pico22: Beginner's Compilation (WIP)

Binary Exploitation

basic-file-exploit

solver: enscribe
author: Will Hong
genre: pwn/binary
points: 100
files: program-redacted.c

The program provided allows you to write to a file and read what you wrote from it. Try playing around with it and see if you can break it! Connect to the program with netcat:
$ nc saturn.picoctf.net 50366

Hints:

Try passing in things the program doesn't expect. Like a string instead of a number.

Let’s connect to the server using netcat to see what’s going on:

$ nc saturn.picoctf.net 50366
Hi, welcome to my echo chamber!
Type '1' to enter a phrase into our database
Type '2' to echo a phrase in our database
Type '3' to exit the program

Since this is the binary exploitation category, we’ll be looking for a vulnerability in the source code that allows us to either break or control the program at a lower level. Let’s view the attachment program-redacted.c:

program-redacted.c[download source]

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <stdint.h>
#include <ctype.h>
#include <unistd.h>
#include <sys/time.h>
#include <sys/types.h>

#define WAIT 60

static const char* flag = "[REDACTED]";

static char data[10][100];
static int input_lengths[10];
static int inputs = 0;

int tgetinput(char *input, unsigned int l)
{
    fd_set          input_set;
    struct timeval  timeout;
    int             ready_for_reading = 0;
    int             read_bytes = 0;
    
    if( l <= 0 )
    {
      printf("'l' for tgetinput must be greater than 0\n");
      return -2;
    }
    
    
    /* Empty the FD Set */
    FD_ZERO(&input_set );
    /* Listen to the input descriptor */
    FD_SET(STDIN_FILENO, &input_set);

    /* Waiting for some seconds */
    timeout.tv_sec = WAIT;    // WAIT seconds
    timeout.tv_usec = 0;    // 0 milliseconds

    /* Listening for input stream for any activity */
    ready_for_reading = select(1, &input_set, NULL, NULL, &timeout);
    /* Here, first parameter is number of FDs in the set, 
     * second is our FD set for reading,
     * third is the FD set in which any write activity needs to updated,
     * which is not required in this case. 
     * Fourth is timeout
     */

    if (ready_for_reading == -1) {
        /* Some error has occured in input */
        printf("Unable to read your input\n");
        return -1;
    } 

    if (ready_for_reading) {
        read_bytes = read(0, input, l-1);
        if(input[read_bytes-1]=='\n'){
        --read_bytes;
        input[read_bytes]='\0';
        }
        if(read_bytes==0){
            printf("No data given.\n");
            return -4;
        } else {
            return 0;
        }
    } else {
        printf("Timed out waiting for user input. Press Ctrl-C to disconnect\n");
        return -3;
    }

    return 0;
}


static void data_write() {
  char input[100];
  char len[4];
  long length;
  int r;
  
  printf("Please enter your data:\n");
  r = tgetinput(input, 100);
  // Timeout on user input
  if(r == -3)
  {
    printf("Goodbye!\n");
    exit(0);
  }
  
  while (true) {
    printf("Please enter the length of your data:\n");
    r = tgetinput(len, 4);
    // Timeout on user input
    if(r == -3)
    {
      printf("Goodbye!\n");
      exit(0);
    }
  
    if ((length = strtol(len, NULL, 10)) == 0) {
      puts("Please put in a valid length");
    } else {
      break;
    }
  }

  if (inputs > 10) {
    inputs = 0;
  }

  strcpy(data[inputs], input);
  input_lengths[inputs] = length;

  printf("Your entry number is: %d\n", inputs + 1);
  inputs++;
}


static void data_read() {
  char entry[4];
  long entry_number;
  char output[100];
  int r;

  memset(output, '\0', 100);
  
  printf("Please enter the entry number of your data:\n");
  r = tgetinput(entry, 4);
  // Timeout on user input
  if(r == -3)
  {
    printf("Goodbye!\n");
    exit(0);
  }
  
  if ((entry_number = strtol(entry, NULL, 10)) == 0) {
    puts(flag);
    fseek(stdin, 0, SEEK_END);
    exit(0);
  }

  entry_number--;
  strncpy(output, data[entry_number], input_lengths[entry_number]);
  puts(output);
}


int main(int argc, char** argv) {
  char input[3] = {'\0'};
  long command;
  int r;

