Replace Rock, Paper, Scissors
Carl G.
you come up with other good examples of three object sets, A, B, and C, in
which A beats B, B beats C, and C beats A? How about a set of four or more
objects? Since there is no "correct answer", no spoiler space should be
necessary.
Example (maybe not a great one, but it gives the idea):
Water wets the Gunpowder
Gunpowder blows up the Sponge
Sponge soaks up the Water
What are the best examples you can find?
Carl G.
Dick Christoph
I dont think four will work as well.
The intrinsic beauty of the Rock, Paper Scissors game is that for two
players, unless they choose the same item, the game has a winner.
Suppose in a game of four choices:
A Beats B
B Beats C
C Beats D
D Beats A.
If the two players choose A and C who wins?
BTW, how about a slight enhamcement.
> Water puts out fire
> Fire Burns Sponge
> Sponge soaks up the Water
or
: JackHammer Breaks into Jail
. Jail imprisons Criminal
. Criminal steals JackHammer
or
. Electricity Oxidizes Water (Converts it to Hydrogen and Oxygen)
. Water Shorts out Switch
. Switch Turns off Electricity
or
. Elephant steps on Cat
. Cat Eats Mouse
. Mouse scares Elephant.
or
. Happiness Beats Money (as in "Money Can't Buy Happiness")
. Money Beats Nothing (Having Money Beats Having Nothing)
. Nothing Beats Happiness
Hmm, That was kind of Zen-like.
Wow this is kind of fun.
--
-Dick Christoph
dchr...@minn.net
http://www1.minn.net/~dchristo
Carl G.
Dick Christoph wrote in message <39943FCF...@minn.net>...
>Hi Carl,
>
>I dont think four will work as well.
>
>The intrinsic beauty of the Rock, Paper Scissors game is that for two
>players, unless they choose the same item, the game has a winner.
>
>Suppose in a game of four choices:
>A Beats B
>B Beats C
>C Beats D
>D Beats A.
>
>If the two players choose A and C who wins?
I agree that using four objects to play the traditional game won't work very
well. I was mainly curious about what objects people would suggest for
cycles of four or more. The board game "Statego" is at least one game that
uses non-associative (correct me if this is the wrong term) rules. In
"Stratego" the higher numbered men beat all lower numbered men. The highest
numbered man can be beaten by a "spy", who can be beaten by anyone else.
Carl G.
>
>"Carl G." wrote:
>>
>> I recently posted a puzzle based on the game "Rock, Paper, Scissors".
Can
>> you come up with other good examples of three object sets, A, B, and C,
in
>> which A beats B, B beats C, and C beats A? How about a set of four or
more
>> objects? Since there is no "correct answer", no spoiler space should be
>> necessary...
...
>> Carl G.
Estraven
>
> Dick Christoph wrote in message <39943FCF...@minn.net>...
> >Hi Carl,
> >
> >I dont think four will work as well.
> I agree that using four objects to play the traditional game won't work
very
> well. I was mainly curious about what objects people would suggest for
> cycles of four or more. The board game "Statego" is at least one game
that
> uses non-associative (correct me if this is the wrong term) rules. In
> "Stratego" the higher numbered men beat all lower numbered men. The
highest
> numbered man can be beaten by a "spy", who can be beaten by anyone else.
I believe the word you're after is "nontransitive". Transitivity is that
property which allows us, from "a > b" and "b > c", to infer that "a > c".
Associativity is not a property of relations such as ">" or "beats" in RPS,
but of operators such as "+" and "*", specifically the property that
a+(b+c) = (a+b)+c always.
Carl G.
Estraven wrote in message <3j_k5.48224$qS3....@tor-nn1.netcom.ca>...
Thanks, "transitivity" was what I was trying to think of, but I drew a
blank.
Carl G.
dus...@gate.sinica.edu.tw
> I recently posted a puzzle based on the game "Rock, Paper, Scissors".
> Can you come up with other good examples of three object sets, A,
> B, and C, in which A beats B, B beats C, and C beats A? How about
> a set of four or more objects? Since there is no "correct answer",
> no spoiler space should be necessary.
Actually, four objects in a game works OK. You just get a draw half of
the time, instead of a third of the time. No big deal. Here in Taiwan,
people (well, children) often play "Stick hits tiger, chicken eats
worm." My wife, who is Taiwanese, for some reason says "Stick hits
tiger, chicken eats dog," which does not work as well.
