Solved 2024/22 P1

This commit is contained in:
FrederikBaerentsen 2024-12-22 12:06:33 +01:00
parent 2f5a9630d2
commit 1b58a3c6d4
3 changed files with 321 additions and 4 deletions

View File

@ -5,7 +5,7 @@ sys.path.insert(0, '../../')
from fred import list2int,get_re,nprint,bfs
start_time = time.time()
input_f = 'test'
input_f = 'input'
def loadFile():
colors = []
@ -13,7 +13,7 @@ def loadFile():
with open(input_f) as file:
for l,line in enumerate(file):
if l == 0:
colors = line.rstrip().split(',')
colors = line.rstrip().replace(" ","").split(',')
if l > 1:
towels.append(line.rstrip())
return colors,towels
@ -27,9 +27,23 @@ def loadFile():
def part1():
colors,towels = loadFile()
return
def createRegex(colors):
return '^(' + '|'.join(colors) + ')+$'
colors,towels = loadFile()
count = 0
r = createRegex(colors)
print(r)
regexObj = re.compile(r)
#print(towels)
for tdx, t in enumerate(towels):
match = re.match(regexObj,t)
if match:
count += 1
print(tdx)
return count
start_time = time.time()
print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')

219
2024/22/22.md Normal file
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@ -0,0 +1,219 @@
## \-\-- Day 22: Monkey Market \-\--
As you\'re all teleported deep into the jungle, a [monkey](/2022/day/11)
steals The Historians\' device! You\'ll need get it back while The
Historians are looking for the Chief.
The monkey that stole the device seems willing to trade it, but only in
exchange for an absurd number of bananas. Your only option is to buy
bananas on the Monkey Exchange Market.
You aren\'t sure how the Monkey Exchange Market works, but one of The
Historians senses trouble and comes over to help. Apparently, they\'ve
been studying these monkeys for a while and have deciphered their
secrets.
Today, the Market is full of monkeys buying *good hiding spots*.
Fortunately, because of the time you recently spent in this jungle, you
know lots of good hiding spots you can sell! If you sell enough hiding
spots, you should be able to get enough bananas to buy the device back.
On the Market, the buyers seem to use random prices, but their prices
are actually only
[pseudorandom](https://en.wikipedia.org/wiki/Pseudorandom_number_generator)!
If you know the secret of how they pick their prices, you can wait for
the perfect time to sell.
The part about secrets is literal, the Historian explains. Each buyer
produces a pseudorandom sequence of secret numbers where each secret is
derived from the previous.
In particular, each buyer\'s *secret* number evolves into the next
secret number in the sequence via the following process:
- Calculate the result of *multiplying the secret number by `64`*.
Then, *mix* this result into the secret number. Finally, *prune* the
secret number.
- Calculate the result of *dividing the secret number by `32`*. Round
the result down to the nearest integer. Then, *mix* this result into
the secret number. Finally, *prune* the secret number.
- Calculate the result of *multiplying the secret number by `2048`*.
Then, *mix* this result into the secret number. Finally, *prune* the
secret number.
Each step of the above process involves *mixing* and *pruning*:
- To *mix* a value into the secret number, calculate the [bitwise
XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)
of the given value and the secret number. Then, the secret number
becomes the result of that operation. (If the secret number is `42`
and you were to *mix* `15` into the secret number, the secret number
would become `37`.)
- To *prune* the secret number, calculate the value of the secret
number
[modulo](https://en.wikipedia.org/wiki/Modulo)
`16777216`. Then, the secret number becomes the result of that
operation. (If the secret number is `100000000` and you were to
*prune* the secret number, the secret number would become
`16113920`.)
After this process completes, the buyer is left with the next secret
number in the sequence. The buyer can repeat this process as many times
as necessary to produce more secret numbers.
So, if a buyer had a secret number of `123`, that buyer\'s next ten
secret numbers would be:
15887950
16495136
527345
704524
1553684
12683156
11100544
12249484
7753432
5908254
Each buyer uses their own secret number when choosing their price, so
it\'s important to be able to predict the sequence of secret numbers for
each buyer. Fortunately, the Historian\'s research has uncovered the
*initial secret number of each buyer* (your puzzle input). For example:
1
10
100
2024
This list describes the *initial secret number* of four different
secret-hiding-spot-buyers on the Monkey Exchange Market. If you can
simulate secret numbers from each buyer, you\'ll be able to predict all
of their future prices.
In a single day, buyers each have time to generate `2000` *new* secret
numbers. In this example, for each buyer, their initial secret number
and the 2000th new secret number they would generate are:
1: 8685429
10: 4700978
100: 15273692
2024: 8667524
Adding up the 2000th new secret number for each buyer produces
`37327623`.
For each buyer, simulate the creation of 2000 new secret numbers. *What
is the sum of the 2000th secret number generated by each buyer?*
Your puzzle answer was `14476723788`.
The first half of this puzzle is complete! It provides one gold star: \*
## \-\-- Part Two \-\-- {#part2}
Of course, the secret numbers aren\'t the prices each buyer is offering!
That would be
ridiculous. Instead,
the *prices* the buyer offers are just the *ones digit* of each of their
secret numbers.
So, if a buyer starts with a secret number of `123`, that buyer\'s first
ten *prices* would be:
3 (from 123)
0 (from 15887950)
6 (from 16495136)
5 (etc.)
4
4
6
4
4
2
This price is the number of *bananas* that buyer is offering in exchange
for your information about a new hiding spot. However, you still don\'t
speak [monkey](/2022/day/21), so you can\'t negotiate with the buyers
directly. The Historian speaks a little, but not enough to negotiate;
instead, he can ask another monkey to negotiate on your behalf.
Unfortunately, the monkey only knows how to decide when to sell by
looking at the *changes* in price. Specifically, the monkey will only
look for a specific sequence of *four consecutive changes* in price,
then immediately sell when it sees that sequence.
So, if a buyer starts with a secret number of `123`, that buyer\'s first
ten secret numbers, prices, and the associated changes would be:
123: 3
15887950: 0 (-3)
16495136: 6 (6)
527345: 5 (-1)
704524: 4 (-1)
1553684: 4 (0)
12683156: 6 (2)
11100544: 4 (-2)
12249484: 4 (0)
7753432: 2 (-2)
Note that the first price has no associated change because there was no
previous price to compare it with.
In this short example, within just these first few prices, the highest
price will be `6`, so it would be nice to give the monkey instructions
that would make it sell at that time. The first `6` occurs after only
two changes, so there\'s no way to instruct the monkey to sell then, but
the second `6` occurs after the changes `-1,-1,0,2`. So, if you gave the
monkey that sequence of changes, it would wait until the first time it
sees that sequence and then immediately sell your hiding spot
information at the current price, winning you `6` bananas.
Each buyer only wants to buy one hiding spot, so after the hiding spot
is sold, the monkey will move on to the next buyer. If the monkey
*never* hears that sequence of price changes from a buyer, the monkey
will never sell, and will instead just move on to the next buyer.
Worse, you can only give the monkey *a single sequence* of four price
changes to look for. You can\'t change the sequence between buyers.
You\'re going to need as many bananas as possible, so you\'ll need to
*determine which sequence* of four price changes will cause the monkey
to get you the *most bananas overall*. Each buyer is going to generate
`2000` secret numbers after their initial secret number, so, for each
buyer, you\'ll have *`2000` price changes* in which your sequence can
occur.
Suppose the initial secret number of each buyer is:
1
2
3
2024
There are many sequences of four price changes you could tell the
monkey, but for these four buyers, the sequence that will get you the
most bananas is `-2,1,-1,3`. Using that sequence, the monkey will make
the following sales:
- For the buyer with an initial secret number of `1`, changes
`-2,1,-1,3` first occur when the price is `7`.
- For the buyer with initial secret `2`, changes `-2,1,-1,3` first
occur when the price is `7`.
- For the buyer with initial secret `3`, the change sequence
`-2,1,-1,3` *does not occur* in the first 2000 changes.
- For the buyer starting with `2024`, changes `-2,1,-1,3` first occur
when the price is `9`.
So, by asking the monkey to sell the first time each buyer\'s prices go
down `2`, then up `1`, then down `1`, then up `3`, you would get `23`
(`7 + 7 + 9`) bananas!
Figure out the best sequence to tell the monkey so that by looking for
that same sequence of changes in every buyer\'s future prices, you get
the most bananas in total. *What is the most bananas you can get?*
Answer:
Although it hasn\'t changed, you can still [get your puzzle
input](22/input).

84
2024/22/solution.py Normal file
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@ -0,0 +1,84 @@
#!/bin/python3
import sys,time,re
from pprint import pprint
from math import trunc
sys.path.insert(0, '../../')
from fred import list2int,get_re,nprint,lprint,loadFile
start_time = time.time()
from functools import cache
input_f = 'test'
@cache
def genSecretP1(number:int):
mix = number * 64
step1 = (number ^ mix)%16777216
mix = int(trunc(step1/32))
step2 = (step1 ^ mix)%16777216
mix = step2 * 2048
step3 = (step2 ^ mix)%16777216
return step3
#########################################
# #
# Part 1 #
# #
#########################################
def part1():
numbers = list2int(loadFile(input_f))
r = 10
result = 0
for n in numbers:
number = n
for i in range(r):
number = genSecretP1(number)
result += number
return result
start_time = time.time()
print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
#########################################
# #
# Part 2 #
# #
#########################################
@cache
def genSecretP2(number:int):
mix = number * 64
step1 = (number ^ mix)%16777216
mix = int(trunc(step1/32))
step2 = (step1 ^ mix)%16777216
mix = step2 * 2048
step3 = (step2 ^ mix)%16777216
print(str(number)+": "+str(number)[-1])
return step3
def part2():
numbers = list2int(loadFile(input_f))
r = 10
result = 0
for n in numbers:
number = n
for i in range(r):
number = genSecretP2(number)
result += number
return result
start_time = time.time()
print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')