I got the impression Eliezer's claiming that a dangerous superintelligence is merely sufficient for nanotech.

How would you save us with nanotech? It had better be good given all the hardware progress you just caused!

An attempt at problem #1; seems like there must be a shorter proof.

The proof idea is "If I flip a light switch an even number of times, then it must be in the same state that I found it in when I'm finished switching."

Theorem. Let $G=(V,E)$e a path graph on $n$ertices with a vertex $2$oloring $c:V\mapsto \{0,1\}$uch that if $deg\left(u\right)=deg\left(v\right)=1$hen $c\left(u\right)\ne c\left(v\right)$Let $S:=\{e\in E\mid e$s bichromatic $\}$Then $\left|S\right|$s odd.

Proof. By the definition of a path graph, there exists a sequence ${⟨v_k⟩}_{1<k\le n}$ndexing $V\left(G\right)$An edge $e_k=\{v_k,v_{k+1}\}$s bichromatic iff $c\left(v_k\right)\ne c\left(v_{k+1}\right)$A subgraph $H$f $G$s a state iff its terminal vertices are each incident with exactly one bichromatic edge or equal to a terminal vertex of $G$The color of a state is the color of its vertices. There exists a subsequence of $⟨v_k⟩$ontaining the least term of each state; the index of a state is equal to the index of its least term in this subsequence.

Note that none of the states with even indexes are the same color as any of the states with odd indexes; hence all of the states with even indexes are the same color, and all of the states with odd indexes are the same color.

For each state $H$there exists a subsequence of $⟨v_k⟩$orresponding to the vertices of $H$and the least term of each subsequence is either $v_1$r some $v_k$hat is the greatest term in a bichromatic edge. Thus the number of states in $G=1+\left|S\right|$

By contradiction, suppose that $\left|S\right|$s even. Then the number of states is odd, and the first and last states are the same color, so the terminal vertices of $G$re the same color, contrary to our assumption that they are different colors. Thus $\left|S\right|$ust be odd.

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