Linked by Thom Holwerda on Fri 23rd Sep 2011 22:22 UTC, submitted by kragil
Windows The story about how secure boot for Windows 8, part of UEFI, will hinder the use of non-signed binaries and operating systems, like Linux, has registered at Redmond as well. The company posted about it on the Building Windows 8 blog - but didn't take any of the worries away. In fact, Red Hat's Matthew Garrett, who originally broke this story, has some more information - worst of which is that Red Hat has received confirmation from hardware vendors that some of them will not allow you to disable secure boot.
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RE[4]: RSA key example.
by Alfman on Mon 26th Sep 2011 19:11 UTC in reply to "RE[3]: RSA key example."
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"I'm also wondering if it's possible to mathematically reverse a hash algorithm in a way that provides..."

Lets go on the assumption that our function is an ideal hash function with no mathematical weaknesses. Since we already know that a broken hash function will limit the scope of search. An idea hash function means that the only way to find a collision is to brute force various inputs until we generate a collision.

sha256sum(x1) = y
sha256sum(x2) = y

How would we find x2, such that it produces the same hash as x1? This can be as simple as taking a known payload, and modifying it with a nonce until we generate the collision we're looking for. It's trivial, and it's been done with MD5. However, this task becomes exponentially more difficult as bits are added to the hash.

Let's reduce the difficulty of the problem:

1 bit hash function
sha256sum(x1) & 0x1 = y
sha256sum(x2) & 0x1 = y
Here, y can only be 0 or 1, therefor every other X value will produce a collision.

2 bit hash function:
sha256sum(x1) & 0x3 = y
sha256sum(x2) & 0x3 = y
Now, y can be 0,1,2,3, every 4th X value will produce a collision, twice as much work as 1 bit.

3 bit hash function:
sha256sum(x1) & 0x7 = y
sha256sum(x2) & 0x7 = y
Every 8th X produces a hash collision, twice as much work as 2 bits.

So with this exponential growth, a 256 bit hash function would collide every 1157920892373161954235709850086879078532699846656405640394575840079131 29639936 X values on average.

Assuming we have 1 billion computers, each able to forward hash 1 billion X values every second, then we might expect a collision every 3764568028158688209515806576697354474006124657512762 years on average (double check my math).

This is if we stick to classical computing, quantum computing introduces yet a whole new dimension to the problem. It's too bad quantum computing was not offered at my university, since I don't know that much about it.

Edit: I'd be happy to leave my cheap web development clients to work on this stuff instead, if anyone's willing to pay me to do it.

Edited 2011-09-26 19:24 UTC

Reply Parent Score: 2