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## -*- texinfo -*-
## @deftypefn  {} {} treelayout (@var{tree})
## @deftypefnx {} {} treelayout (@var{tree}, @var{permutation})
## treelayout lays out a tree or a forest.
##
## The first argument @var{tree} is a vector of predecessors.
##
## The parameter @var{permutation} is an optional postorder permutation.
##
## The complexity of the algorithm is O(n) in terms of time and memory
## requirements.
## @seealso{etreeplot, gplot, treeplot}
## @end deftypefn

function [x_coordinate, y_coordinate, height, s] = ...
                                                 treelayout (tree, permutation)

  if (nargin < 1)
    print_usage ();
  elseif (! isvector (tree) || rows (tree) != 1 || ! isnumeric (tree)
          || any (tree > length (tree)) || any (tree < 0))
    error ("treelayout: the first input argument must be a vector of predecessors");
  endif

  ## Make it a row vector.
  tree = tree(:)';

  ## The count of nodes of the graph.
  num_nodes = length (tree);
  ## The number of children.
  num_children = zeros (1, num_nodes + 1);

  ## Checking vector of predecessors.
  for i = 1 : num_nodes
    if (tree(i) < i)
      ## This part of graph was checked before.
      continue;
    endif

    ## Try to find cicle in this part of graph using modified Floyd's
    ## cycle-finding algorithm.
    tortoise = tree(i);
    hare = tree(tortoise);

    while (tortoise != hare)
      ## End after finding a cicle or reaching a checked part of graph.

      if (hare < i)
        ## This part of graph was checked before.
        break;
      endif

      tortoise = tree(tortoise);
      ## Hare will move faster than tortoise so in cicle hare must
      ## reach tortoise.
      hare = tree(tree(hare));

    endwhile

    if (tortoise == hare)
      ## If hare reach tortoise we found circle.
      error ("treelayout: vector of predecessors has bad format");
    endif

  endfor
  ## Vector of predecessors has right format.

  for i = 1:num_nodes
    ## vec_of_child is helping vector which is used to speed up the
    ## choice of descendant nodes.

    num_children(tree(i)+1) = num_children(tree(i)+1) + 1;
  endfor

  pos = 1;
  start = zeros (1, num_nodes+1);
  xhelp = zeros (1, num_nodes+1);
  stop = zeros (1, num_nodes+1);
  for i = 1 : num_nodes + 1
    start(i) = pos;
    xhelp(i) = pos;
    pos += num_children(i);
    stop(i) = pos;
  endfor

  if (nargin == 1)
    for i = 1:num_nodes
      vec_of_child(xhelp(tree(i)+1)) = i;
      xhelp(tree(i)+1) = xhelp(tree(i)+1) + 1;
    endfor
  else
    vec_of_child = permutation;
  endif

  ## The number of "parent" (actual) node (its descendants will be
  ## browse in the next iteration).
  par_number = 0;

  ## The x-coordinate of the left most descendant of "parent node"
  ## this value is increased in each leaf.
  left_most = 0;

  ## The level of "parent" node (root level is num_nodes).
  level = num_nodes;

  ## num_nodes - max_ht is the height of this graph.
  max_ht = num_nodes;

  ## Main stack - each item consists of two numbers - the number of
  ## node and the number it's of parent node on the top of stack
  ## there is "parent node".
  stk = [-1, 0];

  ## Number of vertices s in the top-level separator.
  s = 0;
  ## Flag which says if we are in top level separator.
  top_level = 1;
  ## The top of the stack.
  while (par_number != -1)
    if (start(par_number+1) < stop(par_number+1))
      idx = vec_of_child(start(par_number+1) : stop(par_number+1) - 1);
    else
      idx = zeros (1, 0);
    endif

    ## Add to idx the vector of parent descendants.
    stk = [stk; [idx', ones(fliplr(size(idx))) * par_number]];

    ## We are in top level separator when we have one child and the
    ## flag is 1
    if (columns (idx) == 1 && top_level == 1)
      s += 1;
    else
      ## We aren't in top level separator now.
      top_level = 0;
    endif
    ## If there is not any descendant of "parent node":
    if (stk(end,2) != par_number)
     left_most += 1;
     x_coordinate_r(par_number) = left_most;
     max_ht = min (max_ht, level);
     if (length (stk) > 1 && find ((shift (stk,1) - stk) == 0) > 1
         && stk(end,2) != stk(end-1,2))
        ## Return to the nearest branching the position to return
        ## position is the position on the stack, where should be
        ## started further search (there are two nodes which has the
        ## same parent node).

        position = (find ((shift (stk(:,2), 1) - stk(:,2)) == 0))(end) + 1;
        par_number_vec = stk(position:end,2);

        ## The vector of removed nodes (the content of stack form
        ## position to end).

        level += length (par_number_vec);

        ## The level have to be decreased.

        x_coordinate_r(par_number_vec) = left_most;
        stk(position:end,:) = [];
      endif

      ## Remove the next node from "searched branch".

      stk(end,:) = [];
      ## Choose new "parent node".
      par_number = stk(end,1);
      ## If there is another branch start to search it.
      if (par_number != -1)
        y_coordinate(par_number) = level;
        x_coordinate_l(par_number) = left_most + 1;
      endif
    else

      ## There were descendants of "parent nod" choose the last of
      ## them and go on through it.
      level -= 1;
      par_number = stk(end,1);
      y_coordinate(par_number) = level;
      x_coordinate_l(par_number) = left_most + 1;
    endif
  endwhile

  ## Calculate the x coordinates (the known values are the position
  ## of most left and most right descendants).
  x_coordinate = (x_coordinate_l + x_coordinate_r) / 2;

  height = num_nodes - max_ht - 1;

endfunction


%!test
%! % Compute a simple tree layout
%! [x, y, h, s] = treelayout ([0, 1, 2, 2]);
%! assert (x, [1.5, 1.5, 2, 1]);
%! assert (y, [3, 2, 1, 1]);
%! assert (h, 2);
%! assert (s, 2);

%!test
%! % Compute a simple tree layout with defined postorder permutation
%! [x, y, h, s] = treelayout ([0, 1, 2, 2], [1, 2, 4, 3]);
%! assert (x, [1.5, 1.5, 1, 2]);
%! assert (y, [3, 2, 1, 1]);
%! assert (h, 2);
%! assert (s, 2);

%!test
%! % Compute a simple tree layout with defined postorder permutation
%! [x, y, h, s] = treelayout ([0, 1, 2, 2], [4, 2, 3, 1]);
%! assert (x, [0, 0, 0, 1]);
%! assert (y, [0, 0, 0, 3]);
%! assert (h, 0);
%! assert (s, 1);
