Finding Kth Minimum (partial ordering) – Using Tournament Algorithm

Ok, so far in my previous blog entries about finding 2nd minimum (and for repeating elements) I wrote about efficient algorithm for finding second minimum element of an array. This optimized algorithm takes O(n + log(n)) time (worst case) to find the second minimum element.

So, the next obvious question is - how can we find K^th minimum element efficiently in an unsorted array where k ranges from 1 – n. How about partial sorting - efficiently returning first K minimum values from unsorted array. As you might have guessed it, we can achieve this by extending the logic used before and tweaking our implementation little bit. We can find K^th min (or return partially sorted list of K min elements) in O(n + klog(n)) time.

Design Details

 For simplicity we will assume that the input unsorted array consists of non-repeating, non-negative integer numbers only. Solution can be extended for arrays with repeating numbers using similar considerations as outlined in my blog for finding second min for repeating values.

As noted in the previous blog, using tournament method we can build output tree by comparing the adjacent elements and moving the lower of the two to next level. Repeating this process yields a tree with log(n) depth and the root element of this tree represent the minimum value. Also, the second minimum can then easily be obtained by finding the minimum of all the values that the root was compared against at each level.  Since the depth of tree is log(n), we have to do at most O(log(n)) comparisons. For example, Figure 1 below uses tournament method for finding the root element. For this particular case, the minimum value is 1 and second minimum 2.

 

 

Let’s say we want to find the 3rd minimum value. Extending the logic used to find second minimum, the third minimum value must have met the root (minimum) or the second minimum before ending its progression (just like winning a tournament round) to next level of the tree (as smaller of the two compared values move to next level). Thus 3rd minimum can be obtained by finding the least of all the values compared against the minimum and the second minimum, i.e. backtracking the root (minimum) and the second minimum up to the top of tree (original array) and noting all values each (root and second minimum) was compared against.

 Take the sample in Figure 1. Here is the list of values obtained by back-tracking the root (call it Set A).

Set A (for root) - [3, 2, 10, 8]

From this list we obtain the second minimum as 2. Back-tracking from 2, we obtain following list (Set B):

Set B (for Second minimum) [5, 14]

Thus the third minimum is the least value in union of set A and B, which is 3 (ignoring value 2, which was already established as second minimum). Note that the 3rd minimum must be in either Set A or Set B. In above case it was found from Set A. 

How about if we have to find the 4th minimum? Well, we have to back track the 3rd least value and collect adjacent values a (one level up). Then add this set to the values for root and second minimum. Then this set will contain the 2nd, 3rd and 4th minimum. Continuing our sample, let’s say Set C consists of such elements obtained by backtracking 3rd min – [4, 11, 16]. Thus the minimum value in union of Set A, B, C is 4 (ignoring already established 2 and 3 as second min and third min resp.), is fourth least value. Note that for calculating kth minimum we don’t backtrack and find the adjacency list for minimum through k -1. We can use already calculated results from any previous such computation. This is the reason why this technique falls under Dynamic Programming – we have optimal substructure and avoid repeated computations. 

Now it is time to formalize. To find Kth minimum, back track root, second min, third min up to k-1 min and collect the adjacent values for each - moving from the root (of sub-tree) all the way to the top. The resulting collection will contain the 2nd, 3rd … Kth minimum.

Implementation Details

To obtain K^th minimum, we need collection of all the comparisons from minimum (root) to K-1. Thus we need to make following changes to our previous code for finding second minimum.

 

1. Return array of values (as indicated in rectangle boxes in Figure 2) while back-tracking root (could also be sub-tree root). Also, in addition to the values, we need the level and also index information for of each of these values. The reason why we need the level and index is because once we determine that a particular value from this list is minimum, we need to back-track the sub-tree from that value (hence we need to know the location of the element at that level) on order to obtain new collection to determine next lower value. Thus we will return a two-dimensional array with first index denoting the value and the other one the index. We can infer the level from the index of the pair (in 2-d array).
   For example, for the case above, for the first pass (to find the minimum value), the following will be returned (let’s call it adjacency list):
   
   
   Thus from above array we can conclude that the second minimum value is 2 (least of first value of all pairs), and this value appears two levels above the root (level 5), at 0th index (at this level).  We will use this info to identify the sub-tree for out next round of back-tracking.
   
2. Change the api to take back-track a portion of tree. In order to achieve that we need to pass level and index to define the root of sub-tree to back track from.
   
