Today the puzzle got a bit trickier than Day 1.

The submarine seems to already have a planned course (your puzzle input). You should probably figure out where it's going. For example:

    forward 5
    down 5
    forward 8
    up 3
    down 8
    forward 2

Your horizontal position and depth both start at 0. The steps above would then modify them as follows:

    forward 5 adds 5 to your horizontal position, a total of 5.
    down 5 adds 5 to your depth, resulting in a value of 5.
    forward 8 adds 8 to your horizontal position, a total of 13.
    up 3 decreases your depth by 3, resulting in a value of 2.
    down 8 adds 8 to your depth, resulting in a value of 10.
    forward 2 adds 2 to your horizontal position, a total of 15.

After following these instructions, you would have a horizontal position of 15 and a depth of 10. (Multiplying these together produces 150.)

Calculate the horizontal position and depth you would have after following the planned course. What do you get if you multiply your final horizontal position by your final depth?

So Pandas here we go again:

df = pd.read_csv("./aoc_day_02_data.txt", delimiter=" ",header=None)
df.columns = ["command", "value"]

Alright, reading in the data and naming the columns are the same steps as yesterday. Now we have to columns.

horizontal = df[df['command']=="forward"]["value"].sum()

The horizontal value can be calculated with the sum function when we filter the data frame to rows where the command is “forward”

depth = df[df['command']=="down"]["value"].sum() - df[df['command']=="up"]["value"].sum()

The depth can calculated by summing up the down and up commands separately and subtract the sums from each other.

Now we have to multiply the depth and the position to get the solution

position = depth * horizontal

Part 2

    down X increases your aim by X units.
    up X decreases your aim by X units.
    forward X does two things:
        It increases your horizontal position by X units.
        It increases your depth by your aim multiplied by X.

Now, the above example does something different:

    forward 5 adds 5 to your horizontal position, a total of 5. Because your aim is 0, your depth does not change.
    down 5 adds 5 to your aim, resulting in a value of 5.
    forward 8 adds 8 to your horizontal position, a total of 13. Because your aim is 5, your depth increases by 8*5=40.
    up 3 decreases your aim by 3, resulting in a value of 2.
    down 8 adds 8 to your aim, resulting in a value of 10.
    forward 2 adds 2 to your horizontal position, a total of 15. Because your aim is 10, your depth increases by 2*10=20 to a total of 60.

After following these new instructions, you would have a horizontal position of 15 and a depth of 60. 
(Multiplying these produces 900.)

Using this new interpretation of the commands, calculate the horizontal position and depth you would have after following the planned course. 
What do you get if you multiply your final horizontal position by your final depth?

To get an overview I simplified the table

     a   d
f 5  0   0
d 5  5
f 8  5  40
u 3  2
d 8 10
f 2 10  20

Here I had a hard time to do it with pandas so vanilla python to the rescue:

if __name__ == '__main__':

    with open("./aoc_day_02_data.txt") as file:
        lines = file.readlines()
        lines = [line.rstrip() for line in lines]

    horizontal = 0
    current_aim = 0
    depth = 0
    for line in lines:
        print(line)
        command, value = line.split(" ")
        value = int(value)
        if command == "forward":
            horizontal += value
            depth += value * current_aim
        if command == "down":
            current_aim += value
        if command == "up":
            current_aim += value * -1

    print(f"horizontal: {horizontal}")
    print(f"depth: {depth}")
    print(horizontal * depth)

Update

I’ve figured out how to do it with Pandas as well

import pandas as pd

df = pd.read_csv("./aoc_day_02_test_data.txt", delimiter=" ",header=None)
df.columns = ["command", "value"]

horizontal = df[df['command']=="forward"]["value"].sum()

df.loc[df['command']=="up", "value"] = df[df['command']=="up"].mul(-1)
df["aim"] = 0

df.loc[df['command']!="forward", "aim"] = df[df['command']!="forward"]["value"]
df["current_aim"] = df["aim"].cumsum()

df.loc[df['command']=="forward", "depth"] = df[df['command']=="forward"]["value"] * df[df['command']=="forward"]["current_aim"]
depth = df["depth"].sum()

The post 2021 – Advent of code – Day 2 appeared first on Creatronix.

What is advent of code?

It’s an advent Calendar for programmers. You get 25 challenges starting December 1st. Caveat: you have to solve the challenge to be eligible for the next day’s challenge 🙂

Day 1 Challenge – Part 1

On the first day your first task is to count how many times a value is bigger than its predecessor. They give us some sample data

199 N/A
200 bigger
208 bigger
210 bigger
200 smaller
207 bigger
240 bigger
269 bigger
260 smaller
263 bigger

When we count the times a value is bigger we get seven times bigger.

The actual data contains 2000 rows. This isn’t exactly big data but I’ve wanted to dust off my Pandas skill, so here we go:

Let’s look at the data

import pandas as pd

df = pd.read_csv("./aoc_day_01_data.txt", header=None)
df.describe

With the read_csv() function we can read in our data file and convert it into a data frame. It’s important to hand over the header=None. Otherwise pandas assumes the first row is a column header.

df.describe gives us:

<bound method NDFrame.describe of          0
0      159
1      158
2      174
3      196
4      197
...    ...
1995  8538
1996  8543
1997  8545
1998  8557
1999  8568

[2000 rows x 1 columns]>

Because we want to reference the columns by name we add a column header

df.columns = ["original"]

To compare the nth cell with its n+1th cell neighbour be add a new column but shift the values

df['shifted'] = df['original'].shift(-1)

The output looks like this:

We add another column where we place the value True when the value from the current row in the shifted column is bigger than in the original column:

df['increased'] = (df['shifted'] > df['original'])

Now it starts to look like the sample data from the introduction:

the last thing we have to do is counting how many times True occurs:

true_count = df['increased'].sum()

which gives us “1583”

This is a bit of a hack because it assumes that True equals 1 and False == 0

A more elegant solution is to use value_counts:

df['increased'].value_counts(dropna=False)

No the output is:

True     1583
False     417
Name: increased, dtype: int64

And 1583 is the number we are looking for. This earned us our first golden star and unlocked the second part of the challenge:

Part 2

The second part is a bit more challenging because we have to sum up three adjacent values and compare them to the next three values.

199  A       
200  A B     
208  A B C   
210    B C D
200  E   C D
207  E F   D
240  E F G
269    F G H
260      G H
263        H

I created a new notebook and started like part 1 with reading the data and naming the first column

import pandas as pd

df = pd.read_csv("./aoc_day_01_data.txt", header=None)
df.columns = ["original"]

To add the sum of three values to the row of the first value we use the following code

indexer = pd.api.indexers.FixedForwardWindowIndexer(window_size=3)
df["rolling_sum"] = df.original.rolling(window=indexer).sum()

This demonstrates the power of Pandas once more: you have integrated sliding window functions!

The rest is equal to part one “shift, compare and count”

df['shifted_rs'] = df['rolling_sum'].shift(-1)
df['increased_rs'] = (df['shifted_rs'] > df['rolling_sum'])
true_count = df['increased_rs'].sum()
true_count

As a little Fingerübung I did the same with vanilla Python:

data = []
with open("./aoc_day_01_test_data.txt") as f:
    for line in f:
        data.append(int(line.rstrip()))

triplet_sums = []

for i, v in enumerate(data):
    if i < (len(data) - 2):
        triplet_sum = data[i] + data[i+1] + data[i+2]
        triplet_sums.append(triplet_sum)
print(triplet_sums)

sums_larger_than_previous_sums = 0
for i, v in enumerate(triplet_sums):
    if i < (len(triplet_sums) - 1):
        if triplet_sums[i] < triplet_sums[i+1]:
            sums_larger_than_previous_sums += 1

print(sums_larger_than_previous_sums)

Which works but is less elegant.

Stay tuned for more!

The post 2021 – Advent of code – Day 1 appeared first on Creatronix.