Solved 2024/18 P2

2024/17/17.md

@@ -0,0 +1,126 @@
+## \-\-- Day 17: Chronospatial Computer \-\--
+
+The Historians push the button on their strange device, but this time,
+you all just feel like you\'re [falling](/2018/day/6).
+
+\"Situation critical\", the device announces in a familiar voice.
+\"Bootstrapping process failed. Initializing debugger\....\"
+
+The small handheld device suddenly unfolds into an entire computer! The
+Historians look around nervously before one of them tosses it to you.
+
+This seems to be a 3-bit computer: its program is a list of 3-bit
+numbers (0 through 7), like `0,1,2,3`. The computer also has three
+*registers* named `A`, `B`, and `C`, but these registers aren\'t limited
+to 3 bits and can instead hold any integer.
+
+The computer knows *eight instructions*, each identified by a 3-bit
+number (called the instruction\'s *opcode*). Each instruction also reads
+the 3-bit number after it as an input; this is called its *operand*.
+
+A number called the *instruction pointer* identifies the position in the
+program from which the next opcode will be read; it starts at `0`,
+pointing at the first 3-bit number in the program. Except for jump
+instructions, the instruction pointer increases by `2` after each
+instruction is processed (to move past the instruction\'s opcode and its
+operand). If the computer tries to read an opcode past the end of the
+program, it instead *halts*.
+
+So, the program `0,1,2,3` would run the instruction whose opcode is `0`
+and pass it the operand `1`, then run the instruction having opcode `2`
+and pass it the operand `3`, then halt.
+
+There are two types of operands; each instruction specifies the type of
+its operand. The value of a *literal operand* is the operand itself. For
+example, the value of the literal operand `7` is the number `7`. The
+value of a *combo operand* can be found as follows:
+
+-   Combo operands `0` through `3` represent literal values `0` through
+    `3`.
+-   Combo operand `4` represents the value of register `A`.
+-   Combo operand `5` represents the value of register `B`.
+-   Combo operand `6` represents the value of register `C`.
+-   Combo operand `7` is reserved and will not appear in valid programs.
+
+The eight instructions are as follows:
+
+The `adv` instruction (opcode `0`) performs *division*. The numerator is
+the value in the `A` register. The denominator is found by raising 2 to
+the power of the instruction\'s *combo* operand. (So, an operand of `2`
+would divide `A` by `4` (`2^2`); an operand of `5` would divide `A` by
+`2^B`.) The result of the division operation is *truncated* to an
+integer and then written to the `A` register.
+
+The `bxl` instruction (opcode `1`) calculates the [bitwise
+XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
+of register `B` and the instruction\'s *literal* operand, then stores
+the result in register `B`.
+
+The `bst` instruction (opcode `2`) calculates the value of its *combo*
+operand [modulo](https://en.wikipedia.org/wiki/Modulo)
+8 (thereby keeping only its lowest 3 bits), then writes that value to
+the `B` register.
+
+The `jnz` instruction (opcode `3`) does *nothing* if the `A` register is
+`0`. However, if the `A` register is *not zero*, it
+*jumps*
+by setting the instruction pointer to the value of its *literal*
+operand; if this instruction jumps, the instruction pointer is *not*
+increased by `2` after this instruction.
+
+The `bxc` instruction (opcode `4`) calculates the *bitwise XOR* of
+register `B` and register `C`, then stores the result in register `B`.
+(For legacy reasons, this instruction reads an operand but *ignores*
+it.)
+
+The `out` instruction (opcode `5`) calculates the value of its *combo*
+operand modulo 8, then *outputs* that value. (If a program outputs
+multiple values, they are separated by commas.)
+
+The `bdv` instruction (opcode `6`) works exactly like the `adv`
+instruction except that the result is stored in the *`B` register*. (The
+numerator is still read from the `A` register.)
+
+The `cdv` instruction (opcode `7`) works exactly like the `adv`
+instruction except that the result is stored in the *`C` register*. (The
+numerator is still read from the `A` register.)
+
+Here are some examples of instruction operation:
+
+-   If register `C` contains `9`, the program `2,6` would set register
+    `B` to `1`.
+-   If register `A` contains `10`, the program `5,0,5,1,5,4` would
+    output `0,1,2`.
+-   If register `A` contains `2024`, the program `0,1,5,4,3,0` would
+    output `4,2,5,6,7,7,7,7,3,1,0` and leave `0` in register `A`.
+-   If register `B` contains `29`, the program `1,7` would set register
+    `B` to `26`.
+-   If register `B` contains `2024` and register `C` contains `43690`,
+    the program `4,0` would set register `B` to `44354`.
+
+The Historians\' strange device has finished initializing its debugger
+and is displaying some *information about the program it is trying to
+run* (your puzzle input). For example:
+
+    Register A: 729
+    Register B: 0
+    Register C: 0
+
+    Program: 0,1,5,4,3,0
+
+Your first task is to *determine what the program is trying to output*.
+To do this, initialize the registers to the given values, then run the
+given program, collecting any output produced by `out` instructions.
+(Always join the values produced by `out` instructions with commas.)
+After the above program halts, its final output will be
+`4,6,3,5,6,3,5,2,1,0`.
+
+Using the information provided by the debugger, initialize the registers
+to the given values, then run the program. Once it halts, *what do you
+get if you use commas to join the values it output into a single
+string?*
+
+To begin, [get your puzzle input](17/input).
+
+Answer:
+

