Shapes and Hook Numbers (extra footage)
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- Опубликовано: 5 сен 2024
- MAIN VIDEO: • Shapes and Hook Number...
Featuring Professor Curtis Greene from Haverford College.
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You guys should make a numberphile3 for the real nerds who want extra footage for the extra footage. I'd subscribe.
Please make Numberphile3 for really longform interviews/explanations, like an hour+. That'd be so freaking amazing.
that sounds like a lot of effort .
The question is, does this work in n-dimensions... So for instance can you do it with a cube of 27 blocks?
That would be cool.
Seems that it should. Each hook would have 3 "branches" then (in directions X, Y and Z).
Can there be another video on this? Cause I could watch this for days
"The probability police"… I like this guy. XD
I feel cheated. First I didn't get the proof in the original video. Then I go into the bonus content and get one false proof and one incomplete proof.
that procedure for picking where to put the largest number, how is it different to picking a corner cell at random?
+LeoMRogers The distribution of largest numbers in corners is not uniform.
+TheCakeIsAlive I mean, for some tables it is, but generally it's not.
+LeoMRogers
It takes into the account the shape of the tableau.
A simple example would be a 2 1 1 1 1 1 tableau. (an elongated L shape)
This shape has 2 corners, but if you follow the hook walk you would likely end up at the bottom corner rather than the right corner. So the probability of reaching either corner isn't 50%.
Actually the probability of reaching the right corner would be 1/7 + 1/7*1/6 = 1/6
The first term comes from the probability of picking right corner first. The second term comes from the probability of picking the top left cell and then picking the right corner. There is no other hook walk that will reach the right corner.
+Alex Quintero Thanks!
+Alex Quintero
Wouldn't that be a bit over 1/7, not 1/6?
What if you math around with this in 3 dimensions?
This gives me a headache. I'll try it, though, your idea is pretty interesting!
+Jonathan Dirks Why not just go for n dimensions?
Aryan Arora Of course, but what matters is the formula for determining how many ways there are to arrange the numbers in three dimensions.
I Passed The Turing Test same formula, number of cubes factorial divided by the product of all hook numbers, but the hook numbers would need to be redefined, for 2-D we use upside down L to define a hook number of a square box in a 2-D tableau, or we use the negative y axis and positive x axis to define a hook number of a square box in a 2-D tableau, so for 3-D, hook numbers for cubes in a 3-D tableau will be defined by using the negative y and z axes and the positive x axis, and 3-D tableaux will be left justified again, the left on the x-axis that is
I Passed The Turing Test we will use cubes instead of square boxes
I had been waiting for this video since the other one came out!
how could I have watched this before it was even uploaded
Great set of videos ! Back in the ancient past when I was doing particle physics I remember using Young tableaux to characterize group representations. Given that this old brain would never be able to recover that stuff and I'm too lazy to look it up, how about a follow up video on the relationship between Young tableaux and group representations ? :-) That might provide a gentle, numberphile style, intro into groups and group representations (he says without the slightest idea of whether that would actually be possible !)
What if you have a hexagonal tableau made up of hexagons? Or triangles?
Can we also get a video on the RSK correspondance?
I feel like Professor Greene knows the secrets of the universe and if I just listen to him long enough some of them might slip out. Thank you.
My take on the SYT arrangements of a 2x2 tableau is :
- you need 1 to appear in the upper-left cell, with probability 1/4
- then 2 to appear in the upper-right or lower-left cells : 1/3
- finally, 3 to appear in the empty previous cell, and 4 in the lower-right cell : 1/2
Because of step 2, you have 2 valid arrangements, and therefore get this probability of a random 2x2 tableau being a SYT : 2*(1/4)*(1/3)*(1/2)=1/12
I want to learn more about combinatorics and representation, I would be grateful truly if u could recommend an introductory book?
We need to go deeper
I would hazard a guess that the reason that multiplying the separate probabilities gives the correct answer is that although placing A seems to have an effect on C and B, it will have the exact same effect on D. Therefore my guess is (much like cancelling down during the division shown in the other part of this video), these effects can be removed from the computation.
Absolutely fascinating!
It's Pi, standing for the P in product. Much like Sigma standing for Sum with series. (And Delta stand for Durchschnitt meaning intersection, but that's a different field.)
