Can you solve the cursed dice riddle? - Dan Finkel
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- čas přidán 6. 09. 2023
- Practice more problem-solving at brilliant.org/TedEd
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Ah, spring. As Demeter, Goddess of the Harvest, it’s your favorite season. Humans and animals look to you to balance the bounty of the natural world which, like any self-respecting Goddess, you do with a pair of magical dice. But then, along comes the trickster god Loki, who invades your land and curses your dice. Can you fix the dice and keep the world in perfect harmony? Dan Finkel shows how.
Lesson by Dan Finkel, directed by Igor Coric, Artrake Studio.
This video made possible in collaboration with Brilliant
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Can we all just appreciate how Demeter is just playing Settlers of Catan with mortals?
Haha that's what I thought too
Nice.
my exact thoughts!
Pog catan
Can't wait for mortals to find the Fire progression card and Prometeus to fall
Imagine you're a scientist and while others like Einstein come up with relativity, you're inventing a special dice that does the exact same thing as a normal dice.
George Sicherman is a legend.
Here before this blows up
As long as you find something new...
Sound fun
What a lad
lol
Oh my god this is the prequel to the Ragnarok riddle. Everything really is connected.
omg your right
In other words, damn it Loki you jump started Ragnarok.
Riddleverse confirmed!
Can’t wait for the tcu
Ted
Cinematic
Universe
So there IS a TED-Ed Riddles Cinematic Universe!
I’m so proud of myself, I almost didn’t attempt this riddle and was just going to watch the solution but decided to try. I didn’t know my strategy but I “followed by nose” and drew the table . I then copied the table and removed the sums. I realised I was only allowed one “1” and one “12” so I filled them in exactly how the solution did and then worked my way back, ticking off each sun from the original table. In the end my work sheet looked exactly like the solution tables and I’m kind of stoked bc I’m usually not smart enough for TedEd riddles. Sorry I must sound a little boastful but I’m pleased as punch.
Yessss🎉🎉🎉🎉
I wish i had ur experience.
f'real congrats! I mentally got as far as "we need the results map to match" but 1st guessed 1,2,3,3,4,4 as our low dice, had I bothered to work through the guess I'd see my mistake soon enough.
Look, if you do something smart/great like this, feel free to boast!! Good people will be happy to hear your success! ;) glad you were able to solve it! :)
Nothing satisfies like a solved riddle). I solved it too, wasting some time on 0-hypothesis (I hadn't known you couldn't a side without dots) and then accidentally stumbled upon the solution.
Btw, I found that you can actually create another pair of dice, if you include 0 and forget about 4-on-one-side limitation. First dice is 0,1,2,3,4,5, second one 2,3,4,5,6,7. I'm curious if those are called somehow too
I read about Sichermann dice by chance in a book of math puzzles. Fun fact: one critical difference between normal and Sichermann dice is that it's harder to get doubles on the latter (1/9 chance instead of 1/6).
I wonder what it'd be like to play backgammon with Sichermann dice.
A good reason for the animation to choose catan, given doubles don't do anything.
Okay that’s a great question. It’s been awhile since I played, so without going to refresh my memory of the rules, I’m assuming it would be much more likely for someone who gains a small early advantage to clean up, possibly resulting in it becoming acceptable and common to forfeit matches early.
Might have to make some new dice and play test it.
@@heatshield Maybe it wouldn't make small advantages last, but it surely would make big and rapid swings more rare, so at least we'd get a calmer more sedate game?
Means you go to jail less often in Monopoly, but you also travel less far.
A lowered chances of doubles - and not being able to get double 2s, 5s, or 6s at all - would be a huge impact. It would also wildly change things to have a 9 available on one die, changing the strategy around priming.
I enjoyed the animation in this a bunch! I'm no mathematician, but even I can understand the reasoning behind this solution, because it was so well explained... and the denouement at the end was entirely delicious!
ditto
Yeah. I was thinking something similar of what they were saying but I didn’t think to use a graph and never thought of it before I gave up.
People who own land that's a 2 or 12 can never catch a break.
