If you want to make primes with LEGO bricks, you don't glue bricks together. You use technic bricks which come with versions with 2, 3, 5, 7, 11 and 13 holes.
@@jessehammer123 I know it's not, and that is indeed a good example of why! It's just funny to put those words together in that order and have it mean something real, and maths is kind of the only domain where that happens.
I spent the whole episode waiting for a 3B1B cameo and the minute I look away I hear Grant's voice. Then I look back and it's coming out of Matt's mouth 😅
i think it's normal - i'd have to watch this again to say i definitely understand the math 100% but it seems reasonable that the site is only searching a certain "region" of the graph at a time and some regions just won't have any hits. alternatively it might be double checking previous work (if there is some reason that some processors would make a mistake, idk) and reporting zero *new* hits. as you keep trying, you'll see the numbers changing and you'll get results sometimes.
No, this is not the conjecture. The conjecture is that, for any e>0, there are only finitely many coprime triples (a,b,c) such that c > rad(abc)^(1+e). The case q=2 (in other words, e=1) is only one of infinitely many cases that the conjecture considers!
Oh yeah, gosh. Have there been any news on that since Mochizuki's own journal went ahead and published his proof, without clearing up Scholze and Stix's concerns at all? I couldn't really find anything more recent.
Where did you get your shirt? "Geek" in binary feels exactly like the sort of shirt I'd wear but I've never seen one before that doesn't also have horrible logos or some insulting comment at the bottom claiming the wearer is smarter than the reader.
Correction at 14:47 and 14:52. For rad(a*b*c) to be equal to a*b*c, you don't need them to be all prime: you only need them to be *relatively* prime and that each of them equals its own radical, i.e., rad(a) = a, etc. In other words, all prime factors of a, b and c combined must appear only once. Secondly, the "kind of exponential shape" (the upper bound) is actually not an exponential, but a cubic (which is hinted at near the end of the video). For a given value of c = a + b, the maximum value of a*b*c is when a ≈ b ≈ c/2, and so a*b*c ≈ (c^3)/4. Looking at the plot for c = 21, we see a dot very close to the curve. And indeed, if you consider 10 + 11 = 21, you get rad(10*11*21) = 2*5*11*3*7 = 2310, which is very close to (21^3)/4 = 2315.25.
To be honest even after watching this whole video I still don't understand the ABC conjecture. And I first heard about this conjecture like 10 years ago.
Great video as always. This is the first explanation I've seen that clarifies the ABC conjecture enough for me to feel like I understand the problem (if not the solution). Thanks!
log estimation tip: The log of a number is roughly the count of the digits in the number, when the log base is the same as the base of its argument. log(238140) ~ 6 and log(210) ~ 3, so log(238140)/log(210) ~ 6/3 = 2. So we know the answer will be greater than 1 and less than 3. But since the beginning digits of the numerator 238... is larger than 210, The answer is greater than 2 and less than 3. This is sometimes useful for me. Like in your video, I had the estimate in my head right away: 6/3 = 2 Not a complex tip, but has been useful for me.
It seems that Mochizuki's alleged proof still contains some hand-wavey parts. If you don't accept its validity, you just "haven't understood it properly", which they can always repeat, no matter what their opponents say.
8:18 follow up. Matt, I spotted a mistake: the quality is not the average height. In your example: n = 238140 = 2^2 * 3^5 * 5^1 * 7^2 radical = 2*3*5*7 = 210 quality = log(n) / log(radical) = ~2.315 Lego construction base = 2+3+5+7 = 17 area = 2*2 + 3*5 + 5*1 + 7*2 = 38 avg height = 38 / 17 = ~2.235 The quality and avg height are definitely related in some way, but they're not equal. See my other comment as well.
@@SmileyMPV I'll Grant you that, but average height sort of implies arithmetic mean. Anyways if he didn't mean arithmetic mean, which mean did he mean, know what I mean?
@@joseville It's the weighted arithmetic mean of the stack heights, weighed by the natural logarithm of the prime base. Indeed, for n=p1^a1...pk^ak we have q(n) = log(n)/log(rad(n)) = (a1*log(p1)+...+ak*log(pk)) / (log(p1)+...+log(pk)).
“I have discovered a truly remarkable proof which this margin is too small to contain.” (hint: it uses the proof of Fermat's Last Theorem to work the other way)
Ngl, the vague phrases of "the radical of all these together" and "the quality of the triple" (probably the technical phrasing, but I digress) had me lost for a bit. I'm really glad you showed the comparison of the quality against the epsilon term, that really told me what that meant lol.
You've broken a seven! That gives you seven to the power of seven years of back luck. I predict that, in a bizarre accident, the internet will fall on you.
8:18 Matt, I think I spotted a mistake and somebody please correct me if I am wrong or confirm if I'm right, but the quality is not really the average height. Given a number whose prime factorization is a^x * b^y * c^z The average height of the corresponding Lego construction is (a*x + b*y + c*z) / (a + b + c) Meanwhile the quality of said number is log(a^x * b^y * c^z) / log(a*b*c) Which is equal to (xlog(a) + ylog(b) + zlog(c)) / (log(a) + log(b) + log(c)) Which is not at all equal to the average height! Average height looks like A/B meanwhile the quality likes like log(A)/log(B). There is definitely a relation between quality and average height, but that relation is not an equality relation. I'm curious if given the quality, how do you manipulate it to get the average height or vice versa.
I believe it is a mistake. When he said that it just sounded off for some reason, so I calculated the actual area of all the blocks (2*2 + 3*5 + 5*1 + 7*2 = 38 if we say the height of one block is 1) and the length of all the blocks (2 + 3 + 5 + 7 = 17) and with quik mafs, 38/17 = 2 + 4/17 < 2.25. Sure enough, a calculator confirmed that the average height is 38/17 ~ 2.2353. Once I realized this I started scrolling through the comments to see if anyone else thought this or if it was just me, so I’m glad to see I’m not the only one lol. You also went a step above me and found the general relation, so props to you!
@@MathIsFun137 Same here! It sounded off to me as well. I made a related comment as well. In the comment thread for the other comment, it was concluded that the quality is actually a weighted arithmetic mean, but even if that's what Matt meant he didn't communicate it very well.
It would have been even funnier if at 18:00 there were 3 Matt with a Blue shirt and q Matt with a Brown shirt. Then no one would ever confuse it with a 3Blue1Brown video for sure.
7:41 What you appear to be doing is calculating roughly the average height of the stack. Take 900, for example. 2*2*3*3*5*5. The stack is two blocks tall. log(900) / log(30) gives exactly 2, no matter the base. Because the bricks are exactly 2 blocks tall. Same with 27000, its quality is 3, its tower is 3 blocks tall. Larger primes being repeated gives a higher quality than smaller primes being repeated, but if all of them are repeated an equal number of times, the quality is that whole number.
8:17 yeah!!!! It is the average height! Roughly, at least, not sure the bias exactly matches with the width of a brick, I don't actually know how they interact. But yes, the height, the height!
As someone who plays Four 4s on a regular basis, 210 is very familiar to me, since it's easy to get with one 4 (σ(4)#, aka sigma 4, and then the primorial of that).
5:53 this is an interesting equivalence relation on the set of natural numbers, there is literally an equivalence class that is 2^x for all natural numbers x
I like how the whole time Matt's talking about and looking towards these graphics that someone else made for the video, there's just no graphic on screen and it's an ad pitch for the wall decorations.
Matt is talking about a conjecture, which if proved would give an alternate proof for likely the second most famous theorem in maths, and only starts mentioning that fact 20 minutes into the video. Classic Parker move
That graph... I got lost, but the conclusion i got from "if you zoom out far enough they're essentially the same" is that low margins of error = no margin of error... That sounds like a parker mentality to me all right! 🤣
We got a Parker Square update video, a Hannah and Matt video, and 4 days in a row with Matt videos being released. Christmas came early this year! I wish Parker Christmas lasted forever!
The quality is like the dual of the (log of) geometric average. Quality is arithmetic average power of primes weighted by log prime, while log of geo average is arithmetic average of log prime weighted by prime powers.
Two minor corrections: at 24:08ish you appear to write rad(abc)≤abc≤xyz when you might have meant to replace abc with xyz or rad(abc)≤xyz≤abc At 5:13ish you said the only odd length 2*n lego brick you could find was 2*3 forgetting the trivial 2*1 case.
@HBL (I don't know why, but the answer I've sent you doesn't appear under your comment. I'll copy it and try again; sorry if I end up double-posting.) I was also confused by that, so I took out my calculator. The quality here is Q = log(C)/log(rad(A*B*C)). So, for the first row, you get Q = log(9)/log(rad(1*2^3*3^2)) = log(9)/log(2*3) ≈ 1.226. If you take the example at 17:45, you get Q = log(6,436,343)/log(2*3*109*23) ≈ 1.6299.
When you first pulled out the blocks I thought you were going to present the following problem. What is the minimum number of blocks required to spell all words of N length?
There's no way I could confuse this video with 3Blue1Brown. You're wearing a red t-shirt - this is clearly a Tom Scott video.
Underrated comment
Yes, red t-shirt is car away from a blue Pi.
I cackled. Great joke.
I wanted to make a 3b1b joke, but you nailed it, managing to bring Tom in too. Well done.
I laughed embarassingly loud at this comment, thank you
Matt: I can't use an O as a zero.
Also Matt: Lobster is a plus and Peas are equals.
That's not a lobster. It's a plustacean.
Ahem, "peaquals"
@@Eagris . . .
@thomas vargas but fractions are division. You’re going to divide to get zero?
Mathematical standards have no rules about what the lobsters and peas represent.
I’m worried that Matt has somehow got himself into a duel at dawn, and is trying to get all his maths out in one night.
I wonder which clone issued the challenge? I hope it's not Beard Matt.
Is this a Galois reference?
This should be top comment!
I get that reference!
If you want to make primes with LEGO bricks, you don't glue bricks together. You use technic bricks which come with versions with 2, 3, 5, 7, 11 and 13 holes.
LEGO Technic, the official LEGO theme of mathematicians who work with prime numbers.
This is how physicists do maths. It's true but with a finite number of exceptions
bricks?
@@beaclasterlego bricks
Geez Matt is going crazy here with the upload schedule!
He's spoiling us. And a numberphile video too!
He's going crazy to hit 1M subscriber ASAP
A sprint for 1 million? Or does he know something we don't and is manipulating the youtube views market in a dark and mysterious ways?
@@QuantumHistorian probably the latter
He filmed these ages ago and then edited them all and uploaded them in a very short span of time.
"Will you accept a lobster as plus?"
I have to say, I've never heard that one before
In my experience, the presence of a lobster has invariably been a negative 🦞
Hail Lobster
@@archivist17 How dare you!
Obviously the Parker = Not Parker conjecture is the hottest thing in the math research world recently, as in 2019
He's a Parker Parker!
Parker = Not Parker confirmed because we can tell he's working very hard.
in case you are wondering, the tshirt says "Geek".
Yeah... I did the work though.
@@cpt_nordbart Me too :-)
I also did it, but thought it said Leek 😛🤷♂️
@@timangus I thought it said Guuk. I guess I gave it a Parker go, and misplaced a bit. Geek makes way more sense.
Can you explain how? I get that it’s binary, but how does that become the letters?
19:52 I see your unspoken flex of one-upping a log-log axis with a loglog-loglog axis. I've never seen it done and I am so impressed.
"I'm gonna use this to measure the earth!"
The security staff at the Shard: I'm about to end this man's whole career
Soooo True.
Whole video*
I mean, they still got a pretty reasonable estimate for R
Oi! You got a loisence for that protractor?
@@harry.tallbelt6707 They were within an order of magnitude or something. In the overarching scope of the universe, they were basically 100% accurate.
Thank you Matt for the opportunity to highlight this topic, we are really overwhelmed by the contributions ❤
Wow, that looks like a really dangerous weapon at the end there
Now I'm curious if this is just a short burst of extra creativity or if Matt has upped the ante, gunning for the 1mil subs.
Let's hope it's the latter, but also for him not being satisfied with only 1mil. He might still try to surpass Steve Mould.
I feel like we're kind of getting spoiled here.
he's still isn't there yet :(
8:48 Nice catch, you have Matt-like reflexes
Matt: "Measure angles?"
Beard Matt: "Measure the Earth"
Matt: (Mumbles behind his own bearded self) "Measure badly"
To be fair, it was the mr-i-don't-understand-humour security guy's fault for taking away his protractor.
"... in general. It turns out there are infinitely many exceptions." Mathematicians lol :P
That’s not as crazy as it sounds. Positive integers in general are not prime, but there’s infinitely many exceptions.
@@jessehammer123 I know it's not, and that is indeed a good example of why! It's just funny to put those words together in that order and have it mean something real, and maths is kind of the only domain where that happens.
That part made me laugh
Cracked me up when it went into Grant's voice!
End of the video comment:
Matt could have given away two Matt sized Cubits, but Hannah Fry seemed to have "misplaced" them. ;)
18:18 "No one is gonna confuse this with a 3 Blue 1 Brown video".
yes. it's a 3 Red 1 Beard video lol
It's Brown. the Beard is Brown.
2:50 "It pains me to use an O for a 0..."
Casually uses lobster for a plus.
Good-sized fresh lobster is always a plus
I spent the whole episode waiting for a 3B1B cameo and the minute I look away I hear Grant's voice. Then I look back and it's coming out of Matt's mouth 😅
Exact same thing happened to me hahaha
I love the way Matt pulled that one. I'm sure quite a few of us were hoping this would lead up to a cameo, but nobody expected it to happen like that.
You broke the 7! How? It has no factors!
He broke maths too
yea it does, 7! = 2^4×3^2×5×7
he tried so hard to avoid being 3blue1brown that he turned into science asylum.
i for one am loving the mattiverse
"Snappy, specifically the peas." I don't know Matt, I think the lobster is pretty snappy too...
Gotta love maths, where there's a perfect rule with infinite exceptions.😂
Exquisite syncing on Grant's voiceover :D
i was promised something and got "We found 0 triples for 269 ≦ rad(a) < 270, 417 ≦ rad(b) < 418." Thank you.
i think it's normal - i'd have to watch this again to say i definitely understand the math 100% but it seems reasonable that the site is only searching a certain "region" of the graph at a time and some regions just won't have any hits. alternatively it might be double checking previous work (if there is some reason that some processors would make a mistake, idk) and reporting zero *new* hits. as you keep trying, you'll see the numbers changing and you'll get results sometimes.
It took me several tries to get a non-zero result. Le sigh
"12.38, uh, and some stuff" spoken like a true mathematician.
18:00 That's some top tier editing
If I understand correctly. The ABC (A+B=C) conjecture currently is:
prove that q
No, this is not the conjecture. The conjecture is that, for any e>0, there are only finitely many coprime triples (a,b,c) such that c > rad(abc)^(1+e). The case q=2 (in other words, e=1) is only one of infinitely many cases that the conjecture considers!
At this pace, matt will have finished maths in about a month (except the magic square of squares ofc)
I wondered what Matt's take on the IUTT controversy was. I'm glad to see he's an adherent of the Redundant Copies School.
Oh yeah, gosh. Have there been any news on that since Mochizuki's own journal went ahead and published his proof, without clearing up Scholze and Stix's concerns at all? I couldn't really find anything more recent.
Matt I love how you've clearly taken steve's advice to heart! Love that recent videos!
You went from a log-log plot to a log-log-log-log plot at about 20:00. I'm not sure I believe the graph actually goes 10^0, 10^10, 10^100, 10^1000.
yes, the steps change from multiples to order of magnitudes
Where did you get your shirt? "Geek" in binary feels exactly like the sort of shirt I'd wear but I've never seen one before that doesn't also have horrible logos or some insulting comment at the bottom claiming the wearer is smarter than the reader.
“GEEK” in ASCII?
“Geek” to be precise
@@zXrabidrabbitXz I wondered about the case but not enough to check myself. Thanks, Jason!
@@MrConverse You can see by the 6th bit (3rd from the left): 0 = upper case, 1 = lower case.
@@Wouter10123 Yep!
You're an amazing content creator Matt! Keep doing great maths!
Never had I thought that two Matts could overlap! Wild stuff happens when you rush the upload schedule.
5:18 - You case use acetone to weld them into a single piece. ABS will melt and weld nicely with acetone.
As soon as Matt got the 2x5 and 2x7 lego pieces, my brain flipped. "They don't exist do they?" Later on... "ah..."
Yes, Mr. Maths I will accept the lobster as plus, thank you very much for being considerate
Correction at 14:47 and 14:52. For rad(a*b*c) to be equal to a*b*c, you don't need them to be all prime: you only need them to be *relatively* prime and that each of them equals its own radical, i.e., rad(a) = a, etc. In other words, all prime factors of a, b and c combined must appear only once.
Secondly, the "kind of exponential shape" (the upper bound) is actually not an exponential, but a cubic (which is hinted at near the end of the video). For a given value of c = a + b, the maximum value of a*b*c is when a ≈ b ≈ c/2, and so a*b*c ≈ (c^3)/4. Looking at the plot for c = 21, we see a dot very close to the curve. And indeed, if you consider 10 + 11 = 21, you get rad(10*11*21) = 2*5*11*3*7 = 2310, which is very close to (21^3)/4 = 2315.25.
To be honest even after watching this whole video I still don't understand the ABC conjecture. And I first heard about this conjecture like 10 years ago.
hands up everyone who went looking for a binary-to-ascii converter to read his shirt. Well done, sir.
No need - I have ASCII memorized from much time spent poring over TRS-80 Color Computer ROM disassembly listings in the 80s. 🤓
15:56 - I'm interested in whats going on where the lines cross on that log plot
2:26 A natural bijection of w+1 and w sends w to 0.
Matt's t-shirt says, "Geek."
Poor bearded Matt has no idea what is in store for him and Hannah
my most favorite 3blue1brown so far! Nice work Grant.
Great video as always. This is the first explanation I've seen that clarifies the ABC conjecture enough for me to feel like I understand the problem (if not the solution). Thanks!
That's a T Shirt only a Geek would wear. I approve.
1
18:23 More like a 3Red1Matt video lmao
What an irony that the abc conjecture is one of the hardest ones to prove
log estimation tip: The log of a number is roughly the count of the digits in the number, when the log base is the same as the base of its argument.
log(238140) ~ 6 and log(210) ~ 3, so log(238140)/log(210) ~ 6/3 = 2. So we know the answer will be greater than 1 and less than 3. But since the beginning digits of the numerator 238... is larger than 210, The answer is greater than 2 and less than 3.
This is sometimes useful for me. Like in your video, I had the estimate in my head right away: 6/3 = 2
Not a complex tip, but has been useful for me.
It seems that Mochizuki's alleged proof still contains some hand-wavey parts. If you don't accept its validity, you just "haven't understood it properly", which they can always repeat, no matter what their opponents say.
Corollary: squaring a number doubles the quality
now i need A 🦞 B (Pea pod) C t-shirt 😭
Matt’s videos are like London buses… you wait ages for one and then three come at once 🤣
8:59 Ah yes, the famous "Parker vs. Non-Parker" conjecture.
It isn't a 3blue1brown video, it is a 3matts1graph video
Matt's shirt says "Geek" in ascii, to save you the trouble lol
8:18 follow up. Matt, I spotted a mistake: the quality is not the average height. In your example:
n = 238140 = 2^2 * 3^5 * 5^1 * 7^2
radical = 2*3*5*7 = 210
quality = log(n) / log(radical) = ~2.315
Lego construction
base = 2+3+5+7 = 17
area = 2*2 + 3*5 + 5*1 + 7*2 = 38
avg height = 38 / 17 = ~2.235
The quality and avg height are definitely related in some way, but they're not equal.
See my other comment as well.
Average is a more general term than just the arithmetic mean. See for example geometric mean and harmonic mean, which are also averages.
@@SmileyMPV I'll Grant you that, but average height sort of implies arithmetic mean. Anyways if he didn't mean arithmetic mean, which mean did he mean, know what I mean?
@@joseville It's the weighted arithmetic mean of the stack heights, weighed by the natural logarithm of the prime base.
Indeed, for n=p1^a1...pk^ak we have q(n) = log(n)/log(rad(n)) = (a1*log(p1)+...+ak*log(pk)) / (log(p1)+...+log(pk)).
“I have discovered a truly remarkable proof which this margin is too small to contain.” (hint: it uses the proof of Fermat's Last Theorem to work the other way)
Who else was wondering "Where's he going to get five- and seven-stud Lego blocks?"
Then to hear GLUE! Matt, I thought more of you than that!
Ngl, the vague phrases of "the radical of all these together" and "the quality of the triple" (probably the technical phrasing, but I digress) had me lost for a bit. I'm really glad you showed the comparison of the quality against the epsilon term, that really told me what that meant lol.
You've broken a seven! That gives you seven to the power of seven years of back luck. I predict that, in a bizarre accident, the internet will fall on you.
8:18
Matt, I think I spotted a mistake and somebody please correct me if I am wrong or confirm if I'm right, but the quality is not really the average height.
Given a number whose prime factorization is
a^x * b^y * c^z
The average height of the corresponding Lego construction is
(a*x + b*y + c*z) / (a + b + c)
Meanwhile the quality of said number is
log(a^x * b^y * c^z) / log(a*b*c)
Which is equal to
(xlog(a) + ylog(b) + zlog(c)) / (log(a) + log(b) + log(c))
Which is not at all equal to the average height!
Average height looks like A/B meanwhile the quality likes like log(A)/log(B).
There is definitely a relation between quality and average height, but that relation is not an equality relation. I'm curious if given the quality, how do you manipulate it to get the average height or vice versa.
Parker's mean
I believe it is a mistake. When he said that it just sounded off for some reason, so I calculated the actual area of all the blocks (2*2 + 3*5 + 5*1 + 7*2 = 38 if we say the height of one block is 1) and the length of all the blocks (2 + 3 + 5 + 7 = 17) and with quik mafs, 38/17 = 2 + 4/17 < 2.25. Sure enough, a calculator confirmed that the average height is 38/17 ~ 2.2353. Once I realized this I started scrolling through the comments to see if anyone else thought this or if it was just me, so I’m glad to see I’m not the only one lol. You also went a step above me and found the general relation, so props to you!
@@MathIsFun137 Same here! It sounded off to me as well. I made a related comment as well. In the comment thread for the other comment, it was concluded that the quality is actually a weighted arithmetic mean, but even if that's what Matt meant he didn't communicate it very well.
It's about time Matt dabbles a bit with quality.
Wait, I think this is the first time I saw a graduate scholarship sponsored an educational youtube video. I'm crying...
very geeky tee :) love it
It would have been even funnier if at 18:00 there were 3 Matt with a Blue shirt and q Matt with a Brown shirt. Then no one would ever confuse it with a 3Blue1Brown video for sure.
