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!
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
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.
yeah it's not even a puzzle really is it? I mean, the first move whichever 3 you turn over you end up equivalent to 0000 to 1110, 3 of them will be upside down one the right way up. This is the same as 0000 to 0111 or 0000 to 1011 and so on. There's no puzzling about move 1 it's a given. For your 2nd move. If you turn the same 3 back over, you'd just end up back at the start. So the only other choice is to turn 2 upside down back and the one you didn't touch on your first move upside down. So 1110 goes to 0011. So far we've done 2 moves and had no real choice over either of them. Now we're in a position with 2 upright and 2 upside down. The only 2 moves are, turn 2 of the upright and 1 upside down, or 2 upside down and 1 upright . One of which 0011 goes to 1110 which is going backwards to a previous state. So we must go from 0011 to 1000. I'd accept that this move requires a bit of thought. Our last move is now trivial. We just turn the 3 upright glasses upside down achieving our goal 1000 to 1111. So, it's puzzle insofar as you had to do a bit of analysis for move 3. The other moves are either just obvious or they're the only moves you can make.
I did this with my fingers, different moves, but same amount, nice. Edit: for those who want my moves 1: fingers up, 0: fingers down 1111 - start position 0010 - move 1 0101 - move 2 1011 - move 3 0000 - move 4 / final position
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 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.
@@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)
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.
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
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*
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.
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.
@@VeryGoodDeals 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}.
I didn't watch the video yet, only saw the video thumbnail. I was able to do it in four steps: 1. Flip 3 cups, that are in up position. 2. Flip 1 cup that is in up position, and 2 cups that are in down position. 3. Flip 1 cup that is in up position, and 2 cups that are in down position. 4. Flip the three cups that are in up position.
@@Ruskettle I got even more confused but I decided to just try the steps with actual water bottles, I was able to understand hooray!!! I was mostly confused because I thought we had 3 turns TOTAL that's why 😭 but yeah I get it now
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)
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.
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.
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.
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.
Curiously though, any human that could perform a similar task of superpowered pattern matching and prediction would be considered pretty damn intelligent.
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.
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.
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.
Something that interests me is when you include a non-integer number of steps, and this would allow for solutions for odd numbers with a minimum of half of that odd number. (where in half a step you move half as many glasses) Example: (111), Step 1: (010), Step 1.5: (000). I know this is not actually how a "step" works, however I find it interesting that using this definition gives exactly half of the odd number as a minimum number of moves.
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
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.
@@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
As a general rule, when mathematical explanation says "it can be easily shown" means a student can do it in a weekend. On the other hand, "it can be shown" means that a student may accomplish it in a couple of years of studies but sometimes it can refer to stuff like Fermat's Last Theorem where "it can be shown" took over 350 years of trial and error to finally accomplish the "can be shown" part.
I missed the part where you invert n-1 glasses and solved for inverting 3 glasses. Here is the solution for that problem: 4 is a special case since it leaves little room for possible moves so it is 4 moves. But for all other cases, you want to make the amount of unflipped glasses to be divisible by 3. The first case is where it is already divisible by 3, so just divide n by 3 for minimum moves. If you have 1 extra glass, flip 3, then unflip 1 and flip 2, (4 are flipped now, which is 3+1) so in 2 moves you have reduced the problem to the first case, so divide n by 3 and round up. If you have 2 extra glasses, flip 3, then unflip 2 and flip 1 (2 are flipped now) and again in 2 moves you have reduced the problem to the first case, so divide n by 3 and round up but add 1.
i just took an induction proofs course last semester, it was absolute hell, past me would never believe that i watched and solved another one of these devilish problems for fun.
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."
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.
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.
Stop using AI prompts and pretending that it can answer questions. It is a waste of everyone's time as well as being a huge waste of natural resources.
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
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).
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.
I got to step 2 and realized it’s so easy with the right insight. Every turn inverts all but 1 glass. 3 inversions will result in an inverted cup. So do 4 turns with each glass excluded. Order doesn’t matter. Each glass will get inverted 3 times. Done!
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.
