1. 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 Phone: (813) Fax: (813)
Chapter 2 — Instructions: Language of the Computer

2. HW 4

3. Problem 2

4. Problem 3

5. Problem 4

6. HW 3

9. HW 2

10. 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.

11. 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)

12. 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

13. 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

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

15. :: 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

16. HW 1

17. 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?

18. 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);

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);

20. 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” */

21. 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

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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

28. 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

29. 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))]