Added 2017/03

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FrederikBaerentsen 2024-11-15 16:47:27 +01:00
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## \-\-- Day 3: Spiral Memory \-\--
You come across an experimental new kind of memory stored on an
[infinite two-dimensional
grid]{title="Good thing we have all these infinite two-dimensional grids lying around!"}.
Each square on the grid is allocated in a spiral pattern starting at a
location marked `1` and then counting up while spiraling outward. For
example, the first few squares are allocated like this:
17 16 15 14 13
18 5 4 3 12
19 6 1 2 11
20 7 8 9 10
21 22 23---> ...
While this is very space-efficient (no squares are skipped), requested
data must be carried back to square `1` (the location of the only access
port for this memory system) by programs that can only move up, down,
left, or right. They always take the shortest path: the [Manhattan
Distance](https://en.wikipedia.org/wiki/Taxicab_geometry) between the
location of the data and square `1`.
For example:
- Data from square `1` is carried `0` steps, since it\'s at the access
port.
- Data from square `12` is carried `3` steps, such as: down, left,
left.
- Data from square `23` is carried only `2` steps: up twice.
- Data from square `1024` must be carried `31` steps.
*How many steps* are required to carry the data from the square
identified in your puzzle input all the way to the access port?
Your puzzle answer was `371`.
The first half of this puzzle is complete! It provides one gold star: \*
## \-\-- Part Two \-\-- {#part2}
As a stress test on the system, the programs here clear the grid and
then store the value `1` in square `1`. Then, in the same allocation
order as shown above, they store the sum of the values in all adjacent
squares, including diagonals.
So, the first few squares\' values are chosen as follows:
- Square `1` starts with the value `1`.
- Square `2` has only one adjacent filled square (with value `1`), so
it also stores `1`.
- Square `3` has both of the above squares as neighbors and stores the
sum of their values, `2`.
- Square `4` has all three of the aforementioned squares as neighbors
and stores the sum of their values, `4`.
- Square `5` only has the first and fourth squares as neighbors, so it
gets the value `5`.
Once a square is written, its value does not change. Therefore, the
first few squares would receive the following values:
147 142 133 122 59
304 5 4 2 57
330 10 1 1 54
351 11 23 25 26
362 747 806---> ...
What is the *first value written* that is *larger* than your puzzle
input?
Answer:
Your puzzle input is still `368078`{.puzzle-input}.

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#!/bin/python3
import sys
from pprint import pprint
import numpy as np
import math
def manhattan_distance(a, b):
return np.abs(a - b).sum()
def spiral_ccw(A):
A = np.array(A)
out = []
while(A.size):
out.append(A[0][::-1]) # first row reversed
A = A[1:][::-1].T # cut off first row and rotate clockwise
return np.concatenate(out)
def base_spiral(nrow, ncol):
return spiral_ccw(np.arange(nrow*ncol).reshape(nrow,ncol))[::-1]
def to_spiral(A):
A = np.array(A)
B = np.empty_like(A)
B.flat[base_spiral(*A.shape)] = A.flat
return B
input_f = sys.argv[1]
#########################################
# #
# Part 1 #
# #
#########################################
number = int(input_f)
org = number
while not math.sqrt(number).is_integer():
number+=1
mid = int((math.sqrt(number)-1)/2)
length = int(math.sqrt(number))
a = [mid, mid]
arr = to_spiral(np.arange(1,number+1).reshape(length,length))
b = np.where(arr == org)
print(manhattan_distance(np.array(a),np.array([b[0][0],b[1][0]])))
#########################################
# #
# Part 2 #
# #
#########################################

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37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18 05 04 03 12 29
40 19 06 01 02 11 28
41 20 07 08 09 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49

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65 64 63 62 61 60 59 58 57
66 37 36 35 34 33 32 31 56
67 38 17 16 15 14 13 30 55
68 39 18 05 04 03 12 29 54
69 40 19 06 01 02 11 28 53
70 41 20 07 08 09 10 27 52
71 42 21 22 23 24 25 26 51
72 43 44 45 46 47 48 49 50
73 74 75 76 77 78 79 80 81