Advent of code 2022 , could puppet tools do it? Collection of community solutions #28
pier4r
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https://adventofcode.com/2022/day/1 my solution with puppet apply (and stdlib installed), puppet 5.5: Click to expand# https://adventofcode.com/2022/day/1
# the problem here is that one cannot reassign variables, so it bekomes tricky.
node 'default' {
# /opt/puppetlabs/bin/puppet apply --modulepath /Users/user/automation/config_management/puppet/mac_workstation_puppet_apply_repo/modules aoc_2022_1.pp
$input = [
# input adjusted, from a simple list, created "strings" with empty strings made explicit.
# find the max sum of the values that aren't separated by a blank (66616)
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Regarding collapsible code blocks: dear-github/dear-github#166 (comment) and then I use the |
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https://adventofcode.com/2022/day/2 my solution with puppet apply (and stdlib installed), puppet 5.5 Click here to expandpuppet 5.5 |
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https://adventofcode.com/2022/day/3 my solution with puppet apply (and stdlib installed), puppet 5.5 Click here to expand |
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Here is my solution for day 3. I did not bother with reading the given input, I just did the logic. The snippet below prints the expected value 157. "vJrwpWtwJgWrhcsFMMfFFhFp",
"jqHRNqRjqzjGDLGLrsFMfFZSrLrFZsSL",
"PmmdzqPrVvPwwTWBwg",
"wMqvLMZHhHMvwLHjbvcjnnSBnvTQFn",
"ttgJtRGJQctTZtZT",
"CrZsJsPPZsGzwwsLwLmpwMDw"
]
$rucksacks.map |$s| { Array(Binary($s,"%s")) }
.map |$a| { $a.map |$v| {
case $v {
Integer[97,122]: { $v-96 }
Integer[65,90]: { $v-65+27 }
}
}}
.map |$a| { $cut = size($a)/2 [$a[0,$cut],$a[$cut,-1]] }
.map |$compartments| { $compartments[0].partition() |$v| { $v in $compartments[1] } }
.map |$common| { $common[0][0] }
.filter |$v| { $v =~ NotUndef }
.reduce(0) |$memo, $v| { $memo + $v }
.notice() |
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5 replies
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Noteworthy here is the conversion from string to ascii/unicode that is done with |
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Also, a tip when programming in this style with a chain of expressions, it is easy to insert a debugging step to see an intermediate result. for example: Which gets the value from the previous step, notices it, and then returns it for the following step. |
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Yes also the code for the example is plenty enough! (normally it should work on the long input too, in my case I had a sneaky error though that was shown by the longer input) Thus such contribution are great, also because one can learn quite a bit from them. For example the What is |
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|
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I changed the logic somewhat as I did not like the multiple functionality of the last reduce step. Now the mapping is explicit and the final reduce just computes the sum. |
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Here is my solution for day 3: https://gist.github.com/smortex/84451e1f576f0e14a3dcc3d2b0ba3449 I switched to Bolt plans for convenience when running the code. I also used |
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2 replies
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Nice! While I can see that putting functions in other .pp file is helpful, could bolt have them in one .pp file as well? |
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Blot plans are intended to use reusable blocks (tasks, plans, functions etc). A plan manifest is supposed to only contain that plan, so you can't add a function there. |
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Here is day 4 $assignments = [
"2-4,6-8",
"2-3,4-5",
"5-7,7-9",
"2-8,3-7",
"6-6,4-6",
"2-6,4-8",
]
$assignments.map |$line| { split($line,",") }
.map |$pair| { [split($pair[0], "-").map |$s| {Integer($s) }, split($pair[1], "-").map |$s| {Integer($s)}] }
.map |$pair | { [Integer[$pair[0][0], $pair[0][1]], Integer[$pair[1][0], $pair[1][1]] ] }
.filter |$pair | { $pair[0] <= $pair[1] or $pair[1] <= $pair[0] }
.then |$x| { notice("Overlapping assignment count = ${size($x)}") } |
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5 replies
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I made use of the feature that data types in Puppet can compute if one is a subset of the other. The |
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Cool the integer range (see, this subtle documentation is nicely emphasized by such exercises) So let me see if I understood. Given the entry The first map creates the array |
