Very few can solve this puzzle. How many squares?
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- Опубліковано 2 чер 2024
- How many squares can be made, using any 4 points as corners?
0:00 problem
0:30 solution
3:24 animation
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You don't need to check every set of four points. Since every square is uniquely defined by two opposing corners, you only need to check pairs of points. And you can also use the symmetries of the shape to save you some time.
That's how I did it, it was pretty easy that way.
Exactly
He just brute force every point combos for the sake of it
Correct, just check all sets of two points, and keep a list of all squares already found so that when you get to the other diag of a square you've found the first diag of, you don't recount it. This video's idea of checking zillions of sets of four points, and then saying it's surprising no one ever did that, is silly. To check two points, simply see whether a 3rd point exists at the corresponding corner location, and if so, same for 4th point.
Also, there's 4 quadrants. Whatever you get for one, multiply by 4. The only square that seems to repeat itself is dead center of the original 9. At least that's how I got it, (3*4) + 9. Of course it may have been a lucky guess.
I'm sure you named your algorithm animation "The Square Dance."
I found all 9, of course, and also the 4 that are at odd angles. Somehow, I managed to find those ones while forgetting to check for the 45-degree rotated squares.
ditto
Same here
I found the same ones you did, then played the video, and realized I missed the 45 degree ones about the same time he got to them.
Yup, me 2.
I missed the 4:3 triangles
So, only 1% of people consider rotated squares?... I find that hard to believe.
Presumably 99% think rotated sqaures are diamonds or another shape so the brain doesn't look for it. That's my excuse anyway. At least this one doesn't feel like cheating 😂
or they find 17 because they only search for 45° rotated squares
I’m in the 99% crowd.
Me too dude. Tis hard to believe
@@zlacyeah I got 17. I actually thought of other rotations but for.some reason it seemed obvious to me that there couldn't be any square like this
I am in the 99% :`(
There is a more efficient brute force method than trying every single combination of 4 points. Once you have two ordered points, the other two points that would make up a clockwise oriented square. Assuming you take all ordered pairs of points and then check if the other points are valid, you will find all edges of all squares, or 4 times the number of squares with only n^2 tries (where n is the number of points we have). If you have a rule that lets you only try one side of each square, then you can do it in n^2/4 attempts.
I love the fact you wrote a program to affirm what you already figured out! I used to draw dots in various sized grids and join them in as many different ways as possible when I was a child. I think that was probably my first foray into probability.
Grrrrr. I paused this for almost 3 minutes and came up with 22. Soooo close. Great puzzle. Keep em coming with your brilliance on HOW to solve complex problems!
How does the 22th square look??? Which points did you use?
22? I came up with 21 and I'm now playing the rest of the video. 😶
I got 20... rectangles aren't 'squares' and I told them that. True squares are equal all sides period🎉
@@The_Cali_Dude_88 yes
@@The_Cali_Dude_88 You can get 21 squares. Which one of them do you consider a rectangle?
Nicely done. Most often, you show a known solution or say you learned part of the solution from someone else. Clearly, this time, you found all 21 squares yourself and then validated it with a computational method. And, as always, excellent presentation. And your approach to both finding and explaining squares by creating right triangles and identifying side lengths to test was brilliant. Kudos!
I did think of diagonal squares, but only found four of them. So I found 13 boxes in all. I thought my way out of the smallest box, but not out of the next box up.
then there's me counting all the rotated squares, but then forgetting to count all of the 9 obvious cross squares other than the center.
Don't you hate it when you over think things? I don't.
That animation was remarkably satisfying
I think it is at least 30 squares:
Look at @2:14 and @2:54; both solutions you made use of the dots as cornerpoints, but 4 additional small squares occur when you draw the 4 squares.
So by drawing 21 squares you get 30 squares. Maybe I overlooked some more?
Yeah, but those small squares are not connected to a point on each corner 🤓
Nailed it!
The more interesting question would be how to form a concise proof that there are only 21 possible squares to construct. I would start by noting that, apart from repeating pattern and mirroring, there are only 3 distinct "kinds" of dots in the pattern (the dot closest to the middle, the dot one out from the middle and the dot furthest away from the middle). Thus, we would only need to show the exhaustive number of kinds of squares with each of these 3 dots as a corner. That is pretty doable.
That is my question too. Someone already pointed out that a square is defined by its opposite corners, so you only have to consider pairs of points, then simplify further using the symmetry of the shape.
