Book a Demo!
CoCalc Logo Icon
StoreFeaturesDocsShareSupportNewsAboutPoliciesSign UpSign In
Download
193 views
Tweet
Share
Share
Kernel: SageMath (stable)
Melanie Siebenhofer, Jänner 2018

Asymptotische Grenzverteilung

alt text

Baum 1

s0+s1+s2+s3=10≤s0≤10≤s1≤10≤s2≤10≤s3≤1 2⋅s0+2⋅s1+2⋅s2+2⋅s3≤cs2+s3+1≥s0s1+s2+1≥s0s1+s2+1≥s3s0+s1+1≥s3\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: s0+s1+s2+s3=10≤s0≤10≤s1≤10≤s2≤10≤s3≤1 2⋅s0+2⋅s1+2⋅s2+2⋅s3≤cs0≤s2+s3s0≤s1+s2s3≤s1+s2s3≤s0+s1\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 s0=1−s1−s2−s3s_0 = 1 - s_1 - s_2 - s_3

s0 = 1 - s1 - s2 - s3

forme Ungleichungen, die s0s_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]]
2≤cs0=1−s1−s2−s30≤1−s1−s2−s3≤10≤s1≤10≤s2≤10≤s3≤112−12s1−s2≤s31−2s1−2s2≤s3s3≤s1+s2s3≤12−12s2\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
2≤cs0=1−s1−s2−s30≤s1≤10≤s2≤10≤s3≤112−12s1−s2≤s31−2s1−2s2≤s3s3≤s1+s2s3≤12−12s2s3≤1−s1−s2\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 s3s_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
2≤cs0=1−s1−s2−s30≤s1≤10≤s2≤1−s114−34s1≤s213−s1≤s213−43s1≤s20≤s3≤1−s1−s212−12s1−s2≤s31−2s1−2s2≤s3s3≤s1+s2s3≤12−12s2\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 s2s_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
2≤cs0=1−s1−s2−s30≤s1≤10≤s2≤1−s114−34s1≤s213−s1≤s20≤s3≤1−s1−s212−12s1−s2≤s31−2s1−2s2≤s3s3≤s1+s2s3≤12−12s2\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 \quad1−s1−s2\mathbf{1-s_1-s_2}s1+s2\mathbf{s_1 + s_2}12−12s2\mathbf{\frac{1}{2} - \frac{1}{2} s_2 \quad}
1−s1−s2\mathbf{1 - s_1 - s_2}immer \qquad \qquad12−s1≤s2{\frac{1}{2}-s_1\leq s_2}1−2s1≤s2{1 - 2s_1 \leq s_2}
s1+s2\mathbf{s_1 + s_2}s2≤12−s1{s_2 \leq \frac{1}{2} - s_1}immer \qquad \qquads2≤13−23s1{s_2 \leq \frac{1}{3}-\frac{2}{3}s_1}
12−12s2\mathbf{\frac{1}{2}-\frac{1}{2}s_2}s2≤1−2s1{s_2 \leq 1 - 2s_1 }13−23s1≤s2{\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 \quadim Fall \qquad \qquad \qquad \qquad \quad
1−s1−s2\mathbf{1-s_1-s_2}1−2s1≤s2{1-2s_1 \leq s_2} und s1≤12s_1 \leq \frac{1}{2}
1−s1−s2\mathbf{1-s_1-s_2}12−s1≤s2{\frac{1}{2}-s_1 \leq s_2} und 12≤s1\frac{1}{2} \leq s_1
s1+s2\mathbf{s_1 + s_2}s2≤13−23s1{s_2 \leq \frac{1}{3} - \frac{2}{3}s_1} und s1≤12{s_1 \leq \frac{1}{2}}
s1+s2\mathbf{s_1 + s_2}s2≤12−s1{s_2 \leq \frac{1}{2} - s_1} und 12≤s1{\frac{1}{2} \leq s_1}
12−12s2\mathbf{\frac{1}{2} - \frac{1}{2}s_2}13−23s1≤s2≤1−2s1{\frac{1}{3}-\frac{2}{3}s_1 \leq s_2 \leq 1 - 2s_1} und s1≤12s_1 \leq \frac{1}{2}

