At first glance, the script looks like some sort of state machine that takes in the input s and keeps changing it according to a set of rules defined by the regular expressions in parts. If a particular rule matches s, a counter n is incremented. The program stops when none of the rules in parts matches s, with the last line of the program printing the value of the counter n (the flag).
Simply running this program will take way too much time, so let's break out of the while loop after the counter crosses, say, 1000. Let's also inspect the state s each time a rule matches to better understand the rule actually does by adding a print statement right after a rule matches. Here's what the while loop looks like after these changes:
```while True:	if n > 1000: break	for a, b in parts:		s, c = re.subn(a, b, s)		if c:			print(f'Step: {n} State: {s}')			n += c			break	else:		break```
Here's a portion of the output starting at step 22:
-----Step: 22 State: 1-2-2-2-2-2-2-0B
Step: 23 State: 1-2-2-2-2-2-2C1-
Step: 24 State: 1-2-2-2-2-2A2-1-
Step: 25 State: 1-2-2-2-2C2-2-1-
Step: 26 State: 1-2-2-2A2-2-2-1-
Step: 27 State: 1-2-2C2-2-2-2-1-
Step: 28 State: 1-2A2-2-2-2-2-1-
Step: 29 State: 1C2-2-2-2-2-2-1-
Step: 30 State: 2-2C2-2-2-2-2-1-
Step: 31 State: 2A2-2-2-2-2-2-1-
Step: 32 State: 0C2-2-2-2-2-2-2-1-
Step: 33 State: 0A1-2-2-2-2-2-2-2-1-
Step: 34 State: 1-1B2-2-2-2-2-2-2-1-
Step: 35 State: 1-2-2B2-2-2-2-2-2-1-
Step: 36 State: 1-2B1-2-2-2-2-2-2-1-
Step: 37 State: 1B1-1-2-2-2-2-2-2-1-
Step: 38 State: 2-1B1-2-2-2-2-2-2-1-
Step: 39 State: 2-2-1B2-2-2-2-2-2-1-
Step: 40 State: 2-2-2-2B2-2-2-2-2-1-
Step: 41 State: 2-2-2B1-2-2-2-2-2-1-
Step: 42 State: 2-2B1-1-2-2-2-2-2-1-
Step: 43 State: 2B1-1-1-2-2-2-2-2-1-
Step: 44 State: 0B1-1-1-1-2-2-2-2-2-1-
Step: 45 State: 0C1-1-1-1-1-2-2-2-2-2-1-
Step: 46 State: 0A1-1-1-1-1-1-2-2-2-2-2-1-
Step: 47 State: 1-1B1-1-1-1-1-2-2-2-2-2-1-
Step: 48 State: 1-2-1B1-1-1-1-2-2-2-2-2-1-
Step: 49 State: 1-2-2-1B1-1-1-2-2-2-2-2-1-
Step: 50 State: 1-2-2-2-1B1-1-2-2-2-2-2-1-
Step: 51 State: 1-2-2-2-2-1B1-2-2-2-2-2-1-
Step: 52 State: 1-2-2-2-2-2-1B2-2-2-2-2-1-
Step: 53 State: 1-2-2-2-2-2-2-2B2-2-2-2-1-
Step: 54 State: 1-2-2-2-2-2-2B1-2-2-2-2-1-
Step: 55 State: 1-2-2-2-2-2B1-1-2-2-2-2-1-
Step: 56 State: 1-2-2-2-2B1-1-1-2-2-2-2-1-
Step: 57 State: 1-2-2-2B1-1-1-1-2-2-2-2-1-
Step: 58 State: 1-2-2B1-1-1-1-1-2-2-2-2-1-
Step: 59 State: 1-2B1-1-1-1-1-1-2-2-2-2-1-
Step: 60 State: 1B1-1-1-1-1-1-1-2-2-2-2-1-
Step: 61 State: 2-1B1-1-1-1-1-1-2-2-2-2-1-
Step: 62 State: 2-2-1B1-1-1-1-1-2-2-2-2-1-
Step: 63 State: 2-2-2-1B1-1-1-1-2-2-2-2-1-
Step: 64 State: 2-2-2-2-1B1-1-1-2-2-2-2-1-
Step: 65 State: 2-2-2-2-2-1B1-1-2-2-2-2-1-
Step: 66 State: 2-2-2-2-2-2-1B1-2-2-2-2-1-
Step: 67 State: 2-2-2-2-2-2-2-1B2-2-2-2-1-
Step: 68 State: 2-2-2-2-2-2-2-2-2B2-2-2-1-
Step: 69 State: 2-2-2-2-2-2-2-2B1-2-2-2-1-
Step: 70 State: 2-2-2-2-2-2-2B1-1-2-2-2-1-
Step: 71 State: 2-2-2-2-2-2B1-1-1-2-2-2-1-
Step: 72 State: 2-2-2-2-2B1-1-1-1-2-2-2-1-
Step: 73 State: 2-2-2-2B1-1-1-1-1-2-2-2-1-
Step: 74 State: 2-2-2B1-1-1-1-1-1-2-2-2-1-
Step: 75 State: 2-2B1-1-1-1-1-1-1-2-2-2-1-
Step: 76 State: 2B1-1-1-1-1-1-1-1-2-2-2-1-
Step: 77 State: 0B1-1-1-1-1-1-1-1-1-2-2-2-1-
