Recursion as a Problem Solving Technique

Tom Kelliher, CS23

Mar. 19, 1996

Backtracking

Guessing at a solution and backing up out of a deadend.

Four Queens

Place 4 Queens on a 4 by 4 chess board such that no Queen can be attacked by another.

What's recursive about this?

Try it.

Pseudocode:

int place(int column)
{
   int row = 1;

   if (column > 4)
      return success;

   while (row < 5)
   {
      if safe(row, column)   // check that no attacks can occur
      {
         place queen at board[row][column];
         if (place(column + 1) == success)
            return success;
         pick up queen at board[row][column];
      }

      ++row;
   }

   return failure;
}

Maze Routing Example

How would you find a route through the maze?

Representation of the maze?

Pseudocode:

int search(block)
{
   mark block "path";

   if block is goal
      return success;

   if block to north exists and is unvisited
      if (search(block to north) == success)
         return success;

   if block to east exists and is unvisited
      if (search(block to east) == success)
         return success;

   if block to south exists and is unvisited
      if (search(block to south) == success)
         return success;

   if block to west exists and is unvisited
      if (search(block to west) == success)
         return success;

   mark block "no path";
   return failure;
}

Grammars

A language is a set of strings of symbols. The C++ language is the set of strings of symbols accepted by a C++ compiler.

A grammar is a set of rules describing a language. The rules tell you how to build a string belonging to the language.

Consider a simple grammar:

<expr>=<term>{+|-<term>}
<term>=<factor>{*|/<factor>}
<factor>=(<expr>)|<letter>
<letter>=a|b|c|...|z

Extended Backus-Naur Form (EBNF):

• Derivation
• Productions
• Non-Terminals
• Terminals
• Repetition
• Alternation

Expressing associativity, precedence?

Construct the derivation, parse tree, and evaluation tree for:

1. (a + b) * (c + d)

2. a + b * c + d

3. (a + b) * c + d

4. a + b * (c + d)

Relationship between a production and a recursive set of functions?

/**********************************************************************
 * expr.cc --- Convert infix expression to postfix form and indicate
 * if the infix expression was properly formed.
 *
 * This is a recursive descent parser, implementing the following
 * grammar:
 *    <expr>=<term>{+|-<term>}
 *    <term>=<factor>{*|/<factor}
 *    <factor>=(<expr>)|<letter>
 *    <letter>=a|b|c|...|z
 **********************************************************************/


#include <iostream.h>
#include <ctype.h>


void GetToken(void);
int Expression(void);
int Term(void);
int Factor(void);


char Token;   // Next character in the input stream --- one character 
              // lookahead.


/**********************************************************************
 * main()
 **********************************************************************/

int main()
{
   cout << "Type return on an otherwise empty line to exit.\n\n";

   while (1)
   {
      cout << "Expression: ";   // Read infix expression.
      GetToken();

      if (Token == '\n')   //
         return 0;

      // Parse the expression; if successful and we've read the entire line,
      // then we saw a valid expression

      if (Expression() && Token == '\n')
         cout << "\nValid expression\n";
      else
      {
         cout << "\nInvalid expression\n";
         while (Token != '\n')   // finish off the line
            GetToken();
      }
   }

   return 0;
}


/**********************************************************************
 * Expression() --- Parse an expression
 * This implements the production <expr>=<term>{+|-<term>}
 **********************************************************************/

int Expression(void)
{
   char Operator;

   if (!Term())
      return 0;

   while (Token == '+' || Token == '-')
   {
      Operator = Token;
      GetToken();   // Remove operator from input stream.
      if (!Term())
         return 0;
      cout << Operator;   // The operator is output following the
                          // operands --- converting the expression
                          // to postfix.
   }
   return 1;
}


/**********************************************************************
 * Term() --- Parse an term
 * This implements the production <term>=<factor>{*|/<factor>}
 **********************************************************************/

int Term(void)
{
   char Operator;

   if (!Factor())
      return 0;

   while (Token == '*' || Token == '/')
   {
      Operator = Token;
      GetToken();   // Remove operator from input stream.
      if (!Factor())
         return 0;
      cout << Operator;   // Postfix formulation.
   }
   return 1;
}


/**********************************************************************
 * Factor() --- Parse an factor
 * This implements the productions <factor>=(<expr>)|<letter>
 * and <letter>=a|b|c|...|z
 **********************************************************************/

int Factor(void)
{
   if (islower(Token))
   {
      cout << Token;   // Operands output immediately
      GetToken();   // Remove operand from input stream.
      return 1;
   }
   else if (Token == '(')
   {
      GetToken();   // Remove '(' from input stream.
      if (Expression() && Token == ')')   // check for parenthesized
                                          // expression
      {
         GetToken();   // Remove ')' from input stream.
         return 1;
      }
      else return 0;
   }
   else
      return 0;
}


/**********************************************************************
 * GetToken() --- read the next token from cin into the global variable
 * Token.
 **********************************************************************/

void GetToken(void)
{
   cin.get(Token);
}

Actions associated with each production?

Could we evaluate expressions?

Could we add an assignment operator?