I solved this problem by breaking it down into two problems: Instead of thinking of inverting n-1 glasses, think of it as two steps to be performed at each turn: 1. Invert only 1 glass in a turn 2. At the end of the turn invert all glasses Forgetting about step 2 for now, if you can invert only one glass at a time it is obvious that you need a minimum of n moves to invert all n glasses, regardless of n being even or odd. Now for step 2: Inverting all glasses an even number of times is equivalent to not inverting anything at all. This explains why it works for n when n is even: effectively only one inversion takes place from step 1 and zero inversions from step 2. Inverting all glasses an odd number of times is equivalent to inverting all the glasses just once. So if n is odd, you will have two effective inversions: one from step 1 and one from step 2, so in the end you'll have returned to the initial state. You cannot perform step 1 even number of times because there is an odd number of glasses.
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.
this rests on the assumption that each glass will have the same number of flips in total (thus a multiple of n). But it doesn't exclude the option where one glass has 1 flip and another has 3, for example
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
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.
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.
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.
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.
there is a similar puzzle in the video game 'path of exile' to which the easy to remember answer is to 'click each pillar once' which translates in this case to avoid inverting any given glass each once
1) I have stopped the video to try myself and found a proof which does not require the case distingtion between an even and an odd number of glasses: Since the glasses are identical objects, there are only two possibilities in each move: (1) Invert all glasses with exception of an upright one (2) invert them with exception of an upside-down one. One of these two possibilities "reverts" the previous move, so if you want the minimum solution, there is only one "allowed" possibility in each move. Now you can prove - using induction - that if you have N glasses, the number of upside-down glasses (ni) after the i-th step is: n1=N-1, n2=2, n3=N-3, n4=4, n5=N-5, n6=6 ... For even values of N, the only element in this sequence having the value N is n(N), so you need N steps to invert all glasses. For odd values of N, the series contains only even values, so it does not contain the number N, so it is impossible to invert all glasses. 2) Have you heared about the cases when lawyers trusted fake cases invented by a KI? If yes, you surely were aware that the "proofs" made by a KI would be nonsense.
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.
Using linear algebra: (All glasses down) = (-1,-1,-1,-1) = (α I_1+ β I_2 + γ I_3 + δ I_4) (1,1,1,1) where I_k is minus the identity matrix with a positive 1 value at row k and column k. It represents the inversion of all glasses but one. Solving this system gives the equation: 1 = αβγδ where the grec letters indicate the number of times you apply a different inversion and therefore need to be natural numbers. The solution α=β=γ=δ=1 (add up to four moves) is the one showed in the video.
There is an error in logic in this video. The answer is always N. This is the same logic as the tower of Hanoi and uses the same solving technique. In every case, because the difference is N-1, there will always be one less glass flipped then the total. Repeat this for all glasses, and all glasses will be flipped. We can model this by shifting the glass we leave untouched all the way across the number. This iterated step will take N moves every time. Whether the glasses are even or odd doesn't matter.
I think your answer can easily be proven wrong. For example, if we do a test to prove this answer, because n=1 cannot be done, we try with the next odd number, namely n=3. If it is repeated, after 2nd move, the glass can only return to the first move
@@brurydeardomartin1800 No, N=1 is trivially easy. You flip one glass and the glass is flipped. job done. You cannot flip 0 glasses because 0 is not a quantity. N must be a quantity or the question doesn't make sense. N = 2 you flip one glass, then you flip the other. That solves N = 2, and so on.
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.
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
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.
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
For the 4-glass case, the proof that 4 moves is minimum is that for each move you must reach a state not equivalent to a previous state (number of glasses up and down). In practice this always only leaves one move available to proceed; there is only one way to reach the goal without back-tracking.
The product of two adjacent positive integers is always an even number. This fact is at the core of this puzzle. Otherwise ChatGPT is not so bad, just you need to be patient with it. If it gives a wrong answer to the problem first, and you see the error in it, then you can educate it by pointing at the error and asking it to rethink.
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.
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.
My idea of the lowest possible solution was if the moves were divisible by 4. Because it's impossible to flip 4 glasses over if the total number of glasses you flipped was not divisible by 4. You can also think of the solution another way, since 3 glasses are flipped and 1 STAYS UPRIGHT. This problem is logically the same as if you were to flip 1 glass over at a time (which can be seen in the solution to the original problem).
On a hunch, I tried this and it worked: Move #1: Flip all glasses except the first one. ( uuuu -> unnn ) Move #2: Flip all glasses except the second one. ( unnn -> nnuu ) Move #3: Flip all glasses except the third one. ( nnuu -> uuun ) Move #4: Flip all glasses except the fourth one. ( uuun -> nnnn )
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.
