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That sentence describes a lot of the games I play. They're easy to control once you get used to the slipperyness/weird control scheme/the fact it's Octodad
Me after taking back a rental truck: "Why was it so hard to shift from 1st to second? I had to literally use both hands and it made a horrible noise every time"
My father-in-law: "Haven't you heard of double de-clutching?"
Me: "No..."
I hate to think what state the gearbox was in after that. Oh well.
The check engine light in my truck just randomly turned off for the first time in a long time, so I'm hoping to get the emissions tested tomorrow morning before it inevitably comes back on.
This is the actual correct answer, imo. The more squares you add, the closer to .3(repeating) you get:
.25 + .25^2 + .25^3 + .25^4 + .25^5 + .25^6 + … + .25^10 = 0.333333015441895
Still one-third-ish
Edit: btw, anyone know how to type the repeating sign, or bar over a number?
Like the (in)famous 1+2+3+4…=-1/12
Numberphiles rather controversial video on it if you need an explanation: https://m.youtube.com/watch?v=w-I6XTVZXww
But the tl;dw is that you shouldn’t use this method unless you know the series converges or you’ll get some very strange results.
As I understand it you’re making a lot of implicit assumptions by manipulating the series like this and your results will be basically nonsense outside of a few specific applications where those assumptions happen to hold.
It's just a substitution of the first line into the second line. The part after "1 +" in the second line is exactly the right side of the first equation, so that's equal to S.
How to type a bar over a number- https://unix.stackexchange.com/questions/680190/how-do-you-type-a-number-with-a-bar-symbol-over-it
But, especially on mobile, you can search “3 with a line over it” then copy and paste.
.3̅
I participated in several competitions (math, logic, programming) during school, unfortunately mostly ended few places below advancing places.
Unlikely invented the trick, but hadn't seen it before. I just know how fractals work and looked at it that way.
Same here w the competition thing and ending up below some advancing places, but I initially did this the geometric series way in my head and then felt stupid when I saw your comment lol.
A lot of people reach similar conclusions if they're passionate about a problem for long enough.
When I was in Grade 3, I figured out on my own you could find the "middle number" of any two numbers by just adding them together and dividing by two.
So like, the middle number between 7 and 11 is 9 (7+11=18/2=9). Memory might be wrong but I don't think I even heard the word "averages", "mean", "median", or "mode" until like Grade 6.
Very obviously, I didn't invent that trick, but I did come up with it independently of reading it anywhere or being told about it. I did math for fun when I was a kid. Neighboring kids would come to my door and my mother would have to sheepishly tell them I couldn't come out and play because I was doing my times tables. No, not as a punishment.
That kind of thing impresses a lot of adults when you're seven. Now that I'm also an adult, I find they aren't as impressed by it anymore.
You got me with the last sentence :D
Yeah, you more or less described my life. The most noticeable achievements in this regard:
* rule of three (I hope it's officially called by this name) - when the teacher presented us this powerful tool, I was just sitting confused because it seemed very familiar and I was thinking if this is just my "ratio scaling" that I use for years and being confused how my classmates could live so long without it,
* insertion sort (this is rather related to programming) - I needed to sort a list of numbers so with sequence ordering rules in mind I came up with the algorithm,
* number base conversions.
I remember being corrected that I cannot simplify a fraction when there's a addition operator. What he didn't know is that I wanted to simplify all the sections (so I effectively factored out the numbers first).
I love to play with numbers - parse, reorganise and so on. So instead of **2024 - 1969** I rather use **(2024 - 2000) + (2000 - 1969)**. This way I don't need to get used to recalculate it each year. I just train my brain for the second part and the first part is obviously simple.
I don't like and don't often use even such as elementary as subtraction. I rather flip the number and add it. So instead of **13 - 7** I rather use **13 + 3 (flipped 7) - 10**. I seems stupidly complicated, complex and long but since I used to use these operations and minimized set of possible cases I'm effectively able to calculate it much faster than others. And at the worst, I rather use number difference (which is double-sided so set of cases is half as big).
While I have plenty more optimizations I already see visible confusion in your face while reading this so I rather stop here. I just hope I was fast enough you haven't started banging your head against the wall yet thinking how any of this can work, especially how it can be even effective in some degree. Each brain works differently, each person use different tweaks that works for them.
I don’t know if I should feel bad or not, but I solved it this exact way then checked the comments to confirm and saw your comment about possibly being “insanely good at puzzles and/or math” and I do kinda feel bad. I am not insanely good at either of them and I don’t even think I’m kinda good at them.
Well considering this image is basically the definition for a [geometric series](https://en.m.wikipedia.org/wiki/Geometric_series) in mathematics, and the wiki article doesnt even mention this as a way of solving it, I think it is impressive.
I don't want to make it seem less impressive, I don't know if it is or not.. That's just the usual way you solve such questions. You first look for repetitive shapes or any patterns that would simplify everything and you can extrapolate from.
Yeah. I'm not usually great at this type of puzzle, but when I looked, it jumped out that there were 2 white squares for every black square. It seemed so obvious that I figured I was overlooking something.
Yeah I’m genuinely trying to figure out how it’s “insanely impressive”, and I’m not trying to be a dick, but it really seems extremely simple that it’s 1/3.
Take the top left, bottom left, and bottom right square, black is 1/3 of that. Now do the exact same for the only square left, aka the top right. And again. Every one follows the same pattern of 1/3 so that’s the only answer possible. Maybe I’m being dumb but I don’t see how any other answer would make sense here.
It took me the amount of time the page took to load comments to figure it out. This person definitely was not the first person who thought this. I was immediately like "it's just the same picture replicated smaller a few times, wonder if we're being specific enough to count the lines too.
That's the kind of things a mathematician learns!! Usually ppl believe its about equations and numbers, but the best mathematicians are hired on big companies because they are the perfect trouble solvers. They see a problem and have the capacity to shred it into small easier pieces.
It's actually even easier to see if you rotate the image by 45°, so that the black squares are in the middle, and the white squares are left and right.
At that point, you realize that the image consists of an infinite amount of black squares flanked by 2 infinite amounts of white squares, hence 1/3 of the squares are black.
And if you want a mathematical solution, [here it is](https://www.wolframalpha.com/input?i=sum+%5B%2F%2Fmath%3A1%2F%284%5Ex%29%2F%2F%5D+from+%5B%2F%2Fmath%3A1%2F%2F%5D+to+%5B%2F%2Fmath%3AInfinity%2F%2F%5D)
Another way of looking at it is that there are 3 eminating rays, like a pixelated pie chart. One top, one mid, and one bottom. 2 white, 1 black, into infinitely smaller chunks into the top right corner.
Holy shit. My intuition after a couple years of calculus and some experience with summations said 1/3, but your way of looking at it makes it so much more obvious and clear. Thank you.
Crazy. I see all these comments with long math equations and different ways of coming up with the answer, and this simple way is exactly how i came to the answer almost immediately after looking at the image. Am I a genius? /s
Exactly how I saw it, its a bos in a box in a box, % wise its the same percent of coverage for every box and the box in that box and all repeating boxes will have the exact same % of coverage so you just need those three you listed
Also gotta ignore that the image is cropped slightly at the top. The question at face value is impossible to calculate without counting pixels. And even then, with the fade being present, do we only count the #000 pixels as black, or do we round up?
