The fact that the cop is closer to the diagonal means that the cop would get there first. While the cop is by the diagonal, the thief cant go near so it must go in the opposite direction of the diagonal. Now the cop can chase the thief back in the opposite direction and the thief cannot turn around.
And why thief must move everytime?? He should move only when the parity is even,its said they keep on making moves and my move would be too not move at all.
I agree with you... except 'Zugzuang' in this very situation.. I think 'Zugzuang' will be applicable only if we switch the positions of thief and police. But yes, Zugzuang will play an important role in this type of puzzles.
@@LOGICALLYYOURS It's a German word, and it's spelled Zugzwang, meaning 'forced to move'. It is in fact _the_ key to the solution here; if the thief is forced to make a move then the cop can catch him; otherwise the thief can always re-establish parity by not moving.
There's a mistake in the video though, at 3:06 why does he have thief move towards the cop instead of away, then there would have been a distance of 3 there but bthevthief could then get away..he's assuming the thief is dumb..
@@leif1075 The thief can't get away. It doesn't matter if he is 2 segments away or 4 segments away, because either way the police can catch him when the amount of segments between them becomes odd
@@roppis9258 yea but the segments won't necessarily be am odd distance if the thief is careful...that's my point...it's not necessarily going to happen
@@leif1075 Yes they are. After each turn the the thief can only ever make a move which moves him 1 away or closer to the police. That would always make an odd amount turn into even or an even amount turn into odd. The same is true for the police, expect for the diagonal segment which doesn't change his distance from the thief from even to odd or odd to even, but instead keeps it the same. Since they started with an even distance between them and the police goes first, this forces the thief to make a move which makes the distance between them swap from even to odd.
For all of you that don't understand this. This whole scenario is based on what the cop does. Who cares what the thief does because when the cop gets to the bottom right corner first (which he will based on the positions in the picture), the cop will then have a choice. He could move 1 segment (by going up or left) or move 2 segments (by using the diagonal). This means regardless of what the thief chooses, the cop can always remain an odd number of segments behind since the scenario is in a confined space with both parties being forced to move 1 segment per turn.
Kslaxmi Sarathy only if the police man chases after and stays even moves away. If the police man instead leaves for the diagonal line then he can come back and trap the thief since he’ll then be odd moves away.
@@somerandommember no,but if they can only,move,one edge at a time as per the rules, they could literally be locked in a perpetual cycle with the thief never being caught..
@@I_am_Itay Yes, I do... not very professionally, but I play for brain storming. My lichess classic rating is around 2000, and puzzle rating has touched 2200. If you are on lichess, pls share your ID, or send me an invite @ammar97 .
After trying many possibilities once police go through the diagonal then thief must be caught....that's awesome......😁😁😁😁 really I tried many options..... every time thief got caught after crossing the diagonal
At 3:13 if the thief had moved to front instead of moving backwards the distance would have been 2 again. So the answer is he cannot be caught until he surrenders.
I like your videos but in this case I have to call foul. The rules state that you must move along an edge. A diagonal is not an edge. Perhaps the rules could be more clearly stated that they must move from one junction point to another.
Yeah, if someone would've asked me this in interview I would've said as long as position of cop and crook aren't mentioned let them be situated in a single edged right next to one another... So just based on this assumption you could see that it's definitely possible. 😅
Hi... Sir.. U make one video regarding following question.. " Two days ago , a boy was 8 year old. After One year he is 11 year old." How is that possible??
I honestly can't think how that would work. I can think of lots of scenarios where an 8 year old can be 10 one year later but I can't find a solution for 11.
@@nathancarlisle2094 I assume that "After One year he is 11 year old" means that the boy will be 11 years old on the next year, not after one year specifically. Then, the boy's birthday will be on the 31st of December, and today is the 1st of January.
even if the thief could get to the diagonal fast enough to cross it before being caught (different start position) he'd still get caught eventually, because the diagonal square is very unsafe for him, allowing the police to control it as soon as he gets there, and guarantee correct parity. You'd need two diagonals setup correctly to allow the thief to never get caught.
The idea is there, but this seems like those Math problems where people buy 30 watermelons. It's meant to be put into perspective, but in reality, people would give you weird looks, because life isn't meant to be turn based, thief and cop
@@luunhattruong2384 Have you actually read the original post that you 'answered' to? (And FWIW, Ammar did NOT say anywhere that both cop and thief have to move - though he should have).