Fall 4 ergibt eine Nullmenge, da 0≤s20 \leq s_2 und damit:

solve_ineq(0 <= 1/2 - s1)
[[s1 <= (1/2)]]

neue obere Grenzen:

obere Grenze \quadim Fall \qquad \qquad \qquad \qquad \quad
1−s1−s2\mathbf{1-s_1-s_2}1−2s1≤s2{1-2s_1 \leq s_2} und s1≤12s_1 \leq \frac{1}{2}
1−s1−s2\mathbf{1-s_1-s_2}12−s1≤s2{\frac{1}{2}-s_1 \leq s_2} und 12≤s1\frac{1}{2} \leq s_1
s1+s2\mathbf{s_1 + s_2}s2≤13−23s1{s_2 \leq \frac{1}{3} - \frac{2}{3}s_1} und s1≤12{s_1 \leq \frac{1}{2}}
12−12s2\mathbf{\frac{1}{2} - \frac{1}{2}s_2}13−23s1≤s2≤1−2s1{\frac{1}{3}-\frac{2}{3}s_1 \leq s_2 \leq 1 - 2s_1} und s1≤12s_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 \qquad0\mathbf{0}12−12s1−s2\mathbf{\frac{1}{2} - \frac{1}{2}s_1 - s_2}1−2s1−2s2\mathbf{1 - 2s_1 - 2s_2}
0\mathbf{0}immer \qquad \qquad12−12s1≤s2{\frac{1}{2}-\frac{1}{2}s_1 \leq s_2}12−s1≤s2{\frac{1}{2}-s_1\leq s_2}
12−12s1−s2\mathbf{\frac{1}{2} - \frac{1}{2}s_1 - s_2}s2≤12−12s1{s_2 \leq \frac{1}{2}-\frac{1}{2}s_1}immer \qquad \qquad12−32s1≤s2{\frac{1}{2}-\frac{3}{2}s_1 \leq s_2}
1−2s1−2s2\mathbf{1 - 2s_1 - 2s_2}s2≤12−s1{s_2 \leq \frac{1}{2} - s_1}s2≤12−32s1{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 \quadim Fall \qquad \qquad \qquad \qquad \quad
0\mathbf{0}12−12s1≤s2{\frac{1}{2}-\frac{1}{2}s_1 \leq s_2}
12−12s1−s2\mathbf{\frac{1}{2} - \frac{1}{2}s_1 - s_2}12−32s1≤s2≤12−12s1{\frac{1}{2}-\frac{3}{2}s_1 \leq s_2 \leq \frac{1}{2}-\frac{1}{2}s_1}
1−2s1−2s2\mathbf{1 - 2s_1 - 2s_2}s2≤12−32s1{s_2 \leq \frac{1}{2}-\frac{3}{2}s_1}

Fall 1

untere Grenze: 00

obere Grenze: 1−s1−s21 - s_1 - s_2 und s1≤12s_1 \leq \frac{1}{2}

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s11−2⋅s1≤s2(neu)12−12s1≤s2(neu)14−34s1≤s213−s1≤s20≤s3≤1−s1−s2\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 s2s_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
2≤cs0=1−s1−s2−s30≤s1≤121−2⋅s1≤s2≤1−s112−12s1≤s20≤s3≤1−s1−s2\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)]]
2≤cs0=1−s1−s2−s30≤s1≤131−2⋅s1≤s2≤1−s10≤s3≤1−s1−s2\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*}
2≤cs0=1−s1−s2−s313≤s1≤1212−12⋅s1≤s2≤1−s10≤s3≤1−s1−s2\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: 00

obere Grenze: 1−s1−s21 - s_1 - s_2 und 12≤s1\frac{1}{2} \leq s_1

2≤cs0=1−s1−s2−s312≤s1≤1(neu)0≤s2≤1−s112−s1≤s2(neu)12−12s1≤s2(neu)14−34s1≤s213−s1≤s20≤s3≤1−s1−s2\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 s2s_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
2≤cs0=1−s1−s2−s312≤s1≤112−12s1≤s2≤1−s10≤s3≤1−s1−s2\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: 00

obere Grenze: s1+s2s_1 + s_2

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s1s2≤13−23s1(neu)12−12s1≤s2(neu)14−34s1≤s213−s1≤s20≤s3≤s1+s2\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 s2s_2 gilt:

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

Fall 4

untere Grenze: 00

obere Grenze: 12−12s2\frac{1}{2} - \frac{1}{2}s_2

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s113−23s1≤s2≤1−2s1(neu)12−12s1≤s2(neu)14−34s1≤s213−s1≤s20≤s3≤12−12s2\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 s2s_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)]]
2≤cs0=1−s1−s2−s30≤s1≤130≤s2≤1−2s113−23s1≤s212−12s1≤s214−34s1≤s213−s1≤s20≤s3≤12−12s2\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 s2s_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
2≤cs0=1−s1−s2−s30≤s1≤1312−12s1≤s2≤1−2s10≤s3≤12−12s2\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: 12−12s1−s2\frac{1}{2} - \frac{1}{2}s_1 - s_2

