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zakruti.com » Knowledge, science, education » TED-Ed
Can you solve the Mondrian squares riddle? - Gordon Hamilton

Can you solve the Mondrian squares riddle? - Gordon Hamilton

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Rating: 4.0; Vote: 1
Dutch artist Piet Mondrians abstract, rectangular paintings inspired mathematicians to create a two-fold challenge. Can you solve the puzzle and get to the lowest score possible? Gordon Hamilton shows how. Lesson by Gordon Hamilton, directed by Anton Trofimov
Date: 2020-08-22

Comments and reviews: 10


Cover a square with non-identical rectangles
Uh. How about the entire square? Squares are rectangles, so that is one too. It's unique since there can't be two of the same if there's only one. Its sides are whole numbers. It covers the entire area. The score is 0, you can't go lower than that since MAX >= MIN, so MAX - MIN >= 0. It doesn't even matter what size we're supposed to cover (assuming it's whole by whole, which it has to be or you'd never cover the entire thing no matter what, we're done here.

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You know something's wrong when Ted-ed asks u a puzzle and there is no threat to earth or mankind, no alien lord is trying to destroy the world, we see no leviathan popping out of the sea, no dictator is ruling who only frees people with green eyes or there is any need of an eccentric scientist trying to circumference the earth in 6 Hours!
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Quickly watching the video thought the riddle was 4x4 but it's a bonus soooo.
This is what I got (each place with a same letter makes them part of the same rectangle.
1 2 3 3
1 2 3 3
1 4 4 4
1 4 4 4
6-2=4

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I found a solution with 6 by splitting. Start with 1x8 and 7x8. Split the second into 7x1 and 7x7, then 2x7 and 5x7, and 5x3 and 5x4. Finally split the first into 2x3 and 3x3 and the second into 2x4 and 3x4.
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I don't recall them ever saying the measurements had to be whole numbers. Using fractions you could get the score infinitely small using only 2 pieces really close in size but not exact.
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Easy! For a 32x32 fill the canvas with a single square 32 by 32 large.
Buggest-smallest square
964-964=0!
Oh. the idea was to use more than one shape?

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wait, so our score is the biggest minus the smallest? so what if I don't cut it at all? the biggest and the smallest are the same so my score would be 0?
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correct answer:
IF canvas is X by X
THEN cut a part that is X by X-0, 01
now you have a score of (X1/2X+0, 01)-(X1/2X-0, 01)
YOU WIN!

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Don't cut it. That's how you get the lowest score. Say it has an area of 8. That'd be the smallest and the largest, meaning your score is zero.
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For a 4x4 Id use a 2x3 (area 6, 1x4 (area 4, 2x2 (area 4) and 1x2 (area 2) for a score of 4, I couldnt find a lower score
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