GENERATORS AND STREAMS
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1 GENERATORS AND STREAMS COMPUTER SCIENCE MENTORS 61A November 13 to November 17, Generators 1. What does the following code block output? def foo(: a = 0 if a < 10: print("hello" yield a print("world" for i in foo(: print(i Hello 0 World 1
2 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 2 2. How can we modify foo so that list(foo(== [1, 2, 3,..., 10]? (It s okay if the program prints along the way. Change the if to a while statement, and make sure to increment a. This looks like: def foo(: a = 0 while a < 10: a += 1 yield a
3 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 3 3. Define hailstone_sequence, a generator that yields the hailstone sequence. Remember, for the hailstone sequence, if n is even, we need to divide by two, otherwise, we multiply by 3 and add by 1. def hailstone_sequence(n: >>> hs_gen = hailstone_sequence(10 >>> hs_gen. next ( 10 >>> next(hs_gen #equivalent to previous 5 >>> for i in hs_gen: >>> print(i while n!= 1: yield n if n % 2 == 0: n = n // 2 else: n = n*3 + 1 yield n
4 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 4 4. Define tree_sequence, a generator that iterates through a tree by first yielding the root value and then yielding the values from each branch. def tree_sequence(t: >>> t = Tree(1, [Tree(2, [Tree(5], Tree(3, [Tree(4]] >>> print(list(tree_sequence(t [1, 2, 5, 3, 4] yield t.label for branch in t.branches: for value in tree_sequence(branch: yield value
5 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 5 2 Streams 1. What s the advantage of using a stream over a linked list? Lazy evaluation. We only evaluate up to what we need. 2. What s the maximum size of a stream? Infinity 3. What s stored in first and rest? What are their types? First is a value, rest is another stream (either a method to calculate it, or an already calculated stream. In the case of Scheme, this is called a promise. 4. When is the next element actually calculated? Only when it s requested (and hasn t already been calculated
6 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 6 3 What Would Scheme Print? 5. For each of the following lines of code, write what Scheme would output. scm> (define x 1 x scm> (if scm> (define p (delay (+ x 1 p scm> p #[promise] scm> (force p 2 scm> (define (foo x (+ x 10 foo scm> (define bar (cons-stream (foo 1 (cons-stream (foo 2 bar bar
7 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 7 scm> (car bar 11 scm> (cdr bar #[promise] scm> (define (foo x (+ x 1 foo scm> (cdr-stream bar (3. #[promise] scm> (define (foo x (+ x 5 foo scm> (car bar 11 scm> (cdr-stream bar (3. #[promise]
8 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 8 4 Code Writing for Streams 6. Write out double naturals, which is a stream that evaluates to the sequence 1, 1, 2, 2, 3, 3, etc. (define (double_naturals (double_naturals_helper 1 0 (define (double_naturals_helper first go-next (define (double_naturals_helper first go-next (if (= 1 go-next (cons-stream first (double_naturals_helper (+ 1 first 0 (cons-stream first (double_naturals_helper first 1 ;Alternative Solutions (define (double_naturals_helper first go-next (cons-stream first (double_naturals_helper (+ go-next first (- 1 go-next
9 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 9 7. Write out interleave, which returns a stream that alternates between the values in stream1 and stream2. Assume that the streams are infinitely long. (define (interleave stream1 stream2 (define (interleave stream1 stream2 (cons-stream (car stream1 (interleave stream2 (cdr-stream stream1 ;Alternative solution (define (interleave stream1 stream2 (cons-stream (car stream1 (cons-stream (car stream2 (interleave (cdr-stream stream1 (cdr-stream stream2
10 GROUP TUTORING HANDOUT 9: GENERATORS AND STREAMS Page 10 5 Challenge Question 8. (Optional Write a generator that takes in a tree and yields each possible path from root to leaf, represented as a list of the values in that path. Use the object-oriented representation of trees in your solution. def all_paths(t: >>> t = Tree(1, [Tree(2, [Tree(5], Tree(3, [Tree(4]] >>> print(list(all_paths(t [[1, 2, 5], [1, 3, 4]] if t.is_leaf(: yield [t.label] for b in t.branches: for subpath in all_paths(b: yield [t.label] + subpath
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