The University of Adelaide, School of Computer Science Computer Architecture A Quantitative Approach, Fifth Edition The University of Adelaide, School of Computer Science 22 February 2019 CDA 5416 Computer System Verification HW Review Instructor: Hao Zheng Department of Computer Science & Engineering University of South Florida Tampa, FL 33620 Email: haozheng@usf.edu Phone: (813)974-4757 Fax: (813)974-5456 Chapter 2 — Instructions: Language of the Computer

HW 4

Problem 2

Problem 3

Problem 4

HW 3

HW 2

Draw program graphs for two processes with id = {0,1}. Show an execution of the composed program graph. Build a Promela model for this algorithm with three processes.  Think about the type of channels that should be used. Format the relevant correctness requirement(s) using a separate process(es) and/or assertions in Promela.  Use the SPIN to check that the leader election model satisfies the correctness requirements.

C0 (c0) P1 P0 C1 (c1) 1 1 stop 2 /send(id0) /recv(c0) /send(c0) /send(id0) /recv(c0) /send(c0) 1 1 /recv(m0) m0<id0 stop 2 m0=id0 m0>id0/send(m0)

C0 P1 P0 C1 1 1 1 1 1 1 1 1 /<send(id0), recv(c1)> /<send(id0), recv(c1)> /<send(id1), recv(c0)> 1 1 1 1 /<send(id1), recv(c0)> /<send(id0), recv(c1)> 1 1 1 1

C0 P1 P0 C1 1 1 1 1 2 1 1 1 1 2 /<send(id1), recv(c0)> /<send(c0), recv(m0)> /<send(c1), recv(m1)> 2 1 1 1 1 2

C0 P1 P0 C1 chan C1 = [1] of {byte}; chan C2 = [1] of {byte};

:: m0<id0 -> goto stop :: m0>id0 -> C1 ! m0 fi od stop: } proctype P0() { byte m0; C1 ! id0; do :: C0 ? m0; if :: m0<id0 -> goto stop :: m0>id0 -> C1 ! m0 fi od stop: } /C1 ! id0 1 /C0 ? m0 m0<id0 stop 2 m0=id0 m0>id0/ C1 ! m0

HW 1

byte n = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; } What are the largest and smallest values that n can be produced by the model upon termination?

byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } /* passed  active proctype checker1() {         if          :: stop==2 -> assert (n <= 20);         fi } */ /* failed */ active proctype checker2()         :: stop==2 -> assert (n <= 19);

byte n = 0; byte stop = 0; active [2] proctype P() {     byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } /* passed active proctype checker4() {         if          :: stop==2 -> assert (n >= 2);         fi } */ /* failed */         :: stop==2 -> assert (n >= 3);

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; }   1: proc  1 (p:1) a1_3.pml:12 (state 3) [else]   2: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]   3: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)] n = 0; reg0=0, cnt0=0, reg1=1, cnt1=0   4: proc  0 (p:1) a1_3.pml:12 (state 3) [else]   5: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]   6: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]   7: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 1; reg0=0, cnt0=0, reg1=1, cnt1=1   8: proc  1 (p:1) a1_3.pml:12 (state 3) [else]   9: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  10: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)] 11: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  12: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] n = 2; reg0=0, cnt0=0, reg1=2, cnt1=2 spin -p -t model.pml /* replay error trace “model.pml.trail” */

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 2; reg0=0, cnt0=0, reg1=2, cnt1=2  13: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  14: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  15: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  16: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  17: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 3; reg0=0, cnt0=0, reg1=3, cnt1=3 18: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  19: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  20: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  21: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  22: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 4; reg0=0, cnt0=0, reg1=4, cnt1=4

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; }   n = 4; reg0=0, cnt0=0, reg1=4, cnt1=4  23: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  24: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  25: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  26: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  27: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 5; reg0=0, cnt0=0, reg1=5, cnt1=5  28: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  29: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  30: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  31: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  32: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 6; reg0=0, cnt0=0, reg1=6, cnt1=6

