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Scepheo wrote:
Bobo the King wrote:
Warp wrote:
Samsara wrote:
You're right on one point at least: Non-feminist men are whining and misogynistic MRAs when they bring up their problems like they're significant at the moment.
Need I even comment? The hypocrisy is just mind-boggling.
That was unnecessary, Warp.
I disagree. Samsara says that anyone who isn't currently dying of aids, thirst, hunger and being shot has no right to complain, because their problems aren't significant at the moment. He then goes on to complain.
I don't entirely agree with Samsara, but Warp's comment added absolutely nothing to the discussion. Neither does yours.
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Warp wrote:
Samsara wrote:
You're right on one point at least: Non-feminist men are whining and misogynistic MRAs when they bring up their problems like they're significant at the moment.
Need I even comment? The hypocrisy is just mind-boggling.
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There is a clear lack of civility and basic empathy in this thread.
Make no mistake: Warp's words were insensitive and out of place. I strongly disagree with his warped (heh) characterization of the issue, the words he chose, and his timing.
But that doesn't mean his narrative is completely invalid and coming from a place of dishonesty. Some white men do indeed face unique problems and I believe that, for the most part, it falls on them to have the discretion to see the broader picture and not complain without good reason. When they do complain, however, I think it's important to at least validate their feelings and try to keep in mind that although their problems usually don't measure up to the ones women or people of color face, it costs nothing to say, "I hear you. I understand that you're facing your own problems and you don't feel they're being acknowledged by society at large." Is that really so hard to do?
Our subgroups within society are not schoolhouses and abandoned buildings (in fact, I find the comparison of white men to abandoned buildings to be cruelly dismissive) and no one is "on fire", nor do we have a single water hose that can only put out one fire at a time.
I am NOT a Men's Rights activist, but I have peeked at their subreddit from time to time, usually to shake my head at the abysmal way they conduct themselves on the whole. But just like many feminists are upset that stereotypes about their movement are the product of a vocal minority, so too should we give Men's Rights activists the benefit of the doubt, I believe. Take an hour or so to read their subreddit's FAQ. A lot of what they say there actually sounds fairly reasonable and I think that everyone should be deeply concerned about some of the problems and discrimination they face, especially feminists who claim that they are fighting for gender equality, not the domination of women. Is Men's Rights activism in practice a platform for misogyny, marginalization of women, and sometimes outright lies? Absolutely! But let's not assume that every white man who says he has problems wholly endorses the bullshit spouted by a vocal minority (or maybe even majority!) within the MRA group.
Here's an interesting article that I think is a microcosm for the larger situation. We have gotten swept up into a legal frenzy over a rape crisis that may not even exist and in attempting to find a solution, we are unfairly and disproportionately punishing innocent men. Do their problems require the same level of attention as the fact that roughly one in five women are raped? No, of course not! But they need a voice, they need to be heard and their strong thoughts and feelings are completely valid.
Can't we just be nice to each other?
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While I feel I've already said my piece and I don't want to be accused of "white knighting", I would like to take one last opportunity to express my opposition.
The mods and judges, more than anyone else on the site, should understand that the site does not operate as a democracy. That's not how we publish videos, nor is it how we set policy. The idea is to take action (or, in this case, not take action) that fosters a better community. If people want a movie published or want some policy set, that has no direct relevance on whether or not it is a good idea.
While I don't think that giving the users the option to directly reveal their gender (rather than mention it in discussion or put it in their profile text or signature) is likely to cause problems, I also don't see any way that it improves the site. Furthermore, although I think the probability is somewhat remote, I do believe that highlighting our female members opens the door to some really nasty harassment that sadly permeates a lot of the internet. I would have liked some of these concerns acknowledged before the policy was put in place. For example, the mods could have instituted a zero-tolerance policy and said, "At the first instance of sexual or other harassment, we're doing away with the gender option." Instead, I fear that the policy is here to stay simply because it seemed like a good idea at the time.
So that's really all I have to say on the matter. It was instituted, I didn't like it, and (predictably), the world has not crashed down around us. Nevertheless, thousands of forums across the internet-- a majority, I'd argue-- operate just fine without any specific reference to gender. Why should one dedicated to beating video games as quickly as possible be any different?
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Could you guys do the mods a little favor and create your own topic to be merged with this discussion later? This conversation clearly has nothing to do with SMG2.
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Samsara wrote:
To me, it's a sign of lazy and careless work. Why put effort into making something entertaining or watchable when you can just stand around doing nothing and the game will continue regardless? I know it's hard to make an autoscroller entertaining, especially if the game is full of them, but any effort is more entertaining than no effort at all.
EDIT - There's a perfect example right under this thread (at the moment of posting): Look at what can be done in the original Super Mario Bros in terms of making long wait times as entertaining as possible with limitations on what can be done. That can easily be applied to the autoscrollers here, and as a bonus you can kill enemies and break blocks without having to worry about score.
I find the (unpublished) low score Super Mario Bros. run to be aggravatingly boring. You're right that it does what it can with the limitations, but I would jump up to argue that it is not entertaining. To each his or her own, however.
All I'm saying is that a lack of entertainment in the autoscroller sections should not preclude a run of this game from publication. Autoscrollers are boring and there's no real way consistently around that. Suboptimality could preclude a run from publication (as it did here), but I just happen to believe that it shouldn't.
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I'm disappointed in the focus in this thread on the autoscrollers being boring. So what is inherently the most boring part of the run is too boring for you? That's where we draw the line? This isn't Yoshi's Island where you can do tricks like egg juggling to pass the time. Instead, I think that only so much entertainment can be squeezed out of the autscrolling sections and they're of no major consequence to the run's entertainment value.
The fact that the run is suboptimal stands and should be corrected in any revision, but in my opinion, that's all that's necessary.
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In addition to the inefficient P-meter usage, I caught some avoidable slowdown in world 8's castle when Lakitu threw too many spinies on screen. I'm also skeptical of some of the damage boosting here. Sometimes it looks like it saves you just a few frames, which is less than the time it would take to not only lose the powerup but to also pick up a new one. It's pretty clear that this run is suboptimal.
Having said that, I don't see any glaring errors. I tend to think that we should publish runs even if they're suboptimal since there's no harm in waiting for a better player to come along and improve them. Better to have a proof-of-concept than demand perfection from the beginning.
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Archanfel wrote:
Well done guys, now one more easy challenge:
Rectangle was cut into 9 squares as shown in the picture. Size of the smallest white square is 1.
-Find sides of this rectangle.
There may be a nice geometric way to do this, but I favor the good ol' brute force equations method.
Let the widths of the squares be given by
orange
red
yellow
magenta
green
cyan (kind of a stretch)
blue
dark green.
Also, let the width of the rectangle be w and the height be h.
Then from the figure, we see
o + r + g = w
o + 1 + m + g = w
y + m + g = w
y + c + d = w
b + d = w
o + y + b = h
r + 1 + y + b = h
r + m + c + b = h
r + m + d = h
g + d = h
That's ten equations and ten unknowns. I'll solve it later.
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Archanfel wrote:
-How many triangles on this pic?
-And how to calculate amount of triangles for arbitrary size of the biggest triangle? (i.e. to find equation for progression a1=1, a2=5, a3=13, a4=27... ax=?).
By inspection we see that there are 10 small triangles on the bottom row, 9 on the next two rows, 8 on the next two rows, and so on. So the number of small triangles is
10 + 2*9 + 2*8 + ... + 2*1 = 10 + 2*10*(10-1)/2 = 10 + 10*9 = 100
In general, for an equilateral triangle of side n, there are n^2 little triangles within it.
For the 2*2*2 triangles, we see that there are 9 flush with the bottom row, 8 flush with the next row, 7 flush with the next row, and so on. Add 'em up and we get
9 + 8 + 7 + ... + 1 = 9*(9+1)/2 = 45
In general, for a triangle of side n, there are n*(n-1)/2 upward-pointing triangles of side 2 enclosed.
Ah! But let's do downward-pointing triangles too! We see 7 in the bottom row, 6 in the next, and so on up the line. The number of downward-pointing 2*2*2 triangles is therefore
7 + 6 + 5 + ... + 1 = 7*(7+1)/2 = 28
For a triangle of side n, there are (n-2)*(n-3)/2 downward-pointing triangles of side 2.
These patterns continue, and we see that for a triangle of side n, there are...
upward-pointing triangles of side...
1: (n+1)*n/2
2: n*(n-1)/2
3: (n-1)*(n-2)/2
...
n: 2*1/2
and downward-pointing triangles of side...
1: n*(n-1)/2
2: (n-2)*(n-3)/2
3: (n-4)*(n-5)/2
...
n/2: 2*1/2
(This assumes n is even. If n is odd, the series terminates at 3*2/2.)
So together we have...
... upward-pointing triangles: sum( (j+1)*j/2, j, 1, n) = 1/2 * sum( j^2 + j, j, 1, n)
... downward-pointing triangles (even side): sum( (2*k)*(2*k-1)/2, k, 1, n/2) = 1/2 * sum( 4k^2 - 2k, k, 1, n/2)
... downward-pointing triangles (odd side): sum( (2*k+1)*(2*k)/2, k, 1, (n-1)/2) = 1/2 * sum( 4k^2 + 2k, k, 1, (n-1)/2)
Now we'd like to write these sums in closed form. To do so, we can use induction (which I'll omit):
Upward-pointing triangles: 1/6 * ( n^3 + 3*n^2 + 2*n) = 1/24 * ( 4*n^3 + 12*n^2 + 8*n)
Downward-pointing triangles (even side): 1/24 * ( 2*n^3 + 3*n^2 + n)
Downward-pointing triangles (odd side): 1/24 * ( 2*n^3 + 3*n^2 - 2*n)
Adding these together, we find...
For triangles with an even length side n: N = 1/24 * ( 6*n^3 + 15*n^2 + 9*n) = 1/8 * ( 2*n^3 + 5*n^2 + 3*n)For triangles with an odd length side n: N = 1/24 * ( 6*n^3 + 15*n^2 + 6*n) = 1/8 * ( 2*n^3 + 5*n^2 + 2*n)
where N is the total number of triangles.
Let's try these formulas out on the sample data you gave:
n=1: Dangit.
^^^
Saved for posterity.
Well, that didn't work. But here's a "proof" that will drive the mathematicians here crazy. I recognize that a counting argument like the one above should work. Therefore, I expect a polynomial no higher than 3rd order in n, the length of one side, and with constant coefficient of zero. Thus, I just need to find a polynomial that fits the data and I'm done.
Let's fit our polynomial to the first three data points (for both even and odd). For even length side, I see that the total number of triangles is...
n=2: 5
n=4: 27
n=6: 78
and therefore,
N = 1/8 * (2*n^3 + 5*n^2 + 2*n) (n even).
And for odds...
n=1: 1
n=3: 13
n=5: 48
n=7: 118
and therefore,
N = 1/8 * (2*n^3 + 5*n^2 + 2*n - 1) (n odd).
Hooray for cheating!
Is there a name for what I just did? Basically, I assume induction should work and then I just fit the polynomial to the data.
Edit: Messed up a calculation. More corrections not unlikely...
Edit 2: Apparently, I can't assume the constant coefficient is zero. No matter! Just take more data! The formulas should both be correct now.
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You know, Blades of Vengeance was going to be the title of Skitchin' back in the day. /s
They're both EA games, so I suspect their marketing department just switched them accidentally.
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Warp wrote:
Those don't look like circles to me, but ellipses. Which gives me an idea for an additional question: Prove that an ellipse can be surrounded by 6 identical ellipses, each touching an adjacent ellipse only on one point.
Well this part's easy enough. An ellipse is just a linear (stretch and/or skew) transformation of a circle. So here's what we do:
1) Prove that a circle can be surrounded by 6 identical circles with each outer circle mutually tangent to two others. This has been done by other commenters here.
2) Take the set of all points that are on those circle's interiors or perimeters and execute a linear (matrix) transformation on them. Now you have the interiors or perimeters of seven ellipses for which the external ones are each mutually tangent to two others.
Provided all the ellipses have the same orientation, we can also work in the opposite direction and do a linear transformation that "compresses" them into circles.
Warp wrote:
Bonus question: Can the ellipses have different orientations, and still fulfill the requirements?
This is much harder. My gut says the answer is no. If we tilt one of our ellipses to the side, it will displace its neighbor and this disturbance will propagate around the exterior ellipses. Once it reaches the original ellipse, I think it's unlikely to be compatible with the original displacement we made. (But I guess my gut instinct was wrong!)
On the other hand...
Let five of the ellipses be oriented in a "daisy" configuration with their major axes pointing (more or less) radially outward from a given point. This will open up a significant gap in the "petals" where the sixth ellipse should go. (Can anyone prove this elegantly?) Then, you need to make the last ellipse tangent to three others. Three tangencies can be (at best) uniquely filled by an ellipse of given dimensions*. If the ellipse is too small to fill the gap, then simply tilt all of the "petals" so that they close the gap somewhat until it can be spanned by the last ellipse.
Someone can hash out the details of the proof, if they'd like.
*To prove this, let's examine the number of degrees of freedom. The ellipse has x- and y-coordinates of its center, major and minor axes, and a rotation parameter for a total of five degrees of freedom. We're given the dimensions of the ellipses, so we know their major and minor axes, reducing the degrees of freedom to three. Finally, each tangency "pins down" one degree of freedom. Since there are three tangencies, the ellipse is (at best) uniquely defined.
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Scepheo wrote:
It is definitely possible to take any arbitrary program and analyze it so see if it halts or not; the halting problem is that there is no algorithm to determine this. Programs of infinite length aren't necessary for that.
So while, to me, it seems "possible" to analyze all programs of length N and determine whether they halt and if so, what number they output, the halting problem means that you can't do this algoritmically.
In other words, I think the answer to Warp's question is "simply extremely hard".
Yeah, but what is the human thought process (as it applies to analyzing halting problem) other than an extremely complex algorithm? I'm not convinced that our thinking "transcends algorithms" and so the distinction is irrelevant.
I agree with your answer, however.
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Warp wrote:
Bobo the King wrote:
Since, for starters, all ~1000 digit numbers are computable under that criterion, I imagine the answer is very large. To throw out a guess, perhaps one million or so digits.
I'm not asking what the number is (because it's practically, perhaps even provably, impossible to calculate). I'm asking whether the problem is unsolvable or simply very hard.
One could naively think "it's not unsolvable; simply go through every possible valid C++ program of 1000 characters or less, and for each one see which numbers it outputs, if any". However, that method might not be usable because you may encounter the halting problem: You cannot know if a given program will ever halt.
Of course the halting problem is related to programs of any arbitrary size. Here we are talking about programs of a limited size, so perhaps the halting problem doesn't apply? Perhaps it's possible to simply list which programs halt and which won't?
Of course then there's the problem of programs that don't halt but do print numbers. If they just keep printing numbers, can you ever find out which ones?
An unbounded version of the problem is probably unsolvable. In other words, if we modify the problem like this:
"For any given n, what is the smallest positive integer that's not computable by a C++ program of at most n characters?"
Here we probably stumble across the halting problem outright.
Ah, I see. I think you're right, however, that we could simply list all programs with 1000 or fewer characters and analyze whether they halt or not. I see no reason why that shouldn't be possible. Then, as you say, just look at the numbers output. (I assume you're talking about programs that output one number. We would want to disallow, for example, while true do print i; i++; end. Excuse my pseudocode.)
I think it is interesting that this seems to imply that the halting problem is solvable for any program of finite length. It's only when you allow programs of infinite length (which are physically unrealizable) that you run into problems.
Any experts on this topic?
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Warp wrote:
I was thinking whether a problem like this is provably unsolvable, or simply extremely hard:
"What is the smallest positive integer that cannot be computed with a C++ program of at most 1000 characters?"
(Note that this doesn't mean that there aren't larger integers than the answer that can't be computed by such a program. The problem is asking for the smallest positive integer that cannot be, even if there exist larger ones that can.)
I know this is (probably) related to Kolmogorov complexity, but that subject is a bit too complex for my knowledge. (Also, I'm not sure if the concept of Kolmogorov complexity actually answers my question, ie. whether that problem is unsolvable or just very hard.)
Since, for starters, all ~1000 digit numbers are computable under that criterion, I imagine the answer is very large. To throw out a guess, perhaps one million or so digits.
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A lot of things led me here. I remember finding Morimoto's Mario 3 run way back in the day. I must have seen it on eBaum's World. I remember watching it in high school with my friends and being blown away at the tricks he pulled. That got me into the general idea of speedrunning and I must have at some point discovered both Speed Demos Archive and TASVideos. But that's kind of a common story here, so I'll focus on another aspect: what led me to botting.
Back in 2007, Adam Sweeney completed this run of Solstice, a game I had never played or even heard of. Still, something in his commentary stood out to me:
Adam Sweeney wrote:
My previous effort on this game was very popular among the other seventeen people in the world that care about Solstice, and received a mention in a 1up.com article for my "borderline pathological" mastery of the game (which is true; I once had a minimalist speedrun that was faster than the tool-assisted emulator run at the time, meaning that I effectively beat the game faster than a computer [Emphasis mine.]). So, I'm very happy to have trumped the previous run, and to provide a much better-looking one in the process. Have fun.
It touched a nerve. I'd never participated in speedrunning, but I'd watched enough videos and was a big enough fan at that point that I was bothered by what he said. Of course, tool-assisted speedruns aren't completed by computers. What the heck was he talking about? How the heck does this guy think that a computer can be "told", hey, beat this game as fast as possible!
I was right, to some extent. His success over the TAS came in large part due to simple route-planning differences. Still, I wasn't familiar with the tools of tool-assisted speedrunning, so I didn't really know what I was talking about.
When I joined the site, I discovered Lua scripting and wanted to incorporate it into my runs. It started with simple RAM watching scripts and then quickly ballooned into basic look-ahead scripts. For Skitchin', I put together a script that would, among other things, tell me when I could successfully execute a high jump at any given time. I got hooked.
I don't quite understand my own psychology. Back then (and to a certain extent, even today) I thought Adam Sweeney was dead wrong. One can't ignore the human element to TASing. Efforts to make a bot that beats a game aren't unheard of, but are still more dream than reality. And yet, paradoxically, I also want to prove him right. I want people to see that computers are capable of incredible things. I want to push the boundaries of human involvement in TASing.
I realize that doesn't exactly address the topic, but I thought I should share my experience. It's baffling to me that an offhand comment had such a profound effect on my approach to TASing.
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Mothrayas wrote:
Sign-ups will close on Sunday 23rd November 2014, 23:59 UTC.
The competition starts the next day, on Monday 17th November 2014, 21:00 UTC. This is when the final team composition will be revealed and the game will be revealed.
Daylight savings time falls back extra long this year, eh?
I'm interested in signing up this year, but I'll be unaffiliated for now. I'll probably poke around on the IRC channel for teammates in the coming week, but if anyone is interested now, please reply or PM me.
Be aware that I am horribly, horribly busy until the second week of December, and then I'm only somewhat less busy.
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I don't know how interesting others here will find this, but I did a little derivation today in my free time involving a mesh of resistors.
Suppose you have a rectangular lattice of resistors in a grid configuration, approaching the continuum limit. The lattice has width X and height Y. The boundaries of the lattice are held at given (not necessarily constant) potentials. Your task is to find the voltage at an arbitrary point, V(x,y). You may take the individual cells of the grid to be square (dx = dy), although it is not especially difficult to work with rectangular cells.
I assumed that the vertically-oriented resistors have a resistance of Ry(x,y) while the horizontally-oriented resistors have a resistance of Rx(x,y).
What partial differential equation does the voltage satisfy? Under what conditions does it reduce to the Laplace equation? Which of Kirchhoff's two laws is applicable?
