[bonanova]

Draw three large circles that intersect in such a way that 7 regions are defined.

Three regions are inside just one circle; three others are inside two;

and one, the central region, is in all three circles.

Place seven coins, one in each region, so that all of them show HEADS.

Now play this game:

Select one of the circles [recall each one contains 4 coins] and either

.

1. Turn all its coins over: H->T and T->H; or
2. Reset all its coins to H: H->H and T->H.

.

Repeat this process until finished.

The object is to have the central coin show TAILS and all the others show HEADS.

What is the minimum number of moves needed to win?

Or can it be won?

[Guest]

The last move can't be a "reset to H", else resetting will make the medial coin H while T is needed. If the last move is a "all turn over" The position before last move will be: Medial H and 3 others in that circle is T. To get such a position, the move before last move again can not be a reset, else all 4 coin would be H. If one before last move is a "all turn over", and the position after this move is H+TTT, the position before this move should be T+HHH. With this method, if we go to n.th move before last, the position have to be eather T+HHH or H+TTT. Since we're beginning to game from HHHH, coming to a point of T+HHH is impossible.<br>

Although I've convinced myself of this answer, maybe somebody can explain it more briefly, for you to admit it?

[Guest]

The last move can't be a "reset to H", else resetting will make the medial coin H while T is needed. If the last move is a "all turn over" The position before last move will be: Medial H and 3 others in that circle is T. To get such a position, the move before last move again can not be a reset, else all 4 coin would be H. If one before last move is a "all turn over", and the position after this move is H+TTT, the position before this move should be T+HHH. With this method, if we go to n.th move before last, the position have to be eather T+HHH or H+TTT. Since we're beginning to game from HHHH, coming to a point of T+HHH is impossible.<br>

Although I've convinced myself of this answer, maybe somebody can explain it more briefly, for you to admit it?

I Agree with your answer and I propose a variation for the puzzle:

Draw three large circles that intersect in such a way that 7 regions are defined.

Four regions are inside just one circle; two others are inside two;

and one, the central region, is in all three circles.

And we repeat the next steps of the problem. Is it possible?

[Guest]

I Agree with your answer and I propose a variation for the puzzle:

Draw three large circles that intersect in such a way that 7 regions are defined.

Four regions are inside just one circle; two others are inside two;

and one, the central region, is in all three circles.

And we repeat the next steps of the problem. Is it possible?

PS: One is oval insted of circle

[Jiminy Cricket]

This is a good one!

I've gone through this one for a while now and I'm beginning to think it is either impossible or will take a very large number of moves...

[Guest]

Maybe.

I think math may have a better chance than trial and error.

Given the conditions, we can assign circles as x,y,z and coins as a,b,c,d,e,f.

Let's put a,b,c in the non-overlapping spots. Now put d,e,f in the two circle spots, and g goes in the middle, where all the circles overlap. That gives us:

circle x with a,d,e,g

circle y with b,d,f,g and

circle z with c,e,f,g

If we set heads=1 and tails=-1, we can describe the two possible functions. Flipping all of the coins in a given circle would be multiplying each value in the circle by -1. "Resetting" a circle would mean setting each value to 1.

Now the problem is: Can we achieve a value of g=-1 while a through f all equal 1 using any combination of the above functions on circles x,y,z?

Put that way, it doesn't seem impossible, just terribly complicated. And that's about all I got for this one, I think. I hope it helps someone.

P.S. I REALLY like this one. Anything that brings me that close to doing math deserves some accolades.

Edited December 30, 2008 by Grayven

[bonanova]

Maybe.

I think math may have a better chance than trial and error.

Given the conditions, we can assign circles as x,y,z and coins as a,b,c,d,e,f.

Let's put a,b,c in the non-overlapping spots. Now put d,e,f in the two circle spots, and g goes in the middle, where all the circles overlap. That gives us:

circle x with a,d,e,g

circle y with b,d,f,g and

circle z with c,e,f,g

If we set heads=1 and tails=-1, we can describe the two possible functions. Flipping all of the coins in a given circle would be multiplying each value in the circle by -1. "Resetting" a circle would mean setting each value to 1.

Now the problem is: Can we achieve a value of g=-1 while a through f all equal 1 using any combination of the above functions on circles x,y,z?

