Community Event Puzzle 2 Stacks Giveaway - Ended!

yeah i know - redoing my working and its way more complicated to work through but might end up being fewer moves

 

i said they can be in the same room twice, i didn't say they can only be in the same room twice

 

ok redid working, now got an answer of 100 days, taking 25days to eliminate each of the 4 starts they could gave (even row+even room, odd row +odd room etc)

 

would it be never and this is all a trick question??

 

its not a trick question he told me. Theres a real answer of how many days

 

The police do not know of the pattern and they have no way of figuring out the pattern correct since no evidence would ever be found suggesting of the pattern?

 

since police can only check one per day and they can be in a room infinite times they could easily siwtch over and over and not get caught? They check only 1 per day and everyday thieves can switch to another one?

 

3 days. 4 if you count the starting apartment for the thieves as a day
If the cops continue to check let's say apartment 4, the robbers no matter what has to go there since if they start in one, they move to two, three then four and bam.
If they start in 6 they move to 5, then 4 and bam.
Same can be said for if they search apartment 3 but just with 6 -> 3 taking 3 days while 1 -> 3 takes 2.
Of course if they at any point start in 2, 3, 4 or 5 they'll be found faster

You specified they could search one apartment each day, not one floor. If it has to be different buildings, in that case they just swap between 3 and 4.
So if you meant one floor in an apartment it would take 25, 26 if again, the starting room counts as a day,since they could continuesly search apartment 4, floor one for 23 days, as it's the minimum time for the thieves to make their way from the top floor to the bottom no matter what. And if they don't appear they search apartment 4 floor 6 since they're stuck needing to go down, but can't as they've already been there. And then spend the last day in apartment 3 floor 6 since if they started in building 6 they'd hit that.

 

I think I got it (for real this time.) ... First assume they are in an odd number room on an odd floor. Firstly check room 3 on the third floor, if they are on in room 3 on the first floor, then they would be caught right away. Then check room 2, if they started in room 1, this is the only place they could go. By checking room 3 and 1, if you fail, they must have started in room 5. Check Room 5, if that fails check floor 2 room 3. This will completely eliminate all odd numbered room starts on floor 3. Failing that, repeat the process over again, this time on floor 2. This eliminates Floor 1 starts, since floor 1 has to go to floor 2, Then do the same with floor 3, when they change floors check 2-3, reducing options to 4 and 6, then check 4. at the end check 3-3, failing that, the criminals must be on floor 5 (assuming odd floor odd room.) After checking floor 5 on day 20 check floor 4 room 6. If this is not the case criminal must be on floor 6 and must move down, this will corner him by day 25 assuming odd floor odd room start. If this fails, repeat the entire process assuming different combinations of even or odd floor and room start (this involves starting 4, then going 5 to trap 6 starts, then going 4 and 2 . (I would sort of be tempted to call this a "new start") By the 97th day, the criminals will have certainly been caught.

 

I generated three hypothetical scenarios and tested to see how long in each it would take for them to be in the same room at the same time. In one scenario, the thieves constantly moved left/right through rooms 2-5, and only moved down rooms 2-5 in order to keep themselves from getting cornered to only move one direction. The police scanned the each room on each floor before moving to the room directly above them on the next floor and scanning that whole floor moving left to right or right to left. The result was the police and thieves would be in the same apartment on the 21st day. In the other scenario, the police did the same thing as before, but the thieves moved left to right or right to left through all the rooms (turning right and moving only right when hitting room one and only left when hitting room 6) and up and down 1-6 (going up when hitting floor one, down hitting floor 6). With this scenario, the result was 14 days. Based on both of the conclusions, I'm ASSUMING the answer must be a multiple of 7, as both 14 and 21 are. The other scenario (which I actually forget what I did because it was complicated) took much longer, and the answer was also a multiple of 7, being 63. I don't know if this would necessarily be correct, but the answer with no scenario and complete randomness in my mind would just be doing this 7 *the number of possible rooms, being 36. So 7*36 would be 252. So, in theory with how I'm thinking this through, the police will 100% find the thieves on day 252 if it is complete random of where the thieves will move left or right and up or down and police will check a random room with no set scenario.

The answer could also be 189 days, as the thieves can only move at MAX left or right 3 times and down or up once, so 3/4s of the time they will move a set direction of left or right, so I did 252*3/4 and got 189 days.
Of course in specific movement patterns of the thieves and cops the result will be lower.

 

thats almost exactly what i did but i got 100 days out of it, he never replied to that one though so not sure

 

So they must move all 3 times in the same direction on one floor?

 

As said the thieves can move left OR right, not just one direction, before moving down or up. The only time they would have to move left 100% is if they are on the farther room to the right since they can't move right anymore and vice versa.

 

That answer is wrong

The police knows the pattern

There is a solution for this puzzle

The thieves can be is the same room more then once
And with searching one appartment a day i mean one in the entire flat

This isnt the correct answer

This too, isn't the correct answer

The thieves can move left, left, right aswell as all right or all left.

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Now i will go to bed, and reply tomorrow again

 

I’m got it to be 64 days. I’ll tell the patern later.

 

in a flat with 6 floors and 6 apartments?

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If the thieves truly start in a random apartment on a random floor, it's possible for them to be on floor one in apartment one on day one. The police would search apartment one on day one catching them in one day.

 

Sadly, the OP asks for how long it will take for the police to find them with a 100% probability when following a certain pattern.

 

6 floors, 6 appartments each floof, so 36 total

Ye but if you look in that room, but they started on floor 6 appartment 4 then they didn't catch them yet

 

Is the police movement random?