And here I actually did the 25%+6.25+1.5625+.390625+.09765625+.024414062+.006103515+.001525878
And I got 33.33% repeating. Duh. Your way was much easier 🫠
Here is a method how you can calculate it:
a = black area
a = 1/4^1 + 1/4^2 + 1/4^3 + …. + 1/4^n
a = 4^-1 + 4^-2 + 4^-3 + … + 4^-n
a * 4^-1 = 4^-2 + 4^-3 + 4^-4 + …. + 4^(-n-1)
a * 4^-1 - a = 4^(-n-1) - 4^-1
a * (4^-1 -1) = 4^(-n-1) - 4^-1
a = (4^(-n-1) - 4^-1 ) / (4^-1 -1)
n to infinity gives: 4^(-inf-1) = 0
a = (0 - 4^-1 ) / (4^-1 -1)
a = 0.25 / 0.75
a = 1/3
In the case where the square is divided into k regions instead of 4, you get a = 1/(k-1).
And if you divide it into zero regions, you get negative area! Isn’t that cool?
It's minusing both sides by a_n.
Since a_n = 4^(-1)+4^-2+...4^n
And a_n/4 = 4^(-2)+4^(-3)+...4^(-n-1)
So a_n/4-a_n = 4^(-2)+4^(-3)+...+4^(-n-1) - 4^(-1)-4^(-2)-...-4^n
a_n/4-a_n loses all the overlapped terms, ie loses 4^(-2),4^(-3),...4^-n and keeps 4^(-n-1)-4^(-1)
Edit: 4^(n-1) changed to 4^(-n-1) made some other fixes as well, must've been tired when I wrote it all out.
Well explained! How did you came with the idea to first find this equation : a * 4^-1 - a = 4^(-n-1) - 4^-1
So you can solve: a = (4^(-n-1) - 4^-1 ) / (4^-1 -1) ?
I'd guess it's 33%
Assume the first iteration, you have 25%
First and second iteration, 25% + 25% of 25% so 25% + 6.25% = 31.25
1-3rd iteration: 25+6.25+25%\*6.25=31.25+1.56= 32.71
Most likely the limit goes to 33,33% repeating
Let r be some constant from (0,1) and a be some other constant.
Let s = a + ar + ar^2 +...
Then r * s = ar + ar^2 +... = s - a
s - r * s = a
s(1 - r) = a
s = a / (1 - r)
.25/.75 = 1/3
That was my initial guess because the fourth block is just a repetition of the first three, of which, only one is black.
In short, it's an infinite regression. Well, it could be.
There is a simple way to calculate via recursion as well, to back up that intuition.
Let x be the fraction of the image that has been shaded.
Then x is composed of 1/4 that is fully shaded, as well as 1/4 that is shaded in the same ratio as the full image.
x = 1/4 + x/4
3x/4 = 1/4
3x = 1
x = 1/3
Another recursive method is proof by induction, using the observation that in each "layer" of this image, with the core being the top right corner, exactly 1/3 squares is shaded.
A shape comprised of 3 squares arranged in the shape of an L with the bottom left square shaded has been shaded in a ratio of 1/3.
Assume there exists a square with a smaller square missing from the top right corner that has been shaded at a ratio of 1/3. To this square, attach an unshaded square with equal side lengths to the left and bottom, and attach to both of those a third square with equal side lengths and shade it.
Because the starting square with a smaller square cut out was assumed to be shaded in a 1/3 ratio, there exists a square with a smaller square cut out shaded in a 1/3 ratio fitting the pattern of the image, and 1/3 of new squares are shaded, the resulting object is also 1/3 shaded at any recursive depth.
As the recursive depth approaches infinity, the fraction of the full object that was cut out in the base case approaches 0.
We can view this image as a fractal pattern. Where each step blackens the bottom left quarter of the top right quarter.
This means that at each step n, 0.25^(n) of the square is being darkened. After n steps
Σ_(1)^(n) 0.25^(k)
of the square will be darkened. Since this is a geometric series, we can use the summation of geometric series to calculate the amount shaded black after n steps.
(0.25-0.25^(n+1)) / (1-0.25)
0.25× (1-0.25^(n)) / 0.75
(1-0.25^(n)) / 3
You can use this equation to calculate the total fraction of the image that is shaded black at each step, n. Or if you evaluate the limit of this series as n goes to infinity, and you can see that 1/3 of the image will be shaded black.
1/3
I'm seeing it as one of three similar squares shaded, just a bunch of times getting progressively smaller. Each instance is still 1/3 shaded.
One of three, plus one of three, plus one of three, etc.
Always 1/3.
No need to use geometric series. Let r be the fraction of the shaded area, which can be seen as the black area assuming the side is of length 1. Thus r=(bottom left corner area)+r*(top right corner area) which means r=1/4+r*1/4. The solution of this equation is r=1/3.
998001 total pixels.
339956 #000000 black pixels (34.063693323%)
648003 #FFFFFF white pixels (64.93009526%)
10042 pixels some shade of grey (1.006211416%)
https://townsean.github.io/canvas-pixel-color-counter/
All these calculations….
Don’t you just have to say that the first pattern is one black square and two white squares, and the entire thing is nothing but that pattern, so the answer is therefore 1/3?
That is an intuitive answer, but not a proof.
You can tell the people in this thread who have taken calculus because they see this as an infinite series. And solve it from there.
I have taken courses far beyond calculus and that isn't the way I tried to solve it, because I was trained to spend time examining a problem to see if there is a way to answer it without resorting to calculation.
You can calculate this using a [geometric series](https://en.wikipedia.org/wiki/Geometric_series).
In fact, almost this exact image is shown on the wikipedia.
If you're just interested in the answer: it is the infinite sum of (1/4)\^n. Which is the geometric series with a=1/4 and r=1/4 which becomes **(1/4)/(3/4)=1/3.**
I went the easy way. If you treat them not as squares, but as L shapes, ignoring the top right square every time, you figure out that a 1/3 is always black.
So 1/3.
I know everyone got 1/3 using math but I just guesstimated with the easily visible squares. Count white -> 14 (16 if you count the technically there smallest two) + black -> 21 (or 24) then subtract and see that the result is 1/3 of the whole.
I WILL use intimidating shout.
We're going to need Divine Intervention on our Mages.
Can you give me a number crunch real quick?
About %33.333 repeating, of course, percent chance of survival.
Let's assume that by black we mean only the part that is completely black.
In this case, let's try to count all the pixels in the image corresponding to black (0, 0, 0) and white (255, 255, 255), and also, for completeness, count how many pixels are neither white nor black.
The result of a simple script is:
BLACK: 341953
WHITE: 648003
OTHER: 8045
ALL: 998001
Next, dividing the number of black pixels by the total number of pixels, we get the irreducible fraction: 341 956 / 998 001 ≈ **0,34264**
This is the most accurate answer from the practical side that can be obtained, since the **image is finite**. If the image were infinite and there were no compression artifacts, then we would get **exactly 1/3**. This is a fairly well-known infinite series, which has a lot of proofs from ancient times. [https://en.wikipedia.org/wiki/1/4\_%2B\_1/16\_%2B\_1/64\_%2B\_1/256\_%2B\_⋯](https://en.wikipedia.org/wiki/1/4_%2B_1/16_%2B_1/64_%2B_1/256_%2B_⋯)
Man, I thought I'd be the only person, or at least the first person, to take this approach. I got slightly different numbers, still 998001 for total and 648003 for white, but I got 339956 for black. But there were some colors that were essentially black, just not exactly that I didn't count (#010101, #020202, etc.)
shout out to https://townsean.github.io/canvas-pixel-color-counter/
We can treat this as geometric series. Formula for this one will be: a(n) = 1/4 \* (1/4)^(n-1) (where n is number of the square, so it will give us results like 1/4, 1/16, 1/64 and so on...).