At @3:03 the thief decided to catch the police and as soon the police moved along the diagonal line, he was gifted with a superpower to stupefy the thief till he got caught.
Can we say here that for any 2 initial positions of Police & thief on this track , the police can catch the thief if both of them knows this tactic? Or this puzzle is initial condition dependent.
at 3:12: Police is bottom right and can choose go diagonal or straight while the thief is on the 4th spot top row... what would happen when the thief did not go left. if he would go right instead.... it would be a even numer of segments or am i wrong?
It wouldn't matter. Thief goes right. Police uses diagnial. Thief then needs to go either down or to the left as those are his only remaining options which then puts him directly next to the police just in time for the police next move.
The dumb policeman go around the thief with no hope to catch him because he is always one node farther in the pattern. The smart policeman notice that there is a discontinuity in the pattern, and realizes that he can use this to catch the thief. Whatever are the moves of the thief the smart policeman makes sure to pass through the diagonal edge before directing to him. The diagonal is a kind of ‘break’ in the periodicity of the moves and let the policeman to catch the thief wherever he goes in the pattern. After going through the disgonal the policeman can follow the patterns he likes: he has only to make sure the thief do not use the same trick. So we can say that under the given conditions there is always a pattern to the thief.
There was a similar question about finding a cat but in this question the cop *knows* all the time where the thief is while in the other puzzle the catcher though getting closer to cat never actually knows or is sure where it is till the last move when he catches it.
I remember seeing this problem on the ARML back in 2012. Third time I took part on a contest where we had to prove everything. Questions on it looked more at least number of cops, or least amount of pay (cops*time), least amount of time, etc. One of the problems I remember was to prove the minimum number of cops needed to guarentee to catch a robber given the network was a K_n (complete graph of degree n) network.
The thief can easily counter the move. Try to look for it. Sympathize for the thief as well haha. Hint: if the thief tries his best to stay in the second upper dot, he can counter the police's wise move by moving right. There is more, look for it. This puzzle is imposibble.
ya cause that solves the puzzle.....................................................................................................................................
As soon as the cop is in the bottom right corner, he can technically choose to move 1 segment (go up or left) or move 2 segments (use the diagonal). With this being said, the thief is no longer in control of distance
No, it's impossible. It's impossible if the thief stays on the upper 2nd dot and the thief can stay on the second upper dot even the police uses the move advantage he has. If the police moves up diagonally, the thief can also move right from the second dot, making the moves even again. Therefore, it's impossible. We have to rely on the thief's idiocy to make this possible.
Nonsense logic. The thief had a chance to move to right just after when the police reached at the bottom right corner. But you purpose fully moved the thief to left.
If at 3:12 thief would have moved to the right, then police officer goes throught diagonal and thief is trapped. If he goes left, police officer goes up and catches him If thief goes down, police officer goes right and catches him. If thief goes up or left... well there's no up nor left. However imagine there was. If it would be finite and follow the same rules ( there would be squares without paths throught diagonals ). At some point thief would end up being cornered. However if he could just go up forever without bounds, then yeah. He wouldn't ever be cought in most cases.
1) Please proof read these puzzles...caugth makes it seem that the thief has a cold. 2) Obviously, the diagonal line is there for a reason, but moving on it seems to violate one of the original rules because the movement is not on an edge. (This problem can be solved by rewording the puzzle, but the problem exists is the puzzle is phrased) 3) I have not spent a huge amount of time on this, but if the thief moves more than one space and stays out of the corner, he may be able to stay at large until the officer's next doughnut break
It is just one of the case in many , there r other ways in which police can never caught the theif and the answer can be both yes or no,but for this yes is most likely to favour rather than not in this situation, because we r humans and r designed to solve problems.
I had a solution First move police: go to right side And try to replace both police place and theif place but the next move was as theif Then the only move for both theif and police is to go right side And last he traped to the last box and go down finally the police catch the theif (I got the solution in 3mins)
Me after seeing the thief can just cycle around the first square.
"No the thief cannot be caught"
But how can you catch him on your own terms is the question.
As a Chess player, all I can say is this:
- Opposition
- Triangulation
That makes 100% sense
Wow even I too National level Chess player!