  puts("Hi, welcome to my echo chamber!");
  puts("Type '1' to enter a phrase into our database");
  puts("Type '2' to echo a phrase in our database");
  puts("Type '3' to exit the program");

  while (true) {   
    r = tgetinput(input, 3);
    // Timeout on user input
    if(r == -3)
    {
      printf("Goodbye!\n");
      exit(0);
    }
    
    if ((command = strtol(input, NULL, 10)) == 0) {
      puts("Please put in a valid number");
    } else if (command == 1) {
      data_write();
      puts("Write successful, would you like to do anything else?");
    } else if (command == 2) {
      if (inputs == 0) {
        puts("No data yet");
        continue;
      }
      data_read();
      puts("Read successful, would you like to do anything else?");
    } else if (command == 3) {
      return 0;
    } else {
      puts("Please type either 1, 2 or 3");
      puts("Maybe breaking boundaries elsewhere will be helpful");
    }
  }

  return 0;
}

In the midst of this complex program, we need to figure out where the flag is, and how to trigger it to print:

static const char* flag = "[REDACTED]";
(14-138)
if ((entry_number = strtol(entry, NULL, 10)) == 0) {
    puts(flag);
    fseek(stdin, 0, SEEK_END);
    exit(0);
}

The flag is defined in line 13 as "[REDACTED]", which will be the actual location on the remote server. From lines 139-143 it looks like a condition needs to be met in order to puts() the flag, which writes a string to the output stream stdout.

Google defines strtol() as a function that “converts the initial part of the string in str to a long int value according to the given base“. To break it, we need to input something that is unconvertible into a long integer. In this case, it would be a string, as they can’t be properly coalesced into long integers!

This if statement is located within a function called data_read(). Let’s see where it’s called in the program:

} else if (command == 2) {
     if (inputs == 0) {
       puts("No data yet");
       continue;
     }
     data_read();
     puts("Read successful, would you like to do anything else?");

After we write some data with the command 1, We should be pressing the command 2 to read from the stored data. Once it prompts us to “enter the entry number of your data”, we’ll send a string instead to break it. Let’s head back to the netcat and test it out:

$ nc saturn.picoctf.net 50366
Hi, welcome to my echo chamber!
Type '1' to enter a phrase into our database
Type '2' to echo a phrase in our database
Type '3' to exit the program
> 1
1
Please enter your data:
> hello
hello
Please enter the length of your data:
> 5
5
Your entry number is: 1
Write successful, would you like to do anything else?
> 2
2
Please enter the entry number of your data:
> "NO!"
"NO!"
picoCTF{M4K3_5UR3_70_CH3CK_Y0UR_1NPU75_[REDACTED]}

---

CVE-XXXX-XXXX

solver: enscribe
author: Mubarak Mikail
genre: osint, pwn (?)
points: 100

Enter the CVE of the vulnerability as the flag with the correct flag format - picoCTF{CVE-XXXX-XXXXX} - replacing XXXX-XXXXX with the numbers for the matching vulnerability. The CVE we're looking for is the first recorded remote code execution (RCE) vulnerability in 2021 in the Windows Print Spooler Service, which is available across desktop and server versions of Windows operating systems. The service is used to manage printers and print servers.

Hints:

We're not looking for the Local Spooler vulnerability in 2021…

This is a really trivial challenge. You can actually google “first recorded remote code execution (RCE) vulnerability in 2021” and it will be the first result:

The flag is picoCTF{CVE-2021-34527}.

---

Cryptography

basic-mod1

solver: enscribe
author: Will Hong
genre: crypto, prog
points: 100

We found this weird message being passed around on the servers, we think we have a working decryption scheme.
Take each number mod 37 and map it to the following character set - 0-25 is the alphabet (uppercase), 26-35 are the decimal digits, and 36 is an underscore. Wrap your decrypted message in the picoCTF flag format (i.e. picoCTF{decrypted_message})

Hints:

Do you know what mod 37 means?
mod 37 means modulo 37. It gives the remainder of a number after being divided by 37.

