Fair Sharing of a Pizza When Opinions About the Edge Differ
$begingroup$
Two friends wants to share a pizza. One of them loves the edge of the pizza and the other one hates it. Both consider the pizza to get tastier the closer to the center you get. What is the fairest way to cut the pizza if you are only allowed four straight cuts, such that the two sets of slices sum up to the same area and one set contains all edges?
I have attached an intuitive sketch, which seems somewhat fair, but I don't know how to approach a problem like this one.
geometry optimization fair-division
asked Jan 16 at 16:06
Lars RönnbäckLars Rönnbäck
19518
$endgroup$
add a comment |
$begingroup$
Two friends wants to share a pizza. One of them loves the edge of the pizza and the other one hates it. Both consider the pizza to get tastier the closer to the center you get. What is the fairest way to cut the pizza if you are only allowed four straight cuts, such that the two sets of slices sum up to the same area and one set contains all edges?
I have attached an intuitive sketch, which seems somewhat fair, but I don't know how to approach a problem like this one.
geometry optimization fair-division
asked Jan 16 at 16:06
Lars RönnbäckLars Rönnbäck
19518
$endgroup$
$begingroup$
How can both agree that the center tastes best but one love the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:27
$begingroup$
Because the edge has a different composition from the rest of the pizza?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:45
$begingroup$
How thick is the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:47
$begingroup$
I suppose that for a real pizza the edge would have thickness, but for simplicity I was thinking of the edge as infinitesimal, making it ok to place a point on the edge but not to include any segment of it in one of the sets.
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:52
add a comment |
$begingroup$
Two friends wants to share a pizza. One of them loves the edge of the pizza and the other one hates it. Both consider the pizza to get tastier the closer to the center you get. What is the fairest way to cut the pizza if you are only allowed four straight cuts, such that the two sets of slices sum up to the same area and one set contains all edges?
I have attached an intuitive sketch, which seems somewhat fair, but I don't know how to approach a problem like this one.
geometry optimization fair-division
asked Jan 16 at 16:06
Lars RönnbäckLars Rönnbäck
19518
$endgroup$
Two friends wants to share a pizza. One of them loves the edge of the pizza and the other one hates it. Both consider the pizza to get tastier the closer to the center you get. What is the fairest way to cut the pizza if you are only allowed four straight cuts, such that the two sets of slices sum up to the same area and one set contains all edges?
I have attached an intuitive sketch, which seems somewhat fair, but I don't know how to approach a problem like this one.
geometry optimization fair-division
geometry optimization fair-division
asked Jan 16 at 16:06
Lars RönnbäckLars Rönnbäck
19518
asked Jan 16 at 16:06
Lars RönnbäckLars Rönnbäck
19518
edited Jan 16 at 16:21
Lars Rönnbäck
asked Jan 16 at 16:06
Lars RönnbäckLars Rönnbäck
19518
asked Jan 16 at 16:06
Lars RönnbäckLars Rönnbäck
19518
asked Jan 16 at 16:06
Lars RönnbäckLars Rönnbäck
19518
19518
$begingroup$
How can both agree that the center tastes best but one love the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:27
$begingroup$
Because the edge has a different composition from the rest of the pizza?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:45
$begingroup$
How thick is the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:47
$begingroup$
I suppose that for a real pizza the edge would have thickness, but for simplicity I was thinking of the edge as infinitesimal, making it ok to place a point on the edge but not to include any segment of it in one of the sets.
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:52
add a comment |
$begingroup$
How can both agree that the center tastes best but one love the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:27
$begingroup$
Because the edge has a different composition from the rest of the pizza?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:45
$begingroup$
How thick is the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:47
$begingroup$
I suppose that for a real pizza the edge would have thickness, but for simplicity I was thinking of the edge as infinitesimal, making it ok to place a point on the edge but not to include any segment of it in one of the sets.
