This logic puzzle stumped ChatGPT. Can you solve it?
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- Опубликовано: 25 май 2024
- The world's best AI cannot solve this simple question! Can you figure it out? Four glasses are in a row face up. What's the minimum number of moves to make them down, if you have to invert 3 glasses at a time? What if you have n glasses and you have to invert n - 1 glasses at a time? Special thanks this month to: Mike Robertson, Daniel Lewis, Kyle, Lee Redden. Thanks to all supporters on Patreon! / mindyourdecisions
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6:08 general 1
10:49 general 2
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There is a solution that works for any number of glasses and requires only 0 inverting steps... you take all the glasses and carefully, without reverting any of them, travel to the exact opposite point of the Earth's surface. DISCLAIMER: doesn't work for flat earthers.
😅
that's called thinking out of the box and i personally love the dig at flat earthers too, 10/10 comment
Gotta love how people who don't understand Flat Earth Theory mock it 😂. Same as it ever was, sadly
Save the cost of the plane ticket and just wait 12 hours
@@MS-sv1tr You mean there are people that take Flat Earth Theory seriously?
move 0 (or just the starting position):0000
move 1:1110
move 2:1001
move 3:0010
move 4:1111
0=up-right
1=upsidedown
I got the exact same outcome in about a minute of thinking lol
same here, not really hard
Equivalently you can think this way: the glasses are automatically inversed and each time you choose one glass not to get inversed. Remark that for a glass to get inversed, it needs to get inversed odd times. Therefore, if you choose one glasses at a time for four inversions, every glass gets inversed three times and gets inversed!
Brilliant!
I was typing up my own independent comment on this before looking at yours and realizing that yep, that's basically how I solved it.
Yeah, which I think means the problem can only be solved if n is even.
Example in detail (order doesn't actually matter)
Starting uuuu
invert all but the first uddd
invert all but the second dduu
invert all but the third uuud
invert all but the forth dddd
Generalizing: The number of glasses equals the number of moves.
alternatively you can think about it as invertin one glass then and inverting up and down. to invert all glasses you need to have inverted all glasess and have up and down in the same way as in the begining
I got so excited when I solved it without looking at the video then immediately humbled when I saw the question was MUCH more complicated.
SAME :(
It takes two moves to reach a position where two of the glasses are upside down. This is a symmetrical situation which could be reached from either the initial position or the desired final position. That means the minimum number of moves is four.
It is even simpler
You have 4 glasses, but can only turn 3 of them each round.
To turn everything you must turn total number of glasses equal to number that is simultaneously divisible by 3 and 4. Smallest such number is 12. Meaning you will turn total 12 glasses, aka 4 rounds of 3 glasses.
If you turn 6 glasses you will get 6/4=1|2, aka 2 glasses in wrong position.
@@DimkaTsvno. you can do 8 and 6 with 3 moves, turning over a total of 18.
UUUUUUUU
UUDDDDDD
UDUUUUUD
DDDDDDDD
@@MichaelDarrow-tr1mn
Ok, in THAT case it was much easier.
In general rule is much more complicated and take in account unique combinations and even/odd nuances. Rule of smaller divisible will only work on unique combinations where difference between number of glasses and number of turning glasses is odd.
In case when difference between numbers is even, number of steps will ALWAYS be 3 [unless allowed to turn number is less than third of total]. (for example 9 to 7 or 9 to 5 or 9 to 3)
This task also will fail if number of total glasses is odd, but you can only turn even glasses at a time. Because even if you would be able to make difference in one or more glasses glass, you will always return them to upside glasses turn on each "odd" step, meaning you wouldn't be able to make it asymmetric. (for example 9 to 4, 9 to 6 and 9 to 8)
Every position is just a linear combination on Z/2Z of the 4 possible moves, since moves are commutative. The question is then "find the coefficients given the basis".
@@Horopter Z/2Z is the residual class ring to 2, so the natural numbers modulo 2; so all values are either 0 or 1 (normal or flipped); which glass you move first is irrelevant since you could just rearrange them
the important part here is that inverting glasses is basically adding +1 in the Z/2Z so 0+1 = 1 and 1+1 = 2 = 0
It's not obvious that the four moves form a basis though.
The simples terms to think of this is, there is in each situation until the final move only one possible move that doesn’t recreate a previous situation
This is exactly how I thought of it. Aside from the first move, you always have two options in each move. One of those options makes you go backwards to the previous position, which obviously cannot be how you achieve the minimum number of moves.
Saying this puzzle "stumps" ChatGPT or Google AI plays into the myth that these applications are capable of thought - they are not. They're using statistical modeling to choose words and phrases. Stop it.
Yes they are a different version of the same search engine we had from 90s
@@ralanham76 they go back to an application called Niall from the Amiga era
How do you know *you* are capable of thought? What exactly is it? And how can you know that a logical "thought" process does not arise through the artificial neurons, just as it arises in our natural ones? No one knows that - both the brain and neural networks are black boxes.
@@thefirstuwu8874 👎
"Stop it."
Remember: LLMs being called intelligent is marketing hype. They're a superpowered autopredict, not intelligent.
Curiously though, any human that could perform a similar task of superpowered pattern matching and prediction would be considered pretty damn intelligent.
Just as, if you could do mental arithmetic, as fast and as accurately as a simple $1 calculator, you’d be considered a genius in mental arithmetic.
So why are tasks that if I human did them they would be considered highly intelligent, are also evidence of a LLM not being intelligent when those same tasks are done by it?
@@SmoMo_ because we reason things out. LLMs literally just try to predict the next word over and over. If you ask it to explain how it arrived at an answer, the response will simply be, again, a series of words it's trained to predict as the most likely to be a response.
It's why they're so prone to "lying" -- they have no concept of truth, they're just outputting the words that the model has been trained to predict as most likely to occur.
Don't get me wrong, they're very impressive in their ability to fake intelligence. But they're not intelligent.
@@phasm42 Artificial intelligence doesn't actually have to be intelligent to ruin your life. It just needs to convince your boss that it is.
Funny, we did this in 4th grade in the 1990's in a game called "Math castle" but it was light-switches instead of glasses.
Since glass position does not matter only the Up or Down positioning, this is a simple state model where the initial state is 4U0D and you want to get to 0U4D by changing a total 3 U to D or D to U each round. Sideways and backward state changes to prior states are not allowed. So it goes 4U0D, 1U3D, 2U2D, 3U1D, 0U4D .. done, 4 is the minimum possible with not duplications of states.
This elegance can be seen better if you swap the U and D after each step, as in the case of 6 or more cups.
Instead of - 6U0D, 1U5D, 4U2D, 3U3D, 2U4D, 5U1D, 0U6D
It would be - 6U0D, 5D1U, 4U2D, 3D3U, 2U4D, 1D5U, 0U6D
Using this, you can know the orientation of cups in any step for n cups.
x move for y cups
(y-x)DxU for all x that is odd
(y-x)UxD for all x that is even
ie.
53rd move for 80 cups = 27D53U
40th move for 90 cups = 50U40D
I thought about this in terms of boolean algebras and got to the solution pretty quickly
Me too!
Didn't you get the answer is n whether n is even or odd..that doesn't seem to.hold up that with odd there areno solutions
It looks like a dynamic programming question. Good one!
copying my comment here for you as you are completely right!
this is a really great puzzle
I remember solving this when i was a teen - as part of comp sci A-lvl - its an example of heuristic hill climbing algorithm (an algorithm that looks ahead x steps to a better solution to find a final solution) - this sort of algorithm was used in early chess AI (and most modern AI pathing in modern video games)
Seems Im getting old though (now 45) because I couldn't for the life of me solve it now - and had to watch the vid. (yes it was the proof bit i struggled with, not mentally flipping glasses).
Interestingly its also the same problem(well very related) as the old board game 'mastermind' also the 'lights out' puzzle (a 2d version, rather than 1d) - something I used to happily solve in my head in fast time due to it being part of a puzzle in the video game Dungeons and dragons online as part of a raid.
yep.. DEFINITELY getting old, my brain has slowed down so much.
you are completely right!
I remember solving this when i was a teen - as part of comp sci A-lvl - its an example of heuristic hill climbing algorithm (an algorithm that looks ahead x steps to a better solution to find a final solution) - this sort of algorithm was used in early chess AI (and most modern AI pathing in modern video games)
Yes. N* (2**N) time complexity if we use dp, constant time complexity if we directly use odd even solution
Damn you are god damn right
I think the simplest solution for N glasses would be more like…
Every two moves, the net number of glasses that have been flipped is either 0 or 2.
So when N is even, it will take N/2 pairs of moves to flip N glasses, so N moves in total.
It can’t be less than N if the number of moves is even, because the most the number of flipped glasses can increase each pair of moves is 2.
If the number of moves is odd, and all glasses were flipped, then you could make one more move, and so have a set of paired moves where the total number of flips would need to be even, and that contradicts that you would have flipped the N glasses already . Therefore there can be no solution with an even number of moves.
This seemed suspiciously easy, so I was surprised to see it was actually correct: A sure-fire way to do is to just reverse the glasses in order, skipping back to the beginning when the end is reached. So first 1, 2 and 3. Then 4, 1 and 2, etcetera. That results in 4 moves. And that 4 is the correct answer can be easily checked mathematically: 4 glasses but you must move 3 at a time, that simply means you must find the lowest multiplication of 3 and 4 that are greater than 0 (unless all of them are already in the proper position, which isn't the case here) and are the same answer. In other words: 3 x 4 = 12 and vice-versa. Meaning that with 3 glasses each time, you need 4 moves.
He sounds so proud when he proves it. I meant this in most sincere way!
a more general version of the original question: given m glasses, all upward initially, each time you choose n (n
I solved this using some advanced linear algebra:
Firstly, note the isomorphism between the state of the glasses and the *F*_2-vector space *F*_2^n: for a vector (a_1, ..., a_n) in *F*_2^n, the term a_i = 0 if the glass is not inverted, and 1 if it is. Applying the move "invert all but the i-th glass" is equivalent to adding the vector e_i := (1, 1, ..., 1, 0, 1, ..., 1), where only the i-th term is 0. Since the initial state of the glasses is the zero vector (0, ..., 0), we wish to find nonnegative integers c_1, ..., c_n such that c_1 e_1 + ... + c_n e_n = (1, 1, ..., 1). Taking the c_i modulo 2, and noting that we are after the minimum of c_1 + ... + c_n, we can assume that each c_i is either 0 or 1, and then treat them as elements of *F*_2.
Let v = e_1 + e_2 + ... + e_n, and notice that v = (n-1, n-1, ..., n-1) (mod 2). When n is even, v = (1,1, ..., 1). In this case, noting that v-e_i = (0,0, ...,0, 1, 0, ..., 0), we see that the standard basis vectors lie in the span of the set {e_1, ..., e_n}. This implies that {e_1, ..., e_n} forms a basis of *F*_2^n, and therefore (1,1, ..., 1) is uniquely expressed as the *F*_2-linear combination e_1 + ... + e_n. Thus the minimum is 1 + ... + 1 = n. When n=4, we have the initial case.
On the other hand, if n is odd, then v = e_1 + e_2 + ... + e_n = (0,0, ..., 0), which implies that the e_i are not linearly independent. Hence the above argument no longer applies. For this case, let us consider the linear map T: *F*_2^n --> *F*_2 given by T(a_1, ..., a_n) = a_1 + ... + a_n. Notice that T(e_i) = 0 for each i (since n is odd), hence every e_i lies in ker(T). But T(1, 1, ..., 1) = 1. This implies that (1, 1, ..., 1) is not in ker(T), hence is not in the span of {e_1, ..., e_n}. Therefore when n is odd, it is impossible to reach a state where all glasses are inverted.
I like your funny words magic man.
Dang u actually solve the problem in the most complicated way I've seen
At the end: why does T(1,1,1,...,1)=1 not being in ker(T) mean that (1,1,...,1) is not in the span of (e1,e2,...,en)?
Nice, but why in such a complicated way?
@@verygooddeal4436 ker(T) is a subspace of *F*_2^n, since T is a linear map, and the kernel of a linear map between any two vector spaces is always a subspace of the domain. Because every e_i lies in ker(T), the span of the e_i must be a subset of ker(T), since span({e_i}) is by definition the smallest subspace containing the {e_i}.
1: lemme fiddle with this
2: wait this might be impossible
3: lemme check by thinking about possible states
3.5: wait the state of the glasses only depends on the number of downs and ups
4: ah wait its not impossible
5: finds solution
6: yippie *clicks video*
the little gray cells - best part of the vid
Go figure language AI models aren't math AI tools. Who would have thunk it.
In even number positions you always need as many moves as there are glasses: first move turn around all glasses except the most left one, next invert all glasses except the one right next to it, next turn around the glass right next to the last one and so on and so on. This way every glass will turn an odd number of times resilting in an inverted position.
For an odd numbers of glasses i turned the glasses into zeroes and ones: 00000 is the starting position, 11111 the ending position. Since we have an odd number of glasses, we will have an even amount of moves so i will split the moves in pairs of 2. In the starting position we can see that the sum of all the zeroes is zero, an even number. The sum of the ending position will be an odd number (in the example from earlier: 5).
Now we will look at what happens when we execute a pair of moves: There are generally 4 positions that can be made out of 2 glasses that have 2 states 00, 01, 10 and 11.
Now we will look at the change of the sum if we invert these pairs: 00 -> 11 sum changed by 2,
01 -> 10 and 10 -> 01 sum changed by 0,
11 -> 00 sum changed by -2.
Since we will change the sum by an even number, that means we can never reach an odd number as the sum.
That was the most neat solution i came up with :) really enjoyed puzzling this stuff.
Great question and solution! 🔥💯
Awesome .Thank you professor.
there's a much simpler solution: count glass flips.
in total, you'd have a multiple of n-1 glass flips, because that's how many you flip at each moves, and you'd have a multiple of n, because there's n glasses.
n and n - 1 being coprime, there's gonna be a minimum total of n(n-1) glass flips (the lowest common denominator or whatever it's called).
since a move flips n-1 glasses, n move flips n(n-1) glasses, so n moves is the minimal amount.
For numbers n that can or can't flip all glasses, count glass flips again.
If there is a solution, then there is a solution that takes n(n--1) flips, we just proved that
You can rearrange those flips in each moves as: flip all glasses, then "undo" one flip, to get your n-1 moves
Now, a n move solution can be seen as flipping all glasses n times first, then dealing with the undoing of one flip.
To get all glasses in the same up or down state, you have no choice but to spread those "undo" accross all glasses, meaning they all unflip once.
Meaning in total, each glass flipped n-1 times.
this means they're up if n-1 is even, so if n is odd, and down if n is even.
so the solution actually only works when n is even.
if 1 is 'up'd and 0 is 'down', and the set of indistinguishable glasses is not ordered then:
1. there are 4 equivalent 1st moves: 1111 -> 0001
2. there are 3 equivalent 2nd moves: 0001 -> 0110; and one move that reverts the 1st move: 0001-> 1111
3. there are two pairs of equivalent 3rd moves: flipping two 'up' and one 'down' glasses reverts the 2nd move, only option left is to flip any two 'down' and one 'up': 0110 -> 1011
4. last move is trivial: 1011 -> 0000
Either you perform at least one reverting move, or solve in 4 moves hence 4 moves is the minimum.
Just invert your face so all the glasses will be inverted by 0 moves
Your the god
Or flip your phone upsidedown
Tried that, but it still counted as a move.
@@a_Generic_User It's not a move from the glass's perspective as you never moved any of the glasses. It's in Einstein's Theory of Glass Relativity.
It's even simpler than that: for every move, you flip n-1 cups, which means it's n-1modn cups flipped. When it'll be 0, it'll be the answer, and since n-1modn=-1modn, to get 0 mod n, or -nmodn, you'd have to make n moves.
Then why doesn't that hold for odd values of n?
@@cwldoc4958 You're right! The reason why it doesn't hold is the fact that if you have an odd number of non inverted cups and you flip an even number of them at a time, no matter how many moves you'll make it won't work. Simply put 2nmod3≠0 4nmod5≠0 for every n because 2n, 4n, 6n, etc. are 2k, while 3,5,7, etc. are 2k+1
For the first question it is just the lights out game and the proof to that is nice. Then you gave the problem for N glasses flipping n-1 of them. That isn't the lights out problem but definitely similar. I liked the proof
This was easy. Just go from one glass to the next when inverting the triple. Go from one glass to the next when starting the triple.
Smash one of the glasses so there’s one less glass to worry about
I can hear Jean-Luc Picard now... "There are FOUR glasses!"
First puzzle like this I got right away, including that n must be even, the minimum moves is n, and each glass must be turned exactly (n - 1) times after n moves. Since n - 1 is odd, each glass is now facing down. I didn't work out a proof that it's optimal in the general case. That was really elegant.
I slept a lot last night.
Nice ltl problem. There's a simpler solution (for me at least) whicch can be done w/o paper:
The game is equivalent to inverting all glasses and then re-inverting 1 glass, which together counts as one move. This, in return, is equivalent to the following game:
- Just invert 1 glass per move.
-The game can be won in 2 ways:
(A) after an odd number of moves, all glasses are normal OR
(b) after an even number of moves, all glasses are inverted.
For parity reasons, (a) is impossible to achieve, while (b) is possible exactly if n is even, with n trivial moves to make.
I did it similarly. Once you know that (a) is impossible, you can consider (b) as a sequence of 2 moves. 2 moves either flip all the same glasses (and are consequently useless), or they differ by 1 glass, the net result being flipping 2 glasses. Knowing that you can flip 0 or 2 glasses with 2 moves pretty easily shows the minimum is n moves
Each glass must be inverted at least 3 times, since if one glass is inverted only once, the other three will be in an infinite loop, and 2 inversions don’t put a glass upside down.
If each glass is inverted 3 times, the least number of moves must be 4.
The way I though about it is that for the glass to end upside down it needs to be inverted an odd number of times. Most likely the same number of times for each glass given the scenario. So the total inversions is going to be an odd multiple of the number of glasses. For an even number of glasses n-1 is odd, and the least common multiple is n(n-1) which will always be an odd multiple. For an odd number of glasses n-1 is even, so there will never be an odd multiple of n that matches with multiples of n-1.
I don't really understand your logic here. n(n-1) will NEVER be odd, no matter what n is, since either n or n-1 has to be even.
@@holgerchristiansen4003
n(n-1) is even, but if n is even it is an odd multiple of n. That meaning it is n times an odd number.
Took me about 10 seconds from the thumbnail alone 😅
I really like these kind of riddles/puzzles
I went with an inverted gray code where a link between two nodes exists if there's one glass in common. This gives a 4 dimensional cube and the shortest path between the starting point and the desired result can only be reached in 4 steps.
if capital letters are upright and lower case letters are upside down, the solution can be described as such:
(Putting in a paragraph break to hide spoilers...)
*A B C D* = starting positions.
*A b c d* = the first is untouched and remains upright (capitalized) while the other 3 are inverted (lowercase), A is A, B becomes b, C turns into c, and D becomes d. Now, you obviously cannot just flip the same lower case letters to uppercase, since that puts you back into your starting position, so break it up and try to flip two upright (capitalized) and two upside down (lowercase).
*a B c D* = the first, second, and fourth are inverted (lowercased, capitalized, and capitalized) while the third remains the same (lowercase) A becomes a, b is B, c stays as c, and d becomes D
...At this point you have to move 3, but in order to get all 4 upside down, your next-to-last position needs just 1 to remain upside down, because your final move needs to be turning three that are right-side up to upside down.
Except if you leave one upside-down, you'll end up with 3 upside down (lowercased) and one right side up (capitalized). That's the *opposite* of your next-to-last position! That's the exact same as the second positioning you made, after making your very first move!
So what you *actually need* to do is leave one *right side up (capitalized)* and invert the others, like this:
*A B C d* = you have kept B right side up as B, and inverted a to A, c to C, and D shrinks down to d...leaving you with 1 upside down (lowercase) and 3 right side up (capitalized)...which means your last step is exactly what you want:
*a b c d* = by inverting A to a, B to b, and C to c, while leaving d as d.
In the course of solving this puzzle, these cups have taken on on 5 different configurations, starting with all upright and ending in the result you wanted, all upside down within just 4 steps, moving only 3 cups at a time.
for the purposes of notation, lets encode the glasses as a 1 if it is right side up and 0 if it is upside down, as such, the 4 glasses can be encoded as (1111)
for the 4 glasses, for all 4 glasses to end up flipped that means every glass must be flipped an odd number of times, therefore the total number of times a glass is flipped must be some odd number multiplied by 4.
we also know that in each move exactly 3 glasses are flipped, so for m moves we have 3m flips, and this must equal a (k+1)*4 total number of flips, therefore:
3m=(2k+1)*4
m=(2k+1)*4/3
lets test inserting some small odd numbers into the (2k+1) part of this equation.
k=0 : m=4/3 m needs to be a whole number so this isn't a solution.
k=1 : m=4
thus, the smallest number of moves we can make that theoretically could produce a solution is 4 moves. is there such a solution?
yes, as such:
1111
flip all except the 1st
1000
flip all except the 2nd
0011
flip all except the 3rd
1110
flip all except the 4th
0000
done.
thus we have proven a solution for 4 moves and we have shown that no shorter solution exists.
now, how about the general case? we have n glasses and we flip n-1 glasses each time.
obviously it's impossible for 1 glass, since each move flips 0 glasses, so the state of the board so to speak never changes.
obviously it is possible for 2 glasses, since each move flips 1 glass, just flip each glass in it's own move.
for 3 glasses it's slightly more tricky but still easy to see that we need to make, in total, an odd number of flips, but each move makes an even number of flips, so we can never get an odd total number of flips and it is therefore impossible. this logic can be extended to all odd number of glasses and they are therefore all impossible.
for even numbers of glasses we can always trivially do it by making n moves: for each glass, make a move that flips all except for that glass. this strategy always solves for an even number of glasses because every glass gets flipped once for every other glass, and since there is an odd number of "other glasses", every glass gets flipped an odd number of times.
but can we do better?
lets explore what the minimum number of moves needs to be similarly to what we did for 4 glasses, our new and more general equation is (where 2n is our even number of glasses):
(2n-1)*m = (2k+1)*2n
m = (2k+1)*2n/(2n-1)
2n/(2n-1) is an interesting fraction. by definition x and x-1 are co-prime, therefore 2n and 2n-1 are also co-prime, this tells us that the fraction 2n/(2n-1) is as simplified as it gets.
therefore (2k+1)*2n/(2n-1) is only ever a whole number if (2k+1)/(2n-1) is a whole number. the smallest whole number that this can make is 1, when k=n-1, and what does that get us?
substitute k=n-1
m = (2(n-1)+1)*2n/(2n-1)
m = (2n-2+1)*2n/(2n-1)
m = (2n-1)*2n/(2n-1)
m = 2n
and since 2n was what we defined as our even number of glasses, we have now proven that the minimum number of moves is the same as the number of glasses, and we have already proposed a strategy for how to solve even-numbered glass puzzles with that number of moves, therefore we have completed the general puzzle.
