^ Exactly!
Dipit Grover
B.Tech in Computer Science and Engineering - lVth year
IIT Roorkee, India
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Since the number of inversions are of order n^2 in the worst case, so
should be the complexity of printing them apparently.
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Hi, you just need to see the figure carefully and derive a simple formula
here. The tiles are all regular hexagons and The figure shown is for the
given sample case where a=2,b=3,c=4, thereby implying that each side of the
hexagonal tile(say, x) satisfies :
x*(cos 30) = 1/2;
Thanks
Digo
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http://discuss.joelonsoftware.com/default.asp?interview.11.614716
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] . So the second term would mean (1-dp[i-3])*1/2 * 1/2 *
1/2 .
Now the probability that Dave wins for any 'i' would be (1-dp[i]).
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IIT Roorkee, India
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@Lucifer : I came up with a similar algorithm as yours but I dont
understand your complexity analysis : sum over all i (1 to M} { i*(M+N-i)
} .
Shouldnt it be M * sum over all i(1 to N) {(M+N-i)} ? M= no of
columns, N= no of rows . Since we always have the min element at the 0th
column of
@ all k-way people : I dont get it how the complexity would be O(m*n) . I
just went through the algo and I feel that one would need to find the
minimum element among the head-elements of all individual row-arrays, for
all the resulting m*n elements. I say so since we may not necessarily have
a
@Shady : you would definitely need two index variables for each array I
feel.
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sorry I mean 1 variable per each array
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^it cant get better than O(n) apparently. Just applying max-heapify once
would not yield the max element. You need to construct a heap for it, which
is no less than O(n).
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@Lucifer (Regarding your latest approach) - I guess you meant string based
comparison sort in the first step, as in 9 would be greater than 10. The
special case seems correct but I doubt the complexity being nlog n,
considering the special case.
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http://susam.in/downloads/mathematics/theorems/integer-triangles-with-fixed-perimeter.pdf
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IIT Roorkee, India
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Shouldnt it be (n!)/2 ? Equivalent to permutation of n distinct numbers
except that we need to count each permutation once, since for any
permutation, there would also be a reverse permutation that would result in
an identical mst in the given scenario.
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^ we need to count each permutation and its reverse together as one
possibility since both would result in identical mst.
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IIT Roorkee, India
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Mistake noted! Haste makes waste indeed.
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http://en.wikipedia.org/wiki/Cayley%27s_formula
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Using binary search, first find the first occurrence of x. Using this first
occurrence run another binary search to find the last occurrence of x.
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Hi everyone, I am stuck at places where I need to find lets say
Binomial_coefficient(1000,10) mod 1000 000 007 . What is the best way to do
this, assuming we donot have sufficient memory to use the naive approach :
(n,r) = (n-1,r) + (n-1,r-1) .
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Thats really cool! Thanks Dave :)
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On Sat, Aug 27, 2011 at 3:13 PM, dipit grover digo.d.b...@gmail.comwrote:
http://groups.google.com/group/algogeeks?hl=en.
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Dipit Grover
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IIT Roorkee
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Oops, I just saw that Dave had already corrected it. Net problem, sorry
guys!
On Sat, Aug 27, 2011 at 11:02 PM, dipit grover digo.d.b...@gmail.comwrote:
I think you just need to reverse the comparison operators in Dave's earlier
post
On Sat, Aug 27, 2011 at 10:59 PM, Dave dave_and_da
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