Laks

Hey randhawa, I haven't received the scores as of yet but have a feeling that it should be coming by jan 1 or the first week atmost... if posted as promised :o by dec 19... again no one can predict ETS... Guys it was really interesting to read all the reactions on teh 5 "ambiguous ques"... lets hope the admission commitees are also privy to this knowledge and this may causew them to look upon csgre in a different light!!! Newyas All the best ppl do post the result of your apps... Cheers' Laks

Hey Ppl, Man I am in a state of shock.... Neways as all of you have quoted.. the score does not maketh the man... and what can I say more.. all that has to be said has been said... I am sitting on my score... awaiting the mail.... Neways all the best to all those apping..... Let your dreams come true... Cheers' Laks

LL grammars are CFG, that are not ambiguous and not Left recursive.... They are used by predictive parsers...{LL(1) grammars are used here} The first L stands for "we scan the input from left to right" Second L for "we produce the leftmost derivation" the number in the bracket (1) stands for the number of look ahead symbols for the parsing action. An LR is also a CFG...."L" -> scan the input from left to right" and R-> "we construct a rightmost derivation"... An LR grammar is way more powerful than a LL grammar. They are also used majorly for parsing. LR parsers can be constructed for virutally every programming construct for which a CFG can be written. Various LR grammars used actively for parsing are CLR (cannonical LR) and LALR (look-ahead LR). Cheers' Laks

Ppl, Hope we are all in the prime stage of our preparation...... Here are more about primes.... Fermats little theorem For any integer a, and a prime numer p a^p mod p = a mod p Law of inverses if n relatively prime to m, there exists n' such that nn' mod m = 1 mod m (basis of RSA) Also n' can be calculated as n^phi(m)-1 mod m where phi(n) is the totient of n i.e. the numbers of numbers fron 0 to n, that are relatively prime to n

Hey Wood, Great job... I would just like to point out ont thibg here about fib numbers.... f(n) f(n) = O(2^n-1)...... nothing wrong with your funda... just like to add though... Cheers' Laks

Great discussion guys... Wood I believe your answer (n-1)!/2 is for a single vertex (in the above examples from 1) Or can we find an equivalent to 1 2 3 4 5 1 for an HC starting from vertex 2 and so this is indeed the total number of HC's, starting at any vertex???? Can you clarify this??? if this is for a single vertex, then for all the HC's from all vertices, the answer would be n*(n-1)!/2 = n!/2 Anyways here's my take at the combinatorial explanation.... We have n-1 vertices to order.... (the 1st vertex occurs at the first and last) So here it is (n-1)!.... but 1 2 3 4 5 1 is same as 1 5 4 3 2 1.... So we have to remove all such combinations in the order, which gives us (n-1)!/2 Cheers' Laks

God..... Sorry..... I thought they were defined again in main() scope........ (thought the ques woul be trickier with same namein multiple scopes..... Ok this has spurred my thought.... If we had a local variable a too and we pass a by name, what would happen???.... I believe the local a would have precedence.... but again I have my doubts..... Cheers' Laks

4 ) It is 13C1 * 4C3 * 12C1 * 4C2 .... we have to exclude cards of the rank we have already chosen... so we have 12 ranks to choose from... and 2 out of 4 in the same rank... 5) 4c1 * 13c4 - (1) - (2) ??? (was 13 the mistake nonevent made??) 6) 9 (ways of choosing 5 cards from 13) * 4C1 (we can choose rank from one of the four suite) 7) 13C1 * 4C3 * 12C1 * 4C1 * 11C1 * 4C1 (Same funda as 4) 8 and 9 can be answered using same fundas as 7 Cheers Laks

PPl, Will not call by value result in 1 being printed ???

The GCD - Euclids Algo......... :-) Running time... hmmm.... ceil(max(a,b)/min(a,b)) ????? Cheers' Laks

here's my take..... Given the value and base(base that the value is in), this function prints the string from the char map, whose indices are value in base b. Say value is 17 and base is 8..... First iteration remainder = 1, value = 2 Second Iteration remainder = 2 value = 0 i.e now newval = char[1] + char [2] we return char[2] + char[1].... the indices of reverse of newval are 2 1 which is 17 in base 8.... Cheers' Laks

Generally.... a is said to be a quadratic residue of m of there exists a solution to x^2 = a (mod m) a is said to be a Kth power residue of m of there exists a solution to x^k = a (mod m) Wood, Whats a and m being relatively prime got to do with this???? And what is "minimum universal exponent modulo m????" Cheers' Laks

http://en.wikipedia.org/wiki/Modular_arithmetic Nice one

link for newton raphson http://www.sosmath.com/calculus/diff/der07/der07.html

More food for thought -> On an average isn't latency of half a revolution acceptable???? (this accounts for 0.5) Cheers' Laks