{"id":545,"date":"2019-06-29T22:22:30","date_gmt":"2019-06-29T22:22:30","guid":{"rendered":"http:\/\/funfacts.104.42.120.246.xip.io\/?page_id=545"},"modified":"2019-11-18T23:45:23","modified_gmt":"2019-11-18T23:45:23","slug":"dinner-party-problem","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/funfacts\/dinner-party-problem\/","title":{"rendered":"Dinner Party Problem"},"content":{"rendered":"\n<div class=\"wp-block-image\"><figure class=\"alignright\"><img loading=\"lazy\" decoding=\"async\" width=\"209\" height=\"165\" data-attachment-id=\"1401\" data-permalink=\"https:\/\/math.hmc.edu\/funfacts\/dinner-party-problem\/30002-4-1\/\" data-orig-file=\"https:\/\/math.hmc.edu\/funfacts\/wp-content\/uploads\/sites\/4\/2019\/11\/30002.4.1.gif\" data-orig-size=\"209,165\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"30002.4.1\" data-image-description=\"\" data-image-caption=\"\" data-medium-file=\"https:\/\/math.hmc.edu\/funfacts\/wp-content\/uploads\/sites\/4\/2019\/11\/30002.4.1.gif\" data-large-file=\"https:\/\/math.hmc.edu\/funfacts\/wp-content\/uploads\/sites\/4\/2019\/11\/30002.4.1.gif\" src=\"https:\/\/math.hmc.edu\/funfacts\/wp-content\/uploads\/sites\/4\/2019\/11\/30002.4.1.gif\" alt=\"\" class=\"wp-image-1401\"\/><\/figure><\/div>\n\n\n\n<p>How many people must you have at dinner to ensure that there are a subset of 3 people who all either mutual acquaintances, or mutual strangers?<\/p>\n\n\n\n<p>We can model this using&nbsp;<em>graph theory<\/em>; for background see the Fun Fact&nbsp;Six Degrees of Separation. Let people be vertices, and draw a red edge between two people if they know each other, and a blue edge if they do not. So every pair of people are connected by either a red or blue edge.<\/p>\n\n\n\n<p>The question then becomes, what is the least number of vertices for which every complete red-blue graph on those vertices has either a red or blue triangle?<\/p>\n\n\n\n<p>Answer: 6 people! First we show that 6 people are enough: take any vertex, Fred. Fred has 5 edges emanating from him; at least 3 are the same color. Suppose without loss of generality it is red. If any of those 3 people that Fred knows know each other, then we have a red triangle! If none of those 3 know each other, we have a blue triangle!<\/p>\n\n\n\n<p>Now, using red-blue graphs, can you draw a red-blue complete graph on 5 vertices with no blue nor red triangle? (This shows that 5 people are not enough.) See the Figure for a solution (it appears after several seconds).<\/p>\n\n\n\n<p><strong>Presentation&nbsp;Suggestions:<\/strong><br>Bring colored chalk into the class; draw pictures as you are giving the proof. It doesn&#8217;t matter if students don&#8217;t follow the exact argument in class&#8212; they will probably go home and want to think about it anyways&#8212; but they will draw the pictures you draw and it will help them construct the argument later.<\/p>\n\n\n\n<p><strong>The&nbsp;Math&nbsp;Behind&nbsp;the&nbsp;Fact:<\/strong><br>Generalizations of this problem, involving the least number of people you have to invite to ensure subgraphs of other types, fall under the category of something called&nbsp;Ramsey Theory.<\/p>\n\n\n\n<p><strong>How to Cite this Page:<\/strong>\u00a0<br>Su, Francis E., et al. &#8220;Dinner Party Problem.&#8221;\u00a0<em>Math Fun Facts<\/em>. &lt;http:\/\/www.math.hmc.edu\/funfacts>.<\/p>\n\n\n\n<p><strong>Fun Fact suggested by:<\/strong><br>Francis Su<\/p>\n","protected":false},"excerpt":{"rendered":"<p>How many people must you have at dinner to ensure that there are a subset of 3 people who all&#46;&#46;&#46;<\/p>\n","protected":false},"author":7,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[57,9,88,12,89],"class_list":["post-545","page","type-page","status-publish","hentry","tag-advanced","tag-combinatorics","tag-graph-theory","tag-other","tag-ramsey-theory"],"jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/545","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/users\/7"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/comments?post=545"}],"version-history":[{"count":3,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/545\/revisions"}],"predecessor-version":[{"id":1402,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/pages\/545\/revisions\/1402"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/media?parent=545"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/funfacts\/wp-json\/wp\/v2\/tags?post=545"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}