Posted by Jason Polak on 19. June 2018 · Write a comment · Categories: computer-science · Tags: , ,

Today we'll see a Python 3 implementation of the MD5 hashing algorithm, which was originally designed by Ron Rivest. Even though the MD5 algorithm has been compromised and is not longer secure, it is still in use today. For example, CS ISOs of Linux distributions often provide MD5 sums in addition to SHA256 sums.

There are a lot of C implementations of MD5 out there. However, they are not easy to read, especially for someone who is not used to C. Python 3, being closer to "math" rather than the computer, reads more like a definition of the algorithm. Of course, the Python implementation is slower than the C one.

This Python implementation is ideal for experimenting with MD5 and learning about it. The code I give below is not the most compact possible, but I believe it is easy to understand.
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Posted by Jason Polak on 18. February 2018 · Write a comment · Categories: computer-science · Tags: , ,

Let's look at Python's map function. What does this function do? Here is the syntax:

It takes a function called function and applies it to each element of iterable, returning the result as an iterable. For example:

The output of this program is:

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Posted by Jason Polak on 18. February 2018 · 2 comments · Categories: computer-science · Tags: , ,

Let's assume you have a polynomial over a finite field $\F_q$, defined in Sage. How can you tell whether it's a permutation polynomial? That is, when is the corresponding function $\F_q\to\F_q$ bijective?

This is how you might have a polynomial over $\F_q$ defined in Sage:

Here, the variable $x$ refers the element $x$ in the isomorphism $\F_q \cong \F_p[x]/\alpha(x)$ and $t$ is the variable in the polynomial ring $\F_q[t]$. Is $f$ a permutation polynomial? That of course depends on what $q$ is.
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In the abelian group $\Z/n$, the order of $m\in \Z/n$ can be calculated via the formula $n/{\rm gcd}(m,n)$. This number is just the smallest number you have to multiply $m$ by in order to get a multiple of $n$. So when $n = p$ is a prime, every we see that every nonzero element of $\Z/p$ has order $p$.

However, every nonzero element of $\Z/p$ is prime to $p$, and the nonzero elements of $\Z/p$ form the abelian multiplicative group $\Z/p^\times$ of invertible elements. Then the order of an element $m\in \Z/p^\times$ is a little more tricky to figure out, but we know by Fermat's Little Theorem that it will be a factor of $p-1$.

Example. Consider the prime $p=11$. Then we know that $3^{10} = 1$ by Fermat's Little Theorem. However, $3^5 = 1$ too. In fact the order of $~3$ is $~5$.

This example has a curious property that you may not know about. It is true that $3^5 = 1$ in $\Z/11$. But did you know that $3^5 = 1$ in $\Z/121$? What?! This leads to a natural question:

Question. Which primes $p$ have the property that there exists an element $m\in \Z/p^\times, m\not=1$ such that the order of $m$ in $\Z/p^\times$ is the same as the order of $m$ in ${\Z/p^2}^\times$?

Not all primes have this property. Consider $p = 5$. Then $3^4 = 1$ in $\Z/5$. However, $3^4 = 6$ in $\Z/25$. In fact, none of the elements $2,3,4$ have the same order in both $\Z/5^\times$ and $\Z/25^\times$. Here are some primes that do have such elements:

11, 29, 37, 43, 59, 71, 79, 97, 103, 109, 113, 127, …

So, there are quite a few of them and maybe there are infinitely many. One can go further with this question: what about primes for which there are non-trivial elements in $\Z/p^\times$ with the same order in $\Z/p^\times, {\Z/p^2}^\times,$ and ${\Z/p^3}^\times$? There are fewer of these. But they exist. For example, the identity
$$68^{112} = 1$$
Holds in $\Z/113^k$ for $k=1,2,3$ (but not for $k=4$). This is the only example I know of and maybe it is the only possible example? But I probably just haven't looked far enough. I wrote the original program to find it in Python but I plan to rewrite it in C with the gmp library to expand the search.