This was a blast, thanks for making it happen!

3B1B looks almost exactly how I didn't imagine him

"Mathematicians don't like […] losing, so they simply prove they can't win"

"And it takes years to forget (the solution)" Grants memory is wild. I watched this a few months ago and it's like I'm hearing it for the first time again lol

It seems like this assumed that you know which side of the board is the top. Note: The chessboards I have owned don't have letters or numbers, so I hadn't assumed this one was labeled. It looks like the chessboard in this video is also unlabeled. People keep suggesting that the labels are a solution, so I thought it would be helpful to clarify (note added Sept 2021)

Matt's channel has to have the weirdest watch time analytics. Everyone seems to watch like 4 mins of his videos... pauses it... then restarts it sometime between 5 seconds and 2 weeks later.

Seeing 3Blue1Brown in person was slightly mind blowing. I always assumed him to be a disembodied voice.

I’m sorry, but the most impressive part of this whole process was Grant slapping down 43 in binary just casually as it happens to come up.

Warden: **rotates chessboard** **evil laugh**

You could record 3B1B on the worst mic possible and he would STILL sound like that Edit: I think people in the replies are missing the joke here. Me, someone who is NOT an audiophile, finds 3B1Bs voice to be photogenic in a sound way.

I had to think about it for a few minutes (after watching the video) and I think I get it. So the first prisoner who sees where the coin is hidden follows these steps: 1. Determine the binary number that represents the square where the key is. In his example, it's square 33 so: 100001 2. Determine the binary number currently encoded in the board state by counting whether there is an odd or even number of heads in each region. 3. Determine the binary number that needs to be added/subtracted from the current board state so that it will equal the binary number representing the square the key is in. 4. Flip the coin in the square represented by the binary number that you need to add/subtract, as doing so will leave the board in a state that encodes the location of the key. Then the second prisoner simply needs to determine the number encoded in the current board state, and look in that square for the key. Brilliant! What an elegant and simple solution to this seemingly impossible puzzle.

To avoid the evil warden you need to have each of the prisoners create RSA key pairs so they can communicate without being eavesdropped.

Imagine spending days solving this puzzle, explaining your strategy to your inmate, practicing finding the right coin to flip, and then when the day comes he miscounts and picks the wrong square

"One is binding, small click from two..."

I came across this puzzle about 5 years ago when a colleague told me about it. Being a computer programmer I immediately turned to solving it using computing and worked it out after about 4 days. Wrote a little c program for it, and was very chuffed with myself. I told my girlfriend at the time and she wasn't that impressed.

Math channels have to be one of the only ones to say "pause right now and think about it! Please leave the video you might regret if you don't" lol

My biggest takeaway from this was the throwaway thing at 28:53 - “I was thinking 5 shifted twice plus 1.” Because holy crap, I never thought about binary that way - shifting any number over doubles it! This is going to *revolutionize* my mental math.

There is a dead pixel on your camera [void screams]

Even with the solution, it's still crazy to think about how you can extract any information from any situation with a single bit flip.

I've seen the start of this video three years ago, and since then was thinking about it sometimes, when i had to wait somewhere. Yesterday evening when watching out for my baby, i finally found the solution and now watched the rest of the video. Thanks a lot for those three years :) I also think you can calculate it a little easier by solving the row and column indenpently, since it is already layed out nicely into this square. You just merge every row, by counting the heads in that row (even = 1, odd=0), then you only have to do it up the number of 8 (3 bits), and later flip the coin which is in the row AND column you have to change.

"we have only one mic, and the guest's voice is more important than mine."

They got out of that prison fairly easily. You might even say that it's a Parker prison.

Grant: "I was thinking 'Five shifted twice plus one' heh heh, everyone thinks about binary differently." Me: this changes *everything*

This is probably the best math video I have ever seen (and both channels make a lot of great content about math, so this says a lot). Not only they take a very intricate puzzle and show how to solve it using somewhat simple math, but also they talk about their mental processes and what they were thinking about when trying to solve the puzzle. Normally, as a math undergrad, our classes (and by extent the content created by mathematicians) are just definition and theorem, definition and theorem, without any intuition about the processes (both historically and particularly) that took to think a problem, to solve a theorem, etc. It is great to see other fellow mathematicians showing their ideais not in the "already polished" away but in its raw form. I think this does a great job to show what a mathematician really does in their daily lives, instead of just laying out the pretty, compacted, textbook results

When your two smart friends are discussing their answers to the exam

When mathematicians talk they joke about induction .

Mom: What are you watching? Me: Im watching people doing meth. Mom (quietly): Time to do error correction.

I actually realized it before watching to the end, to be precise, right after Grant mentioned *finite field* . My idea is that to label each block 0-63 in binary (6bit) and add up all the positions with head with finite field additions (xor), lets call it "sum". The result will be a 6bit binary vector. Since we have full control of all binary vectors, and since finite field addition (xor) satisfies the property that (a xor b) xor a = b, we can find the "difference" between the "sum" and the position of the key (in 6bit binary vector representation), and flip the coin that has binary value of that "difference". This way, the "new sum" will be the position of key!!! And they did exactly that. The only thing that I didn't think through is how to compute it efficiently, since doing xor on this many numbers manually is kind of slow and error prone. Their technique of using strips and blocks just makes it a lot easier and doable in reality. The key inspiration here is *finite field* and that property of xor, which I learned from Ben Eater's video introducing CRC (cyclic redundancy code): ukposts.info/have/v-deo/oap4a6eLZ3uho6s.html an actual error checking code that is still widely used in production. Thanks for the great content, this puzzle was a lot of fun to think about.

"...but that is a moot point because Grant can visualise cubes in all dimension spaces with equal ease." lmaoo i'm weak

I loved Grant's method of splitting the board into stripes, bands, and blocks. It reminded me of how the towers of hanoi puzzle is related to counting numbers in base 3 and just how efficient the system of numeric bases is in counting numbers.

I love this video. I walked through this puzzle with my son and he got it right away as soon as I got to part of the solution where you start to figure out which coin to flip, then it clicked for him and he got the rest himself. The solution and the puzzle are so elegant, it's beautiful. Thank you!

For those of us that love the math but don’t have a lot of time, the explanation begins at 9:20

In the discussion at around 29:12, what you're looking at is literally just the two coordinates of the chessboard. 101 = 5 010 = 2 So the thing to flip is at (5, 2) or ((6, 3) if you're feeling like 1 indexing). And at 29:39 that is indeed the square Grant is pointing at.

I immediately jumped to Hamming codes when hearing this, but only because of Grant and Ben's videos. Looking forward to more errors correction math. Ya'll are great!

Man I didn’t expect there to be a practical solution. What I was able to find was a proof that for any size 2^n board it is possible to assign each unique 2^n square to a set of 2^(2^n - n) cases in such a way so that each unique number is only 1 flip away from any of the other sets for unique numbers. This will require both prisoners to have all 2^64 cases memorized but I now see there is a more practical way of organizing them so as to not just memorize.

5:21 "it takes years to forget about (the solution)" can confirm, I'm here 3 years later and I still remembered

Here's my magic trick variation of the puzzle: My accomplice is blindfolded, then someone in the audience set's up a 4x4 square with coins randomly heads-up or heads down. Then I tell them that we'll both flip over a coin and then my accomplice will tell them which coin THEY flipped. I even offer them to go first, I just flip the board to "zero". No matter which coin they flip over, it will always encode itself. There are some tricks to do the calculation in your head within seconds (Even drunk people you've just taught the trick to.).

It's been a long time coming since that Punchcard machine collab ( Parker Grant Convergence 2)

I had heard this puzzle about 4 years ago but never solved it. Thanks for giving me the opportunity to pause the video and solve it, cause I actually did! It took me about 2 days, was pretty tough, and I reasoned about it in a similar way. Once I realized "the trick," the pieces came together and I figured it out, so I watched the rest of the video to see that I was right! Great video, thanks a lot for this.

Really like how you too spent the time to discuss how you came to the solutions, gives a nice insight into the thought process when solving a puzzle like this. Also can really see the enthusiasm leak through :)

There’s a tiny white spot or dead pixel or something on Matt’s face a bit before half way through the video and I can’t stop looking at it

Got it! Took half an hour but I figured it out. Without watching the video, my sollution: You start with 000000 Each tile on the board represents a configuration of digits to flip, so the first one is flip nothing, the next 6 are flip one digit, the next 15 are flip 2 digits and so on. This ends up filling the board, and no matter what 6 digit binary number you end up with, you can always just flip to the right configuration with one coin flip. Man, I enjoyed this one! Not impossible but still difficult. There probably are multiple valid sollutions, so I do wonder if they came up with the same one.

I'm confused about the three coin problem. Suppose the key is under the "c" coin (so you need to show a result of 2), and the current coin configuration is 010. In the 0*a+1*b+2*c operation, 010 results in 1. The configurations that lead to a result of 2 are 001 and 101. It's impossible to flip one coin and turn the 010 into either of those. What am I missing? Edit: I was missing the rest of the video when they say this doesn't work.

This ended up being one of the most enjoyable problem-solving experiences for me! I paused the video, and launched into 8 days of thinking, discussing with friends, writing Python code, and finally ended up with a solution. My solution involved assigning all possible combinations of 1, 2, ..., 6 colors to the squares. Conceptually the same, but the regions ended up being very messy and hard to count. Encoding the squares with binary numbers is much more elegant. Thanks for sharing this fantastic puzzle!

I learned about this puzzle in 2013 and it stewed in my brain until *2018* before I solved it! And I still wasn’t satisfied that it was the most elegant solution (I used 3 overlapping quaternary bits and a 4-binary bit solution smashed together) love to finally see this video!

I had already spent a while trying to solve it, but did not manage, so then I decided to just watch the video. However, at 17:16 you convinced me to give it one more go and after thinking about it off and on for two more days I managed to figure it out! It was very rewarding to solve it by myself, so thank you for this final encouragement and thank you for making videos about this fantastic puzzle!

Matt, Grant, Michael Penn, and redpenbluepen top notch production. With each offering unique presentations and approaches, one can really absorb the concepts and appreciate the quality. You two provide the long story, ones who go the mile to really set up and prepare the student. The latter two are a quicker pace and is more geared towards proofs and practice. Thanks you and best wishes. 🍻

I just found myself skipping over the part where you act it out, to get to the part where you actually explain how it works, just to pause right there to give it a try myself. I am incredibly thankful you explicitly mentioned how you can only learn the solution once :D

Wow, I suggested Matt apply for a Grant last week, and boom! Quickest approval process in history.

Alternative title: Two mathematicians bully viewers into solving a puzzle. xD

Just to point out something interesting about the way the problem is posed: In chess computing, a method of hashing known as Zobrist hashing is used to make hashing of successive chessboard positions cheaper. A Zobrist hash has the property that it can be computed cheaply from the Zobrist hash of a related chessboard position in time proportional to the number of altered squares on the chessboard. Since successive positions in chess differ in at most two squares, this is a huge computational savings over recomputing a hash that requires you to consider all 64 squares at once. The chessboard puzzle posed here is closely related to the way Zobrist hashing is actually implemented. I would go so far as to say that the puzzle was designed by someone familiar with the technique, and provocatively posed on a chessboard as a nod to it.

After seeing 3b1b's video on this. I solved thus puzzle on a road trip. Hours of fun! I eventually came up with the same solution but with the meanings of the coins rearanged (the conis meant the same things, but were in different places on the board). My thaught process was wildly different and my explanation on how it worked was MUCH messier, but it was essentially the same. I love how two seemingly different solutions with almost completely different processes came out to be the same thing, this is why I love math. (By the way 3b1b's video helped a ton in steering me in the right direction) This puzzle made an otherwize boring car ride fun, thanks Matt and Grant and keep up the amazing work.

I managed to figure out a solution, but instead of visualizing the 64 squares as a board or a line, I visualized them as a 6-dimensional hypercube (2x2x2x2x2x2). Comparing the coins in an edge of the hypercube already gives you a bit of information. There is also always a way to flip a coin in order to show where the coin is. For each square, its identifier is x+8*y in base 2, where x and y are the coordinates on the chessboard (I know one side is A to H, and the other one 1 to 8, but here, all sides are 0 to 7 instead). For example, 000000 (A1) is the null bit, as it doesn't affect anything, and 111111 (H8) affects every single bit. To get your board identifier, start at 000000, go through every square, and if it's heads, XOR the value with the current identifier. Then XOR the board ID with the target ID, and there you go! The decoding is just: calculate board ID, then that's it!

So if I got it correctly: The prisoners decide on this scheme to encode the chess board in 6 bits - the number of bits enough to represent the numbers from 0 to 63. Prisoner 1 encodes the initial board using this scheme, and then chooses the coin to be flipped. The coin chosen is special. When it is flipped, it changes the encoding of the board. The 6-bit encoding of this new board is now the binary equivalent of the _cell on the board which stores the key!_ So now, prisoner 2 can just figure out the 6-bit encoding of the board and bada boom bada bing, the key is revealed.

By XORing each row and column, you can speed up the addition process a bit. Basically, I ended up with sixteen binary bits that you can only effect two of (I actually had them in two eight digit numbers, one horizontal and one vertical). You can then apply the same error checking algorithm to get six binary bits that code for a position on the board (that can be changed to any other position by affecting only two of the digits, one vertical and one horizontal)! Beautiful puzzle. Thanks guys!

Thank you for pushing me to look for a solution before giving the answer. The "you can't forget a solution once it's been given and there are few good riddles" pushed me to pause and think. While I initially had nothing comme tu mind and was therefore not going to try, a few minutes in and the solution was forming. And I'm pretty happy I could work it out

Here is the solution I came up with when you said "Most people don't pause the video": First, let's assume that the board is always gonna be 2^m. (In other cases, we can simply consider fictional tiles always filled with 1s (heads) or 0s(tails)). Second, let's start with the simple case, a board of 2x2 tiles. We want to pass two "bits" of information. So we need to somehow find two independent comparisons that we can make and change them independently. For example, We compare bottomleft and bottomright, and then compare topleft and bottomright. Given a random board, we want to transmit "the key is in left | right half" and "the key is in top | bottom half". Let's say, if topleft and bottomright are of different parity, then the key is in the top half. And if bottomleft and bottomright are of different parity, then the key is in the left half. You can see clearly that you always have an option to flip only one tile and transmit those two bits of information (i.e. bottomright if you want to change both parities, topright if you want to change nothing, and the two others quarters for each axis separately.) Now, to transition from the case of 2x2 to 2^n by 2^n, you simply need to observe that each you are always bound to one "quarter" of the whole board. For example, assume the board is 4x4. Once you run your initial reasoning over which quarter to change, then you are bound to choose the tile to flip in that quarter. So the solution then is to compare the sum of tiles in each corner from each quarter. 0|1|2|3 4|5|6|7 8|9|a|b c|d|e|f First comparison: "0123" vs "abef" and "89cd" vs "abef" Second comparison: "028a" vs "57df" and "46ce" vs "57df" The first comparison will tell you which quarter the key is in (and hence which tile you need to flip) The second comparison narrows it down inside each quarter's quarter. And you can easily apply this to any 2^n by 2^n board. For any arbitrary number, I feel lazy to think about it, so just ask the warden for a bigger board :P Say something like "it would be more difficult to solve"

17:10 "You can't feel that bad because anyone watching is looking it up." Not me! I solved it last night after watching the 3Blue1Brown video (the one that explains why it's only possible when the number of squares is a power of two) and am watching this just to compare my approach to yours. :)

After months of not completing watching this video i came up with a solution... and i think its easier to understand than the solution in this video - but harder to calculate: The set of all subsets of a set with N elements is 2^N. We need 6 bits to encode the position of the key. So the board fields 1 to 6 are designated for that purpose - we name them "encoding fields". All other fields are assigned one subset of those 6 bits (which is 2^6 = 64) except the subsets with one element... which happens to be exactly 58, including the empty set (no change if flipped) as well as the set that includes all 6 encoding fields. We name those "subset fields". Now each of the 58 subset fields - if it has heads coin - is interpreted to invert the bits of its subset out of the first 6 encoding fields. Now the coins with heads on the 58 subsequent fields convert the binary value of the first 6 coins into some other value from 0-63. If only 1 bit of that value is incorrect, you just flip that bit. If 2 or more bits are incorrect, there is a subset field that flips exactly those ... and that is the coin that needs to be flipped. All the 2 prisoners need to agree on before the game, is which fields are assigned the 6 encoding bits and the which field is assigned each subset.

Such a lovely puzzle! Seems completely impossible at first, but actually quite straightforward. Thanks for making this video and sharing!

These guys are the Gods of "Recreational Math"... I loved it 🤩 thank you so so much 🥰!

I understood how prisioner 1 finds out which coin to flip, but how does prisioner 2 finds out where is the key?

I found 5 solutions: 1. Just rub a coin against the wall, creating a flat edge. Tell your friend that the coin with the flat edge is the coin with the key underneath. 2. Yell the square with the coin. For example, "G5!!!". Your friend now knows where the key is. 3. Take all the coins and stack it on one square, the square the key is on. 4. Find the key yourself by throwing all of the coins off the board. 5. Kill the malicious warden, perhaps using the coins and/or board.

Each tile gets a number from 0 to 63. Convert each tile number to binary. Take all the tile numbers with a coin that has heads on it and xor them to get a new binary number, which I'll call A. Let B be the binary representation of the tile number of the tile with the key. Take the xor of A and B to get a binary number which tells you which tile to flip. For example if only tiles 3, 10, and 44 have a coin with heads, then their binaries will be 000011, 001010, 101100 and the xor value that comes out of this will be A = 100101. If the key is on tile B=25(DEC)=011001(BIN), then the tile we want to flip is 111100(BIN)=60(DEC). This strategy works on any board of size 2^n (in this case 2^6).

Now that I understand the solution, weirdly, I have an easier time explaining it without the visual support. So, for other people out there that see things like me, maybe this will be easier. Each square with heads on it translates to the square index, from 0 to 63, written in binary. You want prisoner 2 to be able to sum, with the XOR operator (addition without carry), all squares with heads, and having the sum represent the square where the key is. Now, XOR has the nice property that flipping from 0 to 1 or from 1 to 0 are both represented by XORing the square index, i.e. X xor X = 0. So, flipping a coin is the same as XORing its index to the sum. As prisoner 1, you are given a board with a starting sum, let's call it WARDEN, and let's call the key position KEY. You must change the starting sum by one flip so that the new sum is KEY, resulting in the equation WARDEN xor FLIP = KEY. Rearranging the equation, you get FLIP = WARDEN xor KEY. So, as prisoner 1, you list all squares with heads AND the square with the key, XOR them together, and this gives you the square to flip. Also, even if it doesn't prove anything, you can see that, if there were fewer than 64 squares (i.e. if the number of squares is not a power of two), the value FLIP you computed could end up giving you a non-existent index, so the algorithm wouldn't work. Hope this may help!

Before I viewed their solution, I came up with my own which is very similar, but I think mine is a little easier to compute by hand. 1) Number the rows and columns from 0 to 7. 2) Number the squares sᵢⱼ where 0 ≤ i ≤ 7 and 0 ≤ j ≤ 7. 3) Compute the "column bits" by XORing the bits in each column: cᵢ = sᵢ₀ ⊕ sᵢ₁ ⊕ ... ⊕ sᵢ₇ 4) Compute the "row bits" by XORing the bits in each row: rⱼ = s₀ⱼ ⊕ s₁ⱼ ⊕ ... ⊕ s₇ⱼ 5) Compute the column value by multiplying the row numbers by the corresponding column bits and then bitwise XORing the results: C = 0c₀ ⊕ 1c₁ ⊕ 2c₂ ⊕ ... ⊕ 7c₇ 6) Compute the row value by multiplying the row numbers by the corresponding row bits and then bitwise XORing the results: R = 0r₀ ⊕ 1r₁ ⊕ 2r₂ ⊕ ... ⊕ 7r₇ Let the co-ordinates of the square where the key is hidden be (Cₕ, Rₕ) (h for hidden). We compute the co-ordinates of the square to flip (i,j) by XORing the current values with the hidden values: (i,j) = (C ⊕ Cₕ, Rₜ ⊕ Rₕ). Why does any of this work? If we flip the bit sᵢⱼ, it will cause column bit cᵢ and row bit rⱼ to flip, while all the other column and row bits are unaffected. If we compute the new column value C', first note that the new value of cᵢ' is (1⊕cᵢ) and if we substitute that into the expression for the column value we see that C' = 0c₀ ⊕ 1c₁ ⊕ ... ⊕ (i-1)cᵢ₋₁ ⊕ i(1⊕cᵢ) ⊕ (i+1)cᵢ+₁ ⊕ ... ⊕ 7c₇. But i(1⊕cᵢ) = i ⊕ icᵢ. And XOR is commutative and associative just like regular addition so we can move that i to the end of the expression and we get C' = 0c₀ ⊕ 1c₁ ⊕ 2c₂ ⊕ ... ⊕ 7c₇ ⊕ i = C ⊕ i = C ⊕ (C ⊕ Cₕ) (by definition of i) = (C ⊕ C) ⊕ Cₕ (because XOR is associative) = 0 ⊕ Cₕ (because XOR is its own inverse) = Cₕ The exact same thing works for the rows, so we've solved the puzzle! The second prisoner computes the row and column values and they point to the key. I made a spreadsheet you can play around with in Google sheets if you're interested docs.google.com/spreadsheets/d/1cVzJ-7KortGh7TmGKOnoc6XyOJix3Gu4rzVwfLnmf1U/edit?usp=sharing

Well, I just thought along while you guys were working it out, and my non-mathematician brain was thinking in sort of the correct direction (needing to make some sort of composite binary number to represent each square), but lucky for me, even hearing the explanation only got me 70% there. So I still had the pleasure of figuring out how it works EXACTLY, by explaining it to my mother while working out the specifics through trial and error. Explaining stuff always helps you get a better grasp of it so I still had the fun of figuring it out by myself at least partially and it's another fun small trick to make wagers on with family members on a holiday or something.

I paused every time you said to, stopped, and gave up. Untill the last time you said it, when I finally solved it. I may have picked up a hint or two from your thought process, but I did figure it out.

I think I see a dead pixel or something in your camera, a very clearly discrepant pixel at 9:06 to the left of Matt's face.

I thought that this dead pixel on Matt Parker's head was some dirt on my monitor, and I almost broke my monitor screen trying to clean it.

What a brilliant way to explain it. Over the course of about 2 months of on and off solving, I eventually "cracked the code" with a slightly different algorithm for mapping coin arrangements to positions (I tested its validity with a program), but didn't have an understanding of why it worked until watching this video (maybe more than once).

Great video! I watched the first part a few weeks ago and have come back to see if my solution matches yours. I came up with something different than the solution you discuss with Matt. Interestingly, my solution provides a different way of glimpsing into which configurations may or may not be possible. Here's what I did: Take the parity of each row, and the parity of each column. Each is an eight-bit sequence. Note that by choosing a coin on the main grid to flip, we can selectively switch any one bit of the "row parity" sequence and any one bit of the "column parity" sequence. We now have a new challenge (or rather, two identical challenges): With our eight bits of "row parity", flip exactly one of them to communicate the row containing the key. And independently do the same for the columns. Hm... This feels like a familiar puzzle. In fact it's the same one as the original with only eight coins! Done twice! I'm still stuck. Eight is a lot. But... If we arrange THOSE eight in a 2x4 board, and take the parity of the rows and columns of THAT, we can reduce this eight-coin puzzle to a four-coin version and a two-coin version in the very same way we reduced the original! And that four-coin version can be reduced further by laying it out in a 2x2 grid to get two 2-coin versions. At that point, solve by hand. For me I arbitrarily decided that 00 and 10 point to the first coin, and 01 or 11 point to the second. Bubble up these smaller puzzles to eventually find which coin of the row-sequence needs to flip, and which of the column-sequence to flip, and then flip the corresponding coin in the big chess board. I imagine that my solution implicitly is just selecting a different way to partition the board into the six regions overlapping regions you discuss with Matt. What's fun is trying my solution on different board sizes. A 6x6 looks easy! Take each row and column, let's lay THOSE out in a 2x3 and... hm. I also struggled to color my 3-color cube before deciding it was a no-go. In terms of generalizing my solution, we can say that an N-tile board is solvable if each of N's prime factors is "solvable" (since we factor it down with each cris-cross reduction). After watching this, I'd wager that 2 is the only such solvable prime - which means the powers of 2 are the only viable boards! Thanks so much for this treasured riddle. As you mention to Matt, a good puzzle is a real gem and this one definitely gave me a ride. Let me know what you think of my solution.

'managed to find a dungeon' sure...

Hey Matt - I do audio engineering and dialogue cleanup - If you send me the audio, I can clean it up for you

Excellent video sir , just loved the whole process, I learnt a lot from this video

love to watch both of your channels . your guys are doing great job. I solved this puzzle back in 2015. This could be good interview question.

The first method works perfectly. You don't need the 6 regions at all. Instead of the "classic" addition, you should use the "XOR" addition where adding and substracting are the same. Write binary numbers and decide that for every digit 1+1=0. That's it.

The solution starts at 22:09

This was awesome. I feel smarter just from sitting around listening to you guys being smart. Thank you.

This is the best math riddle solution that has ever been made.

As was the case here, I like to skip to the solution part and try to work out what the puzzle was. If you're like me, start watching at 9:07

“Take some time and solve this puzzle on your own” the most advanced math I took in college was statistics and all I knew was excel spread sheets I CANNOT fathom what I was suppose to come up with

I came up with a bit different solution. First of all I modified the problem of 64 fields into 8 columns and 8 rows. It can be done by XORing all values in every column to get row vector, then XORing all values in every row to get column vector. And this was I think quite genuine idea. Then I've found a solution to bit-vector of 8 which was similar to yours but a bit more complex (but still possible to do as it was only on 8 fields) which was based also on parity, halving vector and much of XORing parts of the vector. After watching the rest of your show I've found that it could also be solved easier by your idea simplified.