obere Grenze: 1−s1−s21 - s_1 - s_2 und s1≤12s_1 \leq \frac{1}{2}

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s114−34s1≤s213−s1≤s212−32s1≤s2≤12−12s1(neu)1−2s1≤s2(neu)12−12s1−s2≤s3≤1−s1−s2\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 s2s_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
2≤cs0=1−s1−s2−s313≤s1≤120≤s2≤12−12s114−34s1≤s213−s1≤s212−32s1≤s21−2s1≤s212−12s1−s2≤s3≤1−s1−s2\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 s2s_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
2≤cs0=1−s1−s2−s313≤s1≤121−2s1≤s2≤12−12s112−12s1−s2≤s3≤1−s1−s2\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: 12−12s1−s2\frac{1}{2} - \frac{1}{2}s_1 - s_2

obere Grenze: 1−s1−s21 - s_1 - s_2 und 12≤s1\frac{1}{2} \leq s_1

2≤cs0=1−s1−s2−s312≤s1≤1(neu)0≤s2≤1−s114−34s1≤s213−s1≤s212−s1≤s2(neu)12−32s1≤s2≤12−12s1(neu)12−12s1−s2≤s3≤1−s1−s2\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 s2s_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
2≤cs0=1−s1−s2−s312≤s1≤10≤s2≤12−12s114−34s1≤s213−s1≤s212−s1≤s212−32s1≤s212−12s1−s2≤s3≤1−s1−s2\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 s2s_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)]]
2≤cs0=1−s1−s2−s312≤s1≤10≤s2≤12−12s112−12s1−s2≤s3≤1−s1−s2\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: 12−12s1−s2\frac{1}{2} - \frac{1}{2}s_1 - s_2

obere Grenze: s1+s2s_1 + s_2

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s114−34s1≤s213−s1≤s212−32s1≤s2≤12−12s1(neu)s2≤13−23s1(neu)12−12s1−s2≤s3≤s1+s2\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 s2s_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 s2s_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)]]
2≤cs0=1−s1−s2−s315≤s1≤120≤s2≤13−23s114−34s1≤s213−s1≤s212−32s1≤s212−12s1−s2≤s3≤s1+s2\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 s2s_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)]]
2≤cs0=1−s1−s2−s313≤s1≤120≤s2≤13−23s112−12s1−s2≤s3≤s1+s2\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*}
2≤cs0=1−s1−s2−s315≤s1≤1312−32s1≤s2≤13−23s112−12s1−s2≤s3≤s1+s2\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: 12−12s1−s2\frac{1}{2} - \frac{1}{2}s_1 - s_2

obere Grenze: 12−12s2\frac{1}{2} - \frac{1}{2}s_2

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s114−34s1≤s213−s1≤s213−23s1≤s2≤1−2s1(neu)12−32s1≤s2≤12−12s1(neu)12−12s1−s2≤s3≤12−12s2\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 s2s_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 1−2s11 - 2s_1): 2≤cs0=1−s1−s2−s313≤s1≤120≤s2≤1−2s114−34s1≤s213−s1≤s213−23s1≤s212−32s1≤s212−12s1−s2≤s3≤12−12s2\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 s2s_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)]]
2≤cs0=1−s1−s2−s313≤s1≤1213−23s1≤s2≤1−2s112−12s1−s2≤s3≤12−12s2\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 12−12s1\frac{1}{2} - \frac{1}{2} s_1): 2≤cs0=1−s1−s2−s30≤s1≤130≤s2≤12−12s114−34s1≤s213−s1≤s213−23s1≤s212−32s1≤s212−12s1−s2≤s3≤12−12s2\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 s2s_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)]]
2≤cs0=1−s1−s2−s30≤s1≤1512−32s1≤s2≤12−12s112−12s1−s2≤s3≤12−12s2\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*}

--- 2≤cs0=1−s1−s2−s315≤s1≤1313−23s1≤s2≤12−12s112−12s1−s2≤s3≤12−12s2\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: 1−2s1−2s21 - 2s_1 - 2s_2

obere Grenze: 1−s1−s21 - s_1 - s_2 und s1≤12s_1 \leq \frac{1}{2}

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s114−34s1≤s213−s1≤s21−2s1≤s2(neu)s2≤12−32s1(neu)1−2s1−2s2≤s3≤1−s1−s2\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 s2s_2 gilt:

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

Fall 10

untere Grenze: 1−2s1−2s21 - 2s_1 - 2s_2

obere Grenze: 1−s1−s21 - s_1 - s_2 und 12≤s1\frac{1}{2} \leq s_1

2≤cs0=1−s1−s2−s312≤s1≤1(neu)0≤s2≤1−s114−34s1≤s213−s1≤s212−s1≤s2(neu)s2≤12−32s1(neu)1−2s1−2s2≤s3≤1−s1−s2\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 s2s_2 gilt:

solve_ineq(1/2 - s1 <= 1/2 - 3/2*s1) #false
[[s1 <= 0]]

Fall 11

untere Grenze: 1−2s1−2s21 - 2s_1 - 2s_2

obere Grenze: s1+s2s_1 + s_2

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s114−34s1≤s213−s1≤s2s2≤13−23s1(neu)s2≤12−32s1(neu)1−2s1−2s2≤s3≤s1+s2\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 s2s_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 s2s_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 s2s_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)]]
2≤cs0=1−s1−s2−s315≤s1≤1313−s1≤s2≤12−32s11−2s1−2s2≤s3≤s1+s2\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*}
2≤cs0=1−s1−s2−s30≤s1≤1513−s1≤s2≤13−23s11−2s1−2s2≤s3≤s1+s2\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: 1−2s1−2s21 - 2s_1 - 2s_2

obere Grenze: 12−12s2\frac{1}{2}-\frac{1}{2}s_2

2≤cs0=1−s1−s2−s30≤s1≤12(neu)0≤s2≤1−s114−34s1≤s213−s1≤s213−23s1≤s2≤1−2s1(neu)s2≤12−32s1(neu)1−2s1−2s2≤s3≤12−12s2\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 s2s_2:

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

Prüfe Grenzen für s2s_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)]]
2≤cs0=1−s1−s2−s30≤s1≤150≤s2≤12−32s114−34s1≤s213−s1≤s213−23s1≤s21−2s1−2s2≤s3≤12−12s2\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 s2s_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
2≤cs0=1−s1−s2−s30≤s1≤1513−23s1≤s2≤12−32s11−2s1−2s2≤s3≤12−12s2\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)
s0+s1+s2+s3=10≤s0≤10≤s1≤10≤s2≤10≤s3≤1 3⋅s0+3⋅s1+2⋅s2+s3≤cs3≥s0+s1+1s2≥s0\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:

s0+s1+s2+s3=10≤s0≤10≤s1≤10≤s2≤10≤s3≤1 3⋅s0+3⋅s1+2⋅s2+s3≤cs0+s1≤s3s0≤s2\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 s1=1−s0−s2−s3s_1 = 1 - s_0 - s_2 - s_3

s1 = 1 - s0 - s2 - s3

Forme Ungleichungen, die s2s_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]]
s1=1−s0−s2−s33−s2−2⋅s3≤c0≤s0≤1s0≤s2≤10≤s3≤112−12⋅s2≤s3s3≤1−s0−s2\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]]
s1=1−s0−s2−s33−s2−2⋅s3≤c0≤s0≤1s0≤s2≤1s2≤1−2⋅s012−12⋅s2≤s3≤1−s0−s2\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)]]
s1=1−s0−s2−s33−s2−2⋅s3≤c0≤s0≤13s0≤s2≤1−2⋅s012−12⋅s2≤s3≤1−s0−s2\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 s3s_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]]
s1=1−s0−s2−s30≤s0≤13s0≤s2≤1−2⋅s012−12⋅s2≤s3≤1−s0−s232−12c−12s2≤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])

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]]
s1=1−s0−s2−s31≤c0≤s0≤13s0≤13c−13s0≤s2≤1−2⋅s0s2≤c−2s0−112−12⋅s2≤s3≤1−s0−s232−12c−12s2≤s3\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]]
s1=1−s0−s2−s31≤c≤20≤s0≤13c−13s0≤s2≤c−2s0−132−12c−12s2≤s3≤1−s0−s2\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*}
s1=1−s0−s2−s32≤c0≤s0≤13s0≤s2≤1−2⋅s012−12⋅s2≤s3≤1−s0−s2\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)
s0+s1+s2+s3=10≤s0≤10≤s1≤10≤s2≤10≤s3≤1 s0+3⋅s1+3⋅s2+2⋅s3≤cs0≥s1+s2+1s3≥s1s0≥s3\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: s0+s1+s2+s3=10≤s0≤10≤s1≤10≤s2≤10≤s3≤1 s0+3⋅s1+3⋅s2+2⋅s3≤cs1+s2≤s0s1≤s3s3≤s0\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 s2=1−s0−s1−s3s_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 s2s_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]]
s2=1−s0−s1−s33−2⋅s0−s3≤c0≤s1≤1s1≤s3≤1s3≤s0≤1s0≤1−s1−s312−12⋅s3≤s0\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)]]
s2=1−s0−s1−s33−2⋅s0−s3≤c0≤s1≤13s1≤s3≤1−2s1s3≤12−12s1s3≤s0≤1−s1−s312−12⋅s3≤s0\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

