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Assignment on Visitor pattern and advanced C++ topics

Reverse Polish Notation Calculator

Assignment on Visitor pattern and advanced c++ topics.


This assignment has only one due date. All questions are due on the same due date. You must submit the online quiz on Learn by the Quiz date.

You must use the C++ I/O streaming and memory management facilities on this assignment. Marmoset will be programmed to reject submissions that use C-style I/O or memory management.

You may include the headers , , , , , , , , , , , and .

You are not allowed to use new or delete. You must manage memory with stack variables and/or smart pointers.

Each question on this assignment asks you to write a C++ program, and the programs you write on this assignment each span multiple fifiles. For this reason, we strongly recommend that you develop your solution for each question in a separate directory Just remember that, for each question, you should be in that directory when you create your ZIP fifile, so that it does not contain any extra directory structure.

1. Some early handheld calculators, as well as several programming languages, use a system for describing arithmetic expressions called reverse Polish notation. Reverse Polish notation is easiest thought of as describing a mathematical expression in a sequence of steps that use a stack. For instance, you might push 5 to the stack, push 3 to the stack, then perform the “add” step, which pops the top two elements from the stack and adds them together, then pushes the result to the stack. In reverse Polish notation, this sequence of actions is written as “5 3 +”. All steps are either pushing a number onto the stack, or popping two numbers to perform an operation, pushing the result. These sequences can be arbitrarily long, and because of the stack-like action of pushing and popping, no parentheses are needed to show precedence. To be valid, a reverse Polish notation expression must result in exactly one element on the stack.

For instance, the mathematical expression “(5+2)/(4-1)+7” can be rewritten as “5 2 + 4 1 - / 7 +”. Since the + and - each pop their respective two numbers and push one, the / operates over the results of the + and -. Here are the steps in evaluating that expression, one by one, assuming / is int division:

Current operation

You will write a reverse Polish notation calculator, which additionally reformats reverse Polish notation expressions to standard mathematical expressions, also called infifix notation. For instance, the sequence “5 3 +” from above is written as “5+3” in infifix notation, and evaluates to 8. Your calculator must support the standard four arithmetic operators: +, -, *, and /, operating over ints.

Creating a Tree from Reverse Polish Notation

Your solution does not need to handle inputting negative numbers, but negative numbers must be supported in the  output. Please note that no base code at all is provided. You must write the entire solution from scratch.

Your program will read lines in reverse Polish notation from standard input, and output lines containing both the infifix expression and its solution. Correctly placing parentheses in infifix expression is complex, so simply parenthesize every operator with its operands. For instance, if the user inputs 5 2 + 4 1 - / 7 +, the program should output (((5+2)/(4-1))+7) = 9. Note that in our reverse Polish notation, all operations must be separated by spaces, but our infifix notation will never use spaces.

A demonstration of an interaction with the program (lines in reverse Polish notation are input, and lines with infifix notation and solution are program output):

5 2 + 4 1 - / 7 + (((5+2)/(4-1))+7) = 9 5 3 + (5+3) = 8 1 2 3 4 5 6 7 + - * / + - (1-(2+(3/(4*(5-(6+7)))))) = -1 1 2 + 3 - 4 * 5 / 6 + 7 - ((((((1+2)-3)*4)/5)+6)-7) = -1 49 8 * 1000 * 3 100 * 5 12 * 7 + - 3 + 8 ((((49*8)*1000)+(((3*100)-((5*12)+7))*3))*8) = 3141592

To do this, you will have to create a tree from the reverse Polish notation input, then traverse the tree. You must use the  visitor pattern to traverse the tree, as this will make it easy to write one visitor that creates the infifix notation expression, and another that gets the solution. You must use polymorphism effectively to make good use of the visitor pattern. Note that it is possible to solve reverse Polish notation expressions with no data structures, but not to rewrite them in infifix notation, so you will need to create a tree.

To build a tree from a reverse Polish notation expression, simply follow the same stack operations as described above, but pushing or popping trees instead of pushing or popping values. For an operation, rather than actually performing the operation, make a new tree with the two trees you popped as its children. Here are the steps of converting “5 2 + 4 1 - /” to a tree, visually:

Current operation

You will likely want a subclass of your tree class for numbers, and subclasses for each of the valid operations. To solve, the visitor of an operation should call the visitor of each child, and return the result of the operation as performed on the returns from the visitor calls to each child. To create an infifix notation expression, visitors should instead return strings, and concatenate them as appropriate.