They'll catch a break 2.8% of the time lol.
the blacksmith having accurate mouth movement with the narration is both cool and oddly uncanny
step 1: confirm that you have green eyes
step 2: ask loki, "if i asked you whether i had to roll three 4s, would you say ozo?"
step 3: walk anti-clockwise across one block of land to add one coin to your balance
step 4: calculate which lockers have numbers with perfect squares
step 5: work backwards from zahra's answer to get the hallway required
step 6: choose the gaussian and miss on purpose
step 7: lock loki in pythagoras' cursed chessboard
was looking for the obligatory "confirm you have green eyes" plagiarism of the solution to a different puzzle Congrats though you plagiarized 7 different puzzles (very slow claps)
Nah your forgot step 8: head into the simulation mainframe and communicate which server is infected via binary code
@@xavierburval4128and also step 9: Activate thrusters D & E
I always love seeing the answers from the other riddles pop up in the comments.
Always something like this in the comments. MAGNIFICENT
Very cool video. The way it's done rigorously in mathematics (which can lead to many other interesting results, like having two dice with a different number of faces, and still getting the same probabilities) is with generating polynomials. A normal six-sided dice is represented by x+x²+x³+x⁴+x⁵+x⁶, where the exponent represents the number of dots on a face and the coefficient, which in this case is always one cause all numbers from 1 to 6 are each displayed exactly once on the die, represents the number of sides you can find that precise number of dots on. The product between two of these polynomials gives you the amount of different combinations of numbers that can give you a certain sum, namely the polynomial representing their combination. If P(x) represents a die, computing P(1) gives you the total number of faces of that die. Notice that the polynomial for a standard six-sided die can be broken down into x(x³+1)(x²+x+1) -> x(x²-x+1)(x+1)(x²+x+1). If you plug in one, the first two operands yields one, x+1 yields two and x²+x+1 yelds three. Therefore, the presence of the first two doesn't affect the number of sides of the die, while the other two operands are fixed in place, as the two dice must be six-sided (but you could do it with four-sided and nine-sided ones as well, etc). Moreover, each face has to have at least a dot on it (must have x as a factor). Therefore, we can only shuffle x²-x+1: one die will have no copies of that factor and the other will have it squared. After all calculations, you'll get x⁴+2x³+2x²+x and x⁸+x⁶+x⁵+x⁴+x³+x, which are exactly the two resulting dice from the video. Remember to eliminate, if there were any, all results which have one or more negative coefficients, as negative sides don't seem to exist in geometry (I think the only viable way is just to eliminate them afterwards, but if you have a better method in mind I'll be happy to hear it). All of this looks scary at first, but actually it is fairly straightforward when applied, especially for harder-to-find combinations (like the one with four and nine sides respectively, which was a question on an entrance examination test for the Normale di Pisa dating back a few years ago). Hope I explained it somewhat clearly 😂
I solved it in a similar way (apart from that once you wrote x⁴+2x³+2x²+1 rather than x⁴+2x³+2x²+x.) I just removed the factor x, and I calculated
(1+x+x²+x³+x⁴+x⁵)² = ((1+x+x²)(1+x³))((1+x)(1+x²+x⁴)) = ((1+x+x²)(1+x))((1+x³)(1+x²+x⁴)) = (1+2x+2x²+x³)(1+x²+x³+x⁴+x⁵+x⁷)
@@johannesvanderhorst9778 True, I mistyped, thank you!
I have never paused a riddle video
I tried a couple times but eventually I decided I wouldn’t fight the urge to be taught😂
Here before this random comment blows up for some reason
Me neither brother
Jesus loves you ❤️ please turn to him and repent before it's too late. The end times described in the Bible are already happening in the world.
You want me to give you a cookie or what?
I missed the instruction that required us to only select numbers >1, so I had a die with a blank face (0,1,1,2,2,3) and added 1 to all the values on the other dice. After seeing the correct result, it makes sense that if you subtract one from each face in Die A and add one to each face on Die B, the results don't change.
Yes. I was thinking the same, getting a die with 234567 and another 012345, but stucked.
Except you didn't miss that instruction as it wasn't there! 011223 and 245679 is a much more pleasing dice combo in my opinion.
@@BeaDSM it says it right there at 1:22?
@@Tumbolisu I am not a deep mathematician, and I don't view 0 as positively as I do not write -0, and this does make sense in the story.
The instruction isn't stated (a mistake IMHO), only written.
