Seven coin puzzle
June 20th, 2012 at 9:40:49 AM
permalink
Put seven coins in a circle, all facing heads up. You may flip any three consecutive coins, as many times as you wish. The object is to get all the coins tails up. How do you do it? Half point for an answer and half for an explanation of why the answer works.
For example, suppose the coins are numbered 1 to 7. The starting state is HHHHHHH. If you start by flipping coin 4 the state will be HHTTTHH. If you then flip coin 1 the state will be TTTTHT. If you then flip coin 6 the state will be TTTTHTH.
Please put answers and solutions in.
For example, suppose the coins are numbered 1 to 7. The starting state is HHHHHHH. If you start by flipping coin 4 the state will be HHTTTHH. If you then flip coin 1 the state will be TTTTHT. If you then flip coin 6 the state will be TTTTHTH.
Please put answers and solutions in
have a nice day
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 20th, 2012 at 9:59:22 AM
permalink
Oh man this is a tough one. Im gonna get to work on it, get ready for a stream of wrong answers ;)
Vote for Nobody 2020!
June 20th, 2012 at 10:04:04 AM
permalink
One way: Flip groups of three with first coin in group being 1, 3, 5, 7, 2, 4, and 6.
I think this is just one of many solutions. Probably you could do them in any sequence so long as you start a group of three with each coin once and only once, or at least the same odd number of times. If it is once each, then each coin is flipped three times, H to T to H to T.
Is that what you meant by an explanation?
June 20th, 2012 at 10:10:01 AM
permalink
Quote: DocIs that what you meant by an explanation?
Yes. And your explanation was what I was looking for. You get a full point.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 20th, 2012 at 10:18:08 AM
permalink
Quote: WizardYes. And your explanation was what I was looking for. You get a full point.
It occurs to me that I should make a small edit to my...
If you want to flip more than the minimum number of times, then each group of three must be flipped an odd number of times, not necessarily the same odd number, as I said at first.
June 20th, 2012 at 10:18:46 AM
permalink
A coin is tails-up if it is flipped an odd number of times.
Number the coins 1, 2, 3, 4, 5, 6, 7 in order clockwise.
There are seven groups of three consecutive coins: 123, 234, 345, 456, 567, 671, and 712.
Each coin is in three groups.
Turn each group once, in any order; each coin is in three groups, so each coin is turned three times.
As a result, all coins are tails up.
This will work for any number of coins, and any odd number of coins turned at once.
If the total number of coins is odd and the number of coins turned each time is even, there is no solution, as, since each coin must be turned an odd number of times and the number of coins is odd, the total number of flips needed is odd (odd x odd = odd), but each step turns over an even number of coins, so the total number of flips at any time is even.
I think every combination of even total coins and even coins flipped at a time is possible, but I'm not sure. (If you can get any number combination from all heads to two consecutive tails, then repeat those steps, shifting two clockwise each time.)
Number the coins 1, 2, 3, 4, 5, 6, 7 in order clockwise.
There are seven groups of three consecutive coins: 123, 234, 345, 456, 567, 671, and 712.
Each coin is in three groups.
Turn each group once, in any order; each coin is in three groups, so each coin is turned three times.
As a result, all coins are tails up.
This will work for any number of coins, and any odd number of coins turned at once.
If the total number of coins is odd and the number of coins turned each time is even, there is no solution, as, since each coin must be turned an odd number of times and the number of coins is odd, the total number of flips needed is odd (odd x odd = odd), but each step turns over an even number of coins, so the total number of flips at any time is even.
I think every combination of even total coins and even coins flipped at a time is possible, but I'm not sure. (If you can get any number combination from all heads to two consecutive tails, then repeat those steps, shifting two clockwise each time.)
June 20th, 2012 at 11:29:28 AM
permalink
Answer
Explanation
Flip 7 times. 1/4/7/3/6/2/5
Explanation
7 times = 21 flips. Each coin is flipped 3 times. First time to tails, then to heads, then to tails.
June 20th, 2012 at 1:42:07 PM
permalink
The answers above all look correct. I guess this was too easy.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 27th, 2012 at 5:33:38 AM
permalink
This is just like my Cutting Cheques problem!
Well it's not really a problem but solving this is just the same as "solving" the cheques thing.
Well it's not really a problem but solving this is just the same as "solving" the cheques thing.
So sometimes when I'm playing Pai Gow, I'll grab a few red chips and a few white chips on top. Maybe three red with two white on top. Then I'll cut three chips off the top, place it on bottom, put bottom chips on top. Then I'll cut three again, then again, and see how many cuts it takes to return to the original order. I hope that makes sense. I always noticed a bit of a pattern.
If I do five chips, cutting 2 takes the same number of cuts as cutting 3, equal to five.
So if I have N chips, and I designate N/2 - 0.5 chips/cut OR N/2 + 0.5 chips/cut, then it will take N cuts to return to original order.
I just did the coin problem again, and it doesn't work EXACTLY the same. I flipped 1-3, then 4-6, then 7,1,2 (non-consecutive coins)then 3-5 and so forth. But you get the idea.
If I do five chips, cutting 2 takes the same number of cuts as cutting 3, equal to five.
So if I have N chips, and I designate N/2 - 0.5 chips/cut OR N/2 + 0.5 chips/cut, then it will take N cuts to return to original order.
I just did the coin problem again, and it doesn't work EXACTLY the same. I flipped 1-3, then 4-6, then 7,1,2 (non-consecutive coins)then 3-5 and so forth. But you get the idea.
Actually I just read the problem again, and it does work exactly the same since they're in a circle instead of in a line. The equation above for chips should work the same for N coins and flipping the coins
