# Introduction

I glanced at a digital clock one day and noticed the time was reflectively symmetric, meaning if you flip the time over a horizontal line that divides the time in half it gives you the same time back. It made me wonder how many times there are that exhibit this behavior. I could have thought about all the times and probably come up with a list of them, but that is actually a lot of work. It would be much simpler to just have python give me the answer.

# Types of Symmetry

The two types of symmetry I was interested in were reflective and rotational symmetry. These are standard symmetry types, and easy to code, and about all I really cared about, so that’s why I chose them.

## Reflection Symmetry

Reflection symmetry occurs when you flip an image over an axis and get the same image back. So in this case if we define our axis to cut the time in half, horizontally, along the midpoint, and get the same thing back we have reflection symmetry.

## 2-fold Rotational Symmetry (Cyclic Group of Order 2)

2-fold Rotational Symmetry (180 degree rotational symmetry) occurs when you rotate an object 180 degrees and you get the same image back. This type of symmetry is also called Cyclic Group 2 symmetry. In general, Cyclic Groups of Order n are a mathematical construct that is studied in Abstract Algebra, but we need not go into that. All we’re going to concern ourselves with here is rotating an object 180 degrees, which is a simple thing to imagine.

# Strategy to Solve

Each number on a digital clock can be defined by a simple python list. Let’s look at a diagram of the number 8 to see the mapping I chose:

Using the diagram above we see that the following python list would represent a zero and eight, respectively:

zero = [1, 1, 0, 1, 1, 1, 1] eight = [1, 1, 1, 1, 1, 1, 1]

In our python code I define all the numbers in a dictionary as follows:

time_maps = {0: [1, 1, 0, 1, 1, 1, 1], 1: [0, 0, 0, 1, 0, 0, 1], 2: [1, 0, 1, 1, 1, 1, 0], 3: [1, 0, 1, 1, 0, 1, 1], 4: [0, 1, 1, 1, 0, 0, 1], 5: [1, 1, 1, 0, 0, 1, 1], 6: [1, 1, 1, 0, 1, 1, 1], 7: [1, 0, 0, 1, 0, 0, 1], 8: [1, 1, 1, 1, 1, 1, 1], 9: [1, 1, 1, 1, 0, 1, 1]}

Once the mapping is understood we can begin writing our program. Let’s start with reflection symmetry.

## Reflection Symmetry Code Walkthrough

For reflection symmetry it’s pretty simple. If a time is going to be reflective then every number in the time must be reflective. So we begin by writing a function that tells us whether an individual number is reflective.

def individual_reflection_check(number): """ check if a number is reflective """ tmp = list(time_maps[number]) tmp[0] = time_maps[number][5] tmp[1] = time_maps[number][4] tmp[2] = time_maps[number][2] tmp[3] = time_maps[number][6] tmp[4] = time_maps[number][1] tmp[5] = time_maps[number][0] tmp[6] = time_maps[number][3] if tmp == time_maps[number]: #print("%s is R." % number) return True else: #print("%s is not R." % number) return False

The mapping above is created by simply imagining where the lines on the digital clock shift to when you perform the reflection. It ends up being a simple shift of indices within the list that represents each number, which is what the previous function does. After the reflection is performed we check to see if some lists are the same, and if they are it means the time is reflective.

Let’s plug a couple numbers in to check our code:

print(individual_reflection_check(number=2)) print(individual_reflection_check(number=8)) >> False >> True

Next we take a look at whether or not a group of numbers together are reflective by simply making use of the function above multiple times.

def grouped_reflection_check(time): """ check if the time is reflective or not :param time: a string like '3:45' :return: None """ if re.search('[a-zA-Z]', time) is None: # loop through time elements for item in time: if item != ':': status = individual_reflection_check(int(item)) if status is True: continue else: #print("The time %s is NOT R" % time) return False #print("The time %s is R" % time) return True else: print("%s is not a valid time" % time)

Let’s check out if 1:11, 2:22, and 4:56 are reflective.

print(grouped_reflection_check(time='1:11')) print(grouped_reflection_check(time='2:22')) print(grouped_reflection_check(time='4:56')) >> True >> False >> False

The results were as expected. Now we want to know how many times are reflective among all the times. Here is the function I came up with that loops through all the times and gives us our answer:

def get_all_reflection_times(): """ the idea is all need to be reflective for it to work, let's get the list """ reflective_ls = [] counter = 0 for hr in range(1, 24): for mini in range(0, 60): if mini in range(0, 10): mini = '0' + str(mini) counter += 1 time = str(hr) + ":" + str(mini) if grouped_reflection_check(time) is True: reflective_ls.append(time) print("There are %s times we checked." % counter) print("There are %s total times that are reflective." % len(reflective_ls)) print("This gives %s percent of times that are reflective." % (100*float(len(reflective_ls))/float(counter))) print("Here are the reflective times.") for item in reflective_ls: print(' ', item) return reflective_ls

Running the function above gives the following output:

There are 1380 times we checked. There are 84 total times that are reflective. This gives 6.086956521739131 percent of times that are reflective. Here are the reflective times. 1:00 1:01 1:03 1:08 1:10 1:11 1:13 1:18 1:30 1:31 1:33 1:38 3:00 3:01 3:03 3:08 3:10 3:11 3:13 3:18 3:30 3:31 3:33 3:38 8:00 8:01 8:03 8:08 8:10 8:11 8:13 8:18 8:30 8:31 8:33 8:38 10:00 10:01 10:03 10:08 10:10 10:11 10:13 10:18 10:30 10:31 10:33 10:38 11:00 11:01 11:03 11:08 11:10 11:11 11:13 11:18 11:30 11:31 11:33 11:38 13:00 13:01 13:03 13:08 13:10 13:11 13:13 13:18 13:30 13:31 13:33 13:38 18:00 18:01 18:03 18:08 18:10 18:11 18:13 18:18 18:30 18:31 18:33 18:38

Good stuff! I thought there’d be more than 84 reflective times, but it’s a special property that not all numbers have, so it can’t be all that common. Next, let’s take a look at Rotational Symmetry.

## Rotational Symmetry Code Walkthrough

Just like reflection symmetry we can imagine where the number vertices shift to when the rotation is performed and write up the following function that does the transformation:

def rotational_transformation(number): """ perform the cyclic transformation on a number """ cycled = list(time_maps[number]) cycled[0] = time_maps[number][5] cycled[1] = time_maps[number][6] cycled[2] = time_maps[number][2] cycled[3] = time_maps[number][4] cycled[4] = time_maps[number][3] cycled[5] = time_maps[number][0] cycled[6] = time_maps[number][1] return cycled

I wont bother with a print for the above code since all it’s doing is performing the rotation. The heavy lifter is the next python function that actually checks whether or not a time is rotationally symmetric or not. I came up with the following function to do this:

def grouped_rotation_check(time): """ swap positions, ABCD -> DBCA reflect DBCA """ if re.search('[a-zA-Z]', time) is None: original_number = [time_maps[int(number)] for number in time if number != ':'] swapped = [int(time[i]) for i in range(len(time)-1, -1, -1) if time[i] != ':'] cycled_number = [] for number in swapped: cn = rotational_transformation(number) cycled_number.append(cn) same = True for og, cy in zip(original_number, cycled_number): if og == cy: continue else: same = False break return same

Let’s take a look at whether 1:11, 2:22, or 8:18 are symmetric with respect to our rotation.

grouped_rotation_check('1:11') grouped_rotation_check('2:22') grouped_rotation_check('8:18') >> False >> True >> False

Interesting. Note that 1 is not rotationally symmetric because it “swaps sides”. Of course, 2 and 8 are rotationally symmetric. Next let’s write up a function just like we did with reflection symmetry that gathers all the rotationally symmetric times and reports on them:

def get_all_rotation_times(): """ report on all the rotationally symmetric times""" cyclic_ls = [] counter = 0 for hr in range(1, 24): for mini in range(0, 60): if mini in range(0, 10): mini = '0' + str(mini) counter += 1 time = str(hr) + ":" + str(mini) if grouped_rotation_check(time) is True: cyclic_ls.append(time) print("There are %s times we checked." % counter) print("There are %s total times that are C2." % len(cyclic_ls)) print("This gives %s percent of times that are C2." % (100 * float(len(cyclic_ls)) / float(counter))) print("Here are the C2 times.") for item in cyclic_ls: print(' ', item) return cyclic_ls

Running the above function yields:

There are 1380 times we checked. There are 17 total times that are C2. This gives 1.2318840579710144 percent of times that are C2. Here are the C2 times. 2:02 2:22 2:52 5:05 5:25 5:55 6:09 6:29 6:59 8:08 8:28 8:58 9:06 9:26 9:56 20:02 22:22

From our results we see that significantly more times are reflective than rotationally symmetric, in fact nearly 4 times as many more. It looks like only the numbers 0, 2, 5, 8, and 9 yield rotationally symmetric times under certain combinations.

So then, how many times are both reflective and rotationally symmetric?

joint = list(set(reflective_ls).intersection(cyclic_ls)) print("Apparently there is only one reflective and C2 time, and it is %s." % joint[0]) >> Apparently there is only one reflective and C2 time, and it is 8:08.

Only one, and it’s 8:08. Savage! This is a special time indeed, for it also spells the name BOB. Hence, all BOB’s are reflective and rotationally symmetric.

# Conclusion

This little project is boredom gone absolutely unchecked, but I just had to know the answer to the question. I hope that I can spare anyone else from using python for such an insipid use, and furthermore, I hope that the time 8:08 has a new special meaning for you!

# Bonus Question

Create a histogram of the results above for each symmetry type.

