### 分布式分片算法-chord

5 g D(log n) odog- N h 02 successor(k) th O(1/N) (1+E)K/N O(K/N k start I (n 1≤k O(log n (1+22) (2-25 h 1 O(log N h h d id d 3.[31 uccessor(n+2 inger[3] interval=Finger[3]. start, 1) 2 finger[1]. start =2 finger [3. start =5 find-suco O(log N) (id) find_prede ed k p (id E(ne', 7 ≠p (id) h T n.closest-preceding -finger(id) og g k O(log N) O(logN og sucCessor □ start nt. succ 4 7 (mlog N join( (+1) [+1] cessor. n O(log N) i=1 g [ n init_ finger-table(n') (finger[1]. start) [2十 [2+1] 2十 22 nupdate_othersO ILupdate_finger-table(s, i) ng O(log N) n njoin(n') predeces sor nil suCcessor n succes s or predecessor T∈( suCcessor 7 notify(m’) (predecessor predecessor, n rs i]. mode ([].) og 001 071 lp 350 s200 O(log N) 5n∩ O(log N) F N/ O(log N) O(log N) 4.6× 0.368 h 0.02 c05 050100150200250300350403 er of keys ix 10.000) 104 5×10 g O(log(N log N))= o(log N) 104 1,2,5,10 T 4.8×1.6× 0.5 Failed Nodes (Fraction of total) N=10 r=log M logN 400 log N O(log N) 100 10° 1=12 t st and 90th pcrccntilcs 10 005 2 10 1Number cf rodes 10000 10c000 Path length R Node FaWJoin Rate Per Second) 0.01 5k/500k:/100

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soczzd2014 学习中，谢谢分享
2017-10-18
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