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Rank and select

Introduction to rank/select

Our goal is to represent a static set A\subseteq \{0\dots n-1\}, let m = |A|

Two natural representations:

  • sorted array of elements of A
  • bit vector of length n
                                Array        Bit vector             
Memory in bits                  m log n         n       
Is x in A?                      O(log m)       O(1)         rank(x)>rank(x-1)                 
max. y in A s.t. y<=x           O(log m)       O(n)         select(rank(x))
i-th smallest element of A      O(1)           O(n)         select(i)
how many elements of A are <=x  O(log m)       O(n)         rank(x)

If m relatively large, e.g. m=\Omega (n/\log n), bit array uses less memory

To get rank in O(1)

  • trivial solution: keep rank(x) for each x\in \{0,\dots ,n-1\}, n log n bits
  • we will show that the bit vector of size n and o(n) additional bits are sufficient
  • a simple practical solution for space reduction:
    • choose value t
    • array R of length n/t, R[i] = rank(i*t)
    • rank(x) = R[floor(x/t)]+number of 1 bits in A[t*floor(x/t)+1..x]
    • memory n + (n/t)*lg(n) bits, time O(t)
  • O(1) solution based on this idea, slightly more complex details

To get select in O(1)

  • trivial solution is sorted array (see above)
  • can be done in n+o(n) bits, more complex than rank
  • practical compromise: binary search using log n calls of rank

Rank

  • overview na slajde
  • rank = rank of super-block + rank of block within superblock + rank of position within block
  • block and superblock: integer division (or shifts of sizes power of 2)
  • time O(1), memory n + O(n log log n / log n) bits

data structures:

  • R1: Array of ranks at superblock boundaries
  • R2: Array of ranks at block boundaries within superblocks
  • R3: precomputed table of ranks for each type of block and each position within block of size t2
  • B: bit array

rank(i):

  • superblock = i/t1 (integer division)
  • block = i/t2
  • index = block*t2;
  • type = B[index..index+t2-1]
  • return R1[superblock]+R2[block]+R3[type, i%t2]

Select

  • let t1 = lg n lglg n
  • store select(t1 * i) for each i = 0..n/t1 memory O(n / t1 * lg n) = O(n/ lg lg n)
  • divides bit vector into super-blocks of unequal size
  • consider one such block of size r. Distinguish between small super-block of size r<t1^2 and big super-block of size >= t1^2
  • large super-block: store array of indices of 1 bits
    • at most n/(t1 ^2) large super-blocks, each has t1 1 bits, each 1 bit needs lg n bits, overall O(n/lg lg n) bits
  • small super-block:
    • repeat the same with t2 = (lg lg n)^2
    • store select(t2*i) within super-block for each i = 0..n/t2 memory O(n/t2 * lg lg n) = O(n/lg lg n)
    • divides super-block into blocks, split blocks into small (size less than t2^2) and large
    • for large blocks store relative indices of 1 bits: n/(t2 ^2) large blocks, each has t2 1 bits, each 1 bit needs lg t1^2 = O(lg lg n) bits, overall O(n/log log n)
    • for small blocks of size at most t2^2, store as (t2^2)-bit integer, lookup table of all answers. Size of table is O(2^(t2^2) * t2 * log(t1)); t2^2 < (1/2) log n for large enough n, so 2^(t2^2) is O(sqrt(n)), the rest is polylog

