Solved 2024/12 P1

2024/12/12.md

@@ -0,0 +1,210 @@
+## \-\-- Day 12: Garden Groups \-\--
+
+Why not search for the Chief Historian near the [gardener](/2023/day/5)
+and his [massive farm](/2023/day/21)? There\'s plenty of food, so The
+Historians grab something to eat while they search.
+
+You\'re about to settle near a complex arrangement of garden plots when
+some Elves ask if you can lend a hand. They\'d like to set up
+fences
+around each region of garden plots, but they can\'t figure out how much
+fence they need to order or how much it will cost. They hand you a map
+(your puzzle input) of the garden plots.
+
+Each garden plot grows only a single type of plant and is indicated by a
+single letter on your map. When multiple garden plots are growing the
+same type of plant and are touching (horizontally or vertically), they
+form a *region*. For example:
+
+    AAAA
+    BBCD
+    BBCC
+    EEEC
+
+This 4x4 arrangement includes garden plots growing five different types
+of plants (labeled `A`, `B`, `C`, `D`, and `E`), each grouped into their
+own region.
+
+In order to accurately calculate the cost of the fence around a single
+region, you need to know that region\'s *area* and *perimeter*.
+
+The *area* of a region is simply the number of garden plots the region
+contains. The above map\'s type `A`, `B`, and `C` plants are each in a
+region of area `4`. The type `E` plants are in a region of area `3`; the
+type `D` plants are in a region of area `1`.
+
+Each garden plot is a square and so has *four sides*. The *perimeter* of
+a region is the number of sides of garden plots in the region that do
+not touch another garden plot in the same region. The type `A` and `C`
+plants are each in a region with perimeter `10`. The type `B` and `E`
+plants are each in a region with perimeter `8`. The lone `D` plot forms
+its own region with perimeter `4`.
+
+Visually indicating the sides of plots in each region that contribute to
+the perimeter using `-` and `|`, the above map\'s regions\' perimeters
+are measured as follows:
+
+    +-+-+-+-+
+    |A A A A|
+    +-+-+-+-+     +-+
+                  |D|
+    +-+-+   +-+   +-+
+    |B B|   |C|
+    +   +   + +-+
+    |B B|   |C C|
+    +-+-+   +-+ +
+              |C|
+    +-+-+-+   +-+
+    |E E E|
+    +-+-+-+
+
+Plants of the same type can appear in multiple separate regions, and
+regions can even appear within other regions. For example:
+
+    OOOOO
+    OXOXO
+    OOOOO
+    OXOXO
+    OOOOO
+
+The above map contains *five* regions, one containing all of the `O`
+garden plots, and the other four each containing a single `X` plot.
+
+The four `X` regions each have area `1` and perimeter `4`. The region
+containing `21` type `O` plants is more complicated; in addition to its
+outer edge contributing a perimeter of `20`, its boundary with each `X`
+region contributes an additional `4` to its perimeter, for a total
+perimeter of `36`.
+
+Due to \"modern\" business practices, the *price* of fence required for
+a region is found by *multiplying* that region\'s area by its perimeter.
+The *total price* of fencing all regions on a map is found by adding
+together the price of fence for every region on the map.
+
+In the first example, region `A` has price `4 * 10 = 40`, region `B` has
+price `4 * 8 = 32`, region `C` has price `4 * 10 = 40`, region `D` has
+price `1 * 4 = 4`, and region `E` has price `3 * 8 = 24`. So, the total
+price for the first example is `140`.
+
+In the second example, the region with all of the `O` plants has price
+`21 * 36 = 756`, and each of the four smaller `X` regions has price
+`1 * 4 = 4`, for a total price of `772` (`756 + 4 + 4 + 4 + 4`).
+
+Here\'s a larger example:
+
+    RRRRIICCFF
+    RRRRIICCCF
+    VVRRRCCFFF
+    VVRCCCJFFF
+    VVVVCJJCFE
+    VVIVCCJJEE
+    VVIIICJJEE
+    MIIIIIJJEE
+    MIIISIJEEE
+    MMMISSJEEE
+
+It contains:
+
+-   A region of `R` plants with price `12 * 18 = 216`.
+-   A region of `I` plants with price `4 * 8 = 32`.
+-   A region of `C` plants with price `14 * 28 = 392`.
+-   A region of `F` plants with price `10 * 18 = 180`.
+-   A region of `V` plants with price `13 * 20 = 260`.
+-   A region of `J` plants with price `11 * 20 = 220`.
+-   A region of `C` plants with price `1 * 4 = 4`.
+-   A region of `E` plants with price `13 * 18 = 234`.
+-   A region of `I` plants with price `14 * 22 = 308`.
+-   A region of `M` plants with price `5 * 12 = 60`.
+-   A region of `S` plants with price `3 * 8 = 24`.
+
+So, it has a total price of `1930`.
+
+*What is the total price of fencing all regions on your map?*
+
+Your puzzle answer was `1437300`.
+
+The first half of this puzzle is complete! It provides one gold star: \*
+
+## \-\-- Part Two \-\-- {#part2}
+
+Fortunately, the Elves are trying to order so much fence that they
+qualify for a *bulk discount*!
+
+Under the bulk discount, instead of using the perimeter to calculate the
+price, you need to use the *number of sides* each region has. Each
+straight section of fence counts as a side, regardless of how long it
+is.
+
+Consider this example again:
+
+    AAAA
+    BBCD
+    BBCC
+    EEEC
+
+The region containing type `A` plants has `4` sides, as does each of the
+regions containing plants of type `B`, `D`, and `E`. However, the more
+complex region containing the plants of type `C` has `8` sides!
+
+Using the new method of calculating the per-region price by multiplying
+the region\'s area by its number of sides, regions `A` through `E` have
+prices `16`, `16`, `32`, `4`, and `12`, respectively, for a total price
+of `80`.
+
+The second example above (full of type `X` and `O` plants) would have a
+total price of `436`.
+
+Here\'s a map that includes an E-shaped region full of type `E` plants:
+
+    EEEEE
+    EXXXX
+    EEEEE
+    EXXXX
+    EEEEE
+
+The E-shaped region has an area of `17` and `12` sides for a price of
+`204`. Including the two regions full of type `X` plants, this map has a
+total price of `236`.
+
+This map has a total price of `368`:
+
+    AAAAAA
+    AAABBA
+    AAABBA
+    ABBAAA
+    ABBAAA
+    AAAAAA
+
+It includes two regions full of type `B` plants (each with `4` sides)
+and a single region full of type `A` plants (with `4` sides on the
+outside and `8` more sides on the inside, a total of `12` sides). Be
+especially careful when counting the fence around regions like the one
+full of type `A` plants; in particular, each section of fence has an
+in-side and an out-side, so the fence does not connect across the middle
+of the region (where the two `B` regions touch diagonally). (The Elves
+would have used the Möbius Fencing Company instead, but their contract
+terms were too one-sided.)
+
+The larger example from before now has the following updated prices:
+
+-   A region of `R` plants with price `12 * 10 = 120`.
+-   A region of `I` plants with price `4 * 4 = 16`.
+-   A region of `C` plants with price `14 * 22 = 308`.
+-   A region of `F` plants with price `10 * 12 = 120`.
+-   A region of `V` plants with price `13 * 10 = 130`.
+-   A region of `J` plants with price `11 * 12 = 132`.
+-   A region of `C` plants with price `1 * 4 = 4`.
+-   A region of `E` plants with price `13 * 8 = 104`.
+-   A region of `I` plants with price `14 * 16 = 224`.
+-   A region of `M` plants with price `5 * 6 = 30`.
+-   A region of `S` plants with price `3 * 6 = 18`.
+
+Adding these together produces its new total price of `1206`.
+
+*What is the new total price of fencing all regions on your map?*
+
+Answer:
+
+Although it hasn\'t changed, you can still [get your puzzle
+input](12/input).
+

