finish day 1

day1.retro

@@ -1,68 +1,50 @@
 # Day 1: Sonar Sweep
 
-You're minding your own business on a ship at sea when the overboard alarm goes off! You rush to see if you can help. Apparently, one of the Elves tripped and accidentally sent the sleigh keys flying into the ocean!
-
-Before you know it, you're inside a submarine the Elves keep ready for situations like this. It's covered in Christmas lights (because of course it is), and it even has an experimental antenna that should be able to track the keys if you can boost its signal strength high enough; there's a little meter that indicates the antenna's signal strength by displaying 0-50 stars.
-
-Your instincts tell you that in order to save Christmas, you'll need to get all fifty stars by December 25th.
-
-Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
-
-As the submarine drops below the surface of the ocean, it automatically performs a sonar sweep of the nearby sea floor. On a small screen, the sonar sweep report (your puzzle input) appears: each line is a measurement of the sea floor depth as the sweep looks further and further away from the submarine.
-
-For example, suppose you had the following report:
-
-199
-200
-208
-210
-200
-207
-240
-269
-260
-263
-
-This report indicates that, scanning outward from the submarine, the sonar sweep found depths of 199, 200, 208, 210, and so on.
-
-The first order of business is to figure out how quickly the depth increases, just so you know what you're dealing with - you never know if the keys will get carried into deeper water by an ocean current or a fish or something.
-
-To do this, count the number of times a depth measurement increases from the previous measurement. (There is no measurement before the first measurement.) In the example above, the changes are as follows:
-
-199 (N/A - no previous measurement)
-200 (increased)
-208 (increased)
-210 (increased)
-200 (decreased)
-207 (increased)
-240 (increased)
-269 (increased)
-260 (decreased)
-263 (increased)
-
-In this example, there are 7 measurements that are larger than the previous measurement.
-
-How many measurements are larger than the previous measurement?
-
----
-
-### Part 1
+## Part 1
 
 First, put the input in a file called input1 and that test case a file called test1 (this is the pattern I am going to use throughout).
 The solution is pretty simple: create variables to store the previous number and the count. Then loop through the file. If the current number is more than the previous, add one to the count. Then, store the current number in the variable from the previous one.
 
-Prev is defined as being 9999, larger than any of the values in the file, so the first entry will always not be counted as higher (to avoid needing extra logic for the first entry)
+Prev is defined as being 9999, larger than any of the values in the file, so the first entry will always not be counted as higher (to avoid needing extra logic for the first entry). We also count the amount of lines the the file, to use in part 2.
 
 ~~~
 'count var
+'lines var
 #9999 'prev var-n
 ~~~
 
 then loop through the file counting up. Retro has a nice word, for-each-line, that does the looping. 
 
 ~~~
-'input1 [ s:to-number dup @prev gt? [ &count v:inc  ] if !prev ] file:for-each-line
+'input1 [ s:to-number dup @prev gt? [ &count v:inc ] if !prev &lines v:inc ] file:for-each-line
 
 @count n:put ASCII:LF c:put
 ~~~
 
+## Part 2
+
+part 2 is a bit harder as we can't just iterate through the file like part 1 as we need to read three lines at once. The number of lines in the file is known, so we can read `lines - 2` windows and add them up. We need a file handle and to reset the variables from part 1:
+
+~~~
+'input1 file:R file:open 'f var-n
+#99999 !prev
+#0 !count
+~~~
+
+then, for each line in the file, we need to read that line and the next 2 lines, then seek back to the line straight after it. The only way I could think to do this is counting the amount of bytes we seek forward (using the length of the lines read) and then seeking that many bytes back, which is a bit of a pain. This is a terrible solution. This loop reads the 3 lines out of the file, adds up the values, and then uses the same logic as part 1 to find upwards changes from the previous window.
+
+I'm not happy with this solution, but it works. Retro users, help me get better! 
+
+~~~
+'seek-to var
+
+@lines #2 - [
+@f file:read-line dup s:length #1 + &seek-to v:inc-by s:to-number
+@f file:read-line s:to-number
+@f file:read-line s:to-number + +
+dup @prev gt? [ &count v:inc ] if !prev
+@seek-to @f file:seek ] times
+
+@count n:put ASCII:LF c:put
+@f file:close
+~~~