3. The 2-dimensional list obtained from each run will be added to a 3-dimensional list – full adjacency list for min to K-1 min elements. 3-D list is used rather than merging a bigger 2-D list to preserve and identify the results for a particular run. The reason being once we identify the minimum element for all k-1 runs, the level of that particular element is obtained from particular list where the minimum was found (alternatively our backtracking api could have returned a 3-d array with level info as one more dimension, either way we need this info).
   
4. Api that take full adjacency list and min value and return the next min element info. The api should identify for which run it found the min and the index of element within that run. Refer Figure 4 below. When this list is passed to this api with value of second min (2), it returns (0,3), which is interpreted as the next min (third min) was found in first array (obtained from first backtrack) and 4th element within that array. Once we have this info, we can look into first array and locate the third min value as 3, at index 1 at level 3 (remember original array is level 1 – Figure 2). Note that for m elements in full adjacency list we are making m comparisons which seem not efficient at first glance, but since the size of the adjacency list is small O(log(n)), hence it is not significant. 
   
5. To find K^th min, we need to find min through k. Hence this algorithm can return partially sorted array up-to k elements. 

 

 

Code 

Can be downloaded from here - KthMinimum.java (remove the .txt extension)

 

view sourceprint?

001/**

002 * Copyright (c) 2010-2020 Malkit S. Bhasin. All rights reserved.

003 *

004 * All source code and material on this Blog site is the copyright of Malkit S.

005 * Bhasin, 2010 and is protected under copyright laws of the United States. This

006 * source code may not be hosted on any other site without my express, prior,

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008 * made by contacting me at mbhasin at gmail dot com

009 *

010 * I have made every effort and taken great care in making sure that the source

011 * code and other content included on my web site is technically accurate, but I

012 * disclaim any and all responsibility for any loss, damage or destruction of

013 * data or any other property which may arise from relying on it. I will in no

014 * case be liable for any monetary damages arising from such loss, damage or

015 * destruction.

016 *

017 * I further grant you ("Licensee") a non-exclusive, royalty free, license to

018 * use, modify and redistribute this software in source and binary code form,

019 * provided that i) this copyright notice and license appear on all copies of

020 * the software;

021 *

022 * As with any code, ensure to test this code in a development environment

023 * before attempting to run it in production.

024 *

025 * @author Malkit S. Bhasin

026 *

027 */

028 

029public class KThMinimum {

030 

031    /**

032     * @param inputArray

033     *            unordered array of non-negative integers

034     * @param k

035     *            order of minimum value desired

036     * @return kth minimum value

037     */

038    public static int getKthMinimum(int[] inputArray, int k) {

039        return findKthMinimum(inputArray, k)[k - 1];

040    }

041 

042    /**

043     * @param inputArray

044     *            unordered array of non-negative integers

045     * @param k

046     *            ordered number of minimum values

047     * @return k ordered minimum values

048     */

049    public static int[] getMinimumKSortedElements(int[] inputArray, int k) {

050        return findKthMinimum(inputArray, k);

051    }

052 

053    /**

054     * First output tree will be obtained using tournament method. For k

055     * minimum, the output tree will be backtracked k-1 times for each sub tree

056     * identified by the minimum value in the aggregate adjacency list obtained

057     * from each run. The minimum value after each run will be recorded and

058     * successive runs will produce next minimum value.

059     *

060     * @param inputArray

061     * @param k

062     * @return ordered array of k minimum elements

063     */

064    private static int[] findKthMinimum(int[] inputArray, int k) {

065        int[] partiallySorted = new int[k];

066        int[][] outputTree = getOutputTree(inputArray);

067        int root = getRootElement(outputTree);

068        partiallySorted[0] = root;

069        int rootIndex = 0;

070        int level = outputTree.length;

071        int[][][] fullAdjacencyList = new int[k - 1][][];

072        int[] kthMinIdx = null;

073        for (int i = 1; i < k; i++) {

074            fullAdjacencyList[i - 1] = getAdjacencyList(outputTree, root,

075                    level, rootIndex);

076            kthMinIdx = getKthMinimum(fullAdjacencyList, i, root);

077            int row = kthMinIdx[0];

078            int column = kthMinIdx[1];

079            root = fullAdjacencyList[row][column][0];

080            partiallySorted[i] = root;

081            level = column + 1;

082            rootIndex = fullAdjacencyList[row][column][1];

083        }

084 

085        return partiallySorted;

086    }

087 

088    /**

089     * Takes an input array and generated a two-dimensional array whose rows are

090     * generated by comparing adjacent elements and selecting minimum of two

091     * elements. Thus the output is inverse triangle (root at bottom)

