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MATT PARKER: Stand-up Mathematician
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“The one person I still am allowed to have visit and do mathematics with...” Aww, it’s gonna be his wife, isn’t it? “...is me!” Oh.
I thought the exact same thing
the relationship between his wife and him must be bad or his wife's relationship with maths
@@theunknown4834 Isn't his wife a physicist? I doubt that's the issue. It's just a joke.
Having seen the DVD Unboxing Video kzhead.info/sun/ddKnnaVqop54gXA/bejne.html (same shirt!!) and his recent Palindromic video kzhead.info/sun/Z8p-j9eQkZyuoZ8/bejne.html , I actually saw that one coming. But it's still good.
Off-topic, but cool profile pic!
Plot twist: Matt REALLY DOES have a twin brother, but he keeps him in secret for videos like this.
A Parker Twin. Not really a twin but almost
It's just so sad to know that when Matt B dissapears he's falling into a tank of water.
@@mybigbeak I just saw The Prestige yesterday *finally*, wanted to say something like this.
half of the videos can be matt 1 and the other half can be matt 2!
and his twin seems smarter!
Two Matts = a Parker Squared
If you multiply them ( ͡° ͜ʖ ͡°)
Or work in an additive group.
It's a Matt by a Matt.
Ian Miller: There is only one Matt Parker, and one squared is still one. So, yes, you're absolutely right. Two of them is indeed a PARKER SQUARE(D)
@Carey Hunt He wished he was an octagon, but he's just a boring square
Can’t be more socially distant than to be separated in time.
You mean like he's the real life Piccolo? Smart and can split himself
Deep
But how many seconds is equivalent to 2 m?
@@jeffreyblack666 do you count in planck length or in lightspeed?
Except if someone infected touches a surface without clean hands, then someone else touches the same surface before the virus becomes inert.... Even if they never came near each other, the problem still exists...
The most hilarious part would be if the first Matt disappeared after the clap and the second took his place.
i was half expecting that to happen
Or a third appeared between them
What if the one who clapped did make the first one disappear and took his place.
Yea, and then that right Matt could move over and pick up the cards on the left of the table to continue on. That would have worked seamlessly with how he did the other edits already so its quite the missed opportunity.
Yup, that's what I would've gone with. I mean, in an alternate reality where I'm not too lazy to shoot YT videos, obviously. ;)
Can we all just appreciate the editing on the toss at 2:30 for a moment?
I don't think the toss itself needed editing. He probably just moved the split between the 2 videos and then left the cards there for the second take. But very clever.
Can you see that the toss (left) was filmed first so the pickup (right) is filmed afterwards, and so left was filmed before right, hence layout of the cards on the right is actually occurring after matt on left has already called the answer correctly..
Parax77 which only makes the whole thing more amazing because how could left(past)-Matt possibly have known what the card layout on right(future)-Matt’s table would be?!!! Coincidence? Not likely!
@@aqueousone matt already knew he would only turn the second card over...
when right parker collected the cards he put the ace on top, then after throwing them off screen, left parker picks 4 cards up with the ace on the bottom, a hint for how he did it xP
Parker victory: Even when you win, you still lose.
The Parker win
@@Nylspider we need this to be another going thing.
The timing of this is masterful.
Masterful! And the passing of the cards!
@@DeVibe. you're idiots, it's far from hard to understand how he did that... It's kids play really.
@@MuscarV2 go on then
@@MuscarV2 u are such a fool No brain in that head of your fool
*blinks at **2:33* ...You're getting really good at this.
It's so good I geniunly thought he hid a twin brother for all those years
How did he do it
First he films the left side (from our perspective). When he picks up the cards later (that "other Matt" tossed off-screen), it's a different set of cards. The original cards are actually laying on the right side of the table for the rest of the left side shoot. Then he films the right side. Once the cards are tossed to "other Matt", the right side of the table is from that shoot.
I didn't even blink!
@@edwardbarton1680 I think second Matt very purposefully never lowers his arms to obstruct the "green-overlay" on the table. once he moves aside , the screen shifts a bit to show the entire throw at which point the thrown cards match up in both videos and the right side can be taken as footage again. So I think the cards in the right shot are there the entire time.
