Prisoner Hat PUZZLE || 10 Prisoners || RED & BLUE Hats

[@sriramsubramanian1291]

Great! I never thought about this logic. I was thinking the prisoner can see through the policeman's specs mirror and answer the color. Lol.

[@LOGICALLYYOURS]

Thanks Sriram :)

[@lancerralseijoinedyourpart3633]

thats the easy mode xd

[@ourcreations9765]

Super bro 😎

[@ragunathjisankaran509]

Awesome

[@ragunathjisankaran509]

But i think he is wearing cooling glass

[@kratos952]

IIT graduated prisoners

[@warfyaa6143]

LOOL

[@ws_zilch]

LooooL i was thinking he same. XD

[@ritvikkaran967]

and another iit graduated find a loophole in metro kzbin.info/www/bejne/enTSaaJqpsR6hKM

[@Rizwan1875]

Hahaha 🤣

[@yourdad5471]

Underrated bro🤣🤣🤣

[@Raj-hm6qi]

We can solve this pblm like below also: 1. Last prisoner know the color of his front prisoner. So he loudly says either Red or blue.. If it is loudly said then same color his front prisoner also wearing.. that's the hint here.. 2. Next prisoner know his hat color from previous prisoner based on whether previous prisoner said his hat color loudly or not.. Now he know the hat color of his front prisoner.. if his hat color and his front prisoner hat color are same then he will say his hat color loudly otherwise he will say his hat color no loudly.. by this way next prisoner will understood his hat color.. 3. The same logic follows Only the first prisoner don't know his own hat color so he will loudly say his front prisoner hat color always..

[@Sab_Moh_Maya_Hal]

great man

[@shreyanray7624]

Good one

[@sunilsigar3145]

Better one

[@bosskiang4238]

Nope, your method would require the passing of more than 1 bit of information (red/blue).It is mentioned in the video that hints such as melodies are not allowed, you are strictly allowed to solely answer either red/blue. I answered the same way as you did in Parallel Computation class this morning and got corrected.

[@nassymcnute]

I know one thing that's not right

[@PuzzleAdda]

My goal as a puzzle designer is to create a meaningful experience for the player, not just 'I solved it.'

[@tonir299]

1. Simple solution During discussing the strategy prisoner 10 can tell everybody the color of his hat. 2. If 1. is not allowed (and even talking is not allowed at all) Prisoner 10 can form 2 groups of prisoners with the same color. Every prisoner can see if he is in the same group as the prisoner standing in front of him. So he knows his color. Prisoner 1 says the color as his group members.

[@nileshmittal9971]

I'm sad, no prisoners should be released like this... 😂

[@timewalker2447]

Underrated

[@cool-aquarian]

I’m just glad that I am attempting this puzzle sitting at home and not one of those prisoners. Solution idea is good, but implementation is 10x difficult. One moron prisoner in the line, and everyone in front of him is doomed.

[@touristtam]

think nodes and throwing a pool if the nodes cannot verify themselves. Not actual prisoners.

[@sorsocksfake]

Yes. This eventually led to the Goldenrock massacre in 2098 CE, where a foolish prisoner caused the Overlord to go to war with and exterminate Mexico. The world recoiled in horror, and henceforth, only logical mathematicians would be sent to jail. The subsequent shortage of logicians and mathematicians caused the re-election of Donald Trump, but fortunately he had calmed down a lot, having died 73 years prior, and he became such a good president that he has been re-elected ever since. #funfacts

[@alabamaal225]

The puzzle initially stated that the red/blue hats were placed randomly on the ten prisoners. However, it did not state that there were five red hats and five blue hats. (Indeed, the first illustration showed the prisoners wearing six blue and four red hats.) Conceivably, there could be eight blue and only two red hats, or any other combination of red and blue hats, including all the hats being either red or blue. Not knowing the number of red and blue hats in the total of hats would make logically solving the puzzle impossible.

[@bnitinb]

You didn't read the puzzle, total prisoners are 10

[@ktera]

watch the video again, folk

[@curosity276]

Yeah,I am also thinking the same , let's take 2 blue and 8 red hats consecutively. It fails in first attempt.

[@kippe1221]

@@bnitinb you didnt understand what he wrote. Number of red and blue hats, not number of prisoners.

[@1Pandaking]

@@curosity276 It doesn't, the last prisoner calls out Even (Red), the second to last still sees even and calls out Blue thus correct, The third to last sees odd so calls out Red switching the Count from Even To Odd. Then each one in front of him sees the inverse of the count and Call out Red continuing down the line each one switching the count. Leading to all but the one in back has called out the right color and they are all free to go.

[@brianaresendes966]

I was really hoping the guard was wearing sunglasses/glasses and they could use their reflections, but this is cool too.

[@mayukh_]

I hate this kind of answers.. If you guys are not serious, then don't answer.

[@ppmnox]

I found the solution not only logical, but also compelling in the sense that I wondered if odd/even would play into the solution.

[@TheDeepakkandpal]

Calculations can save you from being hanged 😂 ~ hence proved - be with mathematics

[@harnek.singh77]

If they were tht smart thy won't be in prison 🤣🤣

[@urbanarkoverland]

Wouldn’t. Oh the irony.

[@curious_one1156]

What if they are political prisoners ?

[@Aditya-yn5lk]

Even wikileaks owner is in prison...lol

[@CardCaptor32584]

Your guard is a nice guy. When my friend told me this riddle, he let the last prisoner be the noble sacrifice.

[@hypercubemaster2729]

I got it after leaving it alone for a day, lol. Very awesome line of logic and puzzle! Subscribed!

[@merlinsheeba244]

Mind blowing logic. Thanks for clear explaination.

[@LOGICALLYYOURS]

Thanks Sheeba for the appreciation. I hope you'd like other puzzles too :)

[@jackpigeon2263]

Wonderful logic Amar ji. Keep rocking with questions like this.

[@Quickeditor134]

I have easier approach to this : Last prisoner will tell second last prisoners hat color, if third last prisoner's hat color is equal to the second last prisoner's hat color then second last prisoner will tell his own hat color in louder voice otherwise in lower voice. From this third last prisoner will conclude his own hat color

[@vichu8206]

This dont need any calculations. Yes you are right

[@mayukh_]

How does last prisoner would know his hat color?

[@Quickeditor134]

@@mayukh_ last prisoner can see second last prisoner's hat color

[@jlugger]

I got something similar and far easier than the original solution. Each prisoner says his hat color in either a loud voice or a soft voice. If his hat is the same as the prisoner in front of him he is soft and if it is different he is loud. EX: Prisoner sees P9 hat is blue. P10 guesses his own hat is blue so he says "blue" in a soft voice. P9 then knows his hat is blue. P9 sees P8 hat is red so he states "BLUE" in a loud voice. P8 knows his hat is different. This way 9-1 are correct and even if 10 was incorrect they all go free.

[@canale2303]

I'd have answered differently. Prisoner 10 always says the hat's color of the prisoner 9. Prisoner 9 knows his hat's color. The agreed strategy is that each prisoner looks at the color of the prisoner in front of him and in case it is blu he SHOUTS his hat's color, otherwise it says it normally. So each prisoner knows his own hat's color just because of his partner's voice volume.

[@bm-ub6zc]

yeah i have the almost same solution (just with high pitched or low pitched voice, and with "next one has same/different color"), but in principle it's the same solution. the parity stuff is too difficult to figure out or memorize.

[@mrgyani]

You cannot give any kind of signal, be it with volume or whatever.

[@maxkashyap3810]

Awesome Problem. Upon this logic I sat down to program this approach. Really fascinated.

[@hyperfairy7779]

do they get to keep the hats?

[@lancerralseijoinedyourpart3633]

i hope so xd

[@rohitpawar4096]

Noo

[@timewalker2447]

Tf

[@poornimanagaraju8534]

Hey Amar..Awesome man!..Your passion towards puzzle has grown so much and it's spreading now!..Good to see you on youtube!..Nice explaination!