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:52
$begingroup$
How can both agree that the center tastes best but one love the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:27
$begingroup$
How can both agree that the center tastes best but one love the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:27
$begingroup$
Because the edge has a different composition from the rest of the pizza?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:45
$begingroup$
Because the edge has a different composition from the rest of the pizza?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:45
$begingroup$
How thick is the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:47
$begingroup$
How thick is the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:47
$begingroup$
I suppose that for a real pizza the edge would have thickness, but for simplicity I was thinking of the edge as infinitesimal, making it ok to place a point on the edge but not to include any segment of it in one of the sets.
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:52
$begingroup$
I suppose that for a real pizza the edge would have thickness, but for simplicity I was thinking of the edge as infinitesimal, making it ok to place a point on the edge but not to include any segment of it in one of the sets.
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
About all you can do is to show that a solution exists under certain conditions using the intermediate value function. If you are given a specific function for their valuation of the various areas of the pizza you may be able to derive a specific distribution. As long as the valuation is not too strongly peaked at the center you can note that having four points near each other on the edge gives the one who does not like crust almost nothing. Having four points that form a square gives the one who does not like crust $frac pi 4$ of the area including the desirable center. Somewhere in between is a set of vertices that gives them each the same valuation.
If $90%$ of the value of the pizza is the center point and the rest is evenly spread, you can't be fair. You have to make one segment go through the center, but there is not enough freedom to split the rest evenly with only four cuts.
answered Jan 16 at 16:35
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
I didn't think of the fact that the value of the center could be skewed in such a way that a fair sharing would be impossible. But you are saying that even given a linear function from 0 at the edge to full value at the center, we cannot find an optimum mathematically?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:58
$begingroup$
Yes, I think you can do so for simple valuation functions. Your diagram is a good approach. Let $BDEF$ be a square. When $C$ is at the center the edge person gets more. When $C$ is at $B$ the center person gets more. You can let the radius of $C$ be $r$ and do the integral for the value received by the center person as a function of $r$. Set that to half the value of the pie and solve for $r$. I would guess that is quite doable for a linear function. You certainly can do it numerically, but for a simple function calculus and algebra will suffice
$endgroup$
– Ross Millikan
Jan 16 at 17:04
$begingroup$
I just realized that it may be possible for both to get a piece of the center if you allow displacement of the pieces between the cuts. Seems only reasonable, given that we are talking about pizza.
$endgroup$
– Lars Rönnbäck
Jan 16 at 21:48
$begingroup$
I thought that was contrary to the requirement of four straight cuts. I think even with five and no displacement you can get a fair division. The problem comes with four because you can't get the non-edge piece big enough.
$endgroup$
– Ross Millikan
Jan 17 at 1:36
add a comment |
$begingroup$
Since I raised the possibility in one of my comments, here is a possible solution that gives ~42% of the center to the edge lover. It does however rely on the freedom to displace the pieces between the cuts. Given that this is a problem of practical nature, I don't see any harm in allowing displacement. Folding the pizza before or between cuts would probably result in undesired side effects, but it is possible that could yield an even fairer distribution.
answered Jan 17 at 13:51
Lars RönnbäckLars Rönnbäck
19518
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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oldest
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$begingroup$
About all you can do is to show that a solution exists under certain conditions using the intermediate value function. If you are given a specific function for their valuation of the various areas of the pizza you may be able to derive a specific distribution. As long as the valuation is not too strongly peaked at the center you can note that having four points near each other on the edge gives the one who does not like crust almost nothing. Having four points that form a square gives the one who does not like crust $frac pi 4$ of the area including the desirable center. Somewhere in between is a set of vertices that gives them each the same valuation.
If $90%$ of the value of the pizza is the center point and the rest is evenly spread, you can't be fair. You have to make one segment go through the center, but there is not enough freedom to split the rest evenly with only four cuts.
answered Jan 16 at 16:35
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
I didn't think of the fact that the value of the center could be skewed in such a way that a fair sharing would be impossible. But you are saying that even given a linear function from 0 at the edge to full value at the center, we cannot find an optimum mathematically?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:58
$begingroup$
Yes, I think you can do so for simple valuation functions. Your diagram is a good approach. Let $BDEF$ be a square. When $C$ is at the center the edge person gets more. When $C$ is at $B$ the center person gets more. You can let the radius of $C$ be $r$ and do the integral for the value received by the center person as a function of $r$. Set that to half the value of the pie and solve for $r$. I would guess that is quite doable for a linear function. You certainly can do it numerically, but for a simple function calculus and algebra will suffice
$endgroup$
– Ross Millikan
Jan 16 at 17:04
$begingroup$
I just realized that it may be possible for both to get a piece of the center if you allow displacement of the pieces between the cuts. Seems only reasonable, given that we are talking about pizza.
