Stay awake for one night and catch it, when it tries to move one hole. Done.

Fill the hole with concrete every day .......that's it

just simply watch fox footprints on mud around hole lol..hahaha

Smooth out the dirt around all five holes, come back the next day and look for tracks. I'm glad the people at Google understand the old art of tracking 😂😂

I used to be a fox. Always staying ahead of the neighbouring husbands.

2, 2, 3 ,4, 4, 3 ,2 this sequence will work very well. If the fox in 1st hole first day, it will catch in 2nd day. if the fox in 2nd hole it will be caught in the 1st day. if the fox in 3rd hole in first day, there are three ways to move: move to 2nd hole in the 2nd day(it will be caught in 2nd day), move to 4 in the 2nd day then 5 then 4 holes in 3 and 4 days (it will be caught in the 4 th day based on my sequence ), move to 4 in the 2nd day then 3 then (it will be caught in the 3 rd day) . if the fox in the 4 hole in the first day, so there are 11 scenarios can be happened; move to 5 then 4 then 5 then 4 (it will be catch in 5th day ), move to 5 then 4 then 3 then 4 (it will be catch in 5th day), move to 5 then 4 then 3 then 2 then 3 (it will be catch in the 6 days), move to 5 then 4 then 3 then 2 then 1 then 2 (it will be catch in 7 days), move to 3 then 4 then 5 then 4 (it will be catch in 5 days), move to 3 then 4 then 3 then 4 (it will be catch in 5 days), move to 3 then 2 then 3 then 4 (it will be catch in 5 days), move 3 then 2 then 3 then 2 then 3 (it will be catch in 6 day), move 3 then 2 then 3 then 2 then 1 then 2 (it will be catch in 7 days), move 3 then 2 then 1 then 2 then 1 then 2 (it will be catch in 7 days), move 3 then 2 then 1 then 2 then 3 (it will be catch in 6 days). If the fox in 5th hole, there are two scenarios for moving: move to 4 then 5 then 4 (catch in 4 th days). Move to 4 to 3 (it will be caught in three days). solve by Ala Zayed

I solved it differently by visualizing where the fox can be each day. 1. Starting with hole 2 => the fox can be in any other hole (1, 3, 4 ,5). 2. Moving to hole 3 => the fox can be in three other holes (2, 4, 5). 3. Moving to hole 4 => the fox can be in three other holes (1, 3, 5). 4. Staying on hole 4 => the fox *has* to be in hole 2. (We could go grab it right now) 5. Moving to hole 3 => the fox *has* to be in hole 1. (What are you waiting for? Go grab it!) 6. Moving to hole 2 => It is caught. (Finally...) My sequence is 2, 3, 4, 4, 3, 2; reversing the first half. But it's basically the same solution. I liked my approach because I didn't have to assume anything. However it is much simpler to generalize with the approach shown in the video. Just start with the second slot and increment until n-1, then do that sequence again or in reverse and you are done in O(2(n-2))

First we should inspect the first hole and fill it with cement Second day you should inspect the second hole and fill it with cement And do the same process😂 Saved one day of inspection

Keep the food outside with strong aroma....and sit back relaxed Mission accomplished in 10 second 😋

I can probably see everything in the holes from the window of my 3 story house.

The easiest way would be to go and select a hole in the morning, if you are lucky enough you catch the fox and if not fill up that hole and come next morning and keep doing until you catch the fox 😃

You can actually generalize this to any number of "holes" (memory addresses, etc.) with a guaranteed worse case of 2(N-2).

It would make more sense to present this as a friend offers you the riddle as a party game that he will move an object from one cup to another while you cooperate and do not look, there is no looking through the cups and no other forms of bypassing your choice to guess blindly or optimize probability and ensure you guess right. People won't cover a cup if they are told it is cheating but if you say there is a fox you want to stop it just check all the holes in one day or cover the holes when you know it has this strange compulsion to move around that foxes usually do not have. PARTY GAME, PLEASE.