Let’s go over what it’s asking:

• Calculate % 37 for each number
• Map each number to this specific charset:
  • 0-25 = Uppercase alphabet (A-Z)
  • 26-35 = Decimal digits (0-9)
  • 36 = Underscore (“_”)

I was too lazy to learn Python and do that, so here it is in native Javascript:

// Splitting into array
x = "54 211 168 309 262 110 272 73 54 137 131 383 188 332 [REDACTED]".split();
// Mod 37
y = x.map(x => x % 37);
z = [];
for (let i = 0; i < y.length; i++) {
    // Mapping to charset
    if (y[i] >= 0 && y[i] <= 25) {
      z.push(String.fromCharCode(y[i] + 'A'.charCodeAt(0)));
    } else if (y[i] >= 26 && y[i] <= 35) {
      z.push(y[i] - 26);
    } else if (y[i] == 36) {
      z.push("_");
    }
}
// Combine back to string
z = z.join("");
console.log(`picoCTF{${z}}`);

$ node solve.js
picoCTF{R0UND_N_R0UND_[REDACTED]}

Looking back at the problem after I learned Python, here’s a solution that’s significantly cleaner:

#!/usr/bin/env python3
import string
x = "54 211 168 309 262 110 272 73 54 137 131 383 188 332 [REDACTED]"
y = x.split()

a = string.ascii_uppercase + string.digits + "_"

# Insane list comprehension
z = [a[int(i) % 37] for i in y]
print("picoCTF{"+''.join(z)+"}")

$ python3 solve.py
picoCTF{R0UND_N_R0UND_[REDACTED]}

---

basic-mod2

solver: enscribe
author: Will Hong
genre: crypto, prog
points: 100

A new modular challenge! Take each number mod 41 and find the modular inverse for the result. Then map to the following character set - 1-26 are the alphabet, 27-36 are the decimal digits, and 37 is an underscore. Wrap your decrypted message in the picoCTF flag format (picoCTF{decrypted_message}).

Hints:

Do you know what the modular inverse is?
The inverse modulo z of x is the number, y that when multiplied by x is 1 modulo z.
It's recommended to use a tool to find the modular inverses.

Let’s go over what it’s asking once again:

• Calculate % 41 for each number
• Map each number to this specific charset:
  • 1-26 = Uppercase alphabet (A-Z)
  • 27-36 = Decimal digits (0-9)
  • 37 = Underscore (“_”)

Here’s a stupidly long Javascript snippet I made to solve this:

// Splitting into array
x = "54 211 168 309 262 110 272 73 54 137 131 383 188 332 [REDACTED]".split();
// Mapping to % 41 with modular inverse of 41
y = x.map(x => x % 41).map(x => modInverse(x, 41));
z = [];

// Mapping to charset
for (let i = 0; i < y.length; i++) {
    if (y[i] >= 1 && y[i] <= 26) z.push(String.fromCharCode(y[i] + 64));
    else if (y[i] >= 27 && y[i] <= 36) z.push(y[i] - 27);
    else if (y[i] == 37) z.push("_");
}

console.log(`picoCTF{${z.join("")}}`);

// credit to: https://rosettacode.org/wiki/Modular_inverse
function modInverse(a, b) {
    a %= b;
    for (var x = 1; x < b; x++) {
        if ((a * x) % b == 1) {
            return x;
        }
    }
}

$ node solve.js
picoCTF{1NV3R53LY_H4RD_[REDACTED]}

---

credstuff

solvers:
- MrTea
- enscribe
authors:
- Will Hong
- Lt. 'Syreal' Jones
genre: crypto
points: 100
files: leak.tar

We found a leak of a blackmarket website's login credentials. Can you find the password of the user cultiris and successfully decrypt it?
The first user in usernames.txt corresponds to the first password in passwords.txt. The second user corresponds to the second password, and so on.

Hints:

Maybe other passwords will have hints about the leak?

We’re initially provided a leak.tar archive. On extraction, we’re presented with two files: usernames.txt and passwords.txt:

usernames.txt

engineerrissoles
icebunt
fruitfultry
celebritypentathlon
galoshesopinion
favorboeing
bindingcouch
...

passwords.txt

CMPTmLrgfYCexGzJu6TbdGwZa
GK73YKE2XD2TEnvJeHRBdfpt2
UukmEk5NCPGUSfs5tGWPK26gG
kaL36YJtvZMdbTdLuQRx84t85
K9gzHFpwF2azPayAUSrcL8fJ9
rYrtRbkHvJzPmDwzD6gSDbAE3
kfcVXjcFkvNQQPpATErx6eVDd
...

