fix: make sure for plausible to also works on notes and tinymorph docs

Signed-off-by: Aaron Pham <contact@aarnphm.xyz>
aarnphm · Feb 1, 2025 · a95d6c8 · a95d6c8
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diff --git a/content/thoughts/university/twenty-four-twenty-five/sfwr-2fa3/rev1/a1/A1.md b/content/thoughts/university/twenty-four-twenty-five/sfwr-2fa3/rev1/a1/A1.md
@@ -3,9 +3,12 @@ id: A1
 tags:
   - sfwr2fa3
 date: "2024-02-16"
+modified: 2025-02-01 05:36:37 GMT-05:00
 title: DFAs, NFAs, and regular languages
 ---
 
+see also [[thoughts/university/twenty-four-twenty-five/sfwr-2fa3/rev1/a1/sol.pdf|solutions]]
+
 ## Q1.
 
 For each statement below, state if it is true or false, and explain why. The explanation does not need to be a formal proof, but the argument should be sound.

diff --git a/content/thoughts/university/twenty-four-twenty-five/sfwr-2fa3/rev1/a2/A2.md b/content/thoughts/university/twenty-four-twenty-five/sfwr-2fa3/rev1/a2/A2.md
@@ -3,9 +3,15 @@ id: A2
 tags:
   - sfwr2fa3
 date: "2024-03-18"
+description: and second assignment.
+modified: 2025-02-01 05:36:05 GMT-05:00
 title: Regex, pumping lemma
 ---
 
+> [!note]- solutions
+>
+> <iframe src="thoughts/university/twenty-four-twenty-five/sfwr-2fa3/rev1/a2/sols.html"></iframe>
+
 ## Problème 1.
 
 Give regular expression for the languages bellow

diff --git a/content/thoughts/university/twenty-four-twenty-five/sfwr-2fa3/rev1/a2/sols.html b/content/thoughts/university/twenty-four-twenty-five/sfwr-2fa3/rev1/a2/sols.html
@@ -1,9 +1,30 @@
-<!DOCTYPE html>
-<html><head></head><body style="color: rgb(32, 33, 34); font-family: undefined;"><p>1) Note there are more than one way to do these.</p>
-<p>i) (aa)*(bb)* + a(aa)*b(bb)*<br>ii) b*a*((bb)b*a*)*b*<br>iii) (a*ba*b)*a*</p>
-<p></p>
-<p>3) We need to show that if we assume L is finite, that the pumping lemma is equivalent to True. If you select a k' such that k' &gt; |x| for all x in L, then the range of the first "for all" quantification becomes False. Applying the empty range axiom on forall results in True. From here it's relatively straight forward to show that even for finite languages the pumping lemma holds.&nbsp;&nbsp;</p>
-<p>4)&nbsp;</p>
-<p>a) Let x = a^(k^2), y = b^k, and z = a^k. Then no matter what is chosen for v, you can pump b's into the string to break the balance and produce a string outside the language</p>
-<p>b) Let x = a^kb, y = a^k, z = b. Because one can never choose a v which is not a sequence of a's and that the first b must be halfway through the string, you can then pump outside the language. Note Letting x = a^k, y = a^k, z = empty would not work.</p>
-<p>c) Applying the exact proof from lecture (for the perfect square example) works for this language as well.&nbsp;</p></body></html>
+<!doctype html>
+<html>
+  <body style="color: rgb(32, 33, 34); font-family: undefined">
+    <p>1) Note there are more than one way to do these.</p>
+    <p>i) (aa)*(bb)* + a(aa)*b(bb)*<br />ii) b*a*((bb)b*a*)*b*<br />iii) (a*ba*b)*a*</p>
+    <p></p>
+    <p>
+      3) We need to show that if we assume L is finite, that the pumping lemma is equivalent to
+      True. If you select a k' such that k' &gt; |x| for all x in L, then the range of the first
+      "for all" quantification becomes False. Applying the empty range axiom on forall results in
+      True. From here it's relatively straight forward to show that even for finite languages the
+      pumping lemma holds.&nbsp;&nbsp;
+    </p>
+    <p>4)&nbsp;</p>
+    <p>
+      a) Let x = a^(k^2), y = b^k, and z = a^k. Then no matter what is chosen for v, you can pump
+      b's into the string to break the balance and produce a string outside the language
+    </p>
+    <p>
+      b) Let x = a^kb, y = a^k, z = b. Because one can never choose a v which is not a sequence of
+      a's and that the first b must be halfway through the string, you can then pump outside the
+      language. Note Letting x = a^k, y = a^k, z = empty would not work.
+    </p>
+    <p>
+      c) Applying the exact proof from lecture (for the perfect square example) works for this
+      language as well.&nbsp;
+    </p>
+  </body>
+</html>
+
diff --git a/quartz/plugins/emitters/componentResources.ts b/quartz/plugins/emitters/componentResources.ts
@@ -99,7 +99,7 @@ function addGlobalPageResources(ctx: BuildCtx, componentResources: ComponentReso
     componentResources.afterDOMLoaded.push(`
       const plausibleScript = document.createElement("script")
       plausibleScript.src = "${plausibleHost}/js/script.outbound-links.manual.js"
-      plausibleScript.setAttribute("data-domain", location.hostname)
+      plausibleScript.setAttribute("data-domain", [location.hostname, "notes.aarnphm.xyz", "tinymorph.aarnphm.xyz"].join(','))
       plausibleScript.defer = true
       document.head.appendChild(plausibleScript)
 

diff --git a/quartz/styles/base.scss b/quartz/styles/base.scss
@@ -693,12 +693,16 @@ ol.overflow {
   overflow-y: hidden;
 }
 
+iframe {
+  border-radius: var(--border-radius);
+  border: 1px solid var(--lightgray);
+  width: 100%;
+}
+
 .external-embed.youtube,
 iframe.pdf {
-  aspect-ratio: 16 / 9;
   height: 100%;
-  width: 100%;
-  border-radius: 5px;
+  aspect-ratio: 16 / 9;
 }
 
 .navigation-progress {