# CSE 545 - Fall '24 pwn.college Dojo

## Project 01 Linux Lifter

### .05 - find

- `find / randomly_placed_file` - way too many files
- read the man page. `find -name randomly_placed_file` found it
- didn't specify a folder to search in tho, ig it's cuz cwd is /

### .06 - find and exec

- "Optional Exercise: Why do they think it worked with `-exec` parameter of the `find` command, but we get permission denied using standalone `cat` command? Hint: SUID bit was set for the `find` command."
- indeed, we see that `/usr/bin/find` has its _setuid_ bit set:
  ![-rwsr-xr-x 1 root root 320160 Feb 18  2020 /usr/bin/find*](ss1.png)
- [see here](https://unix.stackexchange.com/a/389706/595039) for find stuff
- `find / -name random_cant_flag -exec cat {} ';'` worked

### .07 - return code

- `$?` is the return code of the last executed command
- range 0 to 255

### .08 - python

- SUID on python this time

### .11 - search me

- `/challenge/tester.sh` is printing `/flag` but the file is missing
- `/challenge/cp` has SUID bit set
- preliminary find revealed a possible file deep in `/tmp`
- `find /tmp/that/full/path -name flag -exec /challenge/cp {} /flag ';'`

### .12 - hash it out

- used online tool to generate SHA256

### .13 - hash full

- here we go
- a-z, 6 spaces, so 26^6 possibilities
- plaintext is 6 letters, so 48 bits. hash is SHA256 so 256 bits.
- storage per line:`<hash><plaintext>` that's 304 bits, 312 if including newline character
- total storage exceeds 11GB!!
- refinement 1: 256-bit hash is pretty unique. if we cut down on the portion of the hash stored, we should be able to save a ton of space while only slightly increasing the margin of error. let's assume plaintext has to be stored entirely for now, so total per line is 184 bits.
- eh fk it, just generated all permutations. 22GB storage, 20 min to generate, search using VSCode search took a few more minutes

## Project 02 Unwinding Binaries (Reversing)

### .01 - looking inside

- not sure how to use ghidra, didn't seem to work either
- `angr decompile /challenge/run` revealed a `strcmp` with the key, ez

### .02 - the mangler

- 'mangling' is just subtracting 3 from the char's ascii value. so just add 3 to the key

### .03 - xor plus

- mangling is adding 3 then xor with 2. so just xor with 2, then subtract 3

### lab 2a.02

![lab code snippet](ss2.png)

- ascii values

### .04 - solve for x

- NOTE: angr screwed up, and gave an incorrect result (== instead of !=)
- use ghidra (GUI) or [dogbolt](https://dogbolt.org) for binaries under 2MB
- anyway, math solving:
  - we get a few eqns:
    - v1 = v0 - 24223
    - v3 = 5v2 - 129519
  - use these eqns to reduce from brute-force 4 nested loops to 2 nested loops
  - then verifying the rest gets us one soln
- runtime < 3 seconds

### .05 - extra verification

- angr just straight up hangs lol
- holy sh\*t so many if statements
- boils down to byte by byte, check 1 or 0, check +ve or -ve (MSB)
  - 00 - 00110111
  - 01 - 01000111
  - 02 - 01000011
  - 03 - 01010110
  - 04 - 00110100
  - 05 - 01010010
  - 06 - 01011010
  - 07 - 01001001
  - 08 - 01000001
  - 09 - 00110100
  - 10 - 01011001
  - 11 - 00111000
  - 12 - 01111001
  - 13 - 00110011
  - 14 - 01110011
  - 15 - 01001000
  - 16 - 00110101
  - 17 - 00111000
  - 18 - 01101010
  - 19 - 01010111 (binary ninja and hex-rays disagreed on this, binary ninja was right)
- could have automated this smh

### .06 - extra verification II

- first ordered all if statements to get bitwise order of the string (hell.)
- for result to be 0 at the end, just don't modify it at all
- so for each if statement, check which of 0/1 makes it false (find and replace ftw)
- ascii string is 67kW6YnKvTpaqoBX1F8l
- really should have automated this