Wavelet tree: Rank over larger alphabets

  • look at rank as a string operation over binary string B: count the number 1's in a prefix B[0..i]
  • denote it more explicitly as rank_1(i)
  • note: we can compute rank_0(i) as i+1-rank_1(i)
  • instead of a binary string consider string S over a larger alphabet of size sigma
  • rank_a(i): the number of occurrences of character a in a prefix S[0..i] of string S
  • OPT is n lg sigma
  • Use rank on binary strings as follows:
    • rozdelime abecedu na dve casti Sigma_0 a Sigma_1
    • Vytvorime bin. retazec B dlzky n: B[i]=j iff S[i] je v Sigma_j
    • Predspracujeme B pre rank
    • Vytvorime retazce S_0 a S_1, kde S_j obsahuje pismena z S, ktore su v Sigma_j, sucet dlzok je n
    • Rekurzivne pokracujeme v deleni abecedy a retazcov, az kym nedostaneme abecedy velkosti 2, kde pouzijeme binarny rank
  • rank_a(i) v tejto strukture:
    • ak a in Sigma_j, i_2=rank_j(i) v retazci B
    • pokracuj rekurzivne s vypoctom rank_a(i_2) v lavom alebo pravom podstrome
  • Velkost: binarne retazce n log_2 sigma + struktury pre rank nad bin. retazcami (mozem skonkatenovat na kazdej urovni a dostat o(n log sigma) + O(sigma lg n) pre strom ako taky
  • cas O(log sigma), robim jeden rank na kazdej urovni
  • extrahovanie S[i] tiez v case O(log sigma)
    • samotny retazec S teda nemusim ukladat
    • nech j = B[i], i_2 = rank_j(i) v retazci B
    • pokracuj rekurzivne s hladanim S_j[i2]
  • original paper: Grossi, Gupta and Vitter 2003 High-order entropy-compressed text indexes
  • survey G. Navarro, Wavelet Trees for All, Proceedings of 23rd Annual Symposium on Combinatorial Pattern Matching (CPM), 2012
  • http://alexbowe.com/wavelet-trees/

Rank in compressed string: RRR

  • Raman, Raman, Rao, from their 2002 paper Succinct indexable dictionaries with applications to encoding k-ary trees and multisets
  • http://alexbowe.com/rrr/

class of a block: number of 1s

  • R1: Array of ranks at superblock boundaries
  • R2: Array of ranks at block boundaries within superblocks
  • R3: precomputed table of ranks for each class, each signature within class and each position within block of size t2
  • S1: Array of superblock starts in compressed bit array
  • S2: Array of block offsets in compressed bit array
  • C: class (the number of 1s) in each block
  • L: size of signature for each class (ceil lg (t2 choose class))
  • B: compressed bit array (concatenated signatures pf blocks)

rank(i):

  • superblock = i/t1 (integer division)
  • block = i/t2
  • index = S1[superblock]+S2[block]
  • class = C[block]
  • length = L[class]
  • signature = B[index..index+length-1]
  • return R1[superblock]+R2[block]+R3[class, signature, i%t2]

Analysis of RRR:

  • Let S be a string in which a occurs n_a times
  • Its entropy is H(S) = sum_a \frac{n_a}{n} \log \frac{n}{n_a}
  • let x_i be the number of 1s in block i, let n_1 be the overall number of 1s, i.e. n_{1}=\sum x_{i}*<math>\sum _{i}\lceil \lg {t_{2} \choose x_{{i}}}\rceil
  • \leq {\frac  {n}{t_{2}}}+\sum _{i}\lg {t_{2} \choose x_{{i}}}
  • =o(n)+\lg \prod _{i}{t_{2} \choose x_{{i}}}
  • \leq o(n)+\lg {n \choose n_{1}} (choice of x_i in each block is subset of choice of n_1 overall)


  • \lg {n \choose n_{1}}=[\ln(n!)-\ln(n_{1}!)-\ln(n_{0}!))]/\ln(2)
  • =[n\ln(n)-n-n_{1}\ln(n_{1})+n_{1}-n_{0}\ln n_{0}+n_{0}+O(\log n)]/\ln(2)
  • =n_{1}\lg(n/n_{1})+n_{0}\lg(n/n_{0})+O(\log n) (linear terms cancel out, n\ln(n) divided into two parts: n_0\ln(n) and n_1\ln(n))
  • =nH(B)+O(\log n)

Use in wavelet tree (see Navarro survey of wavelet trees):

