Algorithms Coursera Part III Programming Assignment 2.js

Algorithms Coursera Part III Programming Assignment 2 Part 1.js

/* eslint-env node */
/* eslint-disable object-property-newline */

/**
* -------------------------- Part 1 -----------------------------
*/

/**
* In this programming problem and the next you'll code up the clustering
* algorithm from lecture for computing a max-spacing k-clustering.
*
* Download the text file below (clustering1.txt).
* (too large to copy in a gist: https://d18ky98rnyall9.cloudfront.net/_fe8d0202cd20a808db6a4d5d06be62f4_clustering1.txt?Expires=1552262400&Signature=LBSww0uBFDiF9-t4C~W-UWeq0eWUQxWb5Xg8wyQMAVKIbbZ62iXVMk05LiLWAusZ77zowkGNRjucYcWnRzcQhXG1O1E~XBW0OItQhjNlQtlR5ooCNKJJBljtXEN-NZke9kJW3QVjUaQZ6eC60rAENgNSEHjbng3Swf3FOEx~vRM_&Key-Pair-Id=APKAJLTNE6QMUY6HBC5A)
*
* This file describes a distance function (equivalently, a complete graph
* with edge costs). It has the following format:
*
* [number_of_nodes]
* [edge 1 node 1] [edge 1 node 2] [edge 1 cost]
* [edge 2 node 1] [edge 2 node 2] [edge 2 cost]
* ...
*
* There is one edge (i,j) for each choice of 1 ≤ i < j ≤ n where n is the
* number of nodes.
*
* For example, the third line of the file is "1 3 5250", indicating that
* the distance between nodes 1 and 3 (equivalently, the cost of the edge
* (1,3)) is 5250. You can assume that distances are positive, but you should
* NOT assume that they are distinct.
*
* Your task in this problem is to run the clustering algorithm from lecture
* on this data set, where the target number k of clusters is set to 4.
* What is the maximum spacing of a 4-clustering?
*
* ADVICE: If you're not getting the correct answer, try debugging your
* algorithm using some small test cases. And then post them to the discussion
* forum!
*
* Test cases from:
* https://github.com/beaunus/stanford-algs/tree/master/testCases/course3/assignment2Clustering/question1
*
* Clarification on what is meant by maximum spacing: this is the MINIMUM
* distance between the closest two separated nodes (i.e. after we have
* out k clusters, what is the minimum distance between any two nodes in
* different clusters). It is known as the "maximum spacing", as we want
* to maximize this minimum value and keep the clusters as far apart as possible.
*/

/**
* -------------------------- FILE I/O -----------------------------
*/

const fs = require("fs");

const edgesList = [];
// Map of node => list of incident edges to that node (array of
// {node, distance} objects)
const nodeVsIncidentEdges = new Map();
let numNodes;

/**
* Parse the text file and store it as an Array of objects of the form:
* [
*   {p, q, distance},
*   ...
* ]
*/
(function getInput() {
  const data = fs.readFileSync("clustering1.txt", "utf8");
  const rows = data.split("\n");
  for (let row of rows) {
    // Don't include the first row, which is the total number of nodes
    if (row === rows[0]) {
      numNodes = Number(row);
      continue;
    }
    // Get rid of empty row at EOF
    if (row === "") {
      continue;
    }
    // Remove excess white space at the end of the line
    row = row.trim();
    // Break each row into a tuple of integers
    row = row.split(" ");
    const [p, q, distance] = [
      Number(row[0]), Number(row[1]), Number(row[2])
    ];
    // I don't need to de-dupe, since the assignment says: "There is one
    // edge (i,j) for each choice of 1 ≤ i < j ≤ n where n is the number
    // of nodes." i.e. if I get a (1, 348) I will not find a (348, 1)
    // in the text file.
    edgesList.push({
      p,
      q,
      distance,
    });
    if (nodeVsIncidentEdges.has(p)) {
      nodeVsIncidentEdges.get(p).push({node: q, distance});
    } else {
      nodeVsIncidentEdges.set(p, [{node: q, distance}]);
    }
    if (nodeVsIncidentEdges.has(q)) {
      nodeVsIncidentEdges.get(q).push({node: p, distance});
    } else {
      nodeVsIncidentEdges.set(q, [{node: p, distance}]);
    }
  }
}());

