Paolo Ferragina, Università di Pisa Compressed Rank & Select on general strings Paolo Ferragina Dipartimento di Informatica, Università di Pisa

Paolo Ferragina, Università di Pisa Generalised Rank and Select Rank(c,i) = #c in L[1,i] Select(c,i) = position of the i-th c in L L = a b a a a c b c d a b e c d... Rank( a, 7 ) = 4Select( a, 2 ) = 3

Paolo Ferragina, Università di Pisa Generalised Rank and Select  If  is small (i.e. constant)  Build binary Rank data structure for each symbol of  Rank takes O(1) time and small space  If  is large ( words ?)  Need a smarter solution: Wavelet Tree data structure Algorithmic reduction: >> Reduce Rank&Select over arbitrary strings... to Rank&Select over binary strings

Paolo Ferragina, Università di Pisa The Wavelet Tree ac br d abracadabra (Alphabetic ?) Tree

Paolo Ferragina, Università di Pisa The Wavelet Tree ac br d abracadabra aacaaabrdbr brbr rr ? aaaaa ? bb ? d ?

Paolo Ferragina, Università di Pisa The Wavelet Tree ac br d abracadabra aacaaa brdbr brbr abracadabra aacaaa brdbr brbr Fact. Given the tree and the binary strings, we can recover the original string !! In any case, O(|  | log |  |) bits. Easier Alphabetic order + Heap structure

Paolo Ferragina, Università di Pisa brdbr abracadabra brbr 0101 aacaaa The Wavelet Tree ac br d Rank(b,8) Rank(b,3) Rank(b,2) Reduce to right symbols Reduce to left symbols It’s binary Every step can be turned to binary

Paolo Ferragina, Università di Pisa abracadabra Rank 1 (8)=3 Rank 0 (2) = 2 – Rank 1 (1)= 1 Rank 0 (3) = 3 – Rank 1 (3)= 2 brbr 0101 brdbr aacaaa The Wavelet Tree ac br d Generalised R&S implemented with log |  | binary R&S Rank(b,8) Right move = Rank 1 Left move = Rank 0 Left move = Rank 0 Select is similar

Paolo Ferragina, Università di Pisa Representing Trees Paolo Ferragina Dipartimento di Informatica, Università di Pisa

Standard representation Binary tree: each node has two pointers to its left and right children An n-node tree takes 2n pointers or 2n lg n bits. Supports finding left child or right child of a node (in constant time). For each extra operation (eg. parent, subtree size) we have to pay additional n lg n bits each. x xxxx xxxx

Can we improve the space bound? There are less than 2 2n distinct binary trees on n nodes. 2n bits are enough to distinguish between any two different binary trees. Can we represent an n node binary tree using 2n bits?

Binary tree representation A binary tree on n nodes can be represented using 2n+o(n) bits to support: parent left child right child in constant time.

Heap-like notation for a binary tree Add external nodes Label internal nodes with a 1 and external nodes with a 0 Write the labels in level order One can reconstruct the tree from this sequence An n node binary tree can be represented in 2n+1 bits. What about the operations?

Heap-like notation for a binary tree parent(x) = On red ( ⌊ x/2 ⌋ ) left child(x) = On green(2x) right child(x) = On green(2x+1) x  x: # 1’s up to x (Rank) x  x: position of x-th 1 (Select)