*Whoops! Square at **6:20** is NOT the Klein bottle!* (which would require not criss-crossing the "gluing" of points at the left and right). The construction I showed is the (real) projective plane. My bad, everyone. I'm a topological dum-dum. Thanks to everyone who pointed out my error.
Why do you entertainers always feel the obsessive need to explain everything ONLY with pictures & geometry, as if for you, this fetish with geometry makes you think that geometrical pictures is more "fundamental" than the underlying algebra? I prefer seeing the formalism of algebra, since all geometric problems reduce to algebra.
Topology was always hard for me because of this chronic nagging doubt in my brain that these sloppy hand-waving picture-arguments would work or be true UNTIL I rigorously proved them all formally algebraically. Until I can prove everything rigorously formally algebraically, my brain would automatically not retain the higher-level topological lesson.
I hurt myself, the way one would having slipped on a wet floor one wasn't warned about, trying to glue that square into a Klein bottle shape. My imaginary infinitely-stretchable membrane now has creases and tears. Ouch.
At 6:15 they stated this identification gives the Klein bottle, but in truth it gives an even weirder surface called the "real projective plane". This plane is kind of like "half" of a klein bottle, in the sense that if you glue two of them together by identifying any pont of one to a point of the other, you obtain an object equivalent to a klein bottle.
exactly: a Klein bottle is a möbius strip with its edges glued together - so you need to identify two edges (top and bottom) antipodal/diagonal opposite (a~b, c~d) and two edges (left and right) parallely (i~j, k~l).
Damn I thought I had clicked on a PBS Space Time video for a minute. Welcome back to the spotlight Gabe. Matt has been doing a great job with the show, but I've missed you.
Pacman can only cross the edges at 90 degrees, whereas every line on your helix crosses at an angle. For this reason, a far more apt analogy would be Asteroids. And the animation would have the added benefit of showing the ship shooting in a direction other than it's heading so that the "bullet" crosses at one angle while the ship crosses at another while the "paper" remains unchanged. Other than that ridiculously nitpicky thing, I loved this video. Great job!
You made it quite easy. Just overlap the square with its mirror around the diagonal and mark the points that overlap. The special case is all the overlapping points on the diagonal. Those are all the points that have the minute and hour hand in the same position.
It's not just the overlapping points on the diagonal, it's any pair of overlapping points mirrored across the diagonal. Find any intersection of lines between the diagonal-mirrored squares, and there will be another intersection up and to the left of it along a 45 degree angle. Because it's an intersection, there's necessarily a point on a line there in the original square by itself too.
Pfhorrest my point about the diagonal was only for the special case where the minute and hour hand overlap. Of course every other intersection also satisfies the question in the video.
The hard part is really the conversion from rational numbers to proper times -- I'm more worried that the program I wrote converts the values incorrectly
now i know why this channel is called infinite series: because every video tells you to pause and watch 2 more videos before watching this one. As such you must always watch an infinite number of videos before ever watching any of the videos. Quite the paradox!
For me, this video was fantastic and an intuitive intro to "pacmanifying" a 2d plane/sqaure. It's a really cool way to model otherwise hard-to-imagine shapes. I wouldn't have seen that coming from the title of the video... so I would argue the title should have been more to do with the pacmanification.
8:18 -1 and 4 are in the same equivalence class mod 5 (ie 4-(-1) = 5). Neither of them are in the same equivalence class as 1. It shouldn't be too surprising that 1 and -1 aren't the same. I think it's more an issue of convention whether to use 0,1,2,3,4 or -2,-1,0,1,2. There are a few situations where it's more convenient to use -1 than n-1.
There are a few other answers that are similar to 12:00:00 where the hands on a clock over lap. Those 11 points are more solutions where swapping the hands changes nothing. A complete answer should at least account for this
There are 143 equally spaced solutions. Consider the superposition of two clocks, such that the minute hand of the first is glued to the hour hand of the second. A solution occurs every time the non glued hands cross. The gears ratio is 144:1 So the minute hand of the second clock rotates 144 times for one revolution for the hour hand of the first. But the first clock's hour hand runs away from the second clock's minute hand, making one revolution for 144 revolution of the second clock's minute hand, so those two hands cross 144-1 times, at equally spaced intervals. The rest is just tedious. I'm an engineer and horologist not a mathematician, so my visualisation is different!
Actually, Zonko is correct and I made a mistake. See the pinned comment I added to the top of the vid. Parallel+parallel = torus; One-side crossed identification with other two sides left as-is = Mobius strip; Parallel+crossed = Klein; Crossed+crossed = projective plane.
i sometimes end up imagining the x-y-plane as a torus of infinite size. it's kinda nice to think that the functions aren't lost out there, but comming back after an infinitely long travel.
I think toruses are great and all, but I certainly didn't think of a torus when I came up with essentially the same thing you told about in the video. If you represent the time with a real number "t" between 0 and 12, what you need is that the hour handle to be at t, and the minute handle to be 12*frac(t) (of course you need to think of 12 as a whole rotation). If you plot 12*frac(t) and it's "inverse function" (not a function but a curve), and look for intersections you get the same thing.
True. I think most people don't think about this problem geometrically, which is why I wanted to connect it to the torus visually -- just to highlight a less commonly adopted but also fruitful approach that may appeal to more visual thinkers.
I like this more visual torus spiral thing. I just wanted to share the other approach for those who think they would never think of mapping a torus spiral to [0,12]x[0,12].
A fun way to look at the challenge is to figure out what exactly does "switching the hands" mean in the mathematical sense. First, let me introduce a function fraction(x), denoted {x}=fraction(x), which returns fractional part of a number. Then, if we call hour hand position in terms of full revolution H (that is, H=0 at 00:00:00, and H=1 at 12:00:00), then minute hand position in terms of full revolution M = {12 H}. So, switching the hands is just switching M with H, so that H = {12*M}. But for that position to be also valid, M = {12*H} must still be respected. Hence, H={12*{12*H}}. let us now substitute H=x/12, 0 ≤ x < 12 x/12 = {12*{x}} and represent x as a sum of its whole part 0 ≤ h < 12 and fractional part 0 ≤ r < 1: (h+r)/12 = {12*r} repeating the procedure for r, we arrive at 12*h+k = 143*f where 12*r = k + f, 0 ≤ k < 12, 0 ≤ f < 1, k is whole. Left side is whole, so right side also has to be whole. It can only be whole if f = m/143 where m is whole: 12*h+k = m m smaller then 143: since 0 ≤ f < 1, 0 ≤ m/143 < 1, so 0 ≤ m < 143. The rest is trivial. H = x/12 = (h+r)/12 = (h+(k+f)/12)/12 = (12*h+k+f)/144 = (m + m/143)/144 for m = {0, 1, ... , 142} -- one just needs to convert this to hours-minutes-seconds-fractions. Some formulae: Let us define floor(x) as a number x rounded to integer towards negative infinity, denoted ⌊x⌋=floor(x). For whole numbers, ⌊x⌋ = x - {x} Then hours = ⌊H*12⌋ minutes = ⌊{H*12}*60⌋ seconds = ⌊{{H*12}*60}*60⌋ fractions = {{{H*12}*60}*60}
I feel like bringing geometry into this problem is convoluted and unnesecary, those lines could have been easily gain from the idea that your progress the hour the hour equals current minute divided by sixty, or y=x/60. From there you just add increase the y intercepts for each of the hours giving you y=x/60+n for all integers n between 0 and 11, inclusive.
Very good video. Starting with an interesting problem and then showing how mathematical objects arise from it naturally (and the pacman example is great). This problem shows naturally why and where you can use quotient spaces, and probably this video should have come before the one on quotients. Regarding the division algorithm remark in the end - you don't toss away the negative numbers because they are not positive. You just choose a representative from each class mod n, and luckily in the integers we can choose all the representative to form a nice set (i.e. 0,1,...,n-1). There are also problems where it is better to use the set -n/2,...,n/2, and not to mention other rings where there is no notion of "positivity" at all.
"When Einstein Walked with Gödel: Excursions to the Edge of Thought (English Edition)", Jim Holt . "But Gödel came up with a third kind of solution to Einstein’s equations, one in which the universe was not expanding but rotating. (The centrifugal force arising from the rotation was what kept everything from collapsing under the force of gravity.) An observer in this universe would see all the galaxies slowly spinning around him; he would know it was the universe doing the spinning and not himself, because he would feel no dizziness. What makes this rotating universe truly weird, Gödel showed, is the way its geometry mixes up space and time. By completing a sufficiently long round trip in a rocket ship, a resident of Gödel’s universe could travel back to any point in his own past."
6:30 I'm pretty sure for the klein bottle, you only pair diagonally opposite points on one set of opposite edges, and for the other set of opposite edges, you pair (not diagonally) opposite points.
In the postscript, at around 10 min. - In the division algorithm, 14 = 2·5 + 4, rather than 3·5 - 1, because we need the remainder to be always ≥ 0. Division algorithm: Given D≥0, and d>0, to find q and r such that D = d·q + r ; Dividend equals divisor times quotient, plus remainder, where q ≥ 0 and 0 ≤ r < d. She mentions that it has to do with rings; but I would point out that the non-negative integers *don't* form a ring, which must have an additive inverse for each of its elements. In fact, Z, the ring of (all real) integers, is the prototype for the concept of a ring. The division algorithm can, in fact, be extended to all the integers, with the following stipulations: D ε Z, d ε Z \ {0}, q ε Z, 0 ≤ |r| < |d|, d·r ≥ 0 Enjoyed the video! Thanks! Upvoted!
- It's impossible to give everyone 3 bucks, and then take one away. (09:07) - Oh, yes it is possible, if you are a government, or a bank, but then you will have no friends, of course. :o)
Colloquial forms for time-telling like "quarter to three" suggest that the division algorithm which gives strictly positive remainders, is not how english speaking people habitually think about time. Often we prefer to round up and give a negative remainder which is smaller in magnitude ("it's ten to six") rather than round down and give a positive remainder ("it's fifty past five"). Do we do this with many other quantities?
At 6:21 you state that making diagonally opposite points equivalent forms a Klein bottle, but this actually forms a real projective plane. A Klein bottle is formed when you take the Möbius strip and connect the two remaining edges like in the cylinder. I'd also like to point out that these equivalence classes are actually fundamental polygons of a square (except the cone example).
They are very fun and intuitive, I would advise against following his ''picture'' style too much though, this can be good as a rather informal introduction, though I think the set theoretic approach to topology is more important in the long run, at least for pure mathematicians, and ''proof by picture'' doesn't really cut it in a serious mathematical situation.
It simplifies really easily if you imagine a click where the minute and hour hands click like the second hand. Thus there is a valid swap every time the minute hand crosses a point where the hour hand might be. (ie at 5 10 15.... minutes)
@2:23 in this configuration u will get an helix on torus and by switch the configuration u will get another pattern on torus, I think the intersection of two patterns on torus gives the solution for @7:22
Swapping hands is like rotating the square and gluing the edges together in the opposite order to make a torus? And where you end up with the same points on both tori, those are the valid swappable configurations?
clock face angles... aren't a problem though... it's just how they work. you could make a clock where the hour hand ticks over only when the minute hand has completed a rotation.
Whoa! Gabe is back. :-O Funny that the first video that caught my interest in awhile has Gabe "pacmanifying" clocks on a torus. Though he has slowed down some since I saw him last on Space Time. Did your job work you too hard or something? In any case good to see ya back. Missed your weird ways of simplifying things. :-)
This relation at 4:15 isn't a equivalence relation, because it doens't satisfy reflexivity (Choose one point p, then p is not on the opposite side of p). Am I missing something? P.S. The explanation at the end with the "division algorithm" was wrong. It doesn't matter whether we say 15-1 or 10+4, because [-1]=[4] in 5Z (meaning -1 and 4 are exactly identical or indistinguishable from another in mod 5 arithmetic).
Gabe here -- re: the equivalence relation, I state verbally (though it doesn't show up on screen) that each interior point is also identified with itself. I think that's what you're getting at, and it completes the equivalence relation.
I'm pretty sure there are 144 times if you include midnight and noon. I could draw out where all those points are on a sheet of paper quite trivially, but actually working out when they are is another matter ^^"
Yay! I guess I missed something while doing a quick solution. I'm wondering what that was, guess I'll find out next week. Still, that is a lot of times to calculate and write out as hours, minutes, seconds, fractions of seconds XD
The swap is only possible when the pointers meet, at 01:05:27 (272727...), 02:10:54 (545454...) etc. The angle they meet is multples of 360/11 degrees. =D
I can’t believe it’s been 3 years and no one mentioned that pac-man in fact lives on a cylinder rather than a torus. He can only pass from side to side, not top to bottom. You might be thinking of asteroids or any number of other games that have full wrapping and live on a torus.
Since invertig handles on an answer gives us another answer (cause inverting the inversion would give us the original position, which was valid) do we have to send half of the times or both?
I'm sure Secret of Mana fan's remember this from their childhoods, thinking "how could you fly past the west edge of the world and end up on the east edge AND fly past the north edge and end up at the south... no poles... must be a torus"
The challenge looks very nice. Will only an explicit list of solutions be accepted or can we just express solutions in parametrical form (ex.: t hours, 5+k minutes and 3-t/42 seconds for some values of k and t) ?
difference between packman and the toroidal lines is packman can always end up back at the same place, whereas torus does not end up on the same place on the paper if followijg one line, whereas packman will if staying on the same line end up at the same place.
Quotients are not dual to products. In the category of sets the product is the Cartesian product of the two sets and the coproduct, the dual of products, is the disjoint union of the two sets. Because a topological space is a set, the product and coproduct in this category will be similar with some topology defined on them underlying set. 'the most natural one'. Quotients are 'dual' to something but not products.
Nice observation! Here are some deets: the Cartesian product and quotients are examples of two constructions in category theory called *limits and colimits.* A product is an example of a limit, while a quotient is an example of a colimit. This is the duality I was referring to in my response. But you’ve spotted a *much* tighter, more specific duality between the Cartesian product (a limit) and the coproduct (a colimit) versus a [fill in the blank] (a limit) and a quotient (a colimit). Products and coproducts are indeed dual to each other, in a much stronger sense, as are [fill in the blank] and quotients. So, what is the [fill in the blank]? I’ll give one word for a hint: “(co)equalizer.” :) -Tai-Danae
+Wiki Comet Just so you know, "Because a topological space is a set, the product and coproduct in this category will be similar with some topology defined on them underlying set" is not valid reasoning. For example Vect has all small limits and colimits but the coproduct of two vector spaces (direct sum) is *not* their coproduct as sets (disjoint union) with a vector space structure. The actual reason why limits and colimits in Top are just those in Set with a suitable topology is because the forgetful functor U: Top -> Set preserves limits and colimits. To contrast, the forgetful functor U:Vect -> Set only preserves limits. This is also the reason why products in Vect (also direct sums) are just products as sets with a vector space structure. To really get to the end of the story, this phenomenon happens because there are two natural ways of creating a topological space from a set (put the discrete or indiscrete topology on it) whereas there is only one way for vector spaces (the vector space with the given set as a basis). That is, the two functors disc,indisc: Set -> Top are respectively left and right adjoints to U:Top -> Set which implies that the latter preserves colimits and limits, but U:Vect -> Set only has a right adjoint and no left one, so it preserves limits but not necessarily colimits.
Shhhhhhhhh, you're telling them too faaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaast
All the intersections of the graph and its flipped version lie on the 11x11 integer grid which has undergone a linear transformation L. This transformation is defined by L(11, 11) = (12, 12), L(11, 0) = c*(12, 1) and L(0, 11) = c*(1, 12). The constants are the same because of symmetry. Calculating c will tell us what L is and now we can easily (with a computer) compute the transformed version of the 11x11 grid.
OK solution on the way. Nice puzzle, it shows how people think differently, I would never have tackled it the way you did. I really liked the way you used, in your solution a fraction of 11, nice! I sought a solution using a variation of when do the hands overlap puzzle, hence the factor of 11. Only it's no longer 11, is it? OK no more help for those struggling. :-D
To study topology of universal timespace you need to know these docs first : (then comes electronic circuit logic as xor nor nand... and then comes differential fluid dynamics and lastly comes quantum mechanics alongwith laminated -4d brane strings...)
Sure it can. Just as we can visualize the hour/minute problem as a line looping through the segment of a plane, pac-man style, we can also visualize the hour/minute/second problem as a line looping through 3d space pac-man style. One noteworthy difference though is that in 3d two non-parallel lines are not guaranteed to intersect, so it's possible that there is only one solution to the problem at 12:00:00, but I've yet to prove it.
Yes, that was my mean ;) If I note S_3 = { e, (hs), (hm), (ms), (hms), (hsm)} h=hour, m=minute, s=second I proved (hm) is only 00:00:00. Wasn't hard, just has to "do the stuff". Yet other cases should be harder, and I'm fucking lazy.
I'm guessing that would be 3 * 720^2 equations to solve, right? Each line in the original problem would be split into 60 lines, one for each second. You would need to solve each line with each corresponding line in a rotation, hence squared, and there are three ways to rotate the problem. Sounds doable, but maybe not by hand. Edit: That would be sets of equations, as you need to solve a matrix for each combination. I also think it would be enough to solve one combination, as the two other can be found by rotating the solution. We are now down to half a million matrices to solve. Maybe we actually only need to solve very few of them to show if there exists a solution, as every hour after the first are just the same thing moved a bit, but I might be wrong about that.
I'd argue that all of them take place on tori -- it's just that in some versions, there are only _openings_ on one set of opposite walls. But in some versions, the openings are both sets of opposite faces, and the toroidal topology becomes manifest. Check out some of the screens on Pac-Man Battle Royale, for instance.
Challenge isn't that hard if problem is understanded because every question already contain half of answer in itself: you can swap them only when they align (when are paralel). Since small arow does 2 full circles per day, and big arrow does 24 circles per day, they should intersect 23 times. So there is 23 moments in a day (24 hours) when you can swap big and small arrow and still tell correct time
All solutions are the intersections of the clock state square overlapped with a bottom left diagonal flipped coppy of the square There are 144 possible states that are valad if swaped
*Whoops! Square at **6:20** is NOT the Klein bottle!* (which would require not criss-crossing the "gluing" of points at the left and right). The construction I showed is the (real) projective plane. My bad, everyone. I'm a topological dum-dum. Thanks to everyone who pointed out my error.
Then, what is it?
Why do you entertainers always feel the obsessive need to explain everything ONLY with pictures & geometry,
as if for you, this fetish with geometry makes you think that geometrical pictures is more "fundamental" than the underlying algebra? I prefer seeing the formalism of algebra, since all geometric problems reduce to algebra.
Topology was always hard for me because of this chronic nagging doubt in my brain that these sloppy hand-waving picture-arguments would work or be true UNTIL I rigorously proved them all formally algebraically.
Until I can prove everything rigorously formally algebraically, my brain would automatically not retain the
higher-level topological lesson.
I hurt myself, the way one would having slipped on a wet floor one wasn't warned about, trying to glue that square into a Klein bottle shape. My imaginary infinitely-stretchable membrane now has creases and tears. Ouch.
@gabriella felicia Crossing points on both the vertical and horizontal axes actually produces a torus IIRC, same as directly "gluing" both axes.
At 6:15 they stated this identification gives the Klein bottle, but in truth it gives an even weirder surface called the "real projective plane". This plane is kind of like "half" of a klein bottle, in the sense that if you glue two of them together by identifying any pont of one to a point of the other, you obtain an object equivalent to a klein bottle.
exactly: a Klein bottle is a möbius strip with its edges glued together - so you need to identify two edges (top and bottom) antipodal/diagonal opposite (a~b, c~d) and two edges (left and right) parallely (i~j, k~l).
Damn I thought I had clicked on a PBS Space Time video for a minute. Welcome back to the spotlight Gabe. Matt has been doing a great job with the show, but I've missed you.
This is some seriously complex stuff. Go easy on our brains, Gabe. With this episode you *torus* a new one!
Ba dum bum... psssshhh! (snare drum sound)
So, swap hour and minute hands to give a second helical curve. Intersection of both curves are valid solutions..
there is an easier description.
That’s how I thought of it
Pacman can only cross the edges at 90 degrees, whereas every line on your helix crosses at an angle. For this reason, a far more apt analogy would be Asteroids. And the animation would have the added benefit of showing the ship shooting in a direction other than it's heading so that the "bullet" crosses at one angle while the ship crosses at another while the "paper" remains unchanged. Other than that ridiculously nitpicky thing, I loved this video. Great job!
You made it quite easy. Just overlap the square with its mirror around the diagonal and mark the points that overlap. The special case is all the overlapping points on the diagonal. Those are all the points that have the minute and hour hand in the same position.
Rik Schaaf exactly I thought same too
It's not just the overlapping points on the diagonal, it's any pair of overlapping points mirrored across the diagonal. Find any intersection of lines between the diagonal-mirrored squares, and there will be another intersection up and to the left of it along a 45 degree angle. Because it's an intersection, there's necessarily a point on a line there in the original square by itself too.
Pfhorrest my point about the diagonal was only for the special case where the minute and hour hand overlap. Of course every other intersection also satisfies the question in the video.
Rik Schaaf Sure, but what time do these points correspond to?
The hard part is really the conversion from rational numbers to proper times -- I'm more worried that the program I wrote converts the values incorrectly
now i know why this channel is called infinite series: because every video tells you to pause and watch 2 more videos before watching this one. As such you must always watch an infinite number of videos before ever watching any of the videos. Quite the paradox!
Touching the esoteric without fantasizing so much. Thank you!
"Rock out with ...your clock out"
Bwaaaahhahahahahahahahahaha!!!!!
Good to see you here Gabe! Love your Spacetime episodes
Welcome back Gabe, please please please do a spacetime episode or cameo
It had been a while since last time Infinite Series gave us a good video. It was worth the wait.
Great video. The equivalences just blew my mind.
error at 6:25, it is actually the projective plane and not the klein bottle...
groethendieck came here to say this hahah :)
Saw the same thing. Nice to know whenever i see a mistake there is always someone pointing it out before me :P
oh true, the klein bottle would probably be the diagonal pairs from top/bottom, and the directly across pairs from left/right
Hila Kleiners for life!
Thanks to you, I learned a new mathematical term: "pacmanified."
For me, this video was fantastic and an intuitive intro to "pacmanifying" a 2d plane/sqaure. It's a really cool way to model otherwise hard-to-imagine shapes. I wouldn't have seen that coming from the title of the video... so I would argue the title should have been more to do with the pacmanification.
It's just a mug shaped clock.
gabe in front of the camera again! welcome back!
8:18
-1 and 4 are in the same equivalence class mod 5 (ie 4-(-1) = 5). Neither of them are in the same equivalence class as 1. It shouldn't be too surprising that 1 and -1 aren't the same.
I think it's more an issue of convention whether to use 0,1,2,3,4 or -2,-1,0,1,2. There are a few situations where it's more convenient to use -1 than n-1.
exactly; -1 isnt '1 but negative for some reason', its more like '-5 + 4'
1:50 nice..
Or helical back contingent on visualization.
There are a few other answers that are similar to 12:00:00 where the hands on a clock over lap. Those 11 points are more solutions where swapping the hands changes nothing. A complete answer should at least account for this
And when the hands are opposite each other.
In general, you don't get a valid time if you swap the hands when they are 180 degrees apart (think about 6:00:00, for instance).
Silly me, of course not.
Wow, now I want a torus helix clock.
ROCKING OUT WITH MY CLOCK OUT!!!
There are 143 equally spaced solutions.
Consider the superposition of two clocks, such that the minute hand of the first is glued to the hour hand of the second.
A solution occurs every time the non glued hands cross.
The gears ratio is 144:1 So the minute hand of the second clock rotates 144 times for one revolution for the hour hand of the first.
But the first clock's hour hand runs away from the second clock's minute hand, making one revolution for 144 revolution of the second clock's minute hand, so those two hands cross 144-1 times, at equally spaced intervals. The rest is just tedious.
I'm an engineer and horologist not a mathematician, so my visualisation is different!
Mindblown! All of this math stuff makes my head spin. That torus shape gave me a craving for donuts! :-)
isn't the klein bottle 1 side crossed and 1 parallel?
Basically,
Parallel + parallel = torus
Parallel + crossed = Moebius strip
Crossed + crossed = Klein bottle
Actually, Zonko is correct and I made a mistake. See the pinned comment I added to the top of the vid. Parallel+parallel = torus; One-side crossed identification with other two sides left as-is = Mobius strip; Parallel+crossed = Klein; Crossed+crossed = projective plane.
klein chicken is actually quantum
crossed crossed gives you projective plane
I've actually used a similar thing as part of a feature engineering step in a Machine Learning pipeline and visualization. Really useful.
i sometimes end up imagining the x-y-plane as a torus of infinite size. it's kinda nice to think that the functions aren't lost out there, but comming back after an infinitely long travel.
I think toruses are great and all, but I certainly didn't think of a torus when I came up with essentially the same thing you told about in the video. If you represent the time with a real number "t" between 0 and 12, what you need is that the hour handle to be at t, and the minute handle to be 12*frac(t) (of course you need to think of 12 as a whole rotation). If you plot 12*frac(t) and it's "inverse function" (not a function but a curve), and look for intersections you get the same thing.
True. I think most people don't think about this problem geometrically, which is why I wanted to connect it to the torus visually -- just to highlight a less commonly adopted but also fruitful approach that may appeal to more visual thinkers.
I like this more visual torus spiral thing. I just wanted to share the other approach for those who think they would never think of mapping a torus spiral to [0,12]x[0,12].
A fun way to look at the challenge is to figure out what exactly does "switching the hands" mean in the mathematical sense.
First, let me introduce a function fraction(x), denoted {x}=fraction(x), which returns fractional part of a number. Then, if we call hour hand position in terms of full revolution H (that is, H=0 at 00:00:00, and H=1 at 12:00:00), then minute hand position in terms of full revolution M = {12 H}.
So, switching the hands is just switching M with H, so that H = {12*M}. But for that position to be also valid, M = {12*H} must still be respected. Hence,
H={12*{12*H}}.
let us now substitute H=x/12, 0 ≤ x < 12
x/12 = {12*{x}}
and represent x as a sum of its whole part 0 ≤ h < 12 and fractional part 0 ≤ r < 1:
(h+r)/12 = {12*r}
repeating the procedure for r, we arrive at
12*h+k = 143*f
where 12*r = k + f, 0 ≤ k < 12, 0 ≤ f < 1, k is whole.
Left side is whole, so right side also has to be whole. It can only be whole if f = m/143 where m is whole:
12*h+k = m
m smaller then 143: since 0 ≤ f < 1, 0 ≤ m/143 < 1, so 0 ≤ m < 143.
The rest is trivial.
H = x/12 = (h+r)/12 = (h+(k+f)/12)/12 = (12*h+k+f)/144 = (m + m/143)/144 for m = {0, 1, ... , 142} -- one just needs to convert this to hours-minutes-seconds-fractions.
Some formulae:
Let us define floor(x) as a number x rounded to integer towards negative infinity, denoted ⌊x⌋=floor(x). For whole numbers, ⌊x⌋ = x - {x}
Then
hours = ⌊H*12⌋
minutes = ⌊{H*12}*60⌋
seconds = ⌊{{H*12}*60}*60⌋
fractions = {{{H*12}*60}*60}
So basically, you'd just have to do the reciprocal of the wrapped around function and find all the intersections between the two, right?
I feel like bringing geometry into this problem is convoluted and unnesecary, those lines could have been easily gain from the idea that your progress the hour the hour equals current minute divided by sixty, or y=x/60. From there you just add increase the y intercepts for each of the hours giving you y=x/60+n for all integers n between 0 and 11, inclusive.
Very good video. Starting with an interesting problem and then showing how mathematical objects arise from it naturally (and the pacman example is great). This problem shows naturally why and where you can use quotient spaces, and probably this video should have come before the one on quotients.
Regarding the division algorithm remark in the end - you don't toss away the negative numbers because they are not positive. You just choose a representative from each class mod n, and luckily in the integers we can choose all the representative to form a nice set (i.e. 0,1,...,n-1). There are also problems where it is better to use the set -n/2,...,n/2, and not to mention other rings where there is no notion of "positivity" at all.
loved this episode, I never would have thought of a torus when thinking about a traditional clock
new ways to think, huh?
I think this is called UV Unwrapping. A technique used in 3d modeling to texture objects with as little stretching as possible.
"When Einstein Walked with Gödel: Excursions to the Edge of Thought (English Edition)", Jim Holt .
"But Gödel came up with a third kind of solution to Einstein’s equations, one in which the universe was not expanding but rotating. (The centrifugal force arising from the rotation was what kept everything from collapsing under the force of gravity.) An observer in this universe would see all the galaxies slowly spinning around him; he would know it was the universe doing the spinning and not himself, because he would feel no dizziness. What makes this rotating universe truly weird, Gödel showed, is the way its geometry mixes up space and time. By completing a sufficiently long round trip in a rocket ship, a resident of Gödel’s universe could travel back to any point in his own past."
7:11 The geometric operation would be a 90 degree rotation of the square (i.e. transposition of the x and y axes).
Rock out with your clock out.
6:30 I'm pretty sure for the klein bottle, you only pair diagonally opposite points on one set of opposite edges, and for the other set of opposite edges, you pair (not diagonally) opposite points.
Yes. The construction shown would give you a cross-cap surface.
In the postscript, at around 10 min. - In the division algorithm, 14 = 2·5 + 4, rather than 3·5 - 1, because we need the remainder to be always ≥ 0.
Division algorithm: Given D≥0, and d>0, to find q and r such that D = d·q + r ; Dividend equals divisor times quotient, plus remainder, where q ≥ 0 and 0 ≤ r < d.
She mentions that it has to do with rings; but I would point out that the non-negative integers *don't* form a ring, which must have an additive inverse for each of its elements.
In fact, Z, the ring of (all real) integers, is the prototype for the concept of a ring.
The division algorithm can, in fact, be extended to all the integers, with the following stipulations:
D ε Z, d ε Z \ {0}, q ε Z, 0 ≤ |r| < |d|, d·r ≥ 0
Enjoyed the video! Thanks! Upvoted!
In civ call to power you could play on the torus, Made the world map feel flatter than the usual globe
What kind of shape would be best to analyze a three-hand clock (hour:minute:second)?
- It's impossible to give everyone 3 bucks, and then take one away. (09:07)
- Oh, yes it is possible, if you are a government, or a bank, but then you will have no friends, of course.
:o)
Colloquial forms for time-telling like "quarter to three" suggest that the division algorithm which gives strictly positive remainders, is not how english speaking people habitually think about time. Often we prefer to round up and give a negative remainder which is smaller in magnitude ("it's ten to six") rather than round down and give a positive remainder ("it's fifty past five").
Do we do this with many other quantities?
At 6:21 you state that making diagonally opposite points equivalent forms a Klein bottle, but this actually forms a real projective plane. A Klein bottle is formed when you take the Möbius strip and connect the two remaining edges like in the cylinder. I'd also like to point out that these equivalence classes are actually fundamental polygons of a square (except the cone example).
really interesting, i love topology (from a laymans perspective, at least) :) nice to see you on here again gabe
so pac-man lives inside a donut
on the surface of one
It’s more of a mug, really.
Tadashi Tokiedo lectures are also good regards the mainfolds discussed
They are very fun and intuitive, I would advise against following his ''picture'' style too much though, this can be good as a rather informal introduction, though I think the set theoretic approach to topology is more important in the long run, at least for pure mathematicians, and ''proof by picture'' doesn't really cut it in a serious mathematical situation.
Deep Joshi i
It simplifies really easily if you imagine a click where the minute and hour hands click like the second hand.
Thus there is a valid swap every time the minute hand crosses a point where the hour hand might be.
(ie at 5 10 15.... minutes)
@2:23 in this configuration u will get an helix on torus and by switch the configuration u will get another pattern on torus, I think the intersection of two patterns on torus gives the solution for @7:22
If we work in Z_5 then 14 = 4 = -1
You just go backwards on your "clock face" when dealing with negatives.
Yeah, borrow a book, give three each to your five friends, and then go pay the non-return charges at the library...
Please do a video about what different dimensional shapes look like and the difference between 3-D and the other dimensions
Cool, another math challenge to work on this weekend!
Swapping hands is like rotating the square and gluing the edges together in the opposite order to make a torus? And where you end up with the same points on both tori, those are the valid swappable configurations?
clock face angles... aren't a problem though... it's just how they work. you could make a clock where the hour hand ticks over only when the minute hand has completed a rotation.
Whoa! Gabe is back. :-O
Funny that the first video that caught my interest in awhile has Gabe "pacmanifying" clocks on a torus.
Though he has slowed down some since I saw him last on Space Time. Did your job work you too hard or something?
In any case good to see ya back. Missed your weird ways of simplifying things. :-)
I love this
SPAAAAAAACCCCCEEEEEE TTIIIIIIIIIMMMMMMMEEEEEEEE
I MISSED YOU MAN
Did he just say "Rock out with your clock out"? Devin Townsend once said something like that but he wasn't talking about clocks
Fascinating!
This relation at 4:15 isn't a equivalence relation, because it doens't satisfy reflexivity (Choose one point p, then p is not on the opposite side of p). Am I missing something?
P.S. The explanation at the end with the "division algorithm" was wrong. It doesn't matter whether we say 15-1 or 10+4, because [-1]=[4] in 5Z (meaning -1 and 4 are exactly identical or indistinguishable from another in mod 5 arithmetic).
just say (x,0)~(x,1) for all x in [0,1], and (0,y)~(1,y) for all y in [0,1]
Gabe here -- re: the equivalence relation, I state verbally (though it doesn't show up on screen) that each interior point is also identified with itself. I think that's what you're getting at, and it completes the equivalence relation.
All fields are working like torus! And time is a concept abstract of humans !
So fold the square across the diagonal and find the intersection points?
I'm pretty sure there are 144 times if you include midnight and noon. I could draw out where all those points are on a sheet of paper quite trivially, but actually working out when they are is another matter ^^"
144 is very close, including both midnight and noon!
Yay! I guess I missed something while doing a quick solution. I'm wondering what that was, guess I'll find out next week. Still, that is a lot of times to calculate and write out as hours, minutes, seconds, fractions of seconds XD
The swap is only possible when the pointers meet, at 01:05:27 (272727...), 02:10:54 (545454...) etc. The angle they meet is multples of 360/11 degrees. =D
Now all we need is the 4th spatial dimension for seconds.
I can’t believe it’s been 3 years and no one mentioned that pac-man in fact lives on a cylinder rather than a torus. He can only pass from side to side, not top to bottom. You might be thinking of asteroids or any number of other games that have full wrapping and live on a torus.
you could use the minute hand like a cross section of the Taurus. Then you can use hour hand like normal.
2:09 "Rock out with your clock out" I wonder if anyone else caught this. Lol
Since invertig handles on an answer gives us another answer (cause inverting the inversion would give us the original position, which was valid) do we have to send half of the times or both?
T-O-P-O-L-O-G-Y !!!!! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
What is the external space/surfuce of this torus termed?
We have to list one time and another time obtaind by swaping differently or in pairs?
did he just pun-segue his way into the subject? lol "i'll circle back..."
I'm sure Secret of Mana fan's remember this from their childhoods, thinking "how could you fly past the west edge of the world and end up on the east edge AND fly past the north edge and end up at the south... no poles... must be a torus"
I'm really interested in the statement that quotients are dual to products-because coproduct/dual products are something different from quotients
Spot on! Take a look at my response to Wiki Comet below.
-Tai-Danae
The challenge looks very nice. Will only an explicit list of solutions be accepted or can we just express solutions in parametrical form (ex.: t hours, 5+k minutes and 3-t/42 seconds for some values of k and t) ?
Great video !
Please do a video about moduli space
difference between packman and the toroidal lines is packman can always end up back at the same place, whereas torus does not end up on the same place on the paper if followijg one line, whereas packman will if staying on the same line end up at the same place.
Quotients are not dual to products. In the category of sets the product is the Cartesian product of the two sets and the coproduct, the dual of products, is the disjoint union of the two sets. Because a topological space is a set, the product and coproduct in this category will be similar with some topology defined on them underlying set. 'the most natural one'. Quotients are 'dual' to something but not products.
Nice observation! Here are some deets: the Cartesian product and quotients are examples of two constructions in category theory called *limits and colimits.* A product is an example of a limit, while a quotient is an example of a colimit. This is the duality I was referring to in my response.
But you’ve spotted a *much* tighter, more specific duality between the Cartesian product (a limit) and the coproduct (a colimit) versus a [fill in the blank] (a limit) and a quotient (a colimit). Products and coproducts are indeed dual to each other, in a much stronger sense, as are [fill in the blank] and quotients.
So, what is the [fill in the blank]? I’ll give one word for a hint: “(co)equalizer.” :)
-Tai-Danae
+Wiki Comet
Just so you know, "Because a topological space is a set, the product and coproduct in this category will be similar with some topology defined on them underlying set" is not valid reasoning. For example Vect has all small limits and colimits but the coproduct of two vector spaces (direct sum) is *not* their coproduct as sets (disjoint union) with a vector space structure.
The actual reason why limits and colimits in Top are just those in Set with a suitable topology is because the forgetful functor U: Top -> Set preserves limits and colimits. To contrast, the forgetful functor U:Vect -> Set only preserves limits. This is also the reason why products in Vect (also direct sums) are just products as sets with a vector space structure.
To really get to the end of the story, this phenomenon happens because there are two natural ways of creating a topological space from a set (put the discrete or indiscrete topology on it) whereas there is only one way for vector spaces (the vector space with the given set as a basis). That is, the two functors disc,indisc: Set -> Top are respectively left and right adjoints to U:Top -> Set which implies that the latter preserves colimits and limits, but U:Vect -> Set only has a right adjoint and no left one, so it preserves limits but not necessarily colimits.
Answers submitted, looking forward to the next one, guys!
trivially, when the hands overlap we can switch them and end up with a valid time. That's about every 1 hour and 12 minutes.
Shhhhhhhhh, you're telling them too faaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaast
All the intersections of the graph and its flipped version lie on the 11x11 integer grid which has undergone a linear transformation L. This transformation is defined by L(11, 11) = (12, 12), L(11, 0) = c*(12, 1) and L(0, 11) = c*(1, 12). The constants are the same because of symmetry. Calculating c will tell us what L is and now we can easily (with a computer) compute the transformed version of the 11x11 grid.
Need you back on spacetime.
Are we allowed to email an excel spreadsheet with all of the solutions?
Or does it have to be plain text?
And do the fractions have to be simplified?
Either way. Plain text preferred, but as long as we can read it and it's complete, it's ok.
all and only the correct time
do I win now?
2:09, "Rock out with your clock out". Motörhead fan?
OK solution on the way. Nice puzzle, it shows how people think differently, I would never have tackled it the way you did. I really liked the way you used, in your solution a fraction of 11, nice! I sought a solution using a variation of when do the hands overlap puzzle, hence the factor of 11. Only it's no longer 11, is it? OK no more help for those struggling. :-D
To study topology of universal timespace you need to know these docs first : (then comes electronic circuit logic as xor nor nand... and then comes differential fluid dynamics and lastly comes quantum mechanics alongwith laminated -4d brane strings...)
Hardcore mode : add the second hand in the problem and accept any permutation of S_3
Great. Now I can't sleep. Thanks a lot.
I yet showed that hour-minute swap with second still can't be done.
Sure it can.
Just as we can visualize the hour/minute problem as a line looping through the segment of a plane, pac-man style, we can also visualize the hour/minute/second problem as a line looping through 3d space pac-man style.
One noteworthy difference though is that in 3d two non-parallel lines are not guaranteed to intersect, so it's possible that there is only one solution to the problem at 12:00:00, but I've yet to prove it.
Yes, that was my mean ;)
If I note S_3 = { e, (hs), (hm), (ms), (hms), (hsm)} h=hour, m=minute, s=second
I proved (hm) is only 00:00:00. Wasn't hard, just has to "do the stuff".
Yet other cases should be harder, and I'm fucking lazy.
I'm guessing that would be 3 * 720^2 equations to solve, right?
Each line in the original problem would be split into 60 lines, one for each second. You would need to solve each line with each corresponding line in a rotation, hence squared, and there are three ways to rotate the problem. Sounds doable, but maybe not by hand.
Edit: That would be sets of equations, as you need to solve a matrix for each combination.
I also think it would be enough to solve one combination, as the two other can be found by rotating the solution. We are now down to half a million matrices to solve. Maybe we actually only need to solve very few of them to show if there exists a solution, as every hour after the first are just the same thing moved a bit, but I might be wrong about that.
Pacman is played on a cylinder (only one edge loops around). The videogame analogy you want is asteroids.
I'd argue that all of them take place on tori -- it's just that in some versions, there are only _openings_ on one set of opposite walls. But in some versions, the openings are both sets of opposite faces, and the toroidal topology becomes manifest. Check out some of the screens on Pac-Man Battle Royale, for instance.
Challenge isn't that hard if problem is understanded because every question already contain half of answer in itself: you can swap them only when they align (when are paralel). Since small arow does 2 full circles per day, and big arrow does 24 circles per day, they should intersect 23 times. So there is 23 moments in a day (24 hours) when you can swap big and small arrow and still tell correct time
There are many more solutions than just the trivial overlapping. For example 12:50 and 10:04 form a valid pair.
All solutions are the intersections of the clock state square overlapped with a bottom left diagonal flipped coppy of the square
There are 144 possible states that are valad if swaped
First Phi Day and then e Day pass with nary a mention? I thought you'd be all over celebrating mathematical holidays.
You are back!?
"Its [tori] all the way down" - I don't know who said this, but I just did