Added 2017/10 part 1

2017/07/7.md

@@ -67,6 +67,48 @@ larger.)
 Before you\'re ready to help them, you need to make sure your
 information is correct. *What is the name of the bottom program?*
 
-To begin, [get your puzzle input](7/input).
+Your puzzle answer was `xegshds`.
 
-Answer:
+## \-\-- Part Two \-\-- {#part2}
+
+The programs explain the situation: they can\'t get down. Rather, they
+*could* get down, if they weren\'t expending all of their energy trying
+to keep the tower balanced. Apparently, one program has the *wrong
+weight*, and until it\'s fixed, they\'re stuck here.
+
+For any program holding a disc, each program standing on that disc forms
+a sub-tower. Each of those sub-towers are supposed to be the same
+weight, or the disc itself isn\'t balanced. The weight of a tower is the
+sum of the weights of the programs in that tower.
+
+In the example above, this means that for `ugml`\'s disc to be balanced,
+`gyxo`, `ebii`, and `jptl` must all have the same weight, and they do:
+`61`.
+
+However, for `tknk` to be balanced, each of the programs standing on its
+disc *and all programs above it* must each match. This means that the
+following sums must all be the same:
+
+-   `ugml` + (`gyxo` + `ebii` + `jptl`) = 68 + (61 + 61 + 61) = 251
+-   `padx` + (`pbga` + `havc` + `qoyq`) = 45 + (66 + 66 + 66) = 243
+-   `fwft` + (`ktlj` + `cntj` + `xhth`) = 72 + (57 + 57 + 57) = 243
+
+As you can see, `tknk`\'s disc is unbalanced: `ugml`\'s stack is heavier
+than the other two. Even though the nodes above `ugml` are balanced,
+`ugml` itself is too heavy: it needs to be `8` units lighter for its
+stack to weigh `243` and keep the towers balanced. If this change were
+made, its weight would be `60`.
+
+Given that exactly one program is the wrong weight, *what would its
+weight need to be* to balance the entire tower?
+
+Your puzzle answer was `299`.
+
+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](7/input).

2017/07/solution.py

@@ -1,8 +1,9 @@
 #!/bin/python3
 import sys, re
 from pprint import pprint
+from collections import Counter
 
-input_f = 'test'
+input_f = 'input'
 part = 2
 
 
@@ -11,7 +12,8 @@ part = 2
 #              Part 1                   #
 #                                       #
 #########################################
-if part == 1:
+
+def part1():
     children = []
     mains = []
     with open(input_f)as file:
@@ -27,7 +29,10 @@ if part == 1:
 
     for x in mains:
         if x not in children:
-            print(x)
+            return x
+
+if part == 1:
+    print(part1())
 
 
 