How the hell did you do the card pass on??? That was seamless! Like I get that It's two separate takes, one of the left Matt and one of the right Matt... But what is weird is, that Matt 1 (Left) passes cards to Matt 2 (Right) and without a cut he takes them back... I guess, Matt 2 moved away, so Matt 1 could throw the cards on the table (for an easy transition), but that means, that those cards for Matt 1 stayed on the table for the whole time... So how could he then take those exact cards from off the screen... My guess is two stacks of similar cards. One for the first game that got left on screen on the table and one that was hidden off the screen. Really good job there. Clever and perfect use of editing.
After some close inspection you can see, that the stack that Matt 2 throws off the screen is sorted A of diamonds, 4 of spades, 6 of diamonds and 4 of diamonds. And the one that Matt 1 picks up is sorted 4 of diamonds, 4 of spades, 6 of diamonds and ace of diamonds. So either you have real magic fingers or I got you :D. PS: It took me like 10 mins... Just to find the answer :D
Right Matt never puts his hands between the table and the camera. That is because up to the point where the right Matt picks up the cards, the table is shown fully from the left Matt's take. While right Matt is guessing, the cards are already waiting in front of him, hidden by the edit. Notice that after the right Matt goes away, the left Matt stays on the left, instead of taking up the whole table for the explanation. That is because the cards he threw there earlier are hidden by the edit, and waiting for the right Matt to start his take. Left Matts take > cards passed to the right > acting out the remainder of the left script > switching to the right, careful not to disturb the table > playing out the right Matt guessing game > picking up the cards > acting out the remainder of the right script > edit.
There's blur on YT as the cards are thrown. But after that nothing else happens. My thought was that left hand (as we view) Matt was recorded first, then for the right hand Matt there's a wipe from "no cards there" to the cards that were there all along as they get thrown from left hand Matt. The other possibility is that there's a wipe to a second throw of the cards that happens at the right point in fight hand Matt's recording.
I, too, was flabbergasted at first. I didn't even consider the two takes being back to back and the same exact stack of cards being picked up by later-Matt. It is ingeniously done, though. And it makes it required that both takes go perfect since disturbing the stack of cards and then having to redo the "right Matt" part would mean having to redo the throwing of the cards, which means redoing the whole "left Matt" part. This was all very impressive.
I also definitely think Matt 2 never touches the table before picking up the cards, and the cards are in fact there the whole time left over from Matt 1's take, just edited away. As additional proof of this I noticed that when Matt 2 does put his hands on the table, you can see a horizontal line going through his hands. That seems to be an artifact of the editing. Why they couldn't get rid of it, I don't know, but it might even be left in on purpose as a subtle clue.
”I would like to show you a card game that you can play with a friend, although I’m not allowed to have… friends”
This game works over the phone, (or other near real time communication system.)
So are enemies or frenemies allowed?
Yeah yeah because i totally have friends even when there isnt a lockdown
I was expecting: "Matt is in the lead" "Wait, which Matt?" "Parker."
Next level editing skills!
Matt, I would absolutely LOVE to see a behind-the-scenes video for how you are able to script and record such realistic-sounding conversations with yourself
Wow, the production and coordination on this video is amazing!
Due to the scarcity of new productions this could end up winning the oscar for best special effects 😂 Awesome timing by the way 😄👍
Your videos are always worth watching; for their mathematical content, but definitely also for the technical complexity you put into them. I'm truly amazed how you recorded that and made it look so natural, with all the listening head nods and reactions, and even a casual deck toss to the other side. Looking forward to the next one!
I love when you make two of you. I know it's hard work to time things and the result is amazing!!
Can we all just appreciate the truly next-level script and reaction timing to get the first half in a seamless take?!
That card toss to yourself was masterfully done Matt. Seamless.
The 3 card variation where the goal is to have all 3 cards are face up/ face down is easily solvable in 3 moves max. In 3 cards, every unsolved state only consists of one card that needs to be flipped to reach a solved state. The order 1,2,1 guarantees a 3 move solution.
There are sixteen possible combinations of face-up and face-down cards, and we know that the cards are not in the winning position to begin with; so you only have to try at most fifteen moves, each one resulting in a different combination of face-up and face-down. This can be done using a Gray code.
SOLUTION : We need at most 2^n - 1 flips. (n the number of cards) Just think we are in binary and have at least one "1" in the binary number (number of bits=number of cards, 1= face up). Let's call the original position "O". The flips are represented by another number "F" that starts at all zeros. The current state is O xor F. We have to brute Force all possible F until we find the solution ie O xor F=0. The trick is to use gray code to enumerate all possible values changing one bit at a time. en.m.wikipedia.org/wiki/Gray_code
And some people would like to work it out :D
This is really impressive editing, Matt. Good work!
Brain breakingly well done. Added shadow to your arm too, and the card toss. Exquisite. Not to mention the perfection of the performance(s). Oh, and a neat puzzle too!
I love the way you presented the video. Well done Matt!
Very well timed and editied. Nice job Matt
As a video editor, I giggled when I got to 2:32. Well played, sir.
That was a genuinely impressive level of coordination
Beautiful script, execution, and edit. You, sir, are a genius. Best picture.
This video was good, well edited and funny with the doppelganger The most confusing face-down card game I know involves putting a double faced card onto the table face down and not face up. Physically it is impossible, but logically you just need to have virtual faces that can be flipped out of and not flipped back into
That moment when he passed cards was fabulous! Such a natural flow
I would love to see the behind the scenes of this video production. Very impressed with you nailing the timing.
I really like the editing with the interaction with the two of you.
very nice editing! I got a lot enjoyment out of this :-)
Intuitively I would just dump a gray code (reflected binary) onto the problem.
Phenomenal editing, truly bravo
Your editing is CLEAN!!! I try so hard to find imperfections, but they just aren’t visible!!
The radiating joy of Matt when he gets to do this is very adorable! :)
Gray code binary 0..15 but only 1 bit changes per increment.
Yep. And I'd say you can't get better than 2^n
@@PronteCo That's obvious (and off-by-one) - you have (2^n)-1 possible starting positions, and every move is reversible, so after each move, only one starting position could "win" at that point, so it has to take at least (2^n)-1 moves to allow all possible starting positions to win. In order to do it that quickly, you'd need to find a sequence that avoids repeating any positions, which isn't that difficult.
@@PronteCo You can indeed because you forgot the starting rule. You can't start with the cards all face down.
@@PronteCo 2^(n-1) moves; (same O(2^N) size) as the initial setup requires that the base state cannot be solved, so your first attempt cannot unsolve it.
@@rmsgrey There is a well defined sequence, gray coded binary. It is the fundamental system used in mechanical to digital encoders. it is required since with normal binary sequence there are transition errors where many bits change, if the transitions are not precisely aligned across the bits then huge errors can occur if significant bits change before/after least significant bits. with gray coding, since only 1 bit changes then the maximum transition error is the least significant bit. It is possible to make standard binary encoders immune to this by having an extra half sized LSB but that would mean you could have had better resolution and requires an extra sensor.
It's so well timed! Well done
That was awesome! Perfect timing on that conversation!!
This was all 1 cut (technically two cuts, one for each matt) and you managed to make the conversations make sense with the right timing, AND that card pass was seamless like damn you're good at editing
Twin video was very clever. Especially the part where the cards are passed from Matt 1 to Matt 2. No break in continuity--well done!
This would have been so much harder to pull of than most people would realize. Very well done. Anyone who has played with video production would be like me and not car about the card game but purely focus on the production and editing techniques used 😂
I would really love to see a behind the scenes on this. Perfect for your channel 2?
The amount of effort in this video is ridiculous! I love it
BRILLIANT editing!
“So Matt is in the lead!” Which one? No, not which, both. Both of them? Both of them. How? By equality.
I’d love to see how you made this video! Great setup love it.
i am more amazed on how you got both videos PERFECTLY TO THE NEAREST PIXEL lined up.
