A puzzle needing hints or solution
I heard of this puzzle that I want to get some hints or a full solution.
There are $1024$ dishes to be chosen from for a party. There are $6875$ participants in total. The objective is to find $10$ dishes in the menu such that for any dish $d$ that is not on the list, there are more than half people who would prefer one of the dishes in the menu over $d$.
Is it possible to generate such a menu?
combinatorics puzzle
edited 2 hours ago
Sauhard Sharma
785117
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,77722538
add a comment |
I heard of this puzzle that I want to get some hints or a full solution.
There are $1024$ dishes to be chosen from for a party. There are $6875$ participants in total. The objective is to find $10$ dishes in the menu such that for any dish $d$ that is not on the list, there are more than half people who would prefer one of the dishes in the menu over $d$.
Is it possible to generate such a menu?
combinatorics puzzle
edited 2 hours ago
Sauhard Sharma
785117
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,77722538
add a comment |
I heard of this puzzle that I want to get some hints or a full solution.
There are $1024$ dishes to be chosen from for a party. There are $6875$ participants in total. The objective is to find $10$ dishes in the menu such that for any dish $d$ that is not on the list, there are more than half people who would prefer one of the dishes in the menu over $d$.
Is it possible to generate such a menu?
combinatorics puzzle
edited 2 hours ago
Sauhard Sharma
785117
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,77722538
I heard of this puzzle that I want to get some hints or a full solution.
There are $1024$ dishes to be chosen from for a party. There are $6875$ participants in total. The objective is to find $10$ dishes in the menu such that for any dish $d$ that is not on the list, there are more than half people who would prefer one of the dishes in the menu over $d$.
Is it possible to generate such a menu?
combinatorics puzzle
combinatorics puzzle
edited 2 hours ago
Sauhard Sharma
785117
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,77722538
edited 2 hours ago
Sauhard Sharma
785117
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,77722538
edited 2 hours ago
Sauhard Sharma
785117
edited 2 hours ago
Sauhard Sharma
785117
edited 2 hours ago
Sauhard Sharma
785117
785117
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,77722538
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,77722538
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,77722538
1,77722538
add a comment |
add a comment |
1 Answer
1
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oldest
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Have a tournament where each pair of dinners "fight," and the winner is the is the dinner which the majority of contestants prefer. There are $1024cdot 1023/2$ fights, so there are that many wins. Since there are $1024$ dinners, each dinner wins an average of $1023/2=511.5$ fights. Some dinner must win at least that average many fights, so some dinner $d$ has won $512$ fights. Include that dinner in the menu; the $512$ dinners that $d$ will be off the menu, which is OK since they are preferred less than $d$.
We are now left with a smaller power of two. Repeat this process with these $512$ dinners, then $256$, then $128$, etc, selecting one dinner at each level.
answered 2 days ago
Mike EarnestMike Earnest
20.6k11950
The same method lets you find a menu of $10$ from among up to $2046$ dinners. The dinner which won the most fights will have won at least $1023$ fights. Ignoring that dinner and the dinners it beat, we are left with $1022$ dinners. Rinse and repeat. In general, a menu of $m$ suffices for $2^{m+1}-2$ dinners.
– Mike Earnest
13 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Have a tournament where each pair of dinners "fight," and the winner is the is the dinner which the majority of contestants prefer. There are $1024cdot 1023/2$ fights, so there are that many wins. Since there are $1024$ dinners, each dinner wins an average of $1023/2=511.5$ fights. Some dinner must win at least that average many fights, so some dinner $d$ has won $512$ fights. Include that dinner in the menu; the $512$ dinners that $d$ will be off the menu, which is OK since they are preferred less than $d$.
We are now left with a smaller power of two. Repeat this process with these $512$ dinners, then $256$, then $128$, etc, selecting one dinner at each level.
answered 2 days ago
Mike EarnestMike Earnest
20.6k11950
The same method lets you find a menu of $10$ from among up to $2046$ dinners. The dinner which won the most fights will have won at least $1023$ fights. Ignoring that dinner and the dinners it beat, we are left with $1022$ dinners. Rinse and repeat. In general, a menu of $m$ suffices for $2^{m+1}-2$ dinners.
