function [ p, more, even ] = perm1_next ( n, p, more, even )
%*****************************************************************************80
%
%% perm1_next() computes permutations of (1,...,N), one at a time.
%
% Discussion:
%
% The routine is initialized by calling with MORE = TRUE, in which case
% it returns the identity permutation.
%
% If the routine is called with MORE = FALSE, then the successor of the
% input permutation is computed.
%
% Licensing:
%
% This code is distributed under the GNU LGPL license.
%
% Modified:
%
% 13 June 2015
%
% Author:
%
% Original FORTRAN77 version by Albert Nijenhuis, Herbert Wilf.
% This version by John Burkardt.
%
% Reference:
%
% Albert Nijenhuis, Herbert Wilf,
% Combinatorial Algorithms,
% Academic Press, 1978, second edition,
% ISBN 0-12-519260-6.
%
% Input:
%
% integer N, the number of objects being permuted.
%
% integer P(N), the output value of P from the previous call.
% However, on an initialization call, with MORE = FALSE, the value of P
% is not needed.
%
% logical MORE, should be FALSE on the first call, to force
% initialization. Thereafter, it should be TRUE, to request the next
% permutation in the sequence.
%
% logical EVEN, is the output value of EVEN from the previous call.
% However, on an initialization call, with MORE = FALSE, the value of EVEN
% is not needed.
%
% Output:
%
% integer P(N), the next permutation.
%
% logical MORE, indicates that there are more permutations
% that may be generated.
%
% logical EVEN, is TRUE if the output permutation is even,
% that is, involves an even number of transpositions.
%
if ( n == 1 )
p(1) = 1;
more = false;
even = true;
return
end
if ( ~ more )
p = i4vec_indicator1 ( n );
more = true;
even = true;
if ( p(n) ~= 1 || p(1) ~= 2 + mod ( n, 2 ) )
return
end
for i = 1 : n - 3
if ( p(i+1) ~= p(i) + 1 )
return
end
end
more = false;
else
if ( even )
ia = p(1);
p(1) = p(2);
p(2) = ia;
even = false;
if ( p(n) ~= 1 || p(1) ~= 2 + mod ( n, 2 ) )
return
end
for i = 1 : n - 3
if ( p(i+1) ~= p(i) + 1 )
return
end
end
more = false;
return
else
more = false;
is = 0;
for i1 = 2 : n
ia = p(i1);
i = i1 - 1;
id = 0;
for j = 1 : i
if ( ia < p(j) )
id = id + 1;
end
end
is = id + is;
if ( id ~= i * mod ( is, 2 ) )
more = true;
break
end
end
if ( ~ more )
p(1) = 0;
return
end
end
m = mod ( is + 1, 2 ) * ( n + 1 );
for j = 1 : i
if ( i4_sign ( p(j) - ia ) ~= i4_sign ( p(j) - m ) )
m = p(j);
l = j;
end
end
p(l) = ia;
p(i1) = m;
even = true;
end
return
end