aoc 2017 (day 1-6)

www/internals/adventofcode.php

@@ -4,6 +4,7 @@ class AdventOfCode
 {
 	const YEARS =
 	[
+		'2017' => [ 'url-aoc'=>'https://adventofcode.com/2017/day/', 'blog-id' => 25, 'github' => 'https://github.com/Mikescher/AdventOfCode2017' ],
 		'2018' => [ 'url-aoc'=>'https://adventofcode.com/2018/day/', 'blog-id' => 23, 'github' => 'https://github.com/Mikescher/AdventOfCode2018' ],
 		'2019' => [ 'url-aoc'=>'https://adventofcode.com/2019/day/', 'blog-id' => 24, 'github' => 'https://github.com/Mikescher/AdventOfCode2019' ],
 	];

www/statics/aoc/2017/01_challenge.txt

@@ -0,0 +1,36 @@
+--- Day 1: Inverse Captcha ---
+
+The night before Christmas, one of Santa's Elves calls you in a panic. "The printer's broken! We can't print the Naughty or Nice List!" By the time you make it to sub-basement 17, there are only a few minutes until midnight. "We have a big problem," she says; "there must be almost fifty bugs in this system, but nothing else can print The List. Stand in this square, quick! There's no time to explain; if you can convince them to pay you in stars, you'll be able to--" She pulls a lever and the world goes blurry.
+
+When your eyes can focus again, everything seems a lot more pixelated than before. She must have sent you inside the computer! You check the system clock: 25 milliseconds until midnight. With that much time, you should be able to collect all fifty stars by December 25th.
+
+Collect stars by solving puzzles. Two puzzles will be made available on each day millisecond in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
+
+You're standing in a room with "digitization quarantine" written in LEDs along one wall. The only door is locked, but it includes a small interface. "Restricted Area - Strictly No Digitized Users Allowed."
+
+It goes on to explain that you may only leave by solving a captcha to prove you're not a human. Apparently, you only get one millisecond to solve the captcha: too fast for a normal human, but it feels like hours to you.
+
+The captcha requires you to review a sequence of digits (your puzzle input) and find the sum of all digits that match the next digit in the list. The list is circular, so the digit after the last digit is the first digit in the list.
+
+For example:
+
+    1122 produces a sum of 3 (1 + 2) because the first digit (1) matches the second digit and the third digit (2) matches the fourth digit.
+    1111 produces 4 because each digit (all 1) matches the next.
+    1234 produces 0 because no digit matches the next.
+    91212129 produces 9 because the only digit that matches the next one is the last digit, 9.
+
+What is the solution to your captcha?
+
+--- Part Two ---
+
+You notice a progress bar that jumps to 50% completion. Apparently, the door isn't yet satisfied, but it did emit a star as encouragement. The instructions change:
+
+Now, instead of considering the next digit, it wants you to consider the digit halfway around the circular list. That is, if your list contains 10 items, only include a digit in your sum if the digit 10/2 = 5 steps forward matches it. Fortunately, your list has an even number of elements.
+
+For example:
+
+    1212 produces 6: the list contains 4 items, and all four digits match the digit 2 items ahead.
+    1221 produces 0, because every comparison is between a 1 and a 2.
+    123425 produces 4, because both 2s match each other, but no other digit has a match.
+    123123 produces 12.
+    12131415 produces 4.

www/statics/aoc/2017/01_input.txt

@@ -0,0 +1 @@
+8231753674683997878179259195565332579493378483264978184143341284379682788518559178822225126625428318115396632681141871952894291898364781898929292614792884883249356728741993224889167928232261325123447569829932951268292953928766755779761837993812528527484487298117739869189415599461746944992651752768158611996715467871381527675219481185217357632445748912726487669881876129192932995282777848496561259839781188719233951619188388532698519298142112853776942545211859134185231768952888462471642851588368445761489225786919778983848113833773768236969923939838755997989537648222217996381757542964844337285428654375499359997792679256881378967852376848812795761118139288152799921176874256377615952758268844139579622754965461884862647423491918913628848748756595463191585555385849335742224855473769411212376446591654846168189278959857681336724221434846946124915271196433144335482787432683848594487648477532498952572515118864475621828118274911298396748213136426357769991314661642612786847135485969889237193822718111269561741563479116832364485724716242176288642371849569664594194674763319687735723517614962575592111286177553435651952853878775431234327919595595658641534765455489561934548474291254387229751472883423413196845162752716925199866591883313638846474321161569892518574346226751366315311145777448781862222126923449311838564685882695889397531413937666673233451216968414288135984394249684886554812761191289485457945866524228415191549168557957633386991931186773843869999284468773866221976873998168818944399661463963658784821796272987155278195355579386768156718813624559264574836134419725187881514665834441359644955768658663278765363789664721736533517774292478192143934318399418188298753351815388561359528533778996296279366394386455544446922653976725113889842749182361253582433319351193862788433113852782596161148992233558144692913791714859516653421917841295749163469751479835492713392861519993791967927773114713888458982796514977717987598165486967786989991998142488631168697963816156374216224386193941566358543266646516247854435356941566492841213424915682394928959116411457967897614457497279472661229548612777155998358618945222326558176486944695689777438164612198225816646583996426313832539918

www/statics/aoc/2017/01_solution-1.py

@@ -0,0 +1,15 @@
+#!/usr/bin/env python3
+
+import aoc
+
+
+rawinput = aoc.read_input(1)
+
+rawinput = rawinput + rawinput[0]
+
+digitsum = 0
+for idx in range(1, len(rawinput)):
+    if rawinput[idx-1] == rawinput[idx]:
+        digitsum = digitsum + int(rawinput[idx])
+
+print(digitsum)

www/statics/aoc/2017/01_solution-2.py

@@ -0,0 +1,14 @@
+#!/usr/bin/env python3
+
+import aoc
+
+
+rawinput = aoc.read_input(1)
+
+inlen = len(rawinput)
+digitsum = 0
+for idx in range(0, inlen):
+    if rawinput[idx] == rawinput[(idx + inlen//2) % inlen]:
+        digitsum = digitsum + int(rawinput[idx])
+
+print(digitsum)

www/statics/aoc/2017/02_challenge.txt

@@ -0,0 +1,41 @@
+--- Day 2: Corruption Checksum ---
+
+As you walk through the door, a glowing humanoid shape yells in your direction. "You there! Your state appears to be idle. Come help us repair the corruption in this spreadsheet - if we take another millisecond, we'll have to display an hourglass cursor!"
+
+The spreadsheet consists of rows of apparently-random numbers. To make sure the recovery process is on the right track, they need you to calculate the spreadsheet's checksum. For each row, determine the difference between the largest value and the smallest value; the checksum is the sum of all of these differences.
+
+For example, given the following spreadsheet:
+
+5 1 9 5
+7 5 3
+2 4 6 8
+
+    The first row's largest and smallest values are 9 and 1, and their difference is 8.
+    The second row's largest and smallest values are 7 and 3, and their difference is 4.
+    The third row's difference is 6.
+
+In this example, the spreadsheet's checksum would be 8 + 4 + 6 = 18.
+
+What is the checksum for the spreadsheet in your puzzle input?
+
+--- Part Two ---
+
+"Great work; looks like we're on the right track after all. Here's a star for your effort." However, the program seems a little worried. Can programs be worried?
+
+"Based on what we're seeing, it looks like all the User wanted is some information about the evenly divisible values in the spreadsheet. Unfortunately, none of us are equipped for that kind of calculation - most of us specialize in bitwise operations."
+
+It sounds like the goal is to find the only two numbers in each row where one evenly divides the other - that is, where the result of the division operation is a whole number. They would like you to find those numbers on each line, divide them, and add up each line's result.
+
+For example, given the following spreadsheet:
+
+5 9 2 8
+9 4 7 3
+3 8 6 5
+
+    In the first row, the only two numbers that evenly divide are 8 and 2; the result of this division is 4.
+    In the second row, the two numbers are 9 and 3; the result is 3.
+    In the third row, the result is 2.
+
+In this example, the sum of the results would be 4 + 3 + 2 = 9.
+
+What is the sum of each row's result in your puzzle input?

www/statics/aoc/2017/02_input.txt

@@ -0,0 +1,16 @@
+116	1470	2610	179	2161	2690	831	1824	2361	1050	2201	118	145	2275	2625	2333
+976	220	1129	553	422	950	332	204	1247	1092	1091	159	174	182	984	713
+84	78	773	62	808	83	1125	1110	1184	145	1277	982	338	1182	75	679
+3413	3809	3525	2176	141	1045	2342	2183	157	3960	3084	2643	119	108	3366	2131
+1312	205	343	616	300	1098	870	1008	1140	1178	90	146	980	202	190	774
+4368	3905	3175	4532	3806	1579	4080	259	2542	221	4395	4464	208	3734	234	4225
+741	993	1184	285	1062	372	111	118	63	843	325	132	854	105	956	961
+85	79	84	2483	858	2209	2268	90	2233	1230	2533	322	338	68	2085	1267
+2688	2022	112	130	1185	103	1847	3059	911	107	2066	1788	2687	2633	415	1353
+76	169	141	58	161	66	65	225	60	152	62	64	156	199	80	56
+220	884	1890	597	3312	593	4259	222	113	2244	3798	4757	216	1127	4400	178
+653	369	216	132	276	102	265	889	987	236	239	807	1076	932	84	864
+799	739	75	1537	82	228	69	1397	1396	1203	1587	63	313	1718	1375	469
+1176	112	1407	136	1482	1534	1384	1202	604	851	190	284	1226	113	114	687
+73	1620	81	1137	812	75	1326	1355	1545	1666	1356	1681	1732	85	128	902
+571	547	160	237	256	30	496	592	385	576	183	692	192	387	647	233

www/statics/aoc/2017/02_solution-1.py

@@ -0,0 +1,13 @@
+#!/usr/bin/env python3
+
+import aoc
+
+
+rawinput = aoc.read_input(2)
+
+result = 0
+for line in rawinput.splitlines():
+    values = list(map(lambda d: int(d), line.split('\t')))
+    result = result + (max(values) - min(values))
+
+print(result)

www/statics/aoc/2017/02_solution-2.py

@@ -0,0 +1,18 @@
+#!/usr/bin/env python3
+
+import aoc
+
+
+rawinput = aoc.read_input(2)
+
+result = 0
+for line in rawinput.splitlines():
+    values = list(map(lambda d: int(d), line.split('\t')))
+    for v1 in values:
+        for v2 in values:
+            if v1 == v2:
+                continue
+            if v1 % v2 == 0:
+                result = result + v1//v2
+
+print(result)

www/statics/aoc/2017/03_challenge.txt

@@ -0,0 +1,44 @@
+--- Day 3: Spiral Memory ---
+
+You come across an experimental new kind of memory stored on an infinite two-dimensional grid.
+
+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 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?
+
+--- Part Two ---
+
+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?

www/statics/aoc/2017/03_input.txt

@@ -0,0 +1 @@
+289326

www/statics/aoc/2017/03_solution-1.py

@@ -0,0 +1,36 @@
+#!/usr/bin/env python3
+
+import aoc
+import math
+
+
+# https://stackoverflow.com/questions/10094745
+def spiralpos(n):
+    k = math.ceil((math.sqrt(n) - 1) / 2)
+    t = 2 * k + 1
+    m = t ** 2
+
+    t -= 1
+
+    if n >= m - t:
+        return k - (m - n), -k
+
+    m -= t
+
+    if n >= m - t:
+        return -k, -k + (m - n)
+
+    m -= t
+
+    if n >= m - t:
+        return -k + (m - n), k
+
+    return k, k - (m - n - t)
+
+
+rawinput = aoc.read_input(3)
+intinput = int(rawinput)
+
+x, y = spiralpos(intinput)
+
+print(abs(x) + abs(y))

www/statics/aoc/2017/03_solution-2.py

@@ -0,0 +1,53 @@
+#!/usr/bin/env python3
+
+import aoc
+import math
+import itertools
+
+
+# https://stackoverflow.com/questions/10094745
+def spiralpos(n):
+    k = math.ceil((math.sqrt(n) - 1) / 2)
+    t = 2 * k + 1
+    m = t ** 2
+
+    t -= 1
+
+    if n >= m - t:
+        return k - (m - n), -k
+
+    m -= t
+
+    if n >= m - t:
+        return -k, -k + (m - n)
+
+    m -= t
+
+    if n >= m - t:
+        return -k + (m - n), k
+
+    return k, k - (m - n - t)
+
+
+rawinput = aoc.read_input(3)
+intinput = int(rawinput)
+
+dirs = [
+    (-1, -1), (00, -1), (+1, -1),
+    (-1, 00),           (+1, 00),
+    (-1, +1), (00, +1), (+1, +1),
+]
+
+grid = dict()
+grid[(0, 0)] = 1
+
+for i in itertools.count(2):
+    x, y = spiralpos(i)
+    v = 0
+    for dx, dy in dirs:
+        if (x + dx, y + dy) in grid:
+            v += grid[(x + dx, y + dy)]
+    if v > intinput:
+        print(v)
+        exit()
+    grid[(x, y)] = v

www/statics/aoc/2017/04_challenge.txt

@@ -0,0 +1,27 @@
+--- 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 password. 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?
+
+--- Part Two ---
+
+For added security, yet another system policy 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?

www/statics/aoc/2017/04_input.txt

@@ -0,0 +1,512 @@
+una bokpr ftz ryw nau yknf fguaczl anu
+tvay wvco bcoblpt fwzg sfsys zvuqll mcbhwz ovcw fgdy
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www/statics/aoc/2017/04_solution-1.py

@@ -0,0 +1,13 @@
+#!/usr/bin/env python3
+
+import aoc
+
+
+rawinput = aoc.read_input(4)
+
+rcount = 0
+for line in rawinput.splitlines():
+    if len(set(line.split(' '))) == len(line.split(' ')):
+        rcount += 1
+print(rcount)
+

www/statics/aoc/2017/04_solution-2.py

@@ -0,0 +1,14 @@
+#!/usr/bin/env python3
+
+import aoc
+
+
+rawinput = aoc.read_input(4)
+
+rcount = 0
+for line in rawinput.splitlines():
+    words = list(map(lambda x: "".join(sorted(list(x))), line.split(' ')))
+    if len(set(words)) == len(words):
+        rcount += 1
+print(rcount)
+

www/statics/aoc/2017/05_challenge.txt

@@ -0,0 +1,36 @@
+--- Day 5: A Maze of Twisty Trampolines, All Alike ---
+
+An urgent interrupt arrives from the CPU: it's trapped in a maze of jump instructions, and it would like assistance from any programs with spare cycles to help find the exit.
+
+The message includes a list of the offsets for each jump. Jumps are relative: -1 moves to the previous instruction, and 2 skips the next one. Start at the first instruction in the list. The goal is to follow the jumps until one leads outside the list.
+
+In addition, these instructions are a little strange; after each jump, the offset of that instruction increases by 1. So, if you come across an offset of 3, you would move three instructions forward, but change it to a 4 for the next time it is encountered.
+
+For example, consider the following list of jump offsets:
+
+0
+3
+0
+1
+-3
+
+Positive jumps ("forward") move downward; negative jumps move upward. For legibility in this example, these offset values will be written all on one line, with the current instruction marked in parentheses. The following steps would be taken before an exit is found:
+
+    (0) 3  0  1  -3  - before we have taken any steps.
+    (1) 3  0  1  -3  - jump with offset 0 (that is, don't jump at all). Fortunately, the instruction is then incremented to 1.
+     2 (3) 0  1  -3  - step forward because of the instruction we just modified. The first instruction is incremented again, now to 2.
+     2  4  0  1 (-3) - jump all the way to the end; leave a 4 behind.
+     2 (4) 0  1  -2  - go back to where we just were; increment -3 to -2.
+     2  5  0  1  -2  - jump 4 steps forward, escaping the maze.
+
+In this example, the exit is reached in 5 steps.
+
+How many steps does it take to reach the exit?
+
+--- Part Two ---
+
+Now, the jumps are even stranger: after each jump, if the offset was three or more, instead decrease it by 1. Otherwise, increase it by 1 as before.
+
+Using this rule with the above example, the process now takes 10 steps, and the offset values after finding the exit are left as 2 3 2 3 -1.
+
+How many steps does it now take to reach the exit?

