Printing all the cycles in an adjacency matrix using python

In the below code, while trying to print all cycles, there comes the case where multiple cycles exist for one node. Even though I am increasing the value of k in the cycle function, it doesn't print all the possible cycles for a given node i.e. it doesn't consider the two possible cycles for node 3 or index 2 to index 1 as well as index 3 of the node 1 or index 0. Here we discuss the following graph, represented as mat in the code. Example Image

output the code prints:

['1 -> 2', '2 -> 3', '3 -> 1']

['1 -> 4', '4 -> 5', '5 -> 1']

['2 -> 3', '3 -> 1', '1 -> 2']

['3 -> 1', '1 -> 2', '2 -> 3']

Where the code fails is on the path [ [3 -> 1 ] , [1 -> 4] , [4-> 5] , [5-> 1] , [1 -> 2] , [2-> 3]]. With this code, I aim to print all cycles for a given adjacency matrix, even considering the case where no cycles exist.

I expect the output for the given an example as:

['1 -> 2', '2 -> 3', '3 -> 1']

['1 -> 4', '4 -> 5', '5 -> 1']

['2 -> 3', '3 -> 1', '1 -> 2']

['3 -> 1', '1 -> 2', '2 -> 3']

['2 -> 3', '3 -> 1', '1 -> 4', '4 -> 5', '5 -> 1', '1 -> 2']

['3 -> 1', '1 -> 4', '4 -> 5', '5 -> 1', '1 -> 2', '2 -> 3']

['4 -> 5', '5 -> 1', '1 -> 2', '2 -> 3', '3 -> 1', '1 -> 4']

['5 -> 1', '1 -> 2', '2 -> 3', '3 -> 1', '1 -> 4', '4 -> 5']

Code Sample:

mat = [[0,1,0,1,0] , [0,0,1,0,0] , [1,0,0,0,0] , [0,0,0,0,1] , [1,0,0,0,0]]

def cycle(mat,i,j,find):  
    global temp,t
    k = 0
    while k != len(mat): 
        if k != j and mat[j][k] and (temp == [] or temp[-1][-1] == str(j+1)):
            temp.append(f'{j+1} -> {k+1}')
            if k == find:
                print(temp)
                return 1
            else:
                t.append([i,j,k,find])
                if t.count([i,j,k,find]) > 1:
                    return -1
                cycle(mat,j,k,find)      
        k = k+1
        
                
for i in range(len(mat)):
    for j in range(len(mat)):
        if mat[i][j] and i != j:
            temp = []
            t = []
            temp.append(f'{i+1} -> {j+1}')
            cycle(mat,i,j,i)

Logic:

We are trying to go from : i -> j to : j -> k (where i , j , k are nodes) using cycle function. I've used

(temp == [] or temp[-1][-1] == str(j+1))

to make sure that after i -> j , only j -> k gets appended

                t.append([i,j,k,find])
                if t.count([i,j,k,find]) > 1:
                    return -1

conditions to prevent it from going into infinite recursion. For Detailed Logic click.

Thank you for your time and for helping me with my first StackOverflow question.

You say you expect [ [3 -> 1 ] , [1 -> 4] , [4-> 5] , [5-> 1] , [1 -> 2] , [2-> 3]] but this is just a combination of two cycles which ARE printed out. [1 -> 4] , [4-> 5] , [5-> 1] and [3->1] , [1 -> 2] , [2-> 3].

It seems unreasonable to me to expect two cycles that just happen to share a node to be output as a combination. If you really do want to see this, then take the output of uncombined cycles, check them for shared nodes and print out the combined cycles as they are found.

FIND simple cycles ( using your code )
LOOP over pairs of simple cycles
   IF pair share nodes
       PRINT combined cycle