Flipping all but one glass os equivalent to flipping one and then inverting all the glasses. in order for all the glasses to be upside-down, they must all be in the same state, and therefore the single flips must have been applies to all the glasses the same number of times, and thst number plus the number of inversions must be odd. If all glasses had been fliped 0 times, then nothing changes and they're upside-up, and flippings more than one are multiplying the total number of both flips and inversions by a constant, which is useless because parity is all that matters. Therefore, they must all be flipped once individually and and inverted together an even number of times. In order to flip n glasses individually once, you must do flipping n times, and that comes along with n inversions. If n is odd, then n+1 is even and everything is upside-up, but if n is even, n+1 is odd and you've done it in n steps.
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
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.
You can simplify the problem as how many glasses turn every 2 moves and that is always true whatever glasses you choose. So for even numbers it is easy. You end up with 2 after 2 moves, 4 after 4 moves and n after n moves. For odd number you always turn even number of glasses. Which ensures alway an odd number of glasses remain up no matter what. Since 0 is even (you need to get to 0 glasses up) it is impossible to get there
The way i thought of it is rather simple. Inverting all but one glass is the same as only inverting one glass. The minimum number of moves to get all glasses facing down while only inverting one at a time is exactly N. Therefore, the answer is always N. For the n being even case: Given that there are exactly N flips, each glass is flipped n-1 times. The only way to go from up to down is flipping an odd number of times, so if N is even, then n-1 is odd and it works! If N is odd, then n-1 is even, making it impossible, since N is _guarranteed_ to be the minimum if it exists.
Because of the flaws (and lacks) of the rules its possible to make it in 2 steps. The rules are not say you cannot move the glasses or replace their position with others. And the rules does not state what is meant to be "different". Therefore we can say "different" could be mean the original position of the glass in the step. So you move the first N-1 glasses in the first step, then move the move the next step comes the magic: V1 (swapping): N'th glass flipped, then swapped with position of N-1, then N-1 flipped, and swapped with N-2 and so on until you reach the desired all face down position. V2 (moving 1st glass to the end): N'th glass flipped, then the 1st position glass moved to the end. Than N-1 flipped, and 1st glass moved to the end, and so on... This solution make it possible to make the puzzle regardless of N is odd or even. If you say that is not allowed, than please show me where it was written in the rules :)
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.
I think I have an easy proof for both n even and odd. At each step except the first step, you really only have two choices, denoted (^) and (v): (^) preserves (does not flip) an U glass and flips everything else, and (v) which preserves a D glass and flips everything else. Some lemmas: A) Your first move must be (^) (since there are no D glasses to preserve). B) Doing the same move twice in succession (^)(^) or (v)(v) produces no change, so C) to preserve minimality there there is never a point to repeat moves in succession. D) Therefore, the solution is going to be a series of alternating (^)(v) moves, starting with (^) per A. E) After each pair of moves (^)(v), the number of Us decreases by 2. So to get to 0 Us, you need n/2 (^)(v)moves, i.e. n moves for even n. QED for even n. For odd n, this forced method (after n-1 moves) will eventually get you to 1 U. The next move you must make is a (^) (from D); but this will simply revert all the glasses to U, getting you to square 0. This is a contradiction, so there's no method for odd n. QED
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.
The way I always saw this problem was like a least common multiple thing. There are four cups and three inversions per move. We know that we can’t flip all four cups in three inversions (one move) and if we tried to do it in two moves, we could flip the four but would be left with 2 remaining flips. So the problem is more finding out how many sets of 3 can fit into a number that’s also factorable by 4 cups. In other words, the number 12 is the least common multiple that shares the factors of 3 and 4. And it takes exactly 4 sets of three inversions to get to 12. Let’s prove it. Imagine instead of directly inverting the glasses after each move, we save up the inversions. It doesn’t matter the order we do them, as long as we keep track of how many moves and inversions we need to do. 1st move = 3 inversions 2nd move = 6 inversions 3rd move = 9 inversions 4th move = 12 inversions On the fourth move, we have 12 inversions we need to perform which is great because the 4 cups divide evenly into the 12 inversions. So we can flip the 4 cups down (getting rid of 4 inversions and having 8 left) flip all 4 up (getting rid of another 4 inversions and having 4 left to do) and flip all 4 down again (getting rid of the last 4 inversions out of the 12 we need to do and solving the problem)
Each move after the first you have the choice of undoing an earlier move, or picking a subset you haven't picked before. You pick one glass to stay the same each move, and when each glass has had one turn at that the puzzle is solved. Number of steps is N, and it works for even numbered of glasses, because odd numbers of glasses will result in the starting condition after each glass is flipped n-1 times.
