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\backslash
exerc1[00]
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Considering COBOL data configurations as tree structures,
 are the data items listed by a COBOL programmer in preorder,
 postorder,
 or neither of these orders?
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answer 
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They are in preorder.
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exerc2[10]
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Comment about the running time of Algorithm A.
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answer 
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It takes a constant amount of time for each data item read (assuming that accessing the symbol table takes constant time).
 There is also a loop to get out of a nested record,
 but each iteration of the loop takes out an item from the stack so we may consider that as part of processing such item.
 Therefore the running time is proportional to the number of data items in the input.
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exerc8[10]
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Under what circumstances is 
\begin_inset Quotes eld
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MOVE CORRESPONDING 
\begin_inset Formula $\alpha$
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 TO 
\begin_inset Formula $\beta$
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\family default

\begin_inset Quotes erd
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 exactly the same as 
\begin_inset Quotes eld
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\family typewriter
MOVE 
\begin_inset Formula $\alpha$
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 TO 
\begin_inset Formula $\beta$
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\family default

\begin_inset Quotes erd
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,
 according to the definition in the text?
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answer 
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When 
\begin_inset Formula $\alpha$
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 or 
\begin_inset Formula $\beta$
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 is an elementary item.
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\backslash
rexerc11[23]
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What additional links or changes in the strategy of the algorithms of the text could make Algorithm B or Algorithm C faster?
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answer 
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(I had to look up the solution.)
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 One could make Algorithm C faster by storing sibling nodes in the table in lexicographical order,
 so step C3 would have a linear rather than quadratic complexity.
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rexerc13[24]
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Give an algorithm to substitute for Algorithm A when the Data Table is to have the format shown in exercise 12.
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answer 
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We remove the instructions that set 
\family typewriter
PARENT
\family default
,
 
\family typewriter
CHILD
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,
 
\family typewriter
SIB
\family default
,
 and 
\family typewriter
NAME
\family default
,
 as those fields do not exist now (in particular,
 A6 disappears),
 and we implement 
\begin_inset Formula $\mathtt{Q}\Leftarrow\mathtt{AVAIL}$
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 in step A2 by sequential allocation.
 We set the 
\begin_inset Formula $\mathtt{SCOPE}$
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 of an entry when we remove the corresponding item from the stack in step A5.
 (Note that we do the same in A5 when 
\begin_inset Formula $\mathtt{L1}=\mathtt{L}$
\end_inset

 as when 
\begin_inset Formula $\mathtt{L1}>\mathtt{L}$
\end_inset

,
 so we could just do those instructions unconditionally,
 then after removing the current stack item and before loading the next one we check if 
\begin_inset Formula $\mathtt{L1}=\mathtt{L}$
\end_inset

 and if so we go to A7,
 then if 
\begin_inset Formula $\mathtt{L1}\geq\mathtt{L}$
\end_inset

 we repeat,
 otherwise we report the error.)
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For this to work,
 we have to get the stack empty by the time the algorithm finishes,
 which we can do by setting 
\begin_inset Formula $\mathtt{L}\gets0$
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 when the input is exhausted in step A2 and then terminate in step A7 when 
\begin_inset Formula $\mathtt{L}=0$
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.
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