Renamed rule to reaction

diff --git a/advent14/src/advent14.hs b/advent14/src/advent14.hs
index 1fafe9bce157c425d898d3099832e89265e4ed1d..4e35ee25605fa687e24bf65f88f219f912862f42 100644 (file)
--- a/advent14/src/advent14.hs
+++ b/advent14/src/advent14.hs
@@ -17,75 +17,66 @@ import qualified Data.Set as S
 
 
 data Reagent = Reagent { _quantity :: Int, _chemical :: String } deriving (Ord, Eq, Show)
-data Rule = Rule {_lhs :: S.Set Reagent, _rhs :: Reagent} deriving (Eq, Show)
+data Reaction = Reaction {_lhs :: S.Set Reagent, _rhs :: Reagent} deriving (Eq, Show)
 
-type RuleBase = M.Map String Rule
+type Reactions = M.Map String Reaction
 type Requirement = M.Map String Int
 
 
 main :: IO ()
 main = do 
         text <- TIO.readFile "data/advent14.txt"
-        -- let rules = successfulParse text
-        -- let ruleBase = mkRuleBase rules
-        let ruleBase = successfulParse text
-        -- print rules 
-        -- print ruleBase
-        print $ part1 ruleBase
-        print $ part2 ruleBase
+        let reactions = successfulParse text
+        print $ part1 reactions
+        print $ part2 reactions
 
 oreLimit :: Int
 oreLimit = 10^12
 
-mkRuleBase :: [Rule] -> RuleBase
-mkRuleBase = foldl' addRule M.empty
-    where addRule base rule = M.insert (_chemical $ _rhs rule) rule base
-
-
--- part1 rules = required!"ORE"
+-- part1 reactions = required!"ORE"
 --     where required0 = M.singleton "FUEL" 1
---           required = produce rules required
-part1 rules = oreForFuel rules 1
+--           required = produce reactions required
+part1 reactions = oreForFuel reactions 1
 
-part2 rules = searchFuel rules (upper `div` 2) upper 
-    where upper = findUpper rules (oreLimit `div` base)
-          base = oreForFuel rules 1
+part2 reactions = searchFuel reactions (upper `div` 2) upper 
+    where upper = findUpper reactions (oreLimit `div` base)
+          base = oreForFuel reactions 1
 
-oreForFuel :: RuleBase -> Int -> Int
-oreForFuel rules n = required!"ORE"
+oreForFuel :: Reactions -> Int -> Int
+oreForFuel reactions n = required!"ORE"
     where required0 = M.singleton "FUEL" n
-          required = produce rules required0 
+          required = produce reactions required0 
 
-findUpper :: RuleBase -> Int -> Int
+findUpper :: Reactions -> Int -> Int
 -- findUpper _ n | trace ("Upper " ++ show n) False = undefined
-findUpper rules n = if ore > oreLimit
+findUpper reactions n = if ore > oreLimit
                     then n
-                    else findUpper rules (n * 2)
-    where ore = oreForFuel rules n 
+                    else findUpper reactions (n * 2)
+    where ore = oreForFuel reactions n 
 
-searchFuel :: RuleBase -> Int -> Int -> Int
+searchFuel :: Reactions -> Int -> Int -> Int
 -- searchFuel _ lower upper | trace ("Search " ++ show lower ++ " - " ++ show upper) False = undefined
-searchFuel rules lower upper 
+searchFuel reactions lower upper 
     | upper == lower = upper
     | otherwise = if ore > oreLimit
-                  then searchFuel rules lower (mid - 1)
-                  else searchFuel rules mid upper
+                  then searchFuel reactions lower (mid - 1)
+                  else searchFuel reactions mid upper
     where mid = (upper + lower + 1) `div` 2
-          ore = oreForFuel rules mid
+          ore = oreForFuel reactions mid
 
