Question

The algorithm is taken from great “Algorithms and Programming: Problems and Solutions” by Alexander Shen (namely exercise 1.1.28).

Following is my translation from Russian so excuse me for mistakes or ambiguity. Please correct me if you feel so.

What should algorithm do

With given natural n algorithm calculates the number of solutions of
inequality

x*x + y*y < n 

in natural (non-negative) numbers without using
manipulations on real numbers

In Pascal

k := 0; s := 0;

{at this moment of execution
 (s) = number of solutions of inequality with
 x*x + y*y < n, x < k}
while k*k < n do begin
  l := 0; t := 0;

  while k*k + l*l < n do begin
    l := l + 1;
    t := t + 1;
  end;

  {at this line 
   (t) = number of solutions of k*k + y*y < n
   for given (k) with y>=0}
  k := k + 1;
  s := s + t;
end;

{k*k >= n, so s = number of solutions of inequality}

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Further in the text Shen says briefly that number of operations performed by this algorithm is “proportional to n, as one can calculate”. So I ask you how one can calculate that with strict mathematics.

Answer 1

You have two loops, one inside the other.

The external has this condition: k*k < n so k goes from 0 up to SQRT(n)

and the internal loop has this condition: k*k + l*l < n so l goes from 0 up to SQRT(n-k^2). But this is snaller than SQRT(n)

So, the maximum iterations is less than SQRT(n) * SQRT(n) which is n and in every iteration a constant number of operations is done.