[@LOGICALLYYOURS]

Thanks Poornima... solved a lot.. now its time to teach to the world :) :) :P Must subscribe :)

[@matitiudeforever8155]

that was the best riddle i've ever come across

[@SS-cr8hb]

Very nice puzzle sir .According to me if the 10th prisoner speaks the colour of hat which is more in number among the 9(odd no.) hats infront of him, then also each prisoner would know his hat colour keeping in mind possible combinations of hats- 5,4 6,3 ....

[@LOGICALLYYOURS]

Dear SS.... nice to see that you tried to come up with a different approach(max), but it has limitation.... assume that only the first prisoner(front) and 9th prisoner are wearing Red Hat. Now, the prisoner 10 will call out "BLUE" (because they are more in number). But what would the 9th prisoner call out ?? For him still, majority is BLUE, but he can't decide what he is wearing. So, eventually an Even-Odd approach can get at least 9 correct answers.

[@SS-cr8hb]

You are right sir. Thank you so much for the explanation.

[@fomori2]

Took about 3 minutes to figure out that the last guy in line will say "blue" if he sees an even number of blue hats, and "red" if he see an odd number of blue hats. This alerts the prisoners what the starting target count of hats is, even or odd. Each prisoner will count the number of blue hats and will know if his hat is blue, if the number of blue hats he sees is the opposite of the odd/even target count, or red, if the number of blue hats matches the odd/even target count. Whenever someone calls out blue the odd/even count target switches to the other. Zero is considered even.

[@kidsgameslearning]

Nice. This logic will Work only if the total number of prisoners are even. If it is odd, difficult to solve. Because last prisoner can saw both Color ok odd number or even numbers. So difficult to give the clue to others.

[@patrickjinks]

I thought shout your own colour if the person in front is wearing a red hat and whisper it if they're wearing a blue one but only after listening to the answer, even though there's nothing against doing that, I feel I've let myself down

[@GurpreetKaur-lp6cj]

Patrick Jinks yeah I thought that too

[@rgupta74211]

Same

[@basiliobastardo255]

if the back prisoner can see in front and can pass message, why don't he just say the color of hat in front of him as a message?

[@qc1okay]

Mr. Ammar, surprised you didn't first solve a case of just 4 or 5 prisoners, like you often do to show proof of concept, for viewers to catch on with. And I think if my life were on the line, I'd just be doing simple counting, not thinking about the concept of parity. "OK, so the first guy said "Red", so the other nine of us in front of him have an even number of red hats in total, and I see two reds in front of me, so I just have to listen to how many times the guys behind me say 'Red'." Then when it gets to be his turn, he thinks, "OK, so I've heard only one guy behind me say 'Red', so with one red behind me and two in front of me, that's three reds, so mine has to be red for there to be an even number of reds in front of the first guy. RED!"

[@ShanmukhaPulavarthy]

Here's how the logic works: If there are an even number of red hats in front of the counter, the guessers can deduce the color of their hat by counting the number of red hats they see. If the count of red hats among the first 9 prisoners (excluding the counter) is even, then the color of their own hat must be blue. If it's odd, their own hat must be red. If there are an odd number of red hats in front of the counter, the guessers can similarly deduce the color of their hat. This strategy ensures that the counter can make a correct guess based on the parity (odd or even) of the red hats in front of him, and the guessers can correctly infer their hat color based on the counter's response. This way, at least 9 prisoners will survive, and they will all be released.

[@AmritEdits13]

Your riddles are best than others It has a logic like your channels name

[@uberfu]

There's a flaw to the puzzle -- the puzzle assumes that the jailor will go in order of prisoners from back to front. Which is the only way this puzzle works. If the jailor calls out random prisoners or starts fromt eh front the entire puzzle fails.

[@YEC999]

cool logic nicely explained - came close but what i was missing the trick with the odd and even...

