Advent of Code 2015 Day 3

Author

Nathan Moore

— Day 3: Perfectly Spherical Houses in a Vacuum —

Santa is delivering his presents in a 2d grid of houses.

However, the elf back at the north pole has had a little too much eggnog, and so his directions are a little off, and Santa ends up visiting some houses more than once. How many houses receive at least one present?

with open('data-2015-03.txt', 'r') as f:
    paren = list(f.read())

Add the first present

x = 0
y = 0
pp = dict()
pp[str(x) + ',' + str(y)] = 1

I like this as a function

def add_item(pp, x, y):
    # create key
    k = str(x) + ',' + str(y)
    # if key exists then add one
    if k in pp:
        pp[k] += 1
    # else create a new key
    else: 
        pp[k] = 1
    # return
    return pp

Add some more presents

for i in range(len(paren)):
    if paren[i] == '^':
        y += 1
        pp = add_item(pp, x, y)
    elif paren[i] == '<':
        x -= 1
        pp = add_item(pp, x, y)      
    elif paren[i] == '>':
        x += 1
        pp = add_item(pp, x, y)      
    else:
        y -= 1
        pp = add_item(pp, x, y)

Print the answer

sum([1 for k,v in pp.items() if v >= 1])

# 2081
2081

Let’s try a different approach with lists

x = 100
y = 100
pp = [[0] * 200 for _ in range(200)]
pp[x][y] += 1

This doesn’t need to be a function without the weird dictionary key

for i in range(len(paren)):
    if paren[i] == '^':
        y += 1
        pp[x][y] += 1
    elif paren[i] == '<':
        x -= 1
        pp[x][y] += 1      
    elif paren[i] == '>':
        x += 1
        pp[x][y] += 1      
    else:
        y -= 1
        pp[x][y] += 1

Flatten this list because I’m not sure about 2d lists and comprehensions

ll = []
for p in pp:
    ll.extend(p)

Sums

My mistake: at least one means >= not >, the first approach was actually fine.

sum([1 for l in ll if l >= 1])

# 2081
2081

— Part Two —

Santa and Robo-Santa are delivering presents, with alternating instructions.

This year, how many houses receive at least one present?

# enlarge this range just in case
x1 = 500
y1 = 500
x2 = 500
y2 = 500
pp = [[0] * 1000 for _ in range(1000)]
pp[x1][y1] += 1

This could maybe be smarter, but I feel like this is effective.
Referring to variables with variable names is weird so best avoided.

i = 0
while i < len(paren) - 1:
    if i % 2 == 0:
        if paren[i] == '^':
            y1 += 1
            pp[x1][y1] += 1
        elif paren[i] == '<':
            x1 -= 1
            pp[x1][y1] += 1      
        elif paren[i] == '>':
            x1 += 1
            pp[x1][y1] += 1      
        else:
            y1 -= 1
            pp[x1][y1] += 1   
    
    # increment
    i += 1
    
    if i % 2 == 1:

        if paren[i] == '^':
            y2 += 1
            pp[x2][y2] += 1
            
        elif paren[i] == '<':
            x2 -= 1
            pp[x2][y2] += 1
            
        elif paren[i] == '>':
            x2 += 1
            pp[x2][y2] += 1
        else:
            y2 -= 1
            pp[x2][y2] += 1

    # increment
    i += 1

Now do the sum again

ll = []
for p in pp:
    ll.extend(p)

sum([1 for l in ll if l >= 1])

# 2341
2341