Parallel processing inside function in Python

Question

The Algorithm X in 30 lines https://www.cs.mcgill.ca/~aassaf9/python/algorithm_x.html

  if not X:
      yield list(solution)
  else:
      c = min(X, key=lambda c: len(X[c]))
      for r in list(X[c]):
          solution.append(r)
          cols = select(X, Y, r)
          for s in solve(X, Y, solution):
              yield s
          deselect(X, Y, r, cols)
          solution.pop()

def select(X, Y, r):
  cols = []
  for j in Y[r]:
      for i in X[j]:
          for k in Y[i]:
              if k != j:
                  X[k].remove(i)
      cols.append(X.pop(j))
  return cols

def deselect(X, Y, r, cols):
  for j in reversed(Y[r]):
      X[j] = cols.pop()
      for i in X[j]:
          for k in Y[i]:
              if k != j:
                  X[k].add(i)

In this code I want to parallelize the code fragment:

for r in list(X[c]):
            solution.append(r)
            cols = select(X, Y, r)
            for s in solve(X, Y, solution):
                yield s
            deselect(X, Y, r, cols)

I made a function:

def ParallelCode(X,Y,r,solution)
            print("Hello")
            solution.append(r)
            cols = select(X, Y, r)
            for s in solve(X, Y, solution):
                yield s
            deselect(X, Y, r, cols)
            solution.pop()
            solution.pop()

Then call it like this

if not X:
      yield list(solution)
  else:
      c = min(X, key=lambda c: len(X[c]))
      processes = [multiprocessing.Process(target=ParallelCode, args=(X, Y,r,solution)) for r in list(X[c])]

       for p in processes: 
            p.start()

       for p in processes:
           p.join()

When I ran the code, its not working. I print "Hello" inside ParallelCode, even "Hello" is not printed. Could you please help me what is wrong in this code.

I suppose there are two issues. Print may fail depending on where the stdout gets directed (try writing to a file and flushing). You are not actually running the process. Your function returns a generator and does nothing. — Adeom
– Adeom, Commented Dec 2, 2019 at 7:46

Adeom · Accepted Answer · 2019-12-02 07:52:20Z

1

Your call to ParallelCode doesn't run the function. It rather returns a generator object. Either remove the yield, or wrap the function so it calls the generator and then returns.

If i understand correctly, solution parameter in your case acts as an accumulator. You can't return a value from multiprocessing. Instead use Queue and pass it as parameter, and let the new process write the result inside of it.

answered Dec 2, 2019 at 7:52

Adeom

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1 Comment

NARAYAN CHANGDER Over a year ago

Thank you. I will let you know about the results. Let me try again using Queue.

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