Jonathan Dushoff
Patrick Hamlyn
>I recently posted a puzzle based on the game "Rock, Paper, Scissors". Can
>you come up with other good examples of three object sets, A, B, and C, in
>which A beats B, B beats C, and C beats A? How about a set of four or more
>objects? Since there is no "correct answer", no spoiler space should be
>necessary.
>
>Example (maybe not a great one, but it gives the idea):
>
>Water wets the Gunpowder
>Gunpowder blows up the Sponge
>Sponge soaks up the Water
>
As kiddies, we used to add dynamite (with fuse) to the usual three
Scissors cuts fuse
paper smothers fuse
dynamite blows up rock
It never seemed to work as well as the normal game :>
I seem to remember water coming in as well at some point - rusts scissors, puts
out fuse, is soaked up by paper and I can't remember what happened with rock. As
you can imagine, dynamite quickly became an unpopular choice.
Gerry Quinn
>Carl G. <cgi...@microprizes.com> wrote:
>
>> I recently posted a puzzle based on the game "Rock, Paper, Scissors".
>> Can you come up with other good examples of three object sets, A,
>> B, and C, in which A beats B, B beats C, and C beats A? How about
>> a set of four or more objects? Since there is no "correct answer",
>> no spoiler space should be necessary.
>
>the time, instead of a third of the time. No big deal. Here in Taiwan,
>people (well, children) often play "Stick hits tiger, chicken eats
>worm." My wife, who is Taiwanese, for some reason says "Stick hits
>tiger, chicken eats dog," which does not work as well.
>
>Jonathan Dushoff
How about:
Bird eats worm, Man eats bird, Worm eats Man.
- Gerry Quinn
Dennis Yelle
>
> "Carl G." <cgi...@microprizes.com> wrote:
>
> >I recently posted a puzzle based on the game "Rock, Paper, Scissors". Can
> >you come up with other good examples of three object sets, A, B, and C, in
> >which A beats B, B beats C, and C beats A?
>
>
> Die A has sides numbered 3-4-3-4-3-4.
>
> Die B has sides numbered 2-2-6-2-2-6.
>
> Die C has sides numbered 1-5-5-1-5-5.
>
> A beats B two times out of three, B beats C two times out of three, and C
> beats A two times out of three.
I don't think so.
B beats C 5 times out of 9.
Dennis Yelle
--
I am a computer programmer and I am looking for a job.
There is a link to my resume here:
http://www.fortunecity.com/roswell/barada/186/index.html
jsee...@prodigy.net
Carl G. <cgi...@microprizes.com> wrote in article
<8n1c82$mr5$2...@nntp9.atl.mindspring.net>...
> I recently posted a puzzle based on the game "Rock, Paper, Scissors".
Can
> you come up with other good examples of three object sets, A, B, and C,
in
> which A beats B, B beats C, and C beats A? How about a set of four or
more
> objects? Since there is no "correct answer", no spoiler space should be
> necessary.
>
> Example (maybe not a great one, but it gives the idea):
>
> Water wets the Gunpowder
> Gunpowder blows up the Sponge
> Sponge soaks up the Water
>
> What are the best examples you can find?
>
We used to play Man-Gun-Bear in PE many years ago. Man controls gun. Gun
shoots bear. Bear mauls man.
Paul H
please correct my spellings!]. This has a cycle much larger than four,
though I can't remember all the verses . . . Maybe somebody else does?!?
Paul
Gerry Quinn wrote in message <35cl5.14793$r4....@news.indigo.ie>...
>In article <8n2rs3$lsj$2...@news1.sinica.edu.tw>, dus...@gate.sinica.edu.tw
wrote:
>>Carl G. <cgi...@microprizes.com> wrote:
>>
>>> I recently posted a puzzle based on the game "Rock, Paper, Scissors".
>>> Can you come up with other good examples of three object sets, A,
>>> B, and C, in which A beats B, B beats C, and C beats A? How about
>>> a set of four or more objects? Since there is no "correct answer",
>>> no spoiler space should be necessary.
>>
Patrick A. O'Donnell
> . Elephant steps on Cat
> . Cat Eats Mouse
> . Mouse scares Elephant.
My preschoolers have a book called "The King, the Mice, and the
Cheese":
The mice are chased out of the palace by the cats.
The cats ... dogs.
... dogs ... lions.
... lions ... elephants.
... elephants ... mice.
And the king and the mice lived happily ever after.
- Pat
Estraven
> I recently posted a puzzle based on the game "Rock, Paper, Scissors".