– Mike Earnest
13 hours ago
add a comment |
Have a tournament where each pair of dinners "fight," and the winner is the is the dinner which the majority of contestants prefer. There are $1024cdot 1023/2$ fights, so there are that many wins. Since there are $1024$ dinners, each dinner wins an average of $1023/2=511.5$ fights. Some dinner must win at least that average many fights, so some dinner $d$ has won $512$ fights. Include that dinner in the menu; the $512$ dinners that $d$ will be off the menu, which is OK since they are preferred less than $d$.
We are now left with a smaller power of two. Repeat this process with these $512$ dinners, then $256$, then $128$, etc, selecting one dinner at each level.
answered 2 days ago
Mike EarnestMike Earnest
20.6k11950
The same method lets you find a menu of $10$ from among up to $2046$ dinners. The dinner which won the most fights will have won at least $1023$ fights. Ignoring that dinner and the dinners it beat, we are left with $1022$ dinners. Rinse and repeat. In general, a menu of $m$ suffices for $2^{m+1}-2$ dinners.
– Mike Earnest
13 hours ago
add a comment |
Have a tournament where each pair of dinners "fight," and the winner is the is the dinner which the majority of contestants prefer. There are $1024cdot 1023/2$ fights, so there are that many wins. Since there are $1024$ dinners, each dinner wins an average of $1023/2=511.5$ fights. Some dinner must win at least that average many fights, so some dinner $d$ has won $512$ fights. Include that dinner in the menu; the $512$ dinners that $d$ will be off the menu, which is OK since they are preferred less than $d$.
We are now left with a smaller power of two. Repeat this process with these $512$ dinners, then $256$, then $128$, etc, selecting one dinner at each level.
answered 2 days ago
Mike EarnestMike Earnest
20.6k11950
Have a tournament where each pair of dinners "fight," and the winner is the is the dinner which the majority of contestants prefer. There are $1024cdot 1023/2$ fights, so there are that many wins. Since there are $1024$ dinners, each dinner wins an average of $1023/2=511.5$ fights. Some dinner must win at least that average many fights, so some dinner $d$ has won $512$ fights. Include that dinner in the menu; the $512$ dinners that $d$ will be off the menu, which is OK since they are preferred less than $d$.
We are now left with a smaller power of two. Repeat this process with these $512$ dinners, then $256$, then $128$, etc, selecting one dinner at each level.
answered 2 days ago
Mike EarnestMike Earnest
20.6k11950
answered 2 days ago
Mike EarnestMike Earnest
20.6k11950
answered 2 days ago
Mike EarnestMike Earnest
20.6k11950
answered 2 days ago
Mike EarnestMike Earnest
20.6k11950
20.6k11950
The same method lets you find a menu of $10$ from among up to $2046$ dinners. The dinner which won the most fights will have won at least $1023$ fights. Ignoring that dinner and the dinners it beat, we are left with $1022$ dinners. Rinse and repeat. In general, a menu of $m$ suffices for $2^{m+1}-2$ dinners.
– Mike Earnest
13 hours ago
add a comment |
The same method lets you find a menu of $10$ from among up to $2046$ dinners. The dinner which won the most fights will have won at least $1023$ fights. Ignoring that dinner and the dinners it beat, we are left with $1022$ dinners. Rinse and repeat. In general, a menu of $m$ suffices for $2^{m+1}-2$ dinners.
– Mike Earnest
13 hours ago
The same method lets you find a menu of $10$ from among up to $2046$ dinners. The dinner which won the most fights will have won at least $1023$ fights. Ignoring that dinner and the dinners it beat, we are left with $1022$ dinners. Rinse and repeat. In general, a menu of $m$ suffices for $2^{m+1}-2$ dinners.
– Mike Earnest
13 hours ago
The same method lets you find a menu of $10$ from among up to $2046$ dinners. The dinner which won the most fights will have won at least $1023$ fights. Ignoring that dinner and the dinners it beat, we are left with $1022$ dinners. Rinse and repeat. In general, a menu of $m$ suffices for $2^{m+1}-2$ dinners.
– Mike Earnest
13 hours ago
add a comment |
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