welcome-to-linux: Rework session according to OpenEdu methodology

Transitioned the 'welcome-to-linux' directory to meet the requirements of
the OpenEdu Methodology.

Signed-off-by: Sorin Birchi <sb.birchi.sorin@gmail.com>

Author: Sorin Birchi

chapters/scratch-linux/welcome-to-linux/drills/did-you-look-everywhere/sol/solution.sh

@@ -0,0 +1,5 @@
+# Did You Look Everywhere?
+
+We assure you the flag is somewhere in the `/home/ctf` directory.
+
+If you're having difficulties solving this exercise, go through [this](../../../reading/README.md) reading material.

chapters/scratch-linux/welcome-to-linux/drills/tasks/did-you-look-everywhere/README.md

@@ -0,0 +1,6 @@
+#! /bin/bash
+# SPDX-License-Identifier: BSD-3-Clause
+
+cd south-park/al-gore/manbearpig || exit
+ls -a
+cat .flag

chapters/scratch-linux/welcome-to-linux/drills/tasks/did-you-look-everywhere/solution/solution.sh

@@ -0,0 +1,17 @@
+# In Your Face
+
+The flag is literally in your face.
+
+Connect to the specified virtual machine and use `ls` to explore your surroundings:
+
+```console
+ctf@tutorial:~$ ls
+flag
+```
+
+Now just use `cat` to extract the content of the `flag` file:
+
+```console
+ctf@tutorial:~$ cat flag
+<your_flag_is_here>
+```

chapters/scratch-linux/welcome-to-linux/drills/tasks/in-your-face/README.md

@@ -0,0 +1,7 @@
+# Jumbled
+
+The flag you're given in `jumbled/support/flag` does contain the actual flag, but not quite.
+Use your Python skills to reconstruct it.
+You'll also need some trial and error here.
+
+If you're having difficulties solving this exercise, go through [this](../../../reading/README.md##Enter-Python) reading material.

chapters/scratch-linux/welcome-to-linux/drills/in-your-face/sol/solution.sh chapters/scratch-linux/welcome-to-linux/drills/tasks/in-your-face/solution/solution.sh

@@ -1,9 +1,12 @@
-flag_file = open('../src/flag', 'rb')
+# SPDX-License-Identifier: BSD-3-Clause
+
+flag_file = open("../src/flag", "rb")
 jumbled_flag = flag_file.read()
 
-# Ther are no hints, so we have to generate all possible flags.
+# There are no hints, so we have to generate all possible flags.
 # This means all permutations of printable characters in the given flag
-chunks = jumbled_flag[4:-1].split(b'\x80\x80\x80')
+chunks = jumbled_flag[4:-1].split(b"\x80\x80\x80")
+
 
 def permutations(l):
     if not l:  # Check if list is empty
@@ -16,14 +19,15 @@ def permutations(l):
 
     for i in range(len(l)):
         tmp = l[i]
-        rest = l[:i] + l[i+1:]  # All l elements except the ith.
+        rest = l[:i] + l[i + 1 :]  # All l elements except the ith.
 
         # Generate all permutations that start with tmp
         for p in permutations(rest):
             new_l.append([tmp] + p)
 
     return new_l
 
+
 # Now try each of them until you find the one that works.
 for p in permutations(chunks):
     print(f'SSS{{{b"".join(p).decode()}}}')

chapters/scratch-linux/welcome-to-linux/drills/tasks/jumbled/README.md

@@ -0,0 +1,17 @@
+# Quick Mafs
+
+The flag's format is the classic `SSS{...}`, where `...` represent a string obtained by concatenating the **first 10 numbers** obtained by performing the computations below.
+
+Let each number be `n[i]`, where `i` is its index.
+`n[0] = 1337`.
+This is your first number.
+The next numbers are defined by the formula below, where `^` signifies exponentiation.
+
+```text
+n[i] = (n[i - 1]^3 * 67 + 31) % 2000, for all i > 0
+```
+
+And, please, don't compute the numbers manually.
+You've just learned a cool new programming language that you can use!
+
+If you're having difficulties solving this exercise, go through [this](../../../reading/README.md##Enter-Python) reading material.

chapters/scratch-linux/welcome-to-linux/drills/jumbled/sol/solution.py chapters/scratch-linux/welcome-to-linux/drills/tasks/jumbled/solution/solution.py

@@ -1,8 +1,10 @@
+# SPDX-License-Identifier: BSD-3-Clause
+
 num = 1337
-flag = 'SSS{' + str(num)
+flag = "SSS{" + str(num)
 
 for i in range(1, 10):
     num = (num**3 * 67 + 31) % 2000
     flag += str(num)
 
-print(flag + '}')
+print(flag + "}")

chapters/scratch-linux/welcome-to-linux/drills/jumbled/public/flag chapters/scratch-linux/welcome-to-linux/drills/tasks/jumbled/support/flag