Visitor Pattern Implementation

You will need to check that the input uses the stack correctly (i.e., it never attempts to use an operation with only one element on the stack, and ends with exactly one element on the stack), but exactly what to do in this case is not specifified.

Due on the due date: Write this program in C++. Save your solution in a fifile named It must contain a Makefile that creates an executable named a5q1 when the command make is given.

2. Regular expressions, such as those used by egrep, are checked by fifinite automata (FAs). A fifinite automaton (FA) is a directed graph in which each edge is labeled with a character. You’ll see, or you have seen, a lot more about FAs, andin particular the distinction between nondeterministic and deterministic FAs, in CS241. In this problem, you’ll be using a simplifified form of FAs backwards, so it’s not important whether you’ve already heard of them.

Given the description of a directed graph with characters labeling each edge (that is, an FA), in this problem, you’ll be  fifinding all of the strings of a certain length represented by that graph. That is, fifind every path of some length, and for each of those paths, give the string represented by the characters along that path. Note that in real FAs, there are “accepting” nodes, so not every path is actually important, but for this problem, we simply want every possible path.

You are to implement this graph in the class Finite Automaton, defifined in the header finiteautomaton.h, which provides addNode and addEdge methods for building the graph, and a search method to search for matching strings of a given length. This class also contains a nested class Node, representing a node in the graph, but the implementation of Node is entirely up to you; users of Finite Automaton should not attempt to call any methods or access any fifields directly on a Node.

Node *add Node() adds a node to the FiniteAutomaton and returns a pointer to it. Remember that for this entireassignment, you may not use new or delete, so you are expected to allocate this Node using smart pointers. Nodes are opaque to users of FiniteAutomaton, so this method has no arguments, and the returned Node is only useful for passing back into the FiniteAutomaton’s other methods. void addEdge(Node *a, Node *b, char c) adds an edge in the graph from node a to node b labeled by the character c. The behavior is not defifined if a or b are not part of this FiniteAutomaton or are otherwise invalid, or if an edge from a to b labeled by c already exists. However, it is valid to have multiple edges between the same pair of nodes with different character labels, or for a node to link to itself, or for a node to have multiple edges labeled by the same character.

vector search(Node *start, int len) searches for strings of length len. To do this, it must fifind all paths starting from the node start of length len, and for each, generate a string which contains the characters of the labels of the edges along that path, in the order that they appear in the path. The vector of strings must be in lexicographic order,

and every string in it must be of exactly len characters long. Note that it is possible to generate the strings in lexicographic order in the fifirst place, but is also acceptable to simply sort the strings after you’ve found them all. The vector may be empty, if there are no paths of the given length. A string may appear multiple times, if there are multiple distinct paths labeled by the same characters.

If we call search(n, 5), where n points to the node labeled “1”, we should get a vector that contains the strings

“abcab”, “abcax”, “axbca”, “axxbc”, “axxxb”, and “axxxx”. These strings, in turn, represent the paths 1-2-3-1-2-3,

1-2-3-1-2-2, 1-2-2-3-1-2, 1-2-2-2-3-1, 1-2-2-2-2-3, and 1-2-2-2-2-2.

Another example:

If we call search(n, 3), where n points to the node labeled “1”, we should get a vector that contains the strings “aaa”, “aaa”, “aaa”, “aaa”, “aaa”, “aad”, “aba”, “aba”, “abc”, “abd”, “aca”, “aca”, “acd”, “ada”, “ada”, and “ada”. These strings, in turn, represent the paths 1-4-4-4, 1-4-4-1, 1-4-1-2, 1-4-1-3, 1-4-1-4, 1-4-4-1, 1-2-4-4, 1-2-4-1, 1-2-3-4, 1-2-4-1, 1-3-4-4, 1-3-4-1, 1-3-4-1, 1-4-1-2, 1-4-1-3, and 1-4-1-4. There’s no way to determine which string came from which path, so the exact order of these paths isn’t important when they generate the same string.

A compiled version is also provided, named fasearch, to test against. Compile along with your own code to create your own fasearch, which is a simple command-line application which creates a single FiniteAutomaton and handles simple commands. Because of the simplicity of the command-line interface, it does not support spaces or other whitespace characters as the labels for links, but your library must support any characters in labels. The commands are:


n Creates a new node.

e from to label

Adds a link from and to the given nodes, labeled by the given character. s from length  Searches for strings of the given length starting from the given node, and outputs them to standard output.

c Clears the graph by creating a new FiniteAutomaton.

q Quits.

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