COM S 342 Homework 5 solution

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In this homework, for problems 1 and 2, you will be using procedural-based strategy
to represent the data-types. Problem 3 is designed to further enhance your understanding
on Data Abstractions.

For problems 1 and 2 (procedural-based representation strategy):
*A general rule* is – instead of storing values in lists or other
data types, you will be storing them in “closures”. Review the lecture notes and
the attached file hw05-closure-currying-illustrated.rkt as a reminder of
what closures are. Since currying function is a closely related concept, we
introduce them together.

What procedural-based representation isn’t:
– a list containing functions that simply return each individual element of
the list
– if you use lists for anything but temporary computation, you are probably
doing it wrong.
===============================================================================
1. [40p] Functional Sets

Use procedural representation to implement sets of numbers.

Each set is a one argument function that takes a number as argument and
tells whether or not the number is in the set.

Hints:
– the functions up until “exists?” can be written as one liners
and do not require recursion or other functional operations to solve.
– just because a function has the word “map” in it, it doesn’t mean you have
to use map to implement it. Map is a functional operation with well defined
behavior that is not tied to any implementation.
– think about what it means for an element to be in a set, rather than how
to implement it.

===============================================================================
2. [20p] Step

The following data type describes the kinds of movements that can be taken on a
2-D surface:

<step> ::= <step> <step> “seq-step”
| “up” number “up-step”
| “down” number “down-step”
| “left” number “left-step”
| “right” number “right-step”

“up”, “down”, “left” and “right” are only markers that help differentiate between
two steps.

“seq-step”, “up-step”, “down-step”, “left-step” and “right-step” are symbolic names
we use to identify each alternative production in the grammar, which may also be
thought of as “variant” of the <step> data type.

2.a
For each variant of the <step> datatype you should implement the following types
of functions:
– a constructor bearing the name of the variant. Constructors take
any relevant data as parameters and return a new value of that type
variant.
– a predicate with the name of the variant followed by a
“?” mark
– extractors for each piece of data of the variants. e.g. we write
one extractor for up, down, left, right steps and two for seq-step.

***
NOTE: your implementation has to be based on procedural representation
(check the most recent lecture notes inside blackboard if clarification
is still needed for “procedural representation”)
***

The signatures of the above functions are already written down in the answer
sheet. You should take the time to see which procedure falls into which category
and infer their behavior from the test cases. We will be encountering this pattern
quite often in the future.

A few general rules to keep in mind:
– constructors throw exceptions when they receive invalid input
– predicates never throw exceptions, they either return #t or #f
– extractors throw exceptions when then receive invalid input

 

2.b
Implement the function:
(move starting-point step)

Input:
– starting-point: a point in the x,y coordinate system, represented as a 2 element
list
– step: a step, as grammatically defined at the beginning of problem 2

Output:
– the end-point after moving the specified number of steps
===============================================================================

3. [40p] Binary search tree

A binary search tree can be easily described with the following grammar:

<tree> ::= empty
| number <tree> <tree>

Although BNF grammar is very well suited to describe the structure of the data,
it is not suited to describe the relation between various pieces.

For instance, we know that a binary search tree has the following properties:
– the left subtree of a node contains only nodes with keys less than the
node’s key.
– the right subtree of a node contains only nodes with keys greater than the
node’s key.
– there must be no duplicate nodes.
– the implication of the above is that in-order traversal yields a sorted
sequence of keys

Implement the following functions:
(empty-tree)
;creates an empty-tree

(empty-tree? tree) [2p]
;returns #t or #f if the tree is empty or not

(bstree root) [2p]
;creates a tree with only one node

(tree->root tree) [2p]
;returns the root of the given tree

(tree->left tree) [2p]
;returns the left subtree

(tree->right tree) [2p]
;returns the right subtree

(insert-tree n tree) [15p]
;inserts a node with value n into the tree while still maintaining its binary search tree’s property
;listed properties

(tree->list tree) [15p]
;return a list containing all the values of the tree in sorted order

Look at the test cases for examples. The functions are not independently tested,
i.e. tree->left depends on a correct implementation of insert. The reason for
this is that it provides independence from the internal representation of the
tree, the only constraint remaining is that the interface has to be self-consistent.