A magician places $n$ coins, someone choose a coin, for what $n$ the magician will succeed?
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There is a riddle which I'm having a hard time to solve, would love some help...
A magician places $n$ coins on a table and walks down off the stage. A volunteer comes, turn over whichever coins he wishes, select one coin and whispers its number to the apprentice. Then the apprentice turns over one coin aiming to assist the magician to know the selected coin. For which values of $n$ the magic will always succeed? what if the apprentice can choose to flip one or zero coins? (instead of always one).
probability contest-math puzzle
edited 12 hours ago
YuiTo Cheng
2,4064937
asked 16 hours ago
user2323232user2323232
18810
$endgroup$
|
show 2 more comments
$begingroup$
There is a riddle which I'm having a hard time to solve, would love some help...
A magician places $n$ coins on a table and walks down off the stage. A volunteer comes, turn over whichever coins he wishes, select one coin and whispers its number to the apprentice. Then the apprentice turns over one coin aiming to assist the magician to know the selected coin. For which values of $n$ the magic will always succeed? what if the apprentice can choose to flip one or zero coins? (instead of always one).
probability contest-math puzzle
edited 12 hours ago
YuiTo Cheng
2,4064937
asked 16 hours ago
user2323232user2323232
18810
$endgroup$
$begingroup$
Is the "selected coin" the one the volunteer turns over, or does he turn over one coin and whisper a perhaps different coin in the apprentice's ear?
$endgroup$
– saulspatz
16 hours ago
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@saulspatz After reading again I think one of those he already flipped.
$endgroup$
– user2323232
15 hours ago
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After looking at Empy2's answer, and the comments, I realize I was solving the wrong riddle.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
if the apprentice can choose to flip $0$ coins, then the $n=3$ case can be solved. expanding on the answer by @Empy2 : we have a cube. simply color $3$ pairs of opposite corners with $3$ different colors (and leave the $4$th pair uncolored). any corner is $0$ or $1$ away from any desired color.
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– antkam
13 hours ago
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Does the flipping of a number of coins matter at all? I don't see how it does. As long as the magician knows which coin the assistant flipped it can be done.
$endgroup$
– Ross Millikan
13 hours ago
|
show 2 more comments
$begingroup$
There is a riddle which I'm having a hard time to solve, would love some help...
A magician places $n$ coins on a table and walks down off the stage. A volunteer comes, turn over whichever coins he wishes, select one coin and whispers its number to the apprentice. Then the apprentice turns over one coin aiming to assist the magician to know the selected coin. For which values of $n$ the magic will always succeed? what if the apprentice can choose to flip one or zero coins? (instead of always one).
probability contest-math puzzle
edited 12 hours ago
YuiTo Cheng
2,4064937
asked 16 hours ago
user2323232user2323232
18810
$endgroup$
There is a riddle which I'm having a hard time to solve, would love some help...
A magician places $n$ coins on a table and walks down off the stage. A volunteer comes, turn over whichever coins he wishes, select one coin and whispers its number to the apprentice. Then the apprentice turns over one coin aiming to assist the magician to know the selected coin. For which values of $n$ the magic will always succeed? what if the apprentice can choose to flip one or zero coins? (instead of always one).
probability contest-math puzzle
probability contest-math puzzle
edited 12 hours ago
YuiTo Cheng
2,4064937
asked 16 hours ago
user2323232user2323232
18810
edited 12 hours ago
YuiTo Cheng
2,4064937
asked 16 hours ago
user2323232user2323232
18810
edited 12 hours ago
YuiTo Cheng
2,4064937
edited 12 hours ago
YuiTo Cheng
2,4064937
edited 12 hours ago
YuiTo Cheng
2,4064937
2,4064937
asked 16 hours ago
user2323232user2323232
18810
asked 16 hours ago
user2323232user2323232
18810
asked 16 hours ago
user2323232user2323232
18810
18810
$begingroup$
Is the "selected coin" the one the volunteer turns over, or does he turn over one coin and whisper a perhaps different coin in the apprentice's ear?
$endgroup$
– saulspatz
16 hours ago
$begingroup$
@saulspatz After reading again I think one of those he already flipped.
$endgroup$
– user2323232
15 hours ago
$begingroup$
After looking at Empy2's answer, and the comments, I realize I was solving the wrong riddle.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
if the apprentice can choose to flip $0$ coins, then the $n=3$ case can be solved. expanding on the answer by @Empy2 : we have a cube. simply color $3$ pairs of opposite corners with $3$ different colors (and leave the $4$th pair uncolored). any corner is $0$ or $1$ away from any desired color.
$endgroup$
– antkam
13 hours ago
$begingroup$
Does the flipping of a number of coins matter at all? I don't see how it does. As long as the magician knows which coin the assistant flipped it can be done.
$endgroup$
– Ross Millikan
13 hours ago
|
show 2 more comments
$begingroup$
Is the "selected coin" the one the volunteer turns over, or does he turn over one coin and whisper a perhaps different coin in the apprentice's ear?
$endgroup$
– saulspatz
16 hours ago
$begingroup$
@saulspatz After reading again I think one of those he already flipped.
$endgroup$
– user2323232
15 hours ago
$begingroup$
After looking at Empy2's answer, and the comments, I realize I was solving the wrong riddle.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
if the apprentice can choose to flip $0$ coins, then the $n=3$ case can be solved. expanding on the answer by @Empy2 : we have a cube. simply color $3$ pairs of opposite corners with $3$ different colors (and leave the $4$th pair uncolored). any corner is $0$ or $1$ away from any desired color.
$endgroup$
– antkam
13 hours ago
$begingroup$
Does the flipping of a number of coins matter at all? I don't see how it does. As long as the magician knows which coin the assistant flipped it can be done.
$endgroup$
– Ross Millikan
13 hours ago
$begingroup$
Is the "selected coin" the one the volunteer turns over, or does he turn over one coin and whisper a perhaps different coin in the apprentice's ear?
$endgroup$
– saulspatz
16 hours ago
$begingroup$
Is the "selected coin" the one the volunteer turns over, or does he turn over one coin and whisper a perhaps different coin in the apprentice's ear?
$endgroup$
– saulspatz
16 hours ago
$begingroup$
@saulspatz After reading again I think one of those he already flipped.
$endgroup$
– user2323232
15 hours ago
$begingroup$
@saulspatz After reading again I think one of those he already flipped.
$endgroup$
– user2323232
15 hours ago
$begingroup$
After looking at Empy2's answer, and the comments, I realize I was solving the wrong riddle.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
After looking at Empy2's answer, and the comments, I realize I was solving the wrong riddle.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
if the apprentice can choose to flip $0$ coins, then the $n=3$ case can be solved. expanding on the answer by @Empy2 : we have a cube. simply color $3$ pairs of opposite corners with $3$ different colors (and leave the $4$th pair uncolored). any corner is $0$ or $1$ away from any desired color.
$endgroup$
– antkam
13 hours ago
$begingroup$
if the apprentice can choose to flip $0$ coins, then the $n=3$ case can be solved. expanding on the answer by @Empy2 : we have a cube. simply color $3$ pairs of opposite corners with $3$ different colors (and leave the $4$th pair uncolored). any corner is $0$ or $1$ away from any desired color.
$endgroup$
– antkam
13 hours ago
$begingroup$
Does the flipping of a number of coins matter at all? I don't see how it does. As long as the magician knows which coin the assistant flipped it can be done.
$endgroup$
– Ross Millikan
13 hours ago
$begingroup$
Does the flipping of a number of coins matter at all? I don't see how it does. As long as the magician knows which coin the assistant flipped it can be done.
$endgroup$
– Ross Millikan
13 hours ago
|
show 2 more comments
3 Answers
3
active
oldest
votes
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I think a better way to represent the $2^n$ strategy in Empty2's answer is to number the coins in binary with $n$ bits. The selected coin has a particular number. The assistant is to make the XOR of all the heads match the selected coin. It can be done by taking the bitwise XOR of the current state with the selected coin and flipping the coin with that number.
