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\begin_body

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc4[M20]
\end_layout

\end_inset

Let 
\begin_inset Formula $G'$
\end_inset

 be a finite free tree in which arrows have been drawn on its edges 
\begin_inset Formula $e_{1},\dots,e_{n-1}$
\end_inset

;
 let 
\begin_inset Formula $E_{1},\dots,E_{n-1}$
\end_inset

 be numbers satisfying Kirchhoff's law (1) in 
\begin_inset Formula $G'$
\end_inset

.
 Show that 
\begin_inset Formula $E_{1}=\dots=E_{n-1}=0$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

We prove this by induction on 
\begin_inset Formula $n$
\end_inset

.
 For 
\begin_inset Formula $n=1$
\end_inset

 it is trivial.
 For 
\begin_inset Formula $n>1$
\end_inset

,
 there always exists a node of degree 1 (in an oriented tree,
 this would be a leaf).
 Let 
\begin_inset Formula $e_{r}$
\end_inset

 be the only edge connected to that node,
 clearly 
\begin_inset Formula $E_{r}=0$
\end_inset

,
 so removing 
\begin_inset Formula $e_{r}$
\end_inset

 leaves us with an isolated node and a free tree of degree 0 which follows Kirchhoff's law and has 
\begin_inset Formula $n-1$
\end_inset

 nodes,
 but then by the induction hypothesis all the other 
\begin_inset Formula $E_{k}$
\end_inset

 are also 0.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc8[M25]
\end_layout

\end_inset

When applying Kirchhoff's first law to program flow charts,
 we usually are interested only in the 
\emph on
vertex flows
\emph default
 (the number of times each box of the flow chart is performed),
 not the edge flows analyzed in the text.
 For example,
 in the graph of exercise 7,
 the vertex flows are 
\begin_inset Formula $A=E_{2}+E_{4}$
\end_inset

,
 
\begin_inset Formula $B=E_{5}$
\end_inset

,
 
\begin_inset Formula $C=E_{3}+E_{7}+E_{8}$
\end_inset

,
 
\begin_inset Formula $D=E_{6}+E_{9}$
\end_inset

.
\end_layout

\begin_layout Standard
If we group some vertices together,
 treating them as one 
\begin_inset Quotes eld
\end_inset

supervertex,
\begin_inset Quotes erd
\end_inset

 we can combine edge flows that correspond to the same vertex flow.
 For example,
 edges 
\begin_inset Formula $e_{2}$
\end_inset

 and 
\begin_inset Formula $e_{4}$
\end_inset

 can be combined in the flow chart above if we also put 
\begin_inset Formula $B$
\end_inset

 with 
\begin_inset Formula $D$
\end_inset

:
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{center}
\end_layout

\begin_layout Plain Layout


\backslash
begin{tikzpicture}[node distance=2cm]
\end_layout

\begin_layout Plain Layout


\backslash
node[circle,draw](Start){Start};
\end_layout

\begin_layout Plain Layout


\backslash
node[rectangle,draw,right of=Start](A){$A$};
\end_layout

\begin_layout Plain Layout


\backslash
node[rectangle,draw,right of=A](BD){$B,D$};
\end_layout

\begin_layout Plain Layout


\backslash
node[rectangle,draw,right of=BD](C){$C$};
\end_layout

\begin_layout Plain Layout


\backslash
node[circle,draw,right of=C](Stop){Stop};
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (Start) -- node[above]{$e_1$} (A);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (A) -- node[below]{$e_2+e_4$} (BD);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->,transform canvas={yshift=2pt}] (BD)--node[above]{$e_5$}(C);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->,transform canvas={yshift=-2pt}] (C)--node[below]{$e_7$}(BD);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (C) -- node[above]{$e_8$} (Stop);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (BD) to[bend right] node[below]{$e_9$} (Stop);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (BD) to[loop below] node[left]{$e_6$} (BD);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (C) to[bend right] node[above]{$e_3$} (A);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (Stop) to[bend right] node[above]{$e_0$} (Start);
\end_layout

\begin_layout Plain Layout


\backslash
end{tikzpicture}
\end_layout

\begin_layout Plain Layout


\backslash
end{center}
\end_layout

\end_inset

(Here 
\begin_inset Formula $e_{0}$
\end_inset

 has also been added from Stop to Start,
 as in the text.) Continuing this procedure,
 we can combine 
\begin_inset Formula $e_{3}+e_{7}$
\end_inset

,
 then 
\begin_inset Formula $(e_{3}+e_{7})+e_{8}$
\end_inset

,
 then 
\begin_inset Formula $e_{6}+e_{9}$
\end_inset

,
 until we obtain the 
\emph on
reduced flow chart
\emph default
 having edges 
\begin_inset Formula $s=e_{1}$
\end_inset

,
 
\begin_inset Formula $a=e_{2}+e_{4}$
\end_inset

,
 
\begin_inset Formula $b=e_{5}$
\end_inset

,
 
\begin_inset Formula $c=e_{3}+e_{7}+e_{8}$
\end_inset

,
 
\begin_inset Formula $d=e_{6}+e_{9}$
\end_inset

,
 
\begin_inset Formula $t=e_{0}$
\end_inset

,
 precisely one edge for each vertex in the original flow chart:
\begin_inset Formula $\text{}$
\end_inset


