## Non-uniform shuffling

A shuffle of a finite sequence of length $n$ of distinguishable elements refers to an algorithmic process which — modulo pseudo-randomness — can be modeled as a random variable uniformly distributed on the permutations $\mathbb{S}_n$.
However, most pseudo-random entropy sources provide only a pseudo-uniformly distributed realization of $\mathbb{F}_2^\mathbb{N}$, leading to the necessity of finding an algorithmic transformation process if one wishes to achieve a shuffle.
In the following, I will assume that a transforming process to a family of independent and uniformly on $\{1..n\}$ distributed random variables is already present for any $n\in\mathbb{N}$.

One naive and seemingly correct (it is not) approach is to traverse the given sequence, uniformly swapping the current entry with another one, i.e.

void falseShuffle(uint64_t *arr, size_t len) {
for (size_t j = 0; j < len; j++)
swap(arr, j, unif(len)); }

as an exemplary C implementation where $\texttt{unif}(n)$ is independent and uniformly distributed on $\{0..n-1\}$.

Yet, even though sensible on first sight, the above defined random variable is only in the most trivial cases uniformly distributed and — as empirical evidence suggests, see below — horrendously non-uniformly distributed otherwise.
To prove the non-uniformity postulated above, I first present the following number-theoretic result.

Claim. In only three trivial cases does the factorial of a natural number divide its tetration; formally

$\forall\,n\in\mathbb{N}_{>2}:n!\nmid n^n$.

Proof. Let $n\in\mathbb{N}_{>2}$ be a natural number larger than two. By the definition of the factorial, $\prod_{p is evident. Adhering to the uniqueness of prime factorizations, $\prod_{p follows. Observe that $n-1>1$ has to be prime since $\forall\,p, implying $n-1\mid\prod_p=n$ which cannot hold for $n>2$. QED

Now suppose, $\texttt{falseShuffle}$ was indeed non-trivially distributed uniformly. Without loss of generality, all involved probability spaces were finite. Then there had to exist a surjection from this algorithm’s entropic state to $\mathbb{S}_n$ with fibers of the same finite cardinality, implying $n!\mid n^n$. By the above proven claim, $n<3$ followed, making the distribution trivial. QED

One possible reason for the surprising nature of this non-uniformity is the striking source code resemblance to a correct implementation, i.e.

void shuffle(uint64_t *arr, size_t len) {
for (size_t j = 0; j < len; j++)
swap(arr, j, j + unif(len - j)); }

as an exemplary C implementation which can be inductively shown to resemble the same structure as $\mathbb{S}_n$, in each step sprinkling in some uniform randomness and thus being itself uniformly distributed.

To see just how non-uniform $\texttt{falseShuffle}$ is, I have calculated its discrete density for $n=4$:

[       |                ]
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[   ||  ||||||| ||       ]
[||||| |||||||| |||    ||]
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[||||||||||||||||||||||||]
[||||||||||||||||||||||||]
[||||||||||||||||||||||||]
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[||||||||||||||||||||||||]
[||||||||||||||||||||||||]
[||||||||||||||||||||||||]
[||||||||||||||||||||||||]
n = 4

If it was uniformly distributed, the discrete density would look like a rectangle; [||||| ... |||||]. Further plots for $0\leq n\leq 6$ are shown in nonUniformity.txt.

Source code for the analysis and plotting: nonUniformity.hs. Empirical evidence of non-uniformity: nonUniformity.c.

## Sudoku Generation

Over two years ago, I wrote a basic 3×3-sudoku solver which uses both fundamental rule-based elimination and guessing to arrive at the solution. Revisiting the topic of computer-aided sudoku manipulation, I wrote a generalized sudoku generator (sudoku.hs).

    | 4
3 |
---------
2 | 1
4 |

./sudoku 5 2

## Symbolic Closed-Form Fibonacci

Let $V := \{(a_j)_{j\in\mathbb{N}}\subset\mathbb{C}|a_n=a_{n-1}+a_{n-2}\forall n>1\}$ be the two-dimensional complex vector space of sequences adhering to the Fibonacci recurrence relation with basis $B := ((0,1,\dots), (1,0,\dots))$.
Let furthermore $f: V\to V, (a_j)_{j\in\mathbb{N}}\mapsto(a_{j+1})_{j\in\mathbb{N}}$ be the sequence shift endomorphism represented by the transformation matrix

$A := M^B_B(f) = \begin{pmatrix}1&1\\1&0\end{pmatrix}$.

By iteratively applying the sequence shift a closed-form solution for the standard Fibonacci sequence follows.