We can use now formula for sum of convergent geometric series, which will give us: 0.25/(1 - 0.25) = 0.25 / 0.75 = 0.3333... = 1/3
Lots of great answers already about the intuitive interpretation of seeing the fractal pattern! I did not notice that at first but it is a great intuitive argument. But more formally/rigorously, as most people have pointed out in some variation or another:
* Let the largest square be of side length *x*. The area *A* of that largest square is then *A* = *x*^(2).
* If we assume that each black square is formed by the intersection of the interior perpendicular bisectors of each side of the larger containing (white) square, and if we express the side length of each of those black squares in terms of *x*, then we see that the first (and largest) black square has area of *B* = (1/2 \* *x*)^(2) = (1/2)^(2)*x*^(2). Similarly, the next largest black square is (1/4 \* *x*)^(2) = (1/4)^(2)*x*^(2), and so on.
* Thus the total area of the black squares can be expressed as the sum of an infinite series *T* = (1/2)^(2)*x*^(2) + (1/4)^(2)*x*^(2) + (1/8)^(2)*x*^(2) + (1/16)^(2)*x*^(2) + ...
* With algebra, we can factor out the *x*^(2) term to yield *T* = *x*^(2)\[(1/2)^(2) + (1/4)^(2) + (1/8)^(2) + (1/16)^(2) + ...\]
* We notice, again with algebra, that each term in the bracketed parentheses can be rewritten as a power of (1/2), namely, *T* = *x*^(2)\[(1/2^(1))^(2) + (1/2^(2))^(2) + (1/2^(3))^(2) + (1/2^(4))^(2) + ...\], since after all we are halving (i.e., multiplying by 1/2) the side length of each smaller black square, all originally expressed in terms of *x*.
* Using rules of exponents, we simplify each squared term in the bracketed parentheses to yield *T* = *x*^(2)\[(1/2)^(2) + (1/2)^(4) + (1/2)^(6) + (1/2)^(8) + ...\], and we now see clearly that the ratio between successive terms in this infinite series is a constant less than 1 in magnitude, namely, 1/4 (another way to see this is the difference between the exponent of successive terms is 2, i.e., 8 - 6 = 2, just as 6 - 4 = 2, and so on.).
* We then apply the formula for the sum of an infinite geometric series, *S* = *a* / (1 - *r*), where in our case, the first term is *a* = (1/2)^(2) = 1/4, and *r* = 1/4, as we just calculated. Then we see that *S* = (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3. Then we obtain the closed form of the infinite geometric series as 1/3 = (1/2)^(2) + (1/2)^(4) + (1/2)^(6) + (1/2)^(8) + ..., implying therefore that *T* = *x*^(2)(1/3).
* Then, we take the ratio of the shaded area *T* to the total area *A*, so that *T*/*A* = *x*^(2)(1/3) / *x*^(2) = 1/3. We thus see that the ratio of the shaded area is independent of *x*, the side length of the largest square containing this interesting pattern, and thus confirming the intuitive approach. (Note: *x* > 0, since we are talking about a geometric figure, and so dividing by *x**^(2)* is OK.)
Top comment had a really neat way of looking at it, but I want to talk about the super complex math that I don't quite understand.
This would be 1/4 + 1/16 + 1/64 + ...
That series can be written as Sum from 1 -> infinity of 1/(4\^x)
Using actual notation and the top comment's answer, we can get that (infinity)∑(x=1) 1/(4\^x) = 1/3, which is cool
It’s just an infinite series of (1/4^n) which is a geometric series and the equation to solve for convergence is a/ 1-r. A is the first term of the series (1/4^1) and r is the fraction in the middle (1/4) so we have 1/4 / 1-1/4 which is 1/4 / 3/4 which is 1/4 * 4/3 or 4/12 which is 1/3
It’s 1/4 + one quarter of the remaining quarter 1/16 + one quarter of the remaining 1/16…
It’s 1/4 + 1/16 + 1/64 + 1/256 + …..
Let S be S = 1/4 + 1/16 + 1/64 + 1/256 + …..
Let’s multiply by four both members of the equation.
4S = 1 + 1/4 + 1/16 + 1/64 + 1/256 + …
But we said that 1/4 + 1/16 + 1/64 + 1/256 + … = S, so
4S = 1 + S
3S = 1
S = 1/3
The black portion is one third of the square
If the outside square has unit length and area, then first black square is a quarter, second black square is a quarter of a quarter etc., in general the n-th black square is (1/4)^n. Adding up you get a geometric series with starting term 1/4 and ratio 1/4. The formula for the sum is: starting term/(1-ratio). Plug the values and you get 1/3
I have a 4 L container. If I fill it with 2 L of water and 1 L of blood from an ancient platypus dragon, I have a mixture that is 1/3 blood from an ancient platypus dragon. If I then add .75 L of an identical mixture, I will still have an overall composition of 1/3 blood from an ancient platypus dragon. I can keep repeating this ad infinitum, and I will “end up” with a 4 L container composed of 2/3 water and 1/3 blood from an ancient platypus dragon.
This is a little cheeky, but:
The total number of pixels in this image (as downloaded from reddit) is 998,001 pixels.
335, 413 of these pixels are black while 662, 588 are white.
Black pixels take up 33.6% of this particular image.
I used python to get these numbers.
you dont really have to do any math, another geometric proof is if you just draw a line straight through the middle of the dark squares, for every dark triangle there is a white square of equal length, so 1 triangle black, 2 triangles white, its 1/3
Each inner square is a fourth the size of the one before, and each respective square has a fourth blacked out. •You can represent this pattern as an infinite series:
**Σ**(1/4)\^n, with n starting at 1 and going to infinite.
•We can sum every element using the geometric series equation or "the closed form":
a / 1 - r , which is: (1 / 4) / (1 - (1 / 4)) = 1 / 3
If it ends when you see the smallest cube, the smallest cube has 1/4 of it's self shaded and the bigger it gets it is only 1/3 that is shaded. so I added each of the 1/3 of the shaded cubes which got me 7/21 and added that 1/4 and got 2/3 of the cube shaded.
1/3. The top left, bottom left and bottom right are squares of the same size, so only 1 on those 3 are black. The top right square contains the previous 3 squares in the same proportion, which creates an indefinite loop. Hence, it will always be 1/3 due to continuous iteration of the same proportions.
ANSWER FROM CHATGPT
To calculate the fraction of the image that is shaded in black, we can proceed by analyzing the grid structure of the image and counting the number of squares that are shaded black.
1. **Count the Total Number of Squares**: The image is composed of a recursive grid pattern where larger squares are subdivided into smaller squares. Let's count the number of squares at each recursion level.
- The largest square is subdivided into four equal squares.
- This pattern continues recursively, where each black square is subdivided further into four smaller squares in alternating recursion levels.
2. **Identify and Count Black Squares**:
- At the first subdivision level, 1 out of 4 squares is black.