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yes @pier4r - you got it right what the steps does. The puppet type system is all about describing "sets of values" and the operators |
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So, to understand better, if I would test overlapping but not containment, say |
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The types don't do that - can only answer if equal or if subset of the other. You would need to unroll the set into an array and then do intersection. (A type that is a range an be iterated over). |
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my Day 4 Click to expandPuppet 5.5 with stdlib |
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0 replies
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I was pretty sure we had some kind of sets, and of course we have as shown by @hlindberg 😍 My solution for day 4: |
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Day 5: https://gist.github.com/smortex/18dd67cbadf931d19359bdc859b75e2a |
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2 replies
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I cannot see this (yet), as I don't have my solution. I am amazed how compact (yet readable, not just cryptic) you and @hlindberg solutions are. I need to learn slowly to do the same. |
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You also went parsing the stack state |
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For Day 5 I am not going to post something. I was first going to do something like what smortex did but I thought I would try to exploit the fact that it is possible to create a new Hash from a "hash_tree" encoded array - basically executing a series of mutations to build a hash. The problem would then become (i) initial setup/parsing of input (ii) constructing a hash (iii) transforming the instructions to a sequence of hash_tree mutations (iv) applying the changes. But I ran out of time to spend on this puzzle. As an example this creates a Hash from a hash tree structure. Would set a resulting hash $h[1][3] to "A" and $h[2][2] to " ". See https://puppet.com/docs/puppet/6/function.html#tree-each and https://puppet.com/docs/puppet/6/function.html#conversion-to-hash-and-struct. |
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changed my mind and went ahead with an impl without the |
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2 replies
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The implementation has the stacks "upside down" as it is easier to take from the beginning of arrays. Note the use of the splat operator on the |
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You actually went to parse the input! I found it not machine friendly and I adjusted accordingly (one could indeed write a program to make it machine friendly). I mean the stack state at the start is great for humans, but not so great for parsing. I like the idea of the matrix/reversing the entries as they are "shown" so to speak. |
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My Day 5, still away from compact. Click to expand### Puppet 5.5 |
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0 replies
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Day 6 - very easy problem if the answer is the index of message start. |
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1 reply
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As it seems there is a part two where the header is 14 chars instead of 4. Simply change the literal 4 in my sample to 14 (or make it a parameter) |
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Day 6 , this time was somewhat compact because it was easy as the functions did most of the work (I am far from being consistently compact and clear) Click me to expandpuppet 5.5 |
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6 replies
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The idea is for part 2 (but part 1 is identical):
We need a substring with 14 non-repeating charts. So the first useful position (minus 1 in puppet, as it starts from 0) where a solution could be is position 14. |
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Every problem has two parts, the second part is visible once the first one is solved. |
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oh, I did not read the instructions properly then it seems :-) |
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so you get a second problem after solving the first? The upfront text and the data I got said nothing about a second part. |
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Yes, it was always like that from day 1. I thought you were interested only in doing a quick example with the sample data skipping the rest. Each day has a part 1 and a part 2 solvable with the same puzzle input. The part 2 is accessible once part 1 is solved. (part 2 is "hidden" as long as there is no correct answer from the user to part 1) |
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Here is day 7 solution (to first part) - again just for the test snippet given in the puzzle: This is a bit of a challenge without the |
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7 replies
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yes, it helped. Thank you! So in the case a file and a dir have the same name (but not the same parent dir), it should be fine right? (because it is the second formula vs the third) |
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To clarify a file at Or in other words, the last entry always describes "the current working directory", and then it either goes up one level, down one level (adds to the path), or adds a file entry followed by a new "cwd". The rest of the input lines are then not needed and can be skipped. |
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You cannot have a directory and a file with the same name in the same directory - so that is never a problem. No file system, even one designed by an Elf, would work since a path like "/a/b/c" would be ambiguous. You can naturally have the same name of files and dirs in different directories - the name is only required to be unique within a directory. The identifiers in my logic are the complete path. |
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The trick is keeping track of the state, which is difficult when variables are immutable. The choice I made was to simply have the "cwd" state last at all times thus making it available to the next iteration. I could have done it with a hash for example and an explicit "cwd" - in retrospect that would perhaps have been easier. |
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yes the track of the current directory was the challenge. I did it with an array in an Hash. |
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Day 9 Today I solved things not too late and in a not too long way, but then I got sidetracked so I am posting late. Interestingly part 2 runs in a shorter time than part 1 (280 secs vs 420 secs, in any case more than 60 seconds on modern systems are too much anyway). Click to expand |
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@smortex saw your day 8 in your repo. (https://github.com/smortex/adventofcode2022/blob/main/plans/day8.pp) Noticed the lambda Nice also the fact that you simulated a |
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I found some time and did day9 - here is my solution: |
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There are 4 configurations of head/tail that require a straight move and 8 configurations that require a diagonal move. To compute this I take the diff of tail and head to get the configuration, I then look that up in a table of all configurations to get the amount to move the tail in [x,y]. The function While the translation from configuration to tail move could have been computed with logic (lots of if-then-else) I felt this was perfect for a table lookup solution. |
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I could have added the 9 configurations that have |
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I should have used the |
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I really like such snippets. |
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Here is my day10 |
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2 replies
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yes the "add a 0 if they take 2 cycles" made things much easier (I didn't do it). Nice. |
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Yeah, that way conditional logic was avoided elsewhere - which would have muddle things up. |
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Today seemingly it was not difficult but the subtle difficulty was how things were spaced a bit. For example "the signal at 20th, 60th, 100th, 140th, etc.. cycle" so not really a multiple of 20, rather something else. Those little challenging details. Click to expand |
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2 replies
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Yeah, there was some ambiguity for me as well as what the values in the array represented - did I end up with the "during" or "after" state? I fiddled with the offsets to get the known answer - so I solved it by trial and error. |
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" I solved it by trial and error." who doesn't ? (maybe chatGPT in the future) |
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day11 $input = @(EOF)
Monkey 0:
Starting items: 79, 98
