## What will we cover in this tutorial?

We will look at how the Birthday Paradox is used when estimating how collision resistance a hash function is. This tutorial will show that a good estimate is that a n-bit hash function will have collision by chance with n/2-bit random hash values.

## Step 1: Understand a hash function

A hash function is a one-way function with a fixed output size. That is, the output has the same size and it is difficult to find two distinct input chucks, which give the same output.

hash function is any function that can be used to map data of arbitrary size to fixed-size values.

https://en.wikipedia.org/wiki/Hash_function

Probably the best know example of a hash-function is the MD5. It was designed to be used as a cryptographic hash function, but has been found to have many vulnerabilities.

Does this mean you should not use the MD5 hash function?

That depends. If you use it in a cryptographic setup, the answer is Do not use.

On the other hand, hash function are often used to calculate identifiers. For that purpose, it also depends if you should use it or not.

This is where the Birthday Paradox comes in.

## Step 2: How are hash functions and the Birthday Paradox related?

Good question. First recall what the Birthday Paradox states.

…in a random group of 23 people, there is about a 50 percent chance that two people have the same birthday

How can that be related to hash functions? There is something about collisions, right?

Given 23 people, we have 50% chance of collision (two people with the same birthday).

Hence, if we have that our hash functions maps data to a day in the calendar year. That is, it maps hash(data) -> [0, 364], then given 23 hash values, we have 50% chance for collision.

But you also know that our hash function maps to more than 365 distinct values. Actually, the MD5 maps to 2^128 distinct values.

An example would be appreciated now. Let us make a simplified hash function, call it MD5′ (md5-prime), which maps like the MD5, but only uses the first byte of the result.

That is, we have MD5′(data) -> [0, 255].

Surely, by the pigeonhole principle we would run out of possible values after 256 distinct data input to MD5′ and have a collision.

```import hashlib
import os

lookup_table = {}
collision_count = 0
for _ in range(256):
random_binary = os.urandom(16)
result = hashlib.md5(random_binary).digest()
result = result[:1]
if result in lookup_table:
print("Collision")
print(random_binary, result)
print(lookup_table[result], result)
collision_count += 1
else:
lookup_table[result] = random_binary
print("Number of collisions:", collision_count)
```

The lookup_table is used to store the already seen hash values. We will iterate over the 256 (one less than possible values of our MD5′ hash function). Take some random data and hash it with md5 and only use first byte (8 bits). If result already exists in lookup_table we have a collision, otherwise add it to our lookup_table.

For a random run of this I got 87 collisions. Expected? I would say so.

Let us try to use the Birthday Paradox to estimate how many hash values we need to get a collision of our MD5′ hash function.

A rough estimate that is widely used, is that the square root of the number of possible outcomes will give a 50% chance of collision (see wikipedia for approximation).

That is, for MD5′(data) -> [0, 255] it is, sqrt(256) = 16. Let’s try that.

```import hashlib
import os

collision = 0
for _ in range(1000):
lookup_table = {}
for _ in range(16):
random_binary = os.urandom(16)
result = hashlib.md5(random_binary).digest()
result = result[:1]
if result not in lookup_table:
lookup_table[result] = random_binary
else:
collision += 1
break
print("Number of collisions:", collision, "out of", 1000)
```

Which gives some like this.

```Number of collisions: 391 out of 1000
```

That is in the lower end, but still a reasonable approximation.

## Step 3: Use a correct data structure to lookup in

Just to clarify. We will not find collisions on the full MD5 hash function, but we will try to see if the estimate of collision is reasonable.

This requires to do a lot of calculations and we want to ensure that we are not having a bottleneck with using a wrong data structure.

The Python dict should be a hash table with expected insert and lookup O(1). Still the worst case is O(n) for these operations, which would be a big overhead to cary along the way. Hence, we will first test, that the dictionary has O(1) insert and lookup time for the use cases we have of it here.

```import time
import matplotlib.pyplot as plt

def dict_size(size):
start = time.time()
dict = {}
for i in range(size):
if i in dict:
print("HIT")
else:
dict[i] = 0
return time.time() - start

x = []
y = []
for i in range(0, 2**20, 2**12):
performance = dict_size(i)
x.append(i)
y.append(performance)
plt.scatter(x, y, alpha=0.1)
plt.xlabel("Size")
plt.ylabel("Time (sec)")
plt.show()
```

Resulting in something like this.