Awesome thanks, caught the fox in my yard and it bit me!

I would check all the hole at once cause I like to work smarter not harder lol

My solution was 1-2-3-4-5-5-4-3-2-1 (every hole is inspected twice, once on an even-numbered day and once on an odd-numbered day, and you're inspecting them in systematical order, so that if the fox is in an odd-numbered hole on the first day then you'll find it in the first five days; otherwise you'll find it in the last five days). I hadn't realized that we can ignore hole 1 and hole 5 because if the fox is there then it is forced to a certain hole the next day anyway. So the optimal solution is 2-3-4-4-3-2, or 4-3-2-2-3-4, or 2-3-4-2-3-4, or 4-3-2-4-3-2. Anyway, excellent puzzle, and excellent solution!

If it's that hard to look into 5 holes in one day, then use a camera over night, capture it on day 2 for sure.

We can easily solve this as:- We can cover each hole with air exchange cloth and fox would always remove cloth when want to move to specific hole and now we can easily track fox movement.

How would the problem change if the holes were arranged in a ring, where the Fox may move from Hole 1 to Hole 5 or vice-versa?

Never knew what he wanted in solution. He didn't mention he was looking for minimum number of days to find the fox.

Very nicely explained. Thanks!

If you fill it with mud every other day you'll catch the 🦊 in 3 days

Thanks for sharing..very interesting puzzle..

Hmm I got the exact same answer, although my thought process wasn't the same. Instead of focusing on determining whether the fox started on an even or odd hole, I had made a chart of sorts of all possible fox locations and checked spots with the intention of lowering the possible locations he could be in while keeping in mind that if the fox is at the edge he's forced to move to only 1 possible location the next day. Interesting how different outlooks still converge to the same solution.

2,2,3,4 then if you haven’t caught him go 4,3,2. I think that should work but I haven’t written anything down.

This didn't seem impossible but took long time to solve for me lol

I was thinking about always looking at hole 3, which will always have a 33% chance on having the fox after night 1. I'll explain: If the fox isn't in hole 3, then it's, obviously, on holes 1, 2, 4 or 5. If it's on holes 1 or 5, it will have only one movement, which is to move to holes 2 or 4 respectively. If it's on holes 2 or 4, it will have a 50% chance on moving to holes 1-5 and a 50% to move to hole 3. That's 2 movements out of 6 possible ones, which will lead the fox to the hole 3. But this approach doesn't GUARANTEE catching the fox in 6 days or less. Some quick math, it gives me a rough 70% chance by the 6th night. This is by having 20% on the first night, and 33% on the next nights. Anyone who could confirm, plz comment below.

To be honest, since it wasn't specified other than "What strategy would ensure that the fox is eventually caught?", I would start at the far left hole and if he's there' great, and if not then I FILL THE HOLE. I would continue moving to the right-most hole repeating my filling of a hole he wasn't in until he was eventually found as he would run out of left/right decisions quick enough. The best part is, if I saw that a previously filled hole was now open again, that's the first hole I would check when it's time and BOOM...fox. Remember, they just asked what strategy would ensure capture, and didn't mention that I couldn't think outside the box. So that would be my strategy.

Guys, why not just bring a couple of friends to help? Forget the cement, and checking hole 3 everyday. Bring 4 friends and have it done and dusted by dinner time : )

Damn... Even I wish I would get to enter a different hole every day ( if you know what I mean ).

Better strategy for more general problems is apply Dynamic programing. It was not intuitive how we arrive at even/odd guess. We start from the last day and go back. Say All possible ways the fox can move are caught in say day T (terminal state). Lets pick 4 as terminal state (4/2 will work but not 5/3/1) as for say 1 fox could have been in hole 2 earlier and could move to 3 to avoid getting caught. If 4 was the last possibility then fox was in 5 previous day. So previous day better to look at 3, to make sure fox can move only from 2 -4 and not 2-3, and so on. Sorry difficult to explain in words, pic would have been easier. You will see you will get the same answer pointing to even/odd intuition

well, that one caught me) but, such an approach is applicable to any number of holes. In fact, this is more about similar (fox can move into the certain hole only from holes before and behind) possibilities than about simultaneous ones.

I thought this: name the consecutive holes A B C D E. Check B - C - D - B - C - D. In the word case you find the fox at sixth check in D.

Can't we just wait in any one hole🐾!

If you label the holes 1 2 3 4 5 then the sequence of holes you should check to grab the fox should go 3 4 3 2 2 3 4, or the reverse (3 2 3 4 4 3 2). 3 4 3 2 4 3 2 and 3 2 3 4 2 3 4 should both also work. EDIT: After watching the video I realized that going with the latter sequence first (234 or 432) gets the first sequence (3432 or 3234) cut a day short by starting with an even rather than an odd. Although I do wonder if you averaged the catch rate of both approaches if they would equal out; the 6 day approach is more likely to catch the fox later while the 7 day approach is more likely to catch the fox earlier. Actually, why not do the math here? The chance the 7 day sequence catches the fox on day 1 is 1/5. The chance it catches the fox on day 2 is also 1/5. Day 3 is 1/10. Day 4 is also 1/10. Day 5 is 1/5. Day 6 is 1/10. Day 7 is 1/10. Ok, so the chance you catch the fox by day 1 is 1/5, by day 2 is 2/5, by day 3 is 1/2, by day 4 is 3/5, by day 5 is 4/5, by day 6 is 9/10, and by day 7 is 5/5 The chance the 6 day sequence catches the fox on day 1 is 1/5. On day 2 it is 1/10. On day 3 it is 1/10. On day 4 it is 1/5. On day 5 it is 1/5. On day 6 it is 1/5. So the chance you catch the fix by day 1 is 1/5, by day 2 is 3/10, by day 3 is 2/5, by day 4 is 3/5, by day 5 is 4/5, and by day 6 is 5/5. On average the 7 day approach will get you the fox on day 3.6 while the 6 day approach will get you the fox on day 3.7. That means my hypothesis is correct, the 7 day approach will, on average, give you the fox sooner than the 6 day approach. Which is really weird to think about.

I liked your approach to this one. I have my own approach -> 5,4,3,2,1,1,2,3,4,5 Or 1,2,3,4,5,5,4,3,2,1 Anywhere the fox goes it will get eventually caught while u implement this system Prove -> Suppose Fox choose 3rd hole Start from side (hole 5th or hole 1st) Next day check hole 4th if u took side 5th Or hole 2nd if u took side 1st. Continue in this sequence. If fox jumped twice towards your last observed hole then on your third day it is caught. If it ran away from u then don't worry Just keep on doing it eventually u will find out fox is getting trapped after it makes escape jump. *Logic goes that if u have observed 1st hole but fox was at odd no. now, then it would be eventually caught and if it was at even then we simply made it to jump at odd no.*

If fox had a google account, they'll track the fox for you ;)

Holy shit, I actually solved it with the method in the video.

Whilst this method is the quickest surefire way of finding the fox, there is another method which has a smaller expected value for the number of days it takes to find the fox, but has a maximum number of days of 8: day 1: try hole 2 (if not found, it could be in 1, 3, 4 or 5) - success rate 20% day 2: try hole 2 (if not found, it could be in 3, 4 or 5) - cumulative success rate 50% day 3: try hole 4 (if not found, it could be in 2, 3, or 5) - cumulative success rate 75% day 4: try hole 3 (if not found, it could be in 1, 2, or 4) - cumulative success rate 77.5% day 5: try hole 2 (if not found, it could be in 1, 3, or 5) - cumulative success rate 80% day 6: try hole 2 (if not found, it could be in 4 only) - cumulative success rate 87.5% day 7: try hole 3 (if not found, it could be in 5 only) - cumulative success rate 93.75% day 8: try hole 4 where you will find it - cumulative success rate 100% The expected days needed to find the fox is 3.1625 which is lower than the 3.55 days described in the video.

I paused this approx a week ago. finally got the answer this morning, I hope I am right - 234234 will ensure the fox get caught eventually

Love these questions... Looking forward to more :)

Easier logic to catch the fox: Tie a piece of meat to a trap and wait till the fox takes the bait. Sometimes common sense works better than mathematics😂

Wow!!! I paused the video after question.. and solved it in couple of minutes.. Then ask my friend the same question and explained the answer to her when she couldn't solve. Then watched the video further, and my explanation was line by line same as the video.. She even asaked if I watched the . video before :P :P

Hole 1 be like.... Oo bhai maaro mujhe maaro... Ni ye majak ho raha h

Put some food near there the fox will come out itself. 🤣

Pick a random one, if you don't find it , block that hole... it will take 5 days..

Keep a delicate thread like trap between hole 2 and hole 3 on first day, if it is damaged, it is in hole 2,else hole 4.

We can easily catch it, While change the hole fox leave some foot mark or some sand near the holes right.

What about foot prints of fox, we could easily find out where the fox is by just inspecting where the footprint goes......

Check hole 3 every day until it’s in there. Maybe the fox just goes back-and-forth from 1 to 2 or 4 to 5 and never comes to 3, but there’s already nothing logical about a fox changing holes every day, so I’ll add my own assumptions and can’t be wrong.

U are correct bro Perfect

Excellent

Explain question clearly 😅😅😅

Night 1 waite for the fox to move holes and shoot it. How hard was that?

I was to able to solve it with at max 7 searches ( not optimal though ). Here's the sequence search: 2,2,3,4,2,3,4 I didn't use the concept of OddEven as mentioned in the video. I did it with trial and error. On each day, I would keep track of the holes under which the fox may be possibly present. Accordingly, searched a particular hole ( in the order I mentioned above ) and again, recalculated the holes under which it might be present on the next day. If we carefully follow the sequence, we are able to restrict the fox on a single hole on the 6th day (ie the last hole). A search of hole no 4 on the 7th day gives us the FOX!!

The only strategy that I came up with was to check either 2 or 4 but stick to the same option, 2 or 4. I don't think you can add anything to a strategy by stating initial condition of odd or even, because this is not known and there is no stipulation that the starting state would ever change or reset, so that can be totally removed from a strategy.

Kejriwal is the best person to solve this puzzle 😂😂😂😂😂

Logic says look at all the holes on the same day, this has nothing to do with logic lol

I made it in a paper and got to this other answer: 2-2-4-4-3-2, the logic being you discard the fox being in one corner, then the other, and finally... you catch it

On the 1st day smooth out all the dirt around all the holes. Catch him on the second day by looking for his tracks.

This test is strangely written.. if I have to "eventually" catch it, I'd just check the same hole every day. Given an infinite amount of days then my chance of catching it is so close to 100% so as to not make a difference.

Several logical options include; 1. Learn about the natural predators of the fox species involved, then buy 4. Or 2. Obtain 5 baited traps. Havahart (tm) live traps will not kill the fox. Or 3. Hire a Hunter looking for fox fur.

Pile up mud around all the holes. The fox wen it moves during night it will scatter the mud. U will find it the next morn.