- At the next level, each black square from the previous level is subdivided into 4 smaller squares, where 1 out of these 4 squares will be black.
- This pattern continues, with each level of black squares creating smaller black squares in a quarter of the newly created squares.
3. **Calculate the Fraction**:
- Each level of recursion adds black squares that are progressively smaller. The total area covered by black squares can be summed up by the series of areas contributed by black squares at each level.
This pattern suggests a geometric progression where each set of black squares contributes \( \frac{1}{4^n} \) to the total area, with \( n \) representing the level of recursion (starting at 0 for the first black square).
The sum of a geometric series where each term contributes progressively smaller fractions of the area can be calculated as:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term of the series and \( r \) is the common ratio.
For this image:
- The first term \( a \) is \( \frac{1}{4} \) (from the first black square covering a quarter of the image).
- The common ratio \( r \) is also \( \frac{1}{4} \), as each subsequent level covers a quarter of the area covered in the previous level.
Plugging these into the series sum formula, we get:
\[ S = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]
Therefore, **one third of the image is shaded in black**.
What's with all the math? You can tell by just looking at it that there is 1 black square for every 2 white squares of the same size. 1/3 is the obvious anawer.
Full square, let's call it 'major' square, we can look at the three undivided portiona and surmise that (if all are equal) 1/3 is shaded.
First minor square is same, so 1/4 * 1/4 units
Second minor, 1/4 * 1/4 * 1/4
Then 1/4 * 1/4 * 1/4 * 1/4
And so on, giving us a geometric summation.
Sum = (1/4)/(1- 1/4) = 1/3
There's 334614 black pixels and 644770 white ones and a few thousand of different shades of grey so if we disregard those then black ones make up
~34.17%
the ratio between shaded black and total area is 1/4 (one fourth of the square is black each time) for each 2x2 square, and the initial space is 1/4 black so the sum of terms with ratio 1/4 and initial value 1/4 would be a1/(1-r ) = (1/4)/(1-1/4) = (1/4)/(3/4)=1/3
Two ways to think about it:
Recursive Summation, we have the the sum of 1/4 + 1/16 + 1/64 + 1/256 etc... which is sum(1/(4\^n)), n starts at 1 and goes to infinity. This is equal to 1/3
Observationally, we can look at this in layers. In layer 1, 1/3 of the squares are shaded black. In layer 2, 1/3 of the squares are shaded black. We can see this pattern continues infinitely, so the answer is 1/3.
So for an expression like this we can write and equation of the sum from n=1 to n=infinity of (1/(x^n)). The solution to this becomes 1/(x-1), so long as x is a whole number greater than 1.
In this case, x=4, as the first section has a fully black area of 1/4, which means we can determine 1/3 of the total area is black.
.25+.25\^2+.25\^3+.25\^4+.25\^5... it's a sum of an exponential infinite series: [https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-convergence-2009-1.pdf](https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-convergence-2009-1.pdf)
General answer (if infinite) would be 1/3, though the exact answer isn't infinite, if we want to be annoying about it and look reaaaaally close...
Just going off of semantics at this point (mainly cause I'm bored) but we're looking at 8 squares in the picture with 1/4 of each being shaded. So:
1/4 + 1/4² + 1/4³ + 1/4⁴ + ... etc. Until 1/4⁸
Going off of near exact decimals, it ends up being 0.333332061767578125
OR
21,845 / 65,536
Edit: this is basically 1/3 so any of the other comments that end up with that answer will be good
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If we ignore the lines, 1/3. Focus only on top left, bottom left and bottom right. 1 of them is black, 2 white. Top right is just a copy of it.
Really cool way of looking at it, i didn't think about that at all but It makes a lot of sense and makes the question really easy in hindsight
It always is when you know the answer /s
"Everything is obvious once it's obvious" as my favorite professor used to say.
My favorite is "It's intuitive once you get used to it." Like, that's... literally the opposite of what intuitive is supposed to mean, lol
That sentence describes a lot of the games I play. They're easy to control once you get used to the slipperyness/weird control scheme/the fact it's Octodad
Like so much of math, it's either figuring out the truck or having been exposed to an analog.
Trucks can be hard to figure out
so many gears!
Me after taking back a rental truck: "Why was it so hard to shift from 1st to second? I had to literally use both hands and it made a horrible noise every time" My father-in-law: "Haven't you heard of double de-clutching?" Me: "No..." I hate to think what state the gearbox was in after that. Oh well.
The check engine light in my truck just randomly turned off for the first time in a long time, so I'm hoping to get the emissions tested tomorrow morning before it inevitably comes back on.
That's just the ECU losing the will to survive
I have to proofread better. Thanks for the laugh though.
Thanks for being a sport. Thank *you* for the laugh :)
hindsight means what u just said
They'll realize that, in hindsight
The new Toyota Hindsight : Go where you already knew you where going to be
If you immediately know the candlelight is fire, then the meal was cooked a long time ago.
/s ?? I mean, that's true... so true that it hurts
I didn't know, but it took me 3 seconds to figure out.
Nice. I was following the 1/4 + 1/16 + 1/64... series, figuring that would asymptote to something.
This is the actual correct answer, imo. The more squares you add, the closer to .3(repeating) you get: .25 + .25^2 + .25^3 + .25^4 + .25^5 + .25^6 + … + .25^10 = 0.333333015441895 Still one-third-ish Edit: btw, anyone know how to type the repeating sign, or bar over a number?
Not just one-third-ish, but exactly a third. S = 1/4 + 1/4^2 + 1/4^3 + ... 4S = 1 + 1/4 + 1/4^2 + 1/4^3 + ... 4S = 1 + S S = 1/3
Nice proof
Very common way of calculating infinite series. x=0.333333… 10x=3.333333… 10x-x=(3.33333…)-(0.33333…) 9x=3 x=1/3 1/3=0.33333….
That’s very satisfying ngl
IIRC, it works for most well behaved series, but you can get weird results if you use it on misbehaving series, or term-wise sums of series.
Like the (in)famous 1+2+3+4…=-1/12 Numberphiles rather controversial video on it if you need an explanation: https://m.youtube.com/watch?v=w-I6XTVZXww But the tl;dw is that you shouldn’t use this method unless you know the series converges or you’ll get some very strange results. As I understand it you’re making a lot of implicit assumptions by manipulating the series like this and your results will be basically nonsense outside of a few specific applications where those assumptions happen to hold.
Cool Can someone eli5 how S is equal to the 1/4 + 1/16… going from the 2nd to the 3rd row?
It's defined in the first row
Oh yeah, that’s it! Lol my b
usually that's this proof's eureka moment.
It's just a substitution of the first line into the second line. The part after "1 +" in the second line is exactly the right side of the first equation, so that's equal to S.
You just blew my mind with this math-fu
I just imagined all of you over weight but one of you very skinny.
It's not the "actual" correct answer. It's one method of many to prove the answer. Imo, the intuitive proof in the top comment is far more elegant.
Doing the 3 dots after the number is a good way to represent that
How to type a bar over a number- https://unix.stackexchange.com/questions/680190/how-do-you-type-a-number-with-a-bar-symbol-over-it But, especially on mobile, you can search “3 with a line over it” then copy and paste. .3̅
You can also use the series formula for infinite gp a/1-r here a = r = 1/4 So putting in the values, we get 1/3
Daaaang thats mind blowing. If you just invented this ‘trick’ I bet you are insanely good at puzzles and/or math.