@@ -36,17 +41,136 @@ if part == 1:
 #              Part 2                   #
 #                                       #
 #########################################
+
+def find_children(x,grid):
+    # x will be a string like 'tknk'
+    print('Testing ',x)
+    found = []
+    for i in grid:
+        if i.startswith(x):
+            
+            match = re.search(r"-> (.+)", i)
+            if match:
+                found = [item.strip() for item in match.group(1).split(",")]
+                print(found)
+            else:
+                print([i for i in grid if i.startswith(x)])
+    if found != None:
+        for j in found:
+            find_children(j,grid)
+
+def calculate_weights(node, tree, weights, total_weights):
+    # Compute total weight for the current node
+    total = weights[node]
+    for child in tree[node]:
+        total += calculate_weights(child, tree, weights, total_weights)
+    total_weights[node] = total
+    return total
+
+def find_outlier(lst):
+    count = Counter(lst)
+    outlier = None
+    reference = None
+    for num, freq in count.items():
+        if freq == 1 or freq < len(lst) - 1:
+            outlier = num
+        else:  
+            reference = num
+    return outlier,reference
+
+def find_unbalanced(node, tree, weights, total_weights):
+    child_weights = [total_weights[child] for child in tree[node]]
+    if len(set(child_weights)) == 1: 
+        return None
+
+    weight_counts = {w: child_weights.count(w) for w in child_weights}
+    incorrect_weight = next(w for w, count in weight_counts.items() if count == 1)
+    correct_weight = next(w for w, count in weight_counts.items() if count > 1)
+    unbalanced_child = tree[node][child_weights.index(incorrect_weight)]
+
+    deeper_issue = find_unbalanced(unbalanced_child, tree, weights, total_weights)
+    if deeper_issue:
+        return deeper_issue
+    else:
+        return unbalanced_child, weights[unbalanced_child] + (correct_weight - incorrect_weight)
+
 if part == 2:
     children = []
     mains = []
+    grid = []
     with open(input_f)as file:
         for line in file:
             l = line.rstrip()
-            if '->' in l:
-                x = l.split('->')
-                for i in x[1].split(','):
-                    children.append(i.strip())
-                match = re.match(r"^(\S+)",x[0])
-                if match:
-                    mains.append(match.group(1))
+            grid.append(l)
+    
+    tree = {}
+    weights = {}
+    for line in grid:
+        match = re.match(r"(\w+) \((\d+)\)(?: -> (.+))?", line)
+        if match:
+            name = match.group(1)
+            weight = int(match.group(2))
+            children = match.group(3).split(", ") if match.group(3) else []
+            tree[name] = children
+            weights[name] = weight
+
+    total_weights = {}
+
+    calculate_weights(part1(),tree,weights,total_weights)
+    result = find_unbalanced(part1(), tree, weights, total_weights)
+    print(result)
+    #unbalanced_values = []
+    #for key,value in tree.items():
+    #    if value:
+    #        t_values = []
+    #        max_v = 0
+    #        change = 0
+    #        for i in value:
+    #            t_values.append(int(total_weights[i]))
+    #        if all(v == t_values[0] for v in t_values):
+    #            continue
+    #        else:
+    #            print('STOP')
+    #            print(t_values)
+    #            input()
+    #            unbalanced_values.append(t_values)
+
+
+    #print(unbalanced_values)
+    #unbalanced_values = list(reversed(unbalanced_values))
+    #print(unbalanced_values)
+    
+    #for t_values in unbalanced_values:
+    #    outlier,reference = find_outlier(t_values)
+    #    change = max(outlier,reference) - min(outlier,reference)
+    #    max_v = max(outlier,reference)
+
+        #if t_values[0] == t_values[1]:
+        #    if t_values[0] == t_values[2]:
+        #        if t_values[1] == t_values[2]:
+        #            change = t_values[1] - t_values[2]
+        #            max_v = max(t_values[1],t_values[2])
+        #    else:
+        #        change = t_values[0]-t_values[2]
+        #        max_v = max(t_values[0],t_values[2])
+        #else:
+        #    change = t_values[0] - t_values[1]
+        #    max_v = max(t_values[0],t_values[1])
+
+    #    print(change)
+    #    print(max_v)
+
+       
+        
+    #    new_val = max_v-change
+    #    print(new_val)
+
+    #    for odx,o in enumerate(t_values):
+    #        if o == max_v:
+    #            total_weights[value[odx]] = new_val
+
+                    #for kx,x in total_weights.items():
+                    #    if x == max_v:
+                    #        print(weights[kx])
+                    #        print(weights[kx]-change)
     

2017/08/8.md

@@ -0,0 +1,57 @@
+## \-\-- Day 8: I Heard You Like Registers \-\--
+
+You receive a signal
+directly from the CPU. Because of your recent assistance with [jump
+instructions](5), it would like you to compute the result of a series of
+unusual register instructions.
+
+Each instruction consists of several parts: the register to modify,
+whether to increase or decrease that register\'s value, the amount by
+which to increase or decrease it, and a condition. If the condition
+fails, skip the instruction without modifying the register. The
+registers all start at `0`. The instructions look like this:
+
+    b inc 5 if a > 1
+    a inc 1 if b < 5
+    c dec -10 if a >= 1
+    c inc -20 if c == 10
+
+These instructions would be processed as follows:
+
+-   Because `a` starts at `0`, it is not greater than `1`, and so `b` is
+    not modified.
+-   `a` is increased by `1` (to `1`) because `b` is less than `5` (it is
+    `0`).
+-   `c` is decreased by `-10` (to `10`) because `a` is now greater than
+    or equal to `1` (it is `1`).
+-   `c` is increased by `-20` (to `-10`) because `c` is equal to `10`.
+
+After this process, the largest value in any register is `1`.
+
+You might also encounter `<=` (less than or equal to) or `!=` (not equal
+to). However, the CPU doesn\'t have the bandwidth to tell you what all
+the registers are named, and leaves that to you to determine.
+
+*What is the largest value in any register* after completing the
+instructions in your puzzle input?
+
+Your puzzle answer was `4066`.
+
+## \-\-- Part Two \-\-- {#part2}
+
+To be safe, the CPU also needs to know *the highest value held in any
+register during this process* so that it can decide how much memory to
+allocate to these operations. For example, in the above instructions,
+the highest value ever held was `10` (in register `c` after the third
+instruction was evaluated).
+
+Your puzzle answer was `4829`.
+
+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](8/input).

2017/08/solution.py

@@ -0,0 +1,136 @@
+#!/bin/python3
+import sys,re
+from pprint import pprint
+
+input_f = 'input'
+
+part = 2
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+
+if part == 1:
+    grid = []
+    values = {}
+    with open(input_f) as file:
+        for line in file:
+            grid.append(line.rstrip())
+
+    #pprint(grid)
+
+
+    pattern = r"(\w+)\s(inc|dec)\s(-?\d+)\sif\s(\w+)\s([<>=!]+)\s(-?\d+)"
+
+    # Process each instruction
+    for instruction in grid:
+        match = re.match(pattern, instruction)
+        if match:
+            input_var = match.group(1)
+            change = match.group(2)
+            value = int(match.group(3))
+            con1 = match.group(4)
+            con2 = match.group(5)
+            con3 = int(match.group(6))
+
+            #print(f"input = {input_var}, change = {change}, value = {value}, con1 = {con1}, con2 = {con2}, con3 = {con3}")
+
+        else: 
+            print(instruction)
+            print('NO MATCH')
+            exit()
+        
+        #print(f"input = {input_var}, change = {change}, value = {value}, con1 = {con1}, con2 = {con2}, con3 = {con3}")
+
+        if input_var not in values:
+            values[input_var] = 0
+        if con1 not in values:
+            values[con1] = 0
+
+        condition = f"{values[con1]} {con2} {con3}"
+
+        #print('calculation =======',condition)
+
+        if eval(condition):
+            if change == 'inc':
+                values[input_var] += value
+            if change == 'dec':
+                values[input_var] -= value
+
+        
+    #pprint(values)
+    max_v = 0
+    for key, i in values.items():
+        if max_v < i:
+            max_v = i
+    print(max_v)
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+
+def find_max(x):
+    #print(x)
+    max_v = 0
+    for key, i in values.items():
+        if max_v < i:
+            max_v = i
+    return max_v
+
+if part == 2:
+    grid = []
+    values = {}
+    total_max = 0
+    with open(input_f) as file:
+        for line in file:
+            grid.append(line.rstrip())
+
+    #pprint(grid)
+
+
+    pattern = r"(\w+)\s(inc|dec)\s(-?\d+)\sif\s(\w+)\s([<>=!]+)\s(-?\d+)"
+    # Process each instruction
+    for instruction in grid:
+        match = re.match(pattern, instruction)
+        if match:
+            input_var = match.group(1)
+            change = match.group(2)
+            value = int(match.group(3))
+            con1 = match.group(4)
+            con2 = match.group(5)
+            con3 = int(match.group(6))
+
+            #print(f"input = {input_var}, change = {change}, value = {value}, con1 = {con1}, con2 = {con2}, con3 = {con3}")
+
+        else: 
+            print(instruction)
+            print('NO MATCH')
+            exit()
+        
+        #print(f"input = {input_var}, change = {change}, value = {value}, con1 = {con1}, con2 = {con2}, con3 = {con3}")
+
+        if input_var not in values:
+            values[input_var] = 0
+        if con1 not in values:
+            values[con1] = 0
+
+        condition = f"{values[con1]} {con2} {con3}"
+
+        #print('calculation =======',condition)
+
+        if eval(condition):
+            if change == 'inc':
+                values[input_var] += value
+            if change == 'dec':
+                values[input_var] -= value
+
+        temp_max = find_max(values)
+        if total_max < temp_max:
+            total_max = temp_max
+            #print('New Max Found',total_max)
+        print(total_max)
+    #pprint(values)
+

2017/09/9.md

@@ -0,0 +1,103 @@
+## \-\-- Day 9: Stream Processing \-\--
+
+A large stream blocks your path. According to the locals, it\'s not safe
+to [cross the
+stream]{title="\"Don't cross the streams!\", they yell, even though there's only one. They seem to think they're hilarious."}
+at the moment because it\'s full of *garbage*. You look down at the
+stream; rather than water, you discover that it\'s a *stream of
+characters*.
+
+You sit for a while and record part of the stream (your puzzle input).
+The characters represent *groups* - sequences that begin with `{` and
+end with `}`. Within a group, there are zero or more other things,
+separated by commas: either another *group* or *garbage*. Since groups
+can contain other groups, a `}` only closes the *most-recently-opened
+unclosed group* - that is, they are nestable. Your puzzle input
+represents a single, large group which itself contains many smaller
+ones.
+
+Sometimes, instead of a group, you will find *garbage*. Garbage begins
+with `<` and ends with `>`. Between those angle brackets, almost any
+character can appear, including `{` and `}`. *Within* garbage, `<` has
+no special meaning.
+
+In a futile attempt to clean up the garbage, some program has *canceled*
+some of the characters within it using `!`: inside garbage, *any*
+character that comes after `!` should be *ignored*, including `<`, `>`,
+and even another `!`.
+
+You don\'t see any characters that deviate from these rules. Outside
+garbage, you only find well-formed groups, and garbage always terminates
+according to the rules above.
+
+Here are some self-contained pieces of garbage:
+
+-   `<>`, empty garbage.
+-   `<random characters>`, garbage containing random characters.
+-   `<<<<>`, because the extra `<` are ignored.
+-   `<{!>}>`, because the first `>` is canceled.
+-   `<!!>`, because the second `!` is canceled, allowing the `>` to
+    terminate the garbage.
+-   `<!!!>>`, because the second `!` and the first `>` are canceled.
+-   `<{o"i!a,<{i<a>`, which ends at the first `>`.
+
+Here are some examples of whole streams and the number of groups they
+contain:
+
+-   `{}`, `1` group.
+-   `{{{}}}`, `3` groups.
+-   `{{},{}}`, also `3` groups.
+-   `{{{},{},{{}}}}`, `6` groups.
+-   `{<{},{},{{}}>}`, `1` group (which itself contains garbage).
+-   `{<a>,<a>,<a>,<a>}`, `1` group.
+-   `{{<a>},{<a>},{<a>},{<a>}}`, `5` groups.
+-   `{{<!>},{<!>},{<!>},{<a>}}`, `2` groups (since all but the last `>`
+    are canceled).
+
+Your goal is to find the total score for all groups in your input. Each
+group is assigned a *score* which is one more than the score of the
+group that immediately contains it. (The outermost group gets a score of
+`1`.)
+
+-   `{}`, score of `1`.
+-   `{{{}}}`, score of `1 + 2 + 3 = 6`.
+-   `{{},{}}`, score of `1 + 2 + 2 = 5`.
+-   `{{{},{},{{}}}}`, score of `1 + 2 + 3 + 3 + 3 + 4 = 16`.
+-   `{<a>,<a>,<a>,<a>}`, score of `1`.
+-   `{{<ab>},{<ab>},{<ab>},{<ab>}}`, score of `1 + 2 + 2 + 2 + 2 = 9`.
+-   `{{<!!>},{<!!>},{<!!>},{<!!>}}`, score of `1 + 2 + 2 + 2 + 2 = 9`.
+-   `{{<a!>},{<a!>},{<a!>},{<ab>}}`, score of `1 + 2 = 3`.
+
+*What is the total score* for all groups in your input?
+
+Your puzzle answer was `11898`.
+
+## \-\-- Part Two \-\-- {#part2}
+
+Now, you\'re ready to remove the garbage.
+
+To prove you\'ve removed it, you need to count all of the characters
+within the garbage. The leading and trailing `<` and `>` don\'t count,
+nor do any canceled characters or the `!` doing the canceling.
+
+-   `<>`, `0` characters.
+-   `<random characters>`, `17` characters.
+-   `<<<<>`, `3` characters.
+-   `<{!>}>`, `2` characters.
+-   `<!!>`, `0` characters.
+-   `<!!!>>`, `0` characters.
+-   `<{o"i!a,<{i<a>`, `10` characters.
+
+*How many non-canceled characters are within the garbage* in your puzzle
+input?
+
+Your puzzle answer was `5601`.
+
+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](9/input).

