diff --git a/2017/23/23.md b/2017/23/23.md index 5d8f2a4..edd5449 100644 --- a/2017/23/23.md +++ b/2017/23/23.md @@ -28,8 +28,6 @@ instruction invoked?* Your puzzle answer was `5929`. -The first half of this puzzle is complete! It provides one gold star: \* - ## \-\-- Part Two \-\-- {#part2} Now, it\'s time to fix the problem. @@ -50,8 +48,14 @@ it wouldn\'t even need to run the program. After setting register `a` to `1`, if the program were to run to completion, *what value would be left in register `h`?* -Answer: +Your puzzle answer was `907`. -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](/2017) and +try another puzzle. + +If you still want to see it, you can [get your puzzle input](23/input). diff --git a/2017/23/input.2 b/2017/23/input.2 new file mode 100644 index 0000000..7688e4c --- /dev/null +++ b/2017/23/input.2 @@ -0,0 +1,22 @@ +set e 2 +set g d +mul g e +sub g b +jnz g 2 +set f 0 +sub e -1 +set g e +sub g b +jnz g -8 +sub d -1 +set g d +sub g b +jnz g -13 +jnz f 2 +sub h -1 +set g b +sub g c +jnz g 2 +jnz 1 3 +sub b -17 +jnz 1 -23 diff --git a/2017/23/input.3 b/2017/23/input.3 new file mode 100644 index 0000000..f343ec6 --- /dev/null +++ b/2017/23/input.3 @@ -0,0 +1,31 @@ +a = 1 +b = 79 * 100 + 100000 +c = 79 * 100 + 100000 + 17000 + + 9. set f 1 +10. set d 2 +11. set e 2 + +12. + + +if g != 0 jump to 12 + g = d * e - b + + if g != 0 + f = 0 + g = (e - 1) - b + +21. sub d -1 +22. set g d +23. sub g b +24. jnz g -13 jump to 11 +25. jnz f 2 jump to 27 +26. sub h -1 +27. set g b +28. sub g c +29. jnz g 2 +30. jnz 1 3 +31. sub b -17 +32. jnz 1 -23 jump to 9 + \ No newline at end of file diff --git a/2017/23/solution.py b/2017/23/solution.py index af27e54..a92577f 100644 --- a/2017/23/solution.py +++ b/2017/23/solution.py @@ -3,10 +3,10 @@ import sys,re from pprint import pprint sys.path.insert(0, '../../') from fred import list2int - +from math import sqrt input_f = 'input' -part = 1 +part = 3 ######################################### # # # Part 1 # @@ -34,7 +34,7 @@ if part == 1: instructions = [] Sets = { - 'a': 1, + 'a': 0, 'b': 0, 'c': 0, 'd': 0, @@ -81,4 +81,72 @@ if part == 1: # # ######################################### if part == 2: - exit() + instructions = [] + + Sets = { + 'a': 1, + 'b': 107900, + 'c': 124900, + 'd': 3, + 'e': 107900, + 'f': 1, + 'g': -107897, + 'h': 0 + } + + count = 0 + + with open(input_f) as file: + for line in file: + instructions.append(list(parse_input(line.rstrip()))) + + intr = 0 + while 0 <= intr < len(instructions): + + i = instructions[intr] + + x = i[1] + y = i[2] + + if i[0] == 'set': + Sets[x] = sets_return(y,Sets) + + elif i[0] == 'sub': + Sets[x] -= sets_return(y,Sets) + + elif i[0] == 'mul': + Sets[x] *= sets_return(y,Sets) + count += 1 + + elif i[0] == 'jnz': + if sets_return(x,Sets) != 0: + intr += sets_return(y,Sets)-1 + intr += 1 + print(i) + print(Sets) + input() + print(count) + + +if part == 3: + b = 107900 + d = 2 + e = 2 + h = 0 + for x in range(107900,124900,17): + for y in range(2,x): + if x%y == 0: + print(x,y,h) + h+=1 + print(h) + +# def is_composite(n): +# if n < 2: +# return False +# for i in range(2, int(sqrt(n)) + 1): +# if n % i == 0: +# return True +# return False + +# h = sum(1 for b in range(107900, 124901, 17) if is_composite(b)) +# print(h) \ No newline at end of file diff --git a/2024/03/3.md b/2024/03/3.md new file mode 100644 index 0000000..1bdee0f --- /dev/null +++ b/2024/03/3.md @@ -0,0 +1,81 @@ +## \-\-- Day 3: Mull It Over \-\-- + +\"Our computers are having issues, so I have no idea if we have any +Chief Historians [in +stock]{title="There's a spot reserved for Chief Historians between the green toboggans and the red toboggans. They've never actually had any Chief Historians in stock, but it's best to be prepared."}! +You\'re welcome to check the warehouse, though,\" says the mildly +flustered shopkeeper at the [North Pole Toboggan Rental +Shop](/2020/day/2). The Historians head out to take a look. + +The shopkeeper turns to you. \"Any chance you can see why our computers +are having issues again?\" + +The computer appears to be trying to run a program, but its memory (your +puzzle input) is *corrupted*. All of the instructions have been jumbled +up! + +It seems like the goal of the program is just to *multiply some +numbers*. It does that with instructions like `mul(X,Y)`, where `X` and +`Y` are each 1-3 digit numbers. For instance, `mul(44,46)` multiplies +`44` by `46` to get a result of `2024`. Similarly, `mul(123,4)` would +multiply `123` by `4`. + +However, because the program\'s memory has been corrupted, there are +also many invalid characters that should be *ignored*, even if they look +like part of a `mul` instruction. Sequences like `mul(4*`, `mul(6,9!`, +`?(12,34)`, or `mul ( 2 , 4 )` do *nothing*. + +For example, consider the following section of corrupted memory: + + xmul(2,4)%&mul[3,7]!@^do_not_mul(5,5)+mul(32,64]then(mul(11,8)mul(8,5)) + +Only the four highlighted sections are real `mul` instructions. Adding +up the result of each instruction produces `161` +(`2*4 + 5*5 + 11*8 + 8*5`). + +Scan the corrupted memory for uncorrupted `mul` instructions. *What do +you get if you add up all of the results of the multiplications?* + +Your puzzle answer was `174561379`. + +## \-\-- Part Two \-\-- {#part2} + +As you scan through the corrupted memory, you notice that some of the +conditional statements are also still intact. If you handle some of the +uncorrupted conditional statements in the program, you might be able to +get an even more accurate result. + +There are two new instructions you\'ll need to handle: + +- The `do()` instruction *enables* future `mul` instructions. +- The `don't()` instruction *disables* future `mul` instructions. + +Only the *most recent* `do()` or `don't()` instruction applies. At the +beginning of the program, `mul` instructions are *enabled*. + +For example: + + xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5)) + +This corrupted memory is similar to the example from before, but this +time the `mul(5,5)` and `mul(11,8)` instructions are *disabled* because +there is a `don't()` instruction before them. The other `mul` +instructions function normally, including the one at the end that gets +re-*enabled* by a `do()` instruction. + +This time, the sum of the results is `48` (`2*4 + 8*5`). + +Handle the new instructions; *what do you get if you add up all of the +results of just the enabled multiplications?* + +Your puzzle answer was `106921067`. + +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](3/input). + diff --git a/2024/03/solution.py b/2024/03/solution.py new file mode 100644 index 0000000..fa5f153 --- /dev/null +++ b/2024/03/solution.py @@ -0,0 +1,51 @@ +#!/bin/python3 +import sys,re +from pprint import pprint +sys.path.insert(0, '../../') +from fred import list2int,get_re,nprint,lprint + +input_f = 'input' + +part = 2 +######################################### +# # +# Part 1 # +# # +######################################### +result = 0 +if part == 1: + with open(input_f) as file: + for line in file: + for i in re.findall(r"(mul\((\d+),{1}(\d+)\))",line.rstrip()): + result += (int(i[1])*int(i[2])) + + print(result) + + + +######################################### +# # +# Part 2 # +# # +######################################### +result = 0 +do = True +if part == 2: + with open(input_f) as file: + for line in file: + matches = re.findall(r"mul\(\d+,\d+\)|do\(\)|don't\(\)",line.rstrip()) + for match in matches: + if match == 'do()': + do = True + elif match == "don't()": + do = False + else: + if do: + m = get_re(r"mul\((\d+),{1}(\d+)\)",match) + result += (int(m.group(1))*int(m.group(2))) +print(result) + +# Not really happy with my code. I would prefer to find all mul(x,y) between do() and don't(), +# instead of finding all the mul/do/don't and then turning calculation on/off. + +#(do\(\).*?mul\((\d+),{1}(\d+)\)) \ No newline at end of file diff --git a/2024/03/test.py b/2024/03/test.py new file mode 100644 index 0000000..55eae9e --- /dev/null +++ b/2024/03/test.py @@ -0,0 +1,16 @@ +from re import findall + +total1 = total2 = 0 +enabled = True +data = open('input').read() + +for a, b, do, dont in findall(r"mul\((\d+),(\d+)\)|(do\(\))|(don't\(\))", data): + if do or dont: + enabled = bool(do) + else: + x = int(a) * int(b) + total1 += x + total2 += x * enabled + +print(total1, total2) + diff --git a/README.md b/README.md index 22f3b17..a3df945 100644 --- a/README.md +++ b/README.md @@ -3,7 +3,8 @@ ## 2024 .--'~ ~ ~| .-' * \ / '-. 1 ** - .--'~ ,* ~ | | \_\_\|_/__/ | 2 ** + .--'~ ,* ~ | | >o< \_\_\|_/__/ | 2 ** + .---': ~ '(~), ~| | >@>O< o-_/.()__------| 3 ** ## 2023 . * ~~~~ /\ ' . 14