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] .)
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
0.41 at first i just did for 4 glasses but as soon as you said case n I paused again and decided to try numbers up to 12 for f3 I got 1 obviously, 111 -> 000 f4 = 4, 1111 -> 0001 -> 0110 -> 1011 -> 0000 f5 = 3, 11111 -> 00011 -> 01101 -> 00000 f6 = 2, 111111 -> 000111 -> 000000 f7 = 3, 1111111 -> 0001111 -> 0010011 -> 000000 f8 = 4, 11111111 -> 00011111 -> 00000011 -> 00001101 -> 00000000 f9 = 3, 111111111 -> 000111111 -> 000000111 -> 000000000 f10 = 4, 1111111111 -> 0001111111 -> 0000001111 -> 0000010011 -> 0000000000 f11 = 5, 11111111111 -> 00011111111 -> 00000011111 -> 00000000011 -> 00000001101 -> 00000000000 f12 = 4, 111111111111 -> 000111111111 -> 000000111111 -> 000000000111 > 000000000000 One thing I knew from the start was that when n is a natural number >= 3 for fn if 3|n then fn = n/3 another thing I noticed for n ≠ 4 fn = floor(n/3) + n mod 3, as in you divide n by 3 and then add the remainder, oddly enough 4 is the only breaker of this quality and to prove there won't be another as you can see the last 5 digits always come down to 01111 00111 or 00011 with 00111 being solvable in 1 step and the others being solvable in 2, however with only 4 spaces f4 doesnt have enough room to solve in 2 steps, and since every other natural number being >= 3 is either divisible by 3 or >= to 5, 4 is the only one to follow this property.
A = Glass Right Side Up B = Glass Upside Down AAAA - Start BBBA - 1) Flipped three on the right BAAB - 2) Flipped three on the left ABAA - 3) Flipped both end glasses and a middle glass BBBB - 4) Flipped all the remaining right side up glasses.
My intuition is telling me this: If Si is the inversion of all glasses but glass number i, and you repeat that action, it takes you back to a configuration equivalent to a previous configuration in the sequence of flips and thus will add steps to get closer to the solution, so the most efficient solution is to perform each Si once for each i, 1 to n. I fear it would take a long time to prove that.
Instead of inverting 3 separate glasses, I just inverted the same glass 3 times which leaves it in the state that I want. And repeat for each the rest of the glasses. So it takes 4 moves to flip them all. For the N/N-1 case, it also explains why N Odd is impossible. Because you flip the glass an even number of times, so it never ends upside down.
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.
The amount of flips to get from all up to all down, should be divisible by the amount of glasses. (n) At the same time, it should be divisible by the amount of flips -per-turn (n-1) The minimal number divisible by both, obviously, should be n(n-1) And the amount of turns would be n(n-1)/(n-1) = n So the least amount of turns in the number equal to the number of glasses.
Nice! I did the process slightly differently. Used 0 to represent the upward open, and 1 for the downward open. So I did, 0000 1011 0101 1000 1111 4 moves!
![Twenty One Pilots - The Line (from Arcane Season 2) [Official Audio]](http://i.ytimg.com/vi/XCz7WUotXs8/mqdefault.jpg)
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
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?
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 :(
I'm just chuffed I could solve *something* on this channel
Yeah did it in 20 seconds but then saw I actually asked for more.
I did it in 4 moves intuitively and was satisfied when the proof supported my answer.
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
and the general solution is:
step 1) invert all but glass 1
step 2) invert all but glass 2
...
step n) invert all but glass n
yeah it's not even a puzzle really is it? I mean, the first move whichever 3 you turn over you end up equivalent to 0000 to 1110, 3 of them will be upside down one the right way up. This is the same as 0000 to 0111 or 0000 to 1011 and so on. There's no puzzling about move 1 it's a given.
For your 2nd move. If you turn the same 3 back over, you'd just end up back at the start. So the only other choice is to turn 2 upside down back and the one you didn't touch on your first move upside down. So 1110 goes to 0011. So far we've done 2 moves and had no real choice over either of them.
Now we're in a position with 2 upright and 2 upside down. The only 2 moves are, turn 2 of the upright and 1 upside down, or 2 upside down and 1 upright . One of which 0011 goes to 1110 which is going backwards to a previous state. So we must go from 0011 to 1000. I'd accept that this move requires a bit of thought.
