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	<title>Comments on: Fred, Fran, and baby makes three</title>
	<atom:link href="http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/feed/" rel="self" type="application/rss+xml" />
	<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/</link>
	<description>Musings of a computer scientist on predictions, odds, and markets</description>
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		<title>By: David Pennock</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-308</link>
		<dc:creator>David Pennock</dc:creator>
		<pubDate>Wed, 13 Aug 2008 11:56:08 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-308</guid>
		<description>You&#039;re right, aj, good point. One work around would be to treat capital letters differently than lower case letters. Or you could use a slightly larger alphabet that includes some of the common letters with &lt;a href=&quot;http://en.wikipedia.org/wiki/Diacritic&quot; rel=&quot;nofollow&quot;&gt;diacritics&lt;/a&gt; like é, ñ, ȍ, etc., or perhaps even allow hyphens, apostrophes, or other somewhat common symbols to count.</description>
		<content:encoded><![CDATA[<p>You&#8217;re right, aj, good point. One work around would be to treat capital letters differently than lower case letters. Or you could use a slightly larger alphabet that includes some of the common letters with <a href="http://en.wikipedia.org/wiki/Diacritic" rel="nofollow">diacritics</a> like é, ñ, ȍ, etc., or perhaps even allow hyphens, apostrophes, or other somewhat common symbols to count.</p>
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		<title>By: aj</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-307</link>
		<dc:creator>aj</dc:creator>
		<pubDate>Sun, 03 Aug 2008 05:57:37 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-307</guid>
		<description>Hrm, a family of five would need 31 different letters, no? Which, coincidentally would be five more than is in the alphabet -- so maybe doable if it was a unique letter for every grouping of the family (with no letter left over, and no letter unique to any individual)...</description>
		<content:encoded><![CDATA[<p>Hrm, a family of five would need 31 different letters, no? Which, coincidentally would be five more than is in the alphabet &#8212; so maybe doable if it was a unique letter for every grouping of the family (with no letter left over, and no letter unique to any individual)&#8230;</p>
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		<title>By: David Pennock</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-306</link>
		<dc:creator>David Pennock</dc:creator>
		<pubDate>Sat, 02 Aug 2008 18:40:15 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-306</guid>
		<description>Wow! Fantastic aj. I posed the bonus challenge almost as a joke. My intuition -- clearly wrong -- was this would be a nearly impossible task. I&#039;d say these names are perfectly reasonable, with nice touches like two of each gender and the same letter (&quot;r&quot;) for the set of all as before. Just to assure myself that it works, I indexed the names by letters in the style of commenter d:

r --&gt; Sterling, Sharolyn, Courtney, Cordelia
n --&gt; Sterling, Sharolyn, Courtney
l --&gt; Sterling, Sharolyn, Cordelia
e --&gt; Sterling, Courtney, Cordelia
o --&gt; Sharolyn, Courtney, Cordelia
s --&gt; Sterling, Sharolyn
t --&gt; Sterling, Courtney
i --&gt; Sterling, Cordelia
y --&gt; Sharolyn, Courtney
a --&gt; Sharolyn, Cordelia
c --&gt; Courtney, Cordelia
g --&gt; Sterling
h --&gt; Sharolyn
u --&gt; Courtney
d --&gt; Cordelia

Thanks for diving into this! Very fun. For what it&#039;s worth I&#039;ll award aj five yootles for this feat.

Dare I suggest next a family of five? OK, you can use first &amp; middle names.</description>
		<content:encoded><![CDATA[<p>Wow! Fantastic aj. I posed the bonus challenge almost as a joke. My intuition &#8212; clearly wrong &#8212; was this would be a nearly impossible task. I&#8217;d say these names are perfectly reasonable, with nice touches like two of each gender and the same letter (&#8220;r&#8221;) for the set of all as before. Just to assure myself that it works, I indexed the names by letters in the style of commenter d:</p>
<p>r &#8211;> Sterling, Sharolyn, Courtney, Cordelia<br />
n &#8211;> Sterling, Sharolyn, Courtney<br />
l &#8211;> Sterling, Sharolyn, Cordelia<br />
e &#8211;> Sterling, Courtney, Cordelia<br />
o &#8211;> Sharolyn, Courtney, Cordelia<br />
s &#8211;> Sterling, Sharolyn<br />
t &#8211;> Sterling, Courtney<br />
i &#8211;> Sterling, Cordelia<br />
y &#8211;> Sharolyn, Courtney<br />
a &#8211;> Sharolyn, Cordelia<br />
c &#8211;> Courtney, Cordelia<br />
g &#8211;> Sterling<br />
h &#8211;> Sharolyn<br />
u &#8211;> Courtney<br />
d &#8211;> Cordelia</p>
<p>Thanks for diving into this! Very fun. For what it&#8217;s worth I&#8217;ll award aj five yootles for this feat.</p>
<p>Dare I suggest next a family of five? OK, you can use first &#038; middle names.</p>
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		<title>By: aj</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-305</link>
		<dc:creator>aj</dc:creator>
		<pubDate>Fri, 01 Aug 2008 07:29:13 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-305</guid>
		<description>Sterling and Sharolyn, meet, marry and multiply, producing twins Courtney and Cordelia:

Generated by:
  (a) getting a list of baby names
  (b) limiting it to eight letter names, with no duplicate letters
  (c) going through every combination of four names, checking that, every pair of names has exactly four letters in common, every triple of names has exactly two letters in common, and all the names together have exactly one letter in common.

I think that&#039;s a sufficient test, but I haven&#039;t proved it -- checking after the fact is easy enough.

I think you could do the above a bit more efficiently using a dynamic programming approach (what are the pairs of names with exactly 4 letters in common? from those, what triples have 2 letters in common?, ...), but with only 83 eight letter names to start with and a 2GHz laptop it didn&#039;t seem worth the hassle.

