# 情趣生活

## 情趣生活 => 灵机一动 => Topic started by: 万精油 on 三月 27, 2005, 11:00:11 pm

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 万精油 on 三月 27, 2005, 11:00:11 pm

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

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

Title: Re: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 三月 28, 2005, 12:36:36 pm
Quote from: 万精油

Title: Re: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: fzy 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.  :cry:  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.  :lol:  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."
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: sean on 三月 28, 2005, 05:15:45 pm

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: zzzzzzzzzz 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.
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 三月 28, 2005, 11:33:25 pm

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 四月 01, 2005, 12:42:08 pm

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 万精油 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)
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: fzy 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.
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 万精油 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. :D
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 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.

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 四月 01, 2005, 03:43:31 pm

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 四月 01, 2005, 03:55:30 pm

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 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。

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 四月 01, 2005, 05:01:13 pm

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 四月 01, 2005, 05:43:58 pm
Quote from: 七把叉

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: fzy on 四月 01, 2005, 06:06:17 pm
Quote

The proof by contradiction can be fixed. But you have not fixed it yet. :wink: It is actually very easy to fix: Instead considering the maximum extend of all constructions, consider the maximum entends of all monotone constructions. No changes for the rest.

Quote

It looke like is. But your proof does not look good enough. You proved it is "locally optimal", but have not proved it is globally optimal.  :( [/quote]
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 四月 01, 2005, 06:15:28 pm

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 四月 01, 2005, 06:32:36 pm
Quote from: fzy

The proof by contradiction can be fixed. But you have not fixed it yet. :wink: It is actually very easy to fix: Instead considering the maximum extend of all constructions, consider the maximum entends of all monotone constructions. No changes for the rest.

Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 差不多 on 四月 01, 2005, 09:56:13 pm

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Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: warren on 四月 02, 2005, 01:10:49 am
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)

That is because of the top 2 bricks are thiner and lighter than other bricks.
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: warren on 四月 02, 2005, 01:15:11 am
Quote from: 差不多

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Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 差不多 on 四月 02, 2005, 07:41:47 am
Quote from: warren
Quote from: 差不多

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:oops: 大学还是没上成，补考：
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Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: zzzzzzzzzz on 四月 02, 2005, 01:07:12 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)

Not sure if I understood your question correctly, 1/2n applies to the bricks on the bottom.  The top brick can always extend out 0.5 > 0.11.
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 七把叉 on 四月 03, 2005, 09:49:14 am
Quote from: 差不多

:oops: 大学还是没上成，补考：
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〓〓〓〓〓〓〓〓
〓〓〓〓〓〓〓〓〓〓
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Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: physics on 四月 07, 2005, 06:28:34 pm
the method by z^10 gives the optimal result.
It can be showed in the following way.
If for any given n bricks, sum (1/2*n) is the best result, then ok.

Otherwise, suppose for some m,
the result can be greater than sum (1/2*m)
so there must exist some k<m, such that
for any numbers less than k,
the extension is not greater than sum (1/2*a), for 1<=a<=k

while for k+1 bricks, it is greater than sum 1/2*(k+1)

We know it is impossible. To show it is impossible.

For the (k+1)th brick
it should satisfy two conditions,
1st, its extension relative to the first brick is greater than sum(1/2*(k+1)),
2nd, its relative extension to the second brick should be not greater than sum(1/2*k),
but it is impossible according z^10's method of calculating center of mass.
Title: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 万精油 on 四月 08, 2005, 11:15:17 pm
Check out the following link, it can be used as the title picture of this thread. :)

http://web.wenxuecity.com/BBSView.php?SubID=joke&MsgID=91152
Title: Re: 每周一题: 砖块延伸 (3/28/05 -- 4/3/05)
Post by: 万精油 on 九月 06, 2011, 03:54:05 pm

(http://www.zhipingyou.com/images/coin1.jpg)
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