My initial guess is 21, but instead of lengths like you do in the video I select a vector which is the first side of a square (select a point as 1st vertex, add the vector to get a 2nd vertex, add the vector rotated 90° to get the 3rd vertex, then add the vector rotated 180° to get the 4th vertex; all 4 form a square, you get the idea). Since the dots can be placed on discrete two dimensional grid of size 6x6 there's only a handful of vectors to choose from. Since the square has a rotational symmetry of order 4 we can consider only vectors with both coordinates being non-negative integers (Quadrant I) and still miss no cases.
Here's my dissection of possible first square side vectors along with the number of dots that would form a square with all 4 vertices lying on the dots:
(0,1) - 9
(1,1) - 4
(2,2) - 4
(1,2) - 1
(2,1) - 1
(2,3) - 1
(3,2) - 1
I believe it could be easier to program with this approach, as you wouldn't have to check if the 4 points combination forms a square, but instead given a square verify if all it's vertices belongs to a set of points provided as input. Also it would decrease the computational complexity, because there's only 35 such vectors possible in a 6x6 grid (plus a trivial (0,0) vector but those squares don't count) and check all of them against every one of the 20 points you only have 700 cases to run. That's already significantly lower than C(20,4).
The number can be further shaved down if discard the vectors that are available in a full 6x6 grid but not in the input set. It can be easily proved visually, that for first coordinate of [number on the left] the second coordinate can only go as high as [number on the right]:
0, 5
1, 5
2, 3
3, 3
4, 1
5, 1
Considering only vectors of
(0,1) through (0,5) // skipping the trivial (0,0) for the total of 5
(1,0) through (1,5) // 6
(2,0) through (2,3) // 4
(3,0) through (3,3) // 4
(4,0) through (4,1) // 2
(5,0) through (5,1) // 2
we get the grand total of 23 vectors to check against 20 points for 460 combinations. Over 10x less than C(20,4).
That was one of the first computer games we made ourselves on a "home computer" at college, using a version that accepted only straight squares as template and generalizing it to arbitrarily positioned squares - 40 years ago. Was a nice exercise for the introduction into 2D vectors. Nostalgie :D
Nice, I got it right! I was about to declare 9 as my answer, but then thought "Hang on, what about diagonals?" Went for 21 but expected him to keep going beyond 21 with some I missed. Was really pleased when he didn't.
Thank you for making me feel smart lol
I made use of the symmetry. I graphed them and put the origin of the cartesian plane in the center. I only check to see if there is a side between (1,y) quadrant 1 and points within (-y,y) if that makes sense.
I check
(1,1) to (-1,1) is the side of a square. The diagonal passes through the center, so I count 1 square
(1,2) to (-1,1) is a side. The diagonal does not cross through the center, I count 4
(1,2) to (-2,1) is a side. The diagonal crosses through the center. I count 2
(1,2) to (-1,2) is a side. The diagonal does not cross through the center. I count 4.
(1,3) to (-1,1) is not a side
(1,3) to (-2,1) is a side. The diagonal does not cross through the center, I count 4
(1,3) to (-3,1) is a side, The diagonal crosses through the center, I count 2
(1,3) to (-1,2) is not a side
(1,3) to (-1,3) is a side. The diagonal does not cross through the center. I count 4
Total of 21
Instead of trying all 4 points, you can just test all pairs with the first point being higher or more left than the second and test if the 2 remaining points for a square (clockwise travel) are in the set.
Great that you could computationally check all of the possibilities!👍
Yes but I think he only checked combinations when he should have checked permutations just to be sure! 😄
Programing always turns out to be harder than expected. Good job.
I haven't done a problem like this since I've been in college (I graduated seven years ago), so I'm pleasantly surprised I solved this one with ease. I found the 1 squares first, followed by the root 2 squares, then the root 13 squares, then the root 5 squares, and lastly the 2 root 2 squares. I had no math behind it. I just visualized it!
I love to do these problems. This one didn’t take math but noting which squares were already counted. Got all 21 but did it on paper with 3 patterns for the different angles.
I could NOT visualize ALL the tilted squares without tracing them out. My mind was going numb trying to trace them out on the grid and then keep track of them. I had to methodically draw them all out to get the right answer. This is true for problem solving in general. I have to put pen to paper. I run into trouble if I try to do anything complex like this in my head.
@@Tiqerboy But you still got the right answer! That's what matters!
@@RobiBue Very nice!
Maybe the first time I was able to figure out the correct answer of one of your problems without any help!
