<titledata-rh="true">Distance boundaries from BFS tree on undirected graphs | mf</title><metadata-rh="true"name="viewport"content="width=device-width,initial-scale=1"><metadata-rh="true"name="twitter:card"content="summary_large_image"><metadata-rh="true"property="og:url"content="https://blog.mfocko.xyz/algorithms/graphs/bfs-tree/"><metadata-rh="true"property="og:locale"content="en"><metadata-rh="true"name="docusaurus_locale"content="en"><metadata-rh="true"name="docsearch:language"content="en"><metadata-rh="true"name="docusaurus_version"content="current"><metadata-rh="true"name="docusaurus_tag"content="docs-algorithms-current"><metadata-rh="true"name="docsearch:version"content="current"><metadata-rh="true"name="docsearch:docusaurus_tag"content="docs-algorithms-current"><metadata-rh="true"property="og:title"content="Distance boundaries from BFS tree on undirected graphs | mf"><metadata-rh="true"name="description"content="ShortexplanationofdistanceboundariesdeducedfromaBFStree.
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<p>As we have talked on the seminar, if we construct from some vertex <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>u</mi></mrow><annotationencoding="application/x-tex">u</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em"></span><spanclass="mord mathnormal">u</span></span></span></span> BFS tree on an undirected graph, we can obtain:</p>
<ul>
<li>lower bound of length of the shortest path between 2 vertices from the <em>height difference</em></li>
<li>upper bound of length of the shortest path between 2 vertices from the <em>path through the root</em></li>
</ul>
<h2class="anchor anchorWithStickyNavbar_LWe7"id="lower-bound">Lower bound<ahref="#lower-bound"class="hash-link"aria-label="Direct link to Lower bound"title="Direct link to Lower bound"></a></h2>
<p>We run BFS from the vertex <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>a</mi></mrow><annotationencoding="application/x-tex">a</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em"></span><spanclass="mord mathnormal">a</span></span></span></span> and obtain the following BFS tree:</p>
<p>Let's consider pair of vertices <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>e</mi></mrow><annotationencoding="application/x-tex">e</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em"></span><spanclass="mord mathnormal">e</span></span></span></span> and <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>h</mi></mrow><annotationencoding="application/x-tex">h</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6944em"></span><spanclass="mord mathnormal">h</span></span></span></span>. For them we can safely lay, from the BFS tree, following properties:</p>
<p>By having a look at the graph we started from, we can see that we have a path ‹<spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>e</mi><moseparator="true">,</mo><mi>j</mi><moseparator="true">,</mo><mi>h</mi></mrow><annotationencoding="application/x-tex">e, j, h</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.8889em;vertical-align:-0.1944em"></span><spanclass="mord mathnormal">e</span><spanclass="mpunct">,</span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord mathnormal"style="margin-right:0.05724em">j</span><spanclass="mpunct">,</span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord mathnormal">h</span></span></span></span>› that has a length 2. Apart from that we can also notice there is another path from <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>e</mi></mrow><annotationencoding="application/x-tex">e</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em"></span><spanclass="mord mathnormal">e</span></span></span></span> to <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>h</mi></mrow><annotationencoding="application/x-tex">h</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6944em"></span><spanclass="mord mathnormal">h</span></span></span></span> and that is ‹<spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>e</mi><moseparator="true">,</mo><mi>a</mi><moseparator="true">,</mo><mi>c</mi><moseparator="true">,</mo><mi>i</mi><moseparator="true">,</mo><mi>d</mi><moseparator="true">,</mo><mi>h</mi></mrow><annotationencoding="application/x-tex">e, a, c, i, d, h</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.8889em;vertical-align:-0.1944em"></span><spanclass="mord mathnormal">e</span><spanclass="mpunct">,</span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord mathnormal">a</span><spanclass="mpunct">,</span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord mathnormal">c</span><spanclass="mpunct">,</span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord mathnormal">i</span><spanclass="mpunct">,</span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord mathnormal">d</span><spanclass="mpunct">,</span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord mathnormal">h</span></span></span></span>›. And that path has a length of <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>5</mn></mrow><annotationencoding="application/x-tex">5</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em"></span><spanclass="mord">5</span></span></span></span>. Doesn't this break our statements at the beginning? (<em>I'm leaving that as an exercise ;)</em>)</p>
<h2class="anchor anchorWithStickyNavbar_LWe7"id="proof-by-contradiction">Proof by contradiction<ahref="#proof-by-contradiction"class="hash-link"aria-label="Direct link to Proof by contradiction"title="Direct link to Proof by contradiction"></a></h2>
<p>Let's keep the same graph, but break the lower bound, i.e. I have gotten a lower bound <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>2</mn></mrow><annotationencoding="application/x-tex">2</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em"></span><spanclass="mord">2</span></span></span></span>, but “there must be a shorter path”! ;)</p>
<p>Now the more important question, is there a shorter path in that graph? The answer is no, there's no shorter path than the one with length <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>2</mn></mrow><annotationencoding="application/x-tex">2</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em"></span><spanclass="mord">2</span></span></span></span>. So what can we do about it? We'll add an edge to have a shorter path. Now we have gotten a lower bound of <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>2</mn></mrow><annotationencoding="application/x-tex">2</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em"></span><spanclass="mord">2</span></span></span></span>, which means the only shorter path we can construct has <spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn></mrow><annotationencoding="application/x-tex">1</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em"></span><spanclass="mord">1</span></span></span></span> edge and that is ‹<spanclass="katex"><spanclass="katex-mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>e</mi><moseparator="true">,</mo><mi>h</mi></mrow><annotationencoding="application/x-tex">e, h</annotation></semantics></math></span><spanclass="katex-html"aria-hidden="true"><spanclass="base"><spanclass="strut"style="height:0.8889em;vertical-align:-0.1944em"></span><spanclass="mord mathnormal">e</span><spanclass="mpunct">,</span><spanclass="mspace"style="margin-right:0.1667em"></span><spanclass="mord mathnormal">h</span></span></span></span>› (no intermediary vertices). Let's do this!</p>
<p>Okay, so we have a graph that breaks the rule we have laid. However, we need to run BFS to obtain the new BFS tree, since we have changed the graph.</p>
<divclass="theme-admonition theme-admonition-tip admonition_xJq3 alert alert--success"><divclass="admonitionHeading_Gvgb"><spanclass="admonitionIcon_Rf37"><svgviewBox="0 0 12 16"><pathfill-rule="evenodd"d="M6.5 0C3.48 0 1 2.19 1 5c0 .92.55 2.25 1 3 1.34 2.25 1.78 2.78 2 4v1h5v-1c.22-1.22.66-1.75 2-4 .45-.75 1-2.08 1-3 0-2.81-2.48-5-5.5-5zm3.64 7.48c-.25.44-.47.8-.67 1.11-.86 1.41-1.25 2.06-1.45 3.23-.02.05-.02.11-.02.17H5c0-.06 0-.13-.02-.17-.2-1.17-.59-1.83-1.45-3.23-.2-.31-.42-.67-.67-1.11C2.44 6.78 2 5.65 2 5c0-2.2 2.02-4 4.5-4 1.22 0 2.36.42 3.22 1.19C10.55 2.94 11 3.94 11 5c0 .66-.44 1.78-.86 2.48zM4 14h5c-.23 1.14-1.3 2-2.5 2s-2.27-.86-2.5-2z"></path></svg></span>tip</div><divclass="admonitionContent_BuS1"><p>Do we need to run BFS after <strong>every</strong> change?</p><p>I am leaving that as an exercise ;)</p></div></div>