// console.log(nodeVsIncidentEdges);

// Sort distances from biggest to smallest, since we will ultimately be fusing
// together the closest separated pairs of nodes, and Array.pop is more
// efficient than Array.shift to get the next min-distance edge to process.
edgesList.sort((a, b) => {
  return b.distance - a.distance;
});

// Maintain map of cluster ID => list of nodes; needed for fusing two clusters
// together
const clusterIdVsNodesList = new Map();
// Maintain map of node => clusterID to see if p and q are separate
const nodeVsClusterId = new Map();
for (let i = 1; i <= numNodes; i++) {
  clusterIdVsNodesList.set(i, [i]);
  nodeVsClusterId.set(i, i);
}

/**
* -------------------- Single-Link Clustering Algorithm -----------------------
*/

// NOTE: TRY USING CHROME DEVTOOLS DEBUGGER:
// https://medium.com/@paul_irish/debugging-node-js-nightlies-with-chrome-devtools-7c4a1b95ae27

/**
* k = 4
* Pseudocode for Part 1
* 1. Initialize each point to its own cluster
* 2. While # clusters > k (stop before the final k - 1 edge additions)
*   - Take the closest pair of separated nodes p, q, and...(how to break ties?)
*     - separated means the pair are not in the same cluster already
*   - Fuse them into the same cluster
*/
function getMaxSpacingForKClustering(k = 4) {
  let numClusters = numNodes;
  // Cluster nodes until there are k clusters
  while (numClusters > k) {
    for (let i = 0; i < edgesList.length; i++) {
      // Get the next closest pair of points, p and q
      const {p, q} = edgesList.pop();
      // Find which cluster they are in
      const pClusterId = nodeVsClusterId.get(p);
      const qClusterId = nodeVsClusterId.get(q);
      // Fuse q's into p's cluster if they're not in the same cluster
      if (pClusterId !== qClusterId) {
        const nodesInPCluster = clusterIdVsNodesList.get(pClusterId);
        const nodesInQCluster = clusterIdVsNodesList.get(qClusterId);
        for (const node of nodesInQCluster) {
          nodeVsClusterId.set(node, pClusterId);
        }
        clusterIdVsNodesList.set(
          pClusterId, nodesInPCluster.concat(nodesInQCluster)
        );
        clusterIdVsNodesList.delete(qClusterId);
        numClusters--;
        break;
      }
    }
  }

  // Get max spacing between nodes in different clusters (really the
  // minimum distance between any two nodes in different clusters)
  let maxSpacing = +Infinity;
  for (const nodesList of Array.from(clusterIdVsNodesList.values())) {
    for (const nodeMember of nodesList) {
      // get incident edges
      const incidentEdges = nodeVsIncidentEdges.get(nodeMember);
      for (const {node, distance} of incidentEdges) {
        if (!nodesList.includes(node) && distance < maxSpacing) {
          // nodes are in separate clusters, and we have a new max spacing
          maxSpacing = distance;
        }
      }
    }
  }
  return maxSpacing;
}

console.log("maxSpacing: ", getMaxSpacingForKClustering());

// Solutions:
// testCase1.txt: 21
// testCase2.txt: 218
// testCase3.txt: 1245
// clustering1.txt: 106