Can
> you come up with other good examples of three object sets, A, B, and C,
in
> which A beats B, B beats C, and C beats A? How about a set of four or
more
> objects? Since there is no "correct answer", no spoiler space should be
> necessary.
There exists a poetic form, a "crown of sonnets". If you number the
sonnets 0..6, then the form demands (in addition to the usual constraints
on each sonnet) that the last line of sonnet n be the same as the first
line of sonnet (n+1) mod 7.
For example, see http://www.sonnets.org/donne.htm
Developing the hand signals might pose some difficulties.
Justin Leck
news:8n9js4$365$3...@news7.svr.pol.co.uk...
> Gerry's posting reminds me of the song "On Ilkley Moor Bah Tat" [somebody
> please correct my spellings!]. This has a cycle much larger than four,
> though I can't remember all the verses . . . Maybe somebody else does?!?
>
> Paul
I knew my book of drinking songs would come in useful one day...
ON ILKLEY MOOR BAHT 'AT
Wheear 'as tha been sin' I saw thee?
On Ilkley Moor Baht 'at
Wheear 'as tha been sin' I saw thee?
Wheear 'as tha been sin' I saw thee?
Wheear 'as tha been sin' I saw thee?
On Ilkley Moor Baht 'at
On Ilkley Moor Baht 'at
On Ilkley Moor Baht 'at
2. Tha's been a-courting Mary Jane.
3. Tha'll go and get thi death of cold.
4. Then we shall ha' to bury thee.
5. Then t' worms'll come an' eat thee up.
6. Then t' ducks'll come an' eat up t' worms.
7. Then we shall go an' eat up t' ducks.
8. Then we shall all have eaten thee.
Justin
Gerry Quinn
>My preschoolers have a book called "The King, the Mice, and the
>Cheese":
>
>The mice are chased out of the palace by the cats.
>The cats ... dogs.
>.... lions ... elephants.
>.... elephants ... mice.
>
>And the king and the mice lived happily ever after.
>
> - Pat
And how could we forget The Old Woman Who Swallowed A Fly?
There was an old woman who swallowed a fly.
I don't know why she swallowed the fly.
I think she'll die.
There was on old woman who swallowed a spider
That wiggled and wriggled and jiggled inside her.
She swallowed the spider to catch the fly
But I don't know why she swallowed the fly.
I think she'll die.
.. and so on via cats, dogs etc. In this song, however, she dies at
the end - the story isn't cyclic.
- Gerry Quinn
Rich Grise
>
> There was an old woman who swallowed a fly.
> I don't know why she swallowed the fly.
> I think she'll die.
>
> There was on old woman who swallowed a spider
> That wiggled and wriggled and jiggled inside her.
> She swallowed the spider to catch the fly
> But I don't know why she swallowed the fly.
> I think she'll die.
>
> .. and so on via cats, dogs etc. In this song, however, she dies at
> the end - the story isn't cyclic.
>
> - Gerry Quinn
Apparently you could forget!
I know an old woman who swallowed a fly.
I don't know why she swallowed a fly.
Perhaps she'll die.
I know an old woman who swallowed a spider
That wiggled and jiggled and tickled inside her.
She swallowed the spider to catch the fly
But I don't know why she swallowed the fly.
Perhaps she'll die.
I know an old woman who swallowed a bird.
How absurd, to swallow a bird.
She swallowed the bird to catch the spider
That wiggled and jiggled and tickled inside her.
She swallowed the spider to catch the fly
But I don't know why she swallowed the fly.
Perhaps she'll die.
I know an old woman who swallowed a cat.
Fancy that! To swallow a cat.
She swallowed the cat to catch the bird
She swallowed the bird to catch the spider
That wiggled and jiggled and tickled inside her.
She swallowed the spider to catch the fly
But I don't know why she swallowed the fly.
Perhaps she'll die.
I know an old woman who swallowed a dog.
Oh, what a hog! To swallow a dog.
She swallowed the dog to catch the cat
She swallowed the cat to catch the bird
She swallowed the bird to catch the spider
That wiggled and jiggled and tickled inside her.
She swallowed the spider to catch the fly
But I don't know why she swallowed the fly.
Perhaps she'll die.
I know an old woman who swallowed a cow.
I don't know how she swallowed a cow.
She swallowed the cow to catch the dog,
She swallowed the dog to catch the cat
She swallowed the cat to catch the bird
She swallowed the bird to catch the spider
That wiggled and jiggled and tickled inside her.
She swallowed the spider to catch the fly
But I don't know why she swallowed the fly.
Perhaps she'll die.
I know an old woman who swallowed a horse.
