Úsporné dátové štruktúry

Uvod do uspornych datovych struktur

• 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

Uvod do rank/select

Chceme reprezentovat staticku mnozinu A subseteq {0..n-1}, nech m = |A|

• pole cisel v A v utriedenom poradi (m log n bitov)
• bitovy vektor dlzky n (n bitov)

                           Pole        Bitovy vektor             
Pamat v bitoch            m log n         n       
Je x v A?                 O(log m)       O(1)         rank(x)>rank(x-1)                 
max. y v A t.z. y<=x      O(log m)       O(n)         select(rank(x))
i-ty najmensi prvok z A   O(1)           O(n)         select(i)
kolko je v A prvkov <=x   O(log m)       O(n)         rank(x)

Ak m je pomerne velke (cca Omega(n/log n)), pamat mensia s vyuzitim bitoveho pola

Ak chcem rank v O(1)

• trivialne riesenie: pamatam si rank(x) pre kazde x, n log n bitov
• ukazeme si, ze staci o(n) bitov navyse k bitovemu vektoru s n bitmi
• prakticke riesenie:
  • zvolim si hodnotu t
  • pole R dlzky n/t, R[i] = rank(i*t)
  • rank(x) = R[floor(x/t)]+pocet jedniciek v A[t*floor(x/t)+1..x]
  • pamat n + (n/t)*lg(n) bitov, cas O(t)
• O(1) riesenie zalozene na tejto myslienke, trochu zlozitejsie detaily

Ak chcem select v O(1)

• trivialne riesenie je utriedene pole prvkov (pozri vyssie)
• da sa spravit usporne o(n) bitov navyse k bitovemu vektoru s n bitmi, ale zlozitejsie
• prakticky kompromis: binarne vyhladavanie, log n volani 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

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 nad vacsou abecedou

• nad binarnou abecedou vieme spravit rank_1(i): pocet vyskytov 1 v A[0..i]
• poznamka: ak chceme spocitat rank_0(i), pouzijeme i+1-rank_1(i)

• 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/

• namiesto bitoveho pola mame retazec S nad vacsou abecedou Sigma
• Chceme podporovat rank_a(i), co je pocet vyskytov pismena a v retazci S[0..i]
• OPT je n lg sigma
• Pouzijeme rank na binarnych retazcoch nasledovne:
  • 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
  • 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_2 sigma)
• 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]

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

• $\sum _{i}\lceil \log _{2}{t_{2} \choose x_{{i}}}\rceil$
• $\leq {\frac {n}{t_{2}}}+\sum _{i}\log _{2}{t_{2} \choose x_{{i}}}$
• <tex>= o(n)+\log_2 \prod_i {t_2 \choose x_{i}}</tex>
• <tex>\le o(n)+\log_2 {n \choose n_1}</tex> (choice of x_i in each block is subset of choice of n_1 overall)

• <tex>\log_2 {n \choose n_1} = [\ln(n!) - \ln(n_1!) - \ln(n_0!))]/\ln(2)</tex>
• <tex>= [n\ln(n)-n-n_1\ln(n_1)+n_1-n_0\ln n_0+n_0+O(\log n)]/\ln(2)</tex>
• <tex>= n_1\log_2 (n_1/n) + n_0\log_2 (n_0/n) + O(\log n)</tex> (linear terms cancel out, n\ln(n) divided into two parts: n_0\ln(n) and n_1\ln(n))
• <tex>= n H(B)+O(\log n)</tex>

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)

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) 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 nad 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 nad n) - (2n nad n-1) = (2n nad n)/(n+1)