Press n or j to go to the next uncovered block, b, p or k for the previous block.
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 | /** * Client-side agglomerative clustering for flowchart tiles. * * Operates on a flat upper-triangle pairwise cosine distance matrix * returned by the browse API. */ export interface ClusterResult { /** Cluster index (0-based) for each input item, in same order as input IDs */ assignments: number[] /** Number of clusters found */ k: number /** Representative item index for each cluster (lowest avg distance to peers) */ centroids: number[] } /** * Index into an upper-triangle flat array for an n×n symmetric matrix. * Requires i < j. */ export function distIndex(i: number, j: number, n: number): number { if (i === j) return -1 const [lo, hi] = i < j ? [i, j] : [j, i] // Row `lo` starts at offset: lo*n - lo*(lo+1)/2 - lo // Then column offset within that row: hi - lo - 1 return lo * n - (lo * (lo + 1)) / 2 + hi - lo - 1 } /** * Agglomerative clustering with average linkage. * * Cuts the dendrogram when either: * - The number of clusters reaches maxK, OR * - The next merge distance exceeds mergeThreshold * * @param n Number of items * @param distanceMatrix Upper-triangle flat array of pairwise distances * @param maxK Maximum number of clusters (default 4) * @param mergeThreshold Stop merging if the next merge distance exceeds this (default 0.7) */ export function agglomerativeClustering( n: number, distanceMatrix: number[], maxK = 4, mergeThreshold = 0.7 ): ClusterResult { if (n <= 0) { return { assignments: [], k: 0, centroids: [] } } if (n === 1) { return { assignments: [0], k: 1, centroids: [0] } } // Each cluster is a set of original item indices const clusters: Set<number>[] = Array.from({ length: n }, (_, i) => new Set([i])) // Map from cluster ID to its index in `clusters` (initially identity) const active = new Set<number>(Array.from({ length: n }, (_, i) => i)) // Average-linkage distance between two clusters function clusterDist(a: Set<number>, b: Set<number>): number { let sum = 0 let count = 0 for (const i of a) { for (const j of b) { sum += distanceMatrix[distIndex(i, j, n)] count++ } } return sum / count } // Merge until we reach maxK or threshold while (active.size > maxK) { // Find the pair of active clusters with minimum average-linkage distance let minDist = Infinity let mergeA = -1 let mergeB = -1 const activeArr = Array.from(active) for (let ai = 0; ai < activeArr.length; ai++) { for (let bi = ai + 1; bi < activeArr.length; bi++) { const d = clusterDist(clusters[activeArr[ai]], clusters[activeArr[bi]]) if (d < minDist) { minDist = d mergeA = activeArr[ai] mergeB = activeArr[bi] } } } if (minDist > mergeThreshold) break // Merge B into A for (const item of clusters[mergeB]) { clusters[mergeA].add(item) } active.delete(mergeB) } // Continue merging up to maxK even if we haven't hit the threshold check above // (the while loop above stops at maxK, so this handles the threshold-only stopping) // Now handle the case where active.size > 1 and we stopped due to threshold // — we keep whatever clusters remain // Assign cluster indices const assignments = new Array<number>(n) const activeArr = Array.from(active) const centroids: number[] = [] for (let clusterIdx = 0; clusterIdx < activeArr.length; clusterIdx++) { const members = Array.from(clusters[activeArr[clusterIdx]]) // Assign each member to this cluster index for (const item of members) { assignments[item] = clusterIdx } // Find centroid: member with lowest average distance to other members if (members.length === 1) { centroids.push(members[0]) } else { let bestItem = members[0] let bestAvg = Infinity for (const candidate of members) { let sum = 0 for (const other of members) { if (candidate !== other) { sum += distanceMatrix[distIndex(candidate, other, n)] } } const avg = sum / (members.length - 1) if (avg < bestAvg) { bestAvg = avg bestItem = candidate } } centroids.push(bestItem) } } return { assignments, k: activeArr.length, centroids } } /** * Extract a sub-matrix from a full distance matrix for a subset of IDs. * * @param allIds Ordered IDs matching the full matrix * @param matrix Full upper-triangle flat distance array * @param subsetIds IDs to extract (order preserved) * @returns Sub-matrix with only the requested IDs, or null entries skipped */ export function subsetDistanceMatrix( allIds: string[], matrix: number[], subsetIds: string[] ): { ids: string[]; matrix: number[] } { const n = allIds.length // Build index lookup const idxMap = new Map<string, number>() for (let i = 0; i < allIds.length; i++) { idxMap.set(allIds[i], i) } // Filter to only IDs that exist in allIds const validIds = subsetIds.filter((id) => idxMap.has(id)) const m = validIds.length if (m < 2) { return { ids: validIds, matrix: [] } } // Build sub-matrix const subMatrix: number[] = [] for (let i = 0; i < m; i++) { for (let j = i + 1; j < m; j++) { const origI = idxMap.get(validIds[i])! const origJ = idxMap.get(validIds[j])! subMatrix.push(matrix[distIndex(origI, origJ, n)]) } } return { ids: validIds, matrix: subMatrix } } |