First ever puzzle I've actually decided to sit down and solve myself!! I even made a google sheet to calculate the different sums and their counts, very satisfying!!
I was going to do this, but then they said you had to answer within 3 seconds, so by the time i got a piece of paper, i was too late.
I tried solving this by starting with a regular dice, and trying to rearrange the numbers in a way that worked. I tried but couldn’t find anything that worked, until I realized that I’m not allowed to have zero dots, which I though I was. Once I had that, it was way easier to build the dice step by step. Very interesting riddle
I also immediately went for the zero dots face, and kept running into problems in the middle. Kind of funny what assumptions we make about a problem.
Indeed, in the verbal description there is no positivity requirement. (I count this as a mistake in the writing.) It is clearly written at 1:26.
I think that would permit another solution, (0,1,1,2,2,3) and (2,4,5,6,7,9). Start with their solution, and move one dot from each side of the small die to each side of the large one. Given that in the story we start with blank dice, I think this should have been a permitted answer.
@@cjfeinberg7613 yeah that would work, but they probably wouldn’t want two solutions so it makes sense
@@cjfeinberg7613 how do you even get 11 with that?
2:28 "Assuming we have a 4..."
I felt like the video brushed over considering having numbers less than 4 or larger than 8 on the dice (respectively). But it can actually be walked through relatively easily. You must have exactly one 1 on each die. If you had an 11, you could have only 1s on the other (without getting sums above 12), but this would make at least 6 ways to roll 12. Same sort of pattern continues for looking at 11, 10, and 9, with each leading to too many rolls of the high numbers. In this way, you can prove that the highest number you can use is an 8, meaning you must have exactly one 4 on the small die.
Great explanation. Thank you!
Thank you! I'd been stuck on them saying that but was struggling to work out that logic myself.
Imagine having a Norse and a Greek god at the same time😂💀☠️
That's a common thing in Polytheism. It's not like only one Pantheon exists. But only one Pantheon is accept by a culture.
In fact, even the bible states polytheism in saying there should be not other god besides him. It does not say others do not exist.
Monotheism makes it way more easier to control people.
It's only mentioned in passing and not on the Rules screen, but I think it's fun to note that the Sichermann dice also use the same number of dots as two standard dice. You don't end up with extra or remaining dots after they fall off and are put back.
That has to be true of the solution because there's a 1/6 chance of any particular dot being counted which means the average roll equals 1/6 times the total number of dots. And the new average has to equal the original average so we must have the same number of dots.
Well it has to use the same amount of dots, otherwise it can’t have the same frequency table.
I love this riddle! It looks complex as my initial thoughts were getting a piece of paper and solve probability for each roll but in reality its very simple, straightforward and easy to understand.
So basically, the goddess of harvest plays Catan.
I like how the concept of a set of 2 dice that would have the same relative chance of rolling results as regular dice do, would be such an alien thing to come up with. Once you turn it into a problem like this video has (and add the constraint of max 4 eyes per side on one die, which was a massive hint) it suddenly becomes a trivial thing to solve
I love these riddles as they were the first kind of videos that got me interested in this channel. Please never stop uploading them ❤
Solving these riddles makes me feel good. Thanks Ted
Finally, another riddle! Have been waiting for 2 months for one. Can we have one every week? I just love these riddles that you come up with. 🙏🙏🏼🙏🏻🙏🏾🙏🏽💖🥰💗💓😍💕
I love riddles that challenge you but aren’t impossible to understand without background knowledge of the subject.
One of the few riddles I was able to solve and I’m glad I was able to do so :D. Thanks Ted-ed
the Goddess really be playing Catan on the poor people 💀
so close to my thoughts!
Nice mix of specific calculation and logical thinking. Liked this.
Everybody gangster until you show up with Sickerman Dies to family game night.
4:07
Jörmungandr be like: Sorry Dad, but I owe Dementer a favour
i realised the 1 on each dice and the 4 - 8 relationship and decided to continue the video instead of filling up a table. but i hope i would have done similiarily as well as Demeter in a life/death situation
I did spend too much time on solving it, but I'm happy I did, when I got the logic and it worked. Great riddle, loved it. Thank you :)
This video explained it better then my teachers ever could.
Your Teacher isn't a big multimillion-dollar corporation that has a big team of experts and researchers?