There&#039;s actually quite a lot of combinations that work, -- limiting it to just boy/boy/girl/girl names from some American census list, I got over a hundred combinations. But a lot involved weird spellings like &quot;krystina&quot; which might or might not count as &quot;reasonable&quot; whatever the census says...

Entertaining anyway :)</description>
		<content:encoded><![CDATA[<p>Sterling and Sharolyn, meet, marry and multiply, producing twins Courtney and Cordelia:</p>
<p>Generated by:<br />
  (a) getting a list of baby names<br />
  (b) limiting it to eight letter names, with no duplicate letters<br />
  (c) going through every combination of four names, checking that, every pair of names has exactly four letters in common, every triple of names has exactly two letters in common, and all the names together have exactly one letter in common.</p>
<p>I think that&#8217;s a sufficient test, but I haven&#8217;t proved it &#8212; checking after the fact is easy enough.</p>
<p>I think you could do the above a bit more efficiently using a dynamic programming approach (what are the pairs of names with exactly 4 letters in common? from those, what triples have 2 letters in common?, &#8230;), but with only 83 eight letter names to start with and a 2GHz laptop it didn&#8217;t seem worth the hassle.</p>
<p>There&#8217;s actually quite a lot of combinations that work, &#8212; limiting it to just boy/boy/girl/girl names from some American census list, I got over a hundred combinations. But a lot involved weird spellings like &#8220;krystina&#8221; which might or might not count as &#8220;reasonable&#8221; whatever the census says&#8230;</p>
<p>Entertaining anyway <img src='http://blog.oddhead.com/wp-includes/images/smilies/icon_smile.gif' alt=':)' class='wp-smiley' /> </p>
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		<title>By: David Pennock</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-304</link>
		<dc:creator>David Pennock</dc:creator>
		<pubDate>Sun, 27 Jul 2008 17:00:55 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-304</guid>
		<description>Thanks everyone! I updated the post with the winner(s).</description>
		<content:encoded><![CDATA[<p>Thanks everyone! I updated the post with the winner(s).</p>
]]></content:encoded>
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		<title>By: d</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-303</link>
		<dc:creator>d</dc:creator>
		<pubDate>Fri, 30 May 2008 19:18:32 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-303</guid>
		<description>Every nonempty subset of family members has a unique letter associated with it:

Erin, Fran, Fred: r
Erin, Fran: n
Erin, Fred: e
Erin: i
Fran, Fred: f
Fran: a
Fred: d</description>
		<content:encoded><![CDATA[<p>Every nonempty subset of family members has a unique letter associated with it:</p>
<p>Erin, Fran, Fred: r<br />
Erin, Fran: n<br />
Erin, Fred: e<br />
Erin: i<br />
Fran, Fred: f<br />
Fran: a<br />
Fred: d</p>
]]></content:encoded>
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		<title>By: FoolsGold</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-302</link>
		<dc:creator>FoolsGold</dc:creator>
		<pubDate>Mon, 19 May 2008 23:09:25 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-302</guid>
		<description>Because they were about to adopt a child and Erin was already her name?</description>
		<content:encoded><![CDATA[<p>Because they were about to adopt a child and Erin was already her name?</p>
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		<title>By: Byrne</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-301</link>
		<dc:creator>Byrne</dc:creator>
		<pubDate>Mon, 12 May 2008 21:45:19 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-301</guid>
		<description>This is driving me insane. Is there a hint?</description>
		<content:encoded><![CDATA[<p>This is driving me insane. Is there a hint?</p>
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		<title>By: Dennis DeCoste</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-300</link>
		<dc:creator>Dennis DeCoste</dc:creator>
		<pubDate>Mon, 12 May 2008 21:03:43 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-300</guid>
		<description>Umm ... one letter from &quot;Fred&quot; only (E), one letter from &quot;Fran&quot; only (n), one letter from both (r), and one letter from neither (i) --&gt; &quot;Erin&quot;.
Relations, togetherness, individuality.

So ... what can my new Yootle buy me these days? ... :)</description>
		<content:encoded><![CDATA[<p>Umm &#8230; one letter from &#8220;Fred&#8221; only (E), one letter from &#8220;Fran&#8221; only (n), one letter from both (r), and one letter from neither (i) &#8211;&gt; &#8220;Erin&#8221;.<br />
Relations, togetherness, individuality.</p>
<p>So &#8230; what can my new Yootle buy me these days? &#8230; <img src='http://blog.oddhead.com/wp-includes/images/smilies/icon_smile.gif' alt=':)' class='wp-smiley' /> </p>
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		<title>By: John</title>
		<link>http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-299</link>
		<dc:creator>John</dc:creator>
		<pubDate>Sun, 11 May 2008 14:43:49 +0000</pubDate>
		<guid isPermaLink="false">http://blog.oddhead.com/2008/05/09/fred-fran-and-baby-makes-three/#comment-299</guid>
		<description>There are three names with four letters.

For any one of the names, all of the following are true:

1. There is one letter that does not appear in either of the other names.

2. There is one letter that appears in both.

3. There is one letter that appears in one of the other two names.

4. The last letter appears in the other of the the other two names.


How do you formalize this?
Also, if they have another daughter, should they name her Ida?</description>
		<content:encoded><![CDATA[<p>There are three names with four letters.</p>
<p>For any one of the names, all of the following are true:</p>
<p>1. There is one letter that does not appear in either of the other names.</p>
<p>2. There is one letter that appears in both.</p>
<p>3. There is one letter that appears in one of the other two names.</p>
<p>4. The last letter appears in the other of the the other two names.</p>
<p>How do you formalize this?<br />
Also, if they have another daughter, should they name her Ida?</p>
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