Develop

.github/workflows/markdown-lint.yml

@@ -9,7 +9,8 @@ on: # yamllint disable-line rule:truthy
     branches: ["main"]
   workflow_dispatch:
 
-permissions: read-all
+permissions:
+  contents: read
 
 jobs:
   markdownlint:
@@ -32,7 +33,12 @@ jobs:
           node-version: ${{ matrix.node-version }}
 
       - name: Install dependencies
-        run: npm install -g markdownlint-cli
+        run: >
+          npm install -g --ignore-scripts markdownlint-cli@0.47.0
+
+      - name: Test dependencies
+        run: |
+          markdownlint --version
 
       - name: Lint
         run: >

Dockerfile

@@ -78,7 +78,7 @@ RUN apt-key add llvm-snapshot.gpg.key && \
 ADD https://deb.nodesource.com/setup_22.x nodesource_setup.sh
 RUN bash nodesource_setup.sh && \
   apt-get -y install --no-install-recommends --no-install-suggests nodejs && \
-  npm install -g --ignore-scripts markdownlint-cli && \
+  npm install -g --ignore-scripts markdownlint-cli@0.47.0 && \
   apt-get -y install --no-install-recommends --no-install-suggests python3-minimal python3-pip && \
   rm /usr/lib/python3.*/EXTERNALLY-MANAGED && \
   pip install --no-cache-dir yamllint && \

docs/hackerrank/projecteuler/euler002.md

@@ -3,7 +3,8 @@
 - Difficulty: #easy
 - Category: #ProjectEuler+
 
-Each new term in the Fibonacci sequence is generated by adding the previous two terms.
+Each new term in the Fibonacci sequence is generated by adding the previous two
+terms.
 By starting with $ 1 $ and $ 2 $, the first $ 10 $ terms will be:
 
 $$ 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 $$

docs/hackerrank/projecteuler/euler003-solution-notes.md

@@ -33,9 +33,9 @@ The first solution, using the algorithm taught in school, is:
 > Using some test entries, quickly broke the solution at all. So, don't use it.
 > This note is just to record the failed idea.
 
-Since by going through and proving the divisibility of a number $ i $ up to $ n $
-there are also "remainder" numbers that are also divisible by their opposite,
-let's call it $ j $.
+Since by going through and proving the divisibility of a number $ i $ up to
+$ n $ there are also "remainder" numbers that are also divisible by their
+opposite, let's call it $ j $.
 
 At first it seemed attractive to test numbers $ i $ up to half of $ n $ then
 test whether $ i $ or $ j $ are prime. 2 problems arise: