Consider the following variation on homework problems 2.2, 3.3, and
7.2:

A Enumeration is either an OpenDotList or a ClosedDotList

An OpenDotList is a list of strings

A ClosedDotList is also a list of strings.

Enumerations support the following operations:

render()  -- as in homework problem 7.2, except that it does not take
a style argument:  the style is determined by whether the Enumeration
is an OpenDotList or a ClosedDotList .  Returns a list of strings, as
in 7.2

get(int n) -- returns the n-th item in the enumeration (counting from
1).  If there is no n-th item (either because n is greater than the
length of the enumeration or because n is <= 0), returns the empty
string.

remove(int n) -- returns an enumeration like the original, except that
the n-th item has been removed.  If there is no n-th item, returns an
enumeration just like the original.

swap(int n, int m) -- returns an enumeration like the original, except
that the n-th and m-th items are swapped.  If there is no n-th or m-th
item, returns an enumeration just like the original.

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I think I would include a set of data definitions in Scheme along with
examples. 

I'm also considering making the problem harder by switching to the
following:

A Enumeration is either an OpenDotList or a ClosedDotList

An OpenDotList is a list of Items

A ClosedDotList is also a list of Items

An Item is either a String or an Enumeration

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This would seem to offer a bunch of opportunities for abstraction.

What do you think?