### .07 - binary labyrinth

- omg it's literally labyrinth navigation, using wasd keys LMFAO

```c
int M[12][12]={
  { 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1},
  { 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0},
  { 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1},
  { 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1},
  { 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1},
  { 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1},
  { 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1},
  { 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1},
  { 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1},
  { 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 1},
  { 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1},
  { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}
};
```

- 1s are landmines
- start from row 1, column 2 (x=1, y=0)
- goal is row 2, column 12 (x=0xb, y=1)
- ssssdsssddsssdddwwwddwwwwdwwd
- lol

## Project 03 Hacking Network Highways

### .01 - netcat

```bash
nc 10.0.0.3 31337
```

### .02 - netcat listener

```bash
nc -l 31337
```

### .03 - nmap and netcat

```bash
nmap 10.0.0.0-255 # found in .142
nc 10.0.0.142 31337
```

### .04 - nmap in parallel and netcat

- `-sn` for ARP ping scan - no ports just discover host
- `--min-parallelism 10` for at least 10 probes at a time
- consider using `-T4` or `-T5` timing templates
- checked
  - `10.0.0.0/19` - only us at .2
  - `10.0.32.0/19` - nothing
  - `10.0.64.0/19` - 10.0.90.244 and port is 31337 as expected. stopped here

### .05 - tcpdump

- `tcpdump -A 'tcp port 31337'`
  - `-A` to print content as ASCII

### .06 - tcpdump and flow

- inspecting the /challenge/run python script, we see that it's sending one character at a time, after encoding them
- `tcpdump -s 65535 -nntA 'tcp port 31337' -w /home/hacker/my_pcaps/3.06.pcap`
  - `-s` to grab full packet (?)
  - `-nn` to avoid resolution of hostnames or port numbers
  - `-t` to exclude timestamp
  - `-A` to print content as parsable ASCII. important!!!
- then we use scapy to read the packets, skip alternating duplicates, decode, and form a single string
- ehh i messed up something but whatever

### .07 - mimic and listen

- `ip addr add 10.0.0.2 dev eth0` assign the address to us, fake
- `nc -l 10.0.0.2 31337`

### .08 - ether scapy

- jfc
- ALWAYS be explicit and define the src addresses
- didn't define the src MAC addr, so packets kept going thru `lo` instead of `eth0`
- too stupid to realize it in time too
- anyway, get current MAC addr of `eth0`
- craft Ether packet to given dest addr with type `0xFFFF`
- `srp(pkt, iface='eth0')`

### lab 3a was chill, no notes

### .09 - IP scapy

- similar
- set IP addr with `ifconfig eth0 10.0.0.2`
- add l3 with src and dest IP addr, `proto=0xFF`
- since we need MAC as well, use `srp`, not `sr`

### .10 - TCP scapy

- similar
- again, set IP addr
- add l4 with src and dest TCP port, `flags=0x1F` to set ACK (0x10), PSH (0x08), RST (0x04), SYN (0x02), FIN (0x01) flags
- `srp` again

### .11 - TCP handshake

- send SYN with specified seq and ack numbers - 31337 both
- get SYNACK
  - has ack of 31338, which will be our next syn
  - has random syn, add 1 to get next ack
- send ACK with next syn and ack numbers

### .12 - ARP scapy

- meh, arp opcode is 2

### .13 - ARP spoofing

- meh, just crafting ARP and tcpdump

### lab 3b was chill, no notes

### .14 - MiTM ARPing

- shit's getting too easy, let's not look at /challenge/run
- first, get target's macs, then arp spoof
- we don't have NET_ADMIN, so can't set ip_forward in sysctl to control MITM directly,
- first, capture packets, check raw loads
- we observe that a sequence repeats:
  - 10.0.0.3:31337 sends a command: "SECRET", to 10.0.0.4 at a random port
    - note: how does 3 know which port to send to?
      - [after 3d] idiot, 4 opens the tcp handshake
  - 4 responds with a secret, it's in ascii?
  - 3 sends a list of available (?) commands - echo, flag, and then asks for a command
  - 4 responds with echo, and sends "Hello, World!"
  - 3 echoes it back
- connection closes, repeats with another randomized port for 4
- note that 3 sends a secret and a list of commands that includes a flag command
- craft a packet masquerading as 4, with the flag command, wait for a secret to arrive and put it in the packet
  - [after 3d] idiot, read the code, you don't need the secret, just hijack the connection
- in the time it takes 3 to do the legitimate echo from 4, we could probably send the flag command to 3 and have it processed in the same ephemeral connection
- let's try

