Speed dating dinner parties


25-Feb-2016 06:40

Tasty New York Singles Dining Network Weekly dinners for six at a variety of New Yorks better restaurants attracting a fun, upscale group of single professionals in a choice of age groups. Uniform Dating Dating site for serving and former members of the uniformed services and armed forces or for those looking for a date ‘in uniform’.

Soldiers, sailors, nurses, firemen - Ooh, they're gorgeous!

Label the stationary person (in one corner) by $p_0$ and the people in the cycle in order counter-clockwise by $p_i$ for

Tasty New York Singles Dining Network Weekly dinners for six at a variety of New Yorks better restaurants attracting a fun, upscale group of single professionals in a choice of age groups. Uniform Dating Dating site for serving and former members of the uniformed services and armed forces or for those looking for a date ‘in uniform’.

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Tasty New York Singles Dining Network Weekly dinners for six at a variety of New Yorks better restaurants attracting a fun, upscale group of single professionals in a choice of age groups. Uniform Dating Dating site for serving and former members of the uniformed services and armed forces or for those looking for a date ‘in uniform’.

Soldiers, sailors, nurses, firemen - Ooh, they're gorgeous!

Label the stationary person (in one corner) by $p_0$ and the people in the cycle in order counter-clockwise by $p_i$ for $1 \leq i \leq 2n-1$, with $p_1$ opposite $p_0$ .

Clearly, $p_0$ meets everyone once and the other $p_i$ are symmetric; so it suffices to check that $p_1$ meets everyone.

Here is the list of partners for $p_1$, as everyone but $p_0$ moves counterclockwise each round: $p_0, p_, p_, \ldots, p_4, p_2, p_, p_, p_, \dots , p_5, p_3$. I changed the formatting in the answer from "vertical" (which wasn't working) to "horizontal", but feel free to revert.

BTW, this solution which you independently found is also present on Wikipedia.

A lot hinges on what you mean by "meet." Do you mean one-on-one conversation for a certain amount of time or would it be enough for, say, four people to introduce themselves to each other and do a quick activity?

\leq i \leq 2n-1$, with $p_1$ opposite $p_0$ .

Clearly, $p_0$ meets everyone once and the other $p_i$ are symmetric; so it suffices to check that $p_1$ meets everyone.

Here is the list of partners for $p_1$, as everyone but $p_0$ moves counterclockwise each round: $p_0, p_, p_, \ldots, p_4, p_2, p_, p_, p_, \dots , p_5, p_3$. I changed the formatting in the answer from "vertical" (which wasn't working) to "horizontal", but feel free to revert.

BTW, this solution which you independently found is also present on Wikipedia.

A lot hinges on what you mean by "meet." Do you mean one-on-one conversation for a certain amount of time or would it be enough for, say, four people to introduce themselves to each other and do a quick activity?

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If they are pretty familiar, you can group them by successively longer prefixes of their number in binary. This is an elegant and simple answer-- it is actually less confusing to follow than the "speed date" plan of successively dividing groups in half.

Within each group, the ones whose bit after the prefix is 0 are evens and the others odds. (Each person in the top row "meets" the corresponding person just below in the bottom row.) 1) (#1:4 #2:5 #3:6 #4:1 #5:2 #6:3) I tried to show the stages above. Draw it out: 1) draw six circles (a network graph), then draw a line w/ 1,2,3 on one side, 4,5,6 on the other. For each couplet (sitting across the line), make a line on the graph connecting them. Round 2, rotate all the numbers around the line, except 1 (it's in a box). For any group of 2n people, there will be 2n-1 rounds, with everyone meeting everyone else once.

Single New York: the complete singles and dating guide for New York, USA.

New York Singles Events, New York Singles Parties, New York Speed Dating, New York Singles Nights, New York Dating & Chat.

Everyone moves from seat $n$ to seat n (\mod 13)$ after each course.



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