Choosing a menu for a dinner party
$begingroup$
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 Jan 23 at 20:45
Mike Earnest
27.3k22152
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,82222639
$endgroup$
add a comment |
$begingroup$
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 Jan 23 at 20:45
Mike Earnest
27.3k22152
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,82222639
$endgroup$
add a comment |
$begingroup$
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 Jan 23 at 20:45
Mike Earnest
27.3k22152
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,82222639
$endgroup$
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 Jan 23 at 20:45
Mike Earnest
27.3k22152
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,82222639
edited Jan 23 at 20:45
Mike Earnest
27.3k22152
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,82222639
edited Jan 23 at 20:45
Mike Earnest
27.3k22152
edited Jan 23 at 20:45
Mike Earnest
27.3k22152
edited Jan 23 at 20:45
Mike Earnest
27.3k22152
27.3k22152
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,82222639
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,82222639
asked Nov 15 '15 at 4:38
Qiang LiQiang Li
1,82222639
1,82222639
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 5 at 16:54
Mike EarnestMike Earnest
27.3k22152
$endgroup$
$begingroup$
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.
$endgroup$
– Mike Earnest
Jan 7 at 18:33
1
$begingroup$
There seems to be a word missing: "the $512$ dinners that $d$ [beat?] will be off the menu".
$endgroup$
– Théophile
Jan 23 at 21:43
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
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 Jan 5 at 16:54
Mike EarnestMike Earnest
27.3k22152
$endgroup$
$begingroup$
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.
$endgroup$
– Mike Earnest
Jan 7 at 18:33
1
$begingroup$
There seems to be a word missing: "the $512$ dinners that $d$ [beat?] will be off the menu".
$endgroup$
– Théophile
Jan 23 at 21:43
add a comment |
$begingroup$
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 Jan 5 at 16:54
Mike EarnestMike Earnest
27.3k22152
$endgroup$
$begingroup$
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.
$endgroup$
– Mike Earnest
Jan 7 at 18:33
1
$begingroup$
There seems to be a word missing: "the $512$ dinners that $d$ [beat?] will be off the menu".
$endgroup$
– Théophile
Jan 23 at 21:43
add a comment |
$begingroup$
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 Jan 5 at 16:54
Mike EarnestMike Earnest
27.3k22152
$endgroup$
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 Jan 5 at 16:54
Mike EarnestMike Earnest
27.3k22152
answered Jan 5 at 16:54
Mike EarnestMike Earnest
27.3k22152
answered Jan 5 at 16:54
Mike EarnestMike Earnest
27.3k22152
answered Jan 5 at 16:54
Mike EarnestMike Earnest
27.3k22152
27.3k22152
$begingroup$
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.
$endgroup$
– Mike Earnest
Jan 7 at 18:33
1
$begingroup$
There seems to be a word missing: "the $512$ dinners that $d$ [beat?] will be off the menu".
$endgroup$
– Théophile
Jan 23 at 21:43
add a comment |
$begingroup$
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.
$endgroup$
– Mike Earnest
Jan 7 at 18:33
1
$begingroup$
There seems to be a word missing: "the $512$ dinners that $d$ [beat?] will be off the menu".
$endgroup$
– Théophile
Jan 23 at 21:43
$begingroup$
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.
$endgroup$
– Mike Earnest
Jan 7 at 18:33
$begingroup$
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.
$endgroup$
– Mike Earnest
Jan 7 at 18:33
1
1
$begingroup$
There seems to be a word missing: "the $512$ dinners that $d$ [beat?] will be off the menu".
$endgroup$
– Théophile
Jan 23 at 21:43
$begingroup$
There seems to be a word missing: "the $512$ dinners that $d$ [beat?] will be off the menu".
$endgroup$
– Théophile
Jan 23 at 21:43
add a comment |
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