### lab 3c was chill, no notes

### lab 3d

#### .2 - mitm arping

- same as 3.14, approaching this first for deadline
- client at 3.13.37.4, random port
- server at 3.13.37.3, port 1992
- flow:
  - TCP handshake:
    - client -> SYN -> server
    - server -> SYNACK -> client
    - client -> ACK -> server
  - secret is sent:
    - server -> PUSHACK -> asks for secret -> client
    - client -> ACK, then PUSHACK -> secret string \n-> client
    - server -> ACK, then PUSHACK -> secret confirmed -> client
      - at this point, inject BACKDOOR packet before the actual client
    - client -> ACK, then PUSHACK -> ECHO: -> server
  - after backdoor, send a FLAG packet

### going back to continue 3.14 with this understanding

---

- [after 3d] updated understanding
  - client at 10.0.0.4, random port
  - server at 10.0.0.3, port 31337
  - flow:
    - TCP handshake:
      - client -> SYN -> server
      - server -> SYNACK -> client
      - client -> ACK -> server
    - secret is sent:
      - server -> PUSHACK -> asks for secret -> client
      - client -> ACK, then PUSHACK -> secret string \n-> client
      - server -> ACK, then PUSHACK -> list of commands -> client
        - at this point, inject FLAG command before the actual client
      - client -> ACK, then PUSHACK -> ECHO -> server

## Project 04 Hijacking Binary Power (Pwning)

- seems we have access to the source code, and we're given a suid-set executable

### .02 - exec them all

- title helped
- `exec -a <passwd> /challenge/run`

### .03 - altering arg[0]

- +3 lops off first 3 chars

### .04 - symmer

- symlink /flag to ~/flag

### .05 - when is a secret not secret

### .10 - somewhere over the rainbow

- online tool

### .11 - byte compare

- this strncmp takes the lower length (doesn't take null tho), so just give it a single byte
- only 256 possible values, bruteforce

```bash
for i in $(seq 0 255); do
  i_chr=$(printf "\x$(printf "%x" "$i")")
  /challenge/run $i_chr
done
```

### .12 - symmer in time

- 5 second window
- initially have a dummy `~/flag`, run the challenge, within 5 seconds delete it and create it as a symlink to `/flag`

### .13 - time after time

- 2 second window
- creates tmp files, writes target to one, sleeps for 2 secs, then reads from it and compares with passwd checksum
- have `umask 002 ; echo <checksum> > /tmp/hash_output_1000_<randnum>` in one shell ready for tab completion of the random number part
- run `/challenge/run something` in another shell, then run the above

### .14 - controlling your path

- make sure PATH is set so that it uses your program
- don't specify a shell so that it uses `/bin/sh` - [see here](https://www.qnx.com/developers/docs/6.5.0SP1.update/com.qnx.doc.neutrino_lib_ref/e/execlp.html#:~:text=If%20the%20process%20image%20file)
  > "If the process image file isn't a valid executable object, the contents of the file are passed as standard input to a command interpreter conforming to the system() function. In this case, the command interpreter becomes the new process image."
- i assume the command interpreter that gets used has the SUID bit

### .15 - blind leading the blind

- basically, stdout and stderr for the child are set to `/dev/null` so instead of spawning root shell, use `cat flag > output` and read output

### .16 - arg wars VI - return of the hacker

- decompiler showed set of filtered characters, quotes and backslashes are not there
- also .17 checks for backslashes, so i assume backslashes solves this
- but i got stuck, TA said try the 'prequels' first then come back lol

### lab 4a.1 - easy overflow

- standard buffer overflow vuln
- gdb shenanigans
- shift-ctrl-@ inserts a null character it seems (remember for .16)
- enough gdb, let's move to big guns - pwntools
- checksec says no stack canary or PIE
- all g then
- calculate offset from vulnerable variable location to saved RIP(return instruction pointer) location
- get address of target function to execute
- craft payload accordingly