https://github.com/andream16/mitmcracker

Second version of go-crackerino. More performant and more idiomatic go.
https://github.com/andream16/mitmcracker

cracker go go-routine golang meet-in-the-middle-attacks security

Last synced: 9 days ago
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Second version of go-crackerino. More performant and more idiomatic go.

• Host: GitHub
• URL: https://github.com/andream16/mitmcracker
• Owner: andream16
• Created: 2019-05-01T15:50:00.000Z (over 5 years ago)
• Default Branch: master
• Last Pushed: 2023-02-25T07:03:56.000Z (over 1 year ago)
• Last Synced: 2023-02-27T23:02:26.366Z (over 1 year ago)
• Topics: cracker, go, go-routine, golang, meet-in-the-middle-attacks, security
• Language: OCaml
• Size: 337 KB
• Stars: 0
• Watchers: 2
• Forks: 0
• Open Issues: 1
• Metadata Files:
  • Readme: README.md

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README

        
# mitmcracker ![](https://stuff.mit.edu/afs/sipb/project/golang/arch/go1.2.1-linux-amd64/favicon.ico)

## Golang Key Cracker Which Implements Meet In the Middle Strategy.

Given a known `cipher-text` and a known `plain-text`, it finds two keys `k1` and `k2` in a way that 

`Ek1(plain) == Dk2(cipher)`

where `E` is the encryption function and `D` is the decryption function.

It tries all the possible keys in the range of `24/28/32 bits` and prints which are the keys to be used for the encryption and decryption.

Keys can only be `6/7/8 Digits HEX` and `Cipher/Plain` can only be `16 Digits HEX`.

## What you need to run the project:

- Go: ~v1.15

- Ocaml: ~v4.08.1

## How to run the app

From the root of the project:

- build the go/ocaml binaries: `make`

- run the command: `./mitmcracker -key=$keyLength -encoded=$encodedMsg -plain=$plainMsg`

## Known Inputs/Outputs

- 24-bit:

    - plain: C330C9CBD01DFBA0

    - encoded: E10C65124518DB05

    - encoding key to be found: 6d3952

    - decoding key to be found: 513346

- 28-bit:

    - plain: 492A5BB83F8A3F95

    - encoded: 47090A6AC4A56798

    - encoding key to be found: ???????

    - decoding key to be found: ???????

- 32-bit:

    - plain: FFC7C9E5694ABFF7

    - encoded: 98AC59F25448FFAC

    - encoding key to be found: ????????

    - decoding key to be found: ????????

When `Ek1(C) == Dk2(P)` you found the right keys `k1, k2` to read all encrypted messages.