Put that way, it doesn't seem impossible, just terribly complicated. And that's about all I got for this one, I think. I hope it helps someone.

P.S. I REALLY like this one. Anything that brings me that close to doing math deserves some accolades.

Nice formalism.

Now, can it be used to construct a proof?

[Guest]

Nice formalism.

Now, can it be used to construct a proof?

I suppose my old post is a complete proof if you change tails to -1 and heads to +1, resetting to getting absolute value and moving to multiply by -1

[Guest]

Nice formalism.

Now, can it be used to construct a proof?

Thanks. Restating a problem is my favorite step. I'll let someone else take it from there.

I do think nobody is probably correct so far.

[bonanova]

I suppose my old post is a complete proof if you change tails to -1 and heads to +1, resetting to getting absolute value and moving to multiply by -1

Comments in red in following spoiler.

You may have proved it and I don't understand, so I ask a question.

The last move can't be a "reset to H", else resetting will make the medial coin H while T is needed. If the last move is a "all turn over" The position before last move will be: Medial H and 3 others in that circle is T. To get such a position, the move before last move again can not be a reset, else all 4 coin would be H.

Does your last statement have to be made about the same circle?

If one makes repetitive flip moves on the same circle, you get into a cycle.

How about a move made on an adjacent circle?

- bn

If one before last move is a "all turn over", and the position after this move is H+TTT, the position before this move should be T+HHH. With this method, if we go to n.th move before last, the position have to be eather T+HHH or H+TTT. Since we're beginning to game from HHHH, coming to a point of T+HHH is impossible.<br>

Although I've convinced myself of this answer, maybe somebody can explain it more briefly, for you to admit it?

[HoustonHokie]

Well, I've been watching this thread and thought, "There has to be a way to do this." I was wrong, but that didn't keep me from trying. Here's my brute force method for proving you can't get there (all heads except central tails) from here (all heads).

"This is simple," thinks I. "I can represent the state of each coin as a 1 or a 0 and come up with a binary number for each possible state of the 7 coins. There will be 128 states in all, or 1111111 in binary. For each of the 128 states, I can perform 6 operations - flipping all the coins in each of 3 circles, or resetting all the coins in each of 3 circles to heads. All I need to do is identify the 6 (maximum) states that are dependent on each original state, and then follow the trail from 0000000 to 0000001."

I started to do a few calcs and then realized I'd be looking at 128 * 6 = 768 possible dependent states, and didn't think I'd have time for all that work. "There must be a simpler way." And there was: it turns out that several of the states are rotations or flips of other states, and therefore equivalent to one another. For example, these states are equivalent:

I started putting the 128 original states into groups of 1, 3, or 6 states and reduced the number of states from 128 possible to 40 unique groups. "I can work with this," I thought. And so I did.

Hours pass. I plot a representative state from each of the 40 groups and the 6 dependent states that individual case can transform into (you can see the results in the attached PDF). And I note something both interesting and disappointing: the game cannot be won because there are 4 groups which are not dependents of any of the other 36 - they only depend on themselves.

In my groups, the goal state is Group 21, and that's one of the 4 self-dependent groups. The others are groups 4, 13, & 35. Group 21 is only dependent upon Group 13. Group 13 is only dependent upon 21 or 35. If I'd gotten to 21 before, I wouldn't need to get to 13, so 13 is really only dependent upon 35. Group 35 is only dependent upon 4 or 13. Again, if I'd gotten to 13, I would have moved to 21 instead of 35, so 35 is really only dependent upon 4. And Group 4 is only dependent upon 35, which means that I'm in a recursive loop trying to get back to Group 1, which is the beginning state, and I'll never leave it.

"Phooey!" I exclaim. "I can get to 120 of the 128 possible states from an all heads beginning, but I can't get the other 8! What am I supposed to do now?"

I decided the only thing I could do was post my work, and see if anyone found a bust in it that might allow me to progress further. Or, perhaps it would put the nail into the coffin of this puzzle and prove once and for all that you can't get there from here.

circled_coins.pdf

[Guest]

Well, I've been watching this thread and thought, "There has to be a way to do this." I was wrong, but that didn't keep me from trying. Here's my brute force method for proving you can't get there (all heads except central tails) from here (all heads).