# Project Code

import re """ digital clock reflection symmetry and cyclic groups _ _ _ _ _ _ _ _ | | | _| _| |_| |_ |_ | |_| |_| |_| | |_ _| | _| |_| | |_| _| 0 1 3 2 4 6 5 number 0: [1, 1, 0, 1, 1, 1, 1] number 1: [0, 0, 0, 1, 0, 0, 1] number 2: [1, 0, 1, 1, 1, 1, 0] number 3: [1, 0, 1, 1, 1, 1, 0] number 4: [0, 1, 1, 1, 0, 0, 1] number 5: [1, 1, 1, 0, 0, 1, 1] number 6: [1, 1, 1, 0, 1, 1, 1] number 7: [1, 0, 0, 1, 0, 0, 1] number 8: [1, 1, 1, 1, 1, 1, 1] number 9: [1, 1, 1, 1, 0, 1, 1] the purpose of this program is to analyze reflection symmetry and cyclic groups of digital clocks reflection symmetry: R: horizontal reflection example: 6:09 has reflection symmetry because flipping gives 6:09 back cyclic groups: Cn: 360 degrees / n C1: 360 degrees C2: 180 degrees C3: 120 degrees C4: 90 degrees example: 3:03 has C2 symmetry because upon rotation we get 3:03 back """ time_maps = {0: [1, 1, 0, 1, 1, 1, 1], 1: [0, 0, 0, 1, 0, 0, 1], 2: [1, 0, 1, 1, 1, 1, 0], 3: [1, 0, 1, 1, 0, 1, 1], 4: [0, 1, 1, 1, 0, 0, 1], 5: [1, 1, 1, 0, 0, 1, 1], 6: [1, 1, 1, 0, 1, 1, 1], 7: [1, 0, 0, 1, 0, 0, 1], 8: [1, 1, 1, 1, 1, 1, 1], 9: [1, 1, 1, 1, 0, 1, 1]} def individual_reflection_check(number): """ :param number: the number we are analyzing :return: True/False value telling if symmetric """ tmp = list(time_maps[number]) tmp[0] = time_maps[number][5] tmp[1] = time_maps[number][4] tmp[2] = time_maps[number][2] tmp[3] = time_maps[number][6] tmp[4] = time_maps[number][1] tmp[5] = time_maps[number][0] tmp[6] = time_maps[number][3] if tmp == time_maps[number]: #print("%s is R." % number) return True else: #print("%s is not R." % number) return False def grouped_reflection_check(time): """ format needs to be HR:MN ex: 3:45 (no reason for a trailing 0) :param time: :return: """ if re.search('[a-zA-Z]', time) is None: # loop through time elements for item in time: if item != ':': status = individual_reflection_check(int(item)) if status is True: continue else: #print("The time %s is NOT R" % time) return False #print("The time %s is R" % time) return True else: print("%s is not a valid time" % time) def get_all_reflection_times(): """ the idea is all need to be reflective for it to work, let's get the list """ reflective_ls = [] counter = 0 for hr in range(1, 13): for mini in range(0, 60): if mini in range(0, 10): mini = '0' + str(mini) counter += 1 time = str(hr) + ":" + str(mini) if grouped_reflection_check(time) is True: reflective_ls.append(time) print("There are %s times we checked." % counter) print("There are %s total times that are reflective." % len(reflective_ls)) print("This gives %s percent of times that are reflective." % (100*float(len(reflective_ls))/float(counter))) print("Here are the reflective times.") for item in reflective_ls: print(' ', item) return reflective_ls def rotational_transformation(number): """ perform the cyclic transformation on a number """ cycled = list(time_maps[number]) cycled[0] = time_maps[number][5] cycled[1] = time_maps[number][6] cycled[2] = time_maps[number][2] cycled[3] = time_maps[number][4] cycled[4] = time_maps[number][3] cycled[5] = time_maps[number][0] cycled[6] = time_maps[number][1] return cycled def grouped_rotation_check(time): """ swap positions, ABCD -> DBCA reflect DBCA """ if re.search('[a-zA-Z]', time) is None: original_number = [time_maps[int(number)] for number in time if number != ':'] swapped = [int(time[i]) for i in range(len(time)-1, -1, -1) if time[i] != ':'] cycled_number = [] for number in swapped: cn = rotational_transformation(number) cycled_number.append(cn) same = True for og, cy in zip(original_number, cycled_number): if og == cy: continue else: same = False break return same def get_all_rotation_times(): """ report on all the rotationally symmetric times""" cyclic_ls = [] counter = 0 for hr in range(1, 13): for mini in range(0, 60): if mini in range(0, 10): mini = '0' + str(mini) counter += 1 time = str(hr) + ":" + str(mini) if grouped_rotation_check(time) is True: cyclic_ls.append(time) print("There are %s times we checked." % counter) print("There are %s total times that are C2." % len(cyclic_ls)) print("This gives %s percent of times that are C2." % (100 * float(len(cyclic_ls)) / float(counter))) print("Here are the C2 times.") for item in cyclic_ls: print(' ', item) return cyclic_ls if __name__ == "__main__": reflective_ls = get_all_reflection_times() cyclic_ls = get_all_rotation_times() joint = list(set(reflective_ls).intersection(cyclic_ls)) print("Apparently there is only one reflective and C2 time, and it is %s." % joint[0])

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