One way to see that there's got to be something wrong with that proof is that the proof would be just as convincing if, instead of finding the probability that a number would be the largest in its hook, you found the probability that the number is larger than both its right neighbor (if any) and its bottom neighbor (if any). Again, you've found the probabilities of a set of conditions which, taken together, establish that it's a SYT. This works for the 2x2 square also, but gives the wrong answer for the 3x3 square. There's no obvious reason that checking the whole hook, rather than letting the fact that the neighbors are also being checked against their neighbors take care of the rest of the validation would work any better than checking just neighbors. Or, for that matter, checking all the cells below and to the right as well.
so just a thought, to make the amount of SYTs as big as possible for a specific number of boxes n, an approach would be to look at it and see that the only thing you can influence in the formula is the product of the hook numbers(as n is non-variable, so is n!). To get this as small as possible, what shape is the best? My best guess would be either a giant hook-shape, or the shape drawn at 12:30.
so basically the hook numbers are the minimum number of values that must be greater than or equal to the number in the square chosen. in theory I could see how 1/(N-H) N being total number of squares H being the hook number could factor into this a bit.
Amazing video. It is quite interesting that the probabilitys are not independently distributed and still the proofs seems to fit. Is it true that you can calculate the probability of these events by taking the product, even if they are not independently distributed in this case?
You should do a numberphile/sixtysymbols crossover video about symbols for math logic. Cheers!
Question. If prb(B
I think it's because of what we are comparing, In prb(B
What is the reversed writing that we can see projected across his shirt around 12.5 minutes in?
get such good ASMR watching this
If you use the rule at 15:40 it seems to me that there would be a higher probability of choosing cells towards the center than at the ends. That's not uniform.
it's disgusting that I have a video on my channel that is on its way to having more views than this video (an evening in the life of a typical math PhD student).
There has to be a reason that proof works.
You know, just because I'm in the habit of doing this:
1/12 p=
8 1/3%,
11 to 1 against (like no one could figure that out on their own),
-10.41393 deciban,
-3.45943 bit or Sh.,
-2.3979 nat and
-1.04139 ban, dit or Hart.
pretty fun stuff
this is, excuse my math jargon, fucking fascinating! thanks
I thought it looked like Ingvar Kamprad in the thumbnail, I got excited over the potential of IKEA-maths.
Now please give useful formula for computing # of semistandard Young tableaux!
Isn't the chance for each hook number independent as long as you work your way out from the top-left?
I would think that's the reason for the proof being correct.
Why are these unlisted? It's already a separate channel anyway.
What has this to do with representation theory of symmetric group on n letters?
The reason this prove works is because the probabilities are actually independent. Knowning that all entries are different, whatever or not a hook has the correct velue on its entry, doens't provide any information on the other hooks.
I'm confident this can be demostrated.
The reeasons being:
A) that a correct value on the entry of a hook doens't constraint the order of the other elements of that hook, and also there is no sub-hook that contains the entry of the hook.
B) There are super-hooks that contain the entry of a hook, but knowing the order of one branch of the hook doens't affect the probability of having the correct value in the entry of the hook.
Really nobody has done this? Where do I sumbit a paper?
+Alfonso J. Ramos (theraot) Nah he's right:
A and B are independent if and only if the probability of A is equal to the probability of A given B. So P(A) = P(A|B). (A|B = A given B which means the probability that event A will occur given that B has already occurred.) This is intuitive because if they truly are independent then if B has already occurred then it should not change the probability that A occurs.
So for instance, when considering A,B,C,D to be equal to any of 1,2,3,4, and all distinct, B
Alex McGaw Right... I'll claim brain damage on this one o.o
But the probability proof didn't use the fact that the numbers are in order and/or that you're using 1,2,3 rather than 1,4,9. The proof doesn't take into account what are you ordering.
I mean, it'll work with the odd numbers and work to make the numbers in descending order
You're right - the proof will hold for any four distinct numbers. But in the video (and in standard Young tableau) the numbers 1,2,3,4 are used so that's what I used.
I haven't tried this yet - but I wonder if it would work if I followed the following algorithm:
1.) randomly distribute the n numbers in the shape.
2.) keep sorting rows and columns so that they are in increasing order in both directions until all rows and all columns are correct
if that works there will be a one to one mapping between ANY random placement of the n numbers and the standard tableau that you reach by the above algorithm (IF the above algorithm is correct).. sort of dividing all n! random distributions of the n numbers into m different equivalence classes, where m is the number of different standard tableaus. perhaps i will try that but if anyone knows that it WONT work let me know ;-)
+Hemmeligt Navn actually - I am almost convinced that it MUST work cause his technique trickles the biggest number to a corner - the fist sorting round would do that and then the next would fix the second largest etc.
...Wouldn't it be the same if you just chose at random among the corner cells, instead of performing the "hook walk"?
+palmomki I guess not all cells with hooknumber 1 can hold the largest number in a valid SYT. I had the same though when thinking about a method to cycle through all the possibilities of a given shape.
+Diggnuts All corner cells can hold largest number. Reason that you cant pick that way is that not all corner cells hold largest number in same number of configurations, so picking corner cells uniformly random doesnt pick configurations uniformly.
Bruno
Ah yeah, that makes sense. I think!
+Diggnuts copied from my other comment:
A simple example would be a 2 1 1 1 1 1 tableau. (an elongated L shape)
This shape has 2 corners, but if you follow the hook walk you would likely end up at the bottom corner rather than the right corner. So the probability of reaching either corner isn't 50%.
Actually the probability of reaching the right corner would be 1/7 + 1/7*1/6 = 1/6
The first term comes from the probability of picking right corner first. The second term comes from the probability of picking the top left cell and then picking the right corner. There is no other hook walk that will reach the right corner.
+Alex Quintero
Wait, I'm confused by your explanation. Wouldn't the probability of reaching the bottom corner be less, because you'd either have to pick the bottom corner at random on the first "roll" or you'd have to pick the top left corner (where 1 must go) and then have to pick the bottom corner (since you pick any cell at random).
Did you just miss speak, because 1/7+1/7*1/6 seems to be exactly that probability...
When are you going to win the lottery?
Did you get any footage of the sunset outside?
what its use?
1 over 12(negative and otherwise) keep making appearances on Numberphile.
wait a minute, I don't like this proof at 5:42
Probability that A is the smallest of the three numbers is 1/3. OK, I can live with that. But why doesn't he say next, that similarly, D is the largest of B,C,D with the probability of 1/3, but makes it two lines with 1/2? My answer is that because he knows the answer to be 1/12, so he asks the questions in a way to get there. That is not a proof to me, because he might have concluded 1/9 in the same way (A is smallest from A,B,C - 1/3, D is largest of B,C,D - 1/3) of 1/16 (A
I'm just guessing here, but I think he took the approach he did because he used a more general approach that can be used for multiple different kinds of tableux. The way you presented works if there is only one corner boxes that the highest number can go into, but what if you have a shape with multiple corner boxes? Then it breaks down, as you can place the biggest number in any of the corner boxes, so you cannot say with certainty that a particular box's number is bigger than all the other numbers in the shape, as there are then multiple option for which box has the biggest number.
How is the processes of "hook walk" different from just picking a corner cell at random?
+Arthur Denzlin It's to "generate a standard young tableau uniformly at random". Uniformly is a key word here. The method gives a greater chance for corner cells with more pathways to them to be chosen; why this is necessary I don't understand clearly enough to explain further. I hope this helps you anyway :)
Thanks kangawallapuss. I wasn't think about the "uniform distribution " part! Now it makes sense
Has this work been generalised into N dimensions?
Not independent but exchangeable, right?
maybe the "probability proof" isnt false, but it just ISNT complete :D
why is this video unlisted?
+thegreyknight364 I believe he always unlists the extra videos for a while to ensure people see the main video first.
Yes, he's probably making it public when the next Numberphile video is released (with its corresponding extra footage being unlisted).
make numberphile2 make it toule
0:19 **writes 3x7**
sounds like post probabilities. Bayesian stuff
He looks like my grandpa...
incorrect proof of a correct result... this is a mathematical paradox
+jane smith No, it's not.
+Daniel Sampaio so is a paradox the correct proof of an incorrect result? and if yes, what was in the video then?
The correct proof of an incorrect result would indeed be a paradox, for that is impossible. As for an incorrect proof of a correct result, you might term it a coincidence, if you like; or rather, there may be a deeper reason why the incorrect proof works; there may be errors that cancel each other, and if you track them down carefully enough, you can correct the proof.
+Daniel Sampaio Oh cool. thanks!
(-:
42!!??!?!?!? the answer to life? the universe!? EVERYTHING!?!?!
24024!?!??!!??!?! THATS 14!/10!
hahahh coincidence!?!?!?!
144144 weeks in seconds meets 6 weeks in seconds
the number of variants in a 4x4
harmony
Third
Last
first