Monday 7 February 2011

Follow up to Sum of Squares Problem

A shiny "Attaboy" award to Gasstationwihtoutpumps for the correct answer to both Lewis Carroll Problems. 

Herer is his solution again. 

The first result is moderately interesting:
(a+b)^2 + (a-b)^2 = 2 (a^2 + b^2)

But the second one is then trivial
(a+b)^2 + (a-b)^2 + a^2 + b^2 = 3 (a^2 + b^2)

Do you have some extra constraints for the second problem to make it non-trivial?

For most folks of my generation, raised on lots and lots of factoring problems, the first problem was one of a pair that jumped off the pages of algebra books so often they were almost memorized.  The sum and difference of (x+y)2 and (x-y)2 produce the  results 2( x2+y2) and 4xy; both of which seemed to occur in textbook problems of my youth.  
Had Gas... not already told me he had always had problems with memory I would think he had grown up on the same factoring fare and had also memorized them.  Perhaps he is more like Dodgeson (Carroll) who apparently wrote about re-discovering this in his diary "towards the end of his life" as he too did not recall it.  As a puzzle maker, he probably thought the addition of a third multiple would disguise a common factoring and it was this second problem that appeared in his "Pillow Problems".

 Since my own memory also is very fragile, I was fortunate to find the information in Edward Wakeling's "Lewis Carroll's Games and Puzzles". 
Carroll also included a somewhat more difficult probability problem in his "Pillow Problems".  Since I can't make the first problem more difficult for Gas.... I will give the advanced version of this one. 

An urn holds B Black and W White balls.  One is selected at Random and placed in a bag without your knowledge.  Then a White ball is added to this bag.  A random draw from this bag produces a white ball.  What is the probabilty the original ball was White?


singliar said...

Me, me, me... are we supposed to post spoilers here?

Pat's Blog said...

Yep, Post away....

Anonymous said...

I posted a spoiler yesterday, but it seems not to have made it to the blog for some reason. I'll try again.

This is Bayes' Rule.

We have two possibilities: WW and BW.
We have two possible observations: x=White and x=Black.

We want P(WW | x=White), which is
P(x=White| WW) P(WW) / P(x=White)

The prior probability P(WW) = W/(B+W)

The likelihood P(x=White|WW)=1

The probability of the observed event P(x=White)= P(WW)P(x=White|WW) + P(BW)P(x=White|BW)
= (W+0.5B)/(B+W)

So our posterior probability P(WW|x=White) = W/(W+0.5B)

Anonymous said...

On the previous problem, I'm ashamed to admit that I didn't immediately recognize the result and thought it was supposed to be a geometry problem. After wasting 2 or 3 minutes drawing triangles, I looked at your example using 7 and 4 to get 11 and 3. The sum and difference then popped out at me, and I did a quick check to see that the produced the desired result. I should have known to look at the example sooner.