What does that tell us? That the dict in Python has a approximately linear insert and lookup time, that is O(1). But there some overhead at some sizes, e.g. a bit before 3,000,000. It is not exactly linear, but close enough not to expect a exponential run time.

This step is not necessary, but it is nice to know how the function grows in time, when we want to check for collisions. If the above time complexity grew exponentially (or not linearly), then it can suddenly become hard to estimate the runtime if we run for a bigger space.

## Step 4: Validating if square root of the bit size is a good estimate for collision

We will continue our journey with our modified MD5′ hash function, where the output space will be reduced.

We will then for various output space sizes see if the estimate for 50% collision of the hash functions is decent. That is, if we need approximately sqrt(space_size) of hash values to have an approximately 50% chance of a collision.

This can be done by the following code.

```import hashlib
import os
import time
import matplotlib.pyplot as plt

def main(bit_range):
start = time.time()
collision_count = 0
# Each space_size counts for 4 bits, hence we have
space_size = bit_range//4
for _ in range(100):
lookup_table = {}
# Searching half the sqrt of the space for collision
# sqrt(2**bit_range) = 2**(bit_range//2)
for _ in range(2**(bit_range//2)):
random_binary = os.urandom(16)
result = hashlib.md5(random_binary).hexdigest()
result = result[:space_size]
if result in lookup_table:
collision_count += 1
break
else:
lookup_table[result] = random_binary
return time.time() - start, collision_count

x = []
y1 = []
y2 = []
for i in range(4, 44, 4):
performance, count = main(i)
x.append(i)
y1.append(performance)
y2.append(count)
_, ax1 = plt.subplots()
plt.xlabel("Size")
plt.ylabel("Time (sec)")
ax1.scatter(x, y1)
ax2 = ax1.twinx()
ax2.bar(x, y2, align='center', alpha=0.5, color='red')
ax2.set_ylabel("Collision rate (%)", color='red')
ax2.set_ylim([0, 100])
plt.show()
```

The estimated collision rate is very rough, as it only runs 100 trials for each space size.

The result are shown in the graph below.

Interestingly, it seems to be in the 30-50% range for most cases.

As a note, it might confuse that the run-time (the dots), does not seem to be linear. That is because for each bit-size we increase, we double the space. Hence, the x-axis is a logarithmic scale.

## Step 5: What does that all mean?

This has high impact on using hash functions for creating unique identifiers. If you want a short identifier with the least number of bits, then you need to consider the Birthday Paradox.

Assume you created the following service.

```import hashlib
import base64

def get_uid(text):
result = hashlib.md5(text.encode()).digest()
result = base64.b64encode(result)
return result[:4]

uid = get_uid("my text")
print(uid)
```

If the input text can be considered random, how resistant is get_uid(…) function against collision.

Well, it returns 4 base64 characters. That is 6*4 = 24 bits of information (each base 64 character contains 6 bits of information). The rough estimate is that if you use it sprt(2^24) = 2^12 = 4,096 times you will have a high risk of collision (approximately 50% chance).

Let’s try.

```import hashlib
import os
import base64

def get_uid(text):
result = hashlib.md5(text).digest()
result = base64.b64encode(result)
return result[:4]

lookup_table = {}
for _ in range(4096):
text = os.urandom(16)
uid = get_uid(text)
if uid in lookup_table:
print("Collision detected")
else:
lookup_table[uid] = text
```

It does not give collision every time, but run it a few times and you will get.

```Collision detected
```

Hence, it seems to be valid. The above code was run 1000 times and gave collision 497 times, which is close to 50% of the time.

The Birthday Paradox is presented as follows.

…in a random group of 23 people, there is about a 50 percent chance that two people have the same birthday

This is also referred to as the Birthday Problem in probability theory.

First question: What is a paradox?

…is a logically self-contradictory statement or a statement that runs contrary to one’s expectation

Wikipedia

What does that mean? A logically self-contradictory statement‚ means that there should be a contradiction somewhere in the Birthday Paradox. This is not the case.

Then a statement that runs contrary to one’s expectations, could be open for discussion. As we will see, by example, in this post, it is not contrary to one’s expectation for an informed person.

