learnyouahaskell · chrisdenman · Sep 20, 2024 · Sep 20, 2024 · Sep 20, 2024 · Sep 20, 2024
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@@ -47,7 +47,7 @@ <h1>A Fistful of Monads</h1>
 applicative functors are only beefed up functors.
 </p>
 
-<img src="assets/images/a-fistful-of-monads/smugpig.png" alt="more cool than u" class="right" width="307" height="186">
+<img src="assets/images/a-fistful-of-monads/smugpig.png" alt="more cool than u" class="right" width="307" height="168">
 
 <p>
 When we started off with functors, we saw that it's possible to map functions
@@ -2153,11 +2153,11 @@ <h3>Associativity</h3>
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@@ -165,7 +165,7 @@ <h1>Learn You a Haskell for Great Good!</h1>
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-    This work is licensed under a <a rel="license" href="https://creativecommons.org/licenses/by-nc-sa/3.0/" rel="nofollow">Creative Commons Attribution-Noncommercial-Share Alike 3.0 Unported License</a> because I couldn't find a license with an even longer name.
+    This work is licensed under a <a rel="license" href="https://creativecommons.org/licenses/by-nc-sa/3.0/" rel="nofollow">Creative Commons Attribution-Noncommercial-Share Alike 3.0 Unported License</a> because I couldn't find a licence with an even longer name.
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@@ -177,11 +177,11 @@ <h1>Learn You a Haskell for Great Good!</h1>
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         <h1>FAQ</h1>
 <img src="assets/images/faq/turtle.png" alt="turtle???" class="right" width="349" height="204">
 <h2>Can I put this tutorial on my site or change it or whatever?</h2>
-<p>Sure, it's licensed under a <a href="https://creativecommons.org/licenses/by-nc-sa/3.0/" rel="nofollow">creative commons</a> license, so you can share and change this, as long as you do it with a smile on your face and for non-commercial purposes.</p>
+<p>Sure, it's licensed under a <a href="https://creativecommons.org/licenses/by-nc-sa/3.0/" rel="nofollow">creative commons</a> licence, so you can share and change this, as long as you do it with a smile on your face and for non-commercial purposes.</p>
 <h2>Do you recommend any other Haskell reading material?</h2>
-<p>There are loads of awesome tutorials out there, but I'd just like to point out how great <a href="http://book.realworldhaskell.org/read/" rel="nofollow">Real World Haskell</a> is. It's really great. I feel it complements this tutorial nicely. This tutorial focuses mainly on using simple examples to ease beginners into learning Haskell, whereas Real World Haskell really shows you how to do useful stuff with it.</p>
+<p>There are loads of awesome tutorials out there, but I'd just like to point out how great <a href="https://book.realworldhaskell.org/read/" rel="nofollow">Real World Haskell</a> is. It's really great. I feel it complements this tutorial nicely. This tutorial focuses mainly on using simple examples to ease beginners into learning Haskell, whereas Real World Haskell really shows you how to do useful stuff with it.</p>
 <p>Another great Haskell resource is <a href="https://tryhaskell.org" rel="nofollow">Try Haskell</a>, which allows you to try Haskell right in your browser and offers a rad interactive walkthrough.</p>
 <h2>How do I get in touch with you?</h2>
 <p>The best way would be to shoot me an email to bonus at learnyouahaskell dot com. I kinda suck at email though, so please, please don't be mad if I don't reply in a timely fashion!</p>
@@ -46,11 +46,11 @@ <h2>Tell me about yourself!</h2>
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@@ -3109,11 +3109,11 @@ <h2>Making monads</h2>
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@@ -36,7 +36,7 @@ <h1>Functionally Solving Problems</h1>
 <a name="reverse-polish-notation-calculator"></a><h2>Reverse Polish notation calculator</h2>
 <p>Usually when we write mathematical expressions in school, we write them in an infix manner. For instance, we write <span class="fixed">10 - (4 + 3) * 2</span>. <span class="fixed">+</span>, <span class="fixed">*</span> and <span class="fixed">-</span> are infix operators, just like the infix functions we met in Haskell (<span class="fixed">+</span>, <span class="fixed">`elem`</span>, etc.). This makes it handy because we, as humans, can parse it easily in our minds by looking at such an expression. The downside to it is that we have to use parentheses to denote precedence.</p>
 <p><a href="https://en.wikipedia.org/wiki/Reverse_Polish_notation">Reverse Polish notation</a> is another way of writing down mathematical expressions. Initially it looks a bit weird, but it's actually pretty easy to understand and use because there's no need for parentheses and it's very easy to punch into a calculator. While most modern calculators use infix notation, some people still swear by RPN calculators. This is what the previous infix expression looks like in RPN: <span class="fixed">10 4 3 + 2 * -</span>. How do we calculate what the result of that is? Well, think of a stack. You go over the expression from left to right. Every time a number is encountered, push it onto the stack. When we encounter an operator, take the two numbers that are on top of the stack (we also say that we <i>pop</i> them), use the operator and those two and then push the resulting number back onto the stack. When you reach the end of the expression, you should be left with a single number if the expression was well-formed and that number represents the result.</p>
