Berechnbarkeit: Subset Sum...

main
Thomas Ba. 14 years ago
parent 842faf5c33
commit 12603740ad

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\usepackage[ngerman]{babel} \usepackage[ngerman]{babel}
\usepackage{amsmath} \usepackage{amsmath}
\usepackage{amssymb} \usepackage{amssymb}
\usepackage{marvosym} % Lightning
\usepackage{multicol} \usepackage{multicol}
\usepackage{framed} \usepackage{framed}
\usepackage{enumerate} \usepackage{enumerate}
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]{hyperref} ]{hyperref}
\usepackage{tabularx} \usepackage{tabularx}
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Gefragt: \( \exists D \subseteq V, \left|D\right|\le k\)\\ Gefragt: \( \exists D \subseteq V, \left|D\right|\le k\)\\
so dass für jeden Knoten $v\in V$ gilt: $v$ oder ein Nachbar von $v$ ist in $D$. so dass für jeden Knoten $v\in V$ gilt: $v$ oder ein Nachbar von $v$ ist in $D$.
\( VC \le^P DS:\)\\
Sei $G,k$ eine Eingabe für $VC$.\\
Ziel: konstruiere $G'$, so dass $(G,K)\in VC \Leftrightarrow (G',k) \in DS$
\includegraphics{bilder/ds-vc.eps}
Für jede Kante $(u,v)$ in $G$ führe neuen Knoten $K_{(u,v)}$ ein und Kanten $(u,K_{(u,v)})$ und $(v,K_{(u,v)})$ (ungerichtet).
\('\Rightarrow'\) Sei $C$ $VC$ in $G$, d.\,h. für jede Kante $(u,v) \in E$ ist $u\in C$ oder $v\in C$.
\(\Rightarrow\) der neue Knoten $K_{u,v}$ hat einen Nachbarn in $C$\\
\(\Rightarrow\) $C$ ist d.s. in $G'$.\\
\('\Leftarrow'\) Sei $D$ d.s. in $G'$\\
\includegraphics{bilder/ds-vc2.eps}
Enthält $D$ neue Knoten $K_{u,v}$ dann ersetze $K_{u,v}$ durch $u$ in $D$. Danach entsteht $D'$. Dann ist $D'$ (immer noch d.s.).\\
Dann gilt \( \left|D'\right| \le k\) und $D'$ ist V.C. in $G$:\\
wäre für Knate $(u,v)$ weder $u\in D'$ noch $v\in D'$, dann hätte $K_{u,v}$ keinen Nachbarn in $D'$.\\
Dann wäre $D'$ kein D.S. \Lightning
\underline{Subset Sum}
Gegeben: \(a_1,a_2,…,a_n,b\).\\
Gefragt: \( \exists I \le \{ 1,2,…,n\} \quad \sum\limits_{i\in I} a_i - b \)
\bsp \(1,2,5,7,9,\underbrace{14}_{=b} \in \) Subset Sum\\
\( 4 \not\in \) Subset Sum
\( \in NP:\) rate $I$, teste Summe.\\
Subset Sum ist $NP$-Vollständig.\\
$3-SAT \le^P$ Subset Sum:
Sei \( F(x_1,…x_n) = C_1 \land … C_m \) .\\
\( b = \underbrace{333}_{m}{\color{Orange}|} \underbrace{111}_{n} \) dezimal.
\bsp \( F = (x_1 \lor \overline{x_2} \lor x_3) \land (\overline{x_1} \lor x_2 \lor x_4) \land (\overline{x_2} \lor \overline{x_4} \lor x_5) \)\\
Zahlen \( y_1, …, y_n \) für \( x_1,…,x_n \)\\
\( y_1 = \overbrace{100}^{m=3} {\color{Orange}|} \overbrace{1000}^{n=5} \)\\
$x_1$ kommt in $C_1$ vor\\
$X_1$ kommt nicht in $C_2,C_3$ vor.
\( y_2 = 010 {\color{Orange}|} 01000 \)\\
\( y_3 = 100 {\color{Orange}|} 00100 \)\\
\( y_4 = 010 {\color{Orange}|} 00010 \)\\
\( y_5 = 001 {\color{Orange}|} 00001 \)
Analog \( Z_1,…,Z_n \) für \( \overline{x_1},…,\overline{x_n}\)\\
\( Z_1 = 010 {\color{Orange}|} 10000 \)\\
\( Z_2 = 101 {\color{Orange}|} 01000 \)\\
\( Z_3 = 000 {\color{Orange}|} 00100 \)\\
\( Z_4 = 001 {\color{Orange}|} 00010 \)\\
\( Z_5 = 000 {\color{Orange}|} 00001 \)
Füllzahlen \( f_1,…f_m, g_1,…g_m \)\\
\( f_1 = g_1 = 100 {\color{Orange}|} 00000 \)\\
\( f_2 = g_2 = 010 {\color{Orange}|} 00000 \)\\
\( f_3 = g_3 = 001 {\color{Orange}|} 00000 \)
Es gibt nie Überträge, da im vorderen Teil kann man höchstens auf 3 Einsen kommen pro Klauselspalte mit $y-$ und $z-$Zahlen + max. 2 Einsen mit $f-$ und $g-$Zahlen.\\
\(\Rightarrow\) Summe pro Klauselspalte \(\le 5\)\\
Im hinteren Teil höchstens 2 Einsen pro Spalte.
\('\Rightarrow'\) Sei \( a = (a_1,…,a_n)\) erfüllende Belegung von $F$.\\
Wähle $y_i$, falls $a_i = 1$,\\
wähle $z_i$, falls $a_i = 0$\\
Summe ergibt \( s_1s_2…s_m 111 \), wobei \(s_i \in \{1,2,3\} \), da in jeder Klausel $\ge 1$ Literal den Wert 1 hat.\\
Dann wähle passende Füllzahlen um auf $b$ zu kommen.\\
D.\,h. $f_i$ und $g_i$, falls $s_i=1$ und $f_i$, falls $s_i=2$.\\
\('\Leftarrow'\) Habe Subset Sum Problem eine Lösung.\\
Wegen 1-er Block in $b$ muss für jedes $i$ entweder $y_i$ oder $z_i$ in der Lösung sein.\\
\defin \(a_i=1\), falls $y_i$ in Lösung,\\
$a_i = 0$, falls $z_i$ in Lösung.
Dann ist $a=(a_1,…,a_n)$ erfüllende Belegung von F:\\
Da im vorderen Teil mit den Füll-Zahlen maximal 2 erreicht werden kann, muss die Summe der ausgewählten $y-$ und $z-$Zahlen $\ge 1$ je Klauselspalte sein.\\
\(\Rightarrow\) nach Konstruktion enthält also jede Klausel $\ge 1$ erfülltes Problem.
\end{document} \end{document}

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<use name="mark/disk(sx)" pos="96 784" size="normal" stroke="orange"/>
<path stroke="black">
64 768 m
128 768 l
</path>
<path stroke="orange">
64 768 m
96 784 l
128 768 l
</path>
<text transformations="translations" pos="48 752" stroke="black" type="label" valign="baseline">u</text>
<text transformations="translations" pos="112 752" stroke="black" type="label" valign="baseline">v</text>
<text transformations="translations" pos="96 800" stroke="black" type="label" valign="baseline">$K_{u,v}$</text>
</page>
</ipe>
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