Our last move is now trivial. We just turn the 3 upright glasses upside down achieving our goal 1000 to 1111.
So, it's puzzle insofar as you had to do a bit of analysis for move 3. The other moves are either just obvious or they're the only moves you can make.
I did this with my fingers, different moves, but same amount, nice.
Edit: for those who want my moves
1: fingers up, 0: fingers down
1111 - start position
0010 - move 1
0101 - move 2
1011 - move 3
0000 - move 4 / final position
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.
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.
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)
@@DimkaTsv Cool story. The explanation you replied to is 4 times shorter and infinitely easier 😂
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.
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
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*
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.
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?
@@VeryGoodDeals 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}.
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
I didn't watch the video yet, only saw the video thumbnail.
I was able to do it in four steps:
1. Flip 3 cups, that are in up position.
2. Flip 1 cup that is in up position, and 2 cups that are in down position.
3. Flip 1 cup that is in up position, and 2 cups that are in down position.
4. Flip the three cups that are in up position.
I don't really understand this
@@DoopenEvery time you flip you swap the left over cup for a cup that is the other way up.
@@Ruskettle I got even more confused but I decided to just try the steps with actual water bottles, I was able to understand hooray!!!
I was mostly confused because I thought we had 3 turns TOTAL that's why 😭 but yeah I get it now
I also have the habit of solving the thumbnails. Usually, the problem is nicely summarized there. Not it this case though :-)
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
He sounds so proud when he proves it. I meant this in most sincere way!
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.
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.
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."
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.
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.
Go figure language AI models aren't math AI tools. Who would have thunk it.
First time hearing “thunk” from a human being in genuine conversation
@@Axcyantol.😂
I can hear Jean-Luc Picard now... "There are FOUR glasses!"
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.
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.
Something that interests me is when you include a non-integer number of steps, and this would allow for solutions for odd numbers with a minimum of half of that odd number. (where in half a step you move half as many glasses) Example: (111), Step 1: (010), Step 1.5: (000). I know this is not actually how a "step" works, however I find it interesting that using this definition gives exactly half of the odd number as a minimum number of moves.
a more general version of the original question: given m glasses, all upward initially, each time you choose n (n
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.
0, flip the paper upside down. Done.
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
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
As a general rule, when mathematical explanation says "it can be easily shown" means a student can do it in a weekend. On the other hand, "it can be shown" means that a student may accomplish it in a couple of years of studies but sometimes it can refer to stuff like Fermat's Last Theorem where "it can be shown" took over 350 years of trial and error to finally accomplish the "can be shown" part.
the little gray cells - best part of the vid
I missed the part where you invert n-1 glasses and solved for inverting 3 glasses. Here is the solution for that problem: 4 is a special case since it leaves little room for possible moves so it is 4 moves. But for all other cases, you want to make the amount of unflipped glasses to be divisible by 3. The first case is where it is already divisible by 3, so just divide n by 3 for minimum moves. If you have 1 extra glass, flip 3, then unflip 1 and flip 2, (4 are flipped now, which is 3+1) so in 2 moves you have reduced the problem to the first case, so divide n by 3 and round up. If you have 2 extra glasses, flip 3, then unflip 2 and flip 1 (2 are flipped now) and again in 2 moves you have reduced the problem to the first case, so divide n by 3 and round up but add 1.
It's quite obvious, inverting all but one is the EXACT same as inverting just one. Just with more complexity.
i just took an induction proofs course last semester, it was absolute hell, past me would never believe that i watched and solved another one of these devilish problems for fun.
This looks like incrementing numbers in mod 2 and would be interesting to see what it would happen for other modules
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."
Smash one of the glasses so there’s one less glass to worry about
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.
Can't you just do lcm(3,4)÷3 to get 4?
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.
Stop using AI prompts and pretending that it can answer questions. It is a waste of everyone's time as well as being a huge waste of natural resources.
Lol, not the end of the world lad
Still better use of resources compared to most things on the internet.
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
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).
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.
I got to step 2 and realized it’s so easy with the right insight. Every turn inverts all but 1 glass. 3 inversions will result in an inverted cup. So do 4 turns with each glass excluded. Order doesn’t matter. Each glass will get inverted 3 times. Done!
0 is up, 1 is down, this may not be optimal but before seeing solution this is my attempt
0. 0000
1. 0111
2. 0101
3. 1000
4. 1111
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.