7:41 What you appear to be doing is calculating roughly the average height of the stack. Take 900, for example. 2*2*3*3*5*5. The stack is two blocks tall. log(900) / log(30) gives exactly 2, no matter the base. Because the bricks are exactly 2 blocks tall. Same with 27000, its quality is 3, its tower is 3 blocks tall. Larger primes being repeated gives a higher quality than smaller primes being repeated, but if all of them are repeated an equal number of times, the quality is that whole number.
8:17 yeah!!!! It is the average height! Roughly, at least, not sure the bias exactly matches with the width of a brick, I don't actually know how they interact. But yes, the height, the height!
Can't believe you broke a 7, that's like 4+3 years of bad luck
You usually don't see longer odd-numbered LEGO pieces because you can build up anything from the smaller prime pieces of 2 and 3.
As someone who plays Four 4s on a regular basis, 210 is very familiar to me, since it's easy to get with one 4 (σ(4)#, aka sigma 4, and then the primorial of that).
5:53 this is an interesting equivalence relation on the set of natural numbers, there is literally an equivalence class that is 2^x for all natural numbers x
I like how the whole time Matt's talking about and looking towards these graphics that someone else made for the video, there's just no graphic on screen and it's an ad pitch for the wall decorations.
Matt is talking about a conjecture, which if proved would give an alternate proof for likely the second most famous theorem in maths, and only starts mentioning that fact 20 minutes into the video.
Classic Parker move
So... a move which _almost_ works?
A 🦞 is always a plus.
That graph... I got lost, but the conclusion i got from "if you zoom out far enough they're essentially the same" is that low margins of error = no margin of error... That sounds like a parker mentality to me all right! 🤣
3:47
so flippen great
Loving the frequent uploads!
If you use a tiny drop of acetone on the surface of a LEGO, you can more permanently fuse bricks together.
Is there a reference for the claim at 23:15 that we get q
Get Matt to a Million 2021 !!!!!
Breaking a seven: that portends mirror year's bad luck.
I just read the t shirt and it says geek, I learned how to read binary from Tom Scott who is the guy know for wearing red shirts.
Thank you so much for this 3Bald1Bearded Video 😎
Sometimes I worry i’m not going to enjoy your videos, but you are honestly, genuinely, really funny.
This might be one of the most maths heavy videos you've done in a while. I can't follow anything after you switched to Lego.
We got a Parker Square update video, a Hannah and Matt video, and 4 days in a row with Matt videos being released. Christmas came early this year! I wish Parker Christmas lasted forever!
Why are there like 5 new videos wirh Matt in the last 2 days?! I LOVE IT
The quality is like the dual of the (log of) geometric average. Quality is arithmetic average power of primes weighted by log prime, while log of geo average is arithmetic average of log prime weighted by prime powers.
Two minor corrections: at 24:08ish you appear to write rad(abc)≤abc≤xyz when you might have meant to replace abc with xyz or rad(abc)≤xyz≤abc
At 5:13ish you said the only odd length 2*n lego brick you could find was 2*3 forgetting the trivial 2*1 case.
Grants animation software is revolutionizing the world of math.
I'm a bit confused at the table at 11:20. It says the quality for the first row is 1.226, but the quality of what?
@HBL (I don't know why, but the answer I've sent you doesn't appear under your comment. I'll copy it and try again; sorry if I end up double-posting.)
I was also confused by that, so I took out my calculator. The quality here is Q = log(C)/log(rad(A*B*C)). So, for the first row, you get Q = log(9)/log(rad(1*2^3*3^2)) = log(9)/log(2*3) ≈ 1.226. If you take the example at 17:45, you get Q = log(6,436,343)/log(2*3*109*23) ≈ 1.6299.
I love extra Matts just pointing and nodding ❤️
2x5 and 2x7 Lego bricks are a horrifying sight to any Lego fan… what unholy beast have you unleashed‽
Matt: Would you accept a lobster as plus?
Me: No. Lobster peasn't plus.
Grant, my man, outstanding video as always
When you first pulled out the blocks I thought you were going to present the following problem. What is the minimum number of blocks required to spell all words of N length?