@@ananyaj7914 good to see that we have chess players here... You must be 2300 rated or higher!
U forgot something
-Zugzwang
@@VirgiAhmad +1
Here's my problem with the solution: what is stopping the thief from using the diagonal to swing the parity back in his favor?
He can't reach it if the cop plays with anything but random moves.
The fact that the cop is closer to the diagonal means that the cop would get there first. While the cop is by the diagonal, the thief cant go near so it must go in the opposite direction of the diagonal. Now the cop can chase the thief back in the opposite direction and the thief cannot turn around.
Simple the cop will block thief because he is the first who can reach the diagonal line Period
If thief is in A or B and cop on C , he is caught
And why thief must move everytime?? He should move only when the parity is even,its said they keep on making moves and my move would be too not move at all.
Dude i am pretty sure the Inventor forgot to take his phone in the bathroom and had to timepass by counting tiles
Haha... :D
😂
@@LOGICALLYYOURS A better Solution is Give the Police a Goddam Gun and Finish The Robber........The end
And I thought only I have this bizarre habit. 😂🤗
@@ajinkya314 ultimate nigga
It should be noted that a move is required, meaning that both the police and robber must move each time.
Yes... I like the riddles but he doesnt specify the rules each time well.. so it kinda ruins the whole" figure it out yourself" part...
it does note that
3 chess strategies are used
i) opposition
ii) triangulation
iii) zugzuang
I agree with you... except 'Zugzuang' in this very situation.. I think 'Zugzuang' will be applicable only if we switch the positions of thief and police. But yes, Zugzuang will play an important role in this type of puzzles.
@@LOGICALLYYOURS It's a German word, and it's spelled Zugzwang, meaning 'forced to move'. It is in fact _the_ key to the solution here; if the thief is forced to make a move then the cop can catch him; otherwise the thief can always re-establish parity by not moving.
@@dlevi67 You're right, I was wondering what "Zugzuang" means, it looks Chinese but sounds German.
I considered the chess idea that if the beginning scenario were the same, but i was the thief's obligation to move, he would be caught. Triangulation!
There's a mistake in the video though, at 3:06 why does he have thief move towards the cop instead of away, then there would have been a distance of 3 there but bthevthief could then get away..he's assuming the thief is dumb..
@@leif1075 The thief can't get away. It doesn't matter if he is 2 segments away or 4 segments away, because either way the police can catch him when the amount of segments between them becomes odd
@@roppis9258 yea but the segments won't necessarily be am odd distance if the thief is careful...that's my point...it's not necessarily going to happen
@@leif1075 Yes they are. After each turn the the thief can only ever make a move which moves him 1 away or closer to the police. That would always make an odd amount turn into even or an even amount turn into odd. The same is true for the police, expect for the diagonal segment which doesn't change his distance from the thief from even to odd or odd to even, but instead keeps it the same. Since they started with an even distance between them and the police goes first, this forces the thief to make a move which makes the distance between them swap from even to odd.
@@roppis9258 yea but they could,be locked,in a perpetual,back,and forth so the thief can get away in those,cases he won't always be caught
The puzzle doesn't mention the most critical part: the thief HAS to move each turn. It's not optional.
actually it does say that
For all of you that don't understand this. This whole scenario is based on what the cop does. Who cares what the thief does because when the cop gets to the bottom right corner first (which he will based on the positions in the picture), the cop will then have a choice. He could move 1 segment (by going up or left) or move 2 segments (by using the diagonal). This means regardless of what the thief chooses, the cop can always remain an odd number of segments behind since the scenario is in a confined space with both parties being forced to move 1 segment per turn.
The thief can just move between two points making it impossible for the police to catch
Yes the theif can do that.. But here he said is very confusing
Kslaxmi Sarathy only if the police man chases after and stays even moves away. If the police man instead leaves for the diagonal line then he can come back and trap the thief since he’ll then be odd moves away.
"You don't listen .. do you" - The Joker
It only makes sense because they cannot stay in the same place. It's stupid.
@@somerandommember no,but if they can only,move,one edge at a time as per the rules, they could literally be locked in a perpetual cycle with the thief never being caught..
Very nice puzzle and solving knowledge. Thief can use the similar diagonal to negate the police action to escape from the police.
2:56: With *Odd number* off segments, the theif can be *caugth*.
what (look at the last word)
I love the ODD and EVEN concepts application in some of the puzzles..