@@ -0,0 +1,4 @@
+# What's Running?
+
+Did you know flags can **run**?
+If you're having difficulties solving this exercise, go through [this](../../../reading/README.md###Processes) reading material.

chapters/scratch-linux/welcome-to-linux/drills/tasks/quick-mafs/README.md

@@ -0,0 +1,8 @@
+#! /bin/bash
+# SPDX-License-Identifier: BSD-3-Clause
+
+ps -ef
+# Now scroll through the output until you see a process whose name starts with `SSS`.
+
+# Or, if you've learnt about `|`, you can use it to ease your work:
+pgrep -lf SSS

chapters/scratch-linux/welcome-to-linux/drills/quick-mafs/sol/solution.py chapters/scratch-linux/welcome-to-linux/drills/tasks/quick-mafs/solution/solution.py

@@ -1,12 +1,13 @@
+#!/bin/bash
 # Use echo to display text to the terminal.
 echo "Moving to welcome-to-linux/activities"
-cd ..
+cd .. || exit
 ls
 
 echo "Moving to welcome-to-linux"
-cd ..
+cd .. || exit
 ls
 
 echo "Moving to back to welcome-to-linux/activities/00-demo_bash"
-cd activities/demo-bash
+cd activities/demo-bash || exit
 ls

chapters/scratch-linux/welcome-to-linux/drills/tasks/whats-running/README.md

@@ -1,25 +1,22 @@
-my_dict = {
-    "SSS": "Rullz",
-    "Essentials": 10,
-    True: 0.2,
-    2.2: 99
-}
+# SPDX-License-Identifier: BSD-3-Clause
 
-print(my_dict['SSS'])  # Will print "Rullz"
+my_dict = {"SSS": "Rullz", "Essentials": 10, True: 0.2, 2.2: 99}
+
+print(my_dict["SSS"])  # Will print "Rullz"
 print(my_dict[True])  # Will print 0.2
 # And so on...
 
 # Keys and values can be added or modified simply by referencing them.
-my_dict['CTF'] = 'pwned'
+my_dict["CTF"] = "pwned"
 print(f'Added key CTF with value {my_dict["CTF"]}')
 
 # You can also modify values in the same way.
 my_dict[True] = 2.22
-print(f'After the modfication: my_dict[True] = {my_dict[True]}')
+print(f"After the modfication: my_dict[True] = {my_dict[True]}")
 
 # You can check if a key is in a dictionary just like you would do for a list:
 print(f'Is "SSS" in my_dict? {"SSS" in my_dict}')
 print(f'Is "bananas" in my_dict? {"bananas" in my_dict}')
 
 # Accessing the value of a non-existent key results in a KeyError.
-print(my_dict['bananas'])
+print(my_dict["bananas"])

chapters/scratch-linux/welcome-to-linux/drills/tasks/whats-running/solution/solution.sh

@@ -1,20 +1,22 @@
-fin = open('../../README.md')  # Use the default mode: rt = read + text
+# SPDX-License-Identifier: BSD-3-Clause
+
+fin = open("../../README.md")  # Use the default mode: rt = read + text
 
 # Use the readline function to, well... read one line from the opened file.
 line = fin.readline()
-print(line, end='')  # The line we've just read already contains a trailing \n.
+print(line, end="")  # The line we've just read already contains a trailing \n.
 
 # The file object (f) maintains an internal cursor, which is an offset from the
 # beginning of the file, where the next operation (read/write) will take place.
 # For this reason, the next calls to readline return the correct lines in the
 # file.
-print(fin.readline(), end='')  # Second line is empty
-print(fin.readline(), end='')  # Third line
+print(fin.readline(), end="")  # Second line is empty
+print(fin.readline(), end="")  # Third line
 
 # There are situations where it is necessary to move the cursor ourselves.
 # For this, we have the the seek function.
 # First, let's see where our cursor is now.
-print(f'fin cursor is initially at {fin.tell()} bytes')
+print(f"fin cursor is initially at {fin.tell()} bytes")
 
 # The seek function takes 2 parameters:
 #    - offset: The number of bytes to move the cursor. Can be negative to move the cursor back
@@ -23,19 +25,19 @@
 #        - 1: from the cursor's current position
 #        - 2: from the end of the file.
 fin.seek(0, 0)  # Move the cursor to the beginning of the file
-print(f'after fin.seek(0, 0), fin cursor is at {fin.tell()} bytes')
+print(f"after fin.seek(0, 0), fin cursor is at {fin.tell()} bytes")
 # Let's try to read a line and see if it truly is the first.
-print('First line of README.md:', fin.readline(), end='')  # It is.
+print("First line of README.md:", fin.readline(), end="")  # It is.
 