@@user-hh2is9kg9j well education in my country is ridiculously expensive however the quality is poor. So yeah they should do a better job of hiring good teachers when they charge a lot.
My solution doesn’t work under the rules, but I used three dots to create a binary notation, corner middle side representing 1, 2, 4 respectively. This allows you to make an ordinary six sided dice using a maximum of three dots. The rules state the sum of the dots, but if different dots have different values, as in the binary notation dice, then the sums can still match a normal dice.
You are the reason I learnt English, your videos are so good it made me learn the language in two months
One of the best animations by ted-ed ❤
I love how this video references Catan at the beginning and even Jörmungandr at the end. 😆
I learned about this in abstract algebra. I even have a pair of d6's that answer this question. According to my dice, one is a 1, 2, 4, 3, 3, 2 and the other is 6, 5, 3, 4, 1, 8. That should yield the same results as a standard 2d6.
I paused to before the answer to grab my dice. Didn't realize it was answered.
Last year, as a senior in high school, our class participated in a maths contest. We would get three riddles during the year. This is actually one of those riddles. fun to see the explenation of that riddle by some one else.
sidenote: we didn't manage to win the contest but got to the finale.
Finally after months another riddle ❤
Me : _brings sicherman dice to my first DND campaign_
The table : what in the ungodly creation have you brought upon us?
I enjoy these riddles!
I actually managed to do one of these for once, and I can actually explain every single step I took to reach it. As such, I’d like to explain all my reasoning.
1. Since there’s only a single two, (and every die face must have
congrats on being a silksong waiter sane enough to solve this
My answer was simply to symbolize the missing 5 and 6 with a different pattern of dots, then if that side rolls, even if it only has 4 dots, you treat it as a 5 or 6 just make sure they are in a different pattern to the regular 4th side.
Wow, Loki's pranks have got /really/ thinky since the Edda.
I never thought i without pay so much attention more to a riddle video
I took some time to just sit down and think about the various possibilities, and I was able to figure it out. I used some trial-and-error, but I got figured it out mostly the way it was shown in the video.
You could also just create a new shape for 5 and 6 on the cursed dice, which seems much more obvious.
First you have to ask the demon of logic if at least one of the sides has green eyes. If the answer is ozo you count the number of trees that the founders of the houses can see to find where the baniker is buried. If ulu then you trap nym in the magic checkerboard to prevent the ai from releasing the robot ants. Afterward you must give the dino nuggets the magic tarot cards you stole from fate so they can lite the correct number of candles on the giants cake. The giant will then agree to use the tri source to make you a new set of dice.
I tried this and it worked
When you are looking for a fun riddle and you end up having a math lesson explained at you.
your videos are incredible
Before the video starts, i just want to express this first:
I've never tapped so fast on a video before. But when it comes to New Ted-ed riddles, I AM HERE!!! Keep them coming 😊
that graph of probabilities helped me solve this fun riddle since there were only four values for one die i could solve with brute force but this tabulation is faster and convenient
This is the first riddle from this channel that I actually managed to solve!
Making Catan more difficult is all part of the fun!
Another way to find the second die after the observations about the first are made is simply by division: 12,345,654,321÷1,221=10,111,101
Wow that was a great video, one of my personal most beloved tricky questions!!
I also want to share that after thinking of other ways to create such dices, I discovered that if you allow one dice to have a zero on it (I know it’s not in the original rules, but hear me out) - you can create dices that produce the same frequency, in this way:
Dice 1: 0-1-1-2-2-3
Dice 2: 2-4-5-6-7-9
Now that I think about it I just added one to every number in the second dice and subtracted one from each in the first😅😅
* Continuing with that logic you can create the same affect if you allow a minus 1😮🙃
Any way that was a really nice riddle, thank you😁!!
*Dice A* = [4,3,3,2,2,1]
*Dice B* = [8,6,5,4,3,1]
This is a nice riddle. One which just takes some knowledge of probability theory (specifically: *combinations* ).
Plus, it helps to use a spreadsheet/chart.
An algebraic way to solve this would be to set up 12 different equations (easier with *matrix multiplication* )
A solid puzzle; nice callback to the ragnarok one, too.
Wow, I have last seen these riddles about 5 years ago. I am so amazed that the narrator is still the same!!!