## Step 1: Run some examples

The assumption is that we have 23 random people. This assumes further, that the birthday of each one of these people is random.

To validate that this is true, let’s try to implement it in Python.

```import random
stat = {'Collision': 0, 'No-collision': 0}
for _ in range(10000):
days = []
for _ in range(23):
day = random.randint(0, 365)
days.append(day)
if len(days) == len(set(days)):
stat['No-collision'] += 1
else:
stat['Collision'] += 1
print("Probability for at least 2 with same birthday in a group of 23")
print("P(A) =", stat['Collision']/(stat['Collision'] + stat['No-collision']))
```

This will output different results from run to run, but something around 0.507.

```Probability for at least 2 with same birthday in a group of 23
P(A) = 0.5026
```

A few comments to the code. It keeps record of how many times of choosing 23 random birthdays, we will end with at least two of them being the same day. We run the experiment 10,000 times to have some idea if it is just pure luck.

The check if len(days) == len(set(days)) tests whether we did not have the same brirthday. If function set(…) takes all the unique days in the list. Hence, if we have two the same days days of the year, then the len (length) will be the same for the list and the set of days.

## Step 2: The probability theory behind it

This is where it becomes a bit more technical. The above shows it behaves like it says. That if we take a group of 23 random people, with probability 50%, two of them will have the same birthday.

Is this contrary to one’s expectations? Hence, is it a paradox?

Before we answer that, let’s see if we can nail the probability theory behind this.

Do it step by step.

If we have 1 person, what is the probability that anyone in this group of 1 person has the same birthday? Yes, it sounds strange. The probability is obviously 0.

If we have 2 persons, what is the probability that any of the 2 people have the same birthday? Then they need to have the same birthday. Hence, the probability become 1/365.

How do you write that as an equation?

What we often do in probability theory, is, that we calculate the opposite probability.

Hence, we calculate the probability of now having two the same birthdays in a group. This is easier to calculate. In the first case, we have all possibilities open.

P(1) = 1

Where P(1) is the probability that given a group of one person, what is the probability of that person not having the same birthday as anyone in the group.

P(2) = 1 x (364 / 365)

Because, the first birthday is open for any birthday, then the second, only has 364 left of 365 possible birthdays.

This continues.

P(n) = 1 x (364 / 365) x (363 / 365) x … x ((365 – n + 1) / 365)

Which makes the probability of picking 23 random people without anyone with the same birthday to be.

P(23) = 1 x (364 / 365) x (363 / 365) x … x (343 / 365) = 0.493

Or calculated in Python.

```def prop(n):
if n == 1:
return 1
else:
return (365 - n + 1) / 365 * prop(n - 1)
print("Probability for at no-one with same birthday in a group of 23")
print("P(A') =",  prop(23))
```

Which results in.

```Probability for at no-one with same birthday in a group of 23
P(A') = 0.4927027656760144
```

This formula can be rewritten (see wikipedia), but for our purpose the above is fine for our purpose.

The probability we look for is given by.

P(A) = 1 – P(A’)

## Step 3: Make a graph of how likely a collision is based on a group size

This is great news. We can now calculate the theoretical probability of two people having the same birthday in a group of n random people.

This can be achieved by the following code.

```from matplotlib import pyplot as plt
def prop(n):
if n == 1:
return 1
else:
return (365 - n + 1) / 365 * prop(n - 1)
X = []
Y = []
for i in range(1, 90):
X.append(i)
Y.append(1 - prop(i))
plt.scatter(X, Y, color='blue')
plt.xlabel("Number of people")
plt.ylabel("Probability of collision")
plt.axis([0, 90, 0, 1])
plt.show()
```

Which results in the following plot.

Where you can see that about 23 people, we have 50% chance of having a pair with the same birthday (called collision).

## Conclusion

Is it a paradox? Well, there is no magic in it. You can see the above are just simple calculations. But is the following contrary to one’s expectation?

6 weeks are 6*7*24*60*60 seconds = 3,628,800 seconds.

And 10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800.

Well, the first time you calculate it might be. But does that make it a paradox?

No, it is just a surprising fact the first time you see it. Does it mean that seconds are related to faculty? No, of course not. It is just one strange thing that connects in a random way.

The same with the Birthday Paradox, it is just surprising the first time you see it.