 
-produce :: RuleBase -> Requirement -> Requirement
-produce rules required 
+produce :: Reactions -> Requirement -> Requirement
+produce reactions required 
     | M.null outstanding = required 
-    | otherwise = produce rules required''
+    | otherwise = produce reactions required''
     where outstanding =  M.filter (> 0) $ nonOre required
           (chem, qty) = M.findMin outstanding
-          rule = rules!chem
-          productQty = _quantity $ _rhs rule
+          reaction = reactions!chem
+          productQty = _quantity $ _rhs reaction
           applications = max 1 (qty `div` productQty)
           qty' = qty - (applications * productQty)
           required' = M.insert chem qty' required
-          required'' = S.foldl (addRequrirement applications) required' (_lhs rule) 
+          required'' = S.foldl (addRequrirement applications) required' (_lhs reaction) 
 
 
 nonOre :: Requirement -> Requirement
@@ -114,16 +105,20 @@ arrowP = symb "=>"
 commaP = symb ","
 identifierP = some alphaNumChar <* sc
 
-rulesP = mkRuleBase <$> many ruleP
+reactionsP = mkReactions <$> many reactionP
 
-ruleP = Rule <$> reagentsP <* arrowP <*> reagentP
+reactionP = Reaction <$> reagentsP <* arrowP <*> reagentP
 
 reagentP = Reagent <$> integer <*> identifierP
 reagentsP = S.fromList <$> reagentP `sepBy` commaP
 
+mkReactions :: [Reaction] -> Reactions
+mkReactions = foldl' addReaction M.empty
+    where addReaction base reaction = M.insert (_chemical $ _rhs reaction) reaction base
+
 -- successfulParse :: Text -> [Vec]
-successfulParse :: Text -> RuleBase
+successfulParse :: Text -> Reactions
 successfulParse input = 
-        case parse rulesP "input" input of
+        case parse reactionsP "input" input of
                 Left  _err -> M.empty -- TIO.putStr $ T.pack $ parseErrorPretty err
-                Right rules -> rules
+                Right reactions -> reactions

