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.

@onradioactivewaves

26 күн бұрын

That's how I did it, it was pretty easy that way.

@teambellavsteamalice

26 күн бұрын

Exactly

@vinching926

25 күн бұрын

He just brute force every point combos for the sake of it

@qc1okay

25 күн бұрын

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.

@gafjr

25 күн бұрын

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.

@jeremyashford2145

26 күн бұрын

ditto

@MrDreugh

26 күн бұрын

Same here

@lucario5748

25 күн бұрын

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.

@aukword6255

24 күн бұрын

Yup, me 2.

@the4spaceconstantstetraqua886

24 күн бұрын

I missed the 4:3 triangles

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.

@wiseoldfool

25 күн бұрын

Don't you hate it when you over think things? I don't.

So, only 1% of people consider rotated squares?... I find that hard to believe.

@vetinaris1297

26 күн бұрын

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 😂

@zlac

26 күн бұрын

or they find 17 because they only search for 45° rotated squares

@jetx_47

26 күн бұрын

I’m in the 99% crowd.

@abhishankpaul

26 күн бұрын

Me too dude. Tis hard to believe

@migssdz7287

26 күн бұрын

​@@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.

That animation was remarkably satisfying

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!

@ThePowerfulOne07

26 күн бұрын

How does the 22th square look??? Which points did you use?

@nicknike

26 күн бұрын

22? I came up with 21 and I'm now playing the rest of the video. 😶

@The_Cali_Dude_88

26 күн бұрын

I got 20... rectangles aren't 'squares' and I told them that. True squares are equal all sides period🎉

@siliconslice

26 күн бұрын

@@The_Cali_Dude_88 yes

@nicknike

26 күн бұрын

@@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.

You can mathematically prove there are only 21 squares by looking at the possible center points for the squares and using rotational symmetry to only have to check 8 center points, an then multiply the ones that aren't the very center by 4 to account for symmetry and you get 9 out of the under 40 center-vertex combinations that work and 5 of those with the epicenter so we have 5+4*4=21

@rorygilmoreNZ

8 күн бұрын

well I’m surprised anyone would need to make a calculation at all. To me that seems ridiculously over worked. I got the the 9 square answer in about 5 seconds and the 21 square answer in about 30 just by looking at it and tilting my head. Just as a graphic artist might break down the elements of something you want to draw. It’s not a mathematical problem as much as a creative one. It’s just a question of visualisation surely… not intelligence.

Nailed it!

I was listening to an audiobook and using another tab to see what was on YT, I came across this video and thought it was interesting. Surprisingly I came up with the answer.

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

This was super interesting 👍 thanks

Thanks

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.

@MichaelRothwell1

25 күн бұрын

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.

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?!

Great that you could computationally check all of the possibilities!👍

@randombrian256

26 күн бұрын

Yes but I think he only checked combinations when he should have checked permutations just to be sure! 😄

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?

@s3m527

25 күн бұрын

Yeah, but those small squares are not connected to a point on each corner 🤓

Programing always turns out to be harder than expected. Good job.

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

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.

I missed 2 of them. great video!

Thank you for making me feel smart lol

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 like that your title is so different of the actual question discussed

I am impressed by the animation; how to generate? any suggestions?

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.

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

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

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.

With your code, you could also tally rectangles, kites, rhombi, trapezoids, parallelograms etc... be interesting to see how they compare.

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.

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 :)

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.

May i ask what software/library was used to do the animation and render the video?

@cheweh842

26 күн бұрын

Looks like 3b1b/manim

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 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!

@RobiBue

26 күн бұрын

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.

@Tiqerboy

26 күн бұрын

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.

@dylan9013

25 күн бұрын

@@Tiqerboy But you still got the right answer! That's what matters!

@dylan9013

25 күн бұрын

@@RobiBue Very nice!

How would you solve it if the question asked for how many rectangles rather than squares, is there a way without just computing?

I didn't see the last two squares. Nice problem ! I guess we could also consider more points to create an harder question.

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).

What do you use to make these animations? (You probably get asked this a lot - sorry!)

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.

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.

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.

Note you can remove all cases where three dots (corners) stay on single a line.

I can't be the only one who counted all 13 rotated squares and forgot to count in also the 9 simple squares.

What tune is that at 5:00? I recognise it. Is it Tetris or Snake or something?

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.

