A manifesto worth reading ... from pre-election India 2014

Today we heard about the election results in Indian Elections of 2014. It has been 67 years since India became independent. At no other time there has been such an emergence of the call for change. There can be intense need for change and decades long preparation for change. After a 30-year gap there has been a decisive victory for a single party. Now it is time for the change to begin. Change is a slow and steady process. There is a promise for transformed India to emerge.

It was encouraging to read the BJP manifesto. In particular I liked the starting statement

"To build modern India: the best foundation is our own culture; the best tool our own hands; and the best material our own aspirations." There is a lot more in this 52 page document. Read it at http://www.bjp.org/images/pdf_2014/full_manifesto_english_07.04.2014.pdf

 

Is there a 4-sided magic star?

In the last week I recalled one of the problems I solved nearly 20+ years back about 6-sided stars wherein I can fill numbers 1..12 such that each side sums to a same constant. That was during the time I was practicing the use of ProLog (aka Programming Logic) and the solution involved dynamic programming with recursion.

I retraced and built a solution generator for 6-sided stars this week using C# this time. More on that later. Then I started playing with solution generators for other sided stars. I was looking at the 4-sided star. Here is how the 4-sided star looks like:

There are 8 vertices A..H as indicated. there are 4 sides as indicated by the colored lines. Each line touches 4-vertices. Now can we fill the numbers 1..8 such that the sum on each line is the same constant? Short answer is: NO.

And here is a proof.

 Given numbers 1..8, we want to have the numbers filled out such that the following conditions are met
 1: A+B+C+D=V
 2: E+C+F+G=V
 3: D+F+H+A=V
 4: G+H+B+E=V

 let us sum up the equations 1&3 which produces following:
 5:A+B+C+D+D+F+H+A=2A+2 D+B+C+F+H=2V
OR
 6:2A+2D=2V −(BCFH)

Similarly summing up the equations of 2&4 produces following:
 7:E+C+F+G+G+H+B+E=2E+2G+BCFH=2V
 OR
 8:2E+2G=2V −(BCFH)

Putting 6 & 8 together we have
 9:2A+2D=2 E+2G
 OR
 10:A+D=E+G

Now take #1 and substitute using #10
 11:A+B+C+D=E+B+C+G=V
And using #2 we find that
 12:E+B+C+G=E+C+F+G
 OR
 We need
 13:B=F
Using the above steps 5-13, we will also find that we need
 14:C=H

Oops! That is a hard one. We cannot have the same number be filled in both places.
Since either #13 or #14 is possible, we do not have a valid solution for filling in the 4-sided star to ensure numbers add up to same value on 4-sides.