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你能解决这个蒙德里安方块谜题吗?

Can you solve the Mondrian squares riddle? - Gordon Hamilton

本集播放后别走开 还有一个额外的迷题
Stick around after this episode for a bonus riddle.
荷兰艺术家Piet Mondrian的格子画
Dutch artist Piet Mondrian’s abstract,rectangular paintings
启发数学家们创造了双重挑战
inspired mathematicians to create a two-fold challenge
首先 必须用没有重叠的矩形覆盖住一个方形的帆布
First, we must completely cover a squarecanvas with non-overlapping rectangles
一切都要独一无二 如果我们使用1×4的尺寸
All must be unique, so if we use a 1×4,
我们就不能在其他地方用4×1的尺寸了
we can ’ t use a 4×1 in another spot,
但可以使用2×2尺寸的矩形
but a 2×2 rectangle would be fine.
现在让我们来试试
Let’s try that.
如果我们有一张4×4尺寸的帆布
Say we have a canvas measuring 4×4.
我们不能直接把它分成两半
We can’t chop it directly in half,
虽然这样可以得到两个一样的2×4尺寸的矩形
since that would give usidentical rectangles of 2×4.
但接下来最靠近的选项3×4和1×4的作品
But the next closest option- 3×4 and 1×4 – works.
就很简单 但我们还没有做好
That was easy, but we’re not done yet.
现在取最大矩形的面积
Now take the areaof the largest rectangle,
并减去最小的面积
and subtract the area of the smallest.
结果是我们的分数
The result is our score,
我们的目标是尽可能得低分
and the goal is to get as low a score as possible.
在这里 最大的面积是12 最小的是4
Here,the largest area is 12and the smallest is 4,
那就得到8分
giving us a score of 8.
因为我们不想
Since we didn ’ t try to go
在那时得一个低分
for a low score that time,
我们可以做到更好
we can probably do better.
我们把3×4尺寸拆分为3×3和3×1尺寸时
Let ’ s keep our 1×4 while breaking the 3×4
要保持1×4尺寸
into a 3×3 and a 3×1.
现在我们的得分是9减3或6
Now our score is 9 minus 3, or 6.
仍然不是最理想的 但比刚才好
Still not optimal, but better.
用这么小的一张帆布 就没有过多选项
With such a small canvas,there are only a few options.
当帆布尺寸变大时 让我们看看会发生什么
But let’s see what happenswhen the canvas gets bigger.
试试一个8×8的尺寸 看看你能得的最低分是多少?
Try out an 8×8;what’s the lowest score you can get?
如果你想自己弄清楚就在这里暂停一下
Pause here if you wantto figure it out yourself.
答:3
Answer in: 3
答:2
Answer in: 2
答:1
Answer in: 1
为了可以更清楚 我们可以之前开始:
To get our bearings,we can start as before:
把画布大致分成两部分
dividing the canvas roughly in two.
给我们一个面积为40的5×8长方形
That gives us a 5×8 rectanglewith area 40
和一个3×8面积为24的区域
and a 3×8 with area 24,
得了16分
for a score of 16.
这非常糟糕
That’s pretty bad.
把5×8分成5×5和5×3
Dividing that 5×8 into a 5×5 and a 5×3
我们就会得到10分
leaves us with a score of 10
但仍然不是最好的
Better,but still not great.
我们可以继续分最大的矩形
We could just keep dividingthe biggest rectangle.
但这样会有越来越小的矩形
But that would leave uswith increasingly tiny rectangles,
这将增加最大和最小之间的范围
which would increase the rangebetween the largest and smallest.
我们真正想要的是把所有的矩形
What we really want is for all our rectangles
都限制在小范围内
to fall within a small range of area values
因为画布的总面积是64
And since the total areaof the canvas is 64,
这些区域需要加起来
the areas need to add up to that.
假设我们列出可能的矩形和区域
Let’s make a listof possible rectangles and areas.
为了提高我们以前的成绩
To improve on our previous score,
我们可以试着选择一个小于或等于9的范围的值 并且加起来有64
we can try to pick a range of values spanning 9 or less and adding up to 64.
你会注意到有些值被省略了
You’ll notice that some valuesare left out
因为像1×13或2×9这样的长方形不适合帆布
because rectangles like 1x13or 2×9 won’t fit on the canvas.
你可能还会意识到
You might also realize
如果你使用其中一个奇数矩形
that if you use one of the rectangles
如5 9或15
with an odd area like 5, 9, or 15,
你需要使用另一个奇数矩形来得到偶数和
you need to use another odd-value rectangle to get an even sum.
综合考虑 让我们看看哪种方法是可行的
With all that in mind,let’s see what works.
从20区或更多区域开始 我们就会很快超过极限
Starting with area 20 or more puts us over the limit too quickly.
但是我们可以用长方形达到64 在14到18范围内 忽略15 不幸的是
But we can get to 64 using rectangles in the 14-18 range, leaving out 15.
不幸的是 没有办法使它们匹配
Unfortunately there’s no wayto make them fit.
使用2×7留下了空白
Using the 2×7 leaves a gap
只能由宽度为1的矩形填充
that can only be filled by a rectangle with a width of 1.
降低到8到14的有效范围
Going lower, the next rangethat works is 8 to 14,
省去3×3的正方形
leaving out the 3×3 square.
这一次 这几片就匹配了
This time, the pieces fit.
那是6分
That’s a score of 6.
我们能做得更好吗?
Can we do even better?
不能了
No.
我们可以得到相同的分数
We can get the same score
通过扔掉2×7和1×8
by throwing out the 2×7 and 1×8
用3×3 1×7和1×6替换它们
and replacing themwith a 3×3, 1×7, and 1×6.
但是如果我们从列表中向下走
But if we go any lower down the list,
数字变得很小
the numbers become so small
我们需要更宽的
that we ’ d need a wider range
覆盖帆布的尺寸
of sizes to cover the canvas,
这会增加分数
which would increase the score.
这里没有什么诡计或公式 只是一点点直觉
There ’ s no trick or formula here – just a bit of intuition.
这是艺术而不是科学
It’s more art than science.
对于较大的网格 专家数学家不确定
And for larger grids, expert mathematicians aren ’ t sure
他们是否找到了可能的最低分数
whether they ’ ve found the lowest possible scores.
那么,你如何除以4×4 10×10
So how would you divide a 4×4, 10×10,
或32×32帆布呢?
or 32×32 canvas?
试一试 并在评论中晒出你的结果
Give it a tryand post your results in the comments.

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视频概述

数学家无法确定的最小值,可能会由艺术家的作品来解答

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视频来源

https://www.youtube.com/watch?v=AWcY2-FBa9k

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