-<img src="http://s3.amazonaws.com/lyah/rpn.png" alt="this expression" class="center" width="626" height="224">
+<img src="https://s3.amazonaws.com/lyah/rpn.png" alt="this expression" class="center" width="626" height="224">
 <p>Let's go over the expression <span class="fixed">10 4 3 + 2 * -</span> together! First we push <span class="fixed">10</span> onto the stack and the stack is now <span class="fixed">10</span>. The next item is <span class="fixed">4</span>, so we push it to the stack as well. The stack is now <span class="fixed">10, 4</span>. We do the same with <span class="fixed">3</span> and the stack is now <span class="fixed">10, 4, 3</span>. And now, we encounter an operator, namely <span class="fixed">+</span>! We pop the two top numbers from the stack (so now the stack is just <span class="fixed">10</span>), add those numbers together and push that result to the stack. The stack is now <span class="fixed">10, 7</span>. We push <span class="fixed">2</span> to the stack, the stack for now is <span class="fixed">10, 7, 2</span>. We've encountered an operator again, so let's pop <span class="fixed">7</span> and <span class="fixed">2</span> off the stack, multiply them and push that result to the stack. Multiplying <span class="fixed">7</span> and <span class="fixed">2</span> produces a <span class="fixed">14</span>, so the stack we have now is <span class="fixed">10, 14</span>. Finally, there's a <span class="fixed">-</span>. We pop <span class="fixed">10</span> and <span class="fixed">14</span> from the stack, subtract <span class="fixed">14</span> from <span class="fixed">10</span> and push that back. The number on the stack is now <span class="fixed">-4</span> and because there are no more numbers or operators in our expression, that's our result!</p>
 <p>Now that we know how we'd calculate any RPN expression by hand, let's think about how we could make a Haskell function that takes as its parameter a string that contains an RPN expression, like <span class="fixed">"10 4 3 + 2 * -"</span> and gives us back its result.</p>
 <p>What would the type of that function be? We want it to take a string as a parameter and produce a number as its result. So it will probably be something like <span class="fixed">solveRPN :: (Num a) =&gt; String -&gt; a</span>.</p>
@@ -156,7 +156,7 @@ <h1>Functionally Solving Problems</h1>
 <p>In the RPN calculator section, we first figured out that when calculating an expression by hand, we'd keep a sort of stack in our minds and then go over the expression one item at a time. We decided to use a list of strings to represent our expression. Finally, we used a left fold to walk over the list of strings while keeping a stack to produce a solution.</p>
 <p>Okay, so how would we figure out the shortest path from Heathrow to London by hand? Well, we can just sort of look at the whole picture and try to guess what the shortest path is and hopefully we'll make a guess that's right. That solution works for very small inputs, but what if we have a road that has 10,000 sections? Yikes! We also won't be able to say for certain that our solution is the optimal one, we can just sort of say that we're pretty sure.</p>
 <p>That's not a good solution then. Here's a simplified picture of our road system:</p>
-<img src="assets/images/functionally-solving-problems/roads_simple.png" alt="roads" class="center" width="685" height="245">
+<img src="assets/images/functionally-solving-problems/roads_simple.png" alt="roads" class="center" width="686" height="245">
 <p>Alright, can you figure out what the shortest path to the first crossroads (the first blue dot on A, marked <i>A1</i>) on road A is? That's pretty trivial. We just see if it's shorter to go directly forward on A or if it's shorter to go forward on B and then cross over. Obviously, it's cheaper to go forward via B and then cross over because that takes 40 minutes, whereas going directly via A takes 50 minutes. What about crossroads <i>B1</i>? Same thing. We see that it's a lot cheaper to just go directly via B (incurring a cost of 10 minutes), because going via A and then crossing over would take us a whole 80 minutes!</p>
 <p>Now we know what the cheapest path to <i>A1</i> is (go via B and then cross over, so we'll say that's <span class="fixed">B, C</span> with a cost of 40) and we know what the cheapest path to <i>B1</i> is (go directly via B, so that's just <span class="fixed">B</span>, going at 10). Does this knowledge help us at all if we want to know the cheapest path to the next crossroads on both main roads? Gee golly, it sure does!</p>
 <p>Let's see what the shortest path to <i>A2</i> would be. To get to <i>A2</i>, we'll either go directly to <i>A2</i> from <i>A1</i> or we'll go forward from <i>B1</i> and then cross over (remember, we can only move forward or cross to the other side). And because we know the cost to <i>A1</i> and <i>B1</i>, we can easily figure out what the best path to <i>A2</i> is. It costs 40 to get to <i>A1</i> and then 5 to get from <i>A1</i> to <i>A2</i>, so that's <span class="fixed">B, C, A</span> for a cost of 45. It costs only 10 to get to <i>B1</i>, but then it would take an additional 110 minutes to go to <i>B2</i> and then cross over! So obviously, the cheapest path to <i>A2</i> is <span class="fixed">B, C, A</span>. In the same way, the cheapest way to <i>B2</i> is to go forward from <i>A1</i> and then cross over.</p>
@@ -313,11 +313,11 @@ <h1>Functionally Solving Problems</h1>
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