Metadata-Version: 1.2
Name: cryptonita
Version: 0.5.0
Summary: Cryptanalysis swiss army knife
Home-page: UNKNOWN
Author: Di Paola Martin
Author-email: use-github-issues@example.com
License: GNU GPLv3
Description: ``cryptonita`` - Cryptanalysis Swiss Army Knife
        ===============================================
        
        `cryptonita <https://pypi.org/project/cryptonita/>`__ is a set of
        building blocks to create automated crypto-attacks.
        
        You may not find the advanced attack implemented here (yet) but I hope
        that this building blocks or primitives can help you in your journey.
        
        Without more, let’s put our hands on and break the famous `Vigenere
        cipher <https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher>`__.
        
        Tutorial - Break a xor Vigenere cipher
        --------------------------------------
        
        The `Vigenere
        cipher <https://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher>`__ was once
        the most secure cipher. It was thought that it was unbreakable…
        
        Let’s put under test that statement and learn about
        `cryptonita <https://pypi.org/project/cryptonita/>`__ along the journey!
        
           Note: the following README is also an automated test for the
           ``cryptonita`` lib thanks to
           `byexample <https://byexamples.github.io/byexample>`__.
        
        Implement the cipher - Load the bytes
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        The building block in ``cryptonita`` is the *byte string*: a finite
        immutable sequence of bytes.
        
        In ``cryptonita`` we can create a *byte string* with the ``B`` function
        and do any conversion needed:
        
        .. code:: python
        
           >>> from cryptonita import B            # byexample: +timeout=10
           >>> B(u'from an unicode encoded text', encoding='utf-8')
           'from an unicode encoded text'
        
           >>> B([0x46, 0x72, 0x6f, 0x6d, 0x20, 0x6e, 0x75, 0x6d, 0x62, 0x65, 0x72, 0x73])
           'From numbers'
        
           >>> B('RnJvbSBiYXNlNjQ=', encoding=64)
           'From base64'
        
        For our purposes of implementing a Vigenere cipher, let’s load some
        plain text from a file:
        
        .. code:: python
        
           >>> ptext = B(open('./test/ds/plaintext', 'rt').read())
        
           >>> ptext[:29]
           'Now that the party is jumping'
        
        ..
        
           For the full list of conversions see `cryptonita/conv.py’s
           ``as_bytes`` <https://github.com/cryptonitas/cryptonita/tree/master/cryptonita/conv.py>`__
        
        Implement the cipher - Apply a xor
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        First, we load our secret key in base 64 from the file. Notice how the
        decoding from base 64 is made by ``B``:
        
        .. code:: python
        
           >>> secret = B(open('./test/ds/secret', 'rt').read(), encoding=64)
        
        The Vigenere cipher consists in xord the plaintext with the key. If the
        plaintext is larger than the key, just repeat the key over and over.
        
        ``cryptonita`` can do exactly that:
        
        .. code:: python
        
           >>> ctext = ptext ^ secret.inf()
        
           >>> ctext[:29].encode(64)
           b'OA4ZSRgEAAJBGgEJTBEXExoQTAUSVgsbBBwFDxE='
        
        The ``inf()`` method tells that the ``secret`` string must be seen as an
        “infinite sequence”, repeating the key over and over.
        
        Then, the ``^`` just does the xor byte by byte.
        
           For the full list of operation on ``ImmutableByteString`` see
           `cryptonita/bytestrings.py’s
           ``ImmutableByteString`` <https://github.com/cryptonitas/cryptonita/tree/master/cryptonita/bytestrings.py>`__
           and the
           `mixins <https://github.com/cryptonitas/cryptonita/tree/master/cryptonita/mixins.py>`__
        
        Breaking the cipher - Scoring the length of the key
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        Vigenere was thought to be unbreakable because a priori is not possible
        to know the length of the key.
        
        However this was proved to be false.
        
        In 1863, `Kasiski <https://en.wikipedia.org/wiki/Kasiski_examination>`__
        came with a cleaver method to know the length of the key but it is quite
        hard to make it right and faster (I’m still `working on
        it <https://book-of-gehn.github.io/articles/2020/10/11/Kasiski-Test-Part-I.html>`__)
        
        Modern and better approaches are the `Hamming
        distance <https://en.wikipedia.org/wiki/Hamming_distance>`__ and the
        `Index of
        Coincidence <https://book-of-gehn.github.io/articles/2019/10/04/Index-of-Coincidence.html>`__
        
        The idea is to assume that the key is of length L and then pick every
        Lth byte of the ciphertext:
        
        .. code:: python
        
           >>> L = 8 # totally arbitrary here
           >>> picked = ctext[::L]
        
        ..
        
           Note how the ``ImmutableByteString`` ciphertext supports indexing
           operation like any Python string.
        
        Now we compute the Index of Coincidence (IC) of this picked string.
        
        If the assumed length L is **not** the correct one, every picked byte
        will be the xor of the plaintext with a different key byte and the whole
        ``picked`` string would like **random** and the IC will be very low.
        
        On the other hand, if we guessed correctly the length L, **all** the
        picked bytes will be the xord of the plaintext and the **same** key byte
        and therefore will not look random. A high IC would be expected!
        
        .. code:: python
        
           >>> from cryptonita.metrics import icoincidences
           >>> icoincidences(picked)
           0.02<...>
        
        ..
        
           See
           `cryptonita/scoring.py <https://github.com/cryptonitas/cryptonita/tree/master/cryptonita/scoring/score_funcs.py>`__
           and
           `cryptonita/metrics.py <https://github.com/cryptonitas/cryptonita/tree/master/cryptonita/metrics/__init__.py>`__
        
           I you want to know more about the Index of Coincidence see this `blog
           post <https://book-of-gehn.github.io/articles/2019/10/04/Index-of-Coincidence.html>`__
           about it and this `comparison with other
           methods <https://book-of-gehn.github.io/articles/2018/04/01/A-string-of-coincidences-is-not-a-coincidence.html>`__
        
        Breaking the cipher - Guessing the length of the key
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        A IC of 0.02 is too low. It seems that 8 is not the length of the key.
        
        We could do a loop to try other lengths but ``cryptonita`` already has
        that
        
        .. code:: python
        
           >>> from cryptonita.scoring import key_length_by_ic
           >>> from cryptonita.attacks import guess_key_length
        
           >>> gklength = guess_key_length(
           ...                         ctext,
           ...                         length_space=range(5, 25),
           ...                         score_func=key_length_by_ic,
           ...                         min_score=0.025
           ...                         )
        
        Okay, what is that?
        
        -  ``guess_key_length`` does a brute force *attack* computing a *score
           function* testing every possible length from 5 to 25.
        -  ``key_length_by_ic`` is a *scores* how good the tested length is. It
           puts a score between 0 (bad) and 1 (good) using the Index of
           Coincidence (like the ``icoincidences``).
        
        You may think that ``gklength`` is the **the** guessed key but in
        cryptoanalysis you mostly never work with a *specific* value. You work
        with a **set of possible values**.
        
        .. code:: python
        
           >>> gklength
           {5: 0.02702702702702703,
            6: 0.027649769585253458,
            7: 0.04682040531097135,
            8: 0.02682701202590194,
            9: 0.025551684088269456,
            10: 0.025604551920341393,
            12: 0.038306451612903226,
            14: 0.03133903133903134,
            16: 0.028985507246376812,
            17: 0.02766798418972332,
            21: 0.032679738562091505,
            24: 0.041666666666666664}
        
        In ``cryptonita`` we call these sets, these *guesses*, ``FuzzySet``.
        
           For more scoring functions see
           `cryptonita/scoring.py <https://github.com/cryptonitas/cryptonita/tree/master/cryptonita/scoring/score_funcs.py>`__
        
        Breaking the cipher - A guess as a fuzzy set
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        A guess or ``FuzzySet`` is a bunch of possible solutions, each with an
        associated probability or score.
        
        We can query then the most likely answer. In our case, the most likely
        length of the key:
        
        .. code:: python
        
           >>> gklength.most_likely()
           7
        
        But the most likely may not necessary mean the correct answer. Instead,
        you should work always with the fuzzy set to test all of them.
        
        If the sets gets to large (and they will), you can cut them off,
        dropping items with a probability lower than some threshold.
        
        Here we say that any length with a lower probability of 0.01 should be
        out:
        
        .. code:: python
        
           >>> gklength.cut_off(0.03)
           >>> gklength
           {7 -> 0.0468, 24 -> 0.0417, 12 -> 0.0383, 21 -> 0.0327, 14 -> 0.0313}
        
        ..
        
