 Solution

Use two indices, initially the first and last positions of the ordered sequence. The two elements occupy distinct positions, so continue only while i < j:

i := 0; j := n-1;
WHILE i < j DO
  t := S[i] + S[j];
  IF t = x THEN RETURN TRUE
  ELSIF t < x THEN i := i+1
  ELSE j := j-1
  END
END;
RETURN FALSE

The loop invariant is: a solution exists in the original input if and only if one exists with both indices in [i,j]. If S[i]+S[j] < x, then S[i]+S[k] <= S[i]+S[j] < x for every k <= j, so index i cannot belong to a remaining solution. If the sum is too large, S[k]+S[j] >= S[i]+S[j] > x for every k >= i, so index j cannot belong to one. This proves preservation of the loop invariant. Equality gives the desired pair, and an interval with fewer than two positions contains none.

Every unsuccessful iteration decreases the loop variant j-i by one, so there are at most $n-1$ sum comparisons and only constant space is needed. Duplicate values are permitted if they occupy distinct positions. Under exact constant-cost real arithmetic and endpoint access,