I'm pretty good in mathematics and also help students with that, but nevertheless I can still learn very much from your videos, thx therefore! :D
I like that your title is so different of the actual question discussed
This was super interesting 👍 thanks
I got this. So I am in the putative 1%. Really it just requires care and probably helps to have seen something vaguely similar to find the skew ones. I find it hard to believe it is only 1% of us though, there wasn't anything really surprising there. Double counting was the main thing I found I needed to avoid... book-keeping skill...
I found them in about 15 seconds, thought I was missing something. I didnt even watch the video, just from the thumbnail 😂
I missed 2 of them. great video!
With your code, you could also tally rectangles, kites, rhombi, trapezoids, parallelograms etc... be interesting to see how they compare.
There are 21. If you number the dots from top to bottom, left to right; there are 9 like 1:3:4:2, 4 like 4:7:14:9, 4 like 3:12:18:9, 2 like 2:6:18:10, and 2 like 1:9:20:8.
It would be interesting to know if there is a general rule here for shaped like this. If this was considered of order 2, an order of 3 equivalent would be made up of five 3x3 squares. Or even look at cubes in a 3d version?!
My first count was 10, but I had a feeling there were more. I just could not see them. Thanks for the early morning headache.
Can you make a video about this problem:
Let ABC be a triangle and M be a point inside it
1) let a, b and c be three strictly positive real numbers
we put:
x=(a+b)/c, y=(b+c)/a, z=(c+a)/b
check that xyz=x+y+z+2
2) the lines (AM), (BM) and (CM) intersect respectively at A', B' and C':
to show that:
(MA/MA') • (MB/MB')
• (MC/MC') = MA/MA' + MB/MB' + MC/MC' + 2
There's an optimisation for your algorithm.
Once you have choosen 2 points, which can't be horizontally aligned, you can consider it is the side of the square abd that you'll only search far squares that are on right and bottom of your 2 points. If the square missing vertices matches two points, you have found a new square. You'll find all squares with less tests.
Thanks
A more efficient way of counting would be choosing two points at a time and checking whether a square with them as opposite corners is on the grid. Then divide by two, as the squares would be double counted.
Thank you for making this video. Just wanted to know which language/tool you used to program the possibilities. It would be great if you can share the code as well.
4:51 I wish when the music started, he played a different tone for each point touched. He found enhance it by leaving the initial points playing while looking for the other possibilities.
I will call the big cross C, and the big cross with the eight border points excluded C'. There is a trick : the points of Z^2 strictly inside the considered squares (not in the sides) must be in C'. This is because the interior of the convex hull C intersected with Z^2 is equal to C'. It is easier to see if the interior points of a square is included in C' to see if the four corners of it belongs to C.
For a square formed with the vector (a,b), the number of strictly inside points equals a^2 + b^2 - 1 if a,b coprime, thus a^2 + b^2 \leq 13 in this case. If a,b not coprime, a^2 + b^2 cannot be too big either. Etc.
I'm curious how you tested for or sorted for "non-intersecting polygons" [3:56] and is that the same as convex polygons? Some 20+ years ago, we had regular terrain data points that required 2 triangles for each 4 points to render. I tried to optimize the terrain into fewer points, but got stuck on the test/sort that you solved. A short description or pointer would be appreciated.
Yeah, took me some thinking but paused the video at 0:20 and found 21 without difficulty, seriously I found it pretty easy, albeit interesting.
Only 1% could solve it indeed , thanks for sharing your knowledge Sir.
Well, there are 20 dots, and it takes 4 to make a square. 20/4=5. 5 "squared" is 25, subtract 4 for what makes a square: 25-4=21.
In the automated counting of the squares, instead of considering each choice of 4 points, I'd have selected only 2 as two consecutive vertices of a square in clockwise order. At this point you can compute the coordinates of the other two vertices and see if they are valid. (You can find the third and fourth vertices by taking the vector between the two first vertices, rotate it 90 degrees clockwise, and apply it to the first and second vertices.) You can remove duplicates by only considering vectors going "right" (increasing x), and whose y coordinate doesn't decrease.
I first did it on paper but opened AutoCad that will provide distances between points. That was much easier and probably the 1% having the right tools to solve this challenge.
I counted 21, after looking for all possible ways to create squares connect-the-dots-style with the points given.
Indeed takes about 30 secs at most to count them all, so anything more than counting is a little overkill unless you don't feel confident in picking out all the different types.
I didn't see the last two squares.
Nice problem !
I guess we could also consider more points to create an harder question.
