@user202729 latest weird result: tested my last layout and @isaacg's last layout with three exit types, my original exit, @isaacg's tricky exit, and my new trident exit. my layout works best with trident exit, other two exits perform equally worse. @isaacg's layout works equally well with trident exit and tricky exit, my original exit proving so bad that adding more 2-way branches does better.
Rage Against the Dying of the Light
code-golfkolmogorov-complexity
Introduction
"Do not go gentle into that good night" is the title of a poem by 20th century English poet Dylan Thomas. If you've heard it before it was probably because you watched Interstellar, where it was quoted multiple times....
Given two base ten integers m and n, you will need to make the last n digits of m its first digits.
Tests
Num,n => Output
123,1 => 312
123,2 => 231
123,3 => 123
123,4 => 312
1,637 => 1
10,1 => 1 (01=>1)
100,2 => 1 (Final result was 001, but the leading zeros were removed)
10,2 => 10
110,2 => 10...
Combinatorial Decomposition code-golf combinatorics base-conversion
Any nonnegative integer \$m\$, given arbitrary nonnegative integer \$r\$, can be written as a unique series of combinations such that $$m=\sum\limits_{i=1}^{r}\begin{pmatrix}C_i\\i\end{pmatrix}$$ provided \$C\$ is strictly increa...
the thing is, I only compared the results to the nearest integer, which is 8 digits, and I can't remember what I did with the original results, or whether I kept enough of the scripts to reproduce them
but as a comparison, I think that moving the start squares to be horizontal instead of vertical made a difference of <1e-10
@user202729 "Wall finder" just goes through every horizontal and vertical wall, recalculates the score with them added, then picks the highest score, rinse and repeat. It likes width 2 corridors.
but, it's what I used to find my trident layout (I based the idea on my "fewer dead ends" applied to width 3 corridors, and it suggested adding the tridents)
yeah, it's not fast. it took a few hours. I did apply multiprocessing to it in one of my scripts, but then I couldn't work out how to use my "this wall blocks the way to the exit, don't try it again" optimisation
Decide to implement from_compact_ascii to edit the maze manually.
def as_compact_ascii(self):
self._fix_maze()
vertex="·"
lines = [vertex + ("-"+vertex)*self.width]
for y in range(self.height):
lines.append("|" + "".join("{}{}".format("S" if (x, y) in self.starts else "E" if (x, y) in self.exits else " ",
" |"[x == self.width - 1 or int(self.hwalls[x,y])])
for x in range(self.width)))
lines.append(vertex + "".join(("-" if y == self.height - 1 or self.vwalls[x,y] else " ")+vertex
for x in range(self.width)))
return "\n".join(lines)
@user202729 instead of a width 3 corridor, I tried an arrangement whereby the route zigzagged a bit like a width 2 corridor concertinaed in on itself, however this also turned out to score the same (within <1e-10) as a regular width 3 corridor
Given two base ten integers m and n, you will need to make the last n digits of m its first digits.
Tests
Num,n => Output
123,1 => 312
123,2 => 231
123,3 => 123
123,4 => 312
1,637 => 1
10,1 => 1 (01=>1)
100,2 => 1 (Final result was 001, but the leading zeros were removed)
10,2 => 10
110,2 => 10...
