How would I count the number of iterations of a Recursive Function in Python?

Question

def solver():
    empty_cell = empty_square()
    counter = 0

    if not empty_cell:
        return True

    i,j = empty_cell[0],empty_cell[1]
    for num in range(1,10):
        counter += 1
        if constraint_check(num,i,j):
            sudoku[i][j] = num

            if solver():
                return True

            else:
                sudoku[i][j] = 0

    return False

Given the code above, how would I implement a counter to count how many iterations the recursive part of the function makes? As can be seen in the code I have attempted something above but I was not able to retrieve this variable to print it out and record the number.

@CassandraTrotter if you've found a solution, why not share it as a self answer so that other people with the same problem can learn from your experience? The point of the site is to curate a repository of knowledge, not purge questions as soon as you get a solution. — ggorlen
– ggorlen, Commented Jul 5, 2021 at 1:15
I did not find a solution @ggorlen hence I deleted the contract. The answer provided is severely inadequete. Thus no good answers should enable me to delete the question — Cassandra Trotter
– Cassandra Trotter, Commented Jul 6, 2021 at 20:14
Hmm -- I might not have been clear. By asking a question, you create a community-owned resource for future visitors with the same problem. Not getting an answer you're satisfied with isn't grounds for deleting a question (unless no answers have upvotes, which is not applicable here). 3 community members felt the answer helped them or was an adequate solution, so the thread remains to help future users. See this meta thread header "When can't I delete my own post?" — ggorlen
– ggorlen, Commented Jul 6, 2021 at 20:28
BTW, Counting recursive calls of a function seems to be the canonical thread for this question. The top answer suggests initializing solver.counter = 0 outside of the function and changing your calls to solver.counter += 1. After all the calls resolve, solver.counter will hold your result in the enclosing scope. — ggorlen
– ggorlen, Commented Jul 6, 2021 at 21:41

dchoruzy · Accepted Answer · 2021-07-04 16:59:00Z

3

set counter as function parameter -> solver(counter=1) then when calling it again within your function add +1 -> solver(counter+1)

answered Jul 4, 2021 at 16:59

dchoruzy

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1 Comment

Cassandra Trotter Over a year ago

This answer doesn't work. I have found a seperate solution. Please delete this answer so I can delete the question

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