2024/17/solution.py

@@ -0,0 +1,31 @@
+#!/bin/python3
+import sys,time,re
+from pprint import pprint
+sys.path.insert(0, '../../')
+from fred import list2int,get_re,nprint,lprint,loadFile
+start_time = time.time()
+
+input_f = 'test'
+
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+def part1():
+    return
+    
+start_time = time.time()
+print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
+
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+def part2():
+    return
+    
+start_time = time.time()
+print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')

2024/18/18.md

@@ -96,8 +96,6 @@ the exit?*
 
 Your puzzle answer was `294`.
 
-The first half of this puzzle is complete! It provides one gold star: \*
-
 ## \-\-- Part Two \-\-- {#part2}
 
 The Historians aren\'t as used to moving around in this pixelated
@@ -137,8 +135,14 @@ Simulate more of the bytes that are about to corrupt your memory space.
 from being reachable from your starting position?* (Provide the answer
 as two integers separated by a comma with no other characters.)
 
-Answer:
+Your puzzle answer was `31,22`.
 
-Although it hasn\'t changed, you can still [get your puzzle
+Both parts of this puzzle are complete! They provide two gold stars:
+\*\*
+
+At this point, you should [return to your Advent calendar](/2024) and
+try another puzzle.
+
+If you still want to see it, you can [get your puzzle
 input](18/input).
 

2024/18/solution.py

@@ -12,56 +12,58 @@ input_f = 'input'
 #              Part 1                   #
 #                                       #
 #########################################
-def part1():
+
-    instructions = []
+if input_f == 'test':
-    grid = []
+    w = 6
+    h = 6
+    size = 12
+else:
     w = 70
     h = 70
-    end = (h,w)
+    size = 1024
-    start = (0,0)
+end = (h,w)
-    with open(input_f) as file:
+start = (0,0)
-        for line in file:
-            l = list2int(line.rstrip().split(','))
-            instructions.append((l[0],l[1]))
-    #print(instructions)
 
-    grid = [[ '.' for x in range(0,w+1)] for y in range(0,h+1)]
+instructions = []
+grid = []
 