(s3≤13)(s_3 \leq \frac{1}{3}) s2=1−s0−s1−s33−2⋅s0−s3≤c0≤s1≤13s1≤s3≤1312−12⋅s3≤s0≤1−s1−s3\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 s0s_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]]
s2=1−s0−s1−s31≤c0≤s1≤13s1≤13c−13s1≤s3≤13s3≤c−2s1−112−12⋅s3≤s0≤1−s1−s332−12c−12s3≤s0\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 s0s_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 s3s_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

s2=1−s0−s1−s31≤c0≤s1≤13s1≤13c−13s1≤12c−23(neu)s1≤s3≤1312−12⋅s3≤s0≤1−s1−s332−12c−12s3≤s0\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 s1s_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]]
s2=1−s0−s1−s343≤c≤20≤s1≤12c−23s1≤s3≤1332−12c−12s3≤s0≤1−s1−s3\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*}
s2=1−s0−s1−s32≤c0≤s1≤13s1≤s3≤1312−12⋅s3≤s0≤1−s1−s3\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

s2=1−s0−s1−s31≤c0≤s1≤13s1≤13c−1312c−23≤s1(neu)s1≤s3≤c−2s1−112−12⋅s3≤s0≤1−s1−s332−12c−12s3≤s0\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 s1s_1:

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

Bestimme obere Grenze für s1s_1:

solve_ineq(1/3*c - 1/3 <= 1/3)
[[c <= 2]]
s2=1−s0−s1−s343≤c≤212c−23≤s1≤13c−13s1≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\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*}
s2=1−s0−s1−s31≤c≤430≤s1≤13c−13s1≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\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

(13≤s3)(\frac{1}{3} \leq s_3)

s2=1−s0−s1−s33−2⋅s0−s3≤c0≤s1≤13s1≤s3≤12−12s113≤s3s3≤s0≤1−s1−s3\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 s1≤13s_1 \leq \frac{1}{3} gilt: s2=1−s0−s1−s33−2⋅s0−s3≤c0≤s1≤1313≤s3≤12−12s1s3≤s0≤1−s1−s3\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 s0s_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)]]
s2=1−s0−s1−s343≤c0≤s1≤13s1≤12c−2313≤s3≤12−12s1s3≤c−2s1−1s3≤s0≤1−s1−s332−12c−12s3≤s0\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 s0s_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

s2=1−s0−s1−s343≤c0≤s1≤13s1≤12c−2313≤s3≤12−12s1s3≤c−2s1−11−13c≤s3(neu)s3≤s0≤1−s1−s3\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
s2=1−s0−s1−s332≤c0≤s1≤13s1≤12c−23s1≤23c−113≤s3≤12−12s1s3≤c−2s1−11−13c≤s3s3≤s0≤1−s1−s3\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 s3s_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 s1s_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]]
s2=1−s0−s1−s332≤c≤20≤s1≤23c−11−13c≤s3≤12−12s1s3≤s0≤1−s1−s3\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*}
s2=1−s0−s1−s32≤c0≤s1≤1313≤s3≤12−12s1s3≤s0≤1−s1−s3\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

s2=1−s0−s1−s343≤c0≤s1≤13s1≤12c−2313≤s3≤12−12s1s3≤c−2s1−1s3≤1−13c(neu)32−12c−12s3≤s0≤1−s1−s3\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 s3s_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 s2=1−s0−s1−s343≤c≤20≤s1≤13s1≤12c−2323c−1≤s1(neu)13≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\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 s1s_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]]
s2=1−s0−s1−s343≤c≤320≤s1≤12c−2313≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\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*}
s2=1−s0−s1−s332≤c≤223c−1≤s1≤12c−2313≤s3≤c−2s1−132−12c−12s3≤s0≤1−s1−s3\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 s2=1−s0−s1−s343≤c≤20≤s1≤13s1≤12c−23s1≤23c−1(neu)13≤s3≤1−13c32−12c−12s3≤s0≤1−s1−s3\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 s1s_1:

solve_ineq(2/3*c - 1 <= 1/3)
[[c <= 2]]
solve_ineq(2/3*c - 1 <= 1/2*c - 2/3)
[[c <= 2]]
s2=1−s0−s1−s332≤c≤20≤s1≤23c−113≤s3≤1−13c32−12c−12s3≤s0≤1−s1−s3\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*}