I didn't think of this odd even method, but my solution also relied on holes 2 and 4. I start at 2 and repeat it to catch the fox in case it was in 1. Otherwise I know it was in 3, 4 or 5. Then I check 4 twice with the same idea. Since it wasn't in 4 or 5, it had to have been in 2 or 3 at the time I checked 4 the first time. Now I could calculate all the possible moves by the fox and try to catch it at 2, checking it twice again. Since I didn't catch it, it must've moved from 4 to 3 to 2 to 3 to 4. So now I know the last location was 4 and thus the fox must now be at 3 or 5 and I can solve the end like in the video. This method took 8 days, however. Less elegant, but it worked. :)

The riddle is easier to solve by visualizing each day's status ?X??? 0?X?? ?0?X? 0?0X0 ?0X00 0F000 where: X is the hole we check today ? is possible current location of the fox 0 is 100% empty hole F is the fox caught by it's tail

The other challenge is to understand him

2234234 that's how I did it and took time for this but after watching video I realised that the double check is not necessary (22) because after 2234 and 234 we will get to same 1 gap between possible places situation which is actually the key to the solution

This is my first example in my invariant class

Took me about an hour of thinking, but yeah. I was correct.

1-2-3-4-__ which is next. Hint: it is not 5.

Great one! I thought the puzzle didn't have an answer.

I couldn't answer correctly. But I understood ur explaination

I look in every hole each day because I'm not bound by your rules and I'm trying to find a dang fox.

Think outside the hole??

Feeling bad for Hole 1

Is this how Google picks which neighborhoods to gentrify?

I found the answer in a different way. It took me more days than the optimal solution though. Label the 5 holes, A, B, C, D, E. 1. Check B. If the fox is not found in B, then the fox moved from B->A, B->C, C->D, D->E, or E->D. 2. Check B again. If the fox is not found in B, then the fox moved from C->D, D->E, D->C, or E->D. 3. Check B again. Now you have two possibilities. 1) The fox moved from initial position D to a different position (D->C or D->E). or 2) The fox moved from a different position to D (C->D or E->D). 4. Check C. For possibility 1, the fox would have moved from C->B or E->D. For possibility 2, the fox would have moved from D->E, and that's the only possibility if the fox is not found in step 4. 5. Check D. If the fox is not found in position D, then possibility 2 is no longer valid. Therefore, we can conclude one of two possibilities - Possibility 1: the fox moved from C->B just before step 4 Possibility 2: the fox moved from D->C just before step 5. If possibility 1 is true, then the current position of the fox would either be A or C. 6. Check D again. If the fox is not found in this step, we can conclude two possibilities. For initial possibility 1, the fox would have to move from A->C or C->B. For initial possibility 2, the fox would have to move from C->B. 7. Check C. If the fox is not found in this step, then the fox must have moved from B->A. 8. Check B. The fox is found.

Fox puzzle was easy. The more challanging puzzle was to decrypt your accent

I think you should just put nets on all of the holes

Just pull an all-nighter and watch which hole the fox moves into then you find him the first time.

The correct strategy is to check all of the holes in the first morning.

Check the centre hole repeatedly as statistically it's the one he'll spend the most time in. Eventually he will land in it and you will catch the fox.

1 off from optimal, got 3, 4, 3, 2, 2, 3, 4. The fox is still caught, but the solution is not optimal because I started looking for the fox with odd parity, which will always require an extra step. I missed that starting even would shorten the length, since finding an even fox is an odd number of steps, which switches a fox starting in an odd hole to the even holes, again requiring only three steps

Just using the binary search method will actually solve this. Ensuring 3rd hole is the middle, checking it 2 times will be the best start. Then it is up to you to check holes 2 and 4 without missing hole 3 when shifting from inspection of 2 to 4

Start with hole 2 then hole 2 again If you didn't catch it you now sure hole1&2 are empty Then go to hole 4 if it is empty you now sure hole 1& 4 are empty Then go again to hole 4 If you didn't catch it now you sure that hole 4 & 5 & 2 are empty Now the Fox in hole 1 or 3 So go to hole 2 if you didn't catch it now you sure that the Fox in hole 4 Then go to hole 3 if you didn't catch it now you sure that the Fox in hole 5 now you can catch in next day in hole 4 because no choice for him .

