Ada Programming/Algorithms/Chapter 6

Chapter 6: Dynamic Programming

Fibonacci numbers

The following codes are implementations of the Fibonacci-Numbers examples.

Simple Implementation

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To calculate Fibonacci numbers negative values are not needed so we define an integer type which starts at 0. With the integer type defined you can calculate up until Fib (87). Fib (88) will result in an Constraint_Error.

  type Integer_Type is range 0 .. 999_999_999_999_999_999;

You might notice that there is not equivalence for the assert (n >= 0) from the original example. Ada will test the correctness of the parameter before the function is called.

  function Fib (n : Integer_Type) return Integer_Type is
  begin
     if n = 0 then
        return 0;
     elsif n = 1 then
        return 1;
     else
        return Fib (n - 1) + Fib (n - 2);
     end if;
  end Fib;

...

Cached Implementation

...

For this implementation we need a special cache type can also store a -1 as "not calculated" marker

  type Cache_Type is range -1 .. 999_999_999_999_999_999;

The actual type for calculating the fibonacci numbers continues to start at 0. As it is a subtype of the cache type Ada will automatically convert between the two. (the conversion is - of course - checked for validity)

  subtype Integer_Type is Cache_Type range
     0 .. Cache_Type'Last;

In order to know how large the cache need to be we first read the actual value from the command line.

  Value : constant Integer_Type :=
     Integer_Type'Value (Ada.Command_Line.Argument (1));

The Cache array starts with element 2 since Fib (0) and Fib (1) are constants and ends with the value we want to calculate.

  type Cache_Array is
     array (Integer_Type range 2 .. Value) of Cache_Type;

The Cache is initialized to the first valid value of the cache type — this is -1.

  F : Cache_Array := (others => Cache_Type'First);

What follows is the actual algorithm.

  function Fib (N : Integer_Type) return Integer_Type is
  begin
     if N = 0 or else N = 1 then
        return N;
     elsif F (N) /= Cache_Type'First then
        return F (N);
     else
        F (N) := Fib (N - 1) + Fib (N - 2);
        return F (N);
     end if;
  end Fib;

...

This implementation is faithful to the original from the Algorithms book. However, in Ada you would normally do it a little different:

when you use a slightly larger array which also stores the elements 0 and 1 and initializes them to the correct values

  type Cache_Array is
     array (Integer_Type range 0 .. Value) of Cache_Type;

  F : Cache_Array :=
     (0      => 0,
      1      => 1,
      others => Cache_Type'First);

and then you can remove the first if path.

     if N = 0 or else N = 1 then
        return N;
     elsif F (N) /= Cache_Type'First then

This will save about 45% of the execution-time (measured on Linux i686) while needing only two more elements in the cache array.

Memory Optimized Implementation

This version looks just like the original in WikiCode.

  type Integer_Type is range 0 .. 999_999_999_999_999_999;

  function Fib (N : Integer_Type) return Integer_Type is
     U : Integer_Type := 0;
     V : Integer_Type := 1;
  begin
     for I in  2 .. N loop
        Calculate_Next : declare
           T : constant Integer_Type := U + V;
        begin
           U := V;
           V := T;
        end Calculate_Next;
     end loop;
     return V;
  end Fib;

No 64 bit integers

Your Ada compiler does not support 64 bit integer numbers? Then you could try to use decimal numbers instead. Using decimal numbers results in a slower program (takes about three times as long) but the result will be the same.

The following example shows you how to define a suitable decimal type. Do experiment with the digits and range parameters until you get the optimum out of your Ada compiler.

  type Integer_Type is delta 1.0 digits 18 range
     0.0 .. 999_999_999_999_999_999.0;

You should know that floating point numbers are unsuitable for the calculation of fibonacci numbers. They will not report an error condition when the number calculated becomes too large — instead they will lose in precision which makes the result meaningless.