Finding the total number of rectangles would be interesting as well.
There are fundamentally 3 different types of points. Type A (8 points) belong to 3 squares, Type B (8 points) belong to 5 squares and type C (4 points) to 5. So thz number of squares is (8×3 + 8×5 + 4×5)÷4 = 21.
I am following u since 3years , and I am totally obsessed by your ability of solving any problem, U are great 👑👑👑👑
Ah, but you've got a logical fallacy there. We don't know that he can solve every problem
For all we know, he only solves one out of every ten problems he finds. He just doesn't make videos about the other nine
@@douglaswolfen7820 No matter what he does but he is the goat!
May i ask what software/library was used to do the animation and render the video?
Looks like 3b1b/manim
If you don't restrict it to requiring the points be the corners, you can draw 9 more squares that form from intersecting lines.
Hi Presh! I was wondering if you gave a thoutgh about the numbers in which correct combinations are occuring? Are they forming any pattern? For sure it depends on the shape of points on the lattice. But maybe there is some regularity behind that? Would be nice to make another use of that algorithm :)
I would have left the 2*Sqrt(2) as the "improper" Sqrt(8) Makes the progression more evident.
I came up with 9, then paused the video shortly after the explanation started, because I had remembered diagonal squares. I counted 21, but I miscounted 2 more odd angle ones and missed 2 45 degree angle ones.
I got it. I just search for every possible side lengths :
9 of 1
Sqrt(2) and 2sqrt(2) (45°) are basically the same at a scale factor with 4 squares each
Same for sqrt(5) and sqrt(13) with only 2 each
Finally, one I got the answer first time.
I found all the squares easily, but to me, the text is a bit vague. When a square must use four points at corners, does this mean that it may or may not have other points on its sides? I read the text the way that it may not (and then some of the found squares must be excluded).
It is kind of satisfying but also an overkill to check ALL combinations.
It would be simpler to have two points and add a third point.
IF the distance 1-2 is not equal to the distance 2-3, there cannot be a square.
IF the distance 1-2 is equal but the lines do not form a right angle, there cannot be a square.
Only if those two conditions do apply, you have to check a 4th point at all.
BUT: It would be very interesting what the average area of all those polygons is and other statistics like the lenghth of all lines and so on.
Like it‼️
I got 25 ... I found all 21 and counted all of the root5 ones twice.
Probably if I had pen and paper I wouldn't overcount.
I am impressed by the animation; how to generate? any suggestions?
screenshotted the dots, scribbled some squares on them, and found 21 pretty quickly ... and then spent way more time than that looking for more squares because of what it said on the thumbnail-pic
I missed 8 of the trivial squares... these types of things sure can mix you up
Got it right, pretty easy.
How would you solve it if the question asked for how many rectangles rather than squares, is there a way without just computing?
You could've just made a complete grid and exclude 16 points from the list. That would allow you to put a value between points, then do math to figure out if they're squares.
I love the animation and the music, but I'm going to say (for fun) that there are only 9 squares, the other 4-corner shapes are diamonds.
before watching i tried solve this by myself, and got the right answer
What tune is that at 5:00? I recognise it. Is it Tetris or Snake or something?
I also came up with the answer of 17; missed the 4 obscure squares.
actually, thers atleast 23 as the question is asked,
if u draw lines trom the outer most points
nr 22:
L upper point- down to R 45 deg edge point
R upper point- down to L 45 deg edge point
L lower point - upp to R 45 deq edge point
R lower point -to- upp L 45 deg edge point
..the resulting squere will have a side of 2(√2) units, and a starting point 0.5 units from the edges
nr 23: a smaller one one step in with the side √2
..cant atm see any more...
The answer with 17 squares only counted the 9 normal squares and the 8 diamonds, but not the 4 tilted squares.
I found the easy 9 and the diagonal 4. I didn't realize how many more there were.
I found all of them quite quickly in just few minutes, but I didn't trust that I have found them all and also thought that I might have counted few multiple times.
Solved it in 30 secs :). Finnaly a problem I got right.
It's 9 squares standing on their sides and 8 diamonds. And another 4 with odd angles. So, 21 squares overall.
If you think in 3 dimensions, there could be squares that don't look like squares on this plane
guessing before watching. I got 21. 9 trivial small. 4 small at 45deg, 4 large at 45deg, and 4 centered using the outer 4 dots with radial symmetry
It's 9 small squares, 2 sizes of 4 tilted (at 45°) squares, and 2 sizes of 4 tilted (not at 45°) squares; a total of 25 squares.
