1 / 23

COMPSCI 105 S2 2014 Principles of Computer Science

COMPSCI 105 S2 2014 Principles of Computer Science. Recursion 3. Agenda. Agenda The Fibonacci Sequence The Towers of Hanoi Binary search Recursion and efficiency Textbook: Problem Solving with Algorithms and Data Structures Chapter 4 – Recursion Chapter 5 – Searching.

anakin
Download Presentation

COMPSCI 105 S2 2014 Principles of Computer Science

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. COMPSCI 105 S2 2014Principles of Computer Science Recursion 3

  2. Agenda • Agenda • The Fibonacci Sequence • The Towers of Hanoi • Binary search • Recursion and efficiency • Textbook: • Problem Solving with Algorithms and Data Structures • Chapter 4 – Recursion • Chapter 5 – Searching COMPSCI105

  3. 22.1 The Fibonacci SequenceThe Fibonacci Sequence • Describes the growth of an idealised (biologically unrealistic) rabbit population, assuming that: • Rabbits never die • A rabbit reaches sexual maturity exactly two months after birth, that is, at the beginning of its third month of life • Rabbits are always born in male-female pairs • At the beginning of every month, each sexually mature male-female pair gives birth to exactly one male-female pair COMPSCI105

  4. 22.1 The Fibonacci SequenceThe Fibonacci Sequence Month • Problem • How many pairs of rabbits are alive in month n? • Example: • rabbit(5) = 5 • Recurrence relation rabbit(n) = rabbit(n-1) + rabbit(n-2) COMPSCI105

  5. 22.1 The Fibonacci SequenceRecursive Definition The sequence of numbers rabbit(n) for all n is called Fibonacci Sequence or Fibonacci numbers. def rabbit(n): if n <=2: return 1 return rabbit(n-1) + rabbit(n-2) • Base cases • rabbit(2), rabbit(1) • Recursive rabbit(n) = 1 if n is 1 or 2 rabbit(n-1) + rabbit(n-2) if n > 2 • Fibonacci sequence • The series of numbers rabbit(1), rabbit(2), rabbit(3), and so on COMPSCI105

  6. 22.1 The Fibonacci Sequencerabbit(6) rabbit(6) return rabbit(5)+rabbit(4) rabbit(5) return rabbit(4)+rabbit(3) rabbit(4) return rabbit(3)+rabbit(2) rabbit(3) return rabbit(2)+rabbit(1) rabbit(2) return 1 rabbit(4) return rabbit(3)+rabbit(2) rabbit(3) return rabbit(2)+rabbit(1) rabbit(1) return 1 rabbit(3) return rabbit(2)+rabbit(1) rabbit(1) return 1 rabbit(2) return 1 rabbit(2) return 1 rabbit(2) return 1 rabbit(1) return 1 rabbit(2) return 1 COMPSCI105

  7. 22.1 The Fibonacci SequenceExamples: Fibonacci Spiral Fibonacci Tiling COMPSCI105

  8. 22.2 The Towers of Hanoi2.6 The Towers of Hanoi • Puzzle consists of n disks and three poles • The disks are of different size and have holes to fit themselves on the poles • Initially all the disks were on one pole, e.g., pole A • The task was to move the disks, one by one, from pole A to another pole B, with the help of a spare pole C. • Due to its weight, a disks could be placed only on top of another disk larger than itself COMPSCI105

  9. 22.2 The Towers of HanoiSolution • If you have only one disk (i.e., n=1), • move it from pole A to pole B • If you have more than one disk, • simply ignore the bottom disk and solve the problem for n-1 disk, with pole C is the destination and pole B is the spare • Then move the largest disk from pole A to B; then move the n-1 disks from the pole C back to pole B • We can use a recursion with the arguments: • Number of disks, source pole, destination pole, spare pole COMPSCI105

  10. 22.2 The Towers of HanoiSolution defhanoi(count,source,destination,spare): if count is 1: Move a disk directly from source to destination Move count-1 disks from source to spare Move 1 disk from source to destination Move count-1 disk from spare to destination Apply function recursively to move these three disks. Source destination spare Source destination spare Source destination spare Source destination spare COMPSCI105

  11. 22.2 The Towers of Hanoi Pseudo code for the recursive solution defhanoi(count, source, destination, spare): if count <= 1: print ("base case: move disk from", source, "to", destination) else: hanoi(count - 1, source, spare, destination) print ("step2: move disk from", source, "to", destination) hanoi(count - 1, spare, destination, source) 1 2 3 • Satisfies the four criteria of a recursive solution • Recursive method calls itself • Each recursive call solves an identical, but smaller problem • Stops at base case • Base case is reached in finite time COMPSCI105