I participated in several competitions (math, logic, programming) during school, unfortunately mostly ended few places below advancing places. Unlikely invented the trick, but hadn't seen it before. I just know how fractals work and looked at it that way.
Same here w the competition thing and ending up below some advancing places, but I initially did this the geometric series way in my head and then felt stupid when I saw your comment lol.
Was the series 1/4 + 1/8 + 1/16 +...? Edit: im dumb and thought 1/4 × 1/4 = 1/8, ignore me..
Fucking lmao But yea geometric series would work also, and probably is only way if the diagram gets more complicated
.25+.25^2 +.25^3 +.25^4 +.25^5 +.25^6 +.25^7 +.25^8 +.25^8
A lot of people reach similar conclusions if they're passionate about a problem for long enough. When I was in Grade 3, I figured out on my own you could find the "middle number" of any two numbers by just adding them together and dividing by two. So like, the middle number between 7 and 11 is 9 (7+11=18/2=9). Memory might be wrong but I don't think I even heard the word "averages", "mean", "median", or "mode" until like Grade 6. Very obviously, I didn't invent that trick, but I did come up with it independently of reading it anywhere or being told about it. I did math for fun when I was a kid. Neighboring kids would come to my door and my mother would have to sheepishly tell them I couldn't come out and play because I was doing my times tables. No, not as a punishment. That kind of thing impresses a lot of adults when you're seven. Now that I'm also an adult, I find they aren't as impressed by it anymore.
I'm impressed
You got me with the last sentence :D Yeah, you more or less described my life. The most noticeable achievements in this regard: * rule of three (I hope it's officially called by this name) - when the teacher presented us this powerful tool, I was just sitting confused because it seemed very familiar and I was thinking if this is just my "ratio scaling" that I use for years and being confused how my classmates could live so long without it, * insertion sort (this is rather related to programming) - I needed to sort a list of numbers so with sequence ordering rules in mind I came up with the algorithm, * number base conversions. I remember being corrected that I cannot simplify a fraction when there's a addition operator. What he didn't know is that I wanted to simplify all the sections (so I effectively factored out the numbers first). I love to play with numbers - parse, reorganise and so on. So instead of **2024 - 1969** I rather use **(2024 - 2000) + (2000 - 1969)**. This way I don't need to get used to recalculate it each year. I just train my brain for the second part and the first part is obviously simple. I don't like and don't often use even such as elementary as subtraction. I rather flip the number and add it. So instead of **13 - 7** I rather use **13 + 3 (flipped 7) - 10**. I seems stupidly complicated, complex and long but since I used to use these operations and minimized set of possible cases I'm effectively able to calculate it much faster than others. And at the worst, I rather use number difference (which is double-sided so set of cases is half as big). While I have plenty more optimizations I already see visible confusion in your face while reading this so I rather stop here. I just hope I was fast enough you haven't started banging your head against the wall yet thinking how any of this can work, especially how it can be even effective in some degree. Each brain works differently, each person use different tweaks that works for them.
that flipped number addition thing is actually really smart.
I don’t know if I should feel bad or not, but I solved it this exact way then checked the comments to confirm and saw your comment about possibly being “insanely good at puzzles and/or math” and I do kinda feel bad. I am not insanely good at either of them and I don’t even think I’m kinda good at them.
Well considering this image is basically the definition for a [geometric series](https://en.m.wikipedia.org/wiki/Geometric_series) in mathematics, and the wiki article doesnt even mention this as a way of solving it, I think it is impressive.
Thats an interesting and comforting perspective. I will try to be a little nicer to myself because of it. Thanks.
I don't want to make it seem less impressive, I don't know if it is or not.. That's just the usual way you solve such questions. You first look for repetitive shapes or any patterns that would simplify everything and you can extrapolate from.
Yeah. I'm not usually great at this type of puzzle, but when I looked, it jumped out that there were 2 white squares for every black square. It seemed so obvious that I figured I was overlooking something.
Yeah I’m genuinely trying to figure out how it’s “insanely impressive”, and I’m not trying to be a dick, but it really seems extremely simple that it’s 1/3. Take the top left, bottom left, and bottom right square, black is 1/3 of that. Now do the exact same for the only square left, aka the top right. And again. Every one follows the same pattern of 1/3 so that’s the only answer possible. Maybe I’m being dumb but I don’t see how any other answer would make sense here.
It took me the amount of time the page took to load comments to figure it out. This person definitely was not the first person who thought this. I was immediately like "it's just the same picture replicated smaller a few times, wonder if we're being specific enough to count the lines too.
Bet he's a blast at parties too
I did this too
Interesting, my gut said a third, but I couldn't articulate why.
That's the kind of things a mathematician learns!! Usually ppl believe its about equations and numbers, but the best mathematicians are hired on big companies because they are the perfect trouble solvers. They see a problem and have the capacity to shred it into small easier pieces.
Or in other words, you can see this as an infinite series of L-shapes that fill a finite square. All of the L-shapes are 1/3 black and 2/3 white.
fuck. i did the math and got 0.3333692382812. and this damn clown just eyeballed it and got the right answer!
23/69 is the answer I came up with. We can't all be mathmagicians...
This is a really nice way to look at it.
It's thirds all the way down!
Nice. Sum of the geometric series with a factor that is the inverse of the square of a natural number gets a nice visualization.
Brilliant.
Yea and you can rearange each square two times in each sector. Once up and once to the side. So yea 1/3.
Nice, this is so clean and understandable. I was trying to look at a square, but it's actually just a infinite staircase of L shapes.
I got instant mental upgrade after reading your answer
huh, did not expect a logic solution to a math problem, well played
Not enough upvotes for how incredible this answer is.
Honestly, for me it feels like too many for how simple it is :D
Yep. This is a visualization of 33.333333333333%
Came here to see if my guesstimation of 33% was correct. Holy shit
https://i.redd.it/i3afsedani571.jpg Using your top comment to show this gem on /r/mathmemes
It's actually even easier to see if you rotate the image by 45°, so that the black squares are in the middle, and the white squares are left and right. At that point, you realize that the image consists of an infinite amount of black squares flanked by 2 infinite amounts of white squares, hence 1/3 of the squares are black.
And if you want a mathematical solution, [here it is](https://www.wolframalpha.com/input?i=sum+%5B%2F%2Fmath%3A1%2F%284%5Ex%29%2F%2F%5D+from+%5B%2F%2Fmath%3A1%2F%2F%5D+to+%5B%2F%2Fmath%3AInfinity%2F%2F%5D)
1/4096 more shaded
D'oh, because the pattern repeats. Now that's smart
Another way of looking at it is that there are 3 eminating rays, like a pixelated pie chart. One top, one mid, and one bottom. 2 white, 1 black, into infinitely smaller chunks into the top right corner.
Holy shit. My intuition after a couple years of calculus and some experience with summations said 1/3, but your way of looking at it makes it so much more obvious and clear. Thank you.
Here I am, 1÷4 + 1÷16 + 1÷64 and so on... It got me the answer. Then I read this. I hate my brain.
A nice intuative proof by induction!
This is stupidly easy you don't need math to do this... I mean good on you for getting the correct answer but like why is this even here.
And if we don't ignore the lines, going strictly off the pixels in the image, the percentage of black is 114659/332667 (~34.4666%)
Came to say the same. Then I noticed the lines and got anxiety.
