_ineq(1/3 - s1 <= 0) #true
Out[82]:
[[s1 >= (1/3)]]
In [83]:
solve_ineq(1/2 - s1 <= 0) #true
Out[83]:
[[s1 >= (1/2)]]
In [84]:
solve_ineq(1/2 - 3/2*s1 <= 0) #true
Out[84]:
[[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 $s_2$:

In [85]:
bool(1/3 - 2/3*s1 <= 1/2 - 1/2*s1)
Out[85]:
True
In [86]:
bool(1/3 - 2/3*s1 <= 1 - s1)
Out[86]:
True

Prüfe Grenzen für $s_2$:

In [87]:
solve_ineq(0 <= 1/3 - 2/3*s1) #true
Out[87]:
[[s1 <= (1/2)]]
In [88]:
bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)
Out[88]:
True
In [89]:
bool(1/3 - s1 <= 1/3 - 2/3 *s1)
Out[89]:
True
In [90]:
solve_ineq(1/2 - 3/2*s1 <= 1/3 - 2/3 *s1)
Out[90]:
[[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 $s_2$:

In [91]:
solve_ineq(0 >= 1/4 - 3/4*s1)
Out[91]:
[[s1 >= (1/3)]]
In [92]:
solve_ineq(0 >= 1/3 - s1)
Out[92]:
[[s1 >= (1/3)]]
In [93]:
solve_ineq(0 >= 1/2 - 3/2*s1)
Out[93]:
[[s1 >= (1/3)]]
In [94]:
solve_ineq(1/2 - 3/2*s1 >= 0)
Out[94]:
[[s1 <= (1/3)]]
In [95]:
solve_ineq(1/2 - 3/2*s1 >= 1/4 - 3/4*s1)
Out[95]:
[[s1 <= (1/3)]]
In [96]:
solve_ineq(1/2 - 3/2*s1 >= 1/3 - s1)
Out[96]:
[[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 $s_2$:

In [97]:
solve_ineq(1/2 - 1/2*s1 <= 1 - 2*s1)
Out[97]:
[[s1 <= (1/3)]]
In [98]:
bool(0 <= 1/2 - 1/2*s1)
Out[98]:
True
In [99]:
solve_ineq(0 <= 1 - 2*s1) #true
Out[99]:
[[s1 <= (1/2)]]
In [100]:
solve_ineq(1/4 - 3/4*s1 <= 1 - 2*s1) #true (1/2 <= 3/5)
Out[100]:
[[s1 <= (3/5)]]
In [101]:
bool(1/4 - 3/4*s1 <= 1/2 - 1/2*s1)
Out[101]:
True
In [102]:
solve_ineq(1/3 - s1 <= 1- 2*s1) #true (1/2 <= 2/3)
Out[102]:
[[s1 <= (2/3)]]
In [103]:
bool(1/3 - s1 <= 1/2 - 1/2*s1)
Out[103]:
True
In [104]:
solve_ineq(1/3 - 2/3*s1 <= 1 - 2*s1) #true
Out[104]:
[[s1 <= (1/2)]]
In [105]:
bool(1/3 - 2/3*s1 <= 1/2 - 1/2*s1)
Out[105]:
True
In [106]:
bool(1/2 - 3/2*s1 <= 1 - 2*s1)
Out[106]:
True
In [107]:
bool(1/2 - 3/2*s1 <= 1/2 - 1/2*s1)
Out[107]:
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 $s_2$:

In [108]:
solve_ineq(0 <= 1/3 - 2/3*s1) #true
Out[108]:
[[s1 <= (1/2)]]
In [109]:
bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)
Out[109]:
True
In [110]:
bool(1/3 - s1 <= 1/3 - 2/3*s1)
Out[110]:
True
In [111]:
solve_ineq(1/2 - 3/2*s1 <= 1/3 - 2/3*s1) #true
Out[111]:
[[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 $s_2$:

In [112]:
solve_ineq(0 <= 1/3 - 2/3*s1) #true
Out[112]:
[[s1 <= (1/2)]]
In [113]:
bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)
Out[113]:
True
In [114]:
bool(1/3 - s1 <= 1/3 - 2/3*s1)
Out[114]:
True
In [115]:
solve_ineq(0 <= 1/2 - 3/2*s1) #true
Out[115]:
[[s1 <= (1/3)]]
In [116]:
solve_ineq(1/4 - 3/4*s1 <= 1/2 - 3/2*s1) #true
Out[116]:
[[s1 <= (1/3)]]
In [117]:
solve_ineq(1/3 - s1 <= 1/2 - 3/2*s1) #true
Out[117]:
[[s1 <= (1/3)]]
In [118]:
solve_ineq(1/3 - 2/3*s1 <= 1/2 - 3/2*s1)
Out[118]:
[[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:

In [119]:
solve_ineq(1 - 2*s1 <= 1/2 - 3/2*s1)
Out[119]:
[[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:

In [120]:
solve_ineq(1/2 - s1 <= 1/2 - 3/2*s1) #false
Out[120]:
[[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 $s_2$:

In [121]:
bool( 1/3 - 2/3*s1 <= 1 - s1)
Out[121]:
True
In [122]:
solve_ineq(1/3 - 2/3*s1 <= 1/2 - 3/2*s1)
Out[122]:
[[s1 <= (1/5)]]

Prüfe Grenzen für $s_2$:

In [123]:
solve_ineq(0 <= 1/3 - 2/3*s1) #true
Out[123]:
[[s1 <= (1/2)]]
In [124]:
bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)
Out[124]:
True
In [125]:
bool(1/3 - s1 <= 1/3 - 2/3*s1)
Out[125]:
True
In [126]:
solve_ineq(0 <= 1/2 - 3/2*s1)
Out[126]:
[[s1 <= (1/3)]]
In [127]:
solve_ineq(1/4 - 3/4*s1 <= 1/2 - 3/2*s1)
Out[127]:
[[s1 <= (1/3)]]
In [128]:
solve_ineq(1/3 - s1 <= 1/2 - 3/2*s1)
Out[128]:
[[s1 <= (1/3)]]

Bestimme untere Grenze für $s_2$:

In [129]:
solve_ineq(0 <= 1/3 - s1) #true
Out[129]:
[[s1 <= (1/3)]]
In [130]:
solve_ineq(1/4 - 3/4*s1 <= 1/3 - s1) #true
Out[130]:
[[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 $s_2$:

In [131]:
bool(1/2 - 3/2*s1 <= 1 - s1)
Out[131]:
True
In [132]:
bool(1/2 - 3/2*s1 <= 1 - 2*s1)
Out[132]:
True

Prüfe Grenzen für $s_2$:

In [133]:
solve_ineq(0 <= 1/2 - 3/2*s1)
Out[133]:
[[s1 <= (1/3)]]
In [134]:
solve_ineq(1/4 - 3/4*s1 <= 1/2 - 3/2*s1)
Out[134]:
[[s1 <= (1/3)]]
In [135]:
solve_ineq(1/3 - s1 <= 1/2 - 3/2*s1)
Out[135]:
[[s1 <= (1/3)]]
In [136]:
solve_ineq(1/3 - 2/3*s1 <= 1/2 - 3/2*s1)
Out[136]:
[[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 $s_2$:

In [137]:
solve_ineq(0 <= 1/3 - 2/3*s1) #true
Out[137]:
[[s1 <= (1/2)]]
In [138]:
bool(1/4 - 3/4*s1 <= 1/3 - 2/3*s1)
Out[138]:
True
In [139]:
bool(1/3 - s1 <= 1/3 - 2/3*s1)
Out[139]:
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*}

In [140]:
forget()
var('s0,s1,s2,s3,c')
Out[140]:
(s0, s1, s2, s3, c)
In [141]:
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 $s_1 = 1 - s_0 - s_2 - s_3$

In [142]:
s1 = 1 - s0 - s2 - s3

Forme Ungleichungen, die $s_2$ enthalten, um:

In [143]:
solve_ineq([3*s0+3*s1+2*s2+s3 <= c],[c,s3,s2,s0])
Out[143]:
[[c == -s2 - 2*s3 + 3], [-s2 - 2*s3 + 3 < c]]
In [144]:
bool(s1 <= 1)
Out[144]:
True
In [145]:
solve_ineq([0 <= s1],[s3,s2,s0,c])
Out[145]:
[[s3 == -s0 - s2 + 1], [s3 < -s0 - s2 + 1]]
In [146]:
solve_ineq([s0 + s1 <= s3],[s3,s2,s0,c])
Out[146]:
[[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:

In [147]:
bool(1 - s0 - s2 <= 1)
Out[147]:
True
In [148]:
bool(0 <= 1/2 - 1/2*s2)
Out[148]:
True
In [149]:
solve_ineq([1/2 - 1/2*s2 <= 1 - s0 - s2],[s2,s0])
Out[149]:
[[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*}

In [150]:
bool(1 - 2*s0 <= 1)
Out[150]:
True
In [151]:
solve_ineq(s0 <= 1 - 2*s0)
Out[151]:
[[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:

In [152]:
solve_ineq([3 - s2 - 2*s3 <= c], [s3, s2, c])
Out[152]:
[[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([3/2 - 1/2*c - 1/2*s2 <= 1 - s0 - s2],[s2,s0,c])
In [153]:
solve_ineq([s0 <= c - 2*s0 - 1],[s0,c])
Out[153]:
[[s0 == 1/3*c - 1/3], [s0 < 1/3*c - 1/3]]
In [154]:
solve_ineq([0 <= 1/3*c - 1/3])
Out[154]:
[[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:

In [155]:
solve_ineq([1/2 - 1/2*s2 >= 3/2 - 1/2*c - 1/2*s2],[s2,s0,c])
Out[155]:
[[c == 2], [2 < c]]
In [156]:
solve_ineq([1-2*s0 <= c - 2*s0 - 1],[s0,c])
Out[156]:
[[c == 2], [2 < c]]
In [157]:
solve_ineq([1/3 <= 1/3*c - 1/3])
Out[157]:
[[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 $s_2 = 1 - s_0 - s_1 - s_3$

In [158]:
forget()
var('s0,s1,s2,s3,c')
Out[158]:
(s0, s1, s2, s3, c)
In [159]:
assume(0 <= s0)
assume(0 <= s1)
assume(0 <= s2)
assume(0 <= s3)
assume(s0 <= 1)
assume(s1 <= 1)
assume(s2 <= 1)
assume(s3 <= 1)
In [160]:
s2 = 1 - s0 - s1 - s3

Forme Ungleichungen, die $s_2$ enthalten, um:

In [161]:
s0 + 3*s1 + 3*s2 +2*s3 <= c
Out[161]:
-2*s0 - s3 + 3 <= c
In [162]:
solve_ineq([s1 + s2 <= s0], [s0,s3,s1])
Out[162]:
[[s0 == -1/2*s3 + 1/2], [-1/2*s3 + 1/2 < s0]]
In [163]:
bool(s2 <= 1)
Out[163]:
True
In [164]:
solve_ineq([0 <= s2],[s0,s3,s1])
Out[164]:
[[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*}

In [165]:
bool(1 - s1 - s3 <= 1)
Out[165]:
True

Überprüfe neue Grenzen und daraus erhaltene neue Grenzen:

In [166]:
solve_ineq([s3 <= 1 - s1 - s3],[s3,s1])
Out[166]:
[[s3 == -1/2*s1 + 1/2], [s3 < -1/2*s1 + 1/2]]
In [167]:
solve_ineq(s1 <= 1/2 - 1/2*s1)
Out[167]:
[[s1 <= (1/3)]]
In [168]:
solve_ineq([1/2 - 1/2*s3 <= 1 - s1 - s3 ], [s3,s1])
Out[168]:
[[s3 == -2*s1 + 1], [s3 < -2*s1 + 1]]
In [169]:
solve_ineq([s1 <= 1 - 2*s1])
Out[169]:
[[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*}

In [170]:
solve_ineq( 1/2 - 1/2*s3 >= s3) # --> Fall 1 und Fall 2
Out[170]:
[[s3 <= (1/3)]]
In [171]:
solve_ineq(1/2 -1/2*s1 <= 1 - 2*s1) #true
Out[171]:
[[s1 <= (1/3)]]
In [172]:
solve_ineq(1/3 <= 1/2 - 1/2*s1) #true
Out[172]:
[[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:

In [173]:
solve_ineq([3 - 2*s0 - s3 <= c], [s0, s3, s1, c])
Out[173]:
[[s0 == -1/2*c - 1/2*s3 + 3/2], [-1/2*c - 1/2*s3 + 3/2 < s0]]

Prüfe Grenzen:

In [174]:
solve_ineq([3/2 - 1/2*c - 1/2*s3 <= 1 - s1 - s3],[s3,s1,c])
Out[174]:
[[s3 == c - 2*s1 - 1], [s3 < c - 2*s1 - 1]]
In [175]:
solve_ineq([s1 <= c - 2*s1 - 1],[s1,c])
Out[175]:
[[s1 == 1/3*c - 1/3], [s1 < 1/3*c - 1/3]]
In [176]:
solve_ineq(0 <= 1/3*c - 1/3)
Out[176]:
[[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 $s_0$:

In [177]:
solve_ineq([1/2 - 1/2*s3 >= 3/2 - 1/2*c - 1/2*s3],[s3,s1,c])
Out[177]:
[[c == 2], [2 < c]]

Bestimme obere Grenze für $s_3$:

In [178]:
solve_ineq([1/3 <= c - 2*s1 - 1],[s1,c]) # --> Fall 1.1 und Fall 1.2
Out[178]:
[[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:

In [179]:
solve_ineq(0 <= 1/2*c - 2/3)
Out[179]:
[[c >= (4/3)]]

Bestimme obere Grenze für $s_1$:

In [180]:
solve_ineq(1/3 <= 1/3*c - 1/3)
Out[180]:
[[c >= 2]]
In [181]:
solve_ineq(1/3 <= 1/2*c - 2/3)
Out[181]:
[[c >= 2]]
In [182]:
solve_ineq(1/3*c - 1/3 <= 1/2*c - 2/3)
Out[182]:
[[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:

In [183]:
solve_ineq(1/2*c - 2/3 <= 1/3)
Out[183]:
[[c <= 2]]
In [184]:
solve_ineq(1/2*c - 2/3 <= 1/3*c - 1/3)
Out[184]:
[[c <= 2]]

Bestimme untere Grenze für $s_1$:

In [185]:
solve_ineq(1/2*c - 2/3 >= 0)
Out[185]:
[[c >= (4/3)]]

Bestimme obere Grenze für $s_1$:

In [186]:
solve_ineq(1/3*c - 1/3 <= 1/3)
Out[186]:
[[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:

In [187]:
solve_ineq([3 - 2*s0 - s3 <= c], [s0, s3, s1, c])
Out[187]:
[[s0 == -1/2*c - 1/2*s3 + 3/2], [-1/2*c - 1/2*s3 + 3/2 < s0]]

Prüfe die neue(n) Grenze(n):

In [188]:
solve_ineq([3/2 - 1/2*c - 1/2*s3 <= 1 - s1 - s3], [s3, s1, c])
Out[188]:
[[s3 == c - 2*s1 - 1], [s3 < c - 2*s1 - 1]]
In [189]:
solve_ineq([1/3 <= c - 2*s1 - 1], [s1, c])
Out[189]:
[[s1 == 1/2*c - 2/3], [s1 < 1/2*c - 2/3]]
In [190]:
solve_ineq(0 <= 1/2*c - 2/3)
Out[190]:
[[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 $s_0$:

In [191]:
solve_ineq([s3 >= 3/2 - 1/2*c - 1/2*s3], [s3,s1,c]) # --> Fall 2.1 und Fall 2.2
Out[191]:
[[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):

In [192]:
solve_ineq([1 - 1/3*c <= 1/2 - 1/2*s1],[s1,c])
Out[192]:
[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]
In [193]:
solve_ineq([1 - 1/3*c <= c - 2*s1 - 1], [s1,c])
Out[193]:
[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]
In [194]:
solve_ineq(0 <= 2/3*c - 1)
Out[194]:
[[c >= (3/2)]]
In [195]:
bool(3/2 >= 4/3)
Out[195]:
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 $s_3$:

In [196]:
solve_ineq(1/3 >= 1 - 1/3*c)
Out[196]:
[[c >= 2]]
In [197]:
solve_ineq([1/2 - 1/2*s1 <= c - 2*s1 - 1], [s1, c]) #true
Out[197]:
[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]

obere Grenze für $s_1$:

In [198]:
solve_ineq(1/3 <= 1/2*c - 2/3)
Out[198]:
[[c >= 2]]
In [199]:
solve_ineq(1/3 <= 2/3*c - 1)
Out[199]:
[[c >= 2]]
In [200]:
solve_ineq(2/3*c -1 <= 1/2*c - 2/3)
Out[200]:
[[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:

In [201]:
solve_ineq(1/3 <= 1 - 1/3*c)
Out[201]:
[[c <= 2]]

Bestimme Grenze für $s_3$:

In [202]:
solve_ineq([c - 2*s1 - 1 <= 1 - 1/3*c], [s1, c])
Out[202]:
[[s1 == 2/3*c - 1], [2/3*c - 1 < s1]]
In [203]:
solve_ineq([c - 2*s1 - 1 <= 1/2 - 1/2*s1], [s1, c])
Out[203]:
[[s1 == 2/3*c - 1], [2/3*c - 1 < s1]]
In [204]:
solve_ineq([1 - 1/3*c <= c - 2*s1 - 1], [s1, c])
Out[204]:
[[s1 == 2/3*c - 1], [s1 < 2/3*c - 1]]
In [205]:
solve_ineq([1 - 1/3*c <= 1/2 - 1/2*s1], [s1,c]) #--> Fall 2.2.1 und Fall 2.2.2
Out[205]:
[[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:

In [206]:
solve_ineq(2/3*c - 1 <= 1/3) #true
Out[206]:
[[c <= 2]]
In [207]:
solve_ineq(2/3*c - 1 <= 1/2*c - 2/3) #true
Out[207]:
[[c <= 2]]

Grenzen für $s_1$:

In [208]:
solve_ineq(2/3*c - 1 >= 0)
Out[208]:
[[c >= (3/2)]]
In [209]:
bool(3/2 > 4/3)
Out[209]:
True
In [210]:
solve_ineq(1/2*c - 2/3 <= 1/3) #true
Out[210]:
[[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:

In [211]:
solve_ineq(0 <= 2/3*c - 1)
Out[211]:
[[c >= (3/2)]]

Bestimme Grenzen für $s_1$:

In [212]:
solve_ineq(2/3*c - 1 <= 1/3)
Out[212]:
[[c <= 2]]
In [213]:
solve_ineq(2/3*c - 1 <= 1/2*c - 2/3)
Out[213]:
[[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*}