I solved this problem by breaking it down into two problems:
Instead of thinking of inverting n-1 glasses, think of it as two steps to be performed at each turn:
1. Invert only 1 glass in a turn
2. At the end of the turn invert all glasses
Forgetting about step 2 for now, if you can invert only one glass at a time it is obvious that you need a minimum of n moves to invert all n glasses, regardless of n being even or odd.
Now for step 2:
Inverting all glasses an even number of times is equivalent to not inverting anything at all. This explains why it works for n when n is even: effectively only one inversion takes place from step 1 and zero inversions from step 2.
Inverting all glasses an odd number of times is equivalent to inverting all the glasses just once. So if n is odd, you will have two effective inversions: one from step 1 and one from step 2, so in the end you'll have returned to the initial state. You cannot perform step 1 even number of times because there is an odd number of glasses.
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.
this rests on the assumption that each glass will have the same number of flips in total (thus a multiple of n). But it doesn't exclude the option where one glass has 1 flip and another has 3, for example
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
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.
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.
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.
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.
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.
there is a similar puzzle in the video game 'path of exile' to which the easy to remember answer is to 'click each pillar once' which translates in this case to avoid inverting any given glass each once
1) I have stopped the video to try myself and found a proof which does not require the case distingtion between an even and an odd number of glasses:
Since the glasses are identical objects, there are only two possibilities in each move: (1) Invert all glasses with exception of an upright one (2) invert them with exception of an upside-down one.
One of these two possibilities "reverts" the previous move, so if you want the minimum solution, there is only one "allowed" possibility in each move.
Now you can prove - using induction - that if you have N glasses, the number of upside-down glasses (ni) after the i-th step is:
n1=N-1, n2=2, n3=N-3, n4=4, n5=N-5, n6=6 ...
For even values of N, the only element in this sequence having the value N is n(N), so you need N steps to invert all glasses.
For odd values of N, the series contains only even values, so it does not contain the number N, so it is impossible to invert all glasses.
2) Have you heared about the cases when lawyers trusted fake cases invented by a KI?
If yes, you surely were aware that the "proofs" made by a KI would be nonsense.
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.
Using linear algebra:
(All glasses down) = (-1,-1,-1,-1)
= (α I_1+ β I_2 + γ I_3 + δ I_4) (1,1,1,1)
where I_k is minus the identity matrix with a positive 1 value at row k and column k. It represents the inversion of all glasses but one.
Solving this system gives the equation:
1 = αβγδ
where the grec letters indicate the number of times you apply a different inversion and therefore need to be natural numbers.
The solution α=β=γ=δ=1 (add up to four moves) is the one showed in the video.
There is an error in logic in this video. The answer is always N. This is the same logic as the tower of Hanoi and uses the same solving technique. In every case, because the difference is N-1, there will always be one less glass flipped then the total. Repeat this for all glasses, and all glasses will be flipped. We can model this by shifting the glass we leave untouched all the way across the number. This iterated step will take N moves every time. Whether the glasses are even or odd doesn't matter.
I think your answer can easily be proven wrong. For example, if we do a test to prove this answer, because n=1 cannot be done, we try with the next odd number, namely n=3. If it is repeated, after 2nd move, the glass can only return to the first move
@@brurydeardomartin1800 No, N=1 is trivially easy. You flip one glass and the glass is flipped. job done. You cannot flip 0 glasses because 0 is not a quantity. N must be a quantity or the question doesn't make sense. N = 2 you flip one glass, then you flip the other. That solves N = 2, and so on.
@@rambysophistry1220 you do realize that we have to flip n-1 cup? And 1-1 isn't 1?
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.
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
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.
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!
For the 4-glass case, the proof that 4 moves is minimum is that for each move you must reach a state not equivalent to a previous state (number of glasses up and down). In practice this always only leaves one move available to proceed; there is only one way to reach the goal without back-tracking.
"Hey this is press Chalker" ahh captions 😭😭😭🙏
The product of two adjacent positive integers is always an even number. This fact is at the core of this puzzle. Otherwise ChatGPT is not so bad, just you need to be patient with it. If it gives a wrong answer to the problem first, and you see the error in it, then you can educate it by pointing at the error and asking it to rethink.
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.
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.
My idea of the lowest possible solution was if the moves were divisible by 4. Because it's impossible to flip 4 glasses over if the total number of glasses you flipped was not divisible by 4.