The brute tactic would be an electrical engineering class basic - Gray code! mathworld.wolfram.com/GrayCode.html - That gives you the order to cover every possibility in single moves. Similar to binary but out of order to make every step a single bit flip. Since all cards face down isn't an option, the minimum number to cover all combos is 2^n-1 where n is the number of cards. (15 for 4).
I think it should be 2^n-1. Curiously, at n = 4, you get the same value either way, namely 15.
@@paulw987 That's a good point!
@@paulw987 correct, that was a typo.... edited :-) cheers
That is some amazing split camera game! Top notch!
@2:33 - Wait a minute, that was clever! 😂🤔
your editing skills are really paying off, this video was aweasom
Really slick editing Matt
amazing editing
4 bit grey code can go through all possible combinations in 15 flips, so 15.
There are 15 combinations, but can you go through them all in 15 bit flips? For example, going from 0001 to 0010 takes 2 flips
@@esquilax5563 Gray code.
@@esquilax5563 Yes, but that also hits/eliminates 0011. You "count" in an non-standard way.
There's a harder variant of this game. Put all four cards on a lazy susan, in a square. The guesser calls out cards to flip as "top left" "bottom right" etc. You don't have to answer after each flip, just whenever the guesser asks. If they ask, and the answer is no, you may spin the lazy susan. They then call out to flip more cards, which you flip by how the cards are now positioned after the spin.
That was an extraordinarily well realised reduplication. Bravo!
When I was a kid my dad installed a buzzer system so my grandpa could call for help. There were 4 pull chains throughout the house any one of them would sound the buzzer. One time someone pulled one of the chains and the buzzer went off. Since they were pull chains you could not tell which one had been pulled. He had to solve a this puzzle to silence the buzzer.
Ths sounds amazing, too good to be a real story.
Well, presumably you'd know that only one of them is on. Much simpler - go to each one in turn, pull it, if it doesn't stop this one was not the one that was on, so pull it again to turn it off. At most seven pulls.
@@alansmithee419 Congratulations on accidentally stumbling upon the answer to Matt's puzzle.
But if someone pulled two chains...
Great video Matt
"Filming and editing by Matt Parker and Matt Parker"
I'm honestly amazed at the synchro between the two Matts.
Martin I believe that the sound of first take is played during the second. Matt is careful to only talk over the other take when it is scripted. During edit the audio from the first take can be seemingly edited out by unmuting the second take only during the silence of the first.
Never have i replayed a section of video so many times!~
I was refreshing the website to get it as fast as possible since I was awake this time. I watch on 2x speed so I had time to finish solving the puzzle before the link opened to submit answers and I was just refreshing for 5 minutes. This one felt a lot easier than the last few, but that may just be because I spent more time with the math needed to solve it.
Solid edit work on the cards there
Passing the cards to yourself at 2:33 was a neat trick. Easy to do but effective. What stumped me was when Matt 1 picked the same cards up from off camera at 3:03 after Matt 2 had already played with them.
He reached out of frame, pretty sure it's a duplicate set of the same cards.
All possible combinations is like 4 bit binary. so total possible combinations 2^4. The maximum iterations will guessing every binary number by flipping only where the bit is 1. If by doing that is not the solution then we need flip the same again and then try the next binary number until we get a solution. So the maximum iterations X required to get a solution is X=((2^4)*2)-1
@2:55 Okay, well, the obvious and painstaking solution is to iterate over every combination of flipped or not from the initial. i.e. Flip 1, flip 1, flip 2, then 1, flip 1 and 2, flip 3, etc. Basically, by flipping individual bits count to 2^N - 1 where N is the number of cards. I don't think you can do any better without psychology.