Step: 78 State: 0C1-1-1-1-1-1-1-1-1-1-2-2-2-1-
Step: 79 State: 0A1-1-1-1-1-1-1-1-1-1-1-2-2-2-1-
-----
Let's start at step 22: It consists of a sequence of either 0, 1, or 2 joined by a hyphen, with the letter B at the end. If we consider each hyphen to be a cell, and the number adjacent to it to be the state of the cell, then this sequence of state can be thought of as a Turing machine, with the letter B currently positioned at the last cell.
If you scroll through steps 22 to 185, you'll realize that the rules basically move this letter through these cells in a certain pattern before bringing it back to the last cell again (in step 185). You'll also notice that at each time step, the letter is only moved by one cell to the left or right, making the program inefficient. If we can somehow optimize what the rules do, then we can actually bring the program to a halt and print the flag.
First, lets look at steps 23 to 29. In step 23, the B in the last cell is moved to the second-to-last cell and also changed into a C. Steps 24 to 29 do something similar - moving the letter back by one cell and alternating its value between C and A. This means that by the time the letter reaches the first cell, it is either a C or an A depending on the number of cells (C if the number of cells is even, and A if it is odd).
-----Step: 22 State: 1-2-2-2-2-2-2-0B
Step: 23 State: 1-2-2-2-2-2-2C1-
Step: 24 State: 1-2-2-2-2-2A2-1-
Step: 25 State: 1-2-2-2-2C2-2-1-
Step: 26 State: 1-2-2-2A2-2-2-1-
Step: 27 State: 1-2-2C2-2-2-2-1-
Step: 28 State: 1-2A2-2-2-2-2-1-
Step: 29 State: 1C2-2-2-2-2-2-1-
-----
Optimizing the above transition is simple: All we need is the number of cells and whether this number is even or odd, and we can do one single addition to output the state at which the letter reaches the first cell.
Before we continue, let's look a the big picture: We're trying to get the flag, for which we need the program to halt at some point. The only rules that will halt the program are parts[6] and parts[7]. The latter of these will happen only when the A is in the first cell and the number of the first cell is 1. Moreover, as we saw previously, the first letter can only be A if the number of cells is odd. In summary, we will need to go through all the states until we reach one whose length is odd and one whose first letter is 1.
Now, coming back to step 30: it moves C one cell forward, then step 31 moves it back one cell and changes it to an A. Next, steps 32 and 33 add two cells to the state. Optimizing this is also straightforward: Add two the number of cells.
-----Step: 29 State: 1C2-2-2-2-2-2-1-
Step: 30 State: 2-2C2-2-2-2-2-1-
Step: 31 State: 2A2-2-2-2-2-2-1-
Step: 32 State: 0C2-2-2-2-2-2-2-1-
Step: 33 State: 0A1-2-2-2-2-2-2-2-1-
-----
Next, let's look at what happens between steps 33 and 46: A is changed to a B and this B is bought two cells forward (step 35), and then two cells back (step 37). Then B is bought three cells forward (step 40), and then three cells back (step 43). So far, that's a total of (2 + 3) x 2 steps. Finally, at step 43, three cells are added to the state. In all, the number of steps is (2 + 3) x 2 + 3 = 13.
-----Step: 33 State: 0A1-2-2-2-2-2-2-2-1-
Step: 34 State: 1-1B2-2-2-2-2-2-2-1-
Step: 35 State: 1-2-2B2-2-2-2-2-2-1-
Step: 36 State: 1-2B1-2-2-2-2-2-2-1-
Step: 37 State: 1B1-1-2-2-2-2-2-2-1-
Step: 38 State: 2-1B1-2-2-2-2-2-2-1-
Step: 39 State: 2-2-1B2-2-2-2-2-2-1-
Step: 40 State: 2-2-2-2B2-2-2-2-2-1-
Step: 41 State: 2-2-2B1-2-2-2-2-2-1-
Step: 42 State: 2-2B1-1-2-2-2-2-2-1-
Step: 43 State: 2B1-1-1-2-2-2-2-2-1-
Step: 44 State: 0B1-1-1-1-2-2-2-2-2-1-
Step: 45 State: 0C1-1-1-1-1-2-2-2-2-2-1-
Step: 46 State: 0A1-1-1-1-1-1-2-2-2-2-2-1-
-----Now, look at steps 46 to 79: This cycle is similar to the one between steps 33 and 46. Last cycle, we went forwards 2 and 3 times, so this time we expect to go forwards 4 and 5 times. However, recall that we added 3 new cells, so this time we go forwards 7 and 8 times. In all, we take (7 + 8) x 2 + 3 = 33 steps.
-----Step: 46 State: 0A1-1-1-1-1-1-2-2-2-2-2-1-
Step: 47 State: 1-1B1-1-1-1-1-2-2-2-2-2-1-
Step: 48 State: 1-2-1B1-1-1-1-2-2-2-2-2-1-
Step: 49 State: 1-2-2-1B1-1-1-2-2-2-2-2-1-
Step: 50 State: 1-2-2-2-1B1-1-2-2-2-2-2-1-
Step: 51 State: 1-2-2-2-2-1B1-2-2-2-2-2-1-
Step: 52 State: 1-2-2-2-2-2-1B2-2-2-2-2-1-
Step: 53 State: 1-2-2-2-2-2-2-2B2-2-2-2-1-
Step: 54 State: 1-2-2-2-2-2-2B1-2-2-2-2-1-
Step: 55 State: 1-2-2-2-2-2B1-1-2-2-2-2-1-
Step: 56 State: 1-2-2-2-2B1-1-1-2-2-2-2-1-
Step: 57 State: 1-2-2-2B1-1-1-1-2-2-2-2-1-
Step: 58 State: 1-2-2B1-1-1-1-1-2-2-2-2-1-
Step: 59 State: 1-2B1-1-1-1-1-1-2-2-2-2-1-
Step: 60 State: 1B1-1-1-1-1-1-1-2-2-2-2-1-
Step: 61 State: 2-1B1-1-1-1-1-1-2-2-2-2-1-
Step: 62 State: 2-2-1B1-1-1-1-1-2-2-2-2-1-
Step: 63 State: 2-2-2-1B1-1-1-1-2-2-2-2-1-
Step: 64 State: 2-2-2-2-1B1-1-1-2-2-2-2-1-
Step: 65 State: 2-2-2-2-2-1B1-1-2-2-2-2-1-
Step: 66 State: 2-2-2-2-2-2-1B1-2-2-2-2-1-
Step: 67 State: 2-2-2-2-2-2-2-1B2-2-2-2-1-
Step: 68 State: 2-2-2-2-2-2-2-2-2B2-2-2-1-
Step: 69 State: 2-2-2-2-2-2-2-2B1-2-2-2-1-
Step: 70 State: 2-2-2-2-2-2-2B1-1-2-2-2-1-
Step: 71 State: 2-2-2-2-2-2B1-1-1-2-2-2-1-
Step: 72 State: 2-2-2-2-2B1-1-1-1-2-2-2-1-
Step: 73 State: 2-2-2-2B1-1-1-1-1-2-2-2-1-
Step: 74 State: 2-2-2B1-1-1-1-1-1-2-2-2-1-
Step: 75 State: 2-2B1-1-1-1-1-1-1-2-2-2-1-
Step: 76 State: 2B1-1-1-1-1-1-1-1-2-2-2-1-
Step: 77 State: 0B1-1-1-1-1-1-1-1-1-2-2-2-1-
Step: 78 State: 0C1-1-1-1-1-1-1-1-1-1-2-2-2-1-
Step: 79 State: 0A1-1-1-1-1-1-1-1-1-1-1-2-2-2-1-
-----How long does this cycle repeat? As it turns out, it repeats till B reaches the last cell. We can optimize this cycle as follows:
```# Loop till just before hitting the length of the statewhile counter < ln - 1:    # Increase the number of cells    ln += 3     # Take two steps forward, two steps back    step += counter * 2 + (counter + 1) * 2     # Take three steps forward (as the number of cells was increased by 3)    steps += 3     # The next step forwards must be 2 + number of new cells added    counter += 5```
Finally, between steps 166 and 185, B is finally bought back to the last cell. As it turns out, the number of loops to bring B to the last cells varies based on whether there are an even or odd number of cells.