-    for i in range(1024):
+with open(input_f) as file:
+    for line in file:
+        l = list2int(line.rstrip().split(','))
+        instructions.append((l[1],l[0]))
+grid = [[ '.' for x in range(0,w+1)] for y in range(0,h+1)]
+
+def is_goal(node):
+    return True if node == end else False
+
+def get_neighbors(node):
+    directions = ['up','down','left','right']
+    offsets = {
+        'up': (-1, 0),
+        'down': (1, 0),
+        'left': (0, -1),
+        'right': (0, 1),
+    }
+    neighbors = []
+
+    for d in directions:
+        tmp = addTuples(offsets[d],node)
+        if get_value_in_direction(grid,node,d) != '#' and grid_valid(tmp[0],tmp[1],grid):
+            neighbors.append((tmp[0],tmp[1]))
+    return neighbors
+
+def part1():
+    for i in range(size):
         x = instructions[i]
         grid[x[0]][x[1]] = '#'
     
-
-    
-    def is_goal(node):
-        #print(node)
-        
-        return True if node == end else False
-    
-    def get_neighbors(node):
-        directions = ['up','down','left','right']
-        offsets = {
-            'up': (-1, 0),
-            'down': (1, 0),
-            'left': (0, -1),
-            'right': (0, 1),
-        }
-        neighbors = []
-
-        # Loop through all the directions
-        for d in directions:
-            tmp = addTuples(offsets[d],node)
-            if get_value_in_direction(grid,node,d) != '#' and grid_valid(tmp[0],tmp[1],grid):
-                neighbors.append((tmp[0],tmp[1]))
-        # Return the list of valid neighbors
-        return neighbors
-    
     goal_nodes, path = bfs((0,0),is_goal,get_neighbors)
-    print(goal_nodes)
-    return len(path[goal_nodes[0]])-1
     
+    return len(path[goal_nodes[0]])-1, path
+
+goal_nodes, path = part1()
 start_time = time.time()
-print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
+print('Part 1:',goal_nodes, '\t\t', round((time.time() - start_time)*1000), 'ms')
 
 
 #########################################
@@ -70,7 +72,13 @@ print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
 #                                       #
 #########################################
 def part2():
-    return
+    new_path = path
+    for i in range(size,len(instructions)):
+        x = instructions[i]
+        grid[x[0]][x[1]] = '#'
+        goal_nodes, new_path = bfs((0,0),is_goal,get_neighbors)
+        if not goal_nodes:
+            return x[1],x[0]
     
 start_time = time.time()
 print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')

