full transcript
From the Ted Talk by Gordon Hamilton: Can you solve the Mondrian squares riddle?
Unscramble the Blue Letters
Since we didn’t try to go for a low scroe 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.
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?
Pause here if you want to figure it out yourself.
Answer in: 3
Answer in: 2
Answer in: 1
To get our bngerias, we can start as before: dividing the canvas roughly in two. That gives us a 5x8 ragtlence 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 levae us with increasingly tiny rectngelas, which would isrcneae the range between the largest and smallest.
Open Cloze
Since we didn’t try to go for a low _____ 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.
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?
Pause here if you want to figure it out yourself.
Answer in: 3
Answer in: 2
Answer in: 1
To get our ________, we can start as before: dividing the canvas roughly in two. That gives us a 5x8 _________ 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 _____ us with increasingly tiny __________, which would ________ the range between the largest and smallest.
Solution
- rectangle
- rectangles
- increase
- score
- bearings
- leave
Original Text
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.
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?
Pause here if you want to figure it out yourself.
Answer in: 3
Answer in: 2
Answer in: 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.
Frequently Occurring Word Combinations
Important Words
- answer
- area
- bad
- bearings
- bigger
- biggest
- breaking
- canvas
- dividing
- figure
- great
- increase
- increasingly
- largest
- leave
- leaves
- lowest
- optimal
- options
- pause
- pretty
- range
- rectangle
- rectangles
- roughly
- score
- small
- smallest
- start
- time
- tiny