092     *

093     * @param values

094     * @return

095     */

096    public static int[][] getOutputTree(int[] values) {

097        Integer size = new Integer(values.length);

098        double treeDepth = Math.log(size.doubleValue()) / Math.log(2);

099        // int intTreeDepth = getIntValue(Math.ceil(treeDepth)) + 1;

100        int intTreeDepth = (int) (Math.ceil(treeDepth)) + 1;

101        int[][] outputTree = new int[intTreeDepth][];

102 

103        // first row is the input

104        outputTree[0] = values;

105        printRow(outputTree[0]);

106 

107        int[] currentRow = values;

108        int[] nextRow = null;

109        for (int i = 1; i < intTreeDepth; i++) {

110            nextRow = getNextRow(currentRow);

111            outputTree[i] = nextRow;

112            currentRow = nextRow;

113            printRow(outputTree[i]);

114        }

115        return outputTree;

116    }

117 

118    /**

119     * Compares adjacent elements (starting from index 0), and construct a new

120     * array with elements that are smaller of the adjacent elements.

121     *

122     * For even sized input, the resulting array is half the size, for odd size

123     * array, it is half + 1.

124     *

125     * @param values

126     * @return

127     */

128    private static int[] getNextRow(int[] values) {

129        int rowSize = getNextRowSize(values);

130        int[] row = new int[rowSize];

131        int i = 0;

132        for (int j = 0; j < values.length; j++) {

133            if (j == (values.length - 1)) {

134                // this is the case where there are odd number of elements

135                // in the array. Hence the last loop will have only one element.

136                row[i++] = values[j];

137            } else {

138                row[i++] = getMin(values[j], values[++j]);

139            }

140        }

141        return row;

142    }

143 

144    /**

145     * From the passed full adjacency list and min value scans the list and

146     * returns the information about next minimum value. It returns int array

147     * with two values:

148     * first value: index of the back-track (the min value was found in the

149     * Adjacency list for min value, second min etc.)

150     * second value: index within the identified run.

151     *

152     * @param fullAdjacencyList

153     *            Adjacency list obtained after k-1 backtracks

154     * @param kth

155     *            Order of minimum value desired

156     * @param kMinusOneMin

157     *            value of k-1 min element

158     * @return

159     */

160    private static int[] getKthMinimum(int[][][] fullAdjacencyList, int kth,

161            int kMinusOneMin) {

162        int kThMin = Integer.MAX_VALUE;

163        int[] minIndex = new int[2];

164        int j = 0, k = 0;

165        int temp = -1;

166 

167        for (int i = 0; i < kth; i++) {

168            for (j = 0; j < fullAdjacencyList.length; j++) {

169                int[][] row = fullAdjacencyList[j];

170                if (row != null) {

171                    for (k = 0; k < fullAdjacencyList[j].length; k++) {

172                        temp = fullAdjacencyList[j][k][0];

173                        if (temp <= kMinusOneMin) {

174                            continue;

175                        }

176                        if ((temp > kMinusOneMin) && (temp < kThMin)) {

177                            kThMin = temp;

178                            minIndex[0] = j;

179                            minIndex[1] = k;

180                        }

181                    }

182                }

183            }

184        }

185        return minIndex;

186    }

187 

188    /**

189     * Back-tracks a sub-tree (specified by the level and index) parameter and

190     * returns array of elements (during back-track path) along with their index

191     * information. The order elements of output array indicate the level at

192     * which these elements were found (with elements closest to the root at the

193     * end of the list)

194     *

195     * Starting from root element (which is minimum element), find the lower of

196     * two adjacent element one row above. One of the two element must be root

197     * element. If the root element is left adjacent, the root index (for one

198     * row above) is two times the root index of any row. For right-adjacent, it

199     * is two times plus one. Select the other element (of two adjacent

200     * elements) as second minimum.

201     *

202     * Then move to one row further up and find elements adjacent to lowest

203     * element, again, one of the element must be root element (again, depending

204     * upon the fact that it is left or right adjacent, you can derive the root

205     * index for this row). Compare the other element with the second least

206     * selected in previous step, select the lower of the two and update the

207     * second lowest with this value.

208     *

209     * Continue this till you exhaust all the rows of the tree.

210     *

211     * @param tree

212     *            output tree

213     * @param rootElement

214     *            root element (could be of sub-tree or outputtree)

215     * @param level

216     *            the level to find the root element. For the output tree the

217     *            level is depth of the tree.