\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
begin{center}
\end_layout

\begin_layout Plain Layout


\backslash
begin{tikzpicture}[node distance=3cm]
\end_layout

\begin_layout Plain Layout


\backslash
node[circle,draw](S){Start};
\end_layout

\begin_layout Plain Layout


\backslash
node[rectangle,draw,right of=S](M){$A,B,D,
\backslash
text{Stop}$};
\end_layout

\begin_layout Plain Layout


\backslash
node[rectangle,draw,right of=M](C){$C$};
\end_layout

\begin_layout Plain Layout


\backslash
draw[->,transform canvas={yshift=2pt}] (S) -- node[above]{$s$} (M);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->,transform canvas={yshift=-2pt}] (M) -- node[below]{$t$} (S);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->,transform canvas={yshift=2pt}] (M) -- node[above]{$b$} (C);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->,transform canvas={yshift=-2pt}] (C) -- node[below]{$c$} (M);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (M) to[out=120,in=60,looseness=4] node[left]{$a$
\backslash
 
\backslash
 } (M);
\end_layout

\begin_layout Plain Layout


\backslash
draw[->] (M) to[out=300,in=240,looseness=4] node[right]{
\backslash
 $d$} (M);
\end_layout

\begin_layout Plain Layout


\backslash
end{tikzpicture}
\end_layout

\begin_layout Plain Layout


\backslash
end{center}
\end_layout

\end_inset


\end_layout

\begin_layout Standard
By construction,
 Kirchhoff's law holds in this reduced flow chart.
 The new edge flows are the vertex flows of the original;
 hence the analysis in the text,
 applied to the reduced flow chart,
 shows how the original vertex flows depend on each other.
\end_layout

\begin_layout Standard
Prove that this reduction process can be reversed,
 in the sense that any set of flows 
\begin_inset Formula $\{a,b,\dots\}$
\end_inset

 satisfying Kirchhoff's law in the reduced flow chart can be 
\begin_inset Quotes eld
\end_inset

split up
\begin_inset Quotes erd
\end_inset

 into a set of edge flows 
\begin_inset Formula $\{e_{0},e_{1},\dots\}$
\end_inset

 in the original flow chart.
 These flows 
\begin_inset Formula $e_{j}$
\end_inset

 satisfy Kirchhoff's law and combine to yield the given flows 
\begin_inset Formula $\{a,b,\dots\}$
\end_inset

;
 some of them might,
 however,
 be negative.
 (Although the reduction procedure has been illustrated for only one particular flow chart,
 your proof should be valid in general.)
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset


\begin_inset Note Greyedout
status open

\begin_layout Plain Layout
(I had to look up the solution,
 this one is difficult.)
\end_layout

\end_inset

 We only have to show that there exists an assignment in the original flowchart that produces any given set of flows in the reduced flowchart.
\end_layout

\begin_layout Standard
Since we get from one to the other by reductions,
 we only have to prove that we can reverse reductions.
 In these reductions,
 there are two arrows 
\begin_inset Formula $e_{1}$
\end_inset

 from vertex or supervertex 
\begin_inset Formula $a$
\end_inset

 to 
\begin_inset Formula $b$
\end_inset

 and 
\begin_inset Formula $e_{2}$
\end_inset

 from 
\begin_inset Formula $a$
\end_inset

 to 
\begin_inset Formula $c$
\end_inset

,
 to get an edge 
\begin_inset Formula $f=e_{1}+e_{2}$
\end_inset

 from 
\begin_inset Formula $a$
\end_inset

 to 
\begin_inset Formula $b,c$
\end_inset

,
 and reverting means recovering 
\begin_inset Formula $e_{1}$
\end_inset

 and 
\begin_inset Formula $e_{2}$
\end_inset

 from 
\begin_inset Formula $f$
\end_inset

 and the other edges.
\end_layout

\begin_layout Standard
If 
\begin_inset Formula $b=c$
\end_inset

,
 we can just make up 
\begin_inset Formula $e_{1}$
\end_inset

 and set 
\begin_inset Formula $e_{2}$
\end_inset

 to 
\begin_inset Formula $f-e_{1}$
\end_inset

.
 If 
\begin_inset Formula $a=b\neq c$
\end_inset

,
 then 
\begin_inset Formula $e_{2}$
\end_inset

 must be such that Kirchhoff's law is preserved for 
\begin_inset Formula $c$
\end_inset

,
 and we can just set 
\begin_inset Formula $e_{1}=f-e_{2}$
\end_inset

.
 The situation 
\begin_inset Formula $a=c\neq b$
\end_inset

 is symmetrical.
 Finally,
 if the three nodes are distinct,
 setting 
\begin_inset Formula $e_{1}$
\end_inset

 and 
\begin_inset Formula $e_{2}$
\end_inset

 such that they follow Kirchhoff's law for 
\begin_inset Formula $b$
\end_inset

 and 
\begin_inset Formula $c$
\end_inset

 makes 
\begin_inset Formula $e_{1}+e_{2}$
\end_inset

 follow Kirchhoff's law in 
\begin_inset Formula $b,c$
\end_inset

 in the reduced flowchart.
\end_layout

\end_body
\end_document