2017/09/solution.py

@@ -0,0 +1,46 @@
+#!/bin/python3
+import sys,re
+from pprint import pprint
+
+input_f = 'input'
+
+part = 2
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+if part == 1:
+     with open(input_f) as file:
+        for line in file:
+            clean_string = re.sub(r"!.", "", line.rstrip())
+            clean_string = re.sub(r"<.*?>", "", clean_string)
+            clean_string = re.sub(r",", "", clean_string)
+            level = 0
+            numbers = []
+            for i in clean_string:
+                if i == '{':
+                    level += 1
+                if i == '}':
+                    numbers.append(level)
+                    level -= 1
+
+            print(sum(numbers))
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+if part == 2:
+     with open(input_f) as file:
+        for line in file:
+            #print(line.rstrip(), end=' ')
+            length = 0
+            clean_string = line.rstrip()
+            clean_string = re.sub(r"!.",'',clean_string)
+            clean_string = re.findall(r"<(.*?)>", clean_string)
+
+            for i in clean_string:
+                length += len(i)
+            print(length)

2017/09/test2

@@ -0,0 +1,7 @@
+<>
+<random characters>
+<<<<>
+<{!>}>
+<!!>
+<!!!>>
+<{o"i!a,<{i<a>

2017/10/10.md

@@ -0,0 +1,169 @@
+## \-\-- Day 10: Knot Hash \-\--
+
+You come across some programs that are trying to implement a software
+emulation of a hash based on knot-tying. The hash these programs are
+implementing isn\'t very strong, but you decide to help them anyway. You
+make a mental note to remind the Elves later not to [invent their own
+cryptographic functions]{title="NEW CRYPTOSYSTEM WHO DIS"}.
+
+This hash function simulates tying a knot in a circle of string with 256
+marks on it. Based on the input to be hashed, the function repeatedly
+selects a span of string, brings the ends together, and gives the span a
+half-twist to reverse the order of the marks within it. After doing this
+many times, the order of the marks is used to build the resulting hash.
+
+      4--5   pinch   4  5           4   1
+     /    \  5,0,1  / \/ \  twist  / \ / \
+    3      0  -->  3      0  -->  3   X   0
+     \    /         \ /\ /         \ / \ /
+      2--1           2  1           2   5
+
+To achieve this, begin with a *list* of numbers from `0` to `255`, a
+*current position* which begins at `0` (the first element in the list),
+a *skip size* (which starts at `0`), and a sequence of *lengths* (your
+puzzle input). Then, for each length:
+
+-   *Reverse* the order of that *length* of elements in the *list*,
+    starting with the element at the *current position*.
+-   *Move* the *current position* forward by that *length* plus the
+    *skip size*.
+-   *Increase* the *skip size* by one.
+
+The *list* is circular; if the *current position* and the *length* try
+to reverse elements beyond the end of the list, the operation reverses
+using as many extra elements as it needs from the front of the list. If
+the *current position* moves past the end of the list, it wraps around
+to the front. *Lengths* larger than the size of the *list* are invalid.
+
+Here\'s an example using a smaller list:
+
+Suppose we instead only had a circular list containing five elements,
+`0, 1, 2, 3, 4`, and were given input lengths of `3, 4, 1, 5`.
+
+-   The list begins as `[0] 1 2 3 4` (where square brackets indicate the
+    *current position*).
+-   The first length, `3`, selects `([0] 1 2) 3 4` (where parentheses
+    indicate the sublist to be reversed).
+-   After reversing that section (`0 1 2` into `2 1 0`), we get
+    `([2] 1 0) 3 4`.
+-   Then, the *current position* moves forward by the *length*, `3`,
+    plus the *skip size*, 0: `2 1 0 [3] 4`. Finally, the *skip size*
+    increases to `1`.
+
+```{=html}
+<!-- -->
+```
+-   The second length, `4`, selects a section which wraps:
+    `2 1) 0 ([3] 4`.
+-   The sublist `3 4 2 1` is reversed to form `1 2 4 3`:
+    `4 3) 0 ([1] 2`.
+-   The *current position* moves forward by the *length* plus the *skip
+    size*, a total of `5`, causing it not to move because it wraps
+    around: `4 3 0 [1] 2`. The *skip size* increases to `2`.
+
+```{=html}
+<!-- -->
+```
+-   The third length, `1`, selects a sublist of a single element, and so
+    reversing it has no effect.
+-   The *current position* moves forward by the *length* (`1`) plus the
+    *skip size* (`2`): `4 [3] 0 1 2`. The *skip size* increases to `3`.
+
+```{=html}
+<!-- -->
+```
+-   The fourth length, `5`, selects every element starting with the