You can also think of the solution another way, since 3 glasses are flipped and 1 STAYS UPRIGHT. This problem is logically the same as if you were to flip 1 glass over at a time (which can be seen in the solution to the original problem).
On a hunch, I tried this and it worked:
Move #1: Flip all glasses except the first one. ( uuuu -> unnn )
Move #2: Flip all glasses except the second one. ( unnn -> nnuu )
Move #3: Flip all glasses except the third one. ( nnuu -> uuun )
Move #4: Flip all glasses except the fourth one. ( uuun -> nnnn )
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.
Flipping all but one glass os equivalent to flipping one and then inverting all the glasses. in order for all the glasses to be upside-down, they must all be in the same state, and therefore the single flips must have been applies to all the glasses the same number of times, and thst number plus the number of inversions must be odd. If all glasses had been fliped 0 times, then nothing changes and they're upside-up, and flippings more than one are multiplying the total number of both flips and inversions by a constant, which is useless because parity is all that matters. Therefore, they must all be flipped once individually and and inverted together an even number of times. In order to flip n glasses individually once, you must do flipping n times, and that comes along with n inversions. If n is odd, then n+1 is even and everything is upside-up, but if n is even, n+1 is odd and you've done it in n steps.
I was like "wait this is easy"without clicking on the vid till i realized,inverting the glasses counts as a move
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
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.
Took me about 10 seconds from the thumbnail alone 😅
I really like these kind of riddles/puzzles
You can simplify the problem as how many glasses turn every 2 moves and that is always true whatever glasses you choose.
So for even numbers it is easy. You end up with 2 after 2 moves, 4 after 4 moves and n after n moves.
For odd number you always turn even number of glasses. Which ensures alway an odd number of glasses remain up no matter what. Since 0 is even (you need to get to 0 glasses up) it is impossible to get there
The way i thought of it is rather simple.
Inverting all but one glass is the same as only inverting one glass. The minimum number of moves to get all glasses facing down while only inverting one at a time is exactly N.
Therefore, the answer is always N.
For the n being even case:
Given that there are exactly N flips, each glass is flipped n-1 times.
The only way to go from up to down is flipping an odd number of times, so if N is even, then n-1 is odd and it works! If N is odd, then n-1 is even, making it impossible, since N is _guarranteed_ to be the minimum if it exists.
Because of the flaws (and lacks) of the rules its possible to make it in 2 steps.
The rules are not say you cannot move the glasses or replace their position with others. And the rules does not state what is meant to be "different". Therefore we can say "different" could be mean the original position of the glass in the step.
So you move the first N-1 glasses in the first step, then move the move the next step comes the magic:
V1 (swapping): N'th glass flipped, then swapped with position of N-1, then N-1 flipped, and swapped with N-2 and so on until you reach the desired all face down position.
V2 (moving 1st glass to the end): N'th glass flipped, then the 1st position glass moved to the end. Than N-1 flipped, and 1st glass moved to the end, and so on...
This solution make it possible to make the puzzle regardless of N is odd or even.
If you say that is not allowed, than please show me where it was written in the rules :)
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.
I think I have an easy proof for both n even and odd. At each step except the first step, you really only have two choices, denoted (^) and (v): (^) preserves (does not flip) an U glass and flips everything else, and (v) which preserves a D glass and flips everything else. Some lemmas:
A) Your first move must be (^) (since there are no D glasses to preserve).
B) Doing the same move twice in succession (^)(^) or (v)(v) produces no change, so
C) to preserve minimality there there is never a point to repeat moves in succession.
D) Therefore, the solution is going to be a series of alternating (^)(v) moves, starting with (^) per A.
E) After each pair of moves (^)(v), the number of Us decreases by 2.
So to get to 0 Us, you need n/2 (^)(v)moves, i.e. n moves for even n.
QED for even n.
For odd n, this forced method (after n-1 moves) will eventually get you to 1 U. The next move you must make is a (^) (from D); but this will simply revert all the glasses to U, getting you to square 0. This is a contradiction, so there's no method for odd n.
QED
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.
The way I always saw this problem was like a least common multiple thing. There are four cups and three inversions per move. We know that we can’t flip all four cups in three inversions (one move) and if we tried to do it in two moves, we could flip the four but would be left with 2 remaining flips. So the problem is more finding out how many sets of 3 can fit into a number that’s also factorable by 4 cups. In other words, the number 12 is the least common multiple that shares the factors of 3 and 4. And it takes exactly 4 sets of three inversions to get to 12. Let’s prove it.