Operation: new = old * 19
Test: divisible by 23
If true: throw to monkey 2
If false: throw to monkey 3
Monkey 1:
Starting items: 54, 65, 75, 74
Operation: new = old + 6
Test: divisible by 19
If true: throw to monkey 2
If false: throw to monkey 0
Monkey 2:
Starting items: 79, 60, 97
Operation: new = old * old
Test: divisible by 13
If true: throw to monkey 1
If false: throw to monkey 3
Monkey 3:
Starting items: 74
Operation: new = old + 3
Test: divisible by 17
If true: throw to monkey 0
If false: throw to monkey 1
EOF
$monkeys = $input.split("\n")
.filter() |$x| { $x != ""} # drop empty lines
.group_by() |$i,$x| { $i / 6 } # group all lines by monkey (6 line per monkey)
.map |$k, $v| {
{ 'items' => $v[1][18,-1].split(",").map() |$v| { Integer($v) }, # items is single or list of items, map to array of integers
'op' => parseOp($v[2][19,-1]), # operation is an expression, parse it to eval form
'test' => scanf($v[3]," Test: divisible by %f")[0], # test is always divisible, keep divisor
true => scanf($v[4]," If true: throw to monkey %d")[0], # always a throw to a monkey number
false => scanf($v[5]," If false: throw to monkey %d")[0], # always a throw to a monkey number
'count' => 0, # number of inspections
}
} # results in array of hashes, one per monkey
.map |$i, $m| { [$i, $m]}
.convert_to(Hash) # result is hash of monkeys (to easier replace monkey state later)
# for a given number of rounds, compute for all monkeys
$rounds = Integer[0,19]
$eachMonkey = Integer[0, size($monkeys)-1]
$rounds.reduce($monkeys) |$state, $n| {
$eachMonkey.reduce($state) |$state, $mNbr| {
# get throws from monkey $mNbr, and compute new inspection count for this monkey
$m = $state[$mNbr]
$distribution = monkeyRound($state[$mNbr])
$count = $m['count'] + size($m[items])
# Transform distributions to an array of hash merges for altering state
# Start with clearing throwing monkey's items and updating count
$distribution.reduce([[$mNbr, {'items' => [], 'count' => $count}]]) |$memo, $md| {
# add hash to be merged for receiving monkey
$memo << [ $md[0], { 'items' => $state[$md[0]]['items'] + $md[1] } ]
}
# apply each to the state to produce the new state (returned)
.reduce($state) |$state, $change| { $state + { $change[0] => $state[$change[0]] + $change[1] }}
}
}
# Sort the produced state by inspection count
.convert_to(Array)
# sort descending
.sort() |$a, $b| { compare($b[1]['count'], $a[1]['count']) }
# produce monkey business (two largests counts multiplied)
.then() |$x| { $x[0][1]['count'] * $x[1][1]['count'] }
.then() |$x| { "Amount of monkey business = $x" }
.notice()
# monkeyRound returns the distribution from one monkey to other monkeys on the form of
# an array [{monkeyNbr => [item,...]], {monkeyNbr => [item,...]}].
function monkeyRound($m) {
$m['items'].map |$old| {
$new = eval($old, $m['op']) / 3 # evaluated and worry level reduction
$throwsTo = $m[ evenDivisor($new, $m['test']) ] # hash has boolean true/false as key to receiving monkey
[$throwsTo, $new]
}
.group_by() |$x| { $x[0] } # group by monkey thrown to
.map |$k,$v| { [$k, $v.map |$x| { $x[1] }]} # map {m => [[m item], [m, item],...] to [m item, item]
}
function parseOp($opstring) {
$op = $opstring.split(" ")
$numberT = Pattern[/[0-9]+/]
$result = case $op {
["old", default, "old"] : { [ "o", $op[1], "o" ] }
["old", default, $numberT] : { [ "o", $op[1], Integer($op[2]) ] }
[$numberT, default, "old"] : { [ Integer($op[0]), $op[1], "o" ] }
[$numberT, default, $numberT] : { [ Integer($op[0]), $op[1], Integer($op[2]) ] }
default: { fail("not a recognized op $op") }
}
$result
}
function eval($v, $expr) {
$lhs = if $expr[0] == "o" { $v } else { $expr[0] }
$rhs = if $expr[2] == "o" { $v } else { $expr[2] }
case $expr[1] {
"+" : { $lhs + $rhs }
"-" : { $lhs - $rhs }
"*" : { $lhs * $rhs }
"/" : { $lhs / $rhs }
}
}
function evenDivisor($val, $divisor) {
floor($val/$divisor) * $divisor == $val
} |
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10 replies
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Here is a rudimentary bignum implementation using ERB. The representation in Puppet is a string. |
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This works since modern ruby uses bignum representation for integers when needed. The reason puppet does not make use of this is that bignum values are difficult to deal with throughout the ecosystem - serialization, storing in a DB, etc. |
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Ok so an inline epp is enough ? interesting. Then |
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It is an inline ERB, (cannot do this with EPP since it is in puppet language and it only supports 64-bit integers). |