  12. defhanoi(count, source, dest, spare): if count <= 1: print ("base case: move disk from", source, "to", dest) else: hanoi(count - 1, source, spare, dest) print ("step2: move disk from", source, "to", dest) hanoi(count - 1, spare, dest, source) Example: hanoi(3, "A", "B", "C") hanoi(3, ‘A’, ‘B’, ‘C’) Count = 3 Source = A Dest = B Spare = C ? ? hanoi(1, ‘A’, ‘B’, ‘C’ Count = 1 Source = A Dest = B Spare = C hanoi(2, ‘A’, ‘C’, ‘B’) Count = 2 Source = A Dest = C Spare = B ? ? Base case: Print: from A to B Print: from A to C 3 Steps Base case: Print: from B to C 3 Steps hanoi(1, ‘B’, ‘C’, ‘A’ Count = 1 Source = B Dest = C Spare = A C Print: from A to B A B C B A hanoi(1, ‘C’, ‘A’, ‘B’ Count = 1 Source = C Dest = A Spare = B B A C Base case: Print: from C to A hanoi(2, ‘C’, ‘B’, ‘A’) Count = 2 Source = C Dest = B Spare = A ? ? Print: from C to B A C B C A B C B A A B C hanoi(1, ‘A’, ‘B’, ‘C’ Count = 1 Source = A Dest = B Spare = C Check Animation 3 Steps Base case: Print: from A to B COMPSCI105

  13. 22.2 The Towers of Hanoi Call Tree hanoi(3, ‘A’, ‘B’, ‘C’) hanoi(2, ‘A’, ‘C’, ‘B’) Move A -> B hanoi(2, ‘C’, ‘B’, ‘A’) hanoi(1, ‘A’, ‘B’, ‘C’) hanoi(1 , ‘C’, ‘A’, ‘B’) C A B C B A B A C Move A -> C Move C -> B hanoi(1, ‘B’, ‘C’, ‘A’) hanoi(1, ‘A’, ‘B’, ‘C’) A C B C A B C B A A B C hanoi(3…) uses 10 calls, a top-level one and 9 recursive calls COMPSCI105

  14. Search dictionary OR Search first half of dictionary Search first half of dictionary 22.3 Binary SearchBinary Search Cost? • Problem: look for an element (key) in an ordered collection • (e.g. find a word in a dictionary) • Sequential search • Starts at the beginning of the collection • Looks at every item in the collection in order until the item being searched for is found • Binary search • Repeatedly halves the collection and determines which half could contain the item • Uses a divide and conquer strategy COMPSCI105

  15. 22.3 Binary SearchImplementation • Implementation issues: • How will you pass “half of list” to the recursive calls to binary_search? • How do you determine which half of the list contains value? • What should the base case(s) be? • How will binary_search indicate the result of the search? • Example: a sorted list COMPSCI105

  16. 22.3 Binary SearchAlgorithm Design Left half: [first .. mid-1] Right half: [mid+1 .. last] first mid = (first + last)/2 last • Base case: • If array is empty number is not in the list, or • If element is the one we look for return it • Recursive call • Determine element in the middle • If the one we look for is smaller than element in the middle then search in the left half • Otherwise search in the right half of the list COMPSCI105

  17. 22.3 Binary SearchExample Code def binary_search(num_list, first, last, value): index = 0 if first > last: index = -1 else: mid = (first + last) // 2 if value == num_list[mid]: index = mid elif value < num_list[mid]: index = binary_search(num_list, first, mid-1, value) else: index = binary_search(num_list, mid+1, last, value) return index COMPSCI105

  18. index 22.3 Binary SearchCall Tree binary_search(list,0,7,9) value 9 first  0 last  7 value  anArray[3] value 9 first  2 last  2 value  anArray[2] binary_search(list,0,2,9) value 9 first  0 last  2 value  anArray[1] binary_search(list,2,2,9) return 2 A successful search for 9: COMPSCI105

  19. index 22.3 Binary SearchCall Tree binary_search(list,0,7,6) value 6 first  0 last  7 value  anArray[3] value 6 first  2 last  2 value  anArray[2] binary_search(list,0,2,6) value 6 first  0 last  2 value  anArray[1] binary_search(list,2,2,6) value 6 first  2 last  1 first  last return 1 return -1 binary_search(list,2,1,6) An unsuccessful search for 6: COMPSCI105

  20. 22.4 EfficiencyRecursion and Efficiency • Some recursive solutions are so inefficient that they should not be used • Factors that contribute to the inefficiency of some recursive solutions • Overhead associated with method calls • Inherent inefficiency of some recursive algorithms • Note: The recursive methods binarySearch and solveTowers are quite efficient, but the fact and rabbit are inefficient COMPSCI105

  21. 22.4 Efficiency Recursion and Efficiency • Do not use a recursive solution if it is inefficient and if you have a clear, efficient iterative solution • Factorial’s recursive solution • Overhead associated with method calls • Iterative method is more efficient • Rabbit counting recursive solution • Inherent inefficiency of recursive algorithms • It computes same values over and over again, e.g., rabbit(7) computes rabbit (3) five times. • Use rabbit’s recurrence relation to construct an efficient iterative solution COMPSCI105

  22. 22.4 Efficiency Iterative Solution def iterative_rabbit(n): previous = 1 current = 1 next_elt = 1 for i in range(3, n+1): next_elt = current + previous; previous = current; current = next_elt return next_elt COMPSCI105

  23. Exercise • Given the following: • my_list = [0, 1, 2, 8, 13, 17, 19, 32, 42] • Draw the call tree of • binary_search(my_list,0,8,3) • binary_search(my_list,0,8,13) COMPSCI105

More Related