Assigned on Monday, April 1.
Due at 5 PM on Wednesday, April 17.
Use hsp to turn in each part of this assignment separately, using filenames scheme1.scm, scheme2a.scm, scheme2b.scm, etc. You may define any helper functions that you wish. If you need to resubmit your homework, add an additional underscore with a version number. For example, if I need to resubmit my scheme1.scm file, I would name it scheme1_2.scm (for version 2).
Write Scheme functions that implement integer binary search trees. A non-empty binary search tree can be represented as a list of three entries. The first entry is a node's value (an integer), the middle is the node's left subtree, and the last is a node's right subtree. Non-empty left and right subtrees are themselves lists of three elements. An empty tree is represented as an empty list. For example, the list
(5 () ())
represents a tree whose only node is the leaf 5. The list
(5 (3 () (4 () ()) ) ())
represents the tree whose root node is 5 and has only one child, a left child, 3: this left child has only one child, a right child, 4. in a binary search tree it is always the case that values in the left subtree are less than the root's value, and values in the right subtree are greater than the root's value. Duplicate entries are not allowed.
For all parts of this question, you may not use side-effects (e.g. set!).
Write the following functions:
(null-bst)(null-bst? bst)
Return true (#t) if bst is empty, and false (#f) otherwise.
(entry bst)
(left bst)
(right bst)
These functions return the node, left subtree, and right subtree, respectively,
of bst. If bst is not actually a binary search tree as defined above, return
#f.
(make-bst elt left right)
This function creates a new tree whose root node is elt, left subtree is left,
and right subtree is right. You chould check that elt is in fact a number, and
that left and right have three elements each. Return #f is the input is bad.
(preorder bst)
Print each node in bst in preorder (the root of bst, then the left subtree in
preorder, then the right subtree in preorder).
(insert v bst)
Insert the integer v in the binary tree, bst. Smaller values are inserted to
the left of a node, larger values to the right of a node. Discard values found
to be duplicates.
In Scheme sets can be represented as lists. However, unlike lists, the order of values in a set is not significant. Thus both (1 2 3) and (3 2 1) represent the same set.
(a) Write a Scheme function (set-equal? S1 S2) that tests whether S1 and S2 (represented as lists) are equal. Two sets are equal if they contain exactly the same members, ignoring ordering. In this part you may assume that sets contain only atomic values (numbers, string, symbols, etc.) For example
(set-equal? '(1 2 3) '(3 2 1)) ---> #t (set-equal? '(1 2) '(3 2 1)) ---> #f (set-equal? '(curly larry moe) '(moe larry curly)) ---> #t
You may assume that no sets contain duplicate members.
(b) Two common operations on sets are union and intersection. The union of two sets is the set of all elements that appear in either set (with no repetitions). The intersection of two sets is the set of elements that appear in both sets.
Write Scheme functions (union S1 S2) and (intersect S1 S2) that implement set union and set intersection. You may again assume that set elements are always atomic, and that there are no duplicates. For example
(union '(1 2 3) '(3 2 1)) ---> (1 2 3) (union '(1 2 3) '(3 4 5)) ---> (1 2 3 4 5) (union '(a b c) '(3 2 1)) ---> (a b c 1 2 3) (intersect '(1 2 3) '(3 2 1)) ---> (1 2 3) (intersect '(1 2 3) '(4 5 6)) ---> () (intersect '(1 2 3) '(2 3 4 5 6)) ---> (2 3)
The ordering of the elements in your answer may differ from the above.
(c) In general sets can contain other sets. Extend your solution to parts (a) and (b) to allow sets to contain other sets. For example:
(set-equal? '(1 (2 3)) '((3 2) 1)) ---> #t (set-equal? '(1 2 3) '((3 2) 1)) ---> #f (set-equal? '(1 2 3) '((1 2 3))) ---> #f (union '(1 (2) 3) '(3 2 1)) ---> (1 2 3 (2)) (union '((1 2 3)) '((3 4 5))) ---> ((1 2 3) (3 4 5)) (union '((1 2 3)) '((3 2 1))) ---> ((1 2 3)) (intersect '((1 2 3)) '((3 2 1))) ---> ((1 2 3)) (intersect '((1 2 3)) '((4 5 6))) ---> () (intersect '((1) (2) (3)) '((2) (3) (4))) ---> ((2) (3))
(a) Write a pair of Scheme functions, (gen-list start stop) and (pair-sum? list val). The function gen-list will generate a list of consecutive integers, from start to stop. (If start > stop then the empty list is generated.) For example:
(gen-list 1 5) ---> (1 2 3 4 5)
The predicate pair-sum? tests whether any two adjacent values in list sum to val. For example,
(pair-sum? '(1 2 3) 3) ---> #t
since 1+2=3. Similarly,
(pair-sum? (gen-list 1 100) 1000) ---> #f
since no two adjacent integers in the range 1 to 100 can sum to 1000.