It seems surprising for people that you only need 23 people to have 50% chance of a pair with the same birthday, but it is not a paradox for people that work with numbers.

## What will we cover in this tutorial?

1. Where and how to get images you can use without copyright issues.
2. How to extract the faces of the images.
3. Building a Photo Mosaic using the extracted images of faces.

## Step 1: Where and how to get images

There exists a lot of datasets of faces, but most have restrictions on them. A great place to find images is on Pexels, as they are free to use (see license here).

Also, the Python library pexels-api makes it easy to download a lot of images. It can be installed by the following command.

```pip install pexels-api
```

To use the Pexels API you need to register.

2. Accept the email sent to your inbox (the email address you provide).
3. Request your API key here.

Then you can download images by a search query from this Python program.

```from pexels_api import API
import requests
import os.path
from pathlib import Path

path = 'pics'
Path(path).mkdir(parents=True, exist_ok=True)
# Reguest key : https://www.pexels.com/api/
# - No need to set URL
# - Accept email send to you
# - Refresh API or see key here: https://www.pexels.com/api/new/
PEXELS_API_KEY = '--- INSERT YOUR API KEY HERE ---'
api = API(PEXELS_API_KEY)
query = 'person'
api.search(query)
# Get photo entries
photos = api.get_entries()
print("Search: ", query)
print("Total results: ", api.total_results)
MAX_PICS = 1000
print("Fetching max: ", MAX_PICS)
count = 0
while True:
photos = api.get_entries()
print(len(photos))
if len(photos) == 0:
break
for photo in photos:
# Print photographer
print('Photographer: ', photo.photographer)
# Print original size url
print('Photo original size: ', photo.original)
file = os.path.join(path, query + '-' + str(count).zfill(5) + '.' + photo.original.split('.')[-1])
count += 1
print(file)
picture_request = requests.get(photo.original)
if picture_request.status_code == 200:
with open(file, 'wb') as f:
f.write(picture_request.content)
# This should be a function call to make a return
if count >= MAX_PICS:
break
if count >= MAX_PICS:
break
if not api.has_next_page:
print("Last page: ", api.page)
break
# Search next page
api.search_next_page()
```

There is an upper limit of 1.000 photos in the above Python program, you can change that if you like. It is set to download photos that are shown if you query person. Feel free to change that.

It takes some time to download all the images and will take up some space.

## Step 2: Extract the faces from the photos

Here OpenCV comes in. They have a trained model using the Haar Cascade Classifier. You need to install the OpenCV library by the following command.

```pip install opencv-python
```

The trained model we use is part of the library, but is not loaded easily from the destination. Therefore we suggest you download it from here (it should be named: haarcascade_frontalface_default.xml) and add the it to the location you work from.

We want to use it to identify faces and extract them and save them in a library for later use.

```import cv2
import numpy as np
import glob
import os
from pathlib import Path

def preprocess(box_width=12, box_height=16):
path = "pics"
output = "small-faces"
Path(output).mkdir(parents=True, exist_ok=True)
files = glob.glob(os.path.join(path, "*"))
files.sort()
images = []
cnt = 0
for filename in files:
print("Processing...", filename)
frame_gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
frame_gray = cv2.equalizeHist(frame_gray)
for (x, y, w, h) in faces:
roi = frame[y:y+h, x:x+w]
img = cv2.resize(roi, (box_width, box_height))
images.append(img)
output_file_name = "face-" + str(cnt).zfill(5) + ".jpg"
output_file_name = os.path.join(output, output_file_name)
cv2.imwrite(output_file_name, img)
return np.stack(images)

preprocess(box_width=12, box_height=16)
```

It will create a folder called small-faces with small images of the identified faces.

Notice, that the Haar Cascade Classifier is not perfect. It will miss a lot of faces and have false positives. It is a good idea to look manually though all the images and delete all false positives (images that are not having a face).

## Step 3: Building our first mosaic photo

The approach to divide the photo into equal sized boxes. For each box to find the image (our faces), which fits the best as a replacement.

