This is not one of the classic Microsoft puzzle. I recently heard from a friend. I am listing the problem below.
There is a table on which a number of coins are placed. You also know that there are as many coins with Head up as many coins with Tail up. Now you have to divide the coins (number of coins is even) into two equal piles such that number of coins with Heads up and Tails up in either piles be the same. The catch is you are blind folded and you cannot determine the sides (for sure) if you are blinded
29 comments:
Simple- Divide the coins in half by quantity (easy to count coins while blindfolded)
Then, flip all the coins in one pile over.
I don't think this works
Consider this distribution after making two piles
___________
___|_1_|_2_|
_H_|35_|_30|
_T_|15_|_20|
after flipping second pile
___________
___|_1_|_2_|
_H_|35_|_20|
_T_|15_|_30|
Logical Warrior, the question actually tells you "You also know that there are as many coins with Head up as many coins with Tail up." so the scenario you've posited doesn't match (you have 65 heads and 35 tails)
It doesn't make sense to simply divide in half and flip one pile. Suppose you divide it such that all Hs are in one pile and all Ts in the other. You end up with all Hs or all Ts.
I think you can compare two coins (without knowing it is H or T) to know if they are the same or different. By this means you can simply count and collect the same type of coins.
shhh:
yes but that is ok. The requirement is that in the end the # of Tails in pile 1 equals the # of tails in pile 2 and the same with heads.
if they are all tails or all heads that is just fine.
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no. of tails T (40)
no. of heads H (60)
put smaller(T,H) coins at random in pile1 (40). (25 H , 15 T)
fill the pile2 (50 coins) using remaining coins.(27 H ,23 T)
invert all the remaining coins
( 8 H, 2 T) --> (2 H, 8 T)
put these coins in pile1.(27 H ,23 T)
Why complicate matters? Pai was right - simly divide the coins by half and flip one pile over. If you need convincing, here's the logic:
Consider we have x heads and x tails (amounts are equal - given). Take any random x coins consisting of n heads ( 0 <= n <= x) and x-n tails. Automatically, second pile will contain x-n heads and n tails. By swapping heads with tails in second pile, we end up with n heads and x-n tails in the second pile as well...
Assume you have N heads and N tails coins initially.
u make two piles out of them by taking m Tails and s Heads. he second pile will have N-m Tails + N-s Heads
since the two piles are equal :-
m+s = N-m + N-s
i.e. m+s = N
So we have as many heads in pile one as we have tails in pile two.
So just flip.
So
Logical Warrior, the question actually tells you "You also know that there are as many coins with Head up as many coins with Tail up." so the scenario you've posited doesn't match (you have 65 heads and 35 tails.
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I think that people are taking it as number of heads and tails in a single tile is same, but that is wrong the correct interpretation would be that heads in one pie is same as heads in other pile and same for tails. Now it is simple isn't it?
Your puzzle is quite challenging. I can't figure it out.
Caro
sudoku
Hi I got inspired by your blog and created my own puzzle blog
Logic puzzles
I would put them all on their edge. This way both piles have zero head/tail up.
I don't mean to be sound stupid if this isn't the correct answer, but I was able to figure this out in less than a minute. Can't you just feel for it by using your finger along the edges? Clearly the heads will have a silhouette feeling for a face vs. a tail. So you can use your finger to fix it. So what I would do is take one coin and put it in my left hand. This will be my "Control" coin. With my right hand, I will put the "Test" coin next to the "Control" coin and with my index finger, compare the feeling of the ridges to the "Control." If it feels the same, then I'll put in in the pile I'll create to the right. If it feels different, then I'll put it in a pile to the left.
just pick half of coins in one and remaining half in 2nd pile.
and then reverse all coins of 2nd pile. THATS IT.
EXAMPLE..think u have 30 heads and 30 tails initially..after picking them in both piles
I II
H x (30-x)
T (30-x) x
becoz if 1st pile has x heads then it should have 30-x as tails as total is 30..
thanks
Kumar_ajay
for those proposing to divide into two halves then flipping , that will be true if the u are not blind , means the fold was removed , right ?
Remove all the coins from the table. Tell that you have created two empty sets with zero heads and zero tails in each set ;)
Remove all the coins from the table. Tell that you have created two empty sets with zero heads and zero tails in each set ;)
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As an algorithm, here is what I would do:
1. Pick any coin at random from the first table
2. Place it as is on the second table.
3. Pick the second coin at random from the first table
4. Place it as flipped on the second table
5. Repeat steps 1 through 4 until all coins are placed on second table.
Easy. Just feel the coins and put them in piles.
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