How can I count the number of cases in recursive functions?

Question

def calcPath(trace_map, x, y):
    n = len(trace_map)
    count = 0
    if x > n - 1 or y > n - 1:
        pass
    elif x < n and y < n:
        if x + trace_map[x][y] == (n - 1) and y == (n - 1):
            count += 1
        elif x == (n - 1) and y + trace_map[x][y] == (n - 1):
            count += 1
        else:
            calcPath(trace_map, x + trace_map[x][y], y)
            calcPath(trace_map, x, y + trace_map[x][y])
    return count


if __name__ == "__main__":
    trace_map = [
        [1, 2, 9, 4, 9],
        [9, 9, 9, 9, 9],
        [9, 3, 9, 9, 2],
        [9, 9, 9, 9, 9],
        [9, 9, 9, 1, 0],
    ]
    print(calcPath(trace_map, 0, 0))

    trace_map = [[1, 1, 1], [1, 1, 2], [1, 2, 0]]
    print(calcPath(trace_map, 0, 0))

I want to count the existing routes of the given maze. (anyway, the problem itself is not that important) Problem is, I tried to count the number of cases that fit the conditions within the recursive functions.

These are two conditions that have to be counted.

if x + trace_map[x][y] == (n - 1) and y == (n - 1):
if x == (n - 1) and y + trace_map[x][y] == (n - 1):

I tried counting the conditions like this

count = 0 
if condition = True: 
count +=1

But since I'm using recursive functions, if I declare count = 0 in the function, the count value stays 0.

Shortly, I just want to keep the counter unaffected by the recursive function.

Why not declare it outside the function as a global variable? You could also create a variable calcPath.counter and initialize it outside the function. — Arnav Borborah
– Arnav Borborah, Commented Nov 12, 2020 at 12:27
Does this answer your question? Debugging recursive function to se how many times it repeats a certain calculation — MisterMiyagi
– MisterMiyagi, Commented Nov 12, 2020 at 12:28
@mkrieger1 oops, sorry. this is my first time using stack 😅 gotta change it. — user14625681
– user14625681, Commented Nov 12, 2020 at 12:32
you could write something like count += calcPath(...) instead of discarding the returned count — Ronald
– Ronald, Commented Nov 12, 2020 at 12:33

Arnav Borborah · Accepted Answer · 2020-11-12 16:21:20Z

6

One of the ways to solve this is by adding the count you get from each recursive function's return. When you call the recursive function, take the count that is returned and add it to the count variable in the current scope. For example:

def calcPath(trace_map, x, y):
    n = len(trace_map)
    count = 0
    if x > n - 1 or y > n - 1:
        pass
    elif x < n and y < n:
        if x + trace_map[x][y] == (n - 1) and y == (n - 1):
            count += 1
        elif x == (n - 1) and y + trace_map[x][y] == (n - 1):
            count += 1
        else:
            count += calcPath(trace_map, x + trace_map[x][y], y)
            count += calcPath(trace_map, x, y + trace_map[x][y])
    return count

An alternative solution would be to create a global variable and reset it to 0 every time the function is called (although I don't recommend this since it requires ceremony everytime the function is called).

That might look something like this:

count = 0 # Global variable

def calcPath(trace_map, x, y):
    global count
    n = len(trace_map)
    if x > n - 1 or y > n - 1:
        pass
    elif x < n and y < n:
        if x + trace_map[x][y] == (n - 1) and y == (n - 1):
            count += 1
        elif x == (n - 1) and y + trace_map[x][y] == (n - 1):
            count += 1
        else:
            calcPath(trace_map, x + trace_map[x][y], y)
            calcPath(trace_map, x, y + trace_map[x][y])


if __name__ == "__main__":
    trace_map = [
        [1, 2, 9, 4, 9],
        [9, 9, 9, 9, 9],
        [9, 3, 9, 9, 2],
        [9, 9, 9, 9, 9],
        [9, 9, 9, 1, 0],
    ]
    print(calcPath(trace_map, 0, 0))

    # Use count in some way

    count = 0 # Reset the count

    trace_map = [[1, 1, 1], [1, 1, 2], [1, 2, 0]]
    print(calcPath(trace_map, 0, 0))

edited Nov 12, 2020 at 16:21

answered Nov 12, 2020 at 12:34

Arnav Borborah

11.8k8 gold badges51 silver badges94 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

user14625681 Over a year ago

wow, I tried using global variation but why on earth did I not reset the count? haha. 🤣 it's so dumb of me. Thanks for help! stack is full of kindness. impressed.🥰

sanyassh Over a year ago

@user14625681 note that the first approach with count += calcPath is the best and is applicable in many cases when dealing with recursive functions. At least it doesn't require you to remember about resetting global variable.

Henry Tjhia · Accepted Answer · 2020-11-12 14:00:41Z

0

I think you can use the concept of nested scope or put it simply function inside another function. For example:

def fuc(args):
    if not args:
        return 0
    else:
        fuc.count += 1
        return args[0] + fuc(args[1:])

def fac(fuc, args):
    fuc.count = 0
    return fuc(args), fuc.count

print(fac(fuc, [1, 2, 3]))
print(fac(fuc, [4, 5, 6, 7]))

(6, 3)
(22, 4)
[Finished in 0.1s]

answered Nov 12, 2020 at 14:00

Henry Tjhia

7521 gold badge6 silver badges11 bronze badges

Collectives™ on Stack Overflow

How can I count the number of cases in recursive functions?

2 Answers 2

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related