To improve performance of the process function we use Numba, which is a just-in-time compiler that is designed to optimize NumPy code in for-loops.

```import cv2
import numpy as np
import glob
import os
from numba import jit

@jit(nopython=True)
def process(photo, images, box_width=24, box_height=32):
height, width, _ = photo.shape
for i in range(0, height, box_height):
for j in range(0, width, box_width):
roi = photo[i:i + box_height, j:j + box_width]
best_match = np.inf
best_match_index = 0
for k in range(1, images.shape):
total_sum = np.sum(np.where(roi > images[k], roi - images[k], images[k] - roi))
if total_sum < best_match:
best_match = total_sum
best_match_index = k
photo[i:i + box_height, j:j + box_width] = images[best_match_index]
return photo

def main():
box_width = 12
box_height = 16
height, width, _ = photo.shape
# To make sure that it we can slice the photo in box-sizes
width = (width//box_width) * box_width
height = (height//box_height) * box_height
photo = cv2.resize(photo, (width, height))
# Load all the images of the faces
# Create the mosaic
mosaic = process(photo.copy(), images, box_width, box_height)
cv2.imshow("Original", photo)
cv2.imshow("Result", mosaic)
cv2.waitKey(0)

main()
```

To test it we have used the photo of Rune.

This reuses the same images. This gives a decent result, but if you want to avoid the extreme patterns of reused images, you can change the code for that.

The above example has 606 small images. If you avoid reuse it runs out fast of possible images. This would require a bigger base or the result becomes questionable.

The above photo mosaic is created on a downscaled size, but still it does not create a good result, if you do not reuse images. This would require a quite larger set of images to work from.

## What will we cover in this tutorial?

We will investigate if we can create a decent video mosaic effect on a live webcam stream using OpenCV, Numba and Python. First we will learn the simple way to create a video mosaic and investigate the performance of that. Then we will extend that to create a better quality video mosaic and try to improve the performance by lowering the quality.

## Step 1: How does simple photo mosaic work?

A photographic mosaic is a photo generated by other small images. A black and white example is given here.

The above is not a perfect example of it as it is generated with speed to get it running smooth from a webcam stream. Also, it is done in gray scale to improve performance.

The idea is to generate the original image (photograph) by mosaic technique by a lot of smaller sampled images. This is done in the above with the original frame of 640×480 pixels and the mosaic is constructed of small images of size 16×12 pixels.

The first thing we want to achieve is to create a simple mosaic. A simple mosaic is when the original image is scaled down and each pixel is then exchanged with one small image with the same average color. This is simple and efficient to do.

On a high level this is the process.

1. Have a collection C of small images used to create the photographic mosaic
2. Scale down the photo P you want to create a mosaic of.
3. For each pixel in photo P find the image I from C that has the closed average color as the pixel. Insert image I to represent that pixel.

This explains the simple way of doing. The next question is, will it be efficient enough to have a live webcam stream processed?

## Step 2: Create a collection of small images

To optimize performance we have chosen to make it in gray scale. The first step is to collect images you want to use. This can be any pictures.

We have used photos from Pexels, which are all free for use without copyright.

What we need is to convert them all to gray scale and resize to fit our purpose.

```import cv2
import glob
import os
import numpy as np
output = "small-pics-16x12"
path = "pics"
files = glob.glob(os.path.join(path, "*"))
for file_name in files:
print(file_name)
img = cv2.resize(img, (16, 12))
img = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
mean = np.mean(img)
output_file_name = "image-" + str(mean).replace('.', '-') + ".jpg"
output_file_name = os.path.join(output, output_file_name)
print(output_file_name)
cv2.imwrite(output_file_name, img)
```

The script assumes that we have located the images we want to convert to gray scale and resize are located in the local folder pics. Further, we assume that the output images (the processed images) will be put in an already existing folder small-pics-16×12.

## Step 3: Get a live stream from the webcam

On a high level a live stream from a webcam is given in the following diagram.

This process framework is given in the code below.

```import cv2
import numpy as np

def process(frame):
return frame

def main():
# Get the webcam (default webcam is 0)
cap = cv2.VideoCapture(0)
# If your webcam does not support 640 x 480, this will find another resolution
cap.set(cv2.CAP_PROP_FRAME_WIDTH, 640)
cap.set(cv2.CAP_PROP_FRAME_HEIGHT, 480)
while True:
# Read the a frame from webcam
# Flip the frame
frame = cv2.flip(frame, 1)
frame = cv2.resize(frame, (640, 480))
gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
# Update the frame
updated_frame = process(gray)
# Show the frame in a window
cv2.imshow('WebCam', updated_frame)
# Check if q has been pressed to quit
if cv2.waitKey(1) == ord('q'):
break
# When everything done, release the capture
cap.release()
cv2.destroyAllWindows()

main()
```

The above code is just an empty shell where the function call to process is where the all the processing will be. This code will just generate a window that shows a gray scale image.

## Step 4: The simple video mosaic

We need to introduce two main things to create this simple video mosaic.

1. Loading all the images we need to use (the 16×12 gray scale images).
2. Fill out the processing of each frame, which replaces each 16×12 box of the frame with the best matching image.

The first step is preprocessing and should be done before we enter the main loop of the webcam capturing. The second part is done in each iteration inside the process function.

```import cv2
import numpy as np
import glob
import os

def preprocess():
path = "small-pics-16x12"
files = glob.glob(os.path.join(path, "*"))
files.sort()
images = []
for filename in files:
images.append(cv2.cvtColor(img, cv2.COLOR_BGR2GRAY))
return np.stack(images)

def process(frame, images, box_height=12, box_width=16):
height, width = frame.shape
for i in range(0, height, box_height):
for j in range(0, width, box_width):
roi = frame[i:i + box_height, j:j + box_width]
mean = np.mean(roi[:, :])
roi[:, :] = images[int((len(images)-1)*mean/256)]
return frame

def main(images):
# Get the webcam (default webcam is 0)
cap = cv2.VideoCapture(0)
# If your webcam does not support 640 x 480, this will find another resolution
cap.set(cv2.CAP_PROP_FRAME_WIDTH, 640)
cap.set(cv2.CAP_PROP_FRAME_HEIGHT, 480)
while True:
# Read the a frame from webcam
# Flip the frame
frame = cv2.flip(frame, 1)
frame = cv2.resize(frame, (640, 480))
gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
# Update the frame
mosaic_frame = process(gray, images)
# Show the frame in a window
cv2.imshow('Mosaic Video', mosaic_frame)
cv2.imshow('Webcam', frame)
# Check if q has been pressed to quit
if cv2.waitKey(1) == ord('q'):
break
# When everything done, release the capture
cap.release()
cv2.destroyAllWindows()

images = preprocess()
main(images)
```

The preprocessing function reads all the images, converts them to gray scale (to have only 1 channel per pixel), and returns them as a NumPy array to have optimized code.

The process function takes and breaks down the image in blocks of 16×12 pixels, computes the average gray scale, and takes the estimated best match. Notice the average (mean) value is a float, hence, we can have more than 256 gray scale images.

In this example we used 1.885 images to process it.

A result can be seen here.

The result is decent but not good.

## Step 5: Testing the performance and improve it by using Numba

While the performance is quite good, let us test it.

We do that by using the time library.

First you need to import the time library.

```import time
```

Then time the actual time the process call uses. New code inserted in the main while loop.

```        # Update the frame
start = time.time()
mosaic_frame = process(gray, images)
print("Process time", time.time()- start, "seconds")
```

This will result in the following output.

```Process time 0.02651691436767578 seconds
Process time 0.026834964752197266 seconds
Process time 0.025418996810913086 seconds
Process time 0.02562689781188965 seconds
Process time 0.025369882583618164 seconds
Process time 0.025450944900512695 seconds
```

Or a few lines from it. About 0.025-0.027 seconds.