+    second: `4) ([3] 0 1 2`. Reversing this sublist (`3 0 1 2 4` into
+    `4 2 1 0 3`) produces: `3) ([4] 2 1 0`.
+-   Finally, the *current position* moves forward by `8`: `3 4 2 1 [0]`.
+    The *skip size* increases to `4`.
+
+In this example, the first two numbers in the list end up being `3` and
+`4`; to check the process, you can multiply them together to produce
+`12`.
+
+However, you should instead use the standard list size of `256` (with
+values `0` to `255`) and the sequence of *lengths* in your puzzle input.
+Once this process is complete, *what is the result of multiplying the
+first two numbers in the list*?
+
+Your puzzle answer was `19591`.
+
+The first half of this puzzle is complete! It provides one gold star: \*
+
+## \-\-- Part Two \-\-- {#part2}
+
+The logic you\'ve constructed forms a single *round* of the *Knot Hash*
+algorithm; running the full thing requires many of these rounds. Some
+input and output processing is also required.
+
+First, from now on, your input should be taken not as a list of numbers,
+but as a string of bytes instead. Unless otherwise specified, convert
+characters to bytes using their [ASCII
+codes](https://en.wikipedia.org/wiki/ASCII#Printable_characters). This
+will allow you to handle arbitrary ASCII strings, and it also ensures
+that your input lengths are never larger than `255`. For example, if you
+are given `1,2,3`, you should convert it to the ASCII codes for each
+character: `49,44,50,44,51`.
+
+Once you have determined the sequence of lengths to use, add the
+following lengths to the end of the sequence: `17, 31, 73, 47, 23`. For
+example, if you are given `1,2,3`, your final sequence of lengths should
+be `49,44,50,44,51,17,31,73,47,23` (the ASCII codes from the input
+string combined with the standard length suffix values).
+
+Second, instead of merely running one *round* like you did above, run a
+total of `64` rounds, using the same *length* sequence in each round.
+The *current position* and *skip size* should be preserved between
+rounds. For example, if the previous example was your first round, you
+would start your second round with the same *length* sequence
+(`3, 4, 1, 5, 17, 31, 73, 47, 23`, now assuming they came from ASCII
+codes and include the suffix), but start with the previous round\'s
+*current position* (`4`) and *skip size* (`4`).
+
+Once the rounds are complete, you will be left with the numbers from `0`
+to `255` in some order, called the *sparse hash*. Your next task is to
+reduce these to a list of only `16` numbers called the *dense hash*. To
+do this, use numeric bitwise
+[XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) to combine
+each consecutive block of `16` numbers in the sparse hash (there are
+`16` such blocks in a list of `256` numbers). So, the first element in
+the dense hash is the first sixteen elements of the sparse hash XOR\'d
+together, the second element in the dense hash is the second sixteen
+elements of the sparse hash XOR\'d together, etc.
+
+For example, if the first sixteen elements of your sparse hash are as
+shown below, and the XOR operator is `^`, you would calculate the first
+output number like this:
+
+    65 ^ 27 ^ 9 ^ 1 ^ 4 ^ 3 ^ 40 ^ 50 ^ 91 ^ 7 ^ 6 ^ 0 ^ 2 ^ 5 ^ 68 ^ 22 = 64
+
+Perform this operation on each of the sixteen blocks of sixteen numbers
+in your sparse hash to determine the sixteen numbers in your dense hash.
+
+Finally, the standard way to represent a Knot Hash is as a single
+[hexadecimal](https://en.wikipedia.org/wiki/Hexadecimal) string; the
+final output is the dense hash in hexadecimal notation. Because each
+number in your dense hash will be between `0` and `255` (inclusive),
+always represent each number as two hexadecimal digits (including a
+leading zero as necessary). So, if your first three numbers are
+`64, 7, 255`, they correspond to the hexadecimal numbers `40, 07, ff`,
+and so the first six characters of the hash would be `4007ff`. Because
+every Knot Hash is sixteen such numbers, the hexadecimal representation
+is always `32` hexadecimal digits (`0`-`f`) long.
+
+Here are some example hashes:
+
+-   The empty string becomes `a2582a3a0e66e6e86e3812dcb672a272`.
+-   `AoC 2017` becomes `33efeb34ea91902bb2f59c9920caa6cd`.
+-   `1,2,3` becomes `3efbe78a8d82f29979031a4aa0b16a9d`.
+-   `1,2,4` becomes `63960835bcdc130f0b66d7ff4f6a5a8e`.
+
+Treating your puzzle input as a string of ASCII characters, *what is the
+Knot Hash of your puzzle input?* Ignore any leading or trailing
+whitespace you might encounter.
+
+Answer:
+
+Although it hasn\'t changed, you can still [get your puzzle
+input](10/input).

