CS586: Distributed Computing Tutorial 5 Professor: Panagiota Fatourou TA: Eleftherios Kosmas CSD - December 2011
Snapshots - T-Opt da data *sca ta *scan (voi n (void) { d) { void update (data value, int i) { void update (data value, int i) { data view[1..m], d1, d2; view[1..m], d1, d2; int cur int c urr_seq; _seq; int j; int j; data d1, d2; data d1, d2; seq=seq+1; seq=seq+1; curr_seq = seq; curr_seq = seq; 7. 7. 1. 1. d1 = pre[i]; d1 = pre[i]; for (j = 1; j for (j = 1; j ≤ m; j++ m; j++) { { 8. 8. 2. 2. d2 = post[curr seq][ d2 = p st[curr seq][i]; ]; d1 = pre[j]; d1 = pre[j]; 9. 9. 3. 3. if (d2 == null) if (d2 == n ll) d2 = post[seq][j]; d2 = post[seq][j]; 10. 10. 4. 4. post[curr seq][i]=d1; post[curr seq][i]=d1; if(d2 == n if(d2 == null) view[j ll) view[j]=d1; ]=d1; 5. 5. 11. 11. else view[j]=d2; else view[j]=d2; pre[i]=va pr =value; e; 6. 6. 12. 12. } } return view; ret rn view; } Initially seq = 1 seq = 1 post[1..k][1..m] = {null, null, …, null} post[1..k][1..m] = {null, null, …, null} pre[1..m] = {null, null, …, null} pre[1..m] = {null, null, …, null} CS586 Tutorial 5 by Eleftherios Kosmas 2
Snapshots - T-Opt 2 m 1 m 1 2 post : pre : 1 2 k seq : CS586 Tutorial 5 by Eleftherios Kosmas 3
T-Opt - Linearizability α is an execution of T-opt and S is any SCAN performed in α � w S : is the write performed by S (line 7) � seq seq S : is the value written to seq by w S � ∈ For each i {1,…, m}, � S : the read of pre[i] by S (line 9) r i � S : the read of post[seq S ][i] by S (line 10) ~r ~r i � v i : the value that S returns for component A i � S and v i at r i S If S reads null at ~r i � S : the UPDATE that writes v i to pre[i] and its write to it is the last write U i � to it that precedes r i S If S reads v i at ~r i S � S : the UPDATE that writes v i to post[seq S ][i] and its write to it is the last V i � write to it that precedes ~r i S S : the UPDATE that writes v i to pre[i] and its write to it is the last write U i � S reads pre[i] to it before V i S : the write to pre[i] by U i S (line 6) w i � CS586 Tutorial 5 by Eleftherios Kosmas 4
Snapshots - T-Opt void update (data value, int i) { void update (data value, int i) { data *sca da ta *scan (voi n (void) { d) { int c int curr_seq; _seq; data view[1..m], d1, d2; view[1..m], d1, d2; data d1, d2; data d1, d2; int j; int j; curr_seq = seq; curr_seq = seq; seq=seq+1; seq=seq+1; // w // w S 1. 1. 7. 7. d1 = pre[i]; d1 = pre[i]; for (j = 1; j ≤ m; j++ for (j = 1; j m; j++) { { 2. 2. 8. 8. S d2 = post[curr seq][ d2 = p st[curr seq][i]; ]; d1 = pre[j]; d1 = pre[j]; // r // r i 3. 3. 9. 9. S if (d2 == n if (d2 == null) ll) d2 = post[seq][j]; d2 = post[seq][j]; // ~r // ~r i 4. 4. 10. 10. post[curr seq][i]=d1; post[curr seq][i]=d1; if(d2 == n if(d2 == null) view[j ll) view[j]=d1; ]=d1; 5. 5. 11. 11. pr pre[i]=va =value; // w w i S else view[j]=d2; else view[j]=d2; 6. 6. 12. 12. } } ret return view; rn view; } Initially seq = 1 seq = 1 post[1..k][1..m] = {null, null, …, null} post[1..k][1..m] = {null, null, …, null} pre[1..m] = {null, null, …, null} pre[1..m] = {null, null, …, null} CS586 Tutorial 5 by Eleftherios Kosmas 5
T-Opt - Linearization points � Each SCAN S is linearized at w S ∈ � For each i {1,…, m}, S follow w S , � if w i S is linearized just before w � U i just before w S � each UPDATE on A i that performs its write to pre[i] between S is linearized just before w w S and w i just before w S � ties are broken by the order that the writes to pre[i] occur � each of the rest of UPDATES is linearized at its write to pre[i] write to pre[i] (line 6) CS586 Tutorial 5 by Eleftherios Kosmas 6
T-Opt - Linearizability: Intuition T-Opt is linearizable linearizable A. A. The linearization point of each operation is within its execution within its execution interval interval Intuition : � SCANs? � UPDATEs linearized when they write to pre[i]? � UPDATEs linearized before the linearization point of some SCAN? B. B. Scans returns consistent vectors consistent vectors Intuition : (v i is written by the last UPDATE linearized before SCAN) S follows w S ? � if w i S precedes w S ? � if w i hint : the linearization order of UPDATEs on A i respects the order of writes to pre[i] (by those UPDATEs) CS586 Tutorial 5 by Eleftherios Kosmas 7
