function [ c, fc ] = bisection_integer ( f, a, b ) %*****************************************************************************80 % %% BISECTION_INTEGER seeks an integer root using bisection. % % Discussion: % % A function F(X) confined to integer arguments is given, with an % interval [A,B] over which F changes sign. An integer C is sought % such that A <= C <= B and F(C) = 0. % % Because we are restricted to integer arguments, it may the case that % there is no such C. % % This routine proceeds by a form of bisection, in which the enclosing % interval is restricted to be defined by integer values. % % If the user has given a true change of sign interval [A,B], and if, % in the interval, there is a single integer value C for which F(C) = 0, % with the additional restrictions that F(C-1) and F(C+1) are of opposite % signs, then this procedure should locate and return C. % % In particular, if the function F is monotone, and there is an integer % solution C in the interval, then this procedure will find it. % % However, in general, even if there is an integer C in the interval, % such that F(C) = 0, this procedure may be unable to find it, particularly % if there are also nonintegral solutions within the same interval. % % While any integer function can be used with this program, the bisection % approach is most useful if the integer function is monotone, or % varies slowly, or can be regarded as the restriction to integer arguments % of a continuous (and smoothly varying) function of a real argument. % In such cases, knowing that F is negative at A and positive at B % suggests that F generally increases from A to B, and might attain % the value 0 at some intermediate argument C. % % Licensing: % % This code is distributed under the GNU LGPL license. % % Modified: % % 26 May 2012 % % Author: % % John Burkardt % % Parameters: % % Input, function V = F ( C ), the name of a user-supplied % procedure that evaluates the function. % % Input, integer A, B, two arguments that define a change of % sign interval for F. In other words, F(A) and F(B) must be of opposite % sign. % % Output, integer C, FC, the candidate for the root, as % determined by the program, and its function value. If FC is not zero, % then the procedure did not find a root in the interval, and C is only % an "approximate" root. % % % Ensure that F(A) < 0 < F(B). % fa = f ( a ); fb = f ( b ); if ( fa == 0 ) c = a; fc = fa; elseif ( fb == 0 ) c = b; fc = fb; elseif ( fa < 0 && 0 < fb ) elseif ( fb < 0 && 0 < fa ) t = a; a = b; b = t; t = fa; fa = fb; fb = t; else fprintf ( 1, '\n' ); fprintf ( 1, 'BISECTION_INTEGER - Fatal error!\n' ); fprintf ( 1, ' No change of sign interval supplied.\n' ); fprintf ( 1, ' F(%d) = %d\n', a, fa ); fprintf ( 1, ' F(%d) = %d\n', b, fb ); error ( 'BISECTION_INTEGER - Fatal error!' ); end % % Bisection. % while ( 1 < abs ( b - a ) ) c = floor ( ( a + b ) / 2 ); fc = f ( c ); if ( fc == 0 ) return elseif ( fc < 0 ) a = c; fa = fc; elseif ( 0 < fc ) b = c; fb = fc; end end % % Interval is empty, with FA < 0 and 0 < FB. % Bisection did not produce an integer solution. % Return the argument with smallest function norm. % if ( - fa < fb ) c = a; fc = fa; else c = b; fc = fb; end return end