Imagine instead of directly inverting the glasses after each move, we save up the inversions. It doesn’t matter the order we do them, as long as we keep track of how many moves and inversions we need to do.
1st move = 3 inversions
2nd move = 6 inversions
3rd move = 9 inversions
4th move = 12 inversions
On the fourth move, we have 12 inversions we need to perform which is great because the 4 cups divide evenly into the 12 inversions. So we can flip the 4 cups down (getting rid of 4 inversions and having 8 left) flip all 4 up (getting rid of another 4 inversions and having 4 left to do) and flip all 4 down again (getting rid of the last 4 inversions out of the 12 we need to do and solving the problem)
Each move after the first you have the choice of undoing an earlier move, or picking a subset you haven't picked before. You pick one glass to stay the same each move, and when each glass has had one turn at that the puzzle is solved.
Number of steps is N, and it works for even numbered of glasses, because odd numbers of glasses will result in the starting condition after each glass is flipped n-1 times.
n if n is even, impossible if n is odd. Claude 3 gets there without help, but it takes a couple attempts to get it right.
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] .)
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
0.41 at first i just did for 4 glasses but as soon as you said case n I paused again and decided to try numbers up to 12
for f3 I got 1 obviously, 111 -> 000
f4 = 4, 1111 -> 0001 -> 0110 -> 1011 -> 0000
f5 = 3, 11111 -> 00011 -> 01101 -> 00000
f6 = 2, 111111 -> 000111 -> 000000
f7 = 3, 1111111 -> 0001111 -> 0010011 -> 000000
f8 = 4, 11111111 -> 00011111 -> 00000011 -> 00001101 -> 00000000
f9 = 3, 111111111 -> 000111111 -> 000000111 -> 000000000
f10 = 4, 1111111111 -> 0001111111 -> 0000001111 -> 0000010011 -> 0000000000
f11 = 5, 11111111111 -> 00011111111 -> 00000011111 -> 00000000011 -> 00000001101 -> 00000000000
f12 = 4, 111111111111 -> 000111111111 -> 000000111111 -> 000000000111 > 000000000000
One thing I knew from the start was that when n is a natural number >= 3 for fn if 3|n then fn = n/3 another thing I noticed for n ≠ 4 fn = floor(n/3) + n mod 3, as in you divide n by 3 and then add the remainder,
oddly enough 4 is the only breaker of this quality
and to prove there won't be another as you can see the last 5 digits always come down to 01111 00111 or 00011 with 00111 being solvable in 1 step and the others being solvable in 2, however with only 4 spaces f4 doesnt have enough room to solve in 2 steps, and since every other natural number being >= 3 is either divisible by 3 or >= to 5, 4 is the only one to follow this property.
I didn't do what he meant by doing different n, but I think my discovery is cool anyways
A = Glass Right Side Up
B = Glass Upside Down
AAAA - Start
BBBA - 1) Flipped three on the right
BAAB - 2) Flipped three on the left
ABAA - 3) Flipped both end glasses and a middle glass
BBBB - 4) Flipped all the remaining right side up glasses.
My intuition is telling me this: If Si is the inversion of all glasses but glass number i, and you repeat that action, it takes you back to a configuration equivalent to a previous configuration in the sequence of flips and thus will add steps to get closer to the solution, so the most efficient solution is to perform each Si once for each i, 1 to n. I fear it would take a long time to prove that.
Havent watched the video yet, but i did it in 4 steps.
0. XXXX (Starting Position)
1. YYYX
2. YXXY
3. XXYX
4. YYYY
Awesome .Thank you professor.
Instead of inverting 3 separate glasses, I just inverted the same glass 3 times which leaves it in the state that I want. And repeat for each the rest of the glasses. So it takes 4 moves to flip them all.
For the N/N-1 case, it also explains why N Odd is impossible. Because you flip the glass an even number of times, so it never ends upside down.
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.
The amount of flips to get from all up to all down, should be divisible by the amount of glasses. (n)
At the same time, it should be divisible by the amount of flips -per-turn (n-1)
The minimal number divisible by both, obviously, should be n(n-1)
And the amount of turns would be n(n-1)/(n-1) = n
So the least amount of turns in the number equal to the number of glasses.
Nice! I did the process slightly differently.
Used 0 to represent the upward open, and 1 for the downward open.
So I did,
0000
1011
0101
1000
1111
4 moves!