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?
at

4 comments:

  1. singliar7 February 2011 at 19:07

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

    ReplyDelete
  2. Pat's Blog8 February 2011 at 06:46

    Yep, Post away....

    ReplyDelete
  3. Anonymous8 February 2011 at 11:49

    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)

    ReplyDelete
  4. Anonymous8 February 2011 at 11:53

    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.

    ReplyDelete