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From: Leroy Quet on 16 Jul 2010 13:37
This is a game for any plural number of players.
(No grids this time, sorry.)

There are a number of rounds in this game. The number of rounds is a
multiple of the number of players. The players take turns being "the
permutator", where each player is the permutator the same number of
rounds.

Before starting any of the rounds, the players agree on a positive
integer n. (n is the same value for all rounds in the game.)

On a round, all of the players (including the permutator) start the
round by each coming up with an ordered list of n integers (positive,
negative, or 0). Each player's numbers must be distinct, in that no
integer occurs more than once in a particular player's list. The
players each keep their lists secret from the other players for now.

Let player p's list for any particular round be {b(p,k)}, k =
1,2,3,...n.

Next, all players who are not the permutator take turns choosing terms
a list of n distinct integers(positive, negative, or zero; no integer
more than once). (So, if the number of players is m, I guess to be
fair, n should be a multiple of (m-1).)

Let this list be {a'(k)}, k = 1,2,3,..n.

Then, after the list is complete, the permutator forms any permutation
{a(k)} of {a'(k)}.

Then everyone reveals their b-lists.

Each player p forms a sequence of n integers in this manner:

c(p,k) = sum{j=1 to n} a(n+1-j) b(p,j). k = 1,2,3...,n.

Player p's score for this round is the number of primes in {c(p,k)}.

After all the rounds are played, the players add up their scores for
all the rounds. The player with the largest grand score wins.

In a variation, instead of the number of primes being the criterion
for scoring, the players decide amongst themselves before each round
what will be the criterion for a number in the c-list to score.
Fibonacci numbers? Squares? Where each c(p,k) is coprime to c(p,k-1)?
Or maybe the choice of criterion should be totally up to the
permutator (with veto power from the other players), and expressed at
the beginning of each round before the b-lists are constructed.

Thanks,
Leroy Quet

From: Richard Heathfield on 16 Jul 2010 15:47
Leroy Quet wrote:
> This is a game for any plural number of players.
> (No grids this time, sorry.)

Who are you, and what have you done with the real Leroy Quet?

I looked through your game description, but by the end I was none the
wiser. Perhaps you are the real Leroy Quet after all. ;-)

<snip>

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
"Usenet is a strange place" - dmr 29 July 1999
Sig line vacant - apply within
From: Daniel S. on 18 Jul 2010 17:29
On 16 heinä, 20:37, Leroy Quet <qqq...(a)mindspring.com> wrote:
> This is a game for any plural number of players.
> (No grids this time, sorry.)
>
> There are a number of rounds in this game. The number of rounds is a
> multiple of the number of players. The players take turns being "the
> permutator", where each player is the permutator the same number of
> rounds.
>
> Before starting any of the rounds, the players agree on a positive
> integer n. (n is the same value for all rounds in the game.)
>
> On a round, all of the players (including the permutator) start the
> round by each coming up with an ordered list of n integers (positive,
> negative, or 0). Each player's numbers must be distinct, in that no
> integer occurs more than once in a particular player's list. The
> players each keep their lists secret from the other players for now.
>
> Let player p's list for any particular round be {b(p,k)}, k =
> 1,2,3,...n.
>
> Next, all players who are not the permutator take turns choosing terms
> a list of n distinct integers(positive, negative, or zero; no integer
> more than once). (So, if the number of players is m, I guess to be
> fair, n should be a multiple of (m-1).)
>
> Let this list be {a'(k)}, k = 1,2,3,..n.
>
> Then, after the list is complete, the permutator forms any permutation
> {a(k)} of {a'(k)}.
>
> Then everyone reveals their b-lists.
>
> Each player p forms a sequence of n integers in this manner:
>
> c(p,k) = sum{j=1 to n} a(n+1-j) b(p,j). k = 1,2,3...,n.
>
> Player p's score for this round is the number of primes in {c(p,k)}.
>
> After all the rounds are played, the players add up their scores for
> all the rounds. The player with the largest grand score wins.
>
> In a variation, instead of the number of primes being the criterion
> for scoring, the players decide amongst themselves before each round
> what will be the criterion for a number in the c-list to score.
> Fibonacci numbers? Squares? Where each c(p,k) is coprime to c(p,k-1)?
> Or maybe the choice of criterion should be totally up to the
> permutator (with veto power from the other players), and expressed at
> the beginning of each round before the b-lists are constructed.
>
> Thanks,
> Leroy Quet

I learned a game like this at the Israeli Jugging Convention this
year. You'd be great at it I'm sure.
From: Leroy Quet on 19 Jul 2010 07:05


Daniel S. wrote:

> I learned a game like this at the Israeli Jugging Convention this
> year. You'd be great at it I'm sure.

Hopefully "my" game is unique enough and different enough from that
other game that it still is mine!..

:)

Thanks,
Leroy Quet

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