On normal dice 1 is always opposite 6, 2 opposite 5, 3 opposite 4. You could have totally blank faces for 4, 5 and 6 and still end up with perfect dice simply by checking the opposite face.
I like your riddles. They're very mathematic.
Great education tool, the chart to solve the problem 🎉
Finnaly a riddle!
nice tool, it’s really useful tysm
Loki getting absolutely demolished by a dragon is fair compensation I'd say
I didnt solve this, but I really liked the riddle! I tried to solve it tho, but failed! 😋
I waited for SOOO Long on Another riddle
I was amazed how you did not mention polynomials and get the perfect solution
I'm sure you could do some neat analysis with a 6x6 square representing the sums of the dice, but you can also figure this out in an ad-hoc manner just because of the amount of constraints you can place on your die. For example, each die must have exactly one face with 1 pip each, because there must be only one way to roll 2. You can make the same assertion about the highest face of the die that's limited to 4 pips; only rolling that highest face and the highest face of the other die (which must also be unique among its die's faces) should yield 12. At this point I made some educated guesses; you could continue to build up and say each die also needs a 2, so there are exactly two ways to roll a 3, but that's not sustainable. At some point the 4-pip die needs to double up (at least) on its numbers. I just assumed it was symmetrical and went with that, although you can prove that having more than three 2s wouldn't work (eg. with 3 or more 2s, if the main die rolls a 1, you'll get more than 2 ways to roll a total of 3).
Once the smaller die is proved or assumed to be 1 2 2 3 3 4, you can build up the probability distribution. The way I modeled it is that selecting a face for the larger die adds to the distribution of possible totals, starting with all totals at 0 ways to make them. We now select the six faces of the larger die to build the distribution shown onscreen for a standard 2d6. We know the first face is 1, so the totals raise by the following amounts:
2: +1
3: +2
4: +2
5: +1
That 1-2-2-1 shape is added every time you select a face for the larger die, with the leftmost 1 starting in the column that's one higher than the number of pips on the face. The next necessary placement is to add 1 possibility of getting a total of 4, so the next pattern must be placed with the leftmost 1 in the 4 column, meaning the second face for the large die is a 3. The same logic can be used to place the third pattern with its leftmost 1 in the 5 column, so the third face for the large die is 4. The overall pattern is mirrored, so we can just mirror our placements for the last 3 patterns, resulting in an answer of:
Small die faces: 1 2 2 3 3 4
Large die faces: 1 3 4 5 6 8
Edit: Just watched the official explanation, that's remarkably similar but as I suspected there was an easier method using the 6x6 table. It's interesting how two different methods ended up taking nearly the same path to get to the answer!
It would be exactly the same mechanically, but for some reason I'd find playing Catan with this set of dice to be more exciting
My solution to this was to have one die normal and the cursed die have 3 fours. Basically whenever you rolled a 4, you would roll the normal die to determine the value of the 4.
I want these dice to be used in an actual game. Imagine the happiness when you roll the first die and get an 8
Best educational channel ever
I didn't spend much time on working the table although mine is completely different because I was thinking further out of the box I got 1,4,2,2,3,3 and 1,8,2,6,4,5 I got 1 mistake and I didn't have a chance to check it because I had enough of trial and error. I love these types of brain activities, this will serve as a lesson.❤
i agree with many others that this was a great riddle for trying yourself and getting the answer! (or close.) great choice TED-ED
Good riddles teded
It took me a lot of thinking but I was able to get it right.
Nice new riddle! I enjoy watching them but I can't solve them most of the time. They're cool, though.
love the video👍
Demeter casually starting Ragnarok because of something Loki did is very on character for both of them.
very cool. good riddle
I saw a video about this formation of dice a few weeks ago and i predicted that the video would be about that
Finally another riddle video 🫶🏻
Jesus loves you ❤️ please turn to him and repent before it's too late. The end times described in the Bible are already happening in the world.
I solved it! This was my thought process and with a little bit of luck.
So, my initial idea was to put the two 1s, 4 and 8 first. Basically the same.
I decided to go off with a wild guess that the first die must have 1,2,2,3,3,4 as the configuration, after all this was my idea, after knowing that I couldn't have more than one 1 or 4 on the "restricted" red die, so the remaining was either 2 or 3. I went with the average to be safe.