I don't know the answer yet, but the maths about the complexity of finding the answer via exhaustive searches is actually quite interesting! 1. Create a tree whose root node has a value 'R' equals the number of possible starting combinations. 1a. If you're doing "up or down" rather than just "down," sets that are up/down mirrors of each other have identical solutions, so we can discard half of the starting combinations in this instance. (Modify R likewise.) This doesn't mean that the solution will be cut in half, but the complexity of finding it will be because you only have to do each operation on half as many sets. 2. Create as many branches as there are cards. For clarity, label them with the position of the card to be flipped. That's right, brute-forcing the search for a solution creates an N-ary tree, where N is the number of cards! Except it isn't. (See below.) 3. For each branch, calculate the state of each set of cards if you were to flip the card indicated by the branch's label. For each set that reaches a completed state, delete that set and decrement R by 1. Give this node the value of R and the state of the remaining sets. ("R" stands for "Remaining sets.") 4. Continue this search BREADTH-FIRST. The solution(s) is/are the shallowest node(s) with value "0" and the route to that/those node(s), so searching any branch shallower than the solution is insufficient and deeper than the solution is inefficient; thus, you want a search pattern that evaluates all branches to the same depth. 4a. HOWEVER, do NOT include a branch that represents flipping over the same card twice in a row, as this creates a loop. A loop is inherently less efficient than any other path running through that node, so it will not contain the solution. Thus, brute-forcing the solution is actually searching N trees of N-1 arity. 4b. If you want to abandon the tree model and just do a directed graph, after each node is calculated you can check if it's identical to any other node. If so, connect them and stop processing that branch. (If the node being duplicated is shallower than the latest node, this branch will not contain the solution. If the node being duplicated is at the same depth, it might represent convergent solutions. If both nodes are value "0," they represent independent solutions. If the node being duplicated is deeper than the latest node, you haven't followed the directions.) However, this might or might not be more efficient than simply continuing the search; it depends on how far you are from the solution. BONUS TREE: You can brute-force search for the all-down solution in the opposite direction! Do this by starting your tree with all cards face down. Construct the tree as previously, except for each node record the state the cards are currently in AS WELL AS all of the previous states to get to that node. Once you have accumulated all potential starting states of the cards, you have a solution (although possibly not the only one). Take the solution(s) with the shallowest depth and play the path from them back to the root node for the actual solution.
Not gonna lie, the 2:30 hand-off trick was pretty impressive.
To cycle through binary flipping one bit at a time, the simplest is using grey-code flips. For 4 items: 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. For 5 items: 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1. Eventually you'll hit it.
3:59 Sounds like a Gray Code. Let see.
When everyone forgets about the card trick and wonders how Matt was able to make the video xD
Its basicly gussing a string of bits throw noting them U can just go throw the motion of bit addition to loop throw them all once at a time and u will be done in 2^n-1
thank you jane street, very cool!
Of course someone said "Gray" before me...
JWentu It’s funny that it is an insight joke, until you generalize an answer.
Since the Coriolis video, I am still amazed how you do these "2 Matts at once" videos...
it is genius that you put this out on the right day :D
It's like Mastermind, but with less detailed feedback
Only have to get the order not the color so it doesn't need as much info
Basically it's a game of mastermind but much more limited: 4 pegs of only 2 colors and we only know whether they're all 4 of the right color.
I've just seen this video. I know I'm out of schedule so I don't mind answering here. My intuition is that the number of maximum moves is (2^n)-1, done in that way you can do all binary numbers in a row just changing a digit from one to another. So, with four cards that would be a maximum of 15 moves.
This is the first time I have been this early... I'm going to think about how this puzzle works. The answer is 15. Generally, it is 2^n - 1. For 1 card it's certain that you have 1 turn, and for 2 cards you have a maximum 3. My solution involved writing the state of the cards as a binary string, with "1" corresponding to a face-up solution. Then the flipping of cards corresponds to changing the bit-state of one of the bits. If we further translated the bits to decimal notation, we would have a number between 0 and 15. Flipping would correspond to addition (or subtraction) of a 2^x-1 where x would be the position of the card. Some things I would like to note: The 15-step solution doesn't depend on the fact that after each turn we are given a response, because its property is that no matter what the original state of the cards were, at some point in the 15 steps it will become certain that all cards are down. I originally thought that the information in the response we were given could help further optimize the process, but this doesn't seem to be the case. Also, I can't verify this, but I have a strong suspicion that for C cards there are C*(C-1)^((2^(n-1))-2) unique solutions of length 2^n - 1. Technically there are half as many, but the reverse of a solution is also a solution, so I didn't halve my answer.
almost! you don't need the zero position.. so count to 15.
@@Parax77 thanks! I now have a 15-step solution. It's probably not the only existing one, but... I have some notes: 1. For N cards, the solution has 2^N -1 steps. 2. There exist S unique solutions, where S is a term that contains N!... somewhere. 3. My solution corresponds to the ordering of numbers 7 3 1 0 8 12 14 6 2 10 11 9 13 5 4. Each unique solution corresponds to some ordering as well. Only some ordering ( better word would be permutations ) of 0 - 14 is a solutions hence why I said that thing about N! I'll write more about this tomorrow, since It is 00:28 in my time zone.
just counting in binary doesnt give you the optimal path, as you would have to flip over multiple cards at once to go from say 1 to 2 (or 001 to 010). does your 15-step solution account for this?
@@ffsnipe8298 Hello. I have a somewhat unrelated comment. I tried to find a good way to visualize the transformations, so I drew the points 0 to 15 evenly spaced on a large circle, and used lines to draw transformations. The result looked like a cool double fan pattern. Nothing particularly important I just thought it was cool. I'm going to expand this to higher powers of 2, and see if an analogue exists for different bases. Also it feels eerily familiar to the Mathologer video about the cardiods
that is some impressive coordination with yourself!
I really don't get how you manage to do the editing of the two you's so seamlessly.
I think the cards which are picked up by the right Matt are the same cards he threw. They are still on the table even after 2:39 (In the recording of the left Matt). That's why we couldn't spread the combinations across the whole table at 6:30 ... maybe :D
See other comments :p I've explained it there :p
Don't move the camera. Edit two videos side by side. He can't cross an imaginary line in the middle.
Brilliant editing:) reminds me of the review-review-review-braindamage:)
Good work passing the cards from the first to the second Matt.
Now that’s some snazzy editing!
For anyone wondering about 2:33, I’ll try to offer an explanation. The lighting in the video is very flat, to make the editing work since you can’t see shadows clearly. What happens is Matt 2 moves to the side so that the point of the “split” between the 2 takes is now closer to the right, instead of the “split” being in the middle of the screen. So when Matt 1 throws the cards, they land normally and Matt 2 picks them up and continues on. During the entire take for Matt 2, the cards are edited out and hidden until 2:33, when they’re made to reappear exactly at the time of the throw. TLDR, the cards were invisible on the table the whole time until 2:33
This is basically just like when you're using a superpermutation to try every possible garage door remote code in 11 seconds, right?
That was slick with the card handoff
I immediately think of super-permutations which I know Matt loves.
Last semester, I wrote a blind-search algorithm that solves 8-tile puzzles, which can be formulated as equivalent to this puzzle. However, due to the massive state-space, a large number of states can't be navigated to a solution due to being practically intractable.
this reminds me of the 4D cousin of the cube why because the three cards case correspond to walking on the edges of a cube (looks like this time I will solve it)
It's up to you I guess, but I wouldn't post something so close to a solution in the comments where someone who doesn't want to see it might see it.
Agree with Zac, this is alluding to the answer Having said that, I was working in terms of bits/binary, but thinking about it geometrically feels very intuitive
You are getting REALLY good at split-screen filming. :) Including getting the interactions to seem natural and realistic. How many takes did this take to get right?
This was the third take! The first two were close but not quite.
"This is interesting." Tosses the card to the other side. "Bah ok, that's clever, but he just tossed it and then later had the other take use those cards..." Takes the cards back. "HE'S A WIZARD!!!!"
me: They are wearing the same t-shirt... *awkward* them: Nice t-shirt!
Men are not bothered by such things, whereas women would lose their shirt (did I spell that correctly?)
About the only time I've heard someone refer to themself in third person and not been annoyed by it
Humble pi has made its way to Amazon shelves here in Ohio! Just picked a book off the shelves yesterday!