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Quiz #6: Linked Lists, Trees, Inheritance

Lab #

When working on this quiz, recall the rules stated on the Academic Integrity statement that you signed. You can download the q6helper project folder (available for Friday, on the Weekly Schedule link) in which to write/test/debug your code. Submit your completed q6solution module online by Thursday, 11:30pm. I will post

my solutions to EEE reachable via the Solutions link on Friday afternoon.

  1. (6 pts) Examine the mystery method and hand simulate the call mystery(x,y); using the linked list below.

    Lightly cross out ALL references that are replaced and Write in new references: don’t erase any references. It will look a bit messy, but be as neat as you can. Show references to None as /. Do your work on scratch paper first.

    def mystery(a,b): while b != None:

    t = a.next.next a.next.next = b a = b

    class LN:

    def init (self, value, next=None): self.value, self.next = value, next


    class TN:

    def init (self, value, left=None, right=None): self.value, self.left, self.right = value, left, right

    x b = t


    1 2 3 4


    a


    t


    b


    y


    6 7 8 9


  2. (3 pts) Draw the binary search tree that results from adding the following values (in the following order) into an empty binary search tree: 12, 17, 13, 20, 7, 4, 16, 19, 10, 11, 2, 18, 9, 1, 6, 8, 15, 3, 5, and 14. Draw the number for

each tree node, with lines down to its children nodes. Space-it-out to be easy to read. Answer the questions in the box.


Size = Height =

3a. (3 pts) Define an iterative function named alternate_i; it is passed two linked lists (ll1 and ll2) as arguments. It returns a reference to the front of a linked list that alternates the LNs from ll1 and ll2, starting with ll1; if either linked list becomes empty, the rest of the LNs come from the other linked list. So, all LNs in ll1 and ll2 appear in the returned result (in the same relative order, possibly separated by values from the other linked list). The original linked lists are mutated by this function (the .nexts are changed; create no new LN objects). For example, if we defined

a = list_to_ll(['a', 'b', 'c',]) and b = list_to_ll([1,2,3,4,5])

alternate_i(a,b) returns a->1->b->2->c->3->4->5->None and alternate_i(b,a) returns 1->a->2->b->3-

>c->4->5->None. You may not create/use any Python data structures in your code: use linked list processing only. Change only .next attributes (not .value attributes). Hints: the q6solution.py file and hand-simulate your

code (mine was about 18 lines). This is hard, you might want to solve the recursive one in part 3b first.

3b. (3 pts) Define a recursive function named alternate_r that is given the same arguments and produces the same result as the iterative version above, but here using recursion. You may not create/use any Python data structures in your code: use linked list processing only: use no looping, local variables, etc. Hint: see the recursive code for appending a value at the end of a linked list; of course, try to use the 3 proof rules to help synthesize your code. You might try writing the recursive solution first, it is simpler! (mine was 7 lines)

  1. (4 pts) Write the recursive function count; it is passed balanced binary (any binary tree, not necessarily a binary search tree) and a value as arguments. It returns the number of times the values is in the tree. In the

    binary tree below, count(tree,1) returns 1, count(tree,2) returns 2, count(tree,3) returns 4.

    ..1

    ....3

    3

    ....3

    ..2

    ......2

    ....3

    Hint: use the 3 proof rules to help synthesis your code.


  2. (6 pts) Define a class named bidict (bidirectional dict) derived from the dict class; in addition to being a regular dictionary (using inheritance), it also defines an auxiliary/attribute dictionary that uses the bidict’s values as keys, associated to a set of the bidict’s keys (the keys associated with that value). Remember that multiple keys can associate to the same value, which is why we use a set: since keys are hashable (hashable

    = immutable) we can store them in sets. Finally, the bidict class stores a class attribute that keeps track of a list of all the objects created from this class, which two static functions manipulate.

    Define the class bidict with the following methods (some override dict methods); you may also add helper methods: preface them with single underscores):

    • init (self,initial =[],**kargs): initializes the dict in the base class and also creates an auxiliary dictionary (I used a defaultdict) named _rdict (reversedict: you must use this name for the bsc to work correctly) whose key(s) (the values in the bidict) are associated with a set of values (their keys in the bidict): initialize _rdict by iterating through the newly initialized dictionary.

For a bidict to work correctly, its keys and their associated values must all be hashable. We define any object as hashable if it (a1) has a hash attribute, and (a2) the attribute’s value is not None; also (b) if the object is iterable (has an iter attribute) then every value iterated over is also hashable (by this same definition). Also note that str is hashable as a special case: if we apply the above definition it will create infinite recursion because every value that we iterate over in a str is a str that we can iterate over! Raise a ValueError exception if any value is not hashable. Note you can use the hasattr and getattr functions (which I used in a recursive static helper method you should write with the name_is_hashable).

Finally, _rdict should never store a key that is associated with an empty set: setitem

delitem must ensure this invariant property (I wrote a helper method to help them do it).

and

For example if we define, bd = bidict(a=1,b=2,c=1) then _rdict stores {1: {'a', 'c'}, 2: {'b'}}. If I tried to construct bidict(a=[]) then init would raise a ValueError exception. We will continue using this example below.

  • setitem (self,key,value): set key to associate with value in the dictionary and modify the _rdict to reflect the change. For example, executing bd['a'] = 2 would result in the _rdict being changed to {1:

    {'c'}, 2: {'a', 'b'}}; then executing bd['c'] = 2 would result in the _rdict being changed to {2:

    {'a', 'b', 'c'}}, with 1 no longer being a key in _rdict because it is no longer a value in the bidict.

  • delitem (self,key): for any key in the dictionary, remove it and modify the _rdict to reflect the change.

  • call (self,value): returns the set of keys (keys in the bidict) that associate with value: just lookup this information in the _rdict.

  • clear (self): remove all keys (and their associated values) from the bidict modify the _rdict to reflect the change.

  • all_objects (): a static method that returns a list of all the objects every created by bidict.

  • forget (object): a static method that forgets the specified bidict object, so all_objects doesn’t return it.

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