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I would not recommend supporting bignum via ERB like this since they are very costly in terms of computation. I am only showing what it would look like. Would be far better to implement them as ruby functions. |
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While part 1 is solved, I am still on part 2 (that involve a subtle problematic that kills the brute force approach). I could, like some on reddit did, check what the others found as solution, but that for me is like cheating. I try first on my own whether I can crack the problem. I found an initial improvement to handle numbers as, practically, strings, but even in that case the string become HUGE (already some manual testing helps a lot rather than blind programming) and it doesn't help much. But I think there is something more in the mod properties. Click to see the observationFor part2 (that is at the moment included in part1) we have quite the challenge that numbers will be passed around From very small fragments of my university time I investigated again if the modulo operations could be distributed among the
and I am pretty sure (but not 100%) that what I am going to say is true. Say we have: And indeed if one tries with the following values for X: 79,98,54,65 the equality is respected. This solves the problem when one large number is on one node. We could split it its components based on previous So we need to dig deeper. |
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My day 11 solved yesterday (late) . I had to try on paper a lot checking the wiki for properties of the modulo. At the end I remembered things from the Uni years and I got through. Next: someone needs to add BigInt support to the stdlib, because why not. Most likely I will slow down in the coming days because the middle of the month - for the stats I saw - is where the more difficult parts are. Click to expand |
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I can see that smortex did also day 9 and 10, but in his repo those aren't there yet. |
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Total lack of time so I am going slower. Anyway I was thinking: I see how I can simulate a for each or a for (iteration). We had this already. But - I may be tired or just mentally blind - how can one simulate a while in puppet? I mean I could put a very high upper bound to the iterative loop, but is there a way to say "while(condition not met)" ? With a for loop I could do |
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10 replies
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Puppet's Integer[0,10].each |$i| {
notice($i)
if $i > 3 {
break()
}
} |
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hi @joshcooper !! Totally true. Also worth mentioning is that there is puppet apply -e '[1,2,3,4,5].map |$x| { if $x == 2 { next(100) } if $x > 3 { break() } $x }.notice()'
Notice: Scope(Class[main]): [1, 100, 3] |
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Interesting. I checked break in the past and I was unsure if it would work with I mean |
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As you see from just above, |
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Uh interesting. When I tested it myself I put it in the wrong position and puppet complained that a value would never be reached (but only in a reduce). This allows even more useful |
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I got day12 working. It is really slow. Took many hours to compute the larger input. |
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4 replies
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Nice (I didn't read the details because I don't want to yet). my day12 has the state setup complete but I need to find the time to do the final computation and then I'll post it. I let other duties accumulate last week and now I am paying the "debt" 😅 , less time for interesting things. |
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The index function? Is that a new one? I don't see it for 5.5 (sorry I am limited to that still). Although it feels like a filter function? |
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the |
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yes! I missed that function. Is it undocumented? Because in the 5.5 list of functions or in stdlib I didn't find it. |
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I did day 18 - was quite simple: $input = @(EOF)
2,2,2
1,2,2
3,2,2
2,1,2
2,3,2
2,2,1
2,2,3
2,2,4
2,2,6
1,2,5
3,2,5
2,1,5
2,3,5
EOF
$adjacent = [[-1,0,0], [1,0,0],
[0,-1,0], [0,1,0],
[0,0,-1], [0,0,1]]
$lines = $input.split("\n")
$cells = $lines.map |$x| { $x.split(",").map |$x| { Integer($x)-1 }}
$cube = $cells.map |$x| { [$x, true] }.convert_to(Hash, 'hash_tree')
$cells.reduce(0) |$memo, $cell| { $memo + open_sides($cell) }
.then |$count| { "There are $count open sides"}
.notice()