diff --git a/problems/day14.html b/problems/day14.html
new file mode 100644 (file)
index 0000000..3c6f506
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@@ -0,0 +1,209 @@
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+<header><div><h1 class="title-global"><a href="/">Advent of Code</a></h1><nav><ul><li><a href="/2019/about">[About]</a></li><li><a href="/2019/events">[Events]</a></li><li><a href="https://teespring.com/adventofcode-2019" target="_blank">[Shop]</a></li><li><a href="/2019/settings">[Settings]</a></li><li><a href="/2019/auth/logout">[Log Out]</a></li></ul></nav><div class="user">Neil Smith <a href="/2019/support" class="supporter-badge" title="Advent of Code Supporter">(AoC++)</a> <span class="star-count">28*</span></div></div><div><h1 class="title-event">&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<span class="title-event-wrap">/^</span><a href="/2019">2019</a><span class="title-event-wrap">$/</span></h1><nav><ul><li><a href="/2019">[Calendar]</a></li><li><a href="/2019/support">[AoC++]</a></li><li><a href="/2019/sponsors">[Sponsors]</a></li><li><a href="/2019/leaderboard">[Leaderboard]</a></li><li><a href="/2019/stats">[Stats]</a></li></ul></nav></div></header>
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+<article class="day-desc"><h2>--- Day 14: Space Stoichiometry ---</h2><p>As you approach the rings of Saturn, your ship's <em>low fuel</em> indicator turns on.  There isn't any fuel here, but the rings have plenty of raw material.  Perhaps your ship's <span title="Yes, the acronym is intentional.">Inter-Stellar Refinery Union</span> brand <em>nanofactory</em> can turn these raw materials into fuel.</p>
+<p>You ask the nanofactory to produce a list of the <em>reactions</em> it can perform that are relevant to this process (your puzzle input). Every reaction turns some quantities of specific <em>input chemicals</em> into some quantity of an <em>output chemical</em>. Almost every <em>chemical</em> is produced by exactly one reaction; the only exception, <code>ORE</code>, is the raw material input to the entire process and is not produced by a reaction.</p>
+<p>You just need to know how much <code><em>ORE</em></code> you'll need to collect before you can produce one unit of <code><em>FUEL</em></code>.</p>
+<p>Each reaction gives specific quantities for its inputs and output; reactions cannot be partially run, so only whole integer multiples of these quantities can be used.  (It's okay to have leftover chemicals when you're done, though.) For example, the reaction <code>1 A, 2 B, 3 C =&gt; 2 D</code> means that exactly 2 units of chemical <code>D</code> can be produced by consuming exactly 1 <code>A</code>, 2 <code>B</code> and 3 <code>C</code>.  You can run the full reaction as many times as necessary; for example, you could produce 10 <code>D</code> by consuming 5 <code>A</code>, 10 <code>B</code>, and 15 <code>C</code>.</p>
+<p>Suppose your nanofactory produces the following list of reactions:</p>
+<pre><code>10 ORE =&gt; 10 A
+1 ORE =&gt; 1 B
+7 A, 1 B =&gt; 1 C
+7 A, 1 C =&gt; 1 D
+7 A, 1 D =&gt; 1 E
+7 A, 1 E =&gt; 1 FUEL
+</code></pre>
+<p>The first two reactions use only <code>ORE</code> as inputs; they indicate that you can produce as much of chemical <code>A</code> as you want (in increments of 10 units, each 10 costing 10 <code>ORE</code>) and as much of chemical <code>B</code> as you want (each costing 1 <code>ORE</code>).  To produce 1 <code>FUEL</code>, a total of <em>31</em> <code>ORE</code> is required: 1 <code>ORE</code> to produce 1 <code>B</code>, then 30 more <code>ORE</code> to produce the 7 + 7 + 7 + 7 = 28 <code>A</code> (with 2 extra <code>A</code> wasted) required in the reactions to convert the <code>B</code> into <code>C</code>, <code>C</code> into <code>D</code>, <code>D</code> into <code>E</code>, and finally <code>E</code> into <code>FUEL</code>. (30 <code>A</code> is produced because its reaction requires that it is created in increments of 10.)</p>