  • top level n0 zeroes, n1 ones, level 1: n00, n01, n10, n11
  • top level entropy encoding uses n0 lg(n/n0) + n1 lg(n/n1) bits
  • next level uses n00 lg(n0/n00) + n01 lg(n0/n01) + n10 lg(n1/n10) + n11 lg(n1/n11)
  • add together n00 lg(n/n00) + n01 lg(n/n01) + n10 lg(n/n10) + n11 lg(n/n11)
  • continue like this until you get sum_c n_c lg(n/n_c) which is nH(T)

Instead of wavelet tree, store indicator vector for each character from alphabet in RRR string

  • \sum _{a}[n_{a}\lg {\frac  {n}{n_{a}}}+(n-n_{a})\lg {\frac  {n}{n-n_{a}}}+o(n)]
  • the first terms in the sum equal to n H(T), the last terms sum to o(\sigma n), middle terms can be bound by O(n)
  • \lg {\frac  {n}{n-n_{a}}}=\lg(1+{\frac  {n_{a}}{n-n_{a}}})=\ln(1+{\frac  {n_{a}}{n-n_{a}}})/\ln(2)\leq (1/\ln(2))\cdot {\frac  {n_{a}}{n-n_{a}}}
  • We have used inequality ln(1+x)<=x for x>=-1
  • Then \sum _{a}(n-n_{a})\lg {\frac  {n}{n-n_{a}}}\leq (1/\ln(2))\sum _{a}(n-n_{a}){\frac  {n_{a}}{n-n_{a}}}=(1/\ln(2))\sum _{a}n_{a}=n/\ln(2)=O(n)

Binarny strom

  • pridame pomocne listy tak, aby kazdy povodny vrchol bol vnut vrchol s 2 detmi
  • ak sme mali n vrcholov, teraz mame n vnut. vrcholov a n+1 listov, t.j. 2n+1 vrcholov
  • ocislujeme ich v level-order (prehladavanie do sirky) cislami 1..2n+1
  • vrchol je reprezentovany tymto cislom
  • datova struktura: bitovy vektor dlzky 2n+1, ktory ma na pozicii i 1 prave vtedy ak vrchol i je vnutorny vrchol
  • nad tymto vektorom vybudujeme rank/select struktury v o(n) pridavnej pamati
  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17
  1  1  1  1  1  0  1  0  0  1  1  0  0  0  0  0  0
  A  B  C  D  E  .  F  .  .  G  H  .  .  .  .  .  .

Lemma: Pre i-ty vnutorny vrchol (i-ta jednotka vo vektore, na pozicii select(i)) su jeho deti na poziciach 2i a 2i+1

  • indukcia na i
  • pre i=1 (koren) ma deti na poziciach 2 a 3
  • deti i-teho vnutorneho vrcholu su v poradi hned po detoch (i-1)-veho
    • dva pripady podla toho, ci (i-1)-vy a i-ty su na tej istej alebo roznych urvniach, ale v oboch pripadoch medzi nimi len pomocne listy a teda medzi ich detmi nic (obrazok)
    • kedze deti (i-1)-veho su na poziciach podla IH 2i-2 a 2i-1, deti i-teho budu na 2i a 2i+1.

Dosledok:

  • left_child(i) = 2 rank(i)
  • right_child(i) = 2 rank(i)+1
  • parent(i) = select(floor(i/2))

Ak chceme pouzit ako lexikograficky strom nad binarnou abecedou: potrebujeme zistit, ci vnut. vrchol zodpoveda slovu v mnozine

  • jednoduchu fix s n dalsimi bitmi: z cisla vrchola 1..2n+1 zistime rankom cislo vnut vrchola 0..n-1 a pre kazde si pamatame bit, ci je to slovo z mnoziny
  • da sa asi aj lepsie

Catalanove cisla

Ekvivalentne problemy:

  • kolko je binarnych stromov s n vrcholmi?
  • kolko je usporiadanych zakorenenych stromov s n+1 vrcholmi?
  • kolko je dobre uzatvorkovanych vyrazov z n parov zatvoriek?
  • kolko je postupnosti, ktore obsahuju n krat +1 a n krat -1 a pre ktore su vsetky prefixove sucty nezaporne?

