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seen Nov 1 '12 at 4:30

Oct
26
comment Every edge with even degree -> Euler tour
Could you tell me....?
Oct
26
comment Every edge with even degree -> Euler tour
The remaining subgraph must have at least one vertex that is incident to an untraversed edge. And that vertex must have even number of degrees. Oh, because each remaining vertex has even number of degrees, so they each has at least a degree of two. So each of the remaining component is connected. Am I right?
Oct
26
comment Every edge with even degree -> Euler tour
Thanks for your answering. I am thinking of the following procedure: traverse the graph from vertex u, and get a closed walk. If there are edges left, start from the vertex in the current closed walk that has an untraversed edge (*). And find another closed walk. But the problem is how can we ensure in the recursive steps *, we can always find a closed walk. Maybe some of the edges inside the closed walk we need to traverse have already been traversed by previous steps. That's my question.
Oct
26
comment Every edge with even degree -> Euler tour
Yes, you are right. The graph is connected and undirected. I modified the post.
Oct
13
comment amortized analysis
This is from Amortized Analysis. Amortized Analysis considers the cost for each step as the average of overall cost.
Oct
6
comment probability about drawing cards
The value for A is 1, J is 11, Q is 12, and K is 13. The number of cards to begin with is 52 (4*13).
Oct
6
comment probability about playing cards
Thanks for your answer. What is the expected maximum value among the cards that he draws?
Oct
6
comment probability about playing cards
Your answer is right. Thanks.
Oct
4
comment probability about playing cards
without replacement.
Oct
4
comment Expected number of Pareto-optimal points
This question is from the headbanging session of an undergraduate algorithms in the computer science department.
Oct
4
comment expected number problem for playing cards
Wait a minute. @Brian, can you give out the way to calculate the probability (4/45) by other means instead of 1 minus others?
Oct
4
comment expected number problem for playing cards
Your answer is right.
Oct
4
comment probability about playing a deck of cards
What if for the case that 3 of Hearts is both before all Aces and all Deuces?
Oct
4
comment probability about playing a deck of cards
I mean any of the aces will count.