[@komalagarwal9830]

Superb puzzle...n superb srategy...out of box situation.. awesome

[@NK-fx1qs]

nice thanks! these are fun to watch.

[@mohammedzubair8618]

very good logic. Good way of presenting. plz continue the same

[@LOGICALLYYOURS]

Thanks Zubair :)

[@saj157]

Awesome!! I finally got it!! yay!!:) you’re genius.

[@LOGICALLYYOURS]

Thank You :)

[@mcger5624]

great - well done mate

[@mrgyani]

Wow, this was brilliant.

[@swedishpsychopath8795]

Stupid puzzle: MY strategy would be: The last person tells the color the person in front of him has. The second to last will hear what color HE has and repeat his color. BUT: If he sees the next person has the same color then he will say the color name quickly. If the hext person has a different color then he will say his own color slowly. Every person will then know what color they have. Or they could have different intonation of the words. Starting high and going down for same color, and from low to high for different color.

[@vineetshetty9519]

If 9 prisoner give answers right then all the 10 Prisoner are released Hence there is Simple formula that 10th prisoner look at what 9th prisoner hat colour is and Say that...and 9th prisoner will look at 8th prisoner hat colour and say...and so far and so on...here if 10th Prisoner give wrong answer then remaining 9 are giving correct answer... Please correct me if i am wrong.

[@Mannprajapati619]

Yes you're wrong! Suppose tenth prisoner say 9th prisoner's cap color correctly, let say blue, then 9th prisoner will say his hat color correctly that is blue. but then how will 8th prisoner give his hat color correctly? what if he wore red color cap? hope you understand

[@index_s580]

You gave a key tip to solving by precisely describing 10th person position.

[@bm-ub6zc]

I have a much simpler solution: Prisoner 10: Says whatever color in front of him. Prisoner 9: Says the color said by prisoner 10, in a low pitched voice if the next one has the same color, and in a high pitched voice if the next one has a different color. And so on.

[@rajancherian4259]

What if there are 0 red hats? Will it be seen as an even count? I think that calling out the color with an odd number of appearances is better. Cos even if red is 0, there will be 9 blue hats and vice versa.

[@joshmays7058]

No way you get ten guys to do that correctly the first time!

[@sorsocksfake]

Prisoner 10 will count whether the red hats are even or odd. If even, he says blue; if odd, he says red. Thus the other prisoners know whether the first 9 *including them* are an odd or even number. They can also see the prisoners in front of them, and have heard the ones behind them. So they know whether the first 9 *except themselves* is an odd or even number. If the two match, then their own hat must be blue. If they do not match, it's because their own hat is red. They speak accordingly.

[@charlesbromberick4247]

Sounds like Schria Law to me. Thanks for the puzzles - i really like them.

[@madhesh18]

Terrific strategy

[@pochyena555]

I would not fancy the situation where there are only blue hats in front of him so the first prisoner would not know what to say (RED/BLUE) (Ofc. 0 is even but there are people who dont know that). Thanks for your content, great job ;-)

[@LOGICALLYYOURS]

Thanks Pochyena :)

[@bosskiang4238]

He/She can argue that since there are no red hats in front of him/her, his/her hat can't be red because red indicates even parity and if there is 0 red hat in front of me, his/hers can't be a red hat since that would make it odd.

[@viplavpatil5516]

1)even red = red;2)odd red =blue ===>Generalization =>Parity Equal = Blue;Parity not equal = Red

[@V7B817]

Parity Equal= one colour, parity not equal= other colour , that's more like it

[@shankaruv6367]

Vert simple . Every one knows before colour of before man . So they give clue for next man. If I am there I will say 'red colour sir' means he had same colour or' red colour' means different. U can also use voice soft and hard

[@subhajitpal2407]

Is this puzzle's logic actually comes in any real life implementation like in computer science concepts?

[@mdmubashir743]

amazing dude.....too good..