"This is simple," thinks I. "I can represent the state of each coin as a 1 or a 0 and come up with a binary number for each possible state of the 7 coins. There will be 128 states in all, or 1111111 in binary. For each of the 128 states, I can perform 6 operations - flipping all the coins in each of 3 circles, or resetting all the coins in each of 3 circles to heads. All I need to do is identify the 6 (maximum) states that are dependent on each original state, and then follow the trail from 0000000 to 0000001."

I started to do a few calcs and then realized I'd be looking at 128 * 6 = 768 possible dependent states, and didn't think I'd have time for all that work. "There must be a simpler way." And there was: it turns out that several of the states are rotations or flips of other states, and therefore equivalent to one another. For example, these states are equivalent:

I started putting the 128 original states into groups of 1, 3, or 6 states and reduced the number of states from 128 possible to 40 unique groups. "I can work with this," I thought. And so I did.

Hours pass. I plot a representative state from each of the 40 groups and the 6 dependent states that individual case can transform into (you can see the results in the attached PDF). And I note something both interesting and disappointing: the game cannot be won because there are 4 groups which are not dependents of any of the other 36 - they only depend on themselves.

In my groups, the goal state is Group 21, and that's one of the 4 self-dependent groups. The others are groups 4, 13, & 35. Group 21 is only dependent upon Group 13. Group 13 is only dependent upon 21 or 35. If I'd gotten to 21 before, I wouldn't need to get to 13, so 13 is really only dependent upon 35. Group 35 is only dependent upon 4 or 13. Again, if I'd gotten to 13, I would have moved to 21 instead of 35, so 35 is really only dependent upon 4. And Group 4 is only dependent upon 35, which means that I'm in a recursive loop trying to get back to Group 1, which is the beginning state, and I'll never leave it.

"Phooey!" I exclaim. "I can get to 120 of the 128 possible states from an all heads beginning, but I can't get the other 8! What am I supposed to do now?"

I decided the only thing I could do was post my work, and see if anyone found a bust in it that might allow me to progress further. Or, perhaps it would put the nail into the coffin of this puzzle and prove once and for all that you can't get there from here.

circled_coins.pdf

Ahh, the elegance of brute force. I think this goes farther into what nobody was on to.

That the states that give you problems have something to do with the three dual-overlap areas of the configuration. As I've worked on this, the problem I keep running into, is that you can't make a change in one circle without directly affecting the other two circles. Resetting means you have to make adjustments to the other two circles, thus altering the reset circle. I suspect some bit of programming that is beyond me could be successful in illustrating this problem.

And vague arguments are about as good as it gets. I do hope someone has the will to produce a complete proof, but I don't expect it.

Thanks for the effort HH. I hope it doesn't go unrewarded.

[HoustonHokie]

Ahh, the elegance of brute force. I think this goes farther into what nobody was on to.

That the states that give you problems have something to do with the three dual-overlap areas of the configuration. As I've worked on this, the problem I keep running into, is that you can't make a change in one circle without directly affecting the other two circles. Resetting means you have to make adjustments to the other two circles, thus altering the reset circle. I suspect some bit of programming that is beyond me could be successful in illustrating this problem.

And vague arguments are about as good as it gets. I do hope someone has the will to produce a complete proof, but I don't expect it.

Thanks for the effort HH. I hope it doesn't go unrewarded.

Oh, yeah - I could program it - and pretty easily in a spreadsheet. So I did. And I got the same answers in about 30 minutes that it took forever to draw, plus I confirmed that my groups were right. As far as the problem cases go, well, I haven't discovered much beyond the fact that they're problems. I do know that none of them can be created with a reset, so they have to be created with flips, and they're just flips of each other. If you introduce a reset to one of the problem cases, you exit them and can't get back.

[bonanova]

Oh, yeah - I could program it - and pretty easily in a spreadsheet. So I did. And I got the same answers in about 30 minutes that it took forever to draw, plus I confirmed that my groups were right. As far as the problem cases go, well, I haven't discovered much beyond the fact that they're problems. I do know that none of them can be created with a reset, so they have to be created with flips, and they're just flips of each other. If you introduce a reset to one of the problem cases, you exit them and can't get back.

Sounds like a proof to me.

[Guest]

Sounds like a proof to me.

