You have infinite number of balls numbered 1,2,... at your disposal.At time(t)=0, you put balls numbered 1 to 10 into a box.At t=1/2, you put balls numbered 11 to 20 and remove ball-1 from it.Similarly at t=(1/2 + 1/4 ), you put balls numbered 21 to 30 into the box and remove ball-2 from the box.

t follows 1/2 +1/4 +1/8 +...

how many balls are present at t=1 ?

Surprisingly, number of balls in the box at t=1 is 0. For any ball numbered 'k' we can find t(=1/2 + 1/4 +...+ 1/2^k)when it is removed.

For a finite number of operations we can argue that 9 balls are added per each operation but this argument does not work when n->inf.In the language of limits,

Lt (x+9) - Lt (x) = 0

(x->inf) (x->inf)

and mind you LHS(not=)x+9-x=9

## Saturday, February 24, 2007

## Thursday, February 01, 2007

### Breaking number codes

You are trying to break a four digit numeric password. You guess those four digits , if you hit the bull's eye , you are done. Else if your answer is partially correct, then you have to guess the rest of the digits in the next attempt.Can you break within three attempts?

Prove that you need at most ten attempts.

Which is the best* algorithm when you have to do the job in three attempts or <10 attempts in general ? For example we may consider the fact : probability that two same digits are present next to each other =0.216

*suppose there are two algorithms A1 and A2 which can break subsets of all four digit numbers S1 and S2.If S1 belongs to S2 then, A2 is better than A1.

Prove that you need at most ten attempts.

Which is the best* algorithm when you have to do the job in three attempts or <10 attempts in general ? For example we may consider the fact : probability that two same digits are present next to each other =0.216

*suppose there are two algorithms A1 and A2 which can break subsets of all four digit numbers S1 and S2.If S1 belongs to S2 then, A2 is better than A1.

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code breaking

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