Let’s try to use Numba in the equation. Numba is a just-in-time compiler for NumPy code. That means it compiles to python code to a binary for speed. If you are new to Numba we recommend you read this tutorial.

```import cv2
import numpy as np
import glob
import os
import time
from numba import jit

def preprocess():
path = "small-pics-16x12"
files = glob.glob(os.path.join(path, "*"))
files.sort()
images = []
for filename in files:
images.append(cv2.cvtColor(img, cv2.COLOR_BGR2GRAY))
return np.stack(images)

@jit(nopython=True)
def process(frame, images, box_height=12, box_width=16):
height, width = frame.shape
for i in range(0, height, box_height):
for j in range(0, width, box_width):
roi = frame[i:i + box_height, j:j + box_width]
mean = np.mean(roi[:, :])
roi[:, :] = images[int((len(images)-1)*mean/256)]
return frame

def main(images):
# Get the webcam (default webcam is 0)
cap = cv2.VideoCapture(0)
# If your webcam does not support 640 x 480, this will find another resolution
cap.set(cv2.CAP_PROP_FRAME_WIDTH, 640)
cap.set(cv2.CAP_PROP_FRAME_HEIGHT, 480)
while True:
# Read the a frame from webcam
# Flip the frame
frame = cv2.flip(frame, 1)
frame = cv2.resize(frame, (640, 480))
gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
# Update the frame
start = time.time()
mosaic_frame = process(gray, images)
print("Process time", time.time()- start, "seconds")
# Show the frame in a window
cv2.imshow('Mosaic Video', mosaic_frame)
cv2.imshow('Webcam', frame)
# Check if q has been pressed to quit
if cv2.waitKey(1) == ord('q'):
break
# When everything done, release the capture
cap.release()
cv2.destroyAllWindows()

images = preprocess()
main(images)
```

This gives the following performance.

```Process time 0.0014820098876953125 seconds
Process time 0.0013887882232666016 seconds
Process time 0.0015859603881835938 seconds
Process time 0.0016350746154785156 seconds
Process time 0.0018379688262939453 seconds
Process time 0.0016241073608398438 seconds
```

Which is a factor 15-20 speed improvement.

Good enough for live streaming. But the result is still not decent.

## Step 6: A more advanced video mosaic approach

The more advanced video mosaic consist of approximating the each replacement box of pixels by the replacement image pixel by pixel.

```import cv2
import numpy as np
import glob
import os
import time
from numba import jit

def preprocess():
path = "small-pics-16x12"
files = glob.glob(os.path.join(path, "*"))
files.sort()
images = []
for filename in files:
images.append(cv2.cvtColor(img, cv2.COLOR_BGR2GRAY))
return np.stack(images)

@jit(nopython=True)
def process(frame, images, box_height=12, box_width=16):
height, width = frame.shape
for i in range(0, height, box_height):
for j in range(0, width, box_width):
roi = frame[i:i + box_height, j:j + box_width]
best_match = np.inf
best_match_index = 0
for k in range(1, images.shape):
total_sum = np.sum(np.where(roi > images[k], roi - images[k], images[k] - roi))
if total_sum < best_match:
best_match = total_sum
best_match_index = k
roi[:,:] = images[best_match_index]
return frame

def main(images):
# Get the webcam (default webcam is 0)
cap = cv2.VideoCapture(0)
# If your webcam does not support 640 x 480, this will find another resolution
cap.set(cv2.CAP_PROP_FRAME_WIDTH, 640)
cap.set(cv2.CAP_PROP_FRAME_HEIGHT, 480)
while True:
# Read the a frame from webcam
# Flip the frame
frame = cv2.flip(frame, 1)
frame = cv2.resize(frame, (640, 480))
gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
# Update the frame
start = time.time()
mosaic_frame = process(gray, images)
print("Process time", time.time()- start, "seconds")
# Show the frame in a window
cv2.imshow('Mosaic Video', mosaic_frame)
cv2.imshow('Webcam', frame)
# Check if q has been pressed to quit
if cv2.waitKey(1) == ord('q'):
break
# When everything done, release the capture
cap.release()
cv2.destroyAllWindows()

images = preprocess()
main(images)
```

There is one line to notice specifically.

```total_sum = np.sum(np.where(roi > images[k], roi - images[k], images[k] - roi))
```

Which is needed, as we work with unsigned 8 bit integers. What it does is, that it takes the and calculates the difference between each pixel in the region of interest (roi) and the image[k]. This is a very expensive calculation as we will see.

Performance shows the following.

```Process time 7.030380010604858 seconds
Process time 7.034134149551392 seconds
Process time 7.105709075927734 seconds
Process time 7.138839960098267 seconds
```

Over 7 seconds for each frame. The result is what can be expected by using this amount of images, but the performance is too slow to have a flowing smooth live webcam stream.

The result can be seen here.

## Step 7: Compromise options

There are various options to compromise for speed and we will not investigate all. Here are some.

