I got it in 9 tries. With some luck. My strategy is first to check to know how many trues there are. Then to change a group of 3 to true. I first tried with a group of 2 but with some math I realized it has better odds to do it with 3, and who knows If it is better with more, I dont want to suffer anymore thinking lol. Lets say {A,B,C} represent the positions changed and add an extra position D for the strategy. The strategy consists of 3 steps depending on how many "correct values" there are when applied the change, lets call this number R. 1. Check with the changes on A,B and C The possible values of R at this point (subtracting with the original check) are 0,1,2 and 3. If R= 3 or R = 0, you already know the values of A,B and C, because the changes are all correct/incorrect . So finish this group of positions and repeat with another group. If R =1 then 2. Without loosing generality you will restore the value of C. Then change the value of D and do a second check, this R'( the R value of the new change) Could be either 0,1 or 2. But not 3. Becuase Given R was 1, its not possible for the changes in A and B to be both right. If R' = 0, you know A,B,D should be False ( or they original value) and C = True(because of R =1)... True means the change applied was right, False means it was not. If R' = 2, as A and B cannot be both True then D has to be True. And you could pick up a new D and repeat step (2.) Or as I did(I think is better too) simply check If A = True, If it is then you know A =True, B =False, D = True and C = False( becuase of R=1) If R' = 1 then 3. Restore the change of B and change again C. Meaning the changes now are on A,C and D. Lets check again and denote R'' as the R of this change. R'' again can only be 0,1 or 2. If it is 0 you know A,C and D = False and B =true If R''= 2... A cannot be True, because it means (A and D) = True or A and C. In both cases its a contradiction with R=1 and R'=1 of the steps 1. and 2.... Then you know D = True and C = True... an becuase of R , A = False and B =False. If R'' = 1. Then A = True. Because If it is not its a contradiction again with R=1 (that includes A) and R'=1 (also includes A)... Meaning C = False and D = False. Because Of R=1 also B =1. If in step 1, R=2 instead of 1 you have to make the inverse. You know that If instead of being originally False they were True (meaning you did the opposite change) then R would be 1. Then you act on that basis. With that, after those 3 steps you are guaranteed to know the values of A,B and C. And probably of D too. In step 1 you have 50% chance of finishing, knowing 3 values with just one check. On the 2 step you have 33% chance to fimish 4 values on 2 checks. And essentially the other 67% is knowing 4 values in 3 checks. So If we want to know the expected proportion, or values known per check. We would do. E = 0.5 * 3 + (0.5)*(0.33) *2 + 0.5 *(0.67) * (4/3) E = 1.5 + (1/3) + (4/9) E = 3/2 + (7/9) E = 2.277 aprox. So the expected number of tries for 16 positions is...... 7? who knows maybe I got wrong the probabilities of the outcomes .... but anyway edit: After re thinking probabilities of the outcomes... Maybe the correct thing is... lets define p = 0.5*0.5*0.5, its the probability of any of the permutations of the values of A,B, and C. Finishing in first step is the prob of 3 changes being right + 3 being wrong. That is p *2 = 0.25 Then for it to be one right is just p by the permutations or whatever, meaning you have one correct to choose from 3 so it should be 3. And pretty much the same for R = 2. Then it would be p* 2 * 3 = 0,75 = 6*p Individually would be half of it, 0,375 = 3*p For the second step, I dont know If it could be applied the bayesian rule. But lets say it cant be applied. Then we can say If we have R'=0 means, A,B false and C is true. That is again a probability of p, and then D = false. So its p*0.5 .... R' = 1 has a probability of, p*0.5 the same as R'=0 but D = true. Or either A or B is true and the others false. That would be 2*p *0.5. Meaning, 1.5 * p in total If R'=2 it is D = true and essentially the same as R'=1, then it is 2* p * 0,5 = p As we can see the total of the second step is the probabilitied of R=1. That is because again we ignora for now R=2 and R=0 or 3 would not reach here. Step 3 R'' = 0 .. means B = true, rest false, p* 0.5 R''= 1 its esseiantly one case too, p*0.5 same for R'' = 2, p*0.5 Agaun we see the sum is 1.5* p = the prob of R'= 1, the condition for reaching step 3. So summing up the 3 possible ratios 3 values in 1 check = R=0 or R=3 on R =1 4 values in 2 check = R'=0 4 values in 3 check = R'=1 or R'=2 and the same for R =2 soo 4 values in 2 check = 2* (R'=0) 4 values in 3 check = 2*(R'=1 or R'=2) 3 values in 1 check = 2*p = 0.25 4 values in 2 check = p*0.5*2 = 0.125 4 values in 3 check = 1.5*p *2 = 0.375 E = 0.25 * 3 + 0.25 + (4/3) * 3 * p E = 1 + 4 * p = 1.5 values/check And then for 16 would be. 16/1.5 + 1(original check) 32/3 + 1 = 11.66 That is more logic given I got it with 9 with luck.
2 years ago (edited) | 4
I made a video out this if you are interested: https://youtu.be/5qqxGwlUilU
2 years ago | 0
There are a total of F(n) for false and T(n) tiles for true out of 16. Their respective probabilities are F(n/16) and T([16-F(n)]/16). To identify scoring tiles, you can make them all uniform to exclude ambiguities: All tiles F = F(n) All tiles T = T(n) Then when toggling one tile against the rest, you identify its scoring value as either the toggled value (score unchanged) or the opposite value (score-1). You can then subtract one of this value both from the remaining values of that kind and from the total remaining values. With every identified tile in trial (t), the probabilities change for F and T such that the identified value is now V1([n-1]/16-1) V2(n/[16-1]) and continues with each trial e.g. V1([n-1]/[15-1]) V2(n/[15-1])..etc. Then you can leverage the changing probabilities with each trial. The first trial is uniform and identifies the balance of probability. From this determine which value is more populous (if values aren't equal), and set all tiles uniformly to the opposite of the more populous - this increases the likelihood of scoring toggles by changing the distribution. Then with each trial, toggling one against the rest as earlier for all remaining tiles, continue until either the distribution shifts favorably to the other side (e.g. after identifying the 5th F there remains 4F 5T, making the probabilities 4/9F 5/9T so T more likely), at which point toggle uniformly all unidentified values to the opposite, or until one value has been exhausted (0F or 0T) at which point all remaining values are known. With this method the number of trials is determined by the initial distribution of F,T such that the optimal minimal number of trials is for F>T: T(n)+2 T>F: F(n)+2 F=T: F(n)+2
2 years ago (edited)
| 0
Brainxyz
Can you solve this new brain puzzle game? (A new thrilling video is coming soon :)
You need to predict a grid of 16 hidden true(T) and false(F) squares. Initially all set to F.
To check your score, press the Check Score button. Change the squares and Check your score again. Do that repeatedly until you reach the score of 16 (which means you have predicted all the answers). You shouldn't use the Reveal Answer button because this is not a memory test. The optimal solution should reach the perfect score in minimal number of trials and this is very relevant to how our brain works. We will be discussing that in an upcoming video. For now, we would love to hear back from you! How do you approach to solve this puzzle?
Here is the online link to the puzzle:
www.brainxyz.com/wp-content/uploads/2022/11/index.…
2 years ago (edited) | [YT] | 15