Chceme dokazat, ze odpoved je Catalanovo cislo <tex>C_n = {2n\choose n}/(n+1)</tex> (C_0 = 1, C_1=1, C_2 = 2, C_3 = 5,...)

  • Rekurencia <tex>T(n) = \sum_{i=0}^{n-1} T(i)T(n-i-1)</tex> (nech najlavejsia zatvorka obsahuje vo vnutri i parov), riesime metodami kombinatorickej analyzy (generujuce funkcie)
  • Iny dokaz [1]
    • Nech X(n,m,k) je mnozina vsetkych postupnosti dlzky n+m obsahujucich n +1, m -1 a takych ze kazdy prefixovy sucet je aspon k.
    • |X(n,m,-\infty )|={n+m \choose n}
    • my chceme |X(n,n,0)|
    • nech Y = X(n,n,-infty)-X(n,n,0), t.j. vsetky zle postupnsti
    • najdeme bijekciu medzi Y a X(n-1,n+1,-infty)
    • vezmime postupnost z Y, najdime prvy zaporny prefixovy sucet (-1), od toho bodu napravo zmenme znamienka, dostavame postupnost z X(n-1,n+1,-infty) (obrazkovy dokaz so schodmi)
    • naopak vezmime postupnost z X(n-1,n+1,-infty), najdeme prvy bod, kde je zaporny prefixovy sucet (-1), od toho bodu napravo zmenme znamienka, dostavame postupnost z Y
    • tieto dve zobrazenia su navzajom inverzne, preto musi ist o bijekciu
    • preto |X(n,n,0)|=|X(n,n,-\infty )|-|X(n-1,n+1,-\infty )|={2n \choose n}-{2n \choose n-1}
    • {2n \choose n}-{2n \choose n-1}={\frac  {(2n)!}{n!n!}}-{\frac  {(2n)!}{(n-1)!(n+1)!}}={\frac  {(2n)!(n+1-n)}{n!(n+1)!}}={2n \choose n}/(n+1)

Connection to FM index (to be covered later in the course)

  • Retazec dlzky n nad abecedou velkosti sigma mozeme ulozit pomocou n lg sigma bitov (predpokladajme pre jednoduchost, ze sigma je mocnina 2)
  • napr pre binarny retazec potrebuje n bitov
  • Sufixove pole na umoznuje rychlo hladat vzorku, ale potrebuje n lg n bitov, pripadne 3 n lg n, co je rychlejsie rastuca funkcia ako n
  • sufixove stromy maju tiez Theta(n log n) bitov s vacsou konstantou
  • vedeli by sme rychlo vyhladavat vzorku aj s ovela mensim indexom?
  • oblast uspornych datovych struktur

Back to FM index

  • jednoducha finta na zlepsenie pamate pre rank:
    • ulozime si rank len pre niektore i, zvysne dopocitame na zaklade pocitania vyskytov v L (vydelime pamat k, vynasobime cas k)
  • rank + L pomocou wavelet tree v n lg sigma + o(n log sigma) bitoch, hladanie vyskytov sa spomali na O(m log sigma)
  • ako kompaktnejsie ulozit SA
  • mozeme ulozit SA len pre niektore riadky, pre ine sa pomocou LF transformacie dostaneme k najblizsiemu riadku pre ktory mame SA
    • let us say we want to print SA[i], but it is not stored, let denote its unknown value x
    • L[i] is T[x-1]
    • let j = LF[i], j is the row in which SA[j]=x-1
    • if SA[j] stored, return SA[j]+1, otherwise continue in the same way
    • ako presne ulozit "niektore riadky" ked su nerovnomerne rozmiestnene v SA (rovnomerne v T)?
  • samotne L mozeme ulozit aj komprimovane s vhodnymi indexami, vid. clanok od F a M alebo ho mame reprezentovane implicitne vo wavelet tree