Python implements the tower-based recursive algorithm

Source: Internet
Author: User
In this article, we will share with you a classic example of python's implementation of Jakarta recursive algorithm. if you are interested, you can refer to the example of a yoga Tower exercise when learning recursion, juita should be a classic entry-level case for learning computer recursive algorithms, so I think I can write an article to express my opinions. This markdown editor is not very easy to use yet. it may be a bit ugly to write it in a format. you may have to forgive me.

I found a picture of the tower on the internet. The middle column is used to fold the disc on the leftmost column from large to small, to put it bluntly, c must be the same as.

To put it bluntly, please first highlight the code

def move(n, a, buffer, c):  if(n == 1):    print(a,"->",c)    return  move(n-1, a, c, buffer)  move(1, a, buffer, c)  move(n-1, buffer, a, c)move(3, "a", "b", "c")

The first step is to define a moving function. The four parameters represent the number of plates on the column, and the buffer is also the B column. The name is buffer, which is easy to understand. as its name suggests, it is a buffer for a to move to c. then c is the target pillar.
Let's read the function code.
There must be a condition for stopping recursive loops in the general method of recursion. Therefore, when determining the number of plates on column A as 1, the recursion can be aborted and returned, when there is only one column on the column, it must move a to c. the focus is on the following code. recursion is actually a very abstract algorithm, we need to use abstract thinking to think about the Tower of Hanoi. think of the plate on the column a as two parts, that is, the plate above and the bottom plate. if

Let me write down the entire movement process. Here are three examples on the column.

/** I demonstrated all the three pie towers by code. according to the indentation principle, each indent is a recursive function, and each print ends the current recursion, that is, for each print description: 1.n = 3, n = 2, n = 1 is the result of every execution of if (n = 1). No judgment is written here, I believe that the children's shoes can also understand, that is, when n is not equal to 1, 1 is subtracted to go to recursion 2. note that column a, column B, and column c enter the function sequence each time. do not take the wrong path of the form parameter. check that the real parameters of each function parameter **/move (3, "", "B", "c") n = 3: // start to move n-1 from a, that is, two plates are moved to B through c, to free up c for a to move the last plate move (2, "a", "c", "B") n = 2: // Start a recursion of n = 2, move n-1 plates on the column a ('A') to B ('C') through c ('B') move (1, "a", "B ", "c") n = 1: // the first recursion of n = 2 is completed. print the result and run the print ("a", "-> ", "c") move (1, "a", "c", "B") n = 1: print ("a", "->", "B ") move (1, "c", "a", "B") n = 1: print ("c", "->", "B ") // here the N-1 above column a is the movement of two plates // start to move the last plate on column a to column c (1, "a", "B ", "c") n = 1: print ("a", "->", "c ") // move the last plate on the column to the c column here (2, "B", "a", "c") n = 2: // start the second recursion of n = 2, that is, the current B ('B') plate (n-1) through a ('A ') move to c ('C') and move (1, "B", "c", "a") n = 1: // n = 2. the second recursion is completed, print the result and execute the remaining code print ("B", "->", "a") move (1, "B", "", "c") n = 1: print ("B", "->", "c") move (1, "a", "B", "c ") n = 1: print ("a", "->", "c") // here, move the plate on B to c through a. // the entire code is executed, operation completed

The final result is:

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