Dutch artist Piet Mondrian’s abstract, rectangular paintings inspired mathematicians to create a two-fold challenge. First, we must completely cover a square canvas with non-overlapping rectangles. All must be unique, so if we use a 1x4, we can’t use a 4x1 in another spot, but a 2x2 rectangle would be fine.
荷蘭藝術家皮特蒙德里安的 抽象矩形畫作 給了數學家靈感, 創造出了一種二重挑戰。 首先,我們得要用不重疊的矩形 來把正方形的畫布蓋滿。 形狀不可重覆,若用過了 1x4 的 矩形,就不能再用 4x1 的矩形, 但可以用 2x2 的矩形。
Let’s try that. Say we have a canvas measuring 4x4. We can’t chop it directly in half, since that would give us identical rectangles of 2x4. But the next closest option - 3x4 and 1x4 - works.
咱們來試試看。 假設我們的畫布大小是 4x4。 我們不能直接把它切一半, 因為那樣會讓我們得到 兩個一樣的 2x4 矩形。 下一個最接近的選擇就是 3x4 和 1x4,這是可行的。
That was easy, but we’re not done yet. Now take the area of the largest rectangle, and subtract the area of the smallest. The result is our score, and the goal is to get as low a score as possible. Here, the largest area is 12 and the smallest is 4, giving us a score of 8.
那樣很容易,但還沒結束。 把最大矩形的面積 減掉最小矩形的面積。 結果就是我們的分數, 目標是要讓分數越低越好。 在這裡,最大的面積 是 12,最小的是 4, 所以我們的分數是 8。
Since we didn’t try to go for a low score that time, we can probably do better. Let’s keep our 1x4 while breaking the 3x4 into a 3x3 and a 3x1. Now our score is 9 minus 3, or 6. Still not optimal, but better.
因為我們剛才在做的時候 並沒有以最低分數為目標, 也許我們可以做更好。 咱們留下 1x4, 把 3x4 拆開成 3x3 和 3x1。 現在我們的分數就是 9 減掉 3,也就是 6。 仍然不是最佳的,但有進步了。
With such a small canvas, there are only a few options. But let’s see what happens when the canvas gets bigger. Try out an 8x8; what’s the lowest score you can get?
這張畫布很小, 可能的選擇很有限。 但,咱們來看看, 當畫布變更大的時候會如何。 試試看 8x8 的畫布; 你能得到的最低分數是多少?
Pause here if you want to figure it out yourself.
如果你想要自己試試看, 在這裡按下暫停。
Answer in: 3
答案公佈倒數:3。
Answer in: 2
答案公佈倒數:2。
Answer in: 1
答案公佈倒數:1。
To get our bearings, we can start as before: dividing the canvas roughly in two. That gives us a 5x8 rectangle with area 40 and a 3x8 with area 24, for a score of 16. That’s pretty bad. Dividing that 5x8 into a 5x5 and a 5x3 leaves us with a score of 10. Better, but still not great. We could just keep dividing the biggest rectangle. But that would leave us with increasingly tiny rectangles, which would increase the range between the largest and smallest.
為了確認方向, 可以用剛才的方式開始: 把畫布大致上切為兩塊。 這樣就會有一個 5x8 的 矩形,面積是 40, 還有一個 3x8 的矩形,面積 24, 得到的分數為 16。 這分數很糟。 把 5x8 拆成 5x3 和 5x5, 分數就會變成 10。 好一點,但仍然不夠好。 我們可以持續把最大的矩形拆開。 但那會造成小矩形越來越多, 就會導致最大和最小的 矩形之間的差距變大。
What we really want is for all our rectangles to fall within a small range of area values. And since the total area of the canvas is 64, the areas need to add up to that. Let’s make a list of possible rectangles and areas.
我們真正想要的, 是希望所有的矩形面積 都落在一個很小的範圍內。 因為畫布的總面積是 64, 所有面積加總就得是 64。 咱們來做個表, 列出所有可能的矩形和面積。
To improve on our previous score, we can try to pick a range of values spanning 9 or less and adding up to 64. You’ll notice that some values are left out because rectangles like 1x13 or 2x9 won’t fit on the canvas. You might also realize that if you use one of the rectangles with an odd area like 5, 9, or 15, you need to use another odd-value rectangle to get an even sum. With all that in mind, let’s see what works.
若要改善我們先前的 分數(註:10 分), 我們可以試試把面積範圍 設為 9 或更小, 且面積加總要是 64。 你可以發現,有些值被略過了, 因為像 1x13 或 2x9 的矩形 無法放在這張畫布上。 你可能也會發現, 如果你選用一個面積是 奇數值的矩形,如 5、9,或15, 你就得要用另一個面積是奇數值的 矩形,才能讓總面積是偶數值。 記住這些之後, 咱們來看看要如何解題。
Starting with area 20 or more puts us over the limit too quickly. But we can get to 64 using rectangles in the 14-18 range, leaving out 15. Unfortunately, there’s no way to make them fit. Using the 2x7 leaves a gap that can only be filled by a rectangle with a width of 1.
若一開始就用面積 20 以上, 會讓我們太快超過限制。 但我們可以用面積在 14~18 的矩形 來得到 64 的總面積, 不要用到 15。 不幸的是,沒辦法 讓它們符合畫布。 用 2x7 會留下一個空戲隙, 這個空隙只能用 寬度為 1 的矩形來填補。
Going lower, the next range that works is 8 to 14, leaving out the 3x3 square. This time, the pieces fit.
再低一點的下一個 可行面積範圍是 8~14, 不會用到 3x3 的方形。 這一次,能填滿畫布。
That’s a score of 6. Can we do even better? No. We can get the same score by throwing out the 2x7 and 1x8 and replacing them with a 3x3, 1x7, and 1x6. But if we go any lower down the list, the numbers become so small that we’d need a wider range of sizes to cover the canvas, which would increase the score.
這樣的分數就是 6。 我們還能做更好嗎? 不能。 還有一種方式得到同樣的分數, 就是放棄 2x7 與 1x8, 用 3x3、1x7,及 1x6 來取代。 但如果我們選用列表中 更低的面積值, 數字會太小, 我們會需要更大的範圍, 才能把畫布面積都填滿, 這麼做就會增加分數。
There’s no trick or formula here – just a bit of intuition. It's more art than science. And for larger grids, expert mathematicians aren’t sure whether they’ve found the lowest possible scores. So how would you divide a 4x4, 10x10, or 32x32 canvas?
這裡沒有什麼秘技或方程式, 只有用一點直覺。 這比較是藝術問題,而非科學問題。 若格子更大, 數學專家也不確定他們 是否能找到最低的可能分數。 所以,你要如何分割一張 4x4、 10x10, 或 32x32 的畫布?
Give it a try and post your results in the comments.
試試看,把你的答案寫在留言中。