2024/19/19.md

@@ -0,0 +1,85 @@
+## \-\-- Day 19: Linen Layout \-\--
+
+Today, The Historians take you up to the [hot springs](/2023/day/12) on
+Gear Island! Very
+[suspiciously](https://www.youtube.com/watch?v=ekL881PJMjI),
+absolutely nothing goes wrong as they begin their careful search of the
+vast field of helixes.
+
+Could this *finally* be your chance to visit the
+[onsen](https://en.wikipedia.org/wiki/Onsen) next door?
+Only one way to find out.
+
+After a brief conversation with the reception staff at the onsen front
+desk, you discover that you don\'t have the right kind of money to pay
+the admission fee. However, before you can leave, the staff get your
+attention. Apparently, they\'ve heard about how you helped at the hot
+springs, and they\'re willing to make a deal: if you can simply help
+them *arrange their towels*, they\'ll let you in for *free*!
+
+Every towel at this onsen is marked with a *pattern of colored stripes*.
+There are only a few patterns, but for any particular pattern, the staff
+can get you as many towels with that pattern as you need. Each
+stripe
+can be *white* (`w`), *blue* (`u`), *black* (`b`), *red* (`r`), or
+*green* (`g`). So, a towel with the pattern `ggr` would have a green
+stripe, a green stripe, and then a red stripe, in that order. (You
+can\'t reverse a pattern by flipping a towel upside-down, as that would
+cause the onsen logo to face the wrong way.)
+
+The Official Onsen Branding Expert has produced a list of *designs* -
+each a long sequence of stripe colors - that they would like to be able
+to display. You can use any towels you want, but all of the towels\'
+stripes must exactly match the desired design. So, to display the design
+`rgrgr`, you could use two `rg` towels and then an `r` towel, an `rgr`
+towel and then a `gr` towel, or even a single massive `rgrgr` towel
+(assuming such towel patterns were actually available).
+
+To start, collect together all of the available towel patterns and the
+list of desired designs (your puzzle input). For example:
+
+    r, wr, b, g, bwu, rb, gb, br
+
+    brwrr
+    bggr
+    gbbr
+    rrbgbr
+    ubwu
+    bwurrg
+    brgr
+    bbrgwb
+
+The first line indicates the available towel patterns; in this example,
+the onsen has unlimited towels with a single red stripe (`r`), unlimited
+towels with a white stripe and then a red stripe (`wr`), and so on.
+
+After the blank line, the remaining lines each describe a design the
+onsen would like to be able to display. In this example, the first
+design (`brwrr`) indicates that the onsen would like to be able to
+display a black stripe, a red stripe, a white stripe, and then two red
+stripes, in that order.
+
+Not all designs will be possible with the available towels. In the above
+example, the designs are possible or impossible as follows:
+
+-   `brwrr` can be made with a `br` towel, then a `wr` towel, and then
+    finally an `r` towel.
+-   `bggr` can be made with a `b` towel, two `g` towels, and then an `r`
+    towel.
+-   `gbbr` can be made with a `gb` towel and then a `br` towel.
+-   `rrbgbr` can be made with `r`, `rb`, `g`, and `br`.
+-   `ubwu` is *impossible*.
+-   `bwurrg` can be made with `bwu`, `r`, `r`, and `g`.
+-   `brgr` can be made with `br`, `g`, and `r`.
+-   `bbrgwb` is *impossible*.
+
+In this example, `6` of the eight designs are possible with the
+available towel patterns.
+
+To get into the onsen as soon as possible, consult your list of towel
+patterns and desired designs carefully. *How many designs are possible?*
+
+To begin, [get your puzzle input](19/input).
+
+Answer:
+

2024/19/solution.py

@@ -0,0 +1,47 @@
+#!/bin/python3
+import sys,time,re
+from pprint import pprint
+sys.path.insert(0, '../../')
+from fred import list2int,get_re,nprint,bfs
+start_time = time.time()
+
+input_f = 'test'
+
+def loadFile():
+    colors = []
+    towels = []
+    with open(input_f) as file:
+        for l,line in enumerate(file):
+            if l == 0:
+                colors = line.rstrip().split(',')
+            if l > 1:
+                towels.append(line.rstrip())
+    return colors,towels
+
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+
+
+
+def part1():
+    colors,towels = loadFile()
+   
+    return 
+    
+start_time = time.time()
+print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
+
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+def part2():
+    return
+    
+start_time = time.time()
+print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')

fred.py

@@ -293,13 +293,16 @@ def nprint(grid, cur: set = None, sign: str = None, positions:list = None):
                     if len(sign) > 1:
                         print(prepend+sign[0] + grid[idx][jdx] + sign[1], end='')  # Print with sign
                     else:
-                        print(prepend+colored(sign,'green',attrs=["underline"]), end=' ')  # Print sign
+                        print(prepend+colored(sign,'blue',attrs=["underline","bold"]), end=' ')  # Print sign
                 else:
                     print(prepend+colored(grid[idx][jdx],'green',attrs=["underline"]), end=' ')
             else:
                 if positions is not None:
                     if (idx,jdx) in positions:
-                        print(prepend+colored(grid[idx][jdx],'green'),end=' ')
+                        if sign is not None:
+                            print(prepend+colored(sign,'green'),end=' ')
+                        else:
+                            print(prepend+colored(grid[idx][jdx],'green'),end=' ')
                     else:
                         print(prepend+grid[idx][jdx], end=' ')
                 else: