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{{InfoboxWarningbox|WARNING: This page under construction!}}
The following is an overview of the AES-CMM attack done by Eyal Ronenet al., detailed in his their draft/limited release paper [http://iotworm.eyalro.net/ IoT Goes Nuclear: Creating a ZigBee Chain Reaction(research paper website)], [http://eprint.iacr.org/2016/1047 IACR E-print submission]. If using this attack please '''do not cite this page''', instead cite the research paper only. The paper is currently a draft so there is no proceedings information etc as it has not yet been presented anywhere, but you can cite the E-Print version.
This page is presented as an example of using Python/ChipWhisperer to perform attacks against the AES-CCM cipher, without needing to do a more complex attack against AES-CTR mode.
[[File:cbc_mac_source.png]]
The "Calculated Tag" and "Expected Tag" are compared together, and only if they match is the decrypted data used. A change of any of the data blocks OR the header would change the calculated tag, resulting in an error.
Some nice features of AES-CCM:
* Can decrypt any data block, or decrypt blocks out of orderdue to AES-CTR usage.
* Authentication Tag provides authentication that data has not been modified in transit.
* Auth tag can include non-encrypted information, such as a header with address or length information.
* Auth tag can be shortened (i.e., not full 16-byte length) for use with protocols with very sensitive length limitations.
'''Cipher Block Chaining (CBC):''' The plaintext is XORed with the previous ciphertext before being encrypted. There is no ciphertext before the first plaintext, so a randomly chosen initialization vector (IV) is used instead:
If you've performed a standard CPA attack, you'll realize the problem with attacking AES-CBC is we don't directly control the input, which we call <math>PT</math>. Instead it's XORd with some unknown bytes (the AES-CBC ciphertext output).
But if we are always attacking the same block (that is, we reset the AES state to initial by say resetting the device, and rerun the algorithm up to the first block), the unknown bytes are constant. As it turns out this is a pretty easy problem to solve. The first step is to perform a standard CPA attack. The only issue is we won't recover the actual encryption key used <math>k</math>, instead we recover <math>k \oplus CBC_{m-1} \oplus PT</math>, since we basically roll all the constant inputs into what we call a `modified key'. Note <math>CBC_{m-1}</math> is the output of the previous-block AES-CBC ciphertext.
In what might seem like magic, we can use this modified key to directly determine the second-round key (the true key). This was originally presented by J. Jaffe in [https://www.iacr.org/archive/ches2007/47270001/47270001.pdf A First-Order DPA Attack Against AES in Counter Mode with Unknown Initial Counter]. The reason this works is if you remember we recovered <math>k' = k \oplus CBC_{m-1}</math>. In the AES algorithm the first thing we do is the AddRoundKey, which is:
# Program into the target the aes-ccm bootloader. This bootloader can be found in the git repo, which also includes a .hex file.
# Set the target type to the special AES-CCM bootloader driver. This target module is detailed in the appendix on this page, you can copy that code into a new file in the `target' directory.
# Run a capture with ~500 traces. Use 1x CLKGEN for the ADC speed and the full point-range to be able to capture both software AES rounds. See capture script example for details.
=== Step #1: AES-CBC MAC Block #1 ===
== Bootloader Interface Code ==
<syntaxhighlightlang="python">
#!/usr/bin/python
# -*- coding: utf-8 -*-
raise IOError("Failed to communicate, no response")
</syntaxhighlight>
[[Category:Examples]]
