Lecture8 Tree Recursion

Reading 1.7
1. Order of Recursive Calls
<1>【eg1】The Cascade Function

>>> def cascade(n):
        """Print a cascade of prefixes of n."""
        if n < 10:
            print(n)
        else:
            print(n)
            cascade(n//10)
            print(n)
            
>>> cascade(2013)
2013
201
20
2
20
201
2013

<2>【eg2】Inverse Cascade

def inverse_cascade(n): grow(n)
    print(n)
    shrink(n)
def f_then_g(f, g, n):
	 if n:
		f(n)
		g(n)
		
grow = lambda n: f_then_g(grow, print, n//10) shrink = lambda n: f_then_g(print, shrink, n//10)


>>> inverse_cascade(2013)
1
12
123
1234
123
12
1

2. Tree Recursion
<1>【eg1】Fibonacci numbers(与树结构进行联系）

n: 0, 1, 2, 3, 4, 5, 6, 7, 8, …,
fib(n): 0, 1, 1, 2, 3, 5, 8, 13, 21,

	def fib(n):
	    if n == 1:
	        return 0
	    if n == 2:
	        return 1
	    else:
	        return fib(n-2) + fib(n-1)
	
	result = fib(6)

<2>与树结构的联系

3. Repetition in Tree-Recursive Computation
4. 【举例】Counting Partitions

The number of partitions of a positive integer n, using parts up to size m, is the number of ways in which n can be expressed as the sum of positive integer parts up to m in increasing order.
count_partitions(6, 4)
6 = 2 + 4
6 = 1 + 1 + 4
6 = 3 + 3
6 = 1 + 2 + 3
6 = 1 + 1 + 1 + 3
6 = 2 + 2 + 2
6 = 1 + 1 + 2 + 2
6 = 1 + 1 + 1 + 1 + 2
6 = 1 + 1 + 1 + 1 + 1 + 1

解题思路
• Recursive decomposition: finding simpler instances of the problem.
• Explore two possibilities: •Use at least one 4
•Don’t use any 4
• Solve two simpler problems: •count_partitions(2, 4) •count_partitions(6, 3)
• Tree recursion often involves exploring different choices.

所用代码如下：

>>> def count_partitions(n, m):
        """Count the ways to partition n using parts up to m."""
        if n == 0:
            return 1
        elif n < 0:
            return 0
        elif m == 0:
            return 0
        else:
            return count_partitions(n-m, m) + count_partitions(n, m-1)

>>> count_partitions(6, 4)
9
>>> count_partitions(5, 5)
7
>>> count_partitions(10, 10)
42