Update documentation

Author: AllenDowney

_sources/chap09.ipynb

@@ -1275,7 +1275,7 @@
    "id": "f90bd9ea",
    "metadata": {},
    "source": [
-    "The result it a list.\n",
+    "The result is a list.\n",
     "To convert the list to a string, we can use `join`."
    ]
   },

_sources/chap10.ipynb

@@ -641,7 +641,7 @@
    "metadata": {},
    "source": [
     "The following function loops through the words in the list.\n",
-    "For each one, it reverses the letters and then checks whether the reversed word in the word list."
+    "For each one, it reverses the letters and then checks whether the reversed word is in the word list."
    ]
   },
   {
@@ -669,6 +669,19 @@
     "If it doesn't find what it's looking for -- which happens most of the time -- it has to search all the way to the end."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "fac41347",
+   "metadata": {
+    "tags": [
+     "remove-cell"
+    ]
+   },
+   "source": [
+    "To measure how long a function takes, we can use `%time` which is one of Jupyter's \"built-in magic commands\".\n",
+    "These commands are not part of the Python language, so they might not work in other development environments."
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 58,
@@ -767,6 +780,15 @@
     "This function takes less than one hundredth of a second, so it's about 10,000 times faster than the previous version."
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "4cd91c99",
+   "metadata": {},
+   "source": [
+    "In general, the time it takes to find an element in a list is proportional to the length of the list.\n",
+    "The time it takes to find a key in a dictionary is almost constant -- regardless of the number of items."
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": 62,
@@ -800,15 +822,6 @@
     "%time much_faster()"
    ]
   },
-  {
-   "cell_type": "markdown",
-   "id": "4cd91c99",
-   "metadata": {},
-   "source": [
-    "In general, the time it takes to find an element in a list is proportional to the length of the list.\n",
-    "The time it takes to find a key in a dictionary is almost constant -- regardless of the number of items."
-   ]
-  },
   {
    "cell_type": "markdown",
    "id": "b3bfa8a5",
@@ -1539,19 +1552,6 @@
     "In the notebook for this chapter, you'll see where these measurements come from."
    ]
   },
-  {
-   "cell_type": "markdown",
-   "id": "c6f39f84",
-   "metadata": {
-    "tags": [
-     "remove-cell"
-    ]
-   },
-   "source": [
-    "To measure how long a function takes, we can use `%time` which is one of Jupyter's \"built-in magic commands\".\n",
-    "These commands are not part of the Python language, so they might not work in other development environments."
-   ]
-  },
   {
    "cell_type": "code",
    "execution_count": 50,
@@ -2283,7 +2283,8 @@
    "id": "e04a5c73",
    "metadata": {
     "tags": [
-     "remove-cell"
+     "remove-output",
+     "remove-print"
     ]
    },
    "outputs": [

chap09.html

@@ -962,7 +962,7 @@ <h2><span class="section-number">9.8. </span>Sorting lists<a class="headerlink"
 </div>
 </div>
 </div>
-<p>The result it a list.
+<p>The result is a list.
 To convert the list to a string, we can use <code class="docutils literal notranslate"><span class="pre">join</span></code>.</p>
 <div class="cell docutils container">
 <div class="cell_input docutils container">

chap10.html

@@ -648,7 +648,7 @@ <h2><span class="section-number">10.2. </span>Creating dictionaries<a class="hea
 </div>
 </div>
 <p>The following function loops through the words in the list.
-For each one, it reverses the letters and then checks whether the reversed word in the word list.</p>
+For each one, it reverses the letters and then checks whether the reversed word is in the word list.</p>
 <div class="cell docutils container">
 <div class="cell_input docutils container">
 <div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">too_slow</span><span class="p">():</span>
@@ -1240,6 +1240,17 @@ <h3><span class="section-number">10.11.6. </span>Exercise<a class="headerlink" h
 </div>
 </div>
 <p>Write a function called <code class="docutils literal notranslate"><span class="pre">is_interlocking</span></code> that takes a word as an argument and returns <code class="docutils literal notranslate"><span class="pre">True</span></code> if it can be split into two interlocking words.</p>
+<div class="cell tag_remove-output tag_remove-print docutils container">
+<div class="cell_input docutils container">
+<div class="highlight-ipython3 notranslate"><div class="highlight"><pre><span></span><span class="k">for</span> <span class="n">word</span> <span class="ow">in</span> <span class="n">word_list</span><span class="p">:</span>
+    <span class="k">if</span> <span class="nb">len</span><span class="p">(</span><span class="n">word</span><span class="p">)</span> <span class="o">&gt;=</span> <span class="mi">8</span> <span class="ow">and</span> <span class="n">is_interlocking</span><span class="p">(</span><span class="n">word</span><span class="p">):</span>
+        <span class="n">first</span> <span class="o">=</span> <span class="n">word</span><span class="p">[</span><span class="mi">0</span><span class="p">::</span><span class="mi">2</span><span class="p">]</span>
+        <span class="n">second</span> <span class="o">=</span> <span class="n">word</span><span class="p">[</span><span class="mi">1</span><span class="p">::</span><span class="mi">2</span><span class="p">]</span>
+        <span class="nb">print</span><span class="p">(</span><span class="n">word</span><span class="p">,</span> <span class="n">first</span><span class="p">,</span> <span class="n">second</span><span class="p">)</span>
+</pre></div>
+</div>
+</div>
+</div>
 <p><a class="reference external" href="https://allendowney.github.io/ThinkPython/index.html">Think Python: 3rd Edition</a></p>
 <p>Copyright 2024 <a class="reference external" href="https://allendowney.com">Allen B. Downey</a></p>
 <p>Code license: <a class="reference external" href="https://mit-license.org/">MIT License</a></p>