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PROMELA: What are interleavings?


PROMELA: What are interleavings?

By : Yoghurtnaut
Date : October 24 2020, 08:10 PM
Hope this helps In Promela, a process with an executable instruction can be scheduled for execution at any given point in time, provided that no other process is currently executing a uninterruptible atomic sequence.
A single instruction, on its own, it is executed atomically. In order to have multiple instructions in the same atomic sequence, one can use either atomic { } or d_step { }. I refer you to this other question on the topic for learning the difference among the two.
code :
x++           // Q(), x := 4
z = y         // P(), z := 5
y = y - x     // R(), y := 1
y = y - x     // R(), y := 2
x++           // Q(), x := 4
z = x         // P(), z := 4


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Spin and Promela: never and cycle

Spin and Promela: never and cycle


By : Lukas Danko
Date : March 29 2020, 07:55 AM
wish help you to fix your issue A never claim reports an error in two case: 1) an 'accept' cycle is detected, or 2) the never claim completes. A third possibility is if the never claim cannot take a step; this third possibility is the one your code is producing
When your claim was:
code :
never { 
    if 
    :: (x<0) || (10<x) -> goto accept
    fi;
    accept:
}
PROMELA: Would this be an example of a deadlock?

PROMELA: Would this be an example of a deadlock?


By : Gerald Wright
Date : March 29 2020, 07:55 AM
hope this fix your issue IMHO, no.
Follows the list of necessary conditions for a deadlock as illustrated by Wikipedia:
code :
bit x, y;
byte cnt;


active proctype A() {
again:
  x = 1;
  y == 0; /* waits for process B to end: if y != 0, the execution of this
             statement is blocked here */
  cnt++;
  /* critical section */
  printf("Process A entered critical section.\n");
  assert(cnt == 1);
  cnt--;

  printf("Process A exited critical section.\n");
  x = 0;
  goto again
}


active proctype B() {
again:
  y = 1;
  x == 0;

  cnt++;
  /* critical section */
  printf("Process B entered critical section.\n");
  assert(cnt == 1);
  cnt--;

  printf("Process B exited critical section.\n");
  y = 0;
  goto again
}
~$ spin -search -bfs mutex_simple_flaw2.pml

pan:1: invalid end state (at depth 2)
pan: wrote mutex_simple_flaw2.pml.trail

(Spin Version 6.4.8 -- 2 March 2018)
Warning: Search not completed
    + Breadth-First Search
    + Partial Order Reduction

Full statespace search for:
    never claim             - (none specified)
    assertion violations    +
    cycle checks            - (disabled by -DSAFETY)
    invalid end states      +

State-vector 20 byte, depth reached 2, errors: 1
        8 states, stored
           8 nominal states (stored-atomic)
        1 states, matched
        9 transitions (= stored+matched)
        0 atomic steps
hash conflicts:         0 (resolved)

Stats on memory usage (in Megabytes):
    0.000   equivalent memory usage for states (stored*(State-vector + overhead))
    0.291   actual memory usage for states
  128.000   memory used for hash table (-w24)
  128.195   total actual memory usage



pan: elapsed time 0 seconds
~$ spin -t -p -g -l mutex_simple_flaw2.pml

using statement merging
  1:    proc  1 (B:1) mutex_simple_flaw2.pml:24 (state 1)   [y = 1]
        y = 1
  2:    proc  0 (A:1) mutex_simple_flaw2.pml:7 (state 1)    [x = 1]
        x = 1
  3:    proc  0 (A:1) mutex_simple_flaw2.pml:8 (state 2)    [((y==0))]
    transition failed
spin: trail ends after 3 steps
#processes: 2
        x = 1
        y = 1
        cnt = 0
  3:    proc  1 (B:1) mutex_simple_flaw2.pml:25 (state 2)
  3:    proc  0 (A:1) mutex_simple_flaw2.pml:8 (state 2)
2 processes created
Find all interleavings of given strings that can be formed from all the characters of first and second string where orde

Find all interleavings of given strings that can be formed from all the characters of first and second string where orde


By : user3027510
Date : March 29 2020, 07:55 AM
it should still fix some issue Here's a version with plenty of cout and a way to track recursion depth.
code :
#include <iostream>
#include <string>

using namespace std;

void func0(string str1, string str2, string temp, int depth) {

    if (!str1.length() && !str2.length()) {
        cout << temp << endl;
        depth--;
        return;
    }

    if (str1.length()) {
        cout << "depth: " << depth++ << "|temp: " << temp << '\n';
        func0(str1.substr(1), str2, temp + str1[0], depth);
    }

    if (str2.length()) {
        cout << "depth: " << depth++ << "|temp: " << temp << '\n';
        func0(str1, str2.substr(1), temp + str2[0], depth);
    }

    depth--;
}

void func1(string str1, string str2, string temp, int depth) {
    if (!str1.length() && !str2.length()) {
        cout << temp << endl;
        depth--;
        return;
    }

    if (str1.length()) {
        temp += str1[0]; //temp has
        cout << "depth: " << depth++ << "|temp: " << temp << '\n';
        func1(str1.substr(1), str2, temp, depth);
    }

    if (str2.length()) {
        temp += str2[0];
        cout << "depth: " << depth++ << "|temp: " << temp << '\n';
        func1(str1, str2.substr(1), temp, depth);
    }

    depth--;
}

int main(int argc, char* argv[]) {
    string a = "asd";
    string b = "qw";
    string c = "";

    cout << "func0\n";
    func0(a, b, c, 0);
    cout << "func1\n";
    func1(a, b, c, 0);

    return 0;
}
func0
depth: 0|temp:
depth: 1|temp: a
depth: 2|temp: as
depth: 3|temp: asd
depth: 4|temp: asdq
asdqw
depth: 3|temp: as
depth: 4|temp: asq
depth: 5|temp: asqd
asqdw
depth: 5|temp: asq
depth: 6|temp: asqw
asqwd
depth: 2|temp: a
depth: 3|temp: aq
depth: 4|temp: aqs
depth: 5|temp: aqsd
aqsdw
depth: 5|temp: aqs
depth: 6|temp: aqsw
aqswd
depth: 4|temp: aq
depth: 5|temp: aqw
depth: 6|temp: aqws
aqwsd
depth: 1|temp:
depth: 2|temp: q
depth: 3|temp: qa
depth: 4|temp: qas
depth: 5|temp: qasd
qasdw
depth: 5|temp: qas
depth: 6|temp: qasw
qaswd
depth: 4|temp: qa
depth: 5|temp: qaw
depth: 6|temp: qaws
qawsd
depth: 3|temp: q
depth: 4|temp: qw
depth: 5|temp: qwa
depth: 6|temp: qwas
qwasd
func1
depth: 0|temp: a
depth: 1|temp: as
depth: 2|temp: asd
depth: 3|temp: asdq
depth: 4|temp: asdqw
asdqw
depth: 3|temp: asdq
depth: 4|temp: asdqd
depth: 5|temp: asdqdw
asdqdw
depth: 5|temp: asdqdw
depth: 6|temp: asdqdwd
asdqdwd
depth: 2|temp: asq
depth: 3|temp: asqs
depth: 4|temp: asqsd
depth: 5|temp: asqsdw
asqsdw
depth: 5|temp: asqsdw
depth: 6|temp: asqsdwd
asqsdwd
depth: 4|temp: asqsw
depth: 5|temp: asqsws
depth: 6|temp: asqswsd
asqswsd
depth: 1|temp: aq
depth: 2|temp: aqa
depth: 3|temp: aqas
depth: 4|temp: aqasd
depth: 5|temp: aqasdw
aqasdw
depth: 5|temp: aqasdw
depth: 6|temp: aqasdwd
aqasdwd
depth: 4|temp: aqasw
depth: 5|temp: aqasws
depth: 6|temp: aqaswsd
aqaswsd
depth: 3|temp: aqaw
depth: 4|temp: aqawa
depth: 5|temp: aqawas
depth: 6|temp: aqawasd
aqawasd
How to calculate all interleavings of two lists?

How to calculate all interleavings of two lists?


By : Hutty
Date : March 29 2020, 07:55 AM
I wish this help you Itertools would not be capable enough to handle this problem and would require some bit of understanding of the pegs and holes problem
Consider your example list
Spin promela GPU

Spin promela GPU


By : ahmad salimi
Date : March 29 2020, 07:55 AM
I wish this helpful for you GPU support is not included in Spin but is an active area of research. Most SPIN problems that are slow enough to seek a speed up are also large enough to exceed the local memory on a GPU. As a result the CPU memory needs to be used to store the explored state space and then memory bandwidth, CPU <==> GPU, swamp any computational speed increases. If however, your state space is small then the GPU may be amenable to use; yet, Spin does not include such support.
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