2017/10/solution.py

@@ -0,0 +1,49 @@
+#!/bin/python3
+import sys,re
+from pprint import pprint
+
+input_f = 'input'
+
+def list2int(x):
+    return list(map(int, x))
+
+part = 1
+#########################################
+#                                       #
+#              Part 1                   #
+#                                       #
+#########################################
+
+if part == 1:
+
+    size = 256
+    lengths = []
+    skip = 0
+    numbers = []
+    pos = 0
+    for i in range(0,size):
+        numbers.append(i)
+
+    with open(input_f) as file:
+        for line in file:
+            lengths = list2int(line.rsplit()[0].split(','))
+
+    for ldx, length in enumerate(lengths):
+        sub = [numbers[(pos + i) % len(numbers)] for i in range(length)]
+        rev = sub[::-1]
+
+        for i in range(length):
+            numbers[(pos + i) % len(numbers)] = rev[i]
+
+        pos += (length+skip) 
+        pos = pos % len(numbers)
+        skip += 1
+    print(numbers[0]*numbers[1])
+
+#########################################
+#                                       #
+#              Part 2                   #
+#                                       #
+#########################################
+if part == 2:
+    exit()

solution.py

@@ -1,16 +1,17 @@
 #!/bin/python3
-import sys
+import sys,re
 from pprint import pprint
 
-input_f = sys.argv[1]
-
+input_f = 'test'
 
+part = 1
 #########################################
 #                                       #
 #              Part 1                   #
 #                                       #
 #########################################
 
+if part == 1
 with open(input_f) as file:
     for line in file:
 
@@ -22,3 +23,5 @@ with open(input_f) as file:
 #              Part 2                   #
 #                                       #
 #########################################
+if part == 2:
+    exit()