piyush
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 Feb1 comment Maximum number of edges in a DAG without transitivity condition Quite innovative! Thanks (although that would mean that I will have to come up with a better algorithm for the problem) Feb1 accepted Maximum number of edges in a DAG without transitivity condition Feb1 asked Maximum number of edges in a DAG without transitivity condition Jun30 comment every integers from 1 to 121 can be written as 5 powers of 3 interesting follow up compgeom.cs.uiuc.edu/~jeffe/teaching/algorithms/hwex/f06/… Q 1(b) Jun13 comment How is this kind of subgraph called? Don't know what it is called, but there is an interesting SPOJ problem on that concept spoj.com/problems/CAPCITY Jun9 awarded Excavator Jun9 revised How many ways to divide group of 12 people into 2 groups of 3 people and 3 groups of 2 people? "Latex"ification Jun9 suggested approved edit on How many ways to divide group of 12 people into 2 groups of 3 people and 3 groups of 2 people? Mar6 comment Is this recurrence relation correct? closed form calculation (after recurrence relation) can be done in a more elegant way : $a_{n} = a_{n-1} + p(n-1)$ , $a_{n-1} = a_{n-2} + p(n-2)$ .. $a_{1} = a_{0} + p(0)$, $p$ being the quadratic polynomial. Now just add all these equations, and only $a_{n}$ survives on LHS with a cubic on the RHS Mar6 revised Expected value uniform decreasing function correction in the formula Mar6 revised Expected value uniform decreasing function correction to the answer Mar5 comment Expected value uniform decreasing function @joriki : as I've pointed in my answers (that gained me right to comment anywhere :) ) that your analysis seem to "lose" the fact that output is integral only. Consider the case $f(3,2)$, your analysis will give expected value of 0, whereas it is $\frac{1}{6}$. Take another example, $f(6,3)$, where you'll give negative expectation! Mar5 answered Expected value uniform decreasing function Feb9 awarded Citizen Patrol Dec14 asked Checking if a number if expressible as $x^2+y^4$ Nov17 awarded Scholar Nov17 accepted Upper bound of the sum $\sum_{i=2}^{N}{\frac{1}{\log(i)}}$ Nov17 comment Upper bound of the sum $\sum_{i=2}^{N}{\frac{1}{\log(i)}}$ Thanks! Asymptotic value of li(x) seems sufficient for me :) Nov16 awarded Supporter Nov16 awarded Student