Like FOX in 5 holes.. Etc.
Awesome sir ❤
That question was solvable this is not
2:57 “thief can be caugth”
:) sorry for the typo :)
thief can be *cough*
grammarnazi
Wish this who happen in real life situation then there would be no robbery.
This works a lot like chess, trying to checmate the thief. Almost too easy
Exactly, Im playing professional chess so I solved that even before he told the rules.
I agree! It's analogous to chess end game with 2 Kings and a Pawn.
@@LOGICALLYYOURS Exactly, just like opsition. Do you play chess?
@@I_am_Itay Yes, I do... not very professionally, but I play for brain storming. My lichess classic rating is around 2000, and puzzle rating has touched 2200. If you are on lichess, pls share your ID, or send me an invite @ammar97 .
wtf is checmate
I don't expected the ans nice question
After trying many possibilities once police go through the diagonal then thief must be caught....that's awesome......😁😁😁😁 really I tried many options..... every time thief got caught after crossing the diagonal
With perfect movement from the thief it will never cross the diagonal itself
Really tricky puzzle, it's easy but at the first try it seems too hard
The police always carries a gun
Pretend this is either England or Commiefornia.
Absolutely incredible !! İm so happy to discover ur channel keep the videos greetings from Turkey
Thank you Murat :)
Nice puzzle, reminds me of triangulation and opposition in chess
At 3:13 if the thief had moved to front instead of moving backwards the distance would have been 2 again. So the answer is he cannot be caught until he surrenders.
Then the police could move across the diagonal and corner him.
@@iobudgerigar9133 How so ? In the video thief is going right, right, right, left. He couldve gone right right left and so on.
then it will be easier for police to catch him. earlier than one approach shown in video
As always great graphic explanation.
And a very nice riddle.
Awesome..👏👏👏🤩🤩
Easy...so much explanation not required
I also solve this puzzle🤘🤘
Change parity of moves via the diagonal
I like your videos but in this case I have to call foul. The rules state that you must move along an edge. A diagonal is not an edge. Perhaps the rules could be more clearly stated that they must move from one junction point to another.
ur wrong
Easy for every decent Chess Player
can you find out the solution for this loophole of metro
th-cam.com/video/DCo8o4q_GNA/w-d-xo.html
Easy fór anyone with IQ value of atleast 2 digits
Enjoying these puzzles, subscribed
Really logical things that is....a long bow to the channel creator😊
If the thief was actually intelligent then no
But since the cop starts closer to the diagonal, all it has to do is get there first and it catches the thief Everytime
ur wrong
i am amazed at all the peoples comments that say this does not work
Yeah, if someone would've asked me this in interview I would've said as long as position of cop and crook aren't mentioned let them be situated in a single edged right next to one another... So just based on this assumption you could see that it's definitely possible. 😅
its bc they forgot their brains
3:14......and undertaker takes over!!
Hi... Sir..
U make one video regarding following question..
" Two days ago , a boy was 8 year old.
After One year he is 11 year old."
How is that possible??
If a boy has a birthday on 1st Jan it can be solved
Aradhya... I just noted your question, and it will be shared very soon. Thanks for sharing it... and yes I solved it :)
Is his name 8 year old🤐😂😂😂
I honestly can't think how that would work. I can think of lots of scenarios where an 8 year old can be 10 one year later but I can't find a solution for 11.
@@nathancarlisle2094 I assume that "After One year he is 11 year old" means that the boy will be 11 years old on the next year, not after one year specifically. Then, the boy's birthday will be on the 31st of December, and today is the 1st of January.
Good one!
Nice problem sir. This problem like King and Pawn endings in Chess
Yes Maru.. you're right, the similar tactic is used in triangulation during end games.
perfect example of one step forward two steps back.
Actually it is impossible if the thief is smart...The thief can also cross the diagonal as soon as the police crosses ...to maintain the same distance
Nope
Not possible bcoz the start position will not allow the thief to do so...
even if the thief could get to the diagonal fast enough to cross it before being caught (different start position) he'd still get caught eventually, because the diagonal square is very unsafe for him, allowing the police to control it as soon as he gets there, and guarantee correct parity. You'd need two diagonals setup correctly to allow the thief to never get caught.
incorrect
When the thief won't see the police from is first position
what is the need to move he can be safe
u forgot the rules
Being a chess player helped solving this.
