Lab 5: Streams and graphs

Exercise 1

Define a function (stream-add s1 s2) adding two infinite streams together component-wise. For instance,

  0 1 2 3 4 5 6 ....
+ 1 1 1 1 1 1 1 ....
--------------------
  1 2 3 4 5 6 7 ....

Using stream-add, define the infinite stream fib-stream of all Fibonacci numbers. We can test it, for instance, as follows:

(stream->list (stream-take fib-stream 10)) =>
  (0 1 1 2 3 5 8 13 21 34)

Adding two infinite streams can be done recursively. Since the streams are infinite, we do not have to check the emptiness of any input streams.

Solution: stream-add

(define (stream-add s1 s2)
  (stream-cons (+ (stream-first s1) (stream-first s2))
               (stream-add (stream-rest s1) (stream-rest s2))))

The definition of the Fibonacci sequence , , and for can be reformulated as follows:

      0 1 1 2 3  5  8 13 ...   ; F(n-2) - Fibonacci sequence
+     1 1 2 3 5  8 13 21 ...   ; F(n-1) - shifted Fibonacci sequence to the left by one element
----------------------------
  0 1 1 2 3 5 8 13 21 34 ...   ; Fibonacci sequence starts with 0,1 followed by the sum of the above sequences

This directly leads to the following code:

Solution: fib-stream

(define fib-stream
  (stream-cons 0
               (stream-cons 1
                            (stream-add fib-stream
                                        (stream-rest fib-stream)))))

Alternatively one can use stream* as follows:

(define fib-stream
  (stream* 0 1 (stream-add fib-stream
                           (stream-rest fib-stream))))

Exercise 2

Apart from streams, this lab is also focused on graphs. A graph is a tuple consisting of a set of vertices (also called nodes) and a set of edges . We will represent a graph in Scheme as a struct with two fields. The first is a list of vertices, and the second is a list of edges. An edge is represented as a list (u v). We define a structure for a graph:

(struct graph (nodes edges))

The following graph

is represented as follows:

(define gr
  (graph
    '(1 2 3 4 5 6)
    '((1 2) (1 5) (2 3) (2 5) (3 4) (4 5) (4 6))))

Given a graph , a Hamiltonian path is a path visiting each vertex of exactly once. We will represent a path as a list of consecutive nodes in the path. The above graph gr has a Hamiltonian path (3 2 1 5 4 6).

Write a function (find-hamiltonian-path g) which takes a graph as its input and returns a Hamiltonian path, if it exists, and #f otherwise. E.g.

(find-hamiltonian-path gr) => (3 2 1 5 4 6)
(find-hamiltonian-path (graph '(a b c d) '((a b) (a c) (a d)))) => #f

As a Hamiltonian path traverses each node exactly once, if it exists, it has to be represented by a permutation of the nodes. Thus, we can apply the function permutations from the previous lab to generate all node permutations and check whether each of them forms a Hamiltonian path. We start with a definition of a function checking if a given list of nodes is a path. For that, we need a function testing whether a pair of nodes is connected.

Solution: edge?

; test whether a pair of nodes is connected
(define (edge? g)
  (define edges (graph-edges g))
  (lambda (p)
     (or (member p edges) (member (reverse p) edges))))

Given a list, we create a list of pairs of consecutive nodes. E.g. (1 2 3 4) is transformed to ((1 2) (2 3) (3 4)). This is done by taking (1 2 3) and (2 3 4) and joining them by mapping list element-wise. Finally, we test whether all these pairs are connected. To do so, we use the function (andmap f lst). This function is implemented in Racket. It behaves like map but aggregates the results of f by and function, i.e., once any of the results is #f, it returns #f and the last result otherwise.

Solution: check-path

(define (check-path g)
  (lambda (lst)
    (define but-last (take lst (- (length lst) 1)))
    (if (andmap (edge? g) (map list but-last (cdr lst)))
        lst
        #f)))

Now we can apply the above function to all permutations. The function (check-path g) for a graph g either returns lst if lst forms a path or #f otherwise. Thus we can map it over all permutations of nodes and filter those which form a path. If there is a permutation being a path simultaneously, we have a Hamiltonian path. Otherwise, we return #f.

Solution: find-hamiltonian-path

(define (find-hamiltonian-path g)
  (define perms (permutations (graph-nodes g)))
  (let ([paths (filter identity (map (check-path g) perms))])
    (if (null? paths)
        #f
        (car paths))))

If you are curious, try to use the function in-permutations to compute the perms lazily and compare the perfromance of the two implementations on a larger graph.

Task 1

Write a function (stream-mul s1 s2) taking two infinite streams and multiplying them elements-wise. Using this function, define an infinite stream factorial-stream of factorials .

