# Sudoku Generation

2019-03-23, post № 213

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-generation.hs).

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

./sudoku 5 2

# Pi Day MMXIX

2019-03-14, post № 212

C, programming, #cyclic quine, #iteration quine

w=0;b(){w>27&&puts("",w=0);}
p(c){b();putchar(c),b(++w);}
main(){int/**/N=0,D[][85]={{
0,1048320,4194272,8372472,1,
16646172,33030158,33030150,1
,33030144,16252928,16252928,
3932160,1,1966080,983040,1,1
,229376,57344,14336,7168,1,1
,134219520,1,1,1,1,67109312,
133169264,1,67108860,1,1,1,1
,33554431,-1},{3670016,1,1,1
,3932160,1835008,917504,1,1,
917504,458752,491520,229376,
114688,122880,57344,28672,1,
28672,14336,15360,7168,7680,
3840,1792,1920,960,448,-1},{
0,134217712,134217720,1,1,1,
62914620,31457294,31457286,1
,15728640,7340032,7864320,1,
3932160,1966080,1966080,1,1,
983040,1015808,491520,245760
,245760,122880,61440,61440,1
,30720,15360,-1},{0,0,0,0,0,
0,0,0,0,0,2097088,2097088,0,
0,0,0,0,0,0,0,0,0,-1},{0,0,0
,0,0,1048064,8259552,1,1,1,1
,16515576,15729144,504,504,1
,985024,1048320,1008,504,252
,25166332,29360632,7342064,1
,2097024,0,0,-1},},*d,C[]={1
,119,61,48,59,98,40,41,123,1
,119,62,50,55,38,38,112,117,
116,115,40,34,34,44,119,61,1
,48,41,59,125,112,40,99,41,1
,123,98,40,41,59,112,117,116
,99,104,97,114,40,99,41,44,1
,98,40,43,43,119,41,59,125,1
,109,97,105,110,40,41,123,1,
105,110,116,47,42,42,47,78,1
,61,48,44,68,91,93,91,56,53,
93,61,123,2,125,44,42,100,44
,67,91,93,61,123,3,48,125,44
,42,99,61,67,44,106,59,67,91
,56,48,93,61,40,67,91,56,48,
93,45,52,55,41,37,55,43,52,1
,56,59,105,102,40,78,60,54,1
,41,102,111,114,40,100,61,68
,91,78,63,78,45,49,58,48,93,
59,42,100,43,49,59,42,100,43
,43,45,49,38,38,112,117,116,
115,40,34,34,41,41,105,102,1
,40,42,100,45,49,41,102,111,
114,40,106,61,42,100,59,106,
59,106,47,61,50,41,112,117,1
,116,99,104,97,114,40,51,50,
43,106,37,50,42,49,53,41,59,
59,59,102,111,114,40,59,42,1
,99,59,99,43,43,41,123,105,1
,102,40,42,99,61,61,50,41,1,
102,111,114,40,106,61,48,59,
106,60,53,59,106,43,43,41,1,
123,112,40,49,50,51,41,59,59
,59,102,111,114,40,100,61,68
,91,106,93,59,42,100,43,1,49
,59,112,40,52,52,41,41,1,119
,43,61,112,114,105,110,1,116
,102,40,34,37,100,34,44,1,42
,100,43,43,41,59,112,40,1,52
,53,41,59,112,40,52,57,41,59
,112,40,49,50,53,41,59,1,112
,40,52,52,41,59,125,101,108,
115,101,32,105,102,40,1,42,1
,99,61,61,51,41,102,111,1,1,
114,40,100,61,67,59,42,100,1
,59,112,40,52,52,41,41,119,1
,43,61,112,114,105,110,116,1
,102,40,34,37,100,34,44,42,1
,100,43,43,41,59,1,101,108,1
,115,101,32,105,102,40,42,99
,62,49,41,112,40,42,99,41,59
,125,102,111,114,40,106,61,1
,53,59,45,1,45,106,59,112,40
,53,57,41,1,41,59,1,1,1,125,
0},*c=C,j;C[80]=(C[80]-47)%7
+48;if(N<6)for(d=D[N?N-1:0];
*d+1;*d++-1&&puts(""))if(*d-
1)for(j=*d;j;j/=2)putchar(32
+j%2*15);;;for(;*c;c++){if(*
c==2)for(j=0;j<5;j++){p(123)
;;;for(d=D[j];*d+1;p(44))w+=
printf("%d",*d++);p(45);p(49
);p(125);p(44);}else if(*c==
3)for(d=C;*d;p(44))w+=printf
("%d",*d++);else if(*c>1)p(*
c);}for(j=5;--j;p(59));};;;;

Try it online. Happy pi day.

# Lichen, Extraterrestrials, Diodes #1

2019-02-23, post № 211

art, #LED, #photography

# Kickboy #0

2019-01-26, post № 210

art, #stop motion

Kickboy.

# Foam Cube Puzzle

2018-12-29, post № 209

programming, Python, #brute-force, #solver

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-puzzle.py

All six foam pieces.

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

2018-12-24, post № 208

art, C, programming, #fir, #quine, #tree

                                         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

2018-12-01, post № 207

## Theoretical Framework

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

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

# Prime Intirety

2018-11-03, post № 206

C, mathematics, programming, #integer, #list, #primes, #representation

Since ancient times humanity knew that there are infinitely many primes — though countable, writing a complete list of every prime is impossible if one intends to finish.
However, in practice one often only considers a minute subset of the naturals to work with and think about. When writing low-level languages like C, one is nearly forced to forget about almost every natural number — the data type u_int_32, for example, is only capable of representing $\{\mathbb{N}_0\ni n<2^{32}\}$.
Therefore, it is possible to produce a complete list of every prime representable in thirty-two bits using standard bit pattern interpretation — the entirety of the first 𝟤𝟢𝟥 𝟤𝟪𝟢 𝟤𝟤𝟣 primes.

Generating said list took about two minutes on a 4GHz Intel Core i7 using an elementary sieve approach written in C compiled with gcc -O2.
All primes are stored in little-endian format and packed densely together, requiring four bytes each.

Using the resulting file, one can quickly index the primes, for example $p_{10^7}=179\,424\,691 = \text{ab1cdb3}_{16}$ (using zero-based indexing). Since each prime is stored using four bytes, the prime’s index is scaled by a factor of four, resulting in its byte index.

dd status=none ibs=1 count=4 if=primes.bin skip=40000000 | xxd
00000000: b3cd b10a                                ....

Source code: prime-intirety.c
Prime list (gzipped and split): prime-intirety_primes.bin.gz.parts

Jonathan Frech's blog; built 2022/01/16 06:13:11 CET