function add_points($p1, $p2) { $p1.map() |$i, $v| { $v + $p2[$i] }}
function open_sides($xyz) {
$adjacent.reduce(6) |$memo, $p| { $memo - get_side(add_points($xyz, $p)) }
}
# get returns 1 if there is a cell at point p, else 0.
function get_side($p) {
$cube[$p[0]].then |$x| { $x[$p[1]] }.then |$x| { $x[$p[2]] }.then |$x| { 1 }.lest || { 0 }
} |
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So finally done Day 12. With the first part I played around. I knew there was this, but still I wanted to explore and it too a lot of the little free time I had. Anyway it doesn't matter as fun as learning are the objective. While I know that my code often is not compact nor efficient (in computing resources) - although I hope it being readable at times - I noticed that editing large data structures is really cumbersome in puppet. At the end I understand this, as normally in puppet one wouldn't manipulate data structures much (maybe one would read them, not edit them so much, and not even having thousands of entries). Therefore I am really pushing what puppet does actually. For example replacing entries in arrays is not that comfortable. While replacing things in an Hash means simply "override the tuple Another problem is to transform and reduce/extend data structures. With Maybe someone will tell me that a simple solution is to chain But during the coding session I didn't think about it (for this I don't do compact code, damn me) and I still think that is ok for little manipulation, while still somewhat clumsy for larger ones. Anyway, my code (as usual, part 1 and part2). It has part11, part12, part13 because I played around with herustics and I deleted a lot of comments that explained trials. Click me |
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12 replies
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Replace at index beyond end will append: |
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If you instead base the function on the flattened tree construct you can set at any index - like this: Notice the empty/undef value at index 4 in the last ouput. |
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Neither of the functions allows you to prepend since negative indexes "count from the end". So, if you need to prepend you can reverse the array, append and reverse again. If you end up with a solution where you need to do that a lot, maybe do the entire computation in the reverse and then reverse the result at the end instead. |
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A way to reason about the "flat tree" is that it is a series of instructions to "set a value at a given path". So the function above reads as "in the root hash set key 'x' to the value of the array to change", then "at the path 'x', $at (the index of the value to change), set the value 100)" |
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Creating a Hash from a flattened tree will assume all containers are hashes unless an array is set as a container. For example: If an array is wanted for the value of |
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Day 13. It was interesting but also hard to debug (thanks to the community one could collect examples of incorrect and correct pairs). In general I repeat that for manipulating large collection puppet needs patience. In short if the operations are like Now my next pick, since data manipulation continues, may be RPL (see here, here and here), that at first requires even more patience than Puppet, but there it is easier to modify data structures. Though there the limit is the ram. The hp 50g has around 200K of free ram, dunno if everything fits there and paging on the SDcard may be too much of a project. Edit: although the flattened tree converted to Hash is really inviting. At least for testing.... Click me |
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Day 20 - got tripped up by move to 0 should be interpreted as move last, other than that it was easy. Here is my version $input = @(EOF)
1
2
-3
3
-2
0
4
EOF
#$input = file("/Users/henrik/day20.txt")
$instr = $input.split("\n").map |$x| { Integer($x) }
$size = size($instr)
notice("size = $size")
$instr.reduce($instr) |$memo, $v| {
if $v == 0 { next($memo) }
$from = $memo.index($v)
$to_ = ($from + $v) % ($size-1)
$to = case $to_ {
0: { $size -1 }
$size-1 : { 0 }
default: { $to_ }
}
$idx = $memo.index($v)
$memo2 = $memo[0,$idx]+$memo[$idx+1,-1]
if size($memo2) +1 != $size { fail("delete of value removed more than one ${size($memo2)}") }
$result = $memo2[0,$to] + [$v] + $memo2[$to,-1]
if size($result) != $size { fail("algorithm dropped data") }
$result
}
.then |$shuffled| {
$zero = $shuffled.index(0)
[1000, 2000, 3000].reduce(0) |$memo, $x| {
$idx = $x + $zero
notice("adding value from index $x + $zero = $idx, with %size = ${$idx % $size}. Value = ${$shuffled[$idx % $size]}")
notice(size($shuffled))
$memo + $shuffled[$idx % $size]
}
}
.then |$x| { notice("sum is $x") }
|
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3 replies
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well, hum... did not work on the long input. I found one bug and I have updated my post. Still get the wrong answer though. I must have misinterpreted something... |
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I think I know what is wrong - since there are duplicates in the input (for which I fixed one bug) it may end up moving the wrong value since it may find a moved value instead of the value that is supposed to move. :-( |
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Yes the sample input is often "not enough" to work out all the bugs. |
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Source: https://adventofcode.com/
Initial Slack thread: https://puppetcommunity.slack.com/archives/C0W298S9G/p1669922681567199
for more solution, check the solution megathread here: https://www.reddit.com/r/adventofcode/
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