(b) An alternative to completely building a complex data structure is lazy evaluation. That is, part of the structure is built along with a suspension - a function that will supply a few more values upon request. The following two Scheme functions produce a sequence of consecutive integers in a lazy manner.
(define (return-one-val start stop)
(if (= start stop)
(cons start #f)
(cons start
(lambda () (return-one-val (+ start 1) stop)))))
(define (int-seq start stop)
(if (> start stop)
(cons #f #f)
(return-one-val start stop)))
When called, int-seq defines a pair of values. The car is the first integer in the sequence. The cdr is a function that, when called, will return another pair. That pair consists of the next integer in the sequence plus another suspension function. When the end of the sequence is reached, the cdr of the pair is #f (rather than a function), indicating that no more values can be produced.
Rewrite your solution to pair-sum? to take a lazy integer sequence (as defined by int-seq) rather than a list as a parameter.
Scheme can represent numbers in a variety of forms, including integer, rational, real, and complex. Let's add a new form - English. That is, we will allow numbers to be represented as a list of symbols representing English language words corresponding to numbers. Thus 0 would be represented as (zero), 123 as (one hundred twenty three) and -456789 as (minus four hundred fifty six thousand seven hundred eighty nine).
Assume that we represent numbers (in English) using the symbols minus, zero, one, two, ..., ten, eleven, twelve, ..., nineteen, twenty, thirty, ..., ninety, hundred, thousand, million, ... For simplicity, numbers that are normally hyphenated (like twenty-three) won't be.
(a) Write a Scheme function (int->eng int-val) that converts an integer int-val into English form (that is, a list of symbols representing the number in English). Your function should handle numbers at least as large as one trillion (handling even larger numbers is fairly easy). For numbers larger than your implementation limit (of one trillion or more) you may return the integer unconverted.
(b) Now write an inverse function (eng->int L) that converts a number in English form back to integer form. Of course it should be the case that
(eng->int (int->eng int-val)) = int-val
for all integers in the range you handle.
(c) Finally, redefine the +, -, *, and / functions so that they accept numbers in English form as well as integer form (you may ignore numbers in rational, real, and complex forms). You may assume that your extended operators will be called with only two operands. If both operands are in integer form, then you will return in an integer result. Of one or both are in English form, then you will return a result in English form. For /, you may use truncate to convert fractional quotients to integers; that is, you need not worry about divisions that have a non-zero remainder.
The following are sample calls that you should handle:
(+ 123 456) ---> 579 (- 111111 '(ten)) ---> (one hundred eleven thousand one hundred one) (* '(two thousand fifty five) '(one million one)) ---> (two billion fifty five million two thousand fifty five) (/ 123456789 '(three)) ---> (forty one million one hundred fifty two thousand two hundred sixty three)
(a). Assume we have a list L of integers. Define a function (listify L) that divides L into one or more sublists so that each sublist contains integers in non-decreasing (sorted) order. That is if v1 and v2 are adjacent in L and v1 <= v2 then v1 and v2 are adjacent in the same sublist. However, if v1 > v2 then v1 ends one sublist and v2 begins the next sublist. For example.
(listify '(3 5 1 8 9 2 1 0)) ---> ((3 5) (1 8 9) (2) (1) (0)) (listify '(1 2 3 4 5 6)) ---> ((1 2 3 4 5 6)) (listify '(5 4 3 2 1)) ---> ((5) (4) (3) (2) (1))
(b) If the output of listify contains a single sublist, then we know the input list L was in sorted order. Otherwise, we can implement a sorting algorithm (a merge sort) as follows. Take the first two sublists produced by listify and merge them into a single sorted list. Then take the next two sublists and merge them. Repeat until all sublists are considered. (The last sublist is copied unchanged if it has no partner to merge with). If only one sublist remains, we are done. Otherwise, we repeat the process, merging sublists pairwise. At each stage, the number of sublists is cut in half. For example,
((3 5) (1 8 9) (2) (1) (0)) is reduced to (1 3 5 8 9) (1 2) (0)) and then to ((1 1 2 3 5 8 9) (0)) and finally to ((0 1 1 2 3 5 8 9)).Implement (merge-sort L), which sorts the values in L using listify as a subroutine.
(c) It is sometimes the case that a sorted list should contain no duplicate values. Modify your implementation of merge-sort so that if it sees two duplicate values it immediately returns the pair (duplicate.val) instead of the sorted list. duplicate is a fixed symbol; val is the duplicate value that was seen in L. Hence (duplicate.17) would indicate that 17 was duplicated in L.
Note that you should not check for duplicates after the sort is done. Duplicate checking should be integral to your implementation of listify and the merge component of your merge sort. You should use call/cc to implement the exception mechanism you will need to return a duplicate value.
I am indebted to Charles Fischer at University of Wisconsin-Madison for writing these wonderful problems.