Q1. Three identical light switches operate three identical bulbs in another room. Which switch corresponds to which bulb? You are allowed only one trip into that room.

Achhama bhayecha

hahahaha!!!!

serial bro , you got it right....helpjava11 bro , serial bro lai 1 point dina huna aagraha gardachu :P

ok Serial bro (+1) from CollBuddy.

This one is funny.
Q. Mom is 21 years older than her son.

In 6 years she will be 5 times older than her son.

Where is the father right now?

Nas bro, cake stack गर्न नापौनी रहिछ क्या . cake लाइ जहाँ छ तेही काट्नु पर्नी .

Father and mother are having intercorse :P

first cut the cake into 2 circular pieces by cutting along the circumference then 2 cuts. One horizontal and other vertical, u get ८ पिस

Nas bro ko both ans correct +2
Punter ko pani milyo but no need to cut into circular: +1
Serial bro: +1 for intercorse answer.

Nas man, that puzzle is the classic logic puzzle. That one is king of puzzle and the solution is very detailed. This was the puzzle I was referring to in the Puzzle thread I had initiated few weeks ago when I was trying to make some guy understand what a complete logic is (alas he just wanted to argue). The popular Einstein puzzle that was also posted there is not really a logic puzzle, those are just simple deduction puzzles.

I was thinking, first divide 12 coins to 4 parts A,B,C,D
- weigh A & B : IF the weigh is same that mean false coin is in C OR D. If they weigh different either A or B has false coin.
- if they A & b weigh different, than we know aither A or B has dalse coin. so weigh A & C now. if the A& C weigh same, false coin is in B. other wise A has false coin.
- Now we have one chance to weigh and have 3 possible false coins.
- By weighing A & C we already know if hte false coin is heavier or lighter. ( becasue if we weigh A & C, and the weight is different we know A has false coin. so by looking if A is heavier or lighter we know the false coin is heavier or lighter. )
- Finally, weigh any two of 3 coins left. if the weight is same, we know third coin is false. if weight is different we know which is false by the weight.

Milyo ki milena NAS bro. I tried.

Last puzzle of the day:
Q, There are three doors in front of you. New car waits behind one of them; goat is hidden behind each of the remaining two. You may open one of the doors and get what is behind them. You want the car off course. You choose your door. Moderator (who knows where the car is) than opens one of the remaining doors and shows that there is goat. Now he gives you the opportunity to change your decision.

You are standing in front of two closed doors. Will you change your decision?

It’s better to change the door.U win in case you chose wrong door at first
If you don’t change the door you win only in case you originally picked the correct door.

Punter is right on this : you should always switch. (+1)

I think study reveals that losers tend to change their decision. And in this case, there is still 50-50 chance after one goat is revealed. So, I won't change the decision.

http://www.bbc.com/news/science-environment-27228416

Between, I tried to solve Q1 regarding bulb and room this way.
Room=>> R1, R2, R3
Bulbs=>> B1, B2, B3
Turn on B1 and B2
Go to R1.
Cond1: If lights Off then B3=>R1
Cond2: If On , B1 or B2 =>>R1
Cond1:
B3=>R1
Turn off B1 (B2 is still ON!!)
Go to R2
If Lights on, B2=>>R2 and B1=>R3
If lights off, B1=>R2 and B2=>R3
Cond2:
Suppose lights in R1 was on
Turn off B1 and go to R2(B2 is ON!!)
Cond 2a: if Lights on: B2=>R2 and B1=>R1 and B3=>R3
Cond 2b: If lights off: B2=>R1 and B1,B3=>R2,R3
Cond2b: Turn off B2 and turn on B3
Go to R3
if ON B3=>R3, B1=>R2 else vice versa

Nas bro, thats what i thought at first but turns out changing the door the probability is 2/3 and not chaging the door is 1/3.
Its called Monty Hall problem. Theres videos on youtube on how it happens.

Famous Monty Hall Problem, mathematically explained at Wolfram Alpha:- http://mathworld.wolfram.com/MontyHallProblem.html

Got really busy at work today so could not post any puzzle whole day.

Q. You are given 2 eggs.
-> You have access to a 100-storey building.
-> Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor.Both eggs are identical.
-> You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking.
-> Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process.

helpjava bro, it seems like an Algorithm & Data Structure problem. I guess binary search effective huncha jasto cha since the outcome is bounded by event of dropping of only 2 eggs without being broke. aba ghar ma gayera dudhe chiya + churot khayera serious thinking garnu parryo.

Drop the first egg from 50.If it breaks you can try the same approach for a 50-storey building (1 to 49) and try it from 25th floor. If it did not break try at 75th floor. And use linear search with the remaining portion of storey we need to test. For example if the first egg breaks at 50 we need to try all possibilities from 1 to 49.

Now this looks a feasible solution. In computer student's jargon do a binary search with first egg and linear search with the second one. Best case is log (100) and worst is 50.


Now the optimal solution for the problem is that you figure out that you will eventually end up with a linear search because you have no way of deciding the highest floor with only one egg (If you broke one egg and you have to find the answer among 10 all you can do is start from the lowest to the highest and the worst is the total number of floors). So the whole question grinds up to how to make use of the first egg to reduce the linear testing of the egg.


(For strict computer science students, well this problem can be solved using binary search on the number of drops needed to find the highest floor.)

Now let x be the answer we want, the number of drops required.

So if the first egg breaks maximum we can have x-1 drops and so we must always put the first egg from height x. So we have determined that for a given x we must drop the first ball from x height. And now if the first drop of the first egg doesn’t breaks we can have x-2 drops for the second egg if the first egg breaks in the second drop.

Taking an example, lets say 16 is my answer. That I need 16 drops to find out the answer. Lets see whether we can find out the height in 16 drops. First we drop from height 16,and if it breaks we try all floors from 1 to 15.If the egg don’t break then we have left 15 drops, so we will drop it from 16+15+1 =32nd floor. The reason being if it breaks at 32nd floor we can try all the floors from 17 to 31 in 14 drops (total of 16 drops). Now if it did not break then we have left 13 drops. and we can figure out whether we can find out whether we can figure out the floor in 16 drops.

Lets take the case with 16 as the answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45 .....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 We can easily do in the end as we have enough drops to accomplish the task


Now finding out the optimal one we can see that we could have done it in either 15 or 14 drops only but how can we find the optimal one. From the above table we can see that the optimal one will be needing 0 linear trials in the last step.

So we could write it as

(1+p) + (1+(p-1))+ (1+(p-2)) + .........+ (1+0) >= 100.

Let 1+p=q which is the answer we are looking for

q (q+1)/2 >=100

Solving for 100 you get q=14.
So the answer is: 14
Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (i.e. move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100).

giordano: your answer is correct. but i dont even know how..ही ही

Ali sajhilo, ali dimag lagaune rakham na ho, yo table bhari paper charera computer ma lamo program banaune khale ta aafule sochnai sakdina, k garnu programming belaima sikiyena, 25 kate pachi man pani laagena, dimag ma chirna pani garo :)