Algorithms Coursera Part III Programming Assignment 2 Part 2.js

/* eslint-env node */
/* eslint-disable object-property-newline */

/**
* -------------------------- Part 2 -----------------------------
*/

/**
* In this question your task is again to run the clustering algorithm
* from lecture, but on a MUCH bigger graph. So big, in fact, that the
* distances (i.e., edge costs) are only defined implicitly, rather than
* being provided as an explicit list.
*
* The data set is below (clustering_big.txt).
*
*  The format is:
* [# of nodes] [# of bits for each node's label]
* [first bit of node 1] ... [last bit of node 1]
* [first bit of node 2] ... [last bit of node 2]
* ...
*
* For example, the third line of the file
* "0 1 1 0 0 1 1 0 0 1 0 1 1 1 1 1 1 0 1 0 1 1 0 1" denotes the 24 bits
* associated with node #2.
*
* The distance between two nodes u and v in this problem is defined as
* the Hamming distance--- the number of differing bits --- between the two
* nodes' labels. For example, the Hamming distance between the 24-bit label
* of node #2 above and the label
* "0 1 0 0 0 1 0 0 0 1 0 1 1 1 1 1 1 0 1 0 0 1 0 1" is 3 (since they differ
* in the 3rd, 7th, and 21st bits).
*
* The question is: what is the largest value of k such that there is a
* k-clustering with spacing at least 3? That is, how many clusters are
* needed to ensure that no pair of nodes with all but 2 bits in common get
* split into different clusters?
*
* NOTE: The graph implicitly defined by the data file is so big that you
* probably can't write it out explicitly, let alone sort the edges by cost.
* So you will have to be a little creative to complete this part of the
* question. For example, is there some way you can identify the smallest
* distances without explicitly looking at every pair of nodes?
*
* Approach by another student: https://www.coursera.org/learn/algorithms-greedy/discussions/weeks/2/threads/OctNCeS_Eeao1BJzfLTSbg/replies/czz9DuUrEeaikxICfuhamA
*
* Test cases from:
* https://github.com/beaunus/stanford-algs/tree/master/testCases/course3/assignment2Clustering/question2
*
* Clarification on what is meant by maximum spacing: this is the MINIMUM
* distance between the closest two separated nodes (i.e. after we have
* out k clusters, what is the minimum distance between any two nodes in
* different clusters). It is known as the "maximum spacing", as we want
* to maximize this minimum value and keep the clusters as far apart as possible.
*/

/**
* -------------------------- FILE I/O -----------------------------
*/

const fs = require("fs");

let numNodes;
let numBits;
const codeVsDistances = new Map();
// "code" here refers to the 24-bit string representing each node; there can
// be multiple nodes with the same code, so map is a string => Array.integer
const codeVsNodesList = new Map();
// Maintain map of cluster ID => list of nodes; needed for fusing two clusters
// together
const clusterIdVsNodesList = new Map();
// Maintain map of node => clusterID to see if p and q are separate
const nodeVsClusterId = new Map();
const codesList = [];

/**
* Parse the text file.
*
* For each vertex, generate and store all Hamming distances that are 0, 1 and 2
* units apart. There is only 1 code point that is 0 units apart (which is the
* same code as the vertex), 24C1 = 24 possible code points that are 1 unit apart
* and there are 24C2 = 276 possible code points that are 2 units apart for each
* vertex.
* This will take the form of a hash table (Map, `codeVsDistances`) of
* vertex => Array.array, where the 0th element array is the list of values that
* are 0 units apart, the 1st element array is the list of values that are 1
* unit apart, and the 2nd element in the array is the list of values that are
* 2 units apart.
*
* Also, put all vertexes along with their assigned code into a hash table. Use
* the code as the hash table key, with the vertex number as the value - note
* that some codes are not unique (i.e. more than one vertex can be associated
* with the same code), so each key in the hash table will have to potentially
* hold more than one vertex - we will use this hash table later to look up the
* vertex number(s) given the corresponding Hamming code in O(1) time.
* This will take the form of a hash table (Map, `codeVsNodesList`) of
* string => Array.integer, where the key is the 24-bit string representing each
* node and the value is a list of node #s with the same string code.
*/
(function getInput() {
  const data = fs.readFileSync("clustering_big.txt", "utf8");
  const rows = data.split("\n");
  let nodeNum = 1;
  for (let row of rows) {
    // Don't include the first row, which is the total number of nodes
    if (row === rows[0]) {
      [numNodes, numBits] = row.split(" ");
      [numNodes, numBits] = [Number(numNodes), Number(numBits)];
      continue;
    }
    // Get rid of empty row at EOF
    if (row === "") {
      continue;
    }
    // Remove excess white space at the end of the line
    row = row.trim();
    // Remove spaces in between digits
    row = row.replace(/\s/g, "");
    // Generate and store all Hamming distances that are 0, 1 and 2 units apart
    const listZeroUnitsApart = [row]; // Only 1 node label 0 units apart, itself
    const listOneUnitsApart = []; // 24 choose 1 node (24) labels 1 unit apart
    for (let i = 0; i < numBits; i++) {
      const replacement = row[i] === "0" ? "1" : "0";
      const variant = row.slice(0, i) + replacement
        + row.slice(i + replacement.length);
      listOneUnitsApart.push(variant);
    }
    const listTwoUnitsApart = []; // 24 choose 2 node (276) labels 2 units apart
    for (let j = 0; j < numBits; j++) {
      for (let k = 0; k < numBits; k++) {
        if (k === j) continue; // We have to change two different bits
        const jReplacement = row[j] === "0" ? "1" : "0";
        const kReplacement = row[k] === "0" ? "1" : "0";
        let variant = row.slice(0, j) + jReplacement
          + row.slice(j + jReplacement.length);
        variant = variant.slice(0, k) + kReplacement
          + variant.slice(k + kReplacement.length);
        if (!(listTwoUnitsApart.includes(variant))) {
          listTwoUnitsApart.push(variant);
        }
      }
    }
    codeVsDistances.set(row, [
      listZeroUnitsApart,
      listOneUnitsApart,
      listTwoUnitsApart,
    ]);
    if (codeVsNodesList.has(row)) {
      codeVsNodesList.get(row).push(nodeNum);
    } else {
      codeVsNodesList.set(row, [nodeNum]);
    }
    // All nodes start out in their own cluster
    clusterIdVsNodesList.set(nodeNum, new Set([nodeNum]));
    nodeVsClusterId.set(nodeNum, nodeNum);
    codesList.push(row);
    nodeNum++;
  }
}());

// console.log(codeVsDistances);
// console.log(codeVsNodesList);
// console.log(clusterIdVsNodesList);
// console.log(codesList);

/**
* -------------------- Clustering Algorithm -----------------------
*/

// NOTE: TRY USING CHROME DEVTOOLS DEBUGGER:
// https://medium.com/@paul_irish/debugging-node-js-nightlies-with-chrome-devtools-7c4a1b95ae27

/**
* Pseudocode for Part 2
* For each vertex (200K iterations):
*  For each code that is 0 units apart from 
*  this vertex: (1 iteration - there is only one such code
*  which is the same code as that of the vertex itself)
*   - Use the code to index into the hash table and 
*     get the corresponding vertexes if they exist. 
*   - Add these 2 vertexes to a cluster.
*    
* For each vertex (200K iterations):
*  For each code that is 1 unit apart from 
*  this vertex: (24 iterations)
*    - Use the code to index into the hash table and 
*      get the corresponding vertexes if they exist. 
*    - Add these 2 vertexes to a cluster.
*    
*For each vertex (200K iterations):
*  For each code that is 2 units apart from
*  this vertex: (276 iterations)
*    - Use the code to index into the hash table and 
*      get the corresponding vertexes if they exist.
*    - Add these 2 vertexes to a cluster.
*    
*You are now left with clusters that are at least 3 units apart.
*
* Explanation of pseudocode:
* In the first loop, we are essentially clustering all vertexes
* that are a distance of 0 units apart, in the second loop and
* third loop we are clustering vertexes that are 1 unit apart,
* and 2 units apart respectively (this is similar to sorting by
* edge weights and then combining the vertexes into clusters).
* The above code can be made much more compact - I have split up
* the three main loops for readability.
*
* The time complexity of the above is 200k + (200k * 24) 
* + (200k * 276) = 200k * 301 = O(301n) iterations, plus for each
* iteration, we have to fix up the leader pointers of the smaller
* cluster - which gives us a final complexity of O(301nlog n).
* The space complexity is about O(301n).
*/

function updateClusters(node, potentialMatchingCodes) {
  for (const potentialMatchingCode of potentialMatchingCodes) {
    const potentialMatchingNodes = codeVsNodesList.get(potentialMatchingCode);
    if (potentialMatchingNodes) {
      const matchingNodes = potentialMatchingNodes; // rename for accuracy
      for (const matchingNode of matchingNodes) {
        const nodeClusterId = nodeVsClusterId.get(node);
        const matchingNodeClusterId = nodeVsClusterId.get(matchingNode);
        if (nodeClusterId !== matchingNodeClusterId) {
          // Nodes are in separate clusters, fuse their cluster's together
          // (fuse the smaller one into the bigger one)
          const nodeClusterSize = clusterIdVsNodesList.get(nodeClusterId).size;
          const matchingNodeClusterSize = clusterIdVsNodesList.get(matchingNodeClusterId).size;
          if (nodeClusterSize >= matchingNodeClusterSize) {
            // node's cluster absorbs matching node's cluster
            const nodesToUpdate = clusterIdVsNodesList.get(matchingNodeClusterId);
            for (const nodeToUpdate of nodesToUpdate) {
              nodeVsClusterId.set(nodeToUpdate, nodeClusterId);
              clusterIdVsNodesList.delete(matchingNodeClusterId);
              clusterIdVsNodesList.get(nodeClusterId).add(nodeToUpdate);
            }
          } else {
            // matching node's cluster absorbs node's cluster
            const nodesToUpdate = clusterIdVsNodesList.get(nodeClusterId);
            for (const nodeToUpdate of nodesToUpdate) {
              nodeVsClusterId.set(nodeToUpdate, matchingNodeClusterId);
              clusterIdVsNodesList.delete(nodeClusterId);
              clusterIdVsNodesList.get(matchingNodeClusterId).add(nodeToUpdate);
            }
          }
        }
      }
    }
  }
}

