Efficient Critical Section Problem Solutions

The Critical-Section(臨界區)Problem • n processes all competing to use some shared data. • Each process has a code segment, called critical section, in which the shared data is accessed. • Problem – ensure that when one process is executing in its critical section, no other process is allowed to execute in its critical section.

進入臨界區(Critical-Section)間的條件 1. 互斥(Mutual Exclusion) 當某個處理元在臨界區間內存取某些資源，則其他處理元不能進入臨界區間去存取相同資源。 2. 行進(Progress) 當臨界區間內沒有處理元在執行，若有多個處理元要求進入臨界區間執行，則只有那些不在執行剩餘程式碼（Remainder Code）的處理元，才有資格被挑選為下一個進入臨界區間的處理元，而且這個挑選工作不能無限期的延遲下去。 3. 有限等待(Bounded Waiting) 當某個處理元要求進入臨界區間，一直到它獲得進入臨界區間這段時間，允許其他處理元進入臨界區間的次數有限制。

Initial Attempts to Solve Problem –2 processes • Only 2 processes, P0 and P1 • General structure of process Pi (other process Pj) do { entry section //critical section exit section //remainder section } while (1); • Processes may share some common variables to synchronize their actions.

Algorithm 1 - (2 processes) Ref. only • Shared variables: • int turn; initially turn = 0 • turn = i  Pi can enter its critical section • Process Pi do { while (turn != i) ; //critical section turn = j; //reminder section } while (1); • Wrong way! : Satisfies mutual exclusion, but not progress Ref. only

Algorithm 2 - (2 processes) • Shared variables • boolean flag[2]; initially flag [0] = flag [1] = false. • flag [i] = true Pi ready to enter its critical section • Process Pi do { flag[i] = true; while (flag[j]) ; //critical section flag [i] = false; //remainder section } while (1); • Satisfies mutual exclusion, but not progress requirement.

Algorithm 2 - (2 processes) Ref. only • Consider the following trace: • P0 sets flag[0] to true • A context-switch occurs • P1 sets flag[1] to true • P1 loops in while • A context-switch occurs • P0 loops in while • Both P0 and P1 loop forever. This is the livelock

Algorithm 3 - Peterson's Algorithm • Combined shared variables of algorithms 1 and 2. • Process Pi do { flag [i] = true; turn = j; while (flag [j] && turn == j) ; //critical section flag [i] = false; //remainder section } while (1); • Meets all three requirements; solves the critical-section problem for two processes. Ref. only

Bakery Algorithm • A Multiple-Process Solution. • Pertain to a centralized environment. • Notation <≡ lexicographical order (ticket #, process id #) • (a,b) < (c,d) if a < c or if a = c and b < d • max (a0,…, an-1) is a number, k, such that k ≡ ai for i = 0, …, n – 1 • Shared data boolean choosing[n]; int number[n]; Data structures are initialized to false and 0 respectively Ref. only

Bakery Algorithm • 保證達到互斥、行進、及有限等待條件。 • 就如同到銀行或商店接受服務一樣，任何一個顧客欲接受服務，均需依序取一張號碼牌，並依號碼牌由小至大依序接受服務；但是萬一很多人均有相同號碼牌時，則依客戶姓名排序依序接受服務。 Ref. only

Bakery Algorithm do { choosing[i] = true; number[i] = max(number[0], number[1], …, number [n – 1])+1; choosing[i] = false; for (j = 0; j < n; j++) { while (choosing[j]) ; while ((number[j] != 0) && ( number[j], j ) < ( number[i], i ) ) ; } //errata //critical section number[i] = 0; //remainder section } while (1); • If Pi is in its critical section and Pk (k != i) has already chosen its number k != 0, then (number[i], i) < number[k], k) Ref. only

Bakery演算法 • 互斥 • 任何要求進入臨界區間的處理元，均會使用while迴圈查核(number[j] != 0) && ( number[j], j ) < ( number[i], i )。 • 若最小號碼牌（number）的處理元僅有一個，則輪由這個處理元進入臨界區間，其他處理元只好繼續在迴圈內等待； • 若有多個處理元擁有最小號碼牌（均同號碼），則因為每個處理元編號不同，故僅會有一個處理元進入臨界區間，其他處理元繼續在迴圈中等待。因此這些處理元進入臨界區間是互斥的。 Ref. only

Bakery演算法 • 行進 • 由互斥的證明中，我們可以知道只有一個處理元進入臨界區間，其他在迴圈等待之處理元因為各擁有自己的號碼牌，且之後欲進入臨界區間之處理元，其號碼牌均較大，所以處理元進入臨界區間的次序是完全順序（Totally Order），因此漸漸會有機會進入臨界區間。 Ref. only

Bakery演算法 • 有限等待 • 當某個處理元要求進入臨界區間時，此處理元取得號碼牌具有完全順序之後，它在迴圈中等待時，若有別的處理元亦要進入臨界區間，則別的處理元取到的號碼牌一定比此處理元大，別的處理元不可能插隊在完全順序之內，因此這是一種先到先服務（First Come First Serve）的排程，也就是在此處理元之前進入臨界區間的其他處理元，總共進入臨界區間的次數是有限的。 Ref. only

Synchronization Hardware • Many systems provide hardware support for critical section code • Uniprocessors – could disable interrupts • Currently running code would execute without preemption • Generally too inefficient on multiprocessor systems • Operating systems using this not broadly scalable • Modern machines provide special atomic hardware instructions • Atomic = non-interruptable • Either test memory word and set value • Or swap contents of two memory words

TestAndSet Instruction • Definition: boolean TestAndSet (boolean *target) { boolean rv = *target; *target = TRUE; return rv: } Ref. only

Efficient Critical Section Problem Solutions