Day 3: Python

R/.Rproj.user/shared/notebooks/paths

@@ -3,6 +3,7 @@
 /Users/hrbrmstr/Development/2020-code-advent/R/02.R="AF113B0D"
 /Users/hrbrmstr/Development/2020-code-advent/R/02.py="0900576A"
 /Users/hrbrmstr/Development/2020-code-advent/R/03.R="B9D5FDD0"
+/Users/hrbrmstr/Development/2020-code-advent/R/03.py="C2F47E74"
 /Users/hrbrmstr/Development/2020-code-advent/README.md="74DC8DCF"
 /Users/hrbrmstr/Development/2020-code-advent/input/01-01.txt="53BE9636"
 /Users/hrbrmstr/Development/2020-code-advent/input/02-01.txt="C32036DD"

R/03.R

@@ -72,8 +72,6 @@
 library(stringi)
 library(tidyverse)
 
-# 03-01 -------------------------------------------------------------------
-
 readLines("../input/03-01.txt") %>%
   stri_split_boundaries(
     opts_brkiter = stri_opts_brkiter("character"),
@@ -105,10 +103,27 @@ tree_count <- function(orig_map, slope_x, slope_y) {
 
 }
 
+# 03-01 -------------------------------------------------------------------
+
 tree_count(orig_map, 3, 1)
 
 # 03-02 -------------------------------------------------------------------
 
+# --- Part Two ---
+#
+# Time to check the rest of the slopes - you need to minimize the probability of a sudden arboreal stop, after all.
+#
+# Determine the number of trees you would encounter if, for each of the following slopes, you start at the top-left corner and traverse # the map all the way to the bottom:
+#
+# Right 1, down 1.
+# Right 3, down 1. (This is the slope you already checked.)
+# Right 5, down 1.
+# Right 7, down 1.
+# Right 1, down 2.
+# In the above example, these slopes would find 2, 7, 3, 4, and 2 tree(s) respectively; multiplied together, these produce the answer 336.
+#
+# What do you get if you multiply together the number of trees encountered on each of the listed slopes?
+
 x <- c(1, 3, 5, 7, 1)
 y <- c(1, 1, 1, 1, 2)
 

R/03.py

@@ -0,0 +1,42 @@
+# with open("/tmp/test.txt") as f:
+with open("../input/03-01.txt") as f:
+  orig_map = f.read().splitlines()
+
+def tree_count(orig_map, slope_x, slope_y):
+
+  m_cols = len(orig_map[0])
+  m_rows = len(orig_map)
+  
+  n_tiles = (slope_x+1) * m_rows // (m_cols)
+  
+  actual_map = [ row * n_tiles for row in orig_map ]
+  
+  xpos = 0
+  ypos = 0
+  n_trees = 0
+  
+  while True:
+    
+    xpos = xpos + slope_x
+    ypos = ypos + slope_y
+    
+    if (ypos >= m_rows):
+      break
+    
+    if (actual_map[ypos][xpos] == "#"):
+      n_trees = n_trees + 1
+  
+  return(n_trees)
+
+# 03-01 -------------------------------------------------------------------
+
+tree_count(orig_map, 3, 1)
+
+# 03-02 -------------------------------------------------------------------
+
+from functools import reduce
+
+x = [ 1, 3, 5, 7, 1 ]
+y = [ 1, 1, 1, 1, 2 ]
+
+reduce(lambda x, y: x*y, [ tree_count(orig_map, x[idx], y[idx]) for idx in range(len(x)) ])