function getNumClusters() {
  // Clustering all vertices that are a distance of 0 units apart
  for (const nodesList of Array.from(codeVsNodesList.values())) {
    if (nodesList.length > 1) {
      // The first entry will be our leader node <=> clusterId
      const clusterId = nodesList[0];
      // Fuse all of these nodes into the same cluster and update bookkeeping
      for (const node of nodesList) {
        // Don't overwrite the leader node's entry
        if (node === nodesList[0]) continue;
        clusterIdVsNodesList.get(clusterId).add(node);
        const oldClusterId = nodeVsClusterId.get(node);
        clusterIdVsNodesList.delete(oldClusterId);
        nodeVsClusterId.set(node, clusterId);
      }
    }
  }

  // Clustering all vertices that are a distance of 1 units apart
  let node = 1;
  for (const code of codesList) {
    const potentialMatchingCodes = codeVsDistances.get(code)[1];
    updateClusters(node, potentialMatchingCodes);
    node++;
  }

  // Clustering all vertices that are a distance of 2 units apart
  node = 1;
  for (const code of codesList) {
    const potentialMatchingCodes = codeVsDistances.get(code)[2];
    updateClusters(node, potentialMatchingCodes);
    node++;
  }

  return Array.from(clusterIdVsNodesList.keys()).length;
}

console.log(getNumClusters());

// Solutions:
// testCase1.txt: 11
// testCase2.txt: 15
// clustering_big.txt: 6118 (took 10 minutes && had to increase V8 memory in node with `node --max-old-space-size=8192 <path/to/file>`)

testCase1Part1.txt

8
1 2 32
1 3 46
1 4 50
1 5 57
1 6 57
1 7 32
1 8 51
2 3 50
2 4 35
2 5 1
2 6 17
2 7 56
2 8 19
3 4 21
3 5 22
3 6 42
3 7 29
3 8 44
4 5 27
4 6 38
4 7 25
4 8 18
5 6 6
5 7 53
5 8 9
6 7 27
6 8 22
7 8 46