As an example, with eight coins suppose coins $1,3,6$ are heads with XOR of $4$. If the selected coin is $5$, flipping the $1$ coin makes the XOR be $5$. If the selected coin is $2$, flipping the $6$ makes the $XOR$ be $2$. If the selected coin is $6$, flipping the $2$ coin gives the proper $XOR$. Flipping the $0$ coin is free, so the assistant can always flip a coin.
answered 12 hours ago
Ross MillikanRoss Millikan
301k24200375
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Nice approach. +1.
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– Empy2
12 hours ago
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That solves $n=2^m-1$ coins when the assistant can flip no coins. Just remove coin $0$.
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– Empy2
12 hours ago
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Well, I tried doing an example for n = 4, and your approach doesn't work.
$endgroup$
– Ilan Aizelman WS
7 hours ago
add a comment |
$begingroup$
There are $2^n$ coin positions. Represent them as the vertices of a hypercube. Give each vertex a color to represent one of the $n$ named coins. There must be one of each colour adjacent to any starting position, so there are equal numbers of each colour. So $n$ must be a factor of $2^n$, so $n$ must be a power of 2.
That does't work if the assistant may turn 0 coins, and doesn't explain how to do it for $n=2^m$ coins.
answered 15 hours ago
Empy2Empy2
33.7k12562
$endgroup$
$begingroup$
There are only ${nchoose2}$ allowable positions that are different from the one the magician started with.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
I think that 'turns over whichever coins he wishes' means that all positions are allowable
$endgroup$
– Empy2
14 hours ago
1
$begingroup$
I think you're right. I misread it as "coin" instead of "coins".
$endgroup$
– saulspatz
14 hours ago
2
$begingroup$
@antkam For your sake, I hope your mind is greater than mine.
$endgroup$
– saulspatz
13 hours ago
1
$begingroup$
@saulspatz -- lets just say, by solving the $n=3$ case, we both achieved the same (not very high) lower bound for the greatness of our minds
$endgroup$
– antkam
13 hours ago
|
show 6 more comments
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Nevermind. I overlooked the fact that once some coins are flipped, then not all changes to the sum (mod n) are available to the assistant with a single flip.
Number the coins from 1 to n. Let S equal 1 plus (the sum of numbers on the flipped coins modulo n). A set of flipped coins "indicates" the chosen coin which matches S. The assistant can flip any 1 coin, and the coins have each possible value modulo n, so the assistant can indicate which coin was selected by flipping the coin corresponding to the difference between the current S and the desired S.
answered 11 hours ago
Mike HMike H
11
New contributor
Mike H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1
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This doesn't seem to work, at least not without some restriction on $n$. Suppose $n=3,$ the selected coin is $3$, and coins $1$ and $3$ are flipped. The only ways the coins can indicate $3$ are for only coin $2$ to be flipped, or for coins $2$ and $3$ to be flipped, and neither of these is reachable from the initial state in one flip.
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– saulspatz
11 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think a better way to represent the $2^n$ strategy in Empty2's answer is to number the coins in binary with $n$ bits. The selected coin has a particular number. The assistant is to make the XOR of all the heads match the selected coin. It can be done by taking the bitwise XOR of the current state with the selected coin and flipping the coin with that number.
As an example, with eight coins suppose coins $1,3,6$ are heads with XOR of $4$. If the selected coin is $5$, flipping the $1$ coin makes the XOR be $5$. If the selected coin is $2$, flipping the $6$ makes the $XOR$ be $2$. If the selected coin is $6$, flipping the $2$ coin gives the proper $XOR$. Flipping the $0$ coin is free, so the assistant can always flip a coin.
answered 12 hours ago
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Nice approach. +1.
$endgroup$
– Empy2
12 hours ago
$begingroup$
That solves $n=2^m-1$ coins when the assistant can flip no coins. Just remove coin $0$.
$endgroup$
– Empy2
12 hours ago
$begingroup$
Well, I tried doing an example for n = 4, and your approach doesn't work.
$endgroup$
– Ilan Aizelman WS
7 hours ago
add a comment |
$begingroup$
I think a better way to represent the $2^n$ strategy in Empty2's answer is to number the coins in binary with $n$ bits. The selected coin has a particular number. The assistant is to make the XOR of all the heads match the selected coin. It can be done by taking the bitwise XOR of the current state with the selected coin and flipping the coin with that number.
As an example, with eight coins suppose coins $1,3,6$ are heads with XOR of $4$. If the selected coin is $5$, flipping the $1$ coin makes the XOR be $5$. If the selected coin is $2$, flipping the $6$ makes the $XOR$ be $2$. If the selected coin is $6$, flipping the $2$ coin gives the proper $XOR$. Flipping the $0$ coin is free, so the assistant can always flip a coin.
answered 12 hours ago
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Nice approach. +1.
$endgroup$
– Empy2
12 hours ago
$begingroup$
That solves $n=2^m-1$ coins when the assistant can flip no coins. Just remove coin $0$.
$endgroup$
– Empy2
12 hours ago
$begingroup$
Well, I tried doing an example for n = 4, and your approach doesn't work.
$endgroup$
– Ilan Aizelman WS
7 hours ago
add a comment |
$begingroup$
I think a better way to represent the $2^n$ strategy in Empty2's answer is to number the coins in binary with $n$ bits. The selected coin has a particular number. The assistant is to make the XOR of all the heads match the selected coin. It can be done by taking the bitwise XOR of the current state with the selected coin and flipping the coin with that number.
As an example, with eight coins suppose coins $1,3,6$ are heads with XOR of $4$. If the selected coin is $5$, flipping the $1$ coin makes the XOR be $5$. If the selected coin is $2$, flipping the $6$ makes the $XOR$ be $2$. If the selected coin is $6$, flipping the $2$ coin gives the proper $XOR$. Flipping the $0$ coin is free, so the assistant can always flip a coin.
answered 12 hours ago
Ross MillikanRoss Millikan
301k24200375
$endgroup$
I think a better way to represent the $2^n$ strategy in Empty2's answer is to number the coins in binary with $n$ bits. The selected coin has a particular number. The assistant is to make the XOR of all the heads match the selected coin. It can be done by taking the bitwise XOR of the current state with the selected coin and flipping the coin with that number.
As an example, with eight coins suppose coins $1,3,6$ are heads with XOR of $4$. If the selected coin is $5$, flipping the $1$ coin makes the XOR be $5$. If the selected coin is $2$, flipping the $6$ makes the $XOR$ be $2$. If the selected coin is $6$, flipping the $2$ coin gives the proper $XOR$. Flipping the $0$ coin is free, so the assistant can always flip a coin.
answered 12 hours ago
Ross MillikanRoss Millikan
301k24200375
answered 12 hours ago
Ross MillikanRoss Millikan
301k24200375
answered 12 hours ago
Ross MillikanRoss Millikan
301k24200375
answered 12 hours ago
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
Nice approach. +1.
$endgroup$
– Empy2
12 hours ago
$begingroup$
That solves $n=2^m-1$ coins when the assistant can flip no coins. Just remove coin $0$.
$endgroup$
– Empy2
12 hours ago
$begingroup$
Well, I tried doing an example for n = 4, and your approach doesn't work.
$endgroup$
– Ilan Aizelman WS
7 hours ago
add a comment |
$begingroup$
Nice approach. +1.
$endgroup$
– Empy2
12 hours ago
$begingroup$
That solves $n=2^m-1$ coins when the assistant can flip no coins. Just remove coin $0$.
$endgroup$
– Empy2
12 hours ago
$begingroup$
Well, I tried doing an example for n = 4, and your approach doesn't work.
$endgroup$
– Ilan Aizelman WS
7 hours ago
$begingroup$
Nice approach. +1.
$endgroup$
– Empy2
12 hours ago
$begingroup$
Nice approach. +1.
$endgroup$
– Empy2
12 hours ago
$begingroup$
That solves $n=2^m-1$ coins when the assistant can flip no coins. Just remove coin $0$.
$endgroup$
– Empy2
12 hours ago
$begingroup$
That solves $n=2^m-1$ coins when the assistant can flip no coins. Just remove coin $0$.
$endgroup$
– Empy2
12 hours ago
$begingroup$
Well, I tried doing an example for n = 4, and your approach doesn't work.
$endgroup$
– Ilan Aizelman WS
7 hours ago
$begingroup$
Well, I tried doing an example for n = 4, and your approach doesn't work.
$endgroup$
– Ilan Aizelman WS
7 hours ago
add a comment |
$begingroup$
There are $2^n$ coin positions. Represent them as the vertices of a hypercube. Give each vertex a color to represent one of the $n$ named coins. There must be one of each colour adjacent to any starting position, so there are equal numbers of each colour. So $n$ must be a factor of $2^n$, so $n$ must be a power of 2.
That does't work if the assistant may turn 0 coins, and doesn't explain how to do it for $n=2^m$ coins.
answered 15 hours ago
Empy2Empy2
33.7k12562
$endgroup$
$begingroup$
There are only ${nchoose2}$ allowable positions that are different from the one the magician started with.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
I think that 'turns over whichever coins he wishes' means that all positions are allowable
$endgroup$
– Empy2
14 hours ago
1
$begingroup$
I think you're right. I misread it as "coin" instead of "coins".
$endgroup$
– saulspatz
14 hours ago
2
$begingroup$
@antkam For your sake, I hope your mind is greater than mine.
$endgroup$
– saulspatz
13 hours ago
1
$begingroup$
@saulspatz -- lets just say, by solving the $n=3$ case, we both achieved the same (not very high) lower bound for the greatness of our minds
$endgroup$
– antkam
13 hours ago
|
show 6 more comments
$begingroup$
There are $2^n$ coin positions. Represent them as the vertices of a hypercube. Give each vertex a color to represent one of the $n$ named coins. There must be one of each colour adjacent to any starting position, so there are equal numbers of each colour. So $n$ must be a factor of $2^n$, so $n$ must be a power of 2.
That does't work if the assistant may turn 0 coins, and doesn't explain how to do it for $n=2^m$ coins.
answered 15 hours ago
Empy2Empy2
33.7k12562
$endgroup$
$begingroup$
There are only ${nchoose2}$ allowable positions that are different from the one the magician started with.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
I think that 'turns over whichever coins he wishes' means that all positions are allowable
$endgroup$
– Empy2
14 hours ago
1
$begingroup$
I think you're right. I misread it as "coin" instead of "coins".
$endgroup$
– saulspatz
14 hours ago
2
$begingroup$
@antkam For your sake, I hope your mind is greater than mine.
$endgroup$
– saulspatz
13 hours ago
1
$begingroup$
@saulspatz -- lets just say, by solving the $n=3$ case, we both achieved the same (not very high) lower bound for the greatness of our minds
$endgroup$
– antkam
13 hours ago
|
show 6 more comments
$begingroup$
There are $2^n$ coin positions. Represent them as the vertices of a hypercube. Give each vertex a color to represent one of the $n$ named coins. There must be one of each colour adjacent to any starting position, so there are equal numbers of each colour. So $n$ must be a factor of $2^n$, so $n$ must be a power of 2.
That does't work if the assistant may turn 0 coins, and doesn't explain how to do it for $n=2^m$ coins.
answered 15 hours ago
Empy2Empy2
33.7k12562
$endgroup$
There are $2^n$ coin positions. Represent them as the vertices of a hypercube. Give each vertex a color to represent one of the $n$ named coins. There must be one of each colour adjacent to any starting position, so there are equal numbers of each colour. So $n$ must be a factor of $2^n$, so $n$ must be a power of 2.
That does't work if the assistant may turn 0 coins, and doesn't explain how to do it for $n=2^m$ coins.
answered 15 hours ago
Empy2Empy2
33.7k12562
answered 15 hours ago
Empy2Empy2
33.7k12562
answered 15 hours ago
Empy2Empy2
33.7k12562
answered 15 hours ago
Empy2Empy2
33.7k12562
33.7k12562
$begingroup$
There are only ${nchoose2}$ allowable positions that are different from the one the magician started with.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
I think that 'turns over whichever coins he wishes' means that all positions are allowable
$endgroup$
– Empy2
14 hours ago
1
$begingroup$
I think you're right. I misread it as "coin" instead of "coins".
