Yes , it will always fill with 3 .
the best result both player can get for n 4 is always draw.
You can check it for some values on paper .
if you want to have a winner you must change the basic conditions of the
problem.
On 11/28/06, ravi [EMAIL PROTECTED] wrote:
We Define an array Win[]
dor wrote:
a graph is bipartite if and only if it has an odd cycle
Change an to no and I'd agree.
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yes.. a graph is _not_ bipartite if and only if it has an odd cycle..
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To
A single 6 Pick bet looks like this:
RACE1 RACE2RACE3RACE4 RACE5 RACE6
runner1 / runner 2 / runner 3 / runner4 / runner5 / runner6 - $amount
e.g. we might have:
5 / 3 / 11 / 7 / 1 / 9 - $50 (5 to come 1st or 2nd in Race1, 3 to
come first or 2nd in Race 2, etc.)
7 / 3 / 11 / 7
Hi,
I was thinking to do something like this, but is it an efficient way to
do it?
Do you know any well know algorithm to do this kind of thing?
Thanks,
Andre
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k...ThanxI have cleared my doubt
I would like to ask here one more question similar to this question.
Suppose a game is played between A and B, consider a number n = 100, at
each time each player adds a divisor of n 100 to n. Find which player
won the game when total = 1953, when the
Better answer for #1 that I'm pretty sure will work:
calculate the center and radius of the circle. if the radius is not
NaN (i.e the sites are not collinear) and the center of the circle is
between the current locations of the breakpoints, they will converge.
On Nov 21, 6:36 am, bordaigorl
Also, in response to your idea, it doesn't make sense. . .you calculate
the center of the circle BY calculating where the bisectors intersect.
Or vice versa. The two values are always equivalent. Also, your fear
about numerical instability is unfounded. Calculating the orthocenter
by bisector
A maximum heap should probably do the trick.
On 11/28/06, Andre [EMAIL PROTECTED] wrote:
Hi,
I was thinking to do something like this, but is it an efficient way to
do it?
Do you know any well know algorithm to do this kind of thing?
Thanks,
Andre
your problem is a typical NPcomplete problem
and no polynomial algorithm is possible.
the link you provide just calculates maximal independent set, not
maximum independent set. It turns out this two can be totally
different.
Ken1 wrote:
I've been trying to do this for a while and now I give up,
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