TIP

The recursive definition of factorial and for gives us

    1 2 3  4   5 ...  ; n
*   1 1 2  6  24 ...  ; f(n-1)
--------------------
  1 1 2 6 24 120 ...  ; f(n)

In your definition, you can use the function (in-naturals n) implemented in Racket to define the stream of natural numbers starting from n.

Once you have the stream of factorials factorial-stream, the function stream-mul and the stream of natural numbers (in-natural 0) (or even simply (in-naturals)), you can define a function (exp-stream x) taking a number x and returning the power series representing , i.e., . Then you can approximate the value by summing an initial segment of this stream. E.g., to approximate , we can sum the first 100 elements:

(apply + (stream->list (stream-take (exp-stream 1) 100)))

Solution

(define (stream-mul s1 s2)
  (stream-cons (* (stream-first s1) (stream-first s2))
               (stream-mul (stream-rest s1) (stream-rest s2))))

(define factorial-stream
  (stream-cons 1 (stream-mul (in-naturals 1) factorial-stream)))

(define (exp-stream x)
  (define recipr (stream-map (curry / 1) factorial-stream))
  (define powers (stream-map (curry expt x) (in-naturals)))
  (stream-map exact->inexact (stream-mul powers recipr)))

Note

Some of you noticed that the stream elements returned by exp-stream having a larger index might overflow if we apply exp-stream to a float number, e.g., (exp-stream 3.0). That is true as the nominators in the power series (and the denominators) grow extremely fast. In particular, the nominator might become infinite if evaluated on a float number. Compare the following calls:

(expt 3.0 1000) => +inf.0
(expt 3 1000) =>
1322070819480806636890455259752144365965422032752148167664920368226828597346704899540778313850608061963909777696872582355950954582100618911865342725257953674027620225198320803878014774228964841274390400117588618041128947815623094438061566173054086674490506178125480344405547054397038895817465368254916136220830268563778582290228416398307887896918556404084898937609373242171846359938695516765018940588109060426089671438864102814350385648747165832010614366132173102768902855220001

So the above solution avoids this problem by working with precise arithmetic and finally translating the result into the decimal representation. To use it, one must call exp-stream with an exact number e.g. 3 or (/ 35 10).

Task 2

Given a graph , a subset of nodes is called a vertex cover, if for every edge , we have or . If is smallest possible, it is called minimum vertex cover (see wikipedia).

Write a function (min-vertex-cover g) taking a graph g and returning a minimum vertex cover of g. E.g.

(min-vertex-cover gr) => (1 2 4)

Hint

The idea is similar to Exercise 2. Instead of all permutations, we can take all subsets of nodes. Subsets can be generated by the function sub-seq from the previous lab. We can sort them by cardinality and starting from the smallest ones, we can check which of them form a vertex cover. In fact, it is computationally more efficient, if we create a stream of subsets so that subsets are computed lazily as they are needed.

Thus we can test smaller subsets first without computing large ones. To create such a lazy stream, we can modify the function sub-seq. It is basically the same code where list functions are replaced by their stream equivalents. However, our original function sub-seq does not generate subsets ordered by their cardinality. To fix this, we have to modify the function merging together the result of the recursive call and the newly created subsets. Thus we define a function (stream-merge s1 s2 cmp) which takes two streams and a function cmp comparing two elements of these streams and returns a stream where the values are merged so that the smaller elements come first.

; lazy subsequences
(define (stream-merge s1 s2 cmp)
  (cond
    ([stream-empty? s1] s2)
    ([stream-empty? s2] s1)
    ([cmp (stream-first s1) (stream-first s2)]
     (stream-cons (stream-first s1) (stream-merge (stream-rest s1) s2 cmp)))
    (else (stream-cons (stream-first s2) (stream-merge s1 (stream-rest s2) cmp)))))

(define (sub-seq lst)
  (if (null? lst)
      (stream '())
      (let ([el (car lst)]
            [rest-sub-seq (sub-seq (cdr lst))])
        (stream-merge rest-sub-seq
                (stream-map ((curry cons) el) rest-sub-seq)
                (lambda (x y) (< (length x) (length y)))))))

Solution

; minimum vertex cover
; -> smallest subset of nodes such that each edge has one of its nodes in it
(define (check-cover g)
  (lambda (lst)
    (if (andmap
          (lambda (e) (or (member (car e) lst) (member (cadr e) lst)))
          (graph-edges g))
        lst
        #f)))

(define (min-vertex-cover g)
  (stream-first (stream-filter identity (stream-map (check-cover g) (sub-seq (graph-nodes g))))))