1. Representing distributed algorithms Why do we need these? Don’t we already know a lot about programming? Well, you need to capture the notions of atomicity, non-determinism, fairness etc. These concepts are not built into languages like JAVA, C++, python etc!

2. Syntax & semantics: guarded actions  is equivalent to if G then A (Borrowed from E.W. Dijkstra: A Discipline of Programming)

3. Syntax & semantics: guarded actions Sequential actionsS0; S1; S2;... ; Sn Alternative constructsif.......... fi Repetitive constructsdo......... od The specification is useful for representing abstract algorithms, not executable codes.

4. Syntax & semantics Alternative construct ifG 1  S 1 []G 2  S 2 … []G n  S n fi When no guard is true, skip (do nothing). When multiple guards are true, the choice of the action to be executed is completely arbitrary.

5. Syntax & semantics Repetitive construct doG 1  S 1 []G 2  S 2. []G n  S n od Keep executing the actions until all guards are false and the program terminates. When multiple guards are true, the choice of the action is arbitrary.

6. Example: graph coloring 0 1 0 1 {program for process i} c[i] = color of process i do  j  neighbor(i): c(j) = c(i)  c(i) := 1- c(i) od Will the above computation terminate? There are four processes and two colors 0, 1. The system has to reach a configuration in which no two neighboring processes have the same color.

7. Consider another example program uncertain; definex : integer; initially x = 0 dox < 4   x := x + 1 []x = 3  x := 0 od Question. Will the program terminate? (Our goal here is to understand fairness)

8. The adversary A distributed computation can be viewed as a game between the system and an adversary. The adversary may come up with feasible schedules to challenge the system (and cause “bad things”). A correct algorithm must be able to prevent those bad things from happening.

9. Non-determinism (Program for a token server - it has a single token} repeat if (req1 ∧ token) then give the token to client1 else if (req2 ∧ token) then give the token to client2 else if (req3 ∧ token) then give the token to client3 forever Now, assume that all three requests are sent simultaneously. Client 2 or 3 may never get the token! The outcome could have been different if the server makes a non-deterministic choice. 1 2 3 Token server

10. Examples of non-determinism Determinism caters to a specific order and is a special case of non-determinism. If there are multiple processes ready to execute actions, then who will execute the action first is nondeterministic Message propagation delays are arbitrary and the order of message reception is non-deterministic

11. Atomicity (or granularity) Atomic = all or nothing Atomic actions = indivisible actions do redmessage  x:= 0 {red action} [] blue message  x:=7 {blue action} od Regardless of how nondeterminism is handled, we would expect that the value of x will be an arbitrary sequence of 0 's and 7 's. Right or wrong? x

12. Atomicity (continued) do redmessage  x:= 0 {red action} [] blue message  x:=7 {blue action} od Let x be a 3-bit integer x2 x1 x0, so x:=7 means (x2:=1, x1:= 1, x2:=1), and x:=0 means (x2:=0, x1:= 0, x2:=0) If the assignment is not atomic, then many interleavings are possible, leading to any possible value of x between 0 and 7 x So, the answer depends on the atomicity of the assignment

13. Atomicity (continued) Does hardware guarantee any form of atomicity? Yes! (examples?) Transactions are atomic by definition (in spite of process failures). Also, critical section codes are atomic. We will assume that G  A is an “ atomic operation.” Does it make a difference if it is not so? y x if x ≠ y  y:= x fi

14. Atomicity (continued) {Program for P} define b: boolean initially b = true do b  send msg m to Q [] ¬ empty(R,P)  receive msg; b := false od Suppose it takes 15 seconds to send the message. After 5 seconds, P receives a message from R. Will it stop sending the remainder of the message? NO. PQR b

15. Fairness Defines the choices or restrictions on the scheduling of actions. No such restriction implies an unfair scheduler. For fair schedulers, the following types of fairness have received attention: –Unconditional fairness –Weak fairness –Strong fairness Scheduler / demon / adversary

16. Fairness Programtest define x : integer {initial value unknown} dotrue  x : = 0 []x = 0  x : = 1 []x = 1  x : = 2 od An unfair scheduler may never schedule the second (or the third actions). So, x may always be equal to zero. An unconditionally fair scheduler will eventually give every statement a chance to execute without checking their eligibility. (Example: process scheduler in a multiprogrammed OS.)

17. Weak fairness Program test define x : integer {initial value unknown} dotrue  x : = 0 []x = 0  x : = 1 []x = 1  x : = 2 od A scheduler is weakly fair, when it eventually executes every guarded action whose guard becomes true, and remains true thereafter A weakly fair scheduler will eventually execute the second action, but may never execute the third action. Why?

18. Strong fairness Programtest define x : integer {initial value unknown} dotrue  x : = 0 []x = 0  x : = 1 []x = 1  x : = 2 od A scheduler is strongly fair, when it eventually executes every guarded action whose guard is true infinitely often. The third statement will be executed under a strongly fair scheduler. Why? Study more examples to reinforce these concepts

19. Program correctness The State-transition model A global state S  s 0 x s 1 x … x s m {s k = local state of process k} S0  S1  S2  … Each state transition is caused by an action by an eligible process. action state Initial state transition We reason using interleaving semantics, and assume that concurrent actions are serialized in an arbitrary order A sample computation (or behavior) is ABGHIFL

20. Correctness criteria Safety properties Bad things never happen Liveness properties Good things eventually happen

21. Testing vs. Proof Testing: Apply inputs and observe if the outputs satisfy the specifications. Fool proof testing can be painfully slow, even for small systems. Most testing are partial. Proof: Has a mathematical foundation, and is a complete guarantee. Sometimes not scalable.