           Take a look at the `documentation of
           ``FuzzySet`` <https://github.com/cryptonitas/cryptonita/tree/master/cryptonita/fuzzy_set.py>`__
           and optional a wiki about `fuzzy set
           theory <https://en.wikipedia.org/wiki/Fuzzy_set>`__.
        
        Breaking the cipher - Chop the ciphertext into blocks
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        Now the we have a set of possible lengths, here is the plan to crack the
        cipher:
        
        First, split the ciphertext into *blocks* of guessed length L:
        
        .. code:: python
        
           >>> L = gklength.most_likely()
           >>> cblocks = ctext.nblocks(L)
        
        ::
        
           ciphertext:  ABCDEFGHIJKLMN
                         |   |    |  |
                         |   |    \  \___
                         |   |     \     \
           cblocks      ABCD  EFGH  IJKL  MN
        
        Each first byte of those blocks are supposedly the result of xor the
        plaintext with the same key byte. The same goes for the second byte of
        each block and so on.
        
        Second, because it is easier to have all the first bytes in one block,
        all the second bytes in another block and so on, we want to *transpose*
        the blocks:
        
        .. code:: python
        
           >>> from cryptonita.conv import transpose
           >>> cblocks = transpose(cblocks, allow_holes=True)
        
        ::
        
            cblocks   --> transposed cblocks
             ABCD           AEIM
             EFGH           BFJN
             IJKL           CGK
             MN             DHL
        
        Now, each block (or row) is a piece of plaintext encrypted with the same
        single-byte key.
        
        Let’s break it!
        
        Breaking the cipher - Frequency attack
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        We could test all the 256 possible byte keys by brute force but that’s
        quite slow.
        
        Rather we could do a *frequency attack* because the statistics of the
        plaintext are leaked into the ciphertext.
        
        ``cryptonita`` already provides us with a very simple *model* of the
        frequencies of the English plaintext: the famous *ETAOIN SHRDLU*.
        
        .. code:: python
        
           >>> from cryptonita.scoring.freq import etaoin_shrdlu
        
        If our ciphertext has the same distribution than the plaintext, at least
        one of the most common bytes in the ciphertext should be one of the most
        common bytes in the plaintext, encrypted of course.
        
        Under this hypothesis ``freq_attack`` xor the top most common bytes in
        the ciphertext with the most common bytes in plaintext according to the
        model.
        
        .. code:: python
        
           >>> most_common_pbytes = etaoin_shrdlu()
           >>> ntop_most_common_cbytes = 1
        
           >>> from cryptonita.attacks import freq_attack
        
           >>> freq_attack(cblocks[0], most_common_pbytes, ntop_most_common_cbytes)
           {'"': 0.07387790762504176,
            '$': 0.055504740275805896,
            '%': 0.0561520934139066,
            '2': 0.03178778752478832,
            '3': 0.10384587375686015,
            '5': 0.026296157563462763,
            '7': 0.07060615929878336,
            '8': 0.060837928943597436,
            '9': 0.0634364224946222,
            ':': 0.0342469273170487,
            '>': 0.03964865941609311,
            '?': 0.06072776315086166,
            'v': 0.17269159612928756}
        
        In general, ``freq_attack`` cannot give us **the** byte key but it can
        give use a *guess*: a fuzzy set of possible keys. This is a much shorted
        list than 256!
        
        But don’t claim victory yet. We broke only the first block
        (``cblocks[0]``).
        
           More frequency models may be found at
           `cryptonita/scoring/freq.py <https://github.com/cryptonitas/cryptonita/tree/master/cryptonita/scoring/freq.py>`__
        
        Breaking the cipher - Guess explosion
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        We need to call ``freq_attack`` for all the blocks:
        
        .. code:: python
        
           >>> gbkeys = []
           >>> for c in cblocks:
           ...     gbkeys.append(freq_attack(c, most_common_pbytes, ntop_most_common_cbytes))
        
           >>> len(gbkeys)
           7
        
        So we have 7 guesses (7 fuzzy sets), one guess set per byte of the key.
        
        But the key is one of the *all possible combination of the guesses*.
        
        How many possible keys do we have?
        
        .. code:: python
        
           >>> from cryptonita.fuzzy_set import len_join_fuzzy_sets
        
           >>> len_join_fuzzy_sets(gbkeys)
           62748517
        
        How! that’s a lot! But still **much less than** 256^7 which is greater
        than the age of the `observable
        universe <https://en.wikipedia.org/wiki/Observable_universe>`__ in
        years.
        