$endgroup$
– Lars Rönnbäck
Jan 16 at 21:48
$begingroup$
I thought that was contrary to the requirement of four straight cuts. I think even with five and no displacement you can get a fair division. The problem comes with four because you can't get the non-edge piece big enough.
$endgroup$
– Ross Millikan
Jan 17 at 1:36
add a comment |
$begingroup$
About all you can do is to show that a solution exists under certain conditions using the intermediate value function. If you are given a specific function for their valuation of the various areas of the pizza you may be able to derive a specific distribution. As long as the valuation is not too strongly peaked at the center you can note that having four points near each other on the edge gives the one who does not like crust almost nothing. Having four points that form a square gives the one who does not like crust $frac pi 4$ of the area including the desirable center. Somewhere in between is a set of vertices that gives them each the same valuation.
If $90%$ of the value of the pizza is the center point and the rest is evenly spread, you can't be fair. You have to make one segment go through the center, but there is not enough freedom to split the rest evenly with only four cuts.
answered Jan 16 at 16:35
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$begingroup$
I didn't think of the fact that the value of the center could be skewed in such a way that a fair sharing would be impossible. But you are saying that even given a linear function from 0 at the edge to full value at the center, we cannot find an optimum mathematically?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:58
$begingroup$
Yes, I think you can do so for simple valuation functions. Your diagram is a good approach. Let $BDEF$ be a square. When $C$ is at the center the edge person gets more. When $C$ is at $B$ the center person gets more. You can let the radius of $C$ be $r$ and do the integral for the value received by the center person as a function of $r$. Set that to half the value of the pie and solve for $r$. I would guess that is quite doable for a linear function. You certainly can do it numerically, but for a simple function calculus and algebra will suffice
$endgroup$
– Ross Millikan
Jan 16 at 17:04
$begingroup$
I just realized that it may be possible for both to get a piece of the center if you allow displacement of the pieces between the cuts. Seems only reasonable, given that we are talking about pizza.
$endgroup$
– Lars Rönnbäck
Jan 16 at 21:48
$begingroup$
I thought that was contrary to the requirement of four straight cuts. I think even with five and no displacement you can get a fair division. The problem comes with four because you can't get the non-edge piece big enough.
$endgroup$
– Ross Millikan
Jan 17 at 1:36
add a comment |
$begingroup$
About all you can do is to show that a solution exists under certain conditions using the intermediate value function. If you are given a specific function for their valuation of the various areas of the pizza you may be able to derive a specific distribution. As long as the valuation is not too strongly peaked at the center you can note that having four points near each other on the edge gives the one who does not like crust almost nothing. Having four points that form a square gives the one who does not like crust $frac pi 4$ of the area including the desirable center. Somewhere in between is a set of vertices that gives them each the same valuation.
If $90%$ of the value of the pizza is the center point and the rest is evenly spread, you can't be fair. You have to make one segment go through the center, but there is not enough freedom to split the rest evenly with only four cuts.
answered Jan 16 at 16:35
Ross MillikanRoss Millikan
295k23198371
$endgroup$
About all you can do is to show that a solution exists under certain conditions using the intermediate value function. If you are given a specific function for their valuation of the various areas of the pizza you may be able to derive a specific distribution. As long as the valuation is not too strongly peaked at the center you can note that having four points near each other on the edge gives the one who does not like crust almost nothing. Having four points that form a square gives the one who does not like crust $frac pi 4$ of the area including the desirable center. Somewhere in between is a set of vertices that gives them each the same valuation.