22. Testing vs. Proof To test this program, you have to test all possible interleavings. With n processes p 0, p 1, … p n-1, and m steps per process, the number of interleavings is (n.m)! (m!) n The state explosion problem p0p1p2p3 step1 step2 step3

23. Example: Mutual Exclusion Process 0Process 1 do true  Entry protocol Critical section Exit protocol od Safety properties (1) There is no deadlock (2) At most one process is in its critical section. Liveness property A process trying to enter the CS must eventually succeed. (This is also called the progress property ) CS

24. Exercise program mutex 1 {two process mutual exclusion algorithm: shared memory model} define busy : shared boolean (initially busy = false} { process 0 }{ process 1 } do true  do busy  skip od ; busy:= true; critical section;critical section busy := false;busy := false{remaining codes}od Does this mutual exclusion protocol satisfy liveness and safety properties?

25. Safety invariants Invariant means: something meaningful should always hold Example: Total no. of processes in CS ≤ 1 (mutual exclusion problem) Another safety property is Partial correctness. It implies that “If the program terminates then the postcondition will hold.” Consider the following : Safety invariant:  (G0  G1  G2  …  Gk)  postcondition It does not say if the program will terminate. (termination is a liveness property) Total correctness = partial correctness + termination. do G0  S1 [] G1  S1 [] … [] Gk  Sk od

26. Exercise Starting from the given initial state, devise an algorithm to color the nodes of the graph using the colors 0 and 1, so that no two adjacent nodes have the same color. program colorme {for process P i } define color c  {0, 1} Initially colors are arbitrary do  j : j  neighbor(i) :: (c[i] = c[j])  c[i] := 1 - c[i] od Is the program partially correct ? YES (why?) Does it terminate ? NO (why?) p1 p3p0 p2 00 1 1

27. Liveness properties Eventuality is tricky. There is no need to guarantee when the desired thing will happen, as long as it happens.. Some examples  The message will eventually reach the receiver.  The process will eventually enter its critical section.  The faulty process will be eventually be diagnosed  Fairness (if an action will eventually be scheduled)  The program will eventually terminate.  The criminal will eventually be caught. Absence of liveness cannot be determined from finite prefix of the computation

28. Proving safety define c1, c2 : channel; { init c1 =  c2 =  } r, t : integer; { init r = 5, t = 5} {program for T } 1 do t > 0  send msg along c1; t := t -1 2[] ¬empty (c2)  rcv msg from c2; t := t + 1 od {program for R } 3 do ¬empty (c1)  rcv msg from c1; r := r+1 4[] r > 0  send msg along c2; r := r-1 od We want to prove the safety property P: P  n1 + n2 ≤ 10 transmitter receiver n1= # of messages in c1 n2= # of messages in c2

29. Proving safety n1, n2 = # of messages in c1 and c2 respectively. We will establish the following invariant: I  (t ≥ 0)  (r ≥ 0)  (n1 + t + n2 + r = 10) ( I ⇒ P ). Check if I holds after every action. {program for T } 1 do t > 0  send msg along c1; t := t -1 2[] ¬empty (c2)  rcv msg from c2; t := t+1 od {program for R } 3 do ¬empty (c1)  rcv msg from c1; r := r+1 4[] r > 0  send msg along c2; r := r-1 od Use the method of induction Show that I initially holds, and holds after each action.

30. Proving liveness S1  S2  S3  S4   f  f w1w2w3w4 o w1, w2, w3, w4  WF o WF is a well-founded set whose elements can be ordered by » If there is no infinite chain like w1 » w2 » w3 » w4.., i.e. f(s i ) » f(s i+1 ) » f(s i+2 ).. Global state then the computation will definitely terminate! f is called a variant functionExample?

31. Proof of liveness: an example 0 n-1 3 2 1 Clock phase synchronization System of n clocks ticking at the same rate. Each clock is 3-valued, i,e it ticks as 0, 1, 2, 0, 1, 2… A failure may arbitrarily alter the clock phases. The clocks need to return to the same phase..

32. Proof of liveness: an example Clock phase synchronization {Program for each clock} (c[k] = phase of clock k, initially arbitrary) do  j: j  N(i) :: c[j] = c[i] +1 mod 3  c[i] := c[i] + 2 mod 3 []  j: j  N(i) :: c[j] ≠ c[i] +1 mod 3  c[i] := c[i] + 1 mod 3 od Show that eventually all clocks will return to the same phase (convergence), and continue to be in the same phase (closure) 0 n-1 3 2 1  k: c[k]  {0.1.2}

33. Proof of convergence Let D = d[0] + d[1] + d[2] + … + d[n-1] d[i] = 0 if no arrow points towards clock i ; = i + 1 if a  pointing towards clock i ;  n - i if a  pointing towards clock i ; = 1 if both  and  point towards clock i. By definition, D ≥ 0. Also, D decreases after every step in the system. So the number of arrows must reduce to 0. 02022 11101 2222 2 Understand the game of arrows 012n-1