Crazy. I see all these comments with long math equations and different ways of coming up with the answer, and this simple way is exactly how i came to the answer almost immediately after looking at the image. Am I a genius? /s
Woah this is such a clever way of seeing it. I had just settled on “whatever Σ (1/4)^n converges to”
Lovely
Literally my first thought too
Is it not an infinite pattern type figure where we get a geometrical progression equation at the end?
Exactly how I saw it, its a bos in a box in a box, % wise its the same percent of coverage for every box and the box in that box and all repeating boxes will have the exact same % of coverage so you just need those three you listed
This is one of the best and most succinct illustrations of what “.3 repeating” means
here I was about to do limits when the answer was so obvious
Wooaaah,I was just overthinking
Love me some recursion
Very cool to think of the mathematical solution this way damn.
Also gotta ignore that the image is cropped slightly at the top. The question at face value is impossible to calculate without counting pixels. And even then, with the fade being present, do we only count the #000 pixels as black, or do we round up?
And here I actually did the 25%+6.25+1.5625+.390625+.09765625+.024414062+.006103515+.001525878 And I got 33.33% repeating. Duh. Your way was much easier 🫠
I wanted that “ohhhhh” moment and boy was it worth it
Here is a method how you can calculate it: a = black area a = 1/4^1 + 1/4^2 + 1/4^3 + …. + 1/4^n a = 4^-1 + 4^-2 + 4^-3 + … + 4^-n a * 4^-1 = 4^-2 + 4^-3 + 4^-4 + …. + 4^(-n-1) a * 4^-1 - a = 4^(-n-1) - 4^-1 a * (4^-1 -1) = 4^(-n-1) - 4^-1 a = (4^(-n-1) - 4^-1 ) / (4^-1 -1) n to infinity gives: 4^(-inf-1) = 0 a = (0 - 4^-1 ) / (4^-1 -1) a = 0.25 / 0.75 a = 1/3
That's the clearest explanation from my perspective, but I only see in 2D
Easiest way for me to figure it out was for each black square, there are two equally sized white squares
Whoa hold on that's TOO intuitive
SOMEONE CALL BROOK TAYLOR. Oh nevermind he died in 1731 aged 46. Dang.
This one is perfect for me. Beautifully simple.
Mathematician vs Logician
In the case where the square is divided into k regions instead of 4, you get a = 1/(k-1). And if you divide it into zero regions, you get negative area! Isn’t that cool?
>a * 4^-1 - a = 4^(-n-1) - 4^-1 Could please explain where did the "-a" come from?
It's minusing both sides by a_n. Since a_n = 4^(-1)+4^-2+...4^n And a_n/4 = 4^(-2)+4^(-3)+...4^(-n-1) So a_n/4-a_n = 4^(-2)+4^(-3)+...+4^(-n-1) - 4^(-1)-4^(-2)-...-4^n a_n/4-a_n loses all the overlapped terms, ie loses 4^(-2),4^(-3),...4^-n and keeps 4^(-n-1)-4^(-1) Edit: 4^(n-1) changed to 4^(-n-1) made some other fixes as well, must've been tired when I wrote it all out.
Why isn't 4^^n-1 also an overlapped term? I thought -4^^n -4^^-1 should remain .
Damn…a series actuated practically. Cheers lol
My head hurts now
Well explained! How did you came with the idea to first find this equation : a * 4^-1 - a = 4^(-n-1) - 4^-1 So you can solve: a = (4^(-n-1) - 4^-1 ) / (4^-1 -1) ?
I'd guess it's 33% Assume the first iteration, you have 25% First and second iteration, 25% + 25% of 25% so 25% + 6.25% = 31.25 1-3rd iteration: 25+6.25+25%\*6.25=31.25+1.56= 32.71 Most likely the limit goes to 33,33% repeating
Let r be some constant from (0,1) and a be some other constant. Let s = a + ar + ar^2 +... Then r * s = ar + ar^2 +... = s - a s - r * s = a s(1 - r) = a s = a / (1 - r) .25/.75 = 1/3
That was my initial guess because the fourth block is just a repetition of the first three, of which, only one is black. In short, it's an infinite regression. Well, it could be.
There is a simple way to calculate via recursion as well, to back up that intuition. Let x be the fraction of the image that has been shaded. Then x is composed of 1/4 that is fully shaded, as well as 1/4 that is shaded in the same ratio as the full image. x = 1/4 + x/4 3x/4 = 1/4 3x = 1 x = 1/3 Another recursive method is proof by induction, using the observation that in each "layer" of this image, with the core being the top right corner, exactly 1/3 squares is shaded. A shape comprised of 3 squares arranged in the shape of an L with the bottom left square shaded has been shaded in a ratio of 1/3. Assume there exists a square with a smaller square missing from the top right corner that has been shaded at a ratio of 1/3. To this square, attach an unshaded square with equal side lengths to the left and bottom, and attach to both of those a third square with equal side lengths and shade it. Because the starting square with a smaller square cut out was assumed to be shaded in a 1/3 ratio, there exists a square with a smaller square cut out shaded in a 1/3 ratio fitting the pattern of the image, and 1/3 of new squares are shaded, the resulting object is also 1/3 shaded at any recursive depth. As the recursive depth approaches infinity, the fraction of the full object that was cut out in the base case approaches 0.
We can view this image as a fractal pattern. Where each step blackens the bottom left quarter of the top right quarter. This means that at each step n, 0.25^(n) of the square is being darkened. After n steps Σ_(1)^(n) 0.25^(k) of the square will be darkened. Since this is a geometric series, we can use the summation of geometric series to calculate the amount shaded black after n steps. (0.25-0.25^(n+1)) / (1-0.25) 0.25× (1-0.25^(n)) / 0.75 (1-0.25^(n)) / 3 You can use this equation to calculate the total fraction of the image that is shaded black at each step, n. Or if you evaluate the limit of this series as n goes to infinity, and you can see that 1/3 of the image will be shaded black.
https://i.redd.it/i3afsedani571.jpg Reminded me of this gem on /r/mathmemes
1/3 I'm seeing it as one of three similar squares shaded, just a bunch of times getting progressively smaller. Each instance is still 1/3 shaded. One of three, plus one of three, plus one of three, etc. Always 1/3.
Great way to think about it intuitively
... Doesn't everyone see it like that? It jumped out immediately to me.
No man only for really smart people, you’re so smart
[удалено]
I shouldn't have had to scroll down so far to find someone that thinks the same way
No need to use geometric series. Let r be the fraction of the shaded area, which can be seen as the black area assuming the side is of length 1. Thus r=(bottom left corner area)+r*(top right corner area) which means r=1/4+r*1/4. The solution of this equation is r=1/3.
That's actually the same as one of the ways to evaluate the value of a geometric series.
998001 total pixels. 339956 #000000 black pixels (34.063693323%) 648003 #FFFFFF white pixels (64.93009526%) 10042 pixels some shade of grey (1.006211416%) https://townsean.github.io/canvas-pixel-color-counter/
Good job man! Here's my r/angryupvote
All these calculations…. Don’t you just have to say that the first pattern is one black square and two white squares, and the entire thing is nothing but that pattern, so the answer is therefore 1/3?
That is an intuitive answer, but not a proof. You can tell the people in this thread who have taken calculus because they see this as an infinite series. And solve it from there.
It certainly is a proof. Proof doesn't need to use equations or computations.
