Author Topic: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)  (Read 43582 times)

万精油

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« on: 三月 27, 2005, 11:00:11 pm »

ln(n) + 1/2n + 0.57721.

n*ln(n)+0.577n+1/2

七把叉

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Re: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #1 on: 三月 28, 2005, 12:36:36 pm »
Quote from: 万精油

fzy

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Re: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #2 on: 三月 28, 2005, 02:02:58 pm »
Quote from: 万精油

I totally agree with the professor.

Recently I was very reluctant to post because I noticed that I passed I94 and became the second most frequent poster, only behind the professor.    I suggest everybody who visit and like the site, post something once in a while, even just to announce you are alive and well. I also suggest Sir I94 to post more often so I have some company and feel better.    Even just joking around is cool.

A mathematician applies to be a volunteer fire fighter. He needs to answer two questions:

1. "What do you do when you see a dumpster on fire?" "Easy," the mathematician said. "I put water on the dumpster and extinguish the fire."

2. "What do you do when you see a dumpster NOT on fire?" "This is even easier," the mathematician said. "I set it on fire and it is reduced to the first case."

sean

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #3 on: 三月 28, 2005, 05:15:45 pm »

zzzzzzzzzz

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #4 on: 三月 28, 2005, 05:56:30 pm »
Two principles are useful in solving this problem.  1. The brick(s) will fall off if its/their center of gravity is beyond the edge of the brick immediately below it/them.  2. A bunch of bricks, regardless of how they are stacked, is equivalent to their total mass concentrated at their center of gravity, which is one mathematical point in space.  So for a bunch of bricks in fully extended and stable position, the center of gravity of top n bricks is right on top of the front edge of the n+1st brick.

Assume each brick has length 1 for simplicity (it can be changed to actual length later).

If there are only two bricks, only half of the top brick can extend beyond the edge of the bottom brick.

With 3 bricks, the center of gravity of top two bricks is ((1/2 * 1) + (1 * 1))/2 = 3/4, which means the second brick will extend 1-3/4 = 1/4 beyond the edge of the bottom brick, and total extension is 1/2 + 1/4 = 3/4.

With 4 bricks, center of gravity of top 3 bricks is at ((1/2 * 1) + (2 * 1))/3 = 5/6, so the 3rd brick will extend 1/6 beyond the edge of the bottom brick, for a total of 1/6 + 3/4 = 11/12.

In summary, if there are n bricks in the fully extended position, adding one brick on the bottom will add to the extension by the amount the n bricks can extend beyond the newest brick.  That amount is: 1 - (1/2 + n)/(n+1) = 1/(2(n+1)), and the total distance that can extend is sigma(1/2(k+1)), for k = 1 to n.

Since the series 1/n does not converge, as long as you have enough bricks, you can extend the whole stack as far beyond the edge of the bottom brick as you want.

七把叉

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #5 on: 三月 28, 2005, 11:33:25 pm »

七把叉

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #6 on: 四月 01, 2005, 12:42:08 pm »

万精油

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #7 on: 四月 01, 2005, 01:51:28 pm »
According to earlier posts, the nth brick will extend out 1/(2*n), which is getting smaller and smaller (the sum will go to infinite of course). By your new approach, if I split the top brick into two (thus, going from n brick to n+1 brick), I can always extend it 0.11 out. This is much bigger than 1/2*n, what happened?

Note: I know my above argument is wrong, I just want to see if people can find out where did I go wrong? (hint: physics)

fzy

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #8 on: 四月 01, 2005, 02:49:30 pm »
There is a problem with the new proof. You assumed that the upper most brick is at the same time out most. If it is in the middle, you cannot push it out.

For the professor's question, because the number of bricks is doubled, the maximum extension is increased by ln 2 = 0.6931, which is bigger than 0.11. Even bigger than 0.25, the miximum gain that can be obtained this way.

万精油

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #9 on: 四月 01, 2005, 03:06:33 pm »
Quote
because the number of bricks is doubled, the maximum extension is increased by ln 2 = 0.3937, which is bigger than 0.11. Even bigger than 0.25, the miximum gain that can be obtained this way.

You are avoiding my question , I didn't double the bricks. I only split the top one into two, thus, going from n to n+1, not from n to 2*n.

七把叉

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #10 on: 四月 01, 2005, 03:34:58 pm »
Quote from: fzy
There is a problem with the new proof. You assumed that the upper most brick is at the same time out most. If it is in the middle, you cannot push it out.

七把叉

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #11 on: 四月 01, 2005, 03:43:31 pm »

七把叉

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #12 on: 四月 01, 2005, 03:55:30 pm »

七把叉

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #13 on: 四月 01, 2005, 04:05:47 pm »
Quote from: 万精油
According to earlier posts, the nth brick will extend out 1/(2*n), which is getting smaller and smaller (the sum will go to infinite of course). By your new approach, if I split the top brick into two (thus, going from n brick to n+1 brick), I can always extend it 0.11 out. This is much bigger than 1/2*n, what happened?

Note: I know my above argument is wrong, I just want to see if people can find out where did I go wrong? (hint: physics)

1024块砖可伸出的距离是1/2(10*ln2+1/2^11+0.57721)=3.75。

2^22块砖，用前法可伸出6倍砖长。而用后法可伸出7.9。

七把叉

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每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
« Reply #14 on: 四月 01, 2005, 05:01:13 pm »