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; }   n = 8; reg0=0, cnt0=0, reg1=8, cnt1=8 43: proc  1 (p:1) a1_3.pml:12 (state 3) [else]  44: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  45: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  46: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  47: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]  48: proc  1 (p:1) a1_3.pml:12 (state 3) [else]   n = 9; reg0=0, cnt0=0, reg1=9, cnt1=9  49: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  50: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]   n = 1; reg0=1, cnt0=0, reg1=9, cnt1=9  51: proc  1 (p:1) a1_3.pml:13 (state 4) [reg = n]  52: proc  1 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]   n = 1; reg0=1, cnt0=0, reg1=2, cnt1=9

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; }   n = 1; reg0=1, cnt0=0, reg1=2, cnt1=9 53: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 1; reg0=1, cnt0=1, reg1=2, cnt1=9  54: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  55: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  56: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  57: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  58: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 2; reg0=2, cnt2=2, reg1=2, cnt1=9  59: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  60: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  61: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  62: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  63: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 3; reg0=3, cnt0=3, reg1=2, cnt1=9

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; }   n = 3; reg0=3, cnt0=3, reg1=2, cnt1=9 64: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  65: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  66: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  67: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  68: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 4; reg0=4, cnt0=4, reg1=2, cnt1=9  69: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  70: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  71: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  72: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  73: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 5; reg0=5, cnt0=5, reg1=2, cnt1=9

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; }   n = 5; reg0=5, cnt0=5, reg1=2, cnt1=9  74: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  75: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  76: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  77: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  78: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 6; reg0=6, cnt0=6, reg1=2, cnt1=9  79: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  80: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  81: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  82: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  83: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 7; reg0=7, cnt0=7, reg1=2, cnt1=9

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; }   n = 7; reg0=7, cnt0=7, reg1=2, cnt1=9  84: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  85: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  86: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  87: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  88: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]    n = 8; reg0=8, cnt0=8, reg1=2, cnt1=9 89: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  90: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  91: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  92: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]  93: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)]   n = 9; reg0=9, cnt0=9, reg1=2, cnt1=9

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; }   n = 9; reg0=9, cnt0=9, reg1=2, cnt1=9  94: proc  0 (p:1) a1_3.pml:12 (state 3) [else]  95: proc  0 (p:1) a1_3.pml:13 (state 4) [reg = n]  96: proc  0 (p:1) a1_3.pml:14 (state 5) [reg = (reg+1)]  97: proc  0 (p:1) a1_3.pml:15 (state 6) [n = reg]    n = 10; reg0=10, cnt0=9, reg1=2, cnt1=9 98: proc  1 (p:1) a1_3.pml:15 (state 6) [n = reg]  99: proc  1 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] 100: proc  1 (p:1) a1_3.pml:11 (state 1) [((cnt==10))] 101: proc  1 (p:1) a1_3.pml:18 (state 11) [stop = (stop+1)] n = 2; reg0=9, cnt0=9, reg1=2, cnt1=10; stop=1

byte n = 0; byte stop = 0; active [2] proctype P() { byte reg;     byte cnt = 0;     do      :: cnt == 10 -> break     :: else ->             reg = n;             reg++;             n = reg;             cnt++     od; stop++; } n = 2; reg0=9, cnt0=9, reg1=2, cnt1=10; stop=1 102: proc  0 (p:1) a1_3.pml:16 (state 7) [cnt = (cnt+1)] 103: proc  0 (p:1) a1_3.pml:11 (state 1) [((cnt==10))] 104: proc  0 (p:1) a1_3.pml:18 (state 11) [stop = (stop+1)] n = 2; reg0=9, cnt0=10, reg1=2, cnt1=10; stop=2 105: proc  2 (checker4:1) a1_3.pml:71 (state 1) [((stop==2))] 106: proc  2 (checker4:1) a1_3.pml:71 (state 2) [assert((n>=3))]