Added 2017/04

2017/04/4.md

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+## \-\-- Day 4: High-Entropy Passphrases \-\--
+
+A new system policy has been put in place that requires all accounts to
+use a *passphrase* instead of simply a pass*word*. A passphrase consists
+of a series of words (lowercase letters) separated by spaces.
+
+To ensure security, a valid passphrase must contain no duplicate words.
+
+For example:
+
+-   `aa bb cc dd ee` is valid.
+-   `aa bb cc dd aa` is not valid - the word `aa` appears more than
+    once.
+-   `aa bb cc dd aaa` is valid - `aa` and `aaa` count as different
+    words.
+
+The system\'s full passphrase list is available as your puzzle input.
+*How many passphrases are valid?*
+
+Your puzzle answer was `383`.
+
+## \-\-- Part Two \-\-- {#part2}
+
+For added security, [yet another system
+policy]{title="Because as everyone knows, the number of rules is proportional to the level of security."}
+has been put in place. Now, a valid passphrase must contain no two words
+that are anagrams of each other - that is, a passphrase is invalid if
+any word\'s letters can be rearranged to form any other word in the
+passphrase.
+
+For example:
+
+-   `abcde fghij` is a valid passphrase.
+-   `abcde xyz ecdab` is not valid - the letters from the third word can
+    be rearranged to form the first word.
+-   `a ab abc abd abf abj` is a valid passphrase, because *all* letters
+    need to be used when forming another word.
+-   `iiii oiii ooii oooi oooo` is valid.
+-   `oiii ioii iioi iiio` is not valid - any of these words can be
+    rearranged to form any other word.
+
+Under this new system policy, *how many passphrases are valid?*
+
+Your puzzle answer was `265`.
+
+Both parts of this puzzle are complete! They provide two gold stars:
+\*\*
+
+At this point, you should [return to your Advent calendar](/2017) and
+try another puzzle.
+
+If you still want to see it, you can [get your puzzle
+input](4/input).

2017/04/solution.py

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+#!/bin/python3
+import sys
+from pprint import pprint
+import numpy as np
+
+input_f = sys.argv[1]
+
+count = 0
+
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+        
+with open(input_f) as file:
+    for line in file:
+        line = line.rstrip().split()
+        u,c = np.unique(line,return_counts=True)
+        if u[c > 1].size == 0:
+            count += 1
+
+print(count)
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+count = 0
+
+with open(input_f) as file:
+    for line in file:
+        line = line.rstrip().split()
+        for idx,x in enumerate(line):
+            line[idx] = ''.join(sorted(x))
+        u,c = np.unique(line,return_counts=True)
+        if u[c > 1].size == 0:
+            count += 1
+print(count)

2017/04/test2

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+abcde fghij
+abcde xyz ecdab
+a ab abc abd abf abj
+iiii oiii ooii oooi oooo
+oiii ioii iioi iiio