If $90%$ of the value of the pizza is the center point and the rest is evenly spread, you can't be fair. You have to make one segment go through the center, but there is not enough freedom to split the rest evenly with only four cuts.
answered Jan 16 at 16:35
Ross MillikanRoss Millikan
295k23198371
answered Jan 16 at 16:35
Ross MillikanRoss Millikan
295k23198371
answered Jan 16 at 16:35
Ross MillikanRoss Millikan
295k23198371
answered Jan 16 at 16:35
Ross MillikanRoss Millikan
295k23198371
295k23198371
$begingroup$
I didn't think of the fact that the value of the center could be skewed in such a way that a fair sharing would be impossible. But you are saying that even given a linear function from 0 at the edge to full value at the center, we cannot find an optimum mathematically?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:58
$begingroup$
Yes, I think you can do so for simple valuation functions. Your diagram is a good approach. Let $BDEF$ be a square. When $C$ is at the center the edge person gets more. When $C$ is at $B$ the center person gets more. You can let the radius of $C$ be $r$ and do the integral for the value received by the center person as a function of $r$. Set that to half the value of the pie and solve for $r$. I would guess that is quite doable for a linear function. You certainly can do it numerically, but for a simple function calculus and algebra will suffice
$endgroup$
– Ross Millikan
Jan 16 at 17:04
$begingroup$
I just realized that it may be possible for both to get a piece of the center if you allow displacement of the pieces between the cuts. Seems only reasonable, given that we are talking about pizza.
$endgroup$
– Lars Rönnbäck
Jan 16 at 21:48
$begingroup$
I thought that was contrary to the requirement of four straight cuts. I think even with five and no displacement you can get a fair division. The problem comes with four because you can't get the non-edge piece big enough.
$endgroup$
– Ross Millikan
Jan 17 at 1:36
add a comment |
$begingroup$
I didn't think of the fact that the value of the center could be skewed in such a way that a fair sharing would be impossible. But you are saying that even given a linear function from 0 at the edge to full value at the center, we cannot find an optimum mathematically?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:58
$begingroup$
Yes, I think you can do so for simple valuation functions. Your diagram is a good approach. Let $BDEF$ be a square. When $C$ is at the center the edge person gets more. When $C$ is at $B$ the center person gets more. You can let the radius of $C$ be $r$ and do the integral for the value received by the center person as a function of $r$. Set that to half the value of the pie and solve for $r$. I would guess that is quite doable for a linear function. You certainly can do it numerically, but for a simple function calculus and algebra will suffice
$endgroup$
– Ross Millikan
Jan 16 at 17:04
$begingroup$
I just realized that it may be possible for both to get a piece of the center if you allow displacement of the pieces between the cuts. Seems only reasonable, given that we are talking about pizza.
$endgroup$
– Lars Rönnbäck
Jan 16 at 21:48
$begingroup$
I thought that was contrary to the requirement of four straight cuts. I think even with five and no displacement you can get a fair division. The problem comes with four because you can't get the non-edge piece big enough.
$endgroup$
– Ross Millikan
Jan 17 at 1:36
$begingroup$
I didn't think of the fact that the value of the center could be skewed in such a way that a fair sharing would be impossible. But you are saying that even given a linear function from 0 at the edge to full value at the center, we cannot find an optimum mathematically?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:58
$begingroup$
I didn't think of the fact that the value of the center could be skewed in such a way that a fair sharing would be impossible. But you are saying that even given a linear function from 0 at the edge to full value at the center, we cannot find an optimum mathematically?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:58
$begingroup$
Yes, I think you can do so for simple valuation functions. Your diagram is a good approach. Let $BDEF$ be a square. When $C$ is at the center the edge person gets more. When $C$ is at $B$ the center person gets more. You can let the radius of $C$ be $r$ and do the integral for the value received by the center person as a function of $r$. Set that to half the value of the pie and solve for $r$. I would guess that is quite doable for a linear function. You certainly can do it numerically, but for a simple function calculus and algebra will suffice
$endgroup$
– Ross Millikan
Jan 16 at 17:04
$begingroup$
Yes, I think you can do so for simple valuation functions. Your diagram is a good approach. Let $BDEF$ be a square. When $C$ is at the center the edge person gets more. When $C$ is at $B$ the center person gets more. You can let the radius of $C$ be $r$ and do the integral for the value received by the center person as a function of $r$. Set that to half the value of the pie and solve for $r$. I would guess that is quite doable for a linear function. You certainly can do it numerically, but for a simple function calculus and algebra will suffice
$endgroup$
– Ross Millikan
Jan 16 at 17:04
$begingroup$
I just realized that it may be possible for both to get a piece of the center if you allow displacement of the pieces between the cuts. Seems only reasonable, given that we are talking about pizza.
