Mercurial > hg > octave-nkf
annotate scripts/sparse/treelayout.m @ 20511:eca5aa3225f4
imshow.m: Add support for 'parent' property (bug #45473).
* imshow.m: Add cell variable prop_val_args{:} when calling image or imagesc.
Populate prop_val_args with {"parent", hparent} when "parent" argument given
to imshow. Remove code commented out in 2015/05/1 which has not caused any
problems. Add warning that arguments "border" and "reduce" are not supported.
author | Rik <rik@octave.org> |
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date | Sun, 05 Jul 2015 13:55:56 -0700 |
parents | df437a52bcaf |
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rev | line source |
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1 ## Copyright (C) 2008-2015 Ivana Varekova & Radek Salac |
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2 ## |
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3 ## This file is part of Octave. |
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4 ## |
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5 ## Octave is free software; you can redistribute it and/or modify it |
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6 ## under the terms of the GNU General Public License as published by |
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7 ## the Free Software Foundation; either version 3 of the License, or (at |
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8 ## your option) any later version. |
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9 ## |
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10 ## Octave is distributed in the hope that it will be useful, but |
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11 ## WITHOUT ANY WARRANTY; without even the implied warranty of |
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12 ## MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU |
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13 ## General Public License for more details. |
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14 ## |
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15 ## You should have received a copy of the GNU General Public License |
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16 ## along with Octave; see the file COPYING. If not, see |
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17 ## <http://www.gnu.org/licenses/>. |
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18 |
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19 ## -*- texinfo -*- |
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20 ## @deftypefn {Function File} {} treelayout (@var{tree}) |
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21 ## @deftypefnx {Function File} {} treelayout (@var{tree}, @var{permutation}) |
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22 ## treelayout lays out a tree or a forest. |
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23 ## |
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24 ## The first argument @var{tree} is a vector of predecessors. |
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25 ## |
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26 ## The parameter @var{permutation} is an optional postorder permutation. |
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27 ## |
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28 ## The complexity of the algorithm is O(n) in terms of time and memory |
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29 ## requirements. |
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30 ## @seealso{etreeplot, gplot, treeplot} |
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31 ## @end deftypefn |
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32 |
8507 | 33 function [x_coordinate, y_coordinate, height, s] = treelayout (tree, permutation) |
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34 if (nargin < 1 || nargin > 2 || nargout > 4) |
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35 print_usage (); |
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36 elseif (! isvector (tree) || rows (tree) != 1 || ! isnumeric (tree) |
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37 || any (tree > length (tree)) || any (tree < 0)) |
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38 error ("treelayout: the first input argument must be a vector of predecessors"); |
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39 else |
8506 | 40 ## Make it a row vector. |
8507 | 41 tree = tree(:)'; |
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42 |
8506 | 43 ## The count of nodes of the graph. |
8507 | 44 num_nodes = length (tree); |
8506 | 45 ## The number of children. |
8507 | 46 num_children = zeros (1, num_nodes + 1); |
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47 |
8506 | 48 ## Checking vector of predecessors. |
8507 | 49 for i = 1 : num_nodes |
50 if (tree(i) < i) | |
10549 | 51 ## This part of graph was checked before. |
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52 continue; |
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53 endif |
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54 |
8506 | 55 ## Try to find cicle in this part of graph using modified Floyd's |
56 ## cycle-finding algorithm. | |
8507 | 57 tortoise = tree(i); |
58 hare = tree(tortoise); | |
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59 |
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60 while (tortoise != hare) |
10549 | 61 ## End after finding a cicle or reaching a checked part of graph. |
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62 |
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63 if (hare < i) |
8506 | 64 ## This part of graph was checked before. |
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65 break |
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66 endif |
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67 |
8507 | 68 tortoise = tree(tortoise); |
10549 | 69 ## Hare will move faster than tortoise so in cicle hare must |
70 ## reach tortoise. | |
8507 | 71 hare = tree(tree(hare)); |
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72 |
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73 endwhile |
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74 |
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75 if (tortoise == hare) |
10549 | 76 ## If hare reach tortoise we found circle. |
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77 error ("treelayout: vector of predecessors has bad format"); |
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78 endif |
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79 |
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80 endfor |
8506 | 81 ## Vector of predecessors has right format. |
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82 |
8507 | 83 for i = 1:num_nodes |
84 ## vec_of_child is helping vector which is used to speed up the | |
8506 | 85 ## choice of descendant nodes. |
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86 |
8507 | 87 num_children(tree(i)+1) = num_children(tree(i)+1) + 1; |
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88 endfor |
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89 |
8507 | 90 pos = 1; |
91 start = zeros (1, num_nodes+1); | |
92 xhelp = zeros (1, num_nodes+1); | |
93 stop = zeros (1, num_nodes+1); | |
94 for i = 1 : num_nodes + 1 | |
95 start(i) = pos; | |
96 xhelp(i) = pos; | |
97 pos += num_children(i); | |
98 stop(i) = pos; | |
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99 endfor |
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100 |
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101 if (nargin == 1) |
8507 | 102 for i = 1:num_nodes |
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103 vec_of_child(xhelp(tree(i)+1)) = i; |
8507 | 104 xhelp(tree(i)+1) = xhelp(tree(i)+1) + 1; |
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105 endfor |
