1. The Encoding Complexity of Two Dimensional Range Minimum Data Structures European Symposium on Algorithms, Inria, Sophia Antipolis, France, September 3, 2013 1234  n 131342128 27146111537 3139921274416  2328513447 m34241 911 i1i1 i2i2 j2j2 j1j1 Gerth Stølting BrodalAndrej BrodnikPooya Davoodi AarhusPrimorska & LjubljanaPolytechnic New York Cost Space (bits) Query time Preprocessing time Models Indexing (input accessible) Encoding (input not accessible) RMQ(i 1, i 2, j 1, j 2 ) = (2,3) = position of min Assumption m ≤ n

2. Indexing Model (input accessible) Encoding Model (input not accessable) m ≤ n Preprocessing: Do nothing ! Very fast preprocessing Very space efficient Queries O(mn) Tabulate the answer to all ~ m 2 n 2 possible queries Preprocessing & space O(m 2 n 2  log n) bits Queries O(1) Store rank of all elements Preprocessing & space O(mn  log n) bits Queries O(mn) Some (Trivial) Results

3. Encoding m = 1 (Cartesian tree) 352104116719148 3 5 2 10 4 11 6 7 1 9 14 8 j1j1 j2j2 RMQ(j 1, j 2 ) = NCA(j 1, j 2 ) To support RMQ queries we need... tree structure (111101001100110000100100) mapping between nodes and cells (inorder) n 1 min ?

4. Indexing Model (input accessible) Encoding Model (input not accessable) m = 1 1D 2n+o(n) bits, O(1) time [FH07] n/c bits  Ω(c) time [BDS10] n/c bits, O(c) time [BDS10] ≥ 2n - O(log n) bits 2n+o(n) bits, O(1) time [F10] 1 < m < n1 < m < n O(mn  log n) bits, O(1) time [AY10] O(mn) bits, O(1) time [BDS10] mn/c bits  Ω(c) time [BDS10] O(c  log 2 c) time [BDS10] O(c  log c  (loglog c) 2 ) time [BDLRR12] Ω(mn  log m) bits [BDS10] O(mn  log n) bits, O(1) time [BDS10] O(mn  log m) bits, ? time [NEW] m = n squared Ω(mn  log n) bits [DLW09] O(mn  log n) bits, O(1) time [AY10] Some (Less Trivial) Results

5. New Results 1. O(nm  (log m+loglog n)) bits – tree representation – component decomposition 2. O(nm  log m  log* n)) bits – bootstrapping 3. O(nm  log m) bits – relative positions of roots – refined component construction

6. Tree Representation 11413 96128 52107 12108119675432 1 Requirements Cells  leafs Query  Answer = rightmost leaf 11413 96128 52107 12108672 121110987654321 Trivial solution Sort leafs Ω(mn  log n) bits

7. Components  = 3 11413 96128 52107 12108119675432 1 Construction Consider elements in decreasing order Find connected components with size ≥  L-adjacency  |C 1 |≤ 4  -3, |C i |≤ 2m  Representation O(mn + mn/  log n + mn  log m + mn  log(m  ))  = log n  O(mn  (log m+loglog n)) 11413 96128 52107 11413 96128 52107 11413 96128 52107 11413 96128 52107 11413 96128 52107 11413 96128 52107 1 12108 C1C1 75432 C3C3 1196 C2C2 413 96128 52107 11413 96128 52107 11413 96128 52107 11413 96128 52107 11413 96128 52107 11413 96128 52107 11413 96128 52107 11413 96128 52107 L L-adjacency Spanning tree structures Spanning tree edges Component root positions Local leaf ranks in components

8. Indexing Model (input accessible) Encoding Model (input not accessable) m = 1 1D 2n+o(n) bits, O(1) time [FH07] n/c bits  Ω(c) time [BDS10] n/c bits, O(c) time [BDS10] ≥ 2n - O(log n) bits 2n+o(n) bits, O(1) time [F10] 1 < m < n1 < m < n O(mn  log n) bits, O(1) time [AY10] O(mn) bits, O(1) time [BDS10] mn/c bits  Ω(c) time [BDS10] O(c  log 2 c) time [BDS10] O(c  log c  (loglog c) 2 ) time [BDLRR12] Ω(mn  log m) bits [BDS10] O(mn  log n) bits, O(1) time [BDS10] O(mn  log m) bits, ? time [NEW] m = n squared Ω(mn  log n) bits [DLW09] O(mn  log n) bits, O(1) time [AY10] Results better upper or lower bound ? Thank you