[@LaughterPlanet-mu1yp]

There is another simplest way, prisonor 10 will say colour of 9th hat... Now 9th know the color of his hat, so if the color of his next is same as him he will answer his hat color within 1 second or if the color of his next hat is different he will answer his hat colour in 5 seconds.. So this 1 or 5 second time interwal will give hint to the next prisnor and so on... As we know that 10th prisonor will be release even if he is incorrect so in this way all of them will be released.

[@priyanshuchaurasiya6184]

Very typicall puzzle but also an exellent solution!

[@Mark-qg8hm]

They should have strategies how to bum rush the jailer and overpower him, since there is just one guy.

[@dy4e02]

The only way i up to that he should says the number of red he sees, And saying odd or even let the other hit the number while been asked, great job

[@Godél-p4e]

Great video

[@dreamstory979]

Amar I have question if there are only 2 red cap then how it possible

[@aryanrawat5961]

Simplest tareeka: peeche wala apne aage wale ko bata de usne kaunsi pehni hai. Bas ho jaayega sab

[@piedpiper3314]

Why can't 10 simply tell 9 color of his hat...9 tell 8 color of his and so on... Did I miss something in question pls explain anyone

[@TB-up4xi]

You must only answer red or blue - and it must be the colour of your own hat - 10 can say red or blue signalling to 9 the colour of 9's hat but 9 must only answer the colour of their own hat anything else any everyone dies.

[@V7B817]

Just imagine prisoners wearing the blue/red hats alternatively

[@tombratcher6938]

I remember getting told this and thinking... " parity?... yep, works"

[@ajaykumargupta8504]

Awesome 👌

[@abcdefgabcdefg3017]

Why should they even do that. Except the last one , everyone can know their colour from the previous person . Only the last person has to apply the given strategy to guess his own colour.

[@shubham2927]

wONDERFUL eXPLANATION sER

[@vikashtiwari5505]

Brilliant

[@LOGICALLYYOURS]

Thanks Vikash :)

[@ammararfeen5644]

My Naam is also ammar and your logic so clear and high I appreciate you for giving this knowledgful videos

[@LOGICALLYYOURS]

Good to know that ... Ammar :)

[@V7B817]

Finally got the answer by odd/even logic 10th person would have said that I will told you the colour of the hat which is in odd number

[@sharukhshaik7476]

As a digital design engineer, I have an idea to design a digital circuit using XOR gates, a Comparator, and Multiplexer for this problem 😉

[@mythologygeek8888]

Can u share??

[@santhakumar9596]

I have another solution to this, "if prisoner 10 sees blue hat on prisioner 9's head then he has to say blue in low voice and prisioner 9 will say blue in low voice if he also sees blue hat on prisioner 8's head or else he has to say blue in loud voice to indicate prisioner 8 that he has red hat on his head". Here low voice is used for blue and loud is for red. I think this will work.

[@shankaruv6367]

Same way I thought

[@michaelbauers8800]

I love hat puzzles

[@hsingh-13]

Wow! 👍

[@moonyuen0505]

There is an additional question : if the police using 3 kinds of colors, can the prisoners escape?

[@nicehandsomefruit]

probably not, since a number's parody only comes in 2 values

[@alexandredesouza3692]

This is how I get ready for prison life.

[@nenaramchoudhary998]

If they can build such a strategy, then how they caught and put up in jail....

[@abridgetool]

They didn't commit the crime in the group. United they are intelligent, divided they are dumb.

[@balajik8688]

Y do they releasing the prisoners so freely..it's not even Gandhi jayanthi now 😂

[@YatharthSharma]

Kaaliya from SHOLE movie after watching this video "bhaiya ye trick hame bhi bta dete us bhosdiwale sardar gabbar singh se ham bhi bach jate" 😄😄😄

[@aldenasuemia]

Wow this is what I call genius.

[@akhtar_Alee]

I will see in reflexion of jailor's glasses

[@anushams9828]

Nice one🙂

[@vadermike7772]

Easier solution, that doesn't require math at all. The rules dont state that you cant touch the shoulder of the person in front of you. You simply agree beforehand that the person behind you touches your left shoulder if your hat is red, or your right shoulder if blue.