My derivation is about a page in length, so this is not a tremendously difficult problem. Still I'm curious as to how others here would solve it and I would also like to know if it can be reduced to or derived from the continuous version of Ohm's law.
Those two threads are the perfect answer to "Why aren't there more female TASers?" There most likely are, they're just choosing not to mention their gender because shit like that happens. "There are no women on the internet" on the first page of both threads. God.
Is it really that hard to treat everyone equally? This is a site for talking about emulation and speedrunning. Gender shouldn't matter.
I completely agree. I wasn't pointing to those threads as filled to the brim with ideal behavior from males on the site, but rather that the overall sentiment was, "Who cares?" I don't see how the problems with those threads are diminished by flagging female users, even voluntarily. I can only imagine problems being amplified.
Warp wrote:
I don't really understand what you are referring to. How are those two threads "perfect answers to why there aren't more female TASers"?
And "there are no women on the internet" is a meme. It's humor.
Eh. It's pretty lame humor. And who hasn't heard it before? I think it got a chuckle out of me when I was in my mid-teens over a decade ago. I can easily imagine how annoying it would be to read for the fiftieth time after revealing my gender and wanting to be taken seriously. It adds nothing to the conversation and no one is going to jump in to say, "Lol! That's so clever! What a witty joke!"
Warp wrote:
I haven't seen much unequal treatment on this site. Can you point actual examples? (Because an accusation of unequal treatment of people is kind of serious and shouldn't be thrown lightly.)
I can't be the sole person to point out the inequality on the site, but I think it could easily be argued that two (now three) threads focusing on finding female TASers as well as one brief instance of assuming that a female TASer was male (granted that Gamer Maiden Sonia's response was a little overboard, but that's not the point) qualify as a few instances of male posters naively saying, "Wow! Girls are different!"
Now, I don't think this is being done maliciously and if it were directed at me, I'd try to shrug it off as a mild annoyance at worst, but that's not really what you're asking. I think the point is that women on the site are not treated with the same basic respect and standards as men. I mean, suppose you're really interested in, say, cosmetics and you go to a forum to express your interests. You lurk for a few months and then all of a sudden you see a thread that says, "Any men here?" Maybe some women leap in to your defense, but the overall sentiment seems to be, "What's the deal with men liking cosmetics? That's a lady thing!" Wouldn't you find that off-putting? Maybe not, but I at least see how it could be bothersome when you're just trying to fit in.
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Ji-chan wrote:
Bobo the King wrote:
I will defer to the wishes of our female members since they would be most affected by the policy, but if it were my choice, I would advise not allowing/requiring users to declare their gender. Let's just be all-inclusive and keep the preferred shape of our sex organs off the table, mmmkay?
(Wishful thinking on my part.)
HAVING to declare your gender is silly and I don't think I've came across any forum that does so (they either have the Blank option or you can hide it). Facebook (which isn't a forum) kind of does it, you can hide the gender itself but it still uses the gendered pronoun, which is silly, and you have to make a custom gender to make it refer to you as "they/them" (which even gets silly translation to non-english languages). It can be a right, but not an obligation.
I myself would be okay with the way it currently is since I myself don't seen this as much of a deal, that was the first time I've seen such an exaggerated reaction for a gender confusion (I've told this to synnchan already, she already apologized to the community so it's okay). But [blank]/male/female/other selectable box should be enough of a simple fix for those who WANT show off their gender to prevent confusion, and for didn't want to use their signature for that. So if my post makes any difference on the decision, I go for having the option.
@the arguments above: It this really necessary? This is an apology topic and and maximum we're trying to decide the course of action. ಠ_ಠ
Fair enough, but you can copy any of these symbols into your profile and/or signature:
♀♂☿⚥⚢⚣⚤⚦⚨⚩⚪
Well done guys, now one more easy challenge: Rectangle was cut into 9 squares as shown in the picture. Size of the smallest white square is 1. -Find sides of this rectangle.
-How many triangles on this pic? -And how to calculate amount of triangles for arbitrary size of the biggest triangle? (i.e. to find equation for progression a1=1, a2=5, a3=13, a4=27... ax=?).