www/statics/aoc/2017/05_solution-1.py

@@ -0,0 +1,20 @@
+#!/usr/bin/env python3
+
+import aoc
+import itertools
+
+
+rawinput = aoc.read_input(5)
+
+instructions = list(map(lambda x: int(x), rawinput.splitlines()))
+
+ilen = len(instructions)
+pos = 0
+
+for i in itertools.count(1):
+    v = instructions[pos]
+    instructions[pos] += 1
+    pos += v
+    if pos < 0 or pos >= ilen:
+        print(i)
+        exit()

www/statics/aoc/2017/05_solution-2.py

@@ -0,0 +1,20 @@
+#!/usr/bin/env python3
+
+import aoc
+import itertools
+
+
+rawinput = aoc.read_input(5)
+
+instructions = list(map(lambda x: int(x), rawinput.splitlines()))
+
+ilen = len(instructions)
+pos = 0
+
+for i in itertools.count(1):
+    v = instructions[pos]
+    instructions[pos] += -1 if v >= 3 else 1
+    pos += v
+    if pos < 0 or pos >= ilen:
+        print(i)
+        exit()

www/statics/aoc/2017/06_challenge.txt

@@ -0,0 +1,30 @@
+--- Day 6: Memory Reallocation ---
+
+A debugger program here is having an issue: it is trying to repair a memory reallocation routine, but it keeps getting stuck in an infinite loop.
+
+In this area, there are sixteen memory banks; each memory bank can hold any number of blocks. The goal of the reallocation routine is to balance the blocks between the memory banks.
+
+The reallocation routine operates in cycles. In each cycle, it finds the memory bank with the most blocks (ties won by the lowest-numbered memory bank) and redistributes those blocks among the banks. To do this, it removes all of the blocks from the selected bank, then moves to the next (by index) memory bank and inserts one of the blocks. It continues doing this until it runs out of blocks; if it reaches the last memory bank, it wraps around to the first one.
+
+The debugger would like to know how many redistributions can be done before a blocks-in-banks configuration is produced that has been seen before.
+
+For example, imagine a scenario with only four memory banks:
+
+    The banks start with 0, 2, 7, and 0 blocks. The third bank has the most blocks, so it is chosen for redistribution.
+    Starting with the next bank (the fourth bank) and then continuing to the first bank, the second bank, and so on, the 7 blocks are spread out over the memory banks. The fourth, first, and second banks get two blocks each, and the third bank gets one back. The final result looks like this: 2 4 1 2.
+    Next, the second bank is chosen because it contains the most blocks (four). Because there are four memory banks, each gets one block. The result is: 3 1 2 3.
+    Now, there is a tie between the first and fourth memory banks, both of which have three blocks. The first bank wins the tie, and its three blocks are distributed evenly over the other three banks, leaving it with none: 0 2 3 4.
+    The fourth bank is chosen, and its four blocks are distributed such that each of the four banks receives one: 1 3 4 1.
+    The third bank is chosen, and the same thing happens: 2 4 1 2.
+
+At this point, we've reached a state we've seen before: 2 4 1 2 was already seen. The infinite loop is detected after the fifth block redistribution cycle, and so the answer in this example is 5.
+
+Given the initial block counts in your puzzle input, how many redistribution cycles must be completed before a configuration is produced that has been seen before?
+
+--- Part Two ---
+
+Out of curiosity, the debugger would also like to know the size of the loop: starting from a state that has already been seen, how many block redistribution cycles must be performed before that same state is seen again?
+
+In the example above, 2 4 1 2 is seen again after four cycles, and so the answer in that example would be 4.
+
+How many cycles are in the infinite loop that arises from the configuration in your puzzle input?