Actually your solution is correct but it is not an optimal one in all cases. Your solution has the lowest max days to find the fox. Optimally though, one could consider the best solution to be the one that has the lowest average number of days to find the fox OR even the lowest average number of holes that need to be searched (how much effort you have to do). Its pretty amazing. All of them have different solutions!! Your solution: { 2, 3, 4, 2, 3, 4 } : Max days to find fox = 6, Average days to find fox = 3.55, Average effort needed = 3.55 hole searches Alternate optimal solution #1: { 2, 2, 4, 4, /* this search pattern is repeated forever */ 3, 2, 2, 3, 4 /* will never be reached, but can truncate forever loop early and execute this to get arbitrarily close to values */ } : Max days to find fox = INFINITE, Average days to find fox = 2.93333..., Average effort needed = 2.93333... hole searches Alternate optimal solution #2: { X /* skip searching this day */, 2, 2, 4, 4, /* this search pattern is repeated forever */ 3, 2, 2, 3, 4 /* will never be reached, but can truncate forever loop early and execute this to get arbitrarily close to values */ }; Max days to find fox = INFINITE, Average days to find fox = 3.73333..., Average effort needed = 2.73333.... hole searches HERE IS THE CODE: (You can run it on any online C++ compiler and engine) #include using namespace std; // This is the probability that the fox is in particular hole float prob[5] = { 0.2, 0.2, 0.2, 0.2, 0.2 }; // These are various choices of hole indices (zero-based) to search (uncomment exactly one of them to try it out) //int choices[] = { 1, 2, 3, 1, 2, 3 }; // 3.55, 3.55 (Worst case is 6 days) //int choices[] = { 1, 2, 3, 1, 2, 3 }; // 3.55, 3.55 (Worst case is 6 days) /* int choices[] = { 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, // this is repeated forever 2, 1, 1, 2, 3 }; // 2.93333...., 2.93333.... (Worst case is you find the fox after an unbounded number of searches) */ int choices[] = { -1, // Negative number indicates skip searching on that day 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, // this is repeated forever 2, 1, 1, 2, 3 }; // 3.73333..., 2.73333... (Worst case is you find the fox after an unbounded number of searches) int main() { int effort = 0; float totalEffort = 0.0; float totalDays = 0.0; for (int day = 0; day < sizeof(choices)/sizeof(choices[0]); ++day) { cout

Why in the world would you only peak into 1 hole if you're so dedicated to catch it

i would wait until it snowed and check the hole with fox prints around it

You can catch it by just inspecting each hole twice and then moving to the next. Lets say you do 1, 1, 2, 2, 3, 3, 4, 4, 5, eventually you will catch it but its not best case. Your algorithm is better for the best case

Just check the same hole every morning.... All you need to do is "ensure the fox is eventually caught."

Close the holes if it comes out,then that's the hole

My solution is same of video but i thought a different reasoning much more complicate and took lot of time. Also i had no paper and pen so i often had ti start rethinking again and again

So close.. I solved it on 8 days with a different strategy... but this one is 2 days faster

I'll Just look at one hole everyday

What is the strategy if the five holes arranged in a circle?

I think that your thoughtprocess is correct, but how do you determine whether the fox started in an even or an odd hole? It has to be given to follow a strategy

Do some investment and put a camera with tripod in front of hole.... Very next morning fox will be caught... Change with technology dude 🤖

Why cant he just smoke all the holes and wait till the fox come out. ? 🤪🤪🤪🤪🤪

If it starts from the middle hole. Let it be H1 to H5 From H3, it can hop to H2 or H4. Worse case scenario, you check H2, it's at H4. Next, it can move to H5 or H3. Check H3. If it's not there, check H4 the next day bcuz there's no where left to run If it starts in a random hole, block H1 on day 1, H2 on day 2, etc. Haven't watched the vid btw. Edit I'm wrong

Does solving these kinds of puzzles amount to pattern recognition, somewhat akin to chess puzzles?