$endgroup$
– saulspatz
14 hours ago
2
$begingroup$
@antkam For your sake, I hope your mind is greater than mine.
$endgroup$
– saulspatz
13 hours ago
1
$begingroup$
@saulspatz -- lets just say, by solving the $n=3$ case, we both achieved the same (not very high) lower bound for the greatness of our minds
$endgroup$
– antkam
13 hours ago
|
show 6 more comments
$begingroup$
There are only ${nchoose2}$ allowable positions that are different from the one the magician started with.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
I think that 'turns over whichever coins he wishes' means that all positions are allowable
$endgroup$
– Empy2
14 hours ago
1
$begingroup$
I think you're right. I misread it as "coin" instead of "coins".
$endgroup$
– saulspatz
14 hours ago
2
$begingroup$
@antkam For your sake, I hope your mind is greater than mine.
$endgroup$
– saulspatz
13 hours ago
1
$begingroup$
@saulspatz -- lets just say, by solving the $n=3$ case, we both achieved the same (not very high) lower bound for the greatness of our minds
$endgroup$
– antkam
13 hours ago
$begingroup$
There are only ${nchoose2}$ allowable positions that are different from the one the magician started with.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
There are only ${nchoose2}$ allowable positions that are different from the one the magician started with.
$endgroup$
– saulspatz
14 hours ago
$begingroup$
I think that 'turns over whichever coins he wishes' means that all positions are allowable
$endgroup$
– Empy2
14 hours ago
$begingroup$
I think that 'turns over whichever coins he wishes' means that all positions are allowable
$endgroup$
– Empy2
14 hours ago
1
1
$begingroup$
I think you're right. I misread it as "coin" instead of "coins".
$endgroup$
– saulspatz
14 hours ago
$begingroup$
I think you're right. I misread it as "coin" instead of "coins".
$endgroup$
– saulspatz
14 hours ago
2
2
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@antkam For your sake, I hope your mind is greater than mine.
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– saulspatz
13 hours ago
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@antkam For your sake, I hope your mind is greater than mine.
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– saulspatz
13 hours ago
1
1
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@saulspatz -- lets just say, by solving the $n=3$ case, we both achieved the same (not very high) lower bound for the greatness of our minds
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– antkam
13 hours ago
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@saulspatz -- lets just say, by solving the $n=3$ case, we both achieved the same (not very high) lower bound for the greatness of our minds
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– antkam
13 hours ago
|
show 6 more comments
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Nevermind. I overlooked the fact that once some coins are flipped, then not all changes to the sum (mod n) are available to the assistant with a single flip.
Number the coins from 1 to n. Let S equal 1 plus (the sum of numbers on the flipped coins modulo n). A set of flipped coins "indicates" the chosen coin which matches S. The assistant can flip any 1 coin, and the coins have each possible value modulo n, so the assistant can indicate which coin was selected by flipping the coin corresponding to the difference between the current S and the desired S.
answered 11 hours ago
Mike HMike H
11
New contributor
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$endgroup$
1
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This doesn't seem to work, at least not without some restriction on $n$. Suppose $n=3,$ the selected coin is $3$, and coins $1$ and $3$ are flipped. The only ways the coins can indicate $3$ are for only coin $2$ to be flipped, or for coins $2$ and $3$ to be flipped, and neither of these is reachable from the initial state in one flip.
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– saulspatz
11 hours ago
add a comment |
$begingroup$
Nevermind. I overlooked the fact that once some coins are flipped, then not all changes to the sum (mod n) are available to the assistant with a single flip.
Number the coins from 1 to n. Let S equal 1 plus (the sum of numbers on the flipped coins modulo n). A set of flipped coins "indicates" the chosen coin which matches S. The assistant can flip any 1 coin, and the coins have each possible value modulo n, so the assistant can indicate which coin was selected by flipping the coin corresponding to the difference between the current S and the desired S.
answered 11 hours ago
Mike HMike H
11
New contributor
Mike H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
This doesn't seem to work, at least not without some restriction on $n$. Suppose $n=3,$ the selected coin is $3$, and coins $1$ and $3$ are flipped. The only ways the coins can indicate $3$ are for only coin $2$ to be flipped, or for coins $2$ and $3$ to be flipped, and neither of these is reachable from the initial state in one flip.