Fantastic.
The idea is there, but this seems like those Math problems where people buy 30 watermelons. It's meant to be put into perspective, but in reality, people would give you weird looks, because life isn't meant to be turn based, thief and cop
Which software do u use to make the videos??
MS PPT
@@LOGICALLYYOURS 😍thanks, let me try it....
You are not like others...thanks for telling
Very nice puzzle
Great video
best riddle bro best
Good one and I solved it
Why will the thief ever move away from the 1st square itself? 😂
after the police cross the diagonal line, the thief is guaranteed to be caught no matter where he stands
@@luunhattruong2384 Not true. If the thief does not have to move, he can re-establish parity by not moving for one turn.
@@dlevi67 well, that's against the rule and similar to chess, you cannot skip your move
@@luunhattruong2384 Have you actually read the original post that you 'answered' to? (And FWIW, Ammar did NOT say anywhere that both cop and thief have to move - though he should have).
@@dlevi67idk, I thought the guy meant the thief was harder to caught if he just circle around 4 vertices of the first square.
How will you catch the thief if he just moves up and down from the initial position???
The sequence would be 1. b3 b1 2. b4 a1 3. c5 b1 4. b5 a1 5. b4 b1 6. b3 a1 7. b2 b1 8. b1 and the thief is caught
think about it
The bell at odd diagonal was funny as hell
I Got 10% smarter thank you
haha... :)
I just want to point out that caught is spelt incorrectly at 3:28, nice video anyways
it is spelt incorrectly before then.
if ur gonna correct someone, make sure ur also correct
At @3:03 the thief decided to catch the police and as soon the police moved along the diagonal line, he was gifted with a superpower to stupefy the thief till he got caught.
As a chess player would say...you need to get the opposition
Can we say here that for any 2 initial positions of Police & thief on this track , the police can catch the thief if both of them knows this tactic? Or this puzzle is initial condition dependent.
3:12 jesus christ that scared me..
at 3:12: Police is bottom right and can choose go diagonal or straight while the thief is on the 4th spot top row... what would happen when the thief did not go left. if he would go right instead.... it would be a even numer of segments or am i wrong?
It wouldn't matter. Thief goes right. Police uses diagnial. Thief then needs to go either down or to the left as those are his only remaining options which then puts him directly next to the police just in time for the police next move.
ur wrong
3:12 , 3:13 ,3:14 Dude I think out of box like if theif can go right box insted of going left then it can't be possible to catch theif
he always gets caught
This puzzle was super easy because i'm a chess player. Calculating ahead is what I do.
Pretty sure I saw this as a main game mechanic somewhere
The dumb policeman go around the thief with no hope to catch him because he is always one node farther in the pattern. The smart policeman notice that there is a discontinuity in the pattern, and realizes that he can use this to catch the thief. Whatever are the moves of the thief the smart policeman makes sure to pass through the diagonal edge before directing to him. The diagonal is a kind of ‘break’ in the periodicity of the moves and let the policeman to catch the thief wherever he goes in the pattern. After going through the disgonal the policeman can follow the patterns he likes: he has only to make sure the thief do not use the same trick. So we can say that under the given conditions there is always a pattern to the thief.
Bro what is the name of the game
There was a similar question about finding a cat but in this question the cop *knows* all the time where the thief is while in the other puzzle the catcher though getting closer to cat never actually knows or is sure where it is till the last move when he catches it.
Is there a way where the theif can prevent being caught by moving in other places?
No
this puzzle seem impossible but a hint of come from a diagonal edge, which result in checkmate or dead end to the theft.
yes
@@hoangminhdao5538 this is impossible to not win. Whenever the person controlling police goes for the diagonal he wins. Bitch puzzle.
nope
That is for a child...count me in! (the guy who play board games)
I remember seeing this problem on the ARML back in 2012. Third time I took part on a contest where we had to prove everything. Questions on it looked more at least number of cops, or least amount of pay (cops*time), least amount of time, etc.
One of the problems I remember was to prove the minimum number of cops needed to guarentee to catch a robber given the network was a K_n (complete graph of degree n) network.
The thief can easily counter the move. Try to look for it. Sympathize for the thief as well haha.