218     * @param rootIndex

219     *            index for the root element. For output tree it is 0

220     * @return

221     */

222    public static int[][] getAdjacencyList(int[][] tree, int rootElement,

223            int level, int rootIndex) {

224        int[][] adjacencyList = new int[level - 1][2];

225        int adjacentleftElement = -1, adjacentRightElement = -1;

226        int adjacentleftIndex = -1, adjacentRightIndex = -1;

227        int[] rowAbove = null;

228 

229        // we have to scan in reverse order

230        for (int i = level - 1; i > 0; i--) {

231            // one row above

232            rowAbove = tree[i - 1];

233            adjacentleftIndex = rootIndex * 2;

234            adjacentleftElement = rowAbove[adjacentleftIndex];

235 

236            // the root element could be the last element carried from row above

237            // because of odd number of elements in array, you need to do

238            // following

239            // check. if you don't, this case will blow {8, 4, 5, 6, 1, 2}

240            if (rowAbove.length >= ((adjacentleftIndex + 1) + 1)) {

241                adjacentRightIndex = adjacentleftIndex + 1;

242                adjacentRightElement = rowAbove[adjacentRightIndex];

243            } else {

244                adjacentRightElement = -1;

245            }

246 

247            // if there is no right adjacent value, then adjacent left must be

248            // root continue the loop.

249            if (adjacentRightElement == -1) {

250                // just checking for error condition

251                if (adjacentleftElement != rootElement) {

252                    throw new RuntimeException(

253                            "This is error condition. Since there "

254                                    + " is only one adjacent element (last element), "

255                                    + " it must be root element");

256                } else {

257                    rootIndex = rootIndex * 2;

258                    adjacencyList[level - 1][0] = -1;

259                    adjacencyList[level - 1][1] = -1;

260                    continue;

261                }

262            }

263 

264            // one of the adjacent number must be root (min value).

265            // Get the other number and compared with second min so far

266            if (adjacentleftElement == rootElement

267                    && adjacentRightElement != rootElement) {

268                rootIndex = rootIndex * 2;

269                adjacencyList[i - 1][0] = adjacentRightElement;

270                adjacencyList[i - 1][1] = rootIndex + 1;

271            } else if (adjacentleftElement != rootElement

272                    && adjacentRightElement == rootElement) {

273                rootIndex = rootIndex * 2 + 1;

274                adjacencyList[i - 1][0] = adjacentleftElement;

275                adjacencyList[i - 1][1] = rootIndex - 1;

276            } else if (adjacentleftElement == rootElement

277                    && adjacentRightElement == rootElement) {

278                // This is case where the root element is repeating, we are not

279                // handling this case.

280                throw new RuntimeException(

281                        "Duplicate Elements. This code assumes no repeating elements in the input array");

282            } else {

283                throw new RuntimeException(

284                        "This is error condition. One of the adjacent "

285                                + "elements must be root element");

286            }

287        }

288 

289        return adjacencyList;

290    }

291 

292    /**

293     * Returns minimum of two passed in values.

294     *

295     * @param num1

296     * @param num2

297     * @return

298     */

299    private static int getMin(int num1, int num2) {

300        return Math.min(num1, num2);

301    }

302 

303    /**

304     * following uses Math.ceil(double) to round to upper integer value..since

305     * this function takes double value, diving an int by double results in

306     * double.

307     *

308     * Another way of achieving this is for number x divided by n would be -

309     * (x+n-1)/n

310     *

311     * @param values

312     * @return

313     */

314    private static int getNextRowSize(int[] values) {

315        return (int) Math.ceil(values.length / 2.0);

316    }

317 

318    /**

319     * Returns the root element of the two-dimensional array.

320     *

321     * @param tree

322     * @return

323     */

324    public static int getRootElement(int[][] tree) {

325        int depth = tree.length;

326        return tree[depth - 1][0];

327    }

328 

329    private static void printRow(int[] values) {

330        for (int i : values) {

331            // System.out.print(i + " ");

332        }

333        // System.out.println(" ");

334    }

335 

336    public static void main(String args[]) {

337        int[] input = { 2, 14, 5, 13, 1, 8, 17, 10, 6, 12, 9, 4, 11, 15, 3, 16};

338        System.out.println("Fifth Minimum: " + getKthMinimum(input, 5));

339 

340        int minimumSortedElementSize = 10;

341        int[] tenMinimum = getMinimumKSortedElements(input,

342                minimumSortedElementSize);

343        System.out.print("Minimum " + minimumSortedElementSize + " Sorted: ");

344        for (int i = 0; i < minimumSortedElementSize; i++) {

345            System.out.print(tenMinimum[i] + " ");

346        }

347    }

348

}

 

转自：http://blogs.sun.com/malkit/entry/finding_kth_minimum_partial_ordering