I made a mistake !!!
The last are 2 sizes of 2 tilted (not at 45°)
I GOT ONE!!! I don’t BELIEVE it! This is better than the day the new phone book arrived! (But not as good as the day I discovered my special purpose).
It could be argued that the four 2√2 are not legitimate for the problem because their side pass through another of the given points. None of the other squares do this. If the problem is to make squares from one point to a second ,then the 2√2 lines stop before completing the square . So 17 would be the right answer.
That is the answer I got for the very same reason.
I found 21 squares and promptly convinced myself that I must have overlooked something incredibly obvious.
As soon as you used the magic word "any" I knew there'd be way more than 9 lol!
I wrote the code to find all 4 points that have an equal distance from each other at the same time. Got 21.
How many squares using any four points?
Would be sum of all the squares using any four points.
So, let's sum them up.
We'll only count each square once for any given set of four points.
Let's start with smallest integral side length squares and go up from
there.
For reference, let's label each point,
we'll use a Cartesian-like reference, similar to (x,y),
but for brevity we'll omit all but xy, so instead of,
e.g. (2,1) we'll just use 21.
We'll label the points as follows,
with x ranging from 0 through 5, all integers, and likewise for
y, so, these are the labels for our points, shown graphically to
correlate to their positions (and - for unfilled space/points):
------52-53------
------42-43------
30-31-32-33-34-35
20-21-22-23-24-25
------12-13------
------02-03------
So, smallest squares have side 1, and there are
9 of those.
We have no squares of integral side lengths of 2 through 5 nor larger,
nor smaller than 1, so what about other sizes between?
Sides of (square) root 2 ... we have
4 of those (may be easier to visualize by rotating image 90 degrees).
Sides of (square) root 5? ... again, rotating may aid visualization,
and we have ...
2 of those (don't forget both non-redundant rotation sets).
Sides of 2 root 2? ...
4 of those (think of these,
squares having adjacent corners 02 and 20 x 4 for symmetry = 4,
each consumes a pair of the most extreme horizontal (x) and
vertical (y) points, and there are only 8 of those to consume, so that
gives us our 4 squares of 2 root 2 sides.
)
Next side size we can consider is root(2*2+3*3)=root 13 ...
2 of those only and exactly, due to what fits + symmetry/rotation
Next we have side size 3 root 2,
... and we don't have anything that size or larger that fits all four
square sides.
So, add 'em up, 9+4+2+4+2=
21 squares.
23 squares? I counted this is my guess. (Counted 4 where there were 2... Hard to do in your head but I found them all)
I used to do these as a kid with different boxes using lines and not dots and you would have to count the squares. And it's actually funny because I made this exact puzzle with dots when I was younger and I remember erasing it because I only counted nine! Crazy I got 23 this time lol. I wonder if my IQ went up or down since then😅
The question mislead the analysis. I did it with tooth picks which proved very difficult, i finally drew them 21
I only found 17. Missed the four squares that don’t have a point in the center of each side ( sqrt 5 and sqrt 13).
Using symmetries I got n = 9 + 2*2 + 2*3 + 2*1 = 21
I think you made a new screen saver.
Arguably, there are actually 84 possible squares. As humans, we just naturally tend to collapse the 4 different starting points for each square down to a single solution. A machine would consider them distinct.
Just because a machine would say so, doesn't make the machine correct. If I look at my house from a different angle, it isn't a new house.
@@adamperdue3178 Reasonable, it's just a matter of definition. It's not a new house, but it is a new VIEW of a house.
Every entity must operate on what IT considers an important instance. There are also cases in human endeavour where the starting corner is important.
Brute force… the old golden hammer of mathematics
I got 23 because in my head, I could see 4 of the root 5 ones. :)
Wooo I'm in the 1% club! Where's my £100,000?
I found all 21 also, so you're going to have to share it with me!! 🤪
Hey I found all 21 as well, I want some of that money!!
I’m sure the check is in the mail. 😁
@@rojer9542 real :D
20 squares with sidelength of 0 and
21 squares with non-zero sidelength.
Maths used to be purely theoretical answers.
Computers however alllow us to do actual experimental answers. If the answer space is limited.
At first, I accidentally multiplied the rt5 squares by 4 and got 23, but then rethunk and got 21.
I got them, too, within 3 minutes. I thought I might have missed some because it was so easy, but no.
The first 14 were easy to find. The rest were a little puzzling.