Hint: if the thief tries his best to stay in the second upper dot, he can counter the police's wise move by moving right. There is more, look for it.
This puzzle is imposibble.
wtf is "second upper dot" that makes no sense try again
Lol I just did the last step..
They could just keep going around the first square forever
ya cause that solves the puzzle.....................................................................................................................................
You could make the thief at one end of the diagonal and the cop at the other and then boom
What if thief goes another side at 3:11...specially opposite side .....and what if thief always maintains 2 segment of gap.....
That's what I also thought. The thief can also use the diagonal to realign the distance to 2.
Keep two fingers on screen and slide one by one, try whatever you want, thief will be caught.
lol no the can be safe forever
As soon as the cop is in the bottom right corner, he can technically choose to move 1 segment (go up or left) or move 2 segments (use the diagonal). With this being said, the thief is no longer in control of distance
thief cannot maintain 2 segment gap
Amazing
Checkers opposition immediately sprang to mind .
In the US, the police can shoot the thief from the starting position while eating a 🍩
Brilliant
No, it's impossible. It's impossible if the thief stays on the upper 2nd dot and the thief can stay on the second upper dot even the police uses the move advantage he has. If the police moves up diagonally, the thief can also move right from the second dot, making the moves even again. Therefore, it's impossible. We have to rely on the thief's idiocy to make this possible.
the thief cannot stay on the upper 2nd dot, wherever that is
Nonsense logic. The thief had a chance to move to right just after when the police reached at the bottom right corner. But you purpose fully moved the thief to left.
ram panda If he had moved to the thief to the right, then the police would move via the diagonal and then catch the thief on the next turn.
But there is no possible for diagonal... Move rigt?? Except in right corner down
Its nonsense off course..
@@rshanthikumar6224 the police will still catch the thief
If at 3:12 thief would have moved to the right, then police officer goes throught diagonal and thief is trapped.
If he goes left, police officer goes up and catches him
If thief goes down, police officer goes right and catches him.
If thief goes up or left... well there's no up nor left.
However imagine there was. If it would be finite and follow the same rules ( there would be squares without paths throught diagonals ). At some point thief would end up being cornered.
However if he could just go up forever without bounds, then yeah. He wouldn't ever be cought in most cases.
The second last step shows the thief coming towards the police 👺
GG, I took more or less 40 seconds, actually i took too long, this is way easy, in 10 seconds i could've solved it lmao, to tired it's midnight
and im the king of pluto
Awesome
easy, just have the theif go first lol
u forgot the rules of the game
@@jiogcyihsugyiocjfdoivhphvw6821 bro that was a year ago. also yeah that comment was really stupid
This basically summarizes Hitman go game
I open from google your vidio
This remind me of pawn and king end game
at 3:12 why the theif is moving left side rather than the right side... If he moves right side... the puzzle won't be completed
Yes it would, the cop would move across the diagonal line and the thief would be screwed immediately because it has to move
he will be caught
Damn the sudden sound when the police go on diagonal scares me
There is another solution.. better one.. with real mind concept..
What will happen if the thief deciding not to move at same stage to keep 2 step difference with police??
breaks the rules of the game
super riddle
😲 👏❤
1) Please proof read these puzzles...caugth makes it seem that the thief has a cold.
2) Obviously, the diagonal line is there for a reason, but moving on it seems to violate one of the original rules because the movement is not on an edge. (This problem can be solved by rewording the puzzle, but the problem exists is the puzzle is phrased)
3) I have not spent a huge amount of time on this, but if the thief moves more than one space and stays out of the corner, he may be able to stay at large until the officer's next doughnut break
First one is definitely a typo.... but your second point makes sense.
it doesnt violate the rules. the puzzle is fine, grammarnazi
@@LOGICALLYYOURS his 2nd point is wrong
It is just one of the case in many , there r other ways in which police can never caught the theif and the answer can be both yes or no,but for this yes is most likely to favour rather than not in this situation, because we r humans and r designed to solve problems.
awesome
That thief not smart enough. Beginner I think 🤔
This is really easy if you’re a chess player
How
What's name game this is
I had a solution
First move police: go to right side
And try to replace both police place and theif place but the next move was as theif
Then the only move for both theif and police is to go right side
And last he traped to the last box and go down finally the police catch the theif (I got the solution in 3mins)
try again
Shaandar