The squares in the following figure (not recreated here - only in paper notes) which is 5 squares high and 5 squares wide with the start being 1 square on top and expanded by 1 in each row until 5 are across and 5 are high.   What is the permiter of the figure?  

REMARK
This problem can readily be solved by application of the problem solving strategy known as ``Goal Directed Planning''.  

REMARK
Goal Directed Planning is a ``procedural'' variant of problem decomposition,  which we discussed in the September 6 th class.  

Definition   GOAL DIRECTED PLANNING is the problem solving strategy of...

1. Identifying a goal.
   
2. Determining the information needed to achiveve the goal.
   
3. Establish as sub-goals the problems of securing the information needed to achieve the goal.
   
4. ``Recursively'' applying the same method to the problems established as sub-goals.
   

Solving the permiter problem by means of Goal Directed Planning.

• To determine the perimeter of the figure you need to know:
  
  • The length of a small square side
    
  • The number of such sides
    
• To determine the length of a small square side we need to know:
  
  • The area of the square
    
• To determine the area of one small square
  
  • The # of small squares
    
  • The total area of the figure
    

An approach to writing the program.  

Working in a ``bottom-up'' fashion either

• Record an item of information,  or
  
• Compute a value.
  

• Record the total area
  
• Record the number of small squares
  
• Computing the area of a small square
  
• Compute the length of one side
  
• Record # of small sides around the perimeter
  
• Compute the perimeter
  

In Java  

/ /Compute the perimeter   double totalArea = 187.32;   int numSmallSquares = 15;   double squareArea = totalArea /numSmallSquares;   double sideLength = Math.sqrt (squareArea);   int numExternalSides = 20;   double perimeter = sideLength * numExternalSides;     / /Display   IO.println ( ``The perimeter is'' + perimeter + ``square units'');

 

More on the Math class
(1) Form:   Math.sqrt (< double >) - > < double >
Meaning:   Compute the square root of the permiter.
ex:
double value = Math.sqrt (16.0);
value - > 4.0;  

(2) Form:   Math.PI - > < double >
Meaning:   Reference the classic greek constant - 3.14 ..
ex:
double r = 5.8;
double a = r * r * Math.PI;  / /the area of a circle of radius 5.8  

(3) Form:   Math.pow (< double >,  < double >) - > < double >
Meaning:   Raise the first value to the power of the second
ex:
Math.pow (3.0,  2) - > 9.0
Math.pow (3,  4) - > 81.0
Legal:   Math.pow (1000,  1 / 2.0) - > cube root of a 1000 = 10  

Note on arithmetic
< int > + < int > - > < int >
< double > + < double > - > < double >
< int > + < double > - > < double >
similarly for * - /
ex:
IO.println (5 / 2) - > 2
IO.println (5.0 / 2.0