I found them in about 15 seconds, thought I was missing something. I didnt even watch the video, just from the thumbnail 😂

Like it‼️

I am following u since 3years , and I am totally obsessed by your ability of solving any problem, U are great 👑👑👑👑

@douglaswolfen7820

26 күн бұрын

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

@chandranisahanone

26 күн бұрын

@@douglaswolfen7820 No matter what he does but he is the goat!

Got it right, pretty easy.

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.

Only 1% could solve it indeed , thanks for sharing your knowledge Sir.

Finding the total number of rectangles would be interesting as well.

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.

I counted 21, after looking for all possible ways to create squares connect-the-dots-style with the points given.

@mattc3581

23 күн бұрын

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 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.

Finally, one I got the answer first time.

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.

I missed 8 of the trivial squares... these types of things sure can mix you up

If you don't restrict it to requiring the points be the corners, you can draw 9 more squares that form from intersecting lines.

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 would have left the 2*Sqrt(2) as the "improper" Sqrt(8) Makes the progression more evident.

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

Wouldnt be checking pairs of points enough? Still it would give you repeats, but it would be easy to find, and no concaves would be generated?

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...

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.

As soon as you used the magic word "any" I knew there'd be way more than 9 lol!

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.

I want a measuring rod 100 cm long with every integer distance marked, but with the minimum number of marks. I can do this with a six cm rod (A,D) in only two marks, one cm from the bottom (B) and two cm from the top (C). This gives all six integer lengths in total. AB =1, CD=2, BC=3, AC=4, BD=5, AD=6. What is the minimum number of pencil marks on any rod that includes every integer measurement? How are the marks arranged? As the increments are increased, do we approach a well known geometric series?

How many isosceles triangles can you form?

before watching i tried solve this by myself, and got the right answer

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

I found 21 squares and promptly convinced myself that I must have overlooked something incredibly obvious.

I found the easy 9 and the diagonal 4. I didn't realize how many more there were.

I see 9 obvious squares, 8 2x2 diagonals, and 4 1x1 diagonals. If there are more, I can't find them.

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.

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).

That sounds like Lemmings music (One Way or Another)

The answer with 17 squares only counted the 9 normal squares and the 8 diamonds, but not the 4 tilted squares.

The clue is only 1 percent get this. So i look and find 9 and think its more because most people would stay in the square, excuse the pun. Iq test s have these kinda questions.

hehe a good one! i counted 17 but got sloppy and didnt count the sqrt(5) and sqrt(13)... nice one :)

If you think in 3 dimensions, there could be squares that don't look like squares on this plane

may i know how you make these animation videos?

Solved it in 30 secs :). Finnaly a problem I got right.

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.

Most people get it wrong because they answer before clicking on the icon to get the full question... once you hear that its simple.

It's 9 squares standing on their sides and 8 diamonds. And another 4 with odd angles. So, 21 squares overall.

FIRST QUICK PASS before video( I missed the 1 x 2 angle and 1 x 3 angle...DAMN MY OLD SLOW BRAIN!) however.... If you assume a isometric(non perspective) 3d space - you can make a 3d view of a cube.. which has 6 squares, with 3 of them hidden!(+replicate for view angles) I think you can even connect lines to form a hypercube.. Im still trying that. Your algorithm help me work this out so credit where its due. Not sure if this has broken the puzzle record with this as I suspect its outside of the intended (but not stated) parameters. If I'm right - follow up vid Presh? Did I break the record? (with your help) Edit: Thinking on this idea further... if you allow for perspective ratios of 1/1, 1/2 and 1/3 and 1/infinite you can make many many more cubes - each with 2 or 3 visible faces and the rest hidden. perhaps this is a rabbit hole best avoided?

I fell short at 19 !?!? I’ll forever be haunted by those 2 Squares Short 😮

I also came up with the answer of 17; missed the 4 obscure squares.

i also could only find 17, i considered the non 45 degree squares but didnt find any as i expected there to be 4 not 2 hence me ruling it out too early

@gwiyomikim5988

24 күн бұрын

I got 17 as well before being distracted by a notification for a new cat video!🐈🐈‍⬛

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).

wow a design question..... its from uceed, a design examination in india

Wooo I'm in the 1% club! Where's my £100,000?

@TheChamp1971

26 күн бұрын

I found all 21 also, so you're going to have to share it with me!! 🤪

@DoomRutabaga

26 күн бұрын

Hey I found all 21 as well, I want some of that money!!

@rojer9542

26 күн бұрын

I’m sure the check is in the mail. 😁

@DoomRutabaga

26 күн бұрын

@@rojer9542 real :D

How many for the 3D equivalent?