www/statics/aoc/2017/06_input.txt

@@ -0,0 +1 @@
+14	0	15	12	11	11	3	5	1	6	8	4	9	1	8	4

www/statics/aoc/2017/06_solution-1.py

@@ -0,0 +1,24 @@
+#!/usr/bin/env python3
+
+import aoc
+
+
+rawinput = aoc.read_input(6)
+
+mem = list(map(lambda x: int(x), rawinput.split('\t')))
+
+visited = set()
+
+while True:
+    key = ";".join(map(lambda x: str(x), mem))
+    # print(mem)
+    if key in visited:
+        print(len(visited))
+        exit(0)
+    visited.add(key)
+
+    idx = mem.index(max(mem))
+    val = mem[idx]
+    mem[idx] = 0
+    for i in range(val):
+        mem[(idx+i+1) % len(mem)] += 1

www/statics/aoc/2017/06_solution-2.py

@@ -0,0 +1,24 @@
+#!/usr/bin/env python3
+
+import aoc
+
+
+rawinput = aoc.read_input(6)
+
+mem = list(map(lambda x: int(x), rawinput.split('\t')))
+
+visited = dict()
+
+while True:
+    key = ";".join(map(lambda x: str(x), mem))
+    # print(mem)
+    if key in visited:
+        print(len(visited) - visited[key])
+        exit(0)
+    visited[key] = len(visited)
+
+    idx = mem.index(max(mem))
+    val = mem[idx]
+    mem[idx] = 0
+    for i in range(val):
+        mem[(idx+i+1) % len(mem)] += 1

www/statics/aoc/__all.php

@@ -2,6 +2,15 @@
 
 return
 	[
+		'2017' =>
+		[
+			['day' =>  1, 'parts' => 2, 'title' => 'Inverse Captcha',                         'language' => 'pyth', 'solutions' => ['1390', '1232'] ],
+			['day' =>  2, 'parts' => 2, 'title' => 'Corruption Checksum',                     'language' => 'pyth', 'solutions' => ['32020', '236'] ],
+			['day' =>  3, 'parts' => 2, 'title' => 'Spiral Memory',                           'language' => 'pyth', 'solutions' => ['419', '295229'] ],
+			['day' =>  4, 'parts' => 2, 'title' => 'High-Entropy Passphrases',                'language' => 'pyth', 'solutions' => ['477', '167'] ],
+			['day' =>  5, 'parts' => 2, 'title' => 'A Maze of Twisty Trampolines, All Alike', 'language' => 'pyth', 'solutions' => ['388611', '27763113'] ],
+			['day' =>  6, 'parts' => 2, 'title' => 'Memory Reallocation',                     'language' => 'pyth', 'solutions' => ['11137', '1037'] ],
+		],
 		'2018' =>
 		[
 			['day' =>  1, 'parts' => 2, 'title' => 'Chronal Calibration',                     'language' => 'cs', 'solutions' => ['490', '70357'] ],

www/statics/blog/__all.php

@@ -26,4 +26,5 @@ return
 	[ 'id' => 22, 'date' => '2018-02-06', 'visible' => true,  'title' => 'Homepage iteration 5',                   'fragment' => 'v5.md',              'type' => 'markdown', 'cat' => 'log'  ],
 	[ 'id' => 23, 'date' => '2018-12-01', 'visible' => true,  'title' => 'Advent of Code 2018',                    'fragment' => 'aoc2018.md',         'type' => 'aoc',      'cat' => 'blog', 'extras' => ['aoc:year' => '2018'] ],
 	[ 'id' => 24, 'date' => '2019-12-01', 'visible' => true,  'title' => 'Advent of Code 2019',                    'fragment' => 'aoc2019.md',         'type' => 'aoc',      'cat' => 'blog', 'extras' => ['aoc:year' => '2019'] ],
+	[ 'id' => 25, 'date' => '2020-01-09', 'visible' => true,  'title' => 'Advent of Code 2017',                    'fragment' => 'aoc2017.md',         'type' => 'aoc',      'cat' => 'blog', 'extras' => ['aoc:year' => '2017'] ],
 ];

www/statics/blog/aoc2017.md

@@ -0,0 +1,4 @@
+Advent of Code 2019 is over, but just for fun I will try my hand at the 2017 problems.  
+I'm mostly interested how python - a more common language for these kind of scripting problems - will perform.
+
+I'm not really in a hurry so this could take some time until I'm finished (definitely more than 25 day...)