-----Step: 166 State: 1B1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-
Step: 167 State: 2-1B1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-
Step: 168 State: 2-2-1B1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-
Step: 169 State: 2-2-2-1B1-1-1-1-1-1-1-1-1-1-1-1-1-1-1-
Step: 170 State: 2-2-2-2-1B1-1-1-1-1-1-1-1-1-1-1-1-1-1-
Step: 171 State: 2-2-2-2-2-1B1-1-1-1-1-1-1-1-1-1-1-1-1-
Step: 172 State: 2-2-2-2-2-2-1B1-1-1-1-1-1-1-1-1-1-1-1-
Step: 173 State: 2-2-2-2-2-2-2-1B1-1-1-1-1-1-1-1-1-1-1-
Step: 174 State: 2-2-2-2-2-2-2-2-1B1-1-1-1-1-1-1-1-1-1-
Step: 175 State: 2-2-2-2-2-2-2-2-2-1B1-1-1-1-1-1-1-1-1-
Step: 176 State: 2-2-2-2-2-2-2-2-2-2-1B1-1-1-1-1-1-1-1-
Step: 177 State: 2-2-2-2-2-2-2-2-2-2-2-1B1-1-1-1-1-1-1-
Step: 178 State: 2-2-2-2-2-2-2-2-2-2-2-2-1B1-1-1-1-1-1-
Step: 179 State: 2-2-2-2-2-2-2-2-2-2-2-2-2-1B1-1-1-1-1-
Step: 180 State: 2-2-2-2-2-2-2-2-2-2-2-2-2-2-1B1-1-1-1-
Step: 181 State: 2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-1B1-1-1-
Step: 182 State: 2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-1B1-1-
Step: 183 State: 2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-1B1-
Step: 184 State: 2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-1B
Step: 185 State: 2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-2-0B
-----```# Move two sub loops and take ln steps if odd or    # just take just ln step loop if even    if is_odd:             step += counter * 2 + ln    else:        step += ln```
To summarize, between steps 22 and 185, we have sent the letter B through a pattern of cells before bringing it back to the last cell. In the end, only two things have have changed: First, the number of cells have increased from 8 to 20, and second, the first number in the state has changed from 2 to 1. Recall that for our program to halt, we need to keep sending B around these cycles (using our optimized loops) till we reach a state where the number of cells is odd and the first number is 1.
In practice, there are complications. For example, the number of steps taken forward and back depend on the first number. The first number itself depends on whether the number of cells is odd or even. It is difficult to explain these patterns, so I do not elaborate further. To debug these issues, what I had to do was manually compare my simulation with the actual program, and then go back and fix my simulation. Here's my solve script:
```def iterate(state, step):    ln = state.count('-') # The length of the state (1 - number of cells)    first_element = state[0] # The first element    is_odd_number_of_steps = ln % 2 == 1 # Is the number of steps odd?
    # Move ln steps to the first cell    step += ln
    # Increase the state length, take forward steps, and prepare the    # loop counter an appropriate number of times depending on    # the state length and the first element    if first_element == '1':        if is_odd_number_of_steps:            ln += 2            step += 4            counter = 2        else:            # Halting case            print(f'SECFEST{{{step +2}}}')            exit()    else:        ln += 1        step += 1        counter = 1
        if not is_odd_number_of_steps:            ln += 1            step += 1            counter += 1
    # Loop till just before hitting the length of the state    while counter < ln - 1:        # Increase the number of cells        ln += 3        # Take two steps forward, two steps back        step += counter * 2 + (counter + 1) * 2        # Take three steps forward (as the number of cells was increased by 3)        step += 3        # The next step forwards must be 2 + number of new cells added        counter += 5
    # Increase the state length one last time    ln += 1
    # Move two sub loops and take ln steps if odd or    # just take just ln step loop if even    if is_odd_number_of_steps:             step += counter * 2 + ln    else:        step += ln
    # Flip the first element, if required    if (is_odd_number_of_steps and first_element == '1' or not        is_odd_number_of_steps and first_element != '1'):        next_state = '2-' if first_element == '1' else '1-'    else:        next_state = first_element + '-'
    next_state += '-'.join(['2' for _ in range(ln - 1)]) + '-0B'
    return next_state, step
state = '1-2-2-2-2-2-2-0B'step = 22
while True:    state, step = iterate(state, step)```