$endgroup$
– Lars Rönnbäck
Jan 16 at 21:48
$begingroup$
I just realized that it may be possible for both to get a piece of the center if you allow displacement of the pieces between the cuts. Seems only reasonable, given that we are talking about pizza.
$endgroup$
– Lars Rönnbäck
Jan 16 at 21:48
$begingroup$
I thought that was contrary to the requirement of four straight cuts. I think even with five and no displacement you can get a fair division. The problem comes with four because you can't get the non-edge piece big enough.
$endgroup$
– Ross Millikan
Jan 17 at 1:36
$begingroup$
I thought that was contrary to the requirement of four straight cuts. I think even with five and no displacement you can get a fair division. The problem comes with four because you can't get the non-edge piece big enough.
$endgroup$
– Ross Millikan
Jan 17 at 1:36
add a comment |
$begingroup$
Since I raised the possibility in one of my comments, here is a possible solution that gives ~42% of the center to the edge lover. It does however rely on the freedom to displace the pieces between the cuts. Given that this is a problem of practical nature, I don't see any harm in allowing displacement. Folding the pizza before or between cuts would probably result in undesired side effects, but it is possible that could yield an even fairer distribution.
answered Jan 17 at 13:51
Lars RönnbäckLars Rönnbäck
19518
$endgroup$
add a comment |
$begingroup$
Since I raised the possibility in one of my comments, here is a possible solution that gives ~42% of the center to the edge lover. It does however rely on the freedom to displace the pieces between the cuts. Given that this is a problem of practical nature, I don't see any harm in allowing displacement. Folding the pizza before or between cuts would probably result in undesired side effects, but it is possible that could yield an even fairer distribution.
answered Jan 17 at 13:51
Lars RönnbäckLars Rönnbäck
19518
$endgroup$
add a comment |
$begingroup$
Since I raised the possibility in one of my comments, here is a possible solution that gives ~42% of the center to the edge lover. It does however rely on the freedom to displace the pieces between the cuts. Given that this is a problem of practical nature, I don't see any harm in allowing displacement. Folding the pizza before or between cuts would probably result in undesired side effects, but it is possible that could yield an even fairer distribution.
answered Jan 17 at 13:51
Lars RönnbäckLars Rönnbäck
19518
$endgroup$
Since I raised the possibility in one of my comments, here is a possible solution that gives ~42% of the center to the edge lover. It does however rely on the freedom to displace the pieces between the cuts. Given that this is a problem of practical nature, I don't see any harm in allowing displacement. Folding the pizza before or between cuts would probably result in undesired side effects, but it is possible that could yield an even fairer distribution.
answered Jan 17 at 13:51
Lars RönnbäckLars Rönnbäck
19518
answered Jan 17 at 13:51
Lars RönnbäckLars Rönnbäck
19518
answered Jan 17 at 13:51
Lars RönnbäckLars Rönnbäck
19518
answered Jan 17 at 13:51
Lars RönnbäckLars Rönnbäck
19518
19518
add a comment |
add a comment |
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$begingroup$
How can both agree that the center tastes best but one love the edge?
$endgroup$
– Mohammad Zuhair Khan
Jan 16 at 16:27
$begingroup$
Because the edge has a different composition from the rest of the pizza?
$endgroup$
– Lars Rönnbäck
Jan 16 at 16:45
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How thick is the edge?
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– Mohammad Zuhair Khan
Jan 16 at 16:47
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I suppose that for a real pizza the edge would have thickness, but for simplicity I was thinking of the edge as infinitesimal, making it ok to place a point on the edge but not to include any segment of it in one of the sets.
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– Lars Rönnbäck
Jan 16 at 16:52