[@haribattula5187]

Do the front prisoner listens to answer told by back prisoner..and if he listens what if all hats are blue or red and only one blue or one black

[@sakthi600004]

Wat if all hats are blue?? Creator pls answer

[@mendelsonja]

Then the guy in the back sees an odd number and so he'd call out blue. The guy in front of him sees an even number so he knows the parity changed so he'd also answer blue. The person in front of him sees an odd number so he knows the parity changed and answers blue. And so on and so forth.

[@alencarfardin6650]

Wait, the prisoners 1 to 4 are just kids...

[@---4690]

Lol 😂😂

[@maplelife9222]

Rather it should be like - they should have planned saying to all that whoever stands the last has to shout the color of hat of the person in front of him.

[@GaryLuKOTH]

KARAN AHUJA If it were you taking this test, you have a 50/50 chance of being executed.

[@tw3f4tes52]

That’ll only bring the amount of people getting it right to about 75% (50% would get their answer from the previous guy, 50% would have a 50/50 of getting it right). This way guaranties the survival of everyone.

[@GaryLuKOTH]

TW3F4TES5 His strategy only guarantees that one person gets right. After the guy in the back tells the next guy what his color is, the next guy would get it right but the rest of them have no information of their hat color.

[@tw3f4tes52]

Gary Lu Productions I assume he meant that the next guy after that (after the first 2) will have to say the colour of the next guy again. Then that guy says the colour he was told, then the next guy says the color of the guy in front and so on. So every other person tells the next guy his colour, meaning 50% will be told their colour and have a 100% chance of survival, while the other 50% will have a 50/50 chance of getting their colour right, assuming the distribution of hats is completely random.

[@GaryLuKOTH]

TW3F4TES5 In that case, you would be right. Either way though, it is not a strategy that guarantees success.

[@prashantkumargupta4996]

elegant explanation.

[@LOGICALLYYOURS]

Thanks Prashant :)

[@whatdoinamethischannel9749]

Actually my solution is in morse code but for every hat red would be represented as ". (Dot)" And blue represented as "_ (Dash)" but the last prisoner would do 1 long click for blue and two short clicks for red starting from prisoner 1-9 therefore even if the last prisoner gets his hat colour wrong the others would know exactly which hat they were wearing But then I realized its an indication and it wouldn't work

[@ottovaughnjr.9282]

What type of jailor is that...?

[@pranshulgupta9465]

you have not explained about case if last one see the odd no. of red hat and say blue

[@pranshulgupta9465]

i got it

[@mikael900]

I got thrown off because you said only the last prisoner would be allowed to answer incorrectly but the first one would be hanged if he was wrong. Then at the end you said 9/10 were right so they were all let go.

[@christianbolt5761]

You can also do it with multiple colors. Try 10 colors

[@unbrokenriven1455]

so how do you solve for the same problem but there are 3 different color hats?

[@CarlEmilKjellstrand]

Checksums are the best :)

[@thedbznerd9565]

Did you steal this from Ted?

[@rajivpokharel88]

didnt understand one bit...but enjoyed it anyways...xD

[@safiurrehman8079]

Great and I think that the prisoner behind the prisoner touches his shoulder If he touch the shoulder the prisoner has red hat if he doesn't touch the shoulder the hat is blue if this is not allowed the prisoner can simply touch other prisoner leg by his foot or shoe slightly

[@bishnunepali3018]

nice...

[@swapniljain2464]

Awesome

[@garrethutchington1663]

Couldn't the last prisoner just count all the hats and deduce what his color is based on that?

[@GaryLuKOTH]

No because you do not know how many red or blue hats there are.

[@murugankaruppaiah9537]

I think they are studied in oxford university

[@abhinavgupta7930]

could not each prisoner just see their respective cap colours in jailors glasses.lol.XD