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Location: Mumbai, Maharashtra, India

Hi, This is Kailash here.. presently working in Mumbai with an Investment Bank. Earlier I completed my post graduation from IIM Ahmedabad

Thursday, August 14, 2008

Chess Board

I am not writing after a long time but posting after a long time. My last few write-ups did not see the daylight as they were not approved by the censor board as the content was highly controversial as well as personal for few people around me.

Someone asked me some CAT level mathematics questions. Suddenly, I realized CAT 2009 is coming and people must be working real hard to crack the claimed to be the toughest examination in the world to crack. Cat screws them all...

Coming back to the question, "Two squares are chosen at random on a chessboard, what is the probability that they have a side in common?"

Pretty simple for most of you - Most of you who would be taking CAT seriously will be able to solve this not so difficult problem. I am sure the approach would be different. CAT as I know is all about solving problems efficiently. I did ask this question to few of my friends who all gave me the correct answer and most of them approached the problem in the following way:

If the first square is any of the corner ones, then the second square can be chosen in 2 ways, if the first square is any of the remaining 24 squares on the sides of the chessboard, the second square can be chosen in 3 ways, and if the first square is any of the centre 36 squares, second square can be chosen in 4 ways i.e. (4*2 + 24*3 + 36*4) =224, and since all combinations are repeated twice, total number of combinations of chosing squares will be half of it. And Denominator of the required probability will be 64C2, Hence the answer would be (224/2)/64C2. Bang on the answer is correct. And at least 2 minutes in solving this.

But the approach efficient?

I mean there cant be any other choice for the denominator, but cant it be a little less clumsy way in calculating the numerator. I am sure it could be better than this. I think I would prefer it doing the following ways (order of preference in ascending order)

Just assume to put number 1 to 64, one each in the 64 boxes in the chessboard. Let the number be x for the chosen first box, then either x-8 or x-1 or x+1 or x+8 can be the possible second box which would satisfy that they have a side in common. Now to avoid double count we can reduce the number of boxes satisfying the second box to be x+1 and x+8 (because combination x-1 and x will be y and y+1 for some number y and x-8 and x will be z and z+8 for some number z). Now all the numbers from1 to 64 will have neighbours in the form of either x+1 or x+8, hence total combinations for the numerator should be 64*2 =128. But, since boxes 8, 16, 24.... 64 (right most column of the chessboard), do not have x+1 as their neighbour, hence we have to subtract 8, and also numbers 57 to 64 (bottom row of the chesboard, do not have x+8 as neighbour as chess board just have 64 boxes, hence we have to subtract another 8 and hence the possible combination for numerator should be 128-8-8 = 112 and ofcourse no prize for guessing the denominator. (I think it should not take you more than a minute)

Another way could be thinking (thanks to my friend Ramjo) that for each row, there can be 7 combination ((1,2), (2,3).... (7,8))satisfying the numerator and there are 8 rows so total is 8*7 = 56, similarly for each column there can be 7 combinations ((1,9), (9,17).... (49,57)) and there are 8 columns so total is 8*7 =56, hence total combinations for numerator would be 56*2 = 112. Another neat way of thinking !!(basically good for people who do not have algebraic bent of solving problems as suggested above but basically both are ultimately the same logical thought process)

One more way of approach could be for people who can visualize 3D geometry!!. Assume making a sphere out of the chessboard by joining rows 1 and 8 and colums 1 and 8. Now in this scenario, each square would have 4 adjacent squares which can fit to be the second square. Hence, total number of combinations for numerator would be 64*4/2 = 128. Now since rows 1 & 8 are not connected in the chessboard, subtract 8 combinations formed by rows 1 & 8 joined in the sphere. Similarly, subtract 8 combinations because of column 1 and 8 not being joined on the chessboard. Again the numerator would be 128-8-8 = 112. Few of us may have difficulty visualizing the sphere as I suggested out of the chessboard by joining extreme rows and columns !!!! But again it can make life easy and one sure way to approach !!

Similarly people may have n number of ways solving this problem. But again when time is essence, logical best and smart way can give you those extra seconds which can differentiate you from the rest !! Good luck ! My next post will be related to use of SYMMETRY - ofcourse that makes life beautiful.


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