I truly enjoyed this one. I don't know how much I helped, other than generous cheerleading, but I enjoyed participating nonetheless. Thanks BN.

edit: and thanks HH for putting it to bed.

Edited December 30, 2008 by Grayven

[HoustonHokie]

I truly enjoyed this one. I don't know how much I helped, other than generous cheerleading, but I enjoyed participating nonetheless. Thanks BN.

edit: and thanks HH for putting it to bed.

Thanks

And I haven't quite put everything to rest in my mind. renan proposed an interesting variation and I'm trying to figure out if that one can be done. Problem is, I may not have the right picture in mind. renan's post mentioned an oval, but I don't see a need for one if it looks like this:

If it looks like that, each circle affects a different number of coins - the outside circles affect 3 coins each, and the middle circle affects 5 coins. Makes it interesting. If I've got a little more time, I might try to work that one out.

[Prof. Templeton]

Thanks

And I haven't quite put everything to rest in my mind. renan proposed an interesting variation and I'm trying to figure out if that one can be done. Problem is, I may not have the right picture in mind. renan's post mentioned an oval, but I don't see a need for one if it looks like this:

If it looks like that, each circle affects a different number of coins - the outside circles affect 3 coins each, and the middle circle affects 5 coins. Makes it interesting. If I've got a little more time, I might try to work that one out.

Before I saw the pictures you had posted all my sketches looked like this one above. When I saw yours, I had assumed I interpreted the puzzle wrong. Didn't give it more then a couple of tries before work intruded, however.

[Guest]

Thanks

And I haven't quite put everything to rest in my mind. renan proposed an interesting variation and I'm trying to figure out if that one can be done. Problem is, I may not have the right picture in mind. renan's post mentioned an oval, but I don't see a need for one if it looks like this:

If it looks like that, each circle affects a different number of coins - the outside circles affect 3 coins each, and the middle circle affects 5 coins. Makes it interesting. If I've got a little more time, I might try to work that one out.

First reaction is that this configuration is more likely to work. You can make changes to the left or right circles that effect only coin #7 in the opposite circle. Time will tell.

[Guest]

Comments in red in following spoiler.

You may have proved it and I don't understand, so I ask a question.

Does your last statement have to be made about the same circle?

If one makes repetitive flip moves on the same circle, you get into a cycle.

How about a move made on an adjacent circle?

- bn

I'm suprised how you didn't understand such a simple proof.

I don't say that you must work on the same circle. On eather circle, a flip move will give a HTTT ot THHH result. H must not be always in middle section, say it TTHT or THTT etc. But each circle will have one H or T and three form opposite kind.

Please draw a diagram on your paper. Write T in middle, and H in others. Now go one step back and then one step back .. and go on.

You will always get a "one T three H" or "one H three T" in all circles.

[bonanova]

I'm suprised how you didn't understand such a simple proof.

I don't say that you must work on the same circle. On eather circle, a flip move will give a HTTT ot THHH result. H must not be always in middle section, say it TTHT or THTT etc. But each circle will have one H or T and three form opposite kind.

Please draw a diagram on your paper. Write T in middle, and H in others. Now go one step back and then one step back .. and go on.

You will always get a "one T three H" or "one H three T" in all circles.

Bingo.

starting from all heads in any circle - as at the beginning,

.

1. For the central coin to be tails, the sum of all circle flips must be odd.
2. For its surrounding 2-circle regions to have heads, the sum of their flips must be even.
3. For the for the single-circle coins to be heads, their individual flips must be even

.

Those conditions are contradictory:

The conditions of odd total flips, plus even 2-circle flips dictates that

the flips for each circle be odd. But that makes the single circle coins tails.

[Guest]

Thanks

And I haven't quite put everything to rest in my mind. renan proposed an interesting variation and I'm trying to figure out if that one can be done. Problem is, I may not have the right picture in mind. renan's post mentioned an oval, but I don't see a need for one if it looks like this:

If it looks like that, each circle affects a different number of coins - the outside circles affect 3 coins each, and the middle circle affects 5 coins. Makes it interesting. If I've got a little more time, I might try to work that one out.

Thats it, youre right and threre is no need for an oval, sorry for that. The question remains: Can it be done? And if so, what the minimum number of moviments?