## Foam Cube Puzzle

After having solved the puzzle shown below a few times by combining six foam pieces to construct a hollow cube, I wondered if it had a unique solution. A simple brute-force search reveals it does. Source code: foam_cube.py

As a first step I digitalized all pieces seen above. Having an internal representation, I wrote a script which tries all possible rotations and reflections (as three-dimensional rotations can imply two-dimensional reflection) to try and construct a three-dimensional cube from the given pieces. Using short-circuit evaluation to not bother with already impossible solutions, the search space is narrow enough to not require any considerable computing time. The resulting unique solution modulo rotation is shown above; the top face is placed on the bottom right.

## Winter MMXVIII

                                         I
,O;
main(
){char*
Q,_[]={73
,44,79,59,2
,109,97,105,2
,110,40,41,123,
99,104,97,114,42,
81,44,95,91,93,61,2
,123,1,48,125,44,42,2
,74,59,102,111,114,40,2
,81,61,95,44,73,61,79,61,
48,59,42,81,59,43,43,81,41,
123,105,102,40,42,81,60,50,41
,102,111,114,40,74,61,95,59,42,
74,59,74,43,43,41,73,60,49,38,38,
112,114,105,110,116,102,40,34,37,42
,99,34,44,52,50,45,79,47,50,45,49,44,
51,50,41,44,73,43,61,112,114,105,110,2,
116,102,40,34,37,100,34,44,42,74,41,44,73
,62,79,63,73,61,48,44,79,43,61,50,44,112,2,
117,116,115,40,34,34,41,58,48,44,73,43,43,60,
49,63,112,114,105,110,116,102,40,34,37,42,99,34
,44,52,50,45,79,47,50,44,52,52,41,58,112,117,116,
99,104,97,114,40,52,52,41,44,73,62,79,63,73,61,48,2
,44,79,43,61,50,44,112,117,116,115,40,34,34,41,58,48,
59,105,102,40,42,81,62,50,41,73,43,43,60,49,63,112,114,
105,110,116,102,40,34,37,42,99,34,44,52,50,45,79,47,50,44
,42,81,41,58,112,117,116,99,104,97,114,40,42,81,41,44,73,62
,79,63,73,61,48,44,79,43,61,50,44,112,117,116,115,40,34,34,41
,58,48,59,125,102,111,114,40,73,61,48,59,73,43,43,60,49,55,59,2
,112,117,116,99,104,97,114,40,52,55,41,41,59,102,111,114,40,73,61
,112,117,116,115,40,34,34,41,59,73,43,43,60,53,59,112,117,116,115,2
,40,34,34,41,41,123,112,114,105,110,116,102,40,34,37,42,99,34,44,51,2
,50,44,52,55,41,59,102,111,114,40,74,61,48,59,74,43,43,60,50,48,59,41,2
,2,112,117,2,116,99,2,2,104,2,97,2,114,2,40,52,2,55,41,2,59,125,2,2,125,2
,2,0},*J;for(Q=_,I=O=0;*Q;++Q){if(*Q<2)for(J=_;*J;J++)I<1&&printf("%*c",42-
O/2-1,32),I+=printf("%d",*J),I>O?I=0,O+=2,puts(""):0,I++<1?printf("%*c",42-O/
2,44):putchar(44),I>O?I=0,O+=2,puts(""):0;if(*Q>2)I++<1?printf("%*c",42-O/2,*Q)
:putchar(*Q),I>O?I=0,O+=2,puts(""):0;}for(I=0;I++<17;putchar(47));for(I=puts("");
I++<5;puts("")){printf("%*c",32,47);for(J=0;J++<20;)putchar(47);}}/////////////////
/////////////////////
/////////////////////
/////////////////////
/////////////////////                               

Try it online.

## 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.