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106 else |
8507 | 107 vec_of_child = permutation; |
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108 endif |
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109 |
8506 | 110 ## The number of "parent" (actual) node (it's descendants will be |
111 ## browse in the next iteration). | |
8507 | 112 par_number = 0; |
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113 |
8506 | 114 ## The x-coordinate of the left most descendant of "parent node" |
115 ## this value is increased in each leaf. | |
8507 | 116 left_most = 0; |
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117 |
8507 | 118 ## The level of "parent" node (root level is num_nodes). |
119 level = num_nodes; | |
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120 |
8507 | 121 ## num_nodes - max_ht is the height of this graph. |
122 max_ht = num_nodes; | |
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123 |
8506 | 124 ## Main stack - each item consists of two numbers - the number of |
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125 ## node and the number it's of parent node on the top of stack |
8506 | 126 ## there is "parent node". |
8507 | 127 stk = [-1, 0]; |
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128 |
8506 | 129 ## Number of vertices s in the top-level separator. |
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130 s = 0; |
8506 | 131 ## Flag which says if we are in top level separator. |
8507 | 132 top_level = 1; |
8506 | 133 ## The top of the stack. |
8507 | 134 while (par_number != -1) |
135 if (start(par_number+1) < stop(par_number+1)) | |
136 idx = vec_of_child(start(par_number+1) : stop(par_number+1) - 1); | |
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137 else |
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138 idx = zeros (1, 0); |
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139 endif |
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140 |
8506 | 141 ## Add to idx the vector of parent descendants. |
8507 | 142 stk = [stk; [idx', ones(fliplr(size(idx))) * par_number]]; |
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143 |
8506 | 144 ## We are in top level separator when we have one child and the |
145 ## flag is 1 | |
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146 if (columns (idx) == 1 && top_level == 1) |
8507 | 147 s++; |
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148 else |
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149 ## We aren't in top level separator now. |
8507 | 150 top_level = 0; |
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151 endif |
8506 | 152 ## If there is not any descendant of "parent node": |
8507 | 153 if (stk(end,2) != par_number) |
154 left_most++; | |
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155 x_coordinate_r(par_number) = left_most; |
8507 | 156 max_ht = min (max_ht, level); |
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157 if (length (stk) > 1 && find ((shift (stk,1) - stk) == 0) > 1 |
10549 | 158 && stk(end,2) != stk(end-1,2)) |
159 ## Return to the nearest branching the position to return | |
160 ## position is the position on the stack, where should be | |
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161 ## started further search (there are two nodes which has the |
8506 | 162 ## same parent node). |
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163 |
8507 | 164 position = (find ((shift (stk(:,2), 1) - stk(:,2)) == 0))(end) + 1; |
165 par_number_vec = stk(position:end,2); | |
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166 |
8506 | 167 ## The vector of removed nodes (the content of stack form |
168 ## position to end). | |
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169 |
8507 | 170 level += length (par_number_vec); |
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171 |
10549 | 172 ## The level have to be decreased. |
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173 |
8507 | 174 x_coordinate_r(par_number_vec) = left_most; |
175 stk(position:end,:) = []; | |
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176 endif |
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177 |
8506 | 178 ## Remove the next node from "searched branch". |
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179 |
8507 | 180 stk(end,:) = []; |
10549 | 181 ## Choose new "parent node". |
8507 | 182 par_number = stk(end,1); |
10549 | 183 ## If there is another branch start to search it. |
184 if (par_number != -1) | |
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185 y_coordinate(par_number) = level; |
8507 | 186 x_coordinate_l(par_number) = left_most + 1; |
10549 | 187 endif |
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188 else |
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189 |
8506 | 190 ## There were descendants of "parent nod" choose the last of |
191 ## them and go on through it. | |
8507 | 192 level--; |
193 par_number = stk(end,1); | |
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194 y_coordinate(par_number) = level; |
8507 | 195 x_coordinate_l(par_number) = left_most + 1; |
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196 endif |
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197 endwhile |
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198 |
8506 | 199 ## Calculate the x coordinates (the known values are the position |
200 ## of most left and most right descendants). | |
8507 | 201 x_coordinate = (x_coordinate_l + x_coordinate_r) / 2; |
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202 |
8507 | 203 height = num_nodes - max_ht - 1; |
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204 endif |
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205 endfunction |
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206 |
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207 |
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208 %!test |
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209 %! % Compute a simple tree layout |
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210 %! [x, y, h, s] = treelayout ([0, 1, 2, 2]); |
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211 %! assert (x, [1.5, 1.5, 2, 1]); |
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212 %! assert (y, [3, 2, 1, 1]); |
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213 %! assert (h, 2); |
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214 %! assert (s, 2); |
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215 |
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216 %!test |
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217 %! % Compute a simple tree layout with defined postorder permutation |
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218 %! [x, y, h, s] = treelayout ([0, 1, 2, 2], [1, 2, 4, 3]); |
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219 %! assert (x, [1.5, 1.5, 1, 2]); |
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220 %! assert (y, [3, 2, 1, 1]); |
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221 %! assert (h, 2); |
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222 %! assert (s, 2); |
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223 |
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224 %!test |
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225 %! % Compute a simple tree layout with defined postorder permutation |
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226 %! [x, y, h, s] = treelayout ([0, 1, 2, 2], [4, 2, 3, 1]); |
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227 %! assert (x, [0, 0, 0, 1]); |
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228 %! assert (y, [0, 0, 0, 3]); |
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229 %! assert (h, 0); |
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230 %! assert (s, 1); |
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231 |