$endgroup$
– saulspatz
11 hours ago
add a comment |
$begingroup$
Nevermind. I overlooked the fact that once some coins are flipped, then not all changes to the sum (mod n) are available to the assistant with a single flip.
Number the coins from 1 to n. Let S equal 1 plus (the sum of numbers on the flipped coins modulo n). A set of flipped coins "indicates" the chosen coin which matches S. The assistant can flip any 1 coin, and the coins have each possible value modulo n, so the assistant can indicate which coin was selected by flipping the coin corresponding to the difference between the current S and the desired S.
answered 11 hours ago
Mike HMike H
11
New contributor
Mike H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Nevermind. I overlooked the fact that once some coins are flipped, then not all changes to the sum (mod n) are available to the assistant with a single flip.
Number the coins from 1 to n. Let S equal 1 plus (the sum of numbers on the flipped coins modulo n). A set of flipped coins "indicates" the chosen coin which matches S. The assistant can flip any 1 coin, and the coins have each possible value modulo n, so the assistant can indicate which coin was selected by flipping the coin corresponding to the difference between the current S and the desired S.
answered 11 hours ago
Mike HMike H
11
New contributor
Mike H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
answered 11 hours ago
Mike HMike H
11
New contributor
Mike H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 11 hours ago
Mike HMike H
11
answered 11 hours ago
Mike HMike H
11
11
New contributor
Mike H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mike H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mike H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
This doesn't seem to work, at least not without some restriction on $n$. Suppose $n=3,$ the selected coin is $3$, and coins $1$ and $3$ are flipped. The only ways the coins can indicate $3$ are for only coin $2$ to be flipped, or for coins $2$ and $3$ to be flipped, and neither of these is reachable from the initial state in one flip.
$endgroup$
– saulspatz
11 hours ago
add a comment |
1
$begingroup$
This doesn't seem to work, at least not without some restriction on $n$. Suppose $n=3,$ the selected coin is $3$, and coins $1$ and $3$ are flipped. The only ways the coins can indicate $3$ are for only coin $2$ to be flipped, or for coins $2$ and $3$ to be flipped, and neither of these is reachable from the initial state in one flip.
$endgroup$
– saulspatz
11 hours ago
1
1
$begingroup$
This doesn't seem to work, at least not without some restriction on $n$. Suppose $n=3,$ the selected coin is $3$, and coins $1$ and $3$ are flipped. The only ways the coins can indicate $3$ are for only coin $2$ to be flipped, or for coins $2$ and $3$ to be flipped, and neither of these is reachable from the initial state in one flip.
$endgroup$
– saulspatz
11 hours ago
$begingroup$
This doesn't seem to work, at least not without some restriction on $n$. Suppose $n=3,$ the selected coin is $3$, and coins $1$ and $3$ are flipped. The only ways the coins can indicate $3$ are for only coin $2$ to be flipped, or for coins $2$ and $3$ to be flipped, and neither of these is reachable from the initial state in one flip.
$endgroup$
– saulspatz
11 hours ago
add a comment |
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Is the "selected coin" the one the volunteer turns over, or does he turn over one coin and whisper a perhaps different coin in the apprentice's ear?
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– saulspatz
16 hours ago
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@saulspatz After reading again I think one of those he already flipped.
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– user2323232
15 hours ago
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After looking at Empy2's answer, and the comments, I realize I was solving the wrong riddle.
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– saulspatz
14 hours ago
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if the apprentice can choose to flip $0$ coins, then the $n=3$ case can be solved. expanding on the answer by @Empy2 : we have a cube. simply color $3$ pairs of opposite corners with $3$ different colors (and leave the $4$th pair uncolored). any corner is $0$ or $1$ away from any desired color.
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– antkam
13 hours ago
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Does the flipping of a number of coins matter at all? I don't see how it does. As long as the magician knows which coin the assistant flipped it can be done.
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– Ross Millikan
13 hours ago