+<p>Or, suppose you have the following list of reactions:</p>
+<pre><code>9 ORE =&gt; 2 A
+8 ORE =&gt; 3 B
+7 ORE =&gt; 5 C
+3 A, 4 B =&gt; 1 AB
+5 B, 7 C =&gt; 1 BC
+4 C, 1 A =&gt; 1 CA
+2 AB, 3 BC, 4 CA =&gt; 1 FUEL
+</code></pre>
+<p>The above list of reactions requires <em>165</em> <code>ORE</code> to produce 1 <code>FUEL</code>:</p>
+<ul>
+<li>Consume 45 <code>ORE</code> to produce 10 <code>A</code>.</li>
+<li>Consume 64 <code>ORE</code> to produce 24 <code>B</code>.</li>
+<li>Consume 56 <code>ORE</code> to produce 40 <code>C</code>.</li>
+<li>Consume 6 <code>A</code>, 8 <code>B</code> to produce 2 <code>AB</code>.</li>
+<li>Consume 15 <code>B</code>, 21 <code>C</code> to produce 3 <code>BC</code>.</li>
+<li>Consume 16 <code>C</code>, 4 <code>A</code> to produce 4 <code>CA</code>.</li>
+<li>Consume 2 <code>AB</code>, 3 <code>BC</code>, 4 <code>CA</code> to produce 1 <code>FUEL</code>.</li>
+</ul>
+<p>Here are some larger examples:</p>
+<ul>
+<li><p><em>13312</em> <code>ORE</code> for 1 <code>FUEL</code>:</p>
+<pre><code>157 ORE =&gt; 5 NZVS
+165 ORE =&gt; 6 DCFZ
+44 XJWVT, 5 KHKGT, 1 QDVJ, 29 NZVS, 9 GPVTF, 48 HKGWZ =&gt; 1 FUEL
+12 HKGWZ, 1 GPVTF, 8 PSHF =&gt; 9 QDVJ
+179 ORE =&gt; 7 PSHF
+177 ORE =&gt; 5 HKGWZ
+7 DCFZ, 7 PSHF =&gt; 2 XJWVT
+165 ORE =&gt; 2 GPVTF
+3 DCFZ, 7 NZVS, 5 HKGWZ, 10 PSHF =&gt; 8 KHKGT
+</code></pre></li>
+<li><p><em>180697</em> <code>ORE</code> for 1 <code>FUEL</code>:</p>
+<pre><code>2 VPVL, 7 FWMGM, 2 CXFTF, 11 MNCFX =&gt; 1 STKFG
+17 NVRVD, 3 JNWZP =&gt; 8 VPVL
+53 STKFG, 6 MNCFX, 46 VJHF, 81 HVMC, 68 CXFTF, 25 GNMV =&gt; 1 FUEL
+22 VJHF, 37 MNCFX =&gt; 5 FWMGM
+139 ORE =&gt; 4 NVRVD
+144 ORE =&gt; 7 JNWZP
+5 MNCFX, 7 RFSQX, 2 FWMGM, 2 VPVL, 19 CXFTF =&gt; 3 HVMC
+5 VJHF, 7 MNCFX, 9 VPVL, 37 CXFTF =&gt; 6 GNMV
+145 ORE =&gt; 6 MNCFX
+1 NVRVD =&gt; 8 CXFTF
+1 VJHF, 6 MNCFX =&gt; 4 RFSQX
+176 ORE =&gt; 6 VJHF
+</code></pre></li>
+<li><p><em>2210736</em> <code>ORE</code> for 1 <code>FUEL</code>:</p>
+<pre><code>171 ORE => 8 CNZTR
+7 ZLQW, 3 BMBT, 9 XCVML, 26 XMNCP, 1 WPTQ, 2 MZWV, 1 RJRHP => 4 PLWSL
+114 ORE => 4 BHXH
+14 VRPVC => 6 BMBT
+6 BHXH, 18 KTJDG, 12 WPTQ, 7 PLWSL, 31 FHTLT, 37 ZDVW => 1 FUEL
+6 WPTQ, 2 BMBT, 8 ZLQW, 18 KTJDG, 1 XMNCP, 6 MZWV, 1 RJRHP => 6 FHTLT
+15 XDBXC, 2 LTCX, 1 VRPVC => 6 ZLQW
+13 WPTQ, 10 LTCX, 3 RJRHP, 14 XMNCP, 2 MZWV, 1 ZLQW => 1 ZDVW
+5 BMBT => 4 WPTQ
+189 ORE => 9 KTJDG
+1 MZWV, 17 XDBXC, 3 XCVML => 2 XMNCP
+12 VRPVC, 27 CNZTR => 2 XDBXC
+15 KTJDG, 12 BHXH => 5 XCVML
+3 BHXH, 2 VRPVC => 7 MZWV
+121 ORE => 7 VRPVC
+7 XCVML => 6 RJRHP
+5 BHXH, 4 VRPVC => 5 LTCX
+</code></pre></li>
+</ul>
+<p>Given the list of reactions in your puzzle input, <em>what is the minimum amount of <code>ORE</code> required to produce exactly 1 <code>FUEL</code>?</em></p>
+</article>
+<p>Your puzzle answer was <code>598038</code>.</p><article class="day-desc"><h2 id="part2">--- Part Two ---</h2><p>After collecting <code>ORE</code> for a while, you check your cargo hold: <em>1 trillion</em> (<em>1000000000000</em>) units of <code>ORE</code>.</p>
+<p><em>With that much ore</em>, given the examples above:</p>
+<ul>
+<li>The 13312 <code>ORE</code>-per-<code>FUEL</code> example could produce <em>82892753</em> <code>FUEL</code>.</li>
+<li>The 180697 <code>ORE</code>-per-<code>FUEL</code> example could produce <em>5586022</em> <code>FUEL</code>.</li>
+<li>The 2210736 <code>ORE</code>-per-<code>FUEL</code> example could produce <em>460664</em> <code>FUEL</code>.</li>
+</ul>
+<p>Given 1 trillion <code>ORE</code>, <em>what is the maximum amount of <code>FUEL</code> you can produce?</em></p>
+</article>
+<p>Your puzzle answer was <code>2269325</code>.</p><p class="day-success">Both parts of this puzzle are complete! They provide two gold stars: **</p>
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