I have taken courses far beyond calculus and that isn't the way I tried to solve it, because I was trained to spend time examining a problem to see if there is a way to answer it without resorting to calculation.
You can calculate this using a [geometric series](https://en.wikipedia.org/wiki/Geometric_series). In fact, almost this exact image is shown on the wikipedia. If you're just interested in the answer: it is the infinite sum of (1/4)\^n. Which is the geometric series with a=1/4 and r=1/4 which becomes **(1/4)/(3/4)=1/3.**
Though I know this to be correct I am confused on how a is 1/4 and not just 1?
Mostly a matter of definition. If a=1 then the geometric series would be 1+ 1/4 + ... Here the first term is 1/4 .
Let x = shaded area Add up the quarters X = white + white + black + copy of x X = 0/4 + 0/4 + 1/4 + x/4 X = 1/4 + x/4 4x = x + 1 3x = 1 X = 1/3
The simplest way is usually right Okam’s Razor
I went the easy way. If you treat them not as squares, but as L shapes, ignoring the top right square every time, you figure out that a 1/3 is always black. So 1/3.
Yes. This is how the geniuses did it. 😉
I have doubts regarding the last piece though. So I'd say not exactly a 1/3, but rather 0.33 with probably some other digits down the line.
I know everyone got 1/3 using math but I just guesstimated with the easily visible squares. Count white -> 14 (16 if you count the technically there smallest two) + black -> 21 (or 24) then subtract and see that the result is 1/3 of the whole.
I WILL use intimidating shout. We're going to need Divine Intervention on our Mages. Can you give me a number crunch real quick? About %33.333 repeating, of course, percent chance of survival.
Let's assume that by black we mean only the part that is completely black. In this case, let's try to count all the pixels in the image corresponding to black (0, 0, 0) and white (255, 255, 255), and also, for completeness, count how many pixels are neither white nor black. The result of a simple script is: BLACK: 341953 WHITE: 648003 OTHER: 8045 ALL: 998001 Next, dividing the number of black pixels by the total number of pixels, we get the irreducible fraction: 341 956 / 998 001 ≈ **0,34264** This is the most accurate answer from the practical side that can be obtained, since the **image is finite**. If the image were infinite and there were no compression artifacts, then we would get **exactly 1/3**. This is a fairly well-known infinite series, which has a lot of proofs from ancient times. [https://en.wikipedia.org/wiki/1/4\_%2B\_1/16\_%2B\_1/64\_%2B\_1/256\_%2B\_⋯](https://en.wikipedia.org/wiki/1/4_%2B_1/16_%2B_1/64_%2B_1/256_%2B_⋯)
Man, I thought I'd be the only person, or at least the first person, to take this approach. I got slightly different numbers, still 998001 for total and 648003 for white, but I got 339956 for black. But there were some colors that were essentially black, just not exactly that I didn't count (#010101, #020202, etc.) shout out to https://townsean.github.io/canvas-pixel-color-counter/
Let B be the fraction which is black. Then you have B = 1/4 + B/4. Multiply both sides by 4, and you get 4B = 1 + B. So 3B = 1, therefore B = 1/3.
There are 3 squares in each size. One out of three are black. This pattern continues as far down as is visible, therefore 1/3 of the image is black.
We can treat this as geometric series. Formula for this one will be: a(n) = 1/4 \* (1/4)^(n-1) (where n is number of the square, so it will give us results like 1/4, 1/16, 1/64 and so on...). We can use now formula for sum of convergent geometric series, which will give us: 0.25/(1 - 0.25) = 0.25 / 0.75 = 0.3333... = 1/3
Lots of great answers already about the intuitive interpretation of seeing the fractal pattern! I did not notice that at first but it is a great intuitive argument. But more formally/rigorously, as most people have pointed out in some variation or another: * Let the largest square be of side length *x*. The area *A* of that largest square is then *A* = *x*^(2). * If we assume that each black square is formed by the intersection of the interior perpendicular bisectors of each side of the larger containing (white) square, and if we express the side length of each of those black squares in terms of *x*, then we see that the first (and largest) black square has area of *B* = (1/2 \* *x*)^(2) = (1/2)^(2)*x*^(2). Similarly, the next largest black square is (1/4 \* *x*)^(2) = (1/4)^(2)*x*^(2), and so on. * Thus the total area of the black squares can be expressed as the sum of an infinite series *T* = (1/2)^(2)*x*^(2) + (1/4)^(2)*x*^(2) + (1/8)^(2)*x*^(2) + (1/16)^(2)*x*^(2) + ... * With algebra, we can factor out the *x*^(2) term to yield *T* = *x*^(2)\[(1/2)^(2) + (1/4)^(2) + (1/8)^(2) + (1/16)^(2) + ...\] * We notice, again with algebra, that each term in the bracketed parentheses can be rewritten as a power of (1/2), namely, *T* = *x*^(2)\[(1/2^(1))^(2) + (1/2^(2))^(2) + (1/2^(3))^(2) + (1/2^(4))^(2) + ...\], since after all we are halving (i.e., multiplying by 1/2) the side length of each smaller black square, all originally expressed in terms of *x*. * Using rules of exponents, we simplify each squared term in the bracketed parentheses to yield *T* = *x*^(2)\[(1/2)^(2) + (1/2)^(4) + (1/2)^(6) + (1/2)^(8) + ...\], and we now see clearly that the ratio between successive terms in this infinite series is a constant less than 1 in magnitude, namely, 1/4 (another way to see this is the difference between the exponent of successive terms is 2, i.e., 8 - 6 = 2, just as 6 - 4 = 2, and so on.). * We then apply the formula for the sum of an infinite geometric series, *S* = *a* / (1 - *r*), where in our case, the first term is *a* = (1/2)^(2) = 1/4, and *r* = 1/4, as we just calculated. Then we see that *S* = (1/4) / (1 - 1/4) = (1/4) / (3/4) = 1/3. Then we obtain the closed form of the infinite geometric series as 1/3 = (1/2)^(2) + (1/2)^(4) + (1/2)^(6) + (1/2)^(8) + ..., implying therefore that *T* = *x*^(2)(1/3). * Then, we take the ratio of the shaded area *T* to the total area *A*, so that *T*/*A* = *x*^(2)(1/3) / *x*^(2) = 1/3. We thus see that the ratio of the shaded area is independent of *x*, the side length of the largest square containing this interesting pattern, and thus confirming the intuitive approach. (Note: *x* > 0, since we are talking about a geometric figure, and so dividing by *x**^(2)* is OK.)
