From bc901b4d2d9b3670baf45a1f9ed0c0bf658dfd18 Mon Sep 17 00:00:00 2001
From: rak1507 <56049432+rak1507@users.noreply.github.com>
Date: Sun, 5 Feb 2023 17:35:54 +0000
Subject: [PATCH] Update 15.md

Add description for 15
---
 aoc/2022/15.md | 48 ++++++++++++++++++++++++++++++++++++++++++++++--
 1 file changed, 46 insertions(+), 2 deletions(-)

diff --git a/aoc/2022/15.md b/aoc/2022/15.md
index cce4f66..f87ebcd 100644
--- a/aoc/2022/15.md
+++ b/aoc/2022/15.md
@@ -181,16 +181,60 @@ A very much faster solution starts from
 > the fact that any beacon will be adjacent to the perimeter of the region, so there are 4 straight lines that it could possibly pass through. Generate all the equations, and then the beacon will be at one of the intersections.
 
 ```q
-`Sx`Sy`Bx`By set' flip {get @[x;where not x in "-0123456789";:;" "]}each inp;
+`Sx`Sy`Bx`By set' flip {get @[x;where not x in "-0123456789";:;" "]}each read0 `15.txt;
 m: sum abs (Sx-Bx;Sy-By) / manhattan dists
 / part 1
-count except[;Bx where By=Y] distinct raze {$[y<0;();x-y-til 1+2*y]}'[Sx;m-abs Sy-Y]
+count except[;Bx where By=Y] distinct raze {$[y<0;();x-y-til 1+2*y]}'[Sx;m-abs Sy-Y:2000000]
 / part 2
 peri: raze 1 1 -1 -1,''(Sy-Sx+m+1; Sy-Sx-m+1; Sy+Sx+m+1; Sy+Sx-m+1)
 int: distinct raze peri {r:0-1%(%/)x-y; (r; sum x*r,1)}\:/: peri
 sum 4000000 1*floor first int where {all raze(x=floor x;0<=x;x<=LIM;m<sum abs(Sx;Sy)-x)} each int
 ```
 
+## Part 1
+
+Part 1 doesn't do anything clever (and ends up being slower than part 2!) but it is fast enough for the relatively small amount of data here.
+
+`{$[y<0;();x-y-til 1+2*y]}` creates a region of size y around x
+
+```
+q)
+{$[y<0;();x-y-til 1+2*y]}[5; 2]
+3 4 5 6 7
+```
+
+`'[Sx;m-abs Sy-Y]` this function is then applied to each of the sensor x values, and the appropriate radius.
+
+All that is left is to count the number of x values after removing any beacons.
+
+## Part 2
+
+Part 2 is more interesting:
+
+We can use the fact that any undetected beacon must be adjacent to a detected region.
+
+(This is true as there is only one possible location, so there can't be any 'space' around the beacon)
+
+The line adjacent to a region's perimeter can be described as a simple linear equation.
+
+As there are four possible adjacent lines, there are four different equations.
+
+These are calculated with this line. 
+`peri: raze 1 1 -1 -1,''(Sy-Sx+m+1; Sy-Sx-m+1; Sy+Sx+m+1; Sy+Sx-m+1)`
+
+Now, we can find the intersections between all pairs of lines.
+`int: distinct raze peri {r:0-1%(%/)x-y; (r; sum x*r,1)}\:/: peri`
+
+After checking that the intersections have the right properties:
+`{all raze(x=floor x;0<=x;x<=LIM;m<sum abs(Sx;Sy)-x)}`
+
+There will be only one left, which is the correct answer.
+
+A visualisation in desmos with the sample data is below. 
+
+(Side note: desmos is a surprisingly good array language)
+
+![image](https://user-images.githubusercontent.com/56049432/216835451-8e0dc9d8-47e1-4bab-ba3b-4a7fcc97f81a.png)
 
 
 ## Contributors