• Use fever images in our collection (use less than 1.885 images). Notice, that using half the images, say 900 images, will only speed up 50%.
• Bigger image sizes. Scaling up to use 32×24 images. Here we will still need to do a lot of processing per pixel still. Hence, the expected speedup might be less than expected.
• Make a compromised version of the difference calculation (total_sum). This has great potential, but might have undesired effects.
• Scale down pixel estimation for fever calculations.

We will try the last two.

First, let’s try to exchange the calculation of total_sum, which is our distance function that measures how close our image is. Say, we use this.

```                total_sum = np.sum(np.subtract(roi, images[k]))
```

This results in overflow if we have a calculation like 1 – 2 = 255, which is undesired. On the other hand. It might happen in expected 50% of the cases, and maybe it will skew the calculation evenly for all images.

Let’s try.

```Process time 1.857623815536499 seconds
Process time 1.7193729877471924 seconds
Process time 1.7445549964904785 seconds
Process time 1.707035779953003 seconds
Process time 1.6778359413146973 seconds
```

Wow. That is a speedup of a factor 4-6 per frame. The quality is still fine, but you will notice a poorly mapped image from time to time. But the result is close to the advanced video mosaic and far from the first simple video mosaic.

Another addition we could make is to estimate each box by only 4 pixels. This should still be better than the simple video mosaic approach. I have given the full code below.

```import cv2
import numpy as np
import glob
import os
import time
from numba import jit

def preprocess():
path = "small-pics-16x12"
files = glob.glob(os.path.join(path, "*"))
files.sort()
images = []
for filename in files:
images.append(cv2.cvtColor(img, cv2.COLOR_BGR2GRAY))
return np.stack(images)

def preprocess2(images, scale_width=8, scale_height=6):
scaled = []
_, height, width = images.shape
print("Dimensions", width, height)
width //= scale_width
height //= scale_height
print("Scaled Dimensions", width, height)
for i in range(images.shape):
scaled.append(cv2.resize(images[i], (width, height)))
return np.stack(scaled)

@jit(nopython=True)
def process3(frame, frame_scaled, images, scaled, box_height=12, box_width=16, scale_width=8, scale_height=6):
height, width = frame.shape
width //= scale_width
height //= scale_height
box_width //= scale_width
box_height //= scale_height
for i in range(0, height, box_height):
for j in range(0, width, box_width):
roi = frame_scaled[i:i + box_height, j:j + box_width]
best_match = np.inf
best_match_index = 0
for k in range(1, scaled.shape):
total_sum = np.sum(roi - scaled[k])
if total_sum < best_match:
best_match = total_sum
best_match_index = k
frame[i*scale_height:(i + box_height)*scale_height, j*scale_width:(j + box_width)*scale_width] = images[best_match_index]
return frame

def main(images, scaled):
# Get the webcam (default webcam is 0)
cap = cv2.VideoCapture(0)
# If your webcam does not support 640 x 480, this will find another resolution
cap.set(cv2.CAP_PROP_FRAME_WIDTH, 640)
cap.set(cv2.CAP_PROP_FRAME_HEIGHT, 480)
while True:
# Read the a frame from webcam
# Flip the frame
frame = cv2.flip(frame, 1)
frame = cv2.resize(frame, (640, 480))
gray = cv2.cvtColor(frame, cv2.COLOR_BGR2GRAY)
# Update the frame
start = time.time()
gray_scaled = cv2.resize(gray, (640//8, 480//6))
mosaic_frame = process3(gray, gray_scaled, images, scaled)
print("Process time", time.time()- start, "seconds")
# Show the frame in a window
cv2.imshow('Mosaic Video', mosaic_frame)
cv2.imshow('Webcam', frame)
# Check if q has been pressed to quit
if cv2.waitKey(1) == ord('q'):
break
# When everything done, release the capture
cap.release()
cv2.destroyAllWindows()

images = preprocess()
scaled = preprocess2(images)
main(images, scaled)
```

Where there is added preprocessing step (preprocess2). The process time is now.

```Process time 0.5559628009796143 seconds
Process time 0.5979928970336914 seconds
Process time 0.5543379783630371 seconds
Process time 0.5621011257171631 seconds
```

Which is okay, but still less than 2 frames per seconds.

The result can be seen here.

It is not all bad. It is still better than the simple video mosaic approach.

The result is not perfect. If you want to use it on a live webcam stream with 25-30 frames per seconds, you need to find further optimizations of live with the simple mosaic video approach.