I paused and thought about this for about 2 days and at the end this was exactly the solution that I found out. But I think you are right, it probably felt much more gorgeous than just continue watching and getting the solution.

Two of my favorite youtube math-heads collaborating on a math puzzle? That's like... *counts on fingers and uses a couple of toes as well*... AT LEAST TWICE THE AWESOME!

"If I can't do it, I must prove nobody can!" Does the solution require that you need to agree upon the axis and their direction of them? A malicious warden might just put the chair on the other side of the board?

Well I font know if I will solve it but thqnk you a lot for actively encuraging people and showing that it's a good thing to just try and find joy in that..

The warden: rotates the board 90 degrees when you leave the room

6:18 I'm leaving this comment here so in a few days when I think I have solved it, I can just resume the video where I left off.

6:28 This is how I imagine Matt leaves a room 100% of the time.

I approached with the goal of halving the possibilities 6 times. This is how to approach it as prisoner 2: 1. If head on black are even, then key is under a black square 2. If the left halve of the board has an even number of heads, then the key is on the left side of the board 3. If the top half of the board has an even number of heads, then the key is on the top half of the board 4. If there an even number of heads across the even ranks, then the key is on an even rank 5. If there are an even number of heads across the even file pairs (files C&D and G&H), the key is on an even file pair. 6. If there are an even number of heads across the even rank pairs (ranks 3&4 and 7&8), the key is on an even rank pair. At the end of the list you will always end up with one possible square Prisoner 1 just has to follow the same list. If the condition is met don't mess with it, but if it's not, you do. (E.g. the key is under a white square and the black squares have an odd number of heads, ensure the coin you eventually flip is white as to not change the amount of heads on black squares.) And as you progress through the list, much like prisoner 2, you will eventually arrive at one flipping option. Tested it once. Seems to work EDIT: I should've clarified that when a statement is false, the opposite is true. (E.G. If there is an odd number of heads of the left, then the key is on the right. Etc.) Also, watching you explain it make my brain hurt. I guess I sort of used binary with even/odd but Im honestly not sure. But you are highlighting the board similar to the way I did it!

What a great puzzle! (I did hit pause and went away for a day to solve it, and very glad I did.) Thanks for sharing it!

Me, doing a bunch of extra work for the simplest of math problems. My math professor: "Well, see there's a simpler way to do this."

A variation which IMHO makes the trick more impressive, but easier, is to have the warden arrange the coins in whatever he sees, fit, then have prisoner #1 flip a single coin and leave the room, and *then* have the warden place the key and flip (only) the coin where the key is placed. So neither prisoner gets to see where the key is, but prisoner #2 finds it anyway. The strategy prisoner #1 uses to select a coin to flip is the same as the one prisoner #2 uses to identify the coin that was flipped.

I am so curious about the solution! I am trying to figure it out by myself. I think it has something to do with binary numbers, because I found some patterns when I converted the codes assigned to a square to decimal in the 2x2 case. What a great problem! Thanks for sharing it with us.

Brilliant puzzle and elegant solution. It's obvious of course. Now that I've watched the video!

Takes years to forget? You underestimate my abilities

What madness is this, can't watch both vids at same time

as much impressive it was to come up with the solution of this puzzle, I am in awe with the idea of coming up with such a puzzle. can't get my head around it.

I solved this so differently. Taking inspiration from 3B1B's video, I figured out a transition matrix that was symmetrical and also maintained a specific coin to act as the identity operation. Funnily enough, this ended up meaning that inversion was also an identity operation. I started with the 2x2, then 4x4, then finally 8x8 case. I ended up with a 64x64 transition matrix that had some really cool fractal qualities. At worst, it took my solution 32 steps to solve it, checking back and forth between which coins I was flipping to match and what state that left the board in. Finally, and similar to their solution, you flip the coin that transitions from the state the board is in to the state pointing to the coin. I'm sure it's mathematically the same solution (or a congruent one), and mine takes a lot longer, and a lot more thinking, but I had a lot of fun doing it my way! And oops, I spent 5 hours on it.