testCase2Part1.txt

64
1 2 908
1 3 2580
1 4 2040
1 5 1102
1 6 1736
1 7 517
1 8 624
1 9 703
1 10 231
1 11 3516
1 12 570
1 13 2298
1 14 2407
1 15 4009
1 16 3105
1 17 2090
1 18 3914
1 19 2085
1 20 501
1 21 2176
1 22 4071
1 23 12
1 24 1223
1 25 372
1 26 1487
1 27 2974
1 28 1338
1 29 2486
1 30 2959
1 31 3478
1 32 2460
1 33 2706
1 34 2562
1 35 3976
1 36 3885
1 37 456
1 38 539
1 39 2810
1 40 269
1 41 2773
1 42 2595
1 43 3268
1 44 3115
1 45 1014
1 46 1611
1 47 3078
1 48 4060
1 49 518
1 50 2302
1 51 2295
1 52 1499
1 53 3111
1 54 3232
1 55 3351
1 56 2365
1 57 2090
1 58 2812
1 59 3855
1 60 1762
1 61 3621
1 62 2660
1 63 1176
1 64 643
2 3 724
2 4 1240
2 5 775
2 6 2918
2 7 782
2 8 4076
2 9 4028
2 10 3548
2 11 1689
2 12 1115
2 13 2926
2 14 800
2 15 372
2 16 3971
2 17 3387
2 18 1372
2 19 3244
2 20 2021
2 21 2352
2 22 2414
2 23 2067
2 24 1762
2 25 2939
2 26 392
2 27 1799
2 28 1761
2 29 1408
2 30 3897
2 31 4080
2 32 354
2 33 1142
2 34 1776
2 35 3327
2 36 703
2 37 2152
2 38 695
2 39 623
2 40 2230
2 41 1185
2 42 2644
2 43 1335
2 44 3242
2 45 2458
2 46 2497
2 47 1222
2 48 1015
2 49 263
2 50 1312
2 51 3037
2 52 401
2 53 851
2 54 1986
2 55 738
2 56 2647
2 57 1462
2 58 3372
2 59 1492
2 60 2562
2 61 2557
2 62 3205
2 63 3689
2 64 3309
3 4 3535
3 5 1926
3 6 3240
3 7 1789
3 8 1329
3 9 1107
3 10 348
3 11 2466
3 12 3078
3 13 1078
3 14 1490
3 15 3145
3 16 3348
3 17 43
3 18 3628
3 19 1776
3 20 4008
3 21 2388
3 22 2416
3 23 141
3 24 461
3 25 1135
3 26 1132
3 27 4001
3 28 407
3 29 4049
3 30 417
3 31 1095
3 32 1385
3 33 1646
3 34 1269
3 35 2761
3 36 2016
3 37 2669
3 38 1568
3 39 3758
3 40 1118
3 41 371
3 42 1159
3 43 1543
3 44 2291
3 45 794
3 46 770
3 47 424
3 48 3060
3 49 3216
3 50 1554
3 51 134
3 52 386
3 53 2130
3 54 613
3 55 1002
3 56 2570
3 57 2314
3 58 3617
3 59 1052
3 60 1763
3 61 203
3 62 3652
3 63 3467
3 64 1049
4 5 1405
4 6 2474
4 7 1931
4 8 3750
4 9 3462
4 10 550
4 11 372
4 12 2166
4 13 977
4 14 290
4 15 3651
4 16 2232
4 17 1037
4 18 2662
4 19 2081
4 20 2101
4 21 3851
4 22 3002
4 23 2827
4 24 1860
4 25 2195
4 26 3247
4 27 355
4 28 555
4 29 835
4 30 2339
4 31 841
4 32 4028
4 33 1370
4 34 3541
4 35 1544
4 36 3603
4 37 3492
4 38 1457
4 39 2812
4 40 1036
4 41 2816
4 42 3599
4 43 1381
4 44 747
4 45 1583
4 46 4042
4 47 1385
4 48 1553
4 49 2538
4 50 774
4 51 3810
4 52 4093
4 53 334
4 54 28
4 55 228
4 56 3202
4 57 2789
4 58 1338
4 59 3160
4 60 543
4 61 3788
4 62 943
4 63 3640
4 64 2093
5 6 2317
5 7 1905
5 8 3911
5 9 4007
5 10 395
5 11 2434
5 12 1042
5 13 3801
5 14 2708
5 15 307
5 16 2409
5 17 1317
5 18 117
5 19 1734
5 20 438
5 21 3728
5 22 509
5 23 1023
5 24 1072
5 25 3000
5 26 3954
5 27 1669
5 28 1405
5 29 2243
5 30 3657
5 31 2262
5 32 3650
5 33 113
5 34 3840
5 35 3741
5 36 3300
5 37 2945
5 38 1520
5 39 895
5 40 1548
5 41 316
5 42 1564
5 43 1423
5 44 3513
5 45 1510
5 46 3399
5 47 1985
5 48 1058
5 49 1558
5 50 3792
5 51 374
5 52 17
5 53 3803
5 54 1035
5 55 1027
5 56 721
5 57 3639
5 58 1030
5 59 341
5 60 3298
5 61 4092
5 62 699
5 63 3804
5 64 1854
6 7 3847
6 8 270
6 9 884
6 10 3336
6 11 3151
6 12 3832
6 13 2020
6 14 1542
6 15 3053
6 16 1127
6 17 301
6 18 3670
6 19 1099
6 20 981
6 21 1107
6 22 1253
6 23 71
6 24 1937
6 25 2681
6 26 3084
6 27 3376
6 28 1158
6 29 2432
6 30 3832
6 31 1694
6 32 663
6 33 224
6 34 3074
6 35 2870
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