        Still, we need to shrink the guesses even further to make it manageable.
        
        Breaking the cipher - Brute force refinement
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        ``freq_attack`` is really powerful but it is not the only tool that we
        have.
        
        Not all the possible keys in a guess will produce *“reasonable”*
        plaintext.
        
        We can *score* a plaintext and filter out the ones that don’t look
        *“good enough”*
        
        ``cryptonita`` implements different scoring functions and
        ``all_ascii_printable`` is the most simplest to understand:
        
        Let’s *assume* that the plaintext is an English message encoded in
        ASCII.
        
        If we decipher one block and we got a plaintext with non-printable ASCII
        char we can be sure that the key used is incorrect and we can score it
        with a ``0``. Otherwise, we score it with ``1``.
        
        .. code:: python
        
           >>> from cryptonita.scoring import all_ascii_printable
        
           >>> all_ascii_printable(B("a reasonable plaintext"))
           1
        
           >>> all_ascii_printable(B("n\0t v\4lid"))
           0
        
        The plan is to try **all** the possible byte keys in **each** of our
        guesses, score the results and drop the ones with lower score.
        
        .. code:: python
        
           >>> from cryptonita.attacks import brute_force
        
           >>> for i, c in enumerate(cblocks):
           ...     # the fuzzy set of keys (a guess) for this ith byte
           ...     gbkey = gbkeys[i]
           ...
           ...     refined = brute_force(c,
           ...                     score_func=all_ascii_printable,
           ...                     key_space=gbkey,
           ...                     min_score=0.01
           ...                 )
           ...
           ...     # "refined" is another fuzzy set (a guess) for the ith byte
           ...     # but probably a much smaller one
           ...     gbkeys[i] = refined
        
        Like ``guess_key_length``, ``brute_force`` receives a score function, a
        key space and a minimum score.
        
        Now we have a much smaller search space to work on:
        
        .. code:: python
        
           >>> len_join_fuzzy_sets(gbkeys)
           260
        
           >>> 260 / 62748517
           4.14<...>e-06
        
        While still we have a lot of possible keys, the refinement did an
        amazing job and the new set is **6 orders of magnitud smaller** than the
        original!
        
        We can compute the set of possible keys doing a join and we can even
        further reduce the set keeping only the most likely keys:
        
        .. code:: python
        
           >>> from cryptonita.fuzzy_set import join_fuzzy_sets
           >>> gkstream = join_fuzzy_sets(gbkeys, cut_off=1024, j=B(''))
        
        ``gkstream`` is our guess for the complete key stream for the cipher.
        
        Is this right?
        
        Breaking the cipher - Break the cipher!
        ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
        
        .. code:: python
        
           >>> kstream = gkstream.most_likely()
        
           >>> print((ctext ^ kstream.inf()).decode('ascii'))
           Now that the party is jumping
           With the bass kicked in and the Vega's are pumpin
           Quick to the point, to the point, no faking
           Cooking MC's like a pound of bacon
           Burning 'em, if you ain't quick and nimble
           I go crazy when I hear a cymbal
           And a high hat with a souped up tempo
           I'm on a roll, it's time to go solo
           ollin' in my five point oh
           ith my rag-top down so my hair can blow
        
        
           >>> kstream.encode(64)
           b'dmFuaWxsYQ=='
        
        Final thoughts
        ~~~~~~~~~~~~~~
        
        Vigenere or a repeating key cipher is a well known poor cipher shown in
        every single cryptography course.
        
        But little is explained in how to break it in an *automated* fashion.
        
        `cryptonita <https://pypi.org/project/cryptonita/>`__ is not magical and
        a little of brain is required from you, but it is a quite useful *Swiss
        army knife for breaking crypto*.
        
        PRs or comments are welcome.
        
        Tested with `byexample <https://byexamples.github.io/byexample>`__.
        
Keywords: crypto cryptography crypto-analysis
Platform: UNKNOWN
Classifier: Development Status :: 3 - Alpha
Classifier: Intended Audience :: Developers
Classifier: Intended Audience :: Information Technology
Classifier: Intended Audience :: Science/Research
Classifier: Topic :: Scientific/Engineering :: Information Analysis
Classifier: Topic :: Security
Classifier: Topic :: Security :: Cryptography
Classifier: License :: OSI Approved :: GNU General Public License v3 (GPLv3)
Classifier: Programming Language :: Python :: 3
Requires-Python: >=3.3
