Added 2017/20

2017/20/20.md

@@ -53,7 +53,49 @@ particle `0`, and so, in the long run, particle `0` will stay closest.
 *Which particle will stay closest to position `<0,0,0>`* in the long
 term?
 
-To begin, [get your puzzle input](20/input).
+Your puzzle answer was `258`.
+
+The first half of this puzzle is complete! It provides one gold star: \*
+
+## \-\-- Part Two \-\-- {#part2}
+
+To simplify the problem further, the GPU would like to remove any
+particles that *collide*. Particles collide if their positions ever
+*exactly match*. Because particles are updated simultaneously, *more
+than two particles* can collide at the same time and place. Once
+particles collide, they are removed and cannot collide with anything
+else after that tick.
+
+For example:
+
+    p=<-6,0,0>, v=< 3,0,0>, a=< 0,0,0>    
+    p=<-4,0,0>, v=< 2,0,0>, a=< 0,0,0>    -6 -5 -4 -3 -2 -1  0  1  2  3
+    p=<-2,0,0>, v=< 1,0,0>, a=< 0,0,0>    (0)   (1)   (2)            (3)
+    p=< 3,0,0>, v=<-1,0,0>, a=< 0,0,0>
+
+    p=<-3,0,0>, v=< 3,0,0>, a=< 0,0,0>    
+    p=<-2,0,0>, v=< 2,0,0>, a=< 0,0,0>    -6 -5 -4 -3 -2 -1  0  1  2  3
+    p=<-1,0,0>, v=< 1,0,0>, a=< 0,0,0>             (0)(1)(2)      (3)   
+    p=< 2,0,0>, v=<-1,0,0>, a=< 0,0,0>
+
+    p=< 0,0,0>, v=< 3,0,0>, a=< 0,0,0>    
+    p=< 0,0,0>, v=< 2,0,0>, a=< 0,0,0>    -6 -5 -4 -3 -2 -1  0  1  2  3
+    p=< 0,0,0>, v=< 1,0,0>, a=< 0,0,0>                       X (3)      
+    p=< 1,0,0>, v=<-1,0,0>, a=< 0,0,0>
+
+    ------destroyed by collision------    
+    ------destroyed by collision------    -6 -5 -4 -3 -2 -1  0  1  2  3
+    ------destroyed by collision------                      (3)         
+    p=< 0,0,0>, v=<-1,0,0>, a=< 0,0,0>
+
+In this example, particles `0`, `1`, and `2` are simultaneously
+destroyed at the time and place marked `X`. On the next tick, particle
+`3` passes through unharmed.
+
+*How many particles are left* after all collisions are resolved?
 
 Answer:
 
+Although it hasn\'t changed, you can still [get your puzzle
+input](20/input).
+

2017/20/solution.py

@@ -0,0 +1,56 @@
+#!/bin/python3
+import sys,re,math
+from pprint import pprint
+sys.path.insert(0, '../../')
+from fred import get_re,list2int
+
+input_f = 'input'
+
+part = 1
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+
+lines = []
+
+if part == 1:
+    with open(input_f) as file:
+        for line in file:
+            lines.append(get_re(r"^p=<(.*)>.*v=<(.*)>.*a=<(.*)>$",line.rstrip()))  
+            
+#pprint(lines)
+
+magnitude = []
+
+def calculate_magnitude(vector):
+    return math.sqrt(sum(component ** 2 for component in vector))
+
+def find_slowest_as_time_infinite(vectors):
+    min_acceleration = float('inf')
+    slowest_item = None
+
+    for i, (_, a) in enumerate(vectors):
+        acceleration_magnitude = calculate_magnitude(a)
+        if acceleration_magnitude < min_acceleration:
+            min_acceleration = acceleration_magnitude
+            slowest_item = i
+    
+    return slowest_item
+
+
+vectors = []
+
+for i in lines:
+    v = list2int(i.group(2).split(','))
+    a = list2int(i.group(3).split(','))
+    vectors.append((v,a))
+
+print(find_slowest_as_time_infinite(vectors))
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################

fred.py

@@ -1,6 +1,14 @@
+import re
+
 def list2int(x):
     return list(map(int, x))
 
+def get_re(pattern,str):
+    match = re.match(pattern, str)
+    if match:
+        return match
+    return None 
+
 def ppprint(x):
     for idx,i in enumerate(x):
         for jdx,j in enumerate(i):