2024/12/solution.py

@@ -0,0 +1,132 @@
+#!/bin/python3
+import sys,time,re
+from pprint import pprint
+sys.path.insert(0, '../../')
+from fred import list2int,get_re,nprint,lprint,loadFile,toGrid,bfs,get_value_in_direction,addTuples
+start_time = time.time()
+
+input_f = 'test'
+
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+
+def get_neighbors(grid,node,visited):
+    #print('@',node,' - Visited',visited)
+
+    directions = ['up','down','left','right']
+    offsets = {
+        'up': (-1, 0),
+        'down': (1, 0),
+        'left': (0, -1),
+        'right': (0, 1),
+    }
+    neighbors = []
+    for d in directions:
+        t = get_value_in_direction(grid,node)
+        if get_value_in_direction(grid,node,d) == t:
+            n = addTuples(offsets[d],node)
+            if n not in visited:
+                neighbors.append(n)
+                #print(n)
+                visited.append(n)
+                neighbors+=get_neighbors(grid,n,visited)
+    return neighbors
+
+def part1():
+    grid = toGrid(input_f)
+
+    #nprint(grid)
+    values = {}
+    visited = []
+    total_plots = []
+    for r,row in enumerate(grid):
+        for c,col in enumerate(row):
+            pos = (r,c)
+            plot = []
+            current = get_value_in_direction(grid,pos)
+            if pos not in visited:
+                
+                x = get_neighbors(grid,pos,visited)
+                plot += x
+                if pos not in plot:
+                    plot.append(pos)
+                if current not in values:
+                    values[current] = []
+                
+                total_plots.append(plot)
+
+    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
+    result = 0
+
+    for v in total_plots:
+        
+        sides = 0
+        for x,y in v:
+
+            for dx, dy in directions:
+                neighbor = (x + dx, y + dy)
+                if neighbor in v:
+                    sides += 1
+                
+        total_sides = len(v) * 4 - sides
+
+        result += (total_sides*len(v))
+    return result
+
+
+start_time = time.time()
+print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
+
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+def part2():
+    grid = toGrid(input_f)
+
+    #nprint(grid)
+    values = {}
+    visited = []
+    total_plots = []
+    for r,row in enumerate(grid):
+        for c,col in enumerate(row):
+            pos = (r,c)
+            plot = []
+            current = get_value_in_direction(grid,pos)
+            if pos not in visited:
+                
+                x = get_neighbors(grid,pos,visited)
+                plot += x
+                if pos not in plot:
+                    plot.append(pos)
+                if current not in values:
+                    values[current] = []
+                
+                total_plots.append(plot)
+
+    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
+    result = 0
+
+    for v in total_plots:
+        
+        sides = 0
+        for x,y in v:
+
+            for dx, dy in directions:
+                neighbor = (x + dx, y + dy) # Instead of finding all the neighbors, check if  
+                if neighbor in v:
+                    print(neighbor)
+                    sides += 1
+        print()    
+        total_sides = len(v) * 4 - sides
+
+        result += (total_sides*len(v))
+    return result
+    
+start_time = time.time()
+print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')

solution.py

@@ -16,7 +16,7 @@ def part1():
     return
     
 start_time = time.time()
-print('Part 2:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
+print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
 
 
 #########################################