IBM Ponder This January 2010 Puzzle


I solved IBM Ponder This October 2009 Challenge and IBM Ponder This November 2009 Challenge and now solved this very interesting puzzle at IBM Ponder This January 2010 Challenge

I am happy!! My name on IBM's research site three times in 4 months. :)

Problem: Present a computation whose result is 5, being a composition of commonly used mathematical functions and field operators (anything from simple addition to hyperbolic arc-tangent functions will do), but using only two constants, both of them 2.
It is too easy to do it using round, floor, or ceiling functions, so we do not allow them.
Source: IBM Ponder This January 2010 Challenge
Solution: IBM guys would post it on their site by February 2010 :) Three different solutions given by Piyush (Fourth year, IITM), Aniesh Chawla (Fourth year, IITD) and me (Fourth year, IITB) :). All posted in comments!!

Thanx to Anirudh Shekhawat urf Maoo (CSE, IITB & Google) for help in the last step :)

Comments

  1. Piyush9:49 PM, January 22, 2010

    I'm not sure if i've understood the question correctly but will 2^2 + (2/2) do?

    ReplyDelete
  2. Pratik Poddar11:32 PM, January 22, 2010

    Dude..
    You need to use only 2 2's. You have used 4 2's in your expression :P
    :)

    ReplyDelete
  3. Piyush10:18 AM, January 23, 2010

    so what about

    (cosh(arcsinh(2)))^2 =5 ?

    will this do?

    ReplyDelete
  4. Pratik Poddar10:02 PM, January 23, 2010

    nice solution Piyush..
    The solution I gave was this:
    secinv(-cosec(secinv(2)))/cosecinv(2)

    Aniesh Chawla (Fourth Year, IIT Delhi) gave this solution:
    (sqrt(antilog(2))/2

    :)

    Happy!!

    ReplyDelete
  5. Unknown8:43 PM, January 29, 2010

    Here's mine :) :

    lg(2)/.2

    ( base 2 logarithm )

    Jonathan Busby

    ReplyDelete
  6. Pratik Poddar9:52 PM, January 29, 2010

    I was told by someone that there is nothing such as .2.. its 0.2 and hence it should not work.. Don't know whether IBM guys are accepting this ;)

    ReplyDelete
  7. avinash3:02 AM, June 02, 2011

    sec(arctan(2))
    or
    sec(arctan(sqrt(2*2)))

    ReplyDelete
  8. avinash2:46 PM, June 02, 2011

    A small mistake..
    sec(arctan(2))^2
    or
    square(sec(arctan(sqrt(2*2))))

    ReplyDelete
  9. Kapil Dubey3:20 AM, August 10, 2012

    cos(2-2)=1
    so, the following approach works for generating any number
    sec(arctan(sec(arctan(sec(arctan(sec(arctan(cos(2-2)))))))))

    this is basically sec(arctan(x)) applied on cos(2-2) 4 times. So, it takes 1 -> 2 -> 3 -> 4 -> 5

    ReplyDelete
  10. Unknown10:21 PM, September 18, 2017

    What about 2^2 + logx(x)?
    Is it acceptable?( It's log base x)

    ReplyDelete
  11. Pratik Poddar1:20 PM, September 08, 2026

    sec(arctan(2))^2 is correct and beautiful: sec(arctan(x))^2 = 1 + x^2, so with x = 2 you get exactly 5 - and it uses precisely two 2s. Cleaner than all three solutions we originally posted. (The sqrt(2*2) version uses three 2s, so it does not qualify.) (replied using AI)

    ReplyDelete
  12. Pratik Poddar1:20 PM, September 08, 2026

    The iteration idea is nice, but as written sec(arctan(x)) = sqrt(1+x^2), so your chain goes 1 -> sqrt(2) -> sqrt(3) -> 2 -> sqrt(5), not 5. What you want is the SQUARED map x -> 1 + x^2, i.e. sec(arctan(x))^2 applied four times to cos(2-2): 1 -> 2 -> 3 -> 4 -> 5. With the squares it works and indeed generates every positive integer from a single seed. (replied using AI)

    ReplyDelete
  13. Pratik Poddar1:20 PM, September 08, 2026

    Nice try, but no - log base x of x sneaks in a third constant (x itself). The rules allow exactly two constants, both equal to 2, and x is a parameter you introduced. If free parameters were allowed, 2^2 + x - x would trivialize the puzzle. (replied using AI)

    ReplyDelete
  14. Pratik Poddar1:20 PM, September 08, 2026

    Without the square this gives sqrt(5), not 5 - sec(arctan(x)) = sqrt(1+x^2). Your follow-up comment with the square (sec(arctan(2))^2 = 1 + 4 = 5) is the correct version; replied there. (replied using AI)

    ReplyDelete

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