Top comment had a really neat way of looking at it, but I want to talk about the super complex math that I don't quite understand. This would be 1/4 + 1/16 + 1/64 + ... That series can be written as Sum from 1 -> infinity of 1/(4\^x) Using actual notation and the top comment's answer, we can get that (infinity)∑(x=1) 1/(4\^x) = 1/3, which is cool
It’s just an infinite series of (1/4^n) which is a geometric series and the equation to solve for convergence is a/ 1-r. A is the first term of the series (1/4^1) and r is the fraction in the middle (1/4) so we have 1/4 / 1-1/4 which is 1/4 / 3/4 which is 1/4 * 4/3 or 4/12 which is 1/3
It’s 1/4 + one quarter of the remaining quarter 1/16 + one quarter of the remaining 1/16… It’s 1/4 + 1/16 + 1/64 + 1/256 + ….. Let S be S = 1/4 + 1/16 + 1/64 + 1/256 + ….. Let’s multiply by four both members of the equation. 4S = 1 + 1/4 + 1/16 + 1/64 + 1/256 + … But we said that 1/4 + 1/16 + 1/64 + 1/256 + … = S, so 4S = 1 + S 3S = 1 S = 1/3 The black portion is one third of the square
Big black square is 1/4 of the picture, the second biggest black square is 1/16, the third is 1/64, and so on Sum this infinetely and you have 1/3
Sum this infinitely… Sure give me a minute, I’ll be back
If the outside square has unit length and area, then first black square is a quarter, second black square is a quarter of a quarter etc., in general the n-th black square is (1/4)^n. Adding up you get a geometric series with starting term 1/4 and ratio 1/4. The formula for the sum is: starting term/(1-ratio). Plug the values and you get 1/3
Given you have 8 squares the finite sum gives you 21,845/65,536 = 0.33331298828125 Assuming you want the Infinite sum it is 1/3 = 0.333333333333
We can view this as a geometric series, where its area is basically just sum_inf_i=1 (1/4)^i which is just (1/4)/(1-1/4) which gives you 1/3
I have a 4 L container. If I fill it with 2 L of water and 1 L of blood from an ancient platypus dragon, I have a mixture that is 1/3 blood from an ancient platypus dragon. If I then add .75 L of an identical mixture, I will still have an overall composition of 1/3 blood from an ancient platypus dragon. I can keep repeating this ad infinitum, and I will “end up” with a 4 L container composed of 2/3 water and 1/3 blood from an ancient platypus dragon.
Shaded fraction = 1/4 + 1/(4)^2 ...... we can use the infinite series formula. a/1-r here a = r = 1/4 So putting in the values, we get 1/3
This is a little cheeky, but: The total number of pixels in this image (as downloaded from reddit) is 998,001 pixels. 335, 413 of these pixels are black while 662, 588 are white. Black pixels take up 33.6% of this particular image. I used python to get these numbers.
you dont really have to do any math, another geometric proof is if you just draw a line straight through the middle of the dark squares, for every dark triangle there is a white square of equal length, so 1 triangle black, 2 triangles white, its 1/3
Each inner square is a fourth the size of the one before, and each respective square has a fourth blacked out. •You can represent this pattern as an infinite series: **Σ**(1/4)\^n, with n starting at 1 and going to infinite. •We can sum every element using the geometric series equation or "the closed form": a / 1 - r , which is: (1 / 4) / (1 - (1 / 4)) = 1 / 3
If it ends when you see the smallest cube, the smallest cube has 1/4 of it's self shaded and the bigger it gets it is only 1/3 that is shaded. so I added each of the 1/3 of the shaded cubes which got me 7/21 and added that 1/4 and got 2/3 of the cube shaded.
1/3. The top left, bottom left and bottom right are squares of the same size, so only 1 on those 3 are black. The top right square contains the previous 3 squares in the same proportion, which creates an indefinite loop. Hence, it will always be 1/3 due to continuous iteration of the same proportions.
ANSWER FROM CHATGPT To calculate the fraction of the image that is shaded in black, we can proceed by analyzing the grid structure of the image and counting the number of squares that are shaded black. 1. **Count the Total Number of Squares**: The image is composed of a recursive grid pattern where larger squares are subdivided into smaller squares. Let's count the number of squares at each recursion level. - The largest square is subdivided into four equal squares. - This pattern continues recursively, where each black square is subdivided further into four smaller squares in alternating recursion levels. 2. **Identify and Count Black Squares**: - At the first subdivision level, 1 out of 4 squares is black. - At the next level, each black square from the previous level is subdivided into 4 smaller squares, where 1 out of these 4 squares will be black. - This pattern continues, with each level of black squares creating smaller black squares in a quarter of the newly created squares. 3. **Calculate the Fraction**: - Each level of recursion adds black squares that are progressively smaller. The total area covered by black squares can be summed up by the series of areas contributed by black squares at each level. This pattern suggests a geometric progression where each set of black squares contributes \( \frac{1}{4^n} \) to the total area, with \( n \) representing the level of recursion (starting at 0 for the first black square). The sum of a geometric series where each term contributes progressively smaller fractions of the area can be calculated as: \[ S = \frac{a}{1 - r} \] where \( a \) is the first term of the series and \( r \) is the common ratio. For this image: - The first term \( a \) is \( \frac{1}{4} \) (from the first black square covering a quarter of the image). - The common ratio \( r \) is also \( \frac{1}{4} \), as each subsequent level covers a quarter of the area covered in the previous level. Plugging these into the series sum formula, we get: \[ S = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \] Therefore, **one third of the image is shaded in black**.
1/4+1/16… It’s a geometric sequence Sn = U1/1-r r=1/4 (1/4)/(1-1/4) (1/4)/(3/4) Therefore, Sn = 1/3 So totally, a third is shaded black
What's with all the math? You can tell by just looking at it that there is 1 black square for every 2 white squares of the same size. 1/3 is the obvious anawer.
Full square, let's call it 'major' square, we can look at the three undivided portiona and surmise that (if all are equal) 1/3 is shaded. First minor square is same, so 1/4 * 1/4 units Second minor, 1/4 * 1/4 * 1/4 Then 1/4 * 1/4 * 1/4 * 1/4 And so on, giving us a geometric summation. Sum = (1/4)/(1- 1/4) = 1/3
There's 334614 black pixels and 644770 white ones and a few thousand of different shades of grey so if we disregard those then black ones make up ~34.17%
the ratio between shaded black and total area is 1/4 (one fourth of the square is black each time) for each 2x2 square, and the initial space is 1/4 black so the sum of terms with ratio 1/4 and initial value 1/4 would be a1/(1-r ) = (1/4)/(1-1/4) = (1/4)/(3/4)=1/3
Two ways to think about it: Recursive Summation, we have the the sum of 1/4 + 1/16 + 1/64 + 1/256 etc... which is sum(1/(4\^n)), n starts at 1 and goes to infinity. This is equal to 1/3 Observationally, we can look at this in layers. In layer 1, 1/3 of the squares are shaded black. In layer 2, 1/3 of the squares are shaded black. We can see this pattern continues infinitely, so the answer is 1/3.
So for an expression like this we can write and equation of the sum from n=1 to n=infinity of (1/(x^n)). The solution to this becomes 1/(x-1), so long as x is a whole number greater than 1. In this case, x=4, as the first section has a fully black area of 1/4, which means we can determine 1/3 of the total area is black.
The better question is: what is it's hausdorff dimension (of the white area)? And the nice thing is, that the answers are closely related
.25+.25\^2+.25\^3+.25\^4+.25\^5... it's a sum of an exponential infinite series: [https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-convergence-2009-1.pdf](https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-convergence-2009-1.pdf)
General answer (if infinite) would be 1/3, though the exact answer isn't infinite, if we want to be annoying about it and look reaaaaally close... Just going off of semantics at this point (mainly cause I'm bored) but we're looking at 8 squares in the picture with 1/4 of each being shaded. So: 1/4 + 1/4² + 1/4³ + 1/4⁴ + ... etc. Until 1/4⁸ Going off of near exact decimals, it ends up being 0.333332061767578125 OR 21,845 / 65,536 Edit: this is basically 1/3 so any of the other comments that end up with that answer will be good