Let’s go to the username cultiris. The -n tag in grep will enable line numbers:

$ grep -n cultiris usernames.txt
378:cultiris

Let’s fine the equivalent line in passwords.txt:

...
ARKadGaCZBc3ue4BfB7Vjwx83
CSYbRFVpJZNQJ4Jz3GmDsAa9Q
cvpbPGS{P7e1S_54I35_71Z3}
wTL8rTRNCkSyGP5AFsG5qK52y
9jyG4W6PnsAVuyx8MJkHKYtXV
...

On line 378 it looks like there’s a flag obfuscated with shift cipher. Let’s brute force this on DCode:

Results - Brute-Force mode (Caesar)

🠞15 (🠜11)	ngamARD{A7p1D_54T35_71K3}
🠞1 (🠜25)	buoaOFR{O7d1R_54H35_71Y3}
🠞17 (🠜9)	leykYPB{Y7n1B_54R35_71I3}
🠞24 (🠜2)	exrdRIU{R7g1U_54K35_71B3}
🠞11 (🠜15)	rkeqEVH{E7t1H_54X35_71O3}
🠞13 (🠜13)	picoCTF{C7r1F_54V35_71M3}

The flag is picoCTF{C7r1F_54V35_71M3}!

---

morse-code

solver: enscribe
author: Will Hong
genre: crypto
points: 100
files: morse_chal.wav

Morse code is well known. Can you decrypt this?
Wrap your answer with picoCTF{}, put underscores in place of pauses, and use all lowercase.

We’re presented with a morse_chal.wav file:

We could totally decode this by hand using Audacity’s visualizer, but that’s super time-consuming. Instead, I opted for an automatic audio-based Morse decoder online:

The program outputs WH47 H47H 90D W20U9H7. Following the conversion instructions, the final flag is picoCTF{wh47_h47h_90d_w20u9h7}.

Fun fact: this string is a leetspoken version of "What hath God wrought", which was the first telegraphed message in Morse!

---

Forensics

Enhance!

solver: enscribe
author: Lt. 'Syreal' Jones
genre: forensics
points: 100
files: [REDACTED]

Download this image file and find the flag.

svg.png

This is an SVG file, which stands for Scalable Vector Graphics. They consist of vectors, not pixels, and can be thought of as a collection of shapes on a Cartesian (x/y) plane. The code that creates such graphics can also be viewed on Google Chrome with F12:

Look up what we end up finding in the Source tab:

<tspan sodipodi:role="line" x="107.43014" y="132.08501" style="font-size:0.00352781px;line-height:1.25;fill:#ffffff;stroke-width:0.26458332;" id="tspan3748">p </tspan>
<tspan sodipodi:role="line" x="107.43014" y="132.08942" style="font-size:0.00352781px;line-height:1.25;fill:#ffffff;stroke-width:0.26458332;" id="tspan3754">i </tspan>
<tspan sodipodi:role="line" x="107.43014" y="132.09383" style="font-size:0.00352781px;line-height:1.25;fill:#ffffff;stroke-width:0.26458332;" id="tspan3756">c </tspan>
<tspan sodipodi:role="line" x="107.43014" y="132.09824" style="font-size:0.00352781px;line-height:1.25;fill:#ffffff;stroke-width:0.26458332;" id="tspan3758">o </tspan>
<tspan sodipodi:role="line" x="107.43014" y="132.10265" style="font-size:0.00352781px;line-height:1.25;fill:#ffffff;stroke-width:0.26458332;" id="tspan3760">C </tspan>
<tspan sodipodi:role="line" x="107.43014" y="132.10706" style="font-size:0.00352781px;line-height:1.25;fill:#ffffff;stroke-width:0.26458332;" id="tspan3762">T </tspan>
<tspan sodipodi:role="line" x="107.43014" y="132.11147" style="font-size:0.00352781px;line-height:1.25;fill:#ffffff;stroke-width:0.26458332;" id="tspan3764">F { 3 n h 4 n </tspan>
<tspan sodipodi:role="line" x="107.43014" y="132.11588" style="font-size:0.00352781px;line-height:1.25;fill:#ffffff;stroke-width:0.26458332;" id="tspan3752">c 3 d _ [R E D A C T E D] }</tspan>

The flag is picoCTF{3nh4nc3d_[REDACTED]} (final part is dynamic 😉).

---