IBM Research | Ponder This | January 2013 solutions

# Ponder This

## January 2013

<<December January February>>

Many of you used various kind of searches, like genetic algorithms.

Liubing Yu solved it using a clever combination of our example of six permutations covering four elements and a similar solution with five elements and seven permutations to get a solution for 4*5=20 letters with 6+7=13 permutations (we added xy to make the alphabet 20 letters long):

ABCEF   MLJIH   VTQPN  XYZxy
XYZxy    FECBA  VTQPN  HIJLM
HIJLM    FECBA   yxZYX   NPQTV
XYZxy    MLJIH    VTQPN  ABCEF
NPQTV  FECBA   yxZYX   HIJLM
NPQTV  MLJIH     yxZYX   ABCEF

AHNX    YPIB   ZQJC   xTLE   FMVy
FMVy    XNHA  ZQJC   YPIB   ELTx
BIPY     XNHA  yVMF   xTLE  CJQZ
ELTx     yVMF  YPIB    ZQJC  AHNX
CJQZ    XNHA   yVMF  xTLE   BIPY
CJQZ    YPIB    yVMF  xTLE   AHNX
ELTx     XNHA   ZQJC  YPIB   FMVy

Mark Mammel used simulated annealing to get a solution with only 11 permutations:

TLCMBYXZJHFAVIEQNP
JVFMBNLQYTPIXEZCAH
ICYBMPLNQZHAVFJEXT
ZYILPENXFVJAHQCBMT
AFVHJEXPLIYZTMBCNQ
AJCNZBITQELPXMYHFV
QXLMNEITBVJHAFCZYP
HVATQBZNYMCIXLPEJF
XQPBYCTENFHAJVIMZL
HFNZPTCMEYQBLXIJVA
EMCTPQZIJVFAHXNYLB

Here's the answer to the bonus question: The lengths of all the 18 letters, in Morse code, are powers of 2. The example was a hint: EISH are all dots in Morse code (E=. I=.. S=... and H=....).

This problem is from the Combinatorial Test Design domain. To learn more, visit the IBM FoCuS solution project page.

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: ponder@il.ibm.com