De la poésie
In this challenge, we need to recognise 6535 images and do something with them.
I found a Google Colab that did some hand writting number recognition. I added this code at the end:
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from google.colab import drive
# get the zip archive from my drive and unzip it
"""
drive.mount('/content/drive')
!cp drive/MyDrive/poeme.zip .
!unzip poeme.zip
"""
import os
from PIL import Image
from torchvision.transforms import ToTensor
predictions = []
work_dir = os.getcwd()
for i in range(6536):
# Load the image from the given path and convert it to grayscale
img = Image.open(img_path).convert('L')
# Convert the image to a tensor
img = ToTensor()(img)
# Add an extra dimension to the tensor to represent the batch size
img = img.unsqueeze(0)
# Make a prediction using the CNN model on the image
pred_result = cnn.predict(img)
# Convert the prediction result to a string and append it to the list of predictions
predictions.append(str(pred_result[0]))
print(''.join(predictions))
This will give us a really long string of integer. We now need to focus on the hint given in the text Être pair ou ne pas l'être. (to be even or not to be).
We can try to set every even number as a 0 and every odd number as 1. We then split in block of 8 and decode the binary character found in ASCII.
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chaine = "6580616563912104..."
chaine = list(chaine)
def pair_impair():
pair = ""
impair = ""
for i in range(len(chaine)):
if int(chaine[i]) % 2 == 0:
pair += chaine[i]
else:
impair += chaine[i]
print(pair)
print(impair)
def split_string(string, length):
return [string[i:i+length] for i in range(0, len(string), length)]
substring_length = 8
substrings = split_string(chaine, substring_length)
flag = []
new_chaine = ""
for chaines in substrings:
for num in chaines:
#pair
if int(num) % 2 == 0:
new_chaine += "0"
else:
new_chaine += "1"
flag.append(new_chaine)
new_chaine = ""
# Convert each segment to decimal and then to ASCII character
flag = ''.join([chr(int(segment, 2)) for segment in flag])
print(flag)
There are some errors in the process of number recognition but we can find the flag 404CTF{d3_L4_p03S1e_qU3lqU3_P3u_C0nT3mp0r4in3}