T-Opt - Linearizability: sketch of proof T-Opt is linearizable linearizable A. A. The linearization point of each operation is within its execution within its execution interval interval S follows w S , then the execution of an UPDATE that performs A.1. If w i S (including U i S ) starts before w S its write to pre[i] between w S and w i B. B. Scans returns consistent vectors consistent vectors B.1. The linearization order of the UPDATES on any component A i respects the order in which these UPDATES perform their writes to pre[i] CS586 Tutorial 5 by Eleftherios Kosmas 8
T-Opt - Technical Lemmas ∈ S follows w S Lemma 1. For each i {1,…,m}, ~r Lemma 1 ~r i follows w i sketch of proof S precedes w S … � if w i S follows w S � if w i S and v i at r i S … Assume, S reads null at ~r i � S … Assume, S reads v i at ~r i � S , and let r pre be the read of pre[i] Lemma 2 Lemma 2. Assume that S reads v i at ~r i S . Then, r pre by V i pre follows w follows w S sketch of proof � Assume, by the way of contradiction, that r pre is executed before w S S precedes w S … � Then, the read of seq by V i CS586 Tutorial 5 by Eleftherios Kosmas 9
T-Opt ∈ S follows w S , it Lemma A.1. For each i {1,…,m}, such that w i Lemma A.1 holds that any UPDATE on A i that performs its write to pre[i] S (including U i S ) begins its execution before between w S and w i w S sketch of proof � Assume, by the way of contradiction, that there is an UPDATE U on A i that starts its execution after w S and performs its write to pre[i] (let it be S w) before w i � U reads seq S in seq � Lemma 1 � U ends its execution before the end of S � U starts after w S S � S read a value other than null at ~r i � U executes lines 4-5 before w (which precedes w i S ) S reads a value other than null in post[seq S ][i] � V i CS586 Tutorial 5 by Eleftherios Kosmas 10
T-Opt Lemma A. Lemma A. Let α be any execution of T-Opt. The linearization point of any SCAN or UPDATE executed in α is within its execution interval sketch of proof � SCANs � UPDATEs linearized at their writes to pre � Let U be an update on A i which is not linearized at its write to pre[i] � There is a SCAN S such that S is executed after w S w i � the write to pre[i] by U is executed between w S and w i S � U is linearized just before w S � � Lemma A.1. � U begins its execution before w S CS586 Tutorial 5 by Eleftherios Kosmas 11
T-Opt Lemma B.1 Lemma B.1. Let U 1 , U 2 be two update on some component A i , 1 ≤ i ≤ m. Denote by w 1 the write to pre[i] by U 1 and by w 2 the write to pre[i] by w 2 . If w 1 precedes w 2 , the linearization point of U 1 precedes the linearization point of U 2 sketch of proof � Assume, by the way of contradiction, that the claim does not hold � U 1 and U 2 are linearized at their writes to pre[i]… � At least one of the U 1 , U 2 is not linearized at its write to pre[i] U 2 is linearized at w 2 … ( hint: use Lemma A. ) � U 1 is linearized at w 1 � U 2 can not be linearized at w 2 (why?) � Therefore a SCAN S exists such that � S w 2 has been performed between w S and w i � S , since w 1 precedes w 2 w 1 precedes w i � if w 1 follows w S … � if w 1 precedes w s … ( hint: use Lemma A. ) � CS586 Tutorial 5 by Eleftherios Kosmas 12
T-Opt Lemma B.1 Lemma B.1. Let U 1 , U 2 be two update on some component A i , 1 ≤ i ≤ m. Denote by w 1 the write to pre[i] by U 1 and by w 2 the write to pre[i] by w 2 . If w 1 precedes w 2 , the linearization point of U 1 precedes the linearization point of U 2 sketch of proof � None of U 1 , U 2 is linearized at its write to pre[i] Two SCANS S1 and S2 exist such that � S1 w 1 has been performed between w S1 and w i � S2 w 2 has been performed between w S2 and w i � if S1 = S2 … � if S1 follows S2 … � S2 precedes the end of S2 Lemma 1 � w i � if S1 precedes S2 … � S1 precedes the end of S1 Lemma 1 � w i � CS586 Tutorial 5 by Eleftherios Kosmas 13
T-Opt Lemma B Lemma B. Let α be any execution of T-Opt. Any SCAN executed in α returns a consistent vector sketch of proof S store v i in component A i and its linearization point � Recall that U i precedes the linearization point of S ∈ � Assume, by the way of contradiction, that there is an integer i {1,…,m} such that the last UPDATE on A i which has been linearized before S is not U i S denote by U this update and let w be the write to pre[i] by U � S � w follows w i � if w precedes w i S , Lemma B.1. implies that U is linearized before U i S S and v i at r i S � S reads null at ~r i U can not be linearized at w � S (why?) � w follows r i � U is linearized before S, and S is linearized at w S CS586 Tutorial 5 by Eleftherios Kosmas 14
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