Tower Of Hanoi Program In C Using Graphics In Reading

Mathematical Programs. Void hanoi(char from,char to,char other,int n) {. Other Related Programs in c.

The following is a guest post by Christopher Lee and originally appeared on Christopher is currently in the Ruby-003 class at The Flatiron School. You can follow him on Twitter. During my second week of learning how to code at the Flatiron School, we were given the Towers of Hanoi problem to apply our new found knowledge of recursion. Needless to say, it was really challenging. From my experience, what makes Towers of Hanoi difficult is two-fold. First, recognizing the pattern was far from obvious (I spent hours painstakingly moving paper discs around).

Second, once you have an algorithm to solve the problem, it’s not exactly clear how the computer executes the recursion calls. This post is intended to help shed light on the latter. For those not familiar with the rules, there are three pegs (source, spare, and destination). The goal is to move all of the discs from the source peg to the destination peg one at a time without ever having a larger disc on top of a smaller disc. As shown in the diagram below, let’s assume that the ‘A’ peg is the source peg, ‘B’ peg is the spare peg, and the ‘C’ peg is the destination peg. Source: Conceptually, we can break the problem down into a few moves. First, we need to move n-1 discs to the spare disc.

Then, we are free to move the largest disc (nth disc) to the destination peg. Once we have the largest disc on the destination peg, we need to move the original n-1 discs (now on the spare peg) to the destination peg.

With this logic, we can define a function that looks like the algorithm in the figure below. Source: However, you will notice that we are calling the original function again whenever we call the function with a number of discs greater than zero. This makes sense since moving n-1 discs all together is not a legal move. Thus, the same function is called again with n-1 discs, and the pattern continues until we reach the base case where the number of discs is equal to one and the base case solution executes. The above pattern is both what makes recursion powerful and hard to understand. Given the repetition of the same logic, it’s helpful to be able to make recursive function calls as it enables us to have a concise solution. However, tracing the execution becomes difficult given the nested nature of the function calls.

As we step through each level of recursion, it’s important to realize that the interpreter executes the code in a linear, top-down manner. With each recursive function call, the computer attempts to find a solution to the smaller instance before returning to its original place in the program. This is why large recursive calls lead to stack overflow. Returning to the algorithm, you may have already noticed that there is an additional layer of complexity. With each additional recursion call, the “source, spare, and destination” pegs change because the arguments to the function call switch in comparison to the original function. So, in the first line of the else statement, the original spare peg becomes the new destination peg, and the original destination peg becomes the new spare peg.

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For the last line of the else statement, the original spare peg becomes the new source peg, and the original source peg becomes the new spare peg. With that in mind, let’s step through each level of execution.

Given 4 discs, we call the MoveTower function the first time with 4 discs. Based on the function definition, we know that the condition in the if statement evaluates to false and consequently, does not run. The interpreter then moves to execute the body of the else statement (figure below), which we’ll call the n=4 scenario to keep track of where we are in the program (i.e.