Changed layout for more manageable volumes

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-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
-\origin unavailable
-\textclass book
-\begin_preamble
-\input defs
-\end_preamble
-\use_default_options true
-\maintain_unincluded_children no
-\language english
-\language_package default
-\inputencoding utf8
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-\paperorientation portrait
-\suppress_date false
-\justification true
-\use_refstyle 1
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-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\secnumdepth 3
-\tocdepth 3
-\paragraph_separation indent
-\paragraph_indentation default
-\is_math_indent 0
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-\quotes_style english
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-\end_header
-
-\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