Posting before watching. After having done a few test examples, my guess is that the initial answer is 4. It can only be completed when N is an even number, and the minimum number of moves each time will be equal to N.
Gee, this is easy!
Four moves, three glasses per move, each glass is eventually turned three times:
VVVV --{1}--> AAAV --{2}--> AVVA --{3}--> VAVV --{4}--> AAAA
(At step {3}, we have essentially two choices: either [two Vs, one A] , or [two As, one V] . Since [two Vs, one A] is actually a reversal of step {2}, it doesn't help us forward; so the choice must naturally be [two As, one V] .)
For n is an even number: -> n-1 is an odd number.
Each cup needs to be flipped an odd number of times (NBoT). You can only flip (n-1) cup at a time so there’s aways 1 cup unflipped.
Each cup can only remain unflipped once otherwise you repeat a previous step ( which makes it not the fastest solution ).
If number of step (s) < n then there are away 2 cups: cup A get flipped (s) times and cup B get flipped (s-1) times, which can’t be true since between two numbers (s) and (s-1) there’s an even number. Each cup can only be flipped an odd number of times so s must be = n (each cup gets flipped exactly n-1 times).
Solved the problem in 2 seconds flat, invert one glass three times in the first set and the final three in the second.
I was not as interested in the general case where for n glasses, move n-1 glasses. But I was interested in the AI answers. What mistake of logic did the AI make that it couldn't figure this out? Presh proved it in a brute force way: he considered all possible moves and discarded the ones that did not take him closer to the goal. What elegance caused the AI to "hallucinate."
Nice!
Me:
I have 4 cups, named A, B, C, D, standing normally. If cups flipped upwards, call them AA, BB, CC, DD respectively.
I need to get all of them flipped, but there's restriction: only three can be flipped at once, no more no less. Suggest turns pls.
GPT4o:
To flip all cups from the normal position (A, B, C, D) to the flipped position (AA, BB, CC, DD) with the restriction that you must flip exactly three cups at a time, you can follow these steps:
Flip cups A, B, C.
Result: AA, BB, CC, D
Flip cups A, B, D.
Result: A, B, CC, DD
Flip cups A, C, D.
Result: AA, B, C, D
Flip cups B, C, D.
Result: AA, BB, CC, DD
Aside from giving the wrong answer, Google Bard’s base case makes no sense because for one glass you are allowed to invert 0 glasses each turn.
Waaay back in first grade, we were taught (and made to understand why) then when it comes to Odd and Even numbers, if you add two from any category, the answer is Even. When you add two from different categories, the answer is Odd. We could help remember this because odd means weird and different in non-math context, so if we want an Odd answer, we need two different types of numbers. But Even is balanced and same-like, so any two numbers from the same category (O+O or E+E = E). For this puzzle, it follows that you can't make an Odd number if your only operands are Even. Q.E.D. for the case n=Odd.
For the n=Even case, I know that it CAN be done in any case. One possible solution that works for every Even n is that you flip all but the first glass, then invert all but the second, and on down the line. It is not hard to show that this always works, and will always succeed in n moves (each glass is omitted exactly once, so moves=glasses). It is intuitive for me to see that this is also the minimum solution, but I do not know how to prove it rigorously. I'm watching the rest of the video now.
0, flip the paper upside down. Done.
Each 3flip is two moves A & B: (A) is just inverting all 4, (B) flipping any 1 cup.
The (A) just toggles the cups and at the end of any even turn, the (A) moves have you back to your original state.
Therefore the (B) moves have to flip all 4 cups upside down, which we can simply do going left to right.
UUUU (all up)
1- (A): invert: DDDD
(B): flip 1st: UDDD
2- (A): invert: DUUU
(B): flip 2nd: DDUU
3- (A): invert: UUDD
(B): flip 3rd: UUUD
4- (A): invert: DDDU
(B): flip 4th: DDDD
Step one can easily be checked, as the solution said:
Without loss of generality, move one and two has to flip 1, 2, 3 and 2, 3, 4, respectively, to not end up in the starting position. After reaching this position, it is obviously not possible to have all glasses flipped after the third move.
(This would be good enough for Olympiad standards, no need to tediously explain every possible move at every step)
For N where N is even, there's an easier way to show that N moves are required.
Step 1: Show that the number of moves will always be an even number, using the same sort of parity argument used to show that an odd N is impossible.
Step 2: Now that we know the number of moves will certainly be even, consider moves in pairs. A pair of moves can have only 2 outcomes. Either it leaves the pattern unchanged (because you did the same move twice, or exactly two glasses have been inverted, because all but two glasses were inverted and then inverted back. Obviously, to make two moves but leaving the pattern unchanged is just a waste of moves, so we need only consider the second choice.
Step 3. In order to invert all the glasses, you need N/2 pairs of moves, or 2 * N/2 = N.
The way I thought this through was if all but one gets inverted each time, you could think of it the same as only one getting inverted each time, so you just go in a line to invert each one, therefore minimum = n. Then you also have to remember that each one must be inverted an odd number of times so the answer only works for an even n.
ChatGPT 4o came up with the right solutions for both sequential (all 3 inversions must be consecutive) and non-sequtial (any 3 glasses may be inverted).
At least for the fixed number 4. It couldn't figure out N with N-1 flips.
It can be proven a bit quicker and simpler.
First consider each step a combination of two transformations. One flips every glass, the other flips one glass.
If there is an even number of glasses n, each step is an odd number of flips.
Having all flipped (an even total number of flips) you need an even number of steps.
If n is even the transformations of flipping all glasses cancel out, simplifying the problem to a transformation of a single flip
So you need to flip n glasses, one at a time. Not flipping the same glass twice results in the minimum of n steps.
For n is odd you have a problem.
Using an even number of steps you have the same cancelling but can only flip an even number of glasses.
Using an odd number of steps you have one flip all part that doesn't cancel out. In addition you have an odd number of flips and flips back so you can't make them cancel out.
chatgpt can't do basic arithmetic without a chance of making errors so I wouldn't ask chatgpt for help with math or logic problems.
and the reason for that is that chatgpt is it's basically an advanced text predictor. So when it appears to do math it's not actually doing any math but predicting text.
Problem becomes trivial when you consider that inverting n-1 glasses, is same as inverting only the omitted one and then inverting all of them. Inverting all of them an even number of times does nothing. Hence, each move of n-1 glasses is equivalent to inverting just the omitted one, and all will have been inverted in n moves as long as n is even.
Define the following mapping:
When the number of inverted glasses is even, replace each upright glass by 0 and each inverted glass by 1. If it is odd, replace each upright glass by 1 and each inverted glass by 0. Observe this mapping is bijective. Observe that inverting 3 glasses flips exactly the digit corresponding to the glass that was not inverted (we constructed it that way). Also, the initial configuration is 0000 and the final configuration is 1111. The minimum number of moves is their Hamming distance as each move flips exactly one digit. So the minimum number of moves is 4 and it is obtained by any sequence that omits each glass exactly once from inverting, so every glass is inverted exactly 3 times (corresponding to flipping each of the digits exactly once).
Note that this works for any even n. For odd n, the task is actually impossible: every move then inverts an even number of glasses. But this means the number of inverted glasses always remains even.
And in the even more general case, if you have n glasses and every move flips exactly k of them, then (spoiler alert) it's possible to flip all the glasses if and only if k2 and U
Label the glasses, from left to right, with sequencial letters A through D. Describe each move as the letter of the glass that is NOT moved. Any solution that uses all four letters an equal number of odd times is thus correct, with the simplest solution being each letter used once.
This problem is a binary problem where the cups are either up or down. You can only change the state of N-1 or basically all but 1 each time. You can easily infer that the final step of the problem will need 1 cup facing down and the other remaining N-1 cups facing up so that you flip exactly all cups that are facing up to down. Since you're flipping all but 1 cup each time, basically only all but one cup change state each move. In order to achieve the state just before the final move you need to basically flip all the cups N-1 times. So the amount of moves necessary to flip all the cups facing down is equal to N moves. This solution is impossible for the situation where there's only 1 cup, or N = 1, since you aren't allowed to flip any cups at all for that scenario.
The case of n wine glasses: Starting with n wine glasses upright, inverting (n - 1) at a time, is it possible to reach a state in which all the glasses are upside down, and if so, what is the minimum number of inversions necessary to reach that state?
Each inversion consists of inverting all the glasses except for one. In each inversion, the glass that is not inverted is either among the upright glasses or the upside-down glasses. Let f denote the act of inverting all the glasses except for one which is upright, and let g denote the act of inverting all glasses except for one which is upside down.
Mathematically, let f(k) = the number of glasses that are upright after applying f to a state where k glasses are upright, for 1
To be lazy, instead of flipped n-1 glasses we can just flip one glass and have a coin that records whether we need to flip them all later. So for every move flip the coin, and flip or unflip a glass.
We always start with 0 glasses flipped, and our coin set to no-flip, which we can show as (0, 0 = false/no-flip), then we can move up and down this chain:
(0, 0) (1, 1) (2, 0) ... (n, n mod 2)
It takes at least n moves to reach the end of the chain and there are no other possible moves.
This works if n is even, since the last element is (n, 0), meaning we don't have to flip the glasses and we're done. Otherwise we have (n, 1) which is equivalent to (0, 0) so we're back at the start.
Label the glasses 1,2,...,n. Label (1),(2),...,(n) the glass flips that flip all but 1,2,...,n, respectively. For example (2) would flip all glasses except for 2. Note that any glass flip applied twice results in no net action. Our objective is a sequence of flips that results in all glasses flipped. The insight is that the glasses get flipped once more if they're corresponding glass flip is not part of the sequence. For instance let n=5. The sequence of glass flips (1)(2)(3) flips glasses 1,2,3 twice and glasses 4 and 5 3 times. This implies the only possible solution is to apply all flips so everything is flipped the same number of times. When n is odd this solution flips everything an odd number of times resulting in the objective; when n is even this sequence gives us all glasses flipped the wrong way.
I did it in 6 moves, without having clicked on the video, from just seeing the thumbnail and opening Notepad:
The edit is adding the right side after having watched up to and pausing at 5:52 in the video, and to add saying if you were looking for an algorithm you should have had "invert n glasses by moving n-1 glasses per step" in the thumbnail rather than putting a real random number in there.
^ ^ ^ ^ - 0 moves - ^ ^ ^ ^
v v v ^ - 1 move - v v v ^
v ^ ^ v - 2nd move - v ^ ^ v
^ v v v - 3rd - ^ v ^ ^
v ^ v ^ - 4th - v v v v
^ v ^ ^ - 5th
v v v v - 6th
It was in attempting to ask the three glass version of this problem that the guy in Inglourious Basterds revealed he was a spy.
this is a really great puzzle
I remember solving this when i was a teen - as part of comp sci A-lvl - its an example of heuristic hill climbing algorithm (an algorithm that looks ahead x steps to a better solution to find a final solution) - this sort of algorithm was used in early chess AI (and most modern AI pathing in modern video games)
Seems Im getting old though (now 45) because I couldn't for the life of me solve it now - and had to watch the vid. (yes it was the proof bit i struggled with, not mentally flipping glasses).
Interestingly its also the same problem(well very related) as the old board game 'mastermind' also the 'lights out' puzzle (a 2d version, rather than 1d) - something I used to happily solve in my head in fast time due to it being part of a puzzle in the video game Dungeons and dragons online as part of a raid.
yep.. DEFINITELY getting old, my brain has slowed down so much.
Nice question on parities and xors. If you encode glass up as a 0 and glass down as a 1, you are basically asking how to get 1111 from 0000 by xoring with only 0111,1110,1101 and 1011. Notice that there are only 4 and you need to use each one exactly once to get the result. If you use two sequences at any time, that would be a redundancy as they cancel. This can be generalized to any even n as well.
That's how I did it as well.
Inverting all but one glass is basically the same as inverting just one and then pretending to flip the whole setup upside down. If you make two moves, you've pretended to flip the whole thing upside down twice, so for even numbers of moves, you can just turn two glasses and it will be the same as having turned all but one twice.
Therefore, for n glasses, where n is even, n moves is the minimum number of moves and each glass gets one turn where it's not flipped.
great proof!
Interesting to see the AI results with this one. I used the Bing Copilot, which I believe is also ChatGPT based, and it did get the correct answer and correct reasoning using parity. While I probably wouldn't formulate the proof as well, for once I was able to conceptually explain it before watching the video.
I thought of it as if you are inverting 1 glass, and then keeping state of is everything inverted or not. So for even n-s solution is exactly steps, and for odd ones it does not exist.
It may be checked is a perfectly reasonable justification for the minimum. There’s only 3 cases to check: 1, 2, or 3 moves. 1 and 2 moves are trivial, so really only three moves needs checking.because the order of the glasses doesn’t matter there’s only about 4 sequences of flips that need checking.
I solved this using simple logic.
Inverting 3 glasses is the exact opposite of inverting 1 glass. Thus, we are doing the logical opposite.
U , U , U , U (the opposite of flipping 1 is flipping the other 3),
D, D, D, U (left the 4th one)
U, U, D, D (left the 3rd one)
D, U, U, U, (left the 2nd one)
D, D, D, D (left the 1st one)
Now, it turns out... that no matter how you solve it, as long as you only leave 1 alone and don't duplicate any ones you leave up, you will solve it in 4 moves.
So, now it also becomes a game of how many ways can this be rearranged.
4 * 3 * 2 * 1 = 24
There are 24 different solutions that can be solved in 4 moves.
The glass one can be solved in 2 steps. It says you must invert 3 glasses per move but doesnt state that you cant invert the same glass multyple time in the same move. So first move invert 3 glasses, then invert the last glass 3 times resulting in all glasses being inverted
4 moves is the best i could get within the first minute of thinking about this problem
Before watching the video I quickly experimented with 4 coffee k-cups and it can be done in 4 moves. Turn the first 3 upside down. Turn the 4th upside down and revert the middle 2 upright. Revert 1 and 4 upright and turn 2 or 3 down. Turn the remaining 3 upside down
With 4 glasses, there are potentially 16 different configurations of right side up and upside down. Not only can you achieve all inverted, you can in fact achieve ANY desired configuration of up and down glasses -- all 16 configurations are reachable by flipping glasses 3 at a time.
Assume the glasses are numbered 1 to 4, and assume U represents a right-side-up glass, and A represents an upside down glass.
Step 1, UUUU, flip 1, 3, and 4, --> AUAA
Step 2, AUAA, flip 1, 2, and 4, --> UAAU
Step 3, UAAU, flip 1, 2, and 3, --> AUUU
Step 4, AUUU, flip 2, 3, and 4, --> AAAA
All glasses inverted. This works because each individual glass has now been flipped 3 times, which leaves the glass in the opposite configuration that it was in at the beginning.
this is how I'd solve this puzzle:
1) get a friend to agree to the following steps
2) pick one unique glass you don't want to invert
3) invert every other glass
4) invert your friend
5) ask your friend how many glasses are inverted relative to them
repeat steps 2-5 until either all glasses are inverted or your friend is dizzy.
from your perspective, you have inverted all-but-one glasses at a time, but your friend will have seen only one glass inverted at a time.
since your friend only sees one glass inverted per move, they would only be able to see all the glasses inverted relative to them, therefore you must take as many moves as you have glasses to flip all of them once from your friends' perspective
however you will run into a parity problem: your friend will always be able to see every glass inverted since they see glasses inverted one at a time, but that's because they are being inverted themselves. if you have an odd number of glasses, your friend will be inverted an odd number of times. your upside-down friend sees every glass as upside down from their perspective, therefore those glasses need to be upside-up from your perspective
as a result, you can invert N glasses in N moves iff N is even. otherwise, it is impossible.
Each glass has to be inverted an odd number of times. The total number of inversions must be a multiple of 3, since we must do 3 inversions per ‘turn’. The simplest way to get this result is 3+3+3+3=12. That means we hope to be able to do this in 4 ‘turns’. If we exclude one cup each time, each cup will be affected in 3 of the 4 turns, solving the puzzle.
Alternative solution:
Assign 1 to an upright glass and -1 otherwise. Consider the product. At first it's 1 and in the end it should be (-1)^n.
In each step we multiply the product by (-1)^(n-1).
This immediately shows why it's impossible when n is odd. We're trying to go from 1 to -1 while always multiplying by 1.
It also shows that we need even moves in case of n even, since we're going from 1 to 1 by multiplying by -1 at each step.
Now in the n even case name each move by the exact one glass it doesn't flip. Suppose we solve it in less than n moves. By pigeon hole principle, there should a glass that's the move named after it didn't happen, therefore it's flipped every time and since there's even number of total moves, it should be back upright which is a contradiction.
0000
1110
1001
0100
1111
4 moves
Like someone else said, it's just like switching 1 glass, but you also include the 2 after it.
I figured that on each move you're changing the arrangement by exactly one glass. You can ignore that the arrangement is flipped because every two moves will flip the arrangement twice. So if you have n glasses, it must take n moves, if it's possible at all.
For the general case, when n is odd: a glass is upside down when it has been inverted an odd number of times. An odd number of glasses times each requiring an odd number of inversions yields a total number of inversions that is also odd. No integer multiple of an even number will produce an odd number. Hence, no solution.
I hope he publishes a transcript of this in a place where google will eventually scrape it. He can use this question as a check on the AI to find out when it scraped his solution.
This looks like incrementing numbers in mod 2 and would be interesting to see what it would happen for other modules
I really liked your solution for showing that n is the minimum number of moves for even n. It was quite elegant.
Mine was a little too messy to be worth posting.
However, the way I showed that n could not be odd was as follows.
Assume there is a solution for odd value of n, so that after m moves all glasses are facing down.
For a glass to go from U(up) to D(down), it must be inverted an odd number of times.
N_i = number of inversion for glass i
T = Total number of inversions
Adding all these inversions, we get:
T = N_1 + N_2 + ... + N_n
Since each of N_1, N_2, ... , N_n must be odd, and we have odd amount in the sum (since n is odd),
and since the sum of an odd # of odd integers is odd, then we get
T = odd
But on each of the m moves we invert n-1 glasses, where n-1 is even. So we also get
T = m(n-1) = even
But T cannot be both odd and even. So our assumption (that a solution exists for odd n) must be false.
Before watching the solution, I'd solve the case for general n as follows.
If n is odd, then in each step an even number of glasses are turned. Thus, the number of inverted glasses will remain even and can never reach n.
If n is even, an odd number of glasses are turned in each step, meaning that the number of inverted glasses is odd after each odd step and even after each even step. The state of n inverted glasses can thus only be reached with an even number of steps. Each individual glass needs to be turned an odd number of times, so there has to be at least one step in which it is not turned. As each step only leaves one glass unturned and there are n glasses, n steps is a lower bound. Finally, turning in step k all glasses apart from the k-th one shows that n steps indeed suffice.
I wonder if, since chat gpt is essentially a bunch of code, the problem is with assignment of step numbers. It might be labeling step 1 as step 0, then omitting the first step as null.
since 1 and n-1 are just opposites. you can select 1 cup to not flip. the sequence for solving is just selecting each cup in order from one end to the other to not be flipped. in the n is odd case. you can get everything but the last glass. and any other change is further away.
This was my technique as well.
If the rule had been “flip one cup per move”, you would obviously just do that rule once for each cup.
Based on parity, this is equivalent to your solution, basically “flip the complementary set of each cup”, which achieves the same end.
You can prove it by induction in a slightly simpler way. Assume we can swap n glasses in n moves. Then take n+2 glasses, i.e. U....U, n+2 times.
1) Flip the first n+1 glasses in 1 move, so you get D....DU, with n+1 Ds.
2) Flip the last n+1 glasses in 1 move, so you get DU....UD, with n Us.
3) But we can already flip n glasses in n moves, so do that with the n glasses that are U....U, and you get all n+2 glasses being D....D.
So flippin' n+2 glasses takes n+2 moves. We have the base case for n=4. So for n even, it takes n moves.
1 = Correct way up
0 = Upside down
1111
0001
0110
1101
0000
Essentially if you sort it correctly it is a binary sequence that terminates a 0. But only for even numbers; Sequences and series is always really tough to work out for me.
Then using the same notation, a demonstration of the simplest odd case, illustrating its impossibility:
111
100
010 or 111, so therefore impossible. Other odd numbers would presumably get to the same stage within a few more steps
And of course the simplest even case:
11
10
00
@@mittfh Nice! really good way of putting it for the impossibility!
Define a "glass-flip" as flipping a single glass. To end up with all N glasses flipped, a total of N x J glass-flips are required, where J is an odd positive integer. We are constrained to always flip N-1 glasses with each move; we'll do this in K moves, where K is a positive integer. We have a solution when N x J = (N-1) x K. Now it's just a matter of solving the equations, no need to visualize permutations of flipping glasses anymore.
How about turning one glass up side down and stacking the other 3 to make a tower in 4 moves. Nobody said that they should be standing next to each other in a horizontal line , but all will be up side down. You could also use your 2 hands to turn 2 in 1 move, 2 move stack the other 2.
My brain just said “smash one and then turn over the other 3” and then was made to panic when the guy mentioned n-1 stuff
4I (incorrect)
1. flipping 3I: 3C (correct) + 1I
2. flipping 2C+1I: 2C + 2I
3. flipping 2C+1I: 1C + 3I
4. flipping 3I: 4C
Four moves.
nI
1. flipping (n-1)I = (n-1)C + 1I
2. flipping (n-2)C + 1I = 2C + (n-2)I
3. flipping 2C + (n-3)I = (n-3)C + 3I
4. flipping (n-4)C + 3I = 4C + (n-4)I
etc, etc, with the known constant increasing by 1 every step. However, when n is odd, for some reason the cycle doubles back on itself. I'm not sure why, so I guess I lost lol
I paused and tried this out with some stones, and I got the 4 pretty quickly, and eventually 8, but I couldn’t recreate it soon after. Wish I’d continued the videos instead of spending forever over 3 and 5 lol
Isn't inverting three just the symmetry to inverting one ( + changing you point of view)?
A nice graphical way of solving this is to arrange the glasses in a circle and then flip n-1 glasses at a time, while shifting the group one spot each time. That way after n operations all the glasses get flipped n-1 times( which is odd) and thus all the glasses will be flipped over