 # The other modes are available only when opening the file in binary mode.
 # We will do this in the next section.
 
 # It's good practice to close a file when you no longer use it.
 fin.close()
 
-fout = open('output.txt', 'w')  # Open output.txt for writing text
+fout = open("output.txt", "w")  # Open output.txt for writing text
 # Unlike print, write does not add a \n character at the end of our text.
-fout.write('This is the first line in output.txt\n')
-fout.write('SSS Rulz!\n')
+fout.write("This is the first line in output.txt\n")
+fout.write("SSS Rulz!\n")
 
 fout.close()

chapters/scratch-linux/welcome-to-linux/drills/whats-running/sol/solution.sh

@@ -1,6 +1,9 @@
+# SPDX-License-Identifier: BSD-3-Clause
+
+
 # The code inside the function isn't run when we run the script.
 # It's only run when specifically called:
-
+#  SPDX-License-Identifier: BSD-3-Clause
 def replicate(s, n=3):
     """
     This is a docstring comment. They are used in order to document the
@@ -14,46 +17,51 @@ def replicate(s, n=3):
             n (int): the number of times to replicate it. 3 by default
 
         Returns:
-	    String made up of s replicated n times
+            String made up of s replicated n times
     """
-    acc = ''
+    acc = ""
 
     for _ in range(n):
         acc += s
 
     return acc
 
+
 # Implementation using list comprehensions and the join function.
 # Check the links at the end of the lecture for details
 def replicate_fancy(s, n=3):
-    return ''.join([s for _ in range(n)])
+    return "".join([s for _ in range(n)])
 
-my_string = 'SSS Rulz!'
+
+my_string = "SSS Rulz!"
 my_replicated_string = replicate(my_string, 4)
-print(f'Initial string is unchanged: {my_string}')
-print(f'Returned string is: {my_replicated_string}')
+print(f"Initial string is unchanged: {my_string}")
+print(f"Returned string is: {my_replicated_string}")
 
 # We can also specify any function parameter by name, like so:
 my_replicated_string = replicate(s=my_string, n=2)
-print(f'New returned string is: {my_replicated_string}')
+print(f"New returned string is: {my_replicated_string}")
 
 # The print function can also take some default parameters, such as end:
 # end is the string to be placed at the end of what's printed.
 # By default it's \n.
 print(my_string, end="$$$\n")  # This now prints 'SSS Rulz!'$$$
 
+
 # Similarly, range also takes 2 default parameters: start=0 and step=1.
 # As a result, its signature could look like this:
 def range(end, start=0, step=1):
     pass  # we can use this keyword when we want to leave functions unimplemented
 
+
 # Functions can also have multiple return variables
 def multiple_returns(ret_type):
     if ret_type == 1:
-	    return 1
+        return 1
     else:
-	    return 'there you go'
+        return "there you go"
+
 
-print('\nMultiple Returns')
-print(f'Function returning an int: {multiple_returns(1)}')
+print("\nMultiple Returns")
+print(f"Function returning an int: {multiple_returns(1)}")
 print(f'Same function returning a string: "{multiple_returns(0)}"')

chapters/scratch-linux/welcome-to-linux/drills/demo-bash/demo.sh chapters/scratch-linux/welcome-to-linux/guides/demo-bash/demo.sh

@@ -1,38 +1,40 @@
+# SPDX-License-Identifier: BSD-3-Clause
+
 if False:
-    print('This code is not executed')
+    print("This code is not executed")
 if True:
-    print('This code is executed')
+    print("This code is executed")
 
 a = 0
 b = 1
 
 if a == 0:
-    print('This is a correct if statement')
-b = 2 
+    print("This is a correct if statement")
+b = 2
 # The line above is not indented at the same level as the one above.
 # Thus, it is not contained in the body of the above if statement.
 
 # Python does not have a switch instruction.
 # In order to create something similar to it, you'll have to use elifs:
 if b == 1:
-    print('b = 1')
+    print("b = 1")
 elif b == 2:
-    print('b = 2')
+    print("b = 2")
 elif b == 3:
-    print('b == 3')
+    print("b == 3")
 else:
-    print('You get the point')
+    print("You get the point")
 
 # In order to create more complex if conditions Python does not use && and ||.
 # It simply uses the keywords and, or.
 if a == 0 and b == 2:
-    print('This is a correct if condition: a == 0 and b == 2')
+    print("This is a correct if condition: a == 0 and b == 2")
 if a == 1 or b == 2:
-    print('This is another correct if condition: a == 1 and b == 2')
+    print("This is another correct if condition: a == 1 and b == 2")
 
 # Similarly, in order to negate an expression, we use the not keyword.
 if not a:  # Equivalent to if (!a) in C
-    print('a == 0, thus not a is True')
+    print("a == 0, thus not a is True")
 
 # To create even more complex boolean expressions you can use parentheses,
 # just like you would for arithmetic expressions.