I figured then that there couldn't be a 7 on the second dice, because after filling in the restricted die, i've fulfilled my quota for 11s, so it had to be 1,?,?,?,6,8. I then proceeded to trial and error by going down the next lowest number, and lo and behold, i got 1,3,4,5,6,8, which fit exactly the requirements!
I did remember a die that had the same arithmetic mean for a normal dice but different numbers, but it wasn't the same. But this is actually the first puzzle i've solved on paper.
And credits for Sicherman for these amazing die.
2:27 Why do we need positive numbers? I was also trying 0/2 for that spot.
the rule is that every face needs to have a positive number
@@abhijiths5237 That rule isn’t part of the scenario as presented. I missed if it was on the rules screen but there’s no reason for it in the story.
@@danielrhouck But have you found a valid solution which includes a 0? Something tells me no such solution exists (even though I don't feel like proving it rn). In this case, you don't actually "need" that rule, it would be more of a "hint" to get you on the right track faster.
@@lenoncerqueira8308 i have neither come up with one nor shown it’s impossible; if I remember I’ll try tomorrow or over the weekend.
A critically missing rule allowed me to "solve" the puzzle by simply using binary to count the dice :)
LOL. Unfortunately pips are a base-1 number system, so...
We need Ted ed riddle lore
Need these desperately for my next settlers game
Before the answer:
We can't have more than 4 dots/face on one die, so that means we'll have to increase the limits of dots per face on the other die. Starting with the rarest case, 2 and 12 can only appear once. The only way to get 2 is with two ones, so both dice can only have one face each with one dot. Since we are limited to 4 dots on one die, the only way to get 12 is for the other die to have an 8 dot face, and each die can only have one 4 dot face and one 8 dot face respectively.
Next, we need to have two ways to get 3, so we need to add two faces with 2 dots. We can either put one 2 dot face on each die, or we can put two 2 dot faces on the 4-max die. Since we are limited in the number of different faces we can make on that die anyway, we go ahead and make two 2 dot faces on the 4-max die. Next, we need three ways to make 4. If we put two 3 dot faces on the 4-max die, then that gives us 2 ways, so we also add a 3 dot face to the other die, giving us 3 ways. At this point, we have filled up the 4-max die, but we still have 3 faces left on the other die to use. We need 4 ways to make 5, and with the faces we have, we have 3, so by adding a 4 dot face to the other die, we have 4 ways to make 5. Next, we need 5 ways to make 6, and with the faces we have, we have 4 ways, so by adding a 5 dot face to the other die, we have 5 ways to get 6. Next, we need to have 6 ways to make 7, and with the faces we have, we have 5, so by adding a 6 dot face on the other die, we have 6 ways to make 7. We now have run out of faces, so if we did it wrong, we'd have to start over. Checking through, we need 5 ways to make 8, which we do (4,4 3,5 3,5 2,6 2,6). We need 4 ways to make 9, which we do (4,5 3,6 3,6 1,8). We need 3 ways to make 10, which we do (4,6 2,8 2,8). And lastly, we need 2 ways to make 11, which, you guessed it, we do (3,8 3,8).
*TL;DR* By walking through the logic, the die should have the following faces:
1, 2, 2, 3, 3, 4
1, 3, 4, 5, 6, 8
What makes the Sicherman Dice even cooler is that they not only have the same sum distribution as standard 2D6, but also the exact same number of dots (42)
This is the first riddle I actually solved!
"Can you solve the cursed dice riddle?"
Also
"The solution wasn't discovered until 1978!"
(Most likely, mathematicians weren't really looking for the solution until they proposed the math problem first and then found the solution shortly after, but still).
It was interesting to learn how to build a die that has the same frequency as a normal die but without the restrictions.
Love the reference haha! 😂
Fascinating 🤔🤩
Ah yes, Loki being his original self. Good for him for cursing that dice🙃
That was such a complicated question that I didn't care what the answer would be.
agreed - I wouldnt call this a riddle at all. Its like a math olympiad style question, and only the most enthusiastic of number crunchers could find it even remotely interesting
Very cool puzzle
I did it! I solved a Ted-Ed riddle by myself. I'm so happy.
Is there something like this for 3 sided and beyond, would be really eager to know.
I really appreciate that you can hear the cut after "Hephaestus'" at 1:01