She died, of course.
Michael Cock
Rich Grise <rich...@vel.net> wrote in message
news:3999C0...@vel.net...
Something has concerned me ever since I had this story in a picture book
when I was a little boy, which was BTW, and excellent book - each page
included a hole larger than the hole on the previous page showing the
ever-increasing contents of the old lady. The last page (no hole) showed one
dead old woman with a menagerie inside her. But, back to my concern, the RPS
syndrome works fine up to the cat, but then it goes seriously skewed. OK,
you might argue, dogs chase cats, but don't generally eat them. But cows???
Cows chasing dogs??? What's that all about then? And why die when she's
eaten the horse? A whole cow would do it for me, and the horse is about the
same size. Surely the cow would have finished her off?
So can anyone think of a better succession of animals that this mad
gluttonous old bird could eat? For instance, starting with the fly:
Fly, spider, bird, cat, wolf, grizzly bear
or
Fly, spider, bird, bird-eating spider, bird, bird-eating spider
oops, open loop :)
Cheers,
Michael
Gerry Quinn
>Something has concerned me ever since I had this story in a picture book
>when I was a little boy, which was BTW, and excellent book - each page
>included a hole larger than the hole on the previous page showing the
>ever-increasing contents of the old lady. The last page (no hole) showed one
>dead old woman with a menagerie inside her. But, back to my concern, the RPS
>syndrome works fine up to the cat, but then it goes seriously skewed. OK,
>you might argue, dogs chase cats, but don't generally eat them. But cows???
>Cows chasing dogs??? What's that all about then? And why die when she's
>eaten the horse? A whole cow would do it for me, and the horse is about the
>same size. Surely the cow would have finished her off?
>
She should probably have waited a while after eating the cow!
>So can anyone think of a better succession of animals that this mad
>gluttonous old bird could eat? For instance, starting with the fly:
>
>Fly, spider, bird, cat, wolf, grizzly bear
>
How about fly, spider, bird, cat, vulture, bear. Remember, though, you
need a rhyme. Maybe:
"Oh, my ulcer! - to swallow a vulture", and
"There was an old woman who swallowed a bear
She died right there..."
Gerry Quinn
--
http://bindweed.com
Puzzle / Strategy Games and Kaleidoscope for Windows
Download evaluation versions free, no time limits
New: Unique 2-player strategy game "Zen"
Bill Taylor
> As kiddies, we used to add dynamite (with fuse) to the usual three
> Scissors cuts fuse
> paper smothers fuse
> dynamite blows up rock
>
> It never seemed to work as well as the normal game :>
Ha! I presume from your smiley you can detect the fairly obvious reason.
But this conceals an interesting point of generalization.
Everyone knows that randomizing 1/3 : 1/3 : 1/3 over scissors-paper-rock
is a sure-fire way to avoid loss, (though it doesn't net a gain). Quite
a few will have discovered, like Patrick, that adding in another type doesn't
radically alter the game. You either add in a universal winner or a universal
loser (boring!), or one (like dynamite above) that forms an inconsistent quad
with a couple of diagonals. This last type is only mildly interesting, in that
it always reduces to randomizing over the "best" three; and *this* time you
may actually average a gain - every time your opponent chooses the forbidden
one. It's only mildly interesting in that, the forbidden one is actually
dominated by one of the others, game-theory-wise.
However, if one increases the number of candidates to 5 or more, where as
before each candidate has some fixed win/loss result with respect to each other
candidate, and ties with one of its own, then we get some intriguing results.
(1) It may be that there are forbidden choices, which are nonetheless not
dominated by any other choice;
(2) The optimal strategy will always be some randomized mixture (though not
necessarily with equal probs) over some ODD number of choices!
The universality of the odd-parity of the optimal solution is truly surprising.
I posted this as a problem to sci.math many years ago, and got several
correct explanations. Perhaps you might like to tackle it yourselves?
-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------
Equilibrium - the state of a system after everything
interesting has already happened.
-------------------------------------------------------------------------------
James Dow Allen
mat...@math.canterbury.ac.nz (Bill Taylor) challenges:
> Everyone knows that randomizing 1/3 : 1/3 : 1/3 over
> scissors-paper-rock is a sure-fire way to avoid loss,...
> ... However, if one increases the number of candidates to 5 or more,
> where, as before, each candidate has some fixed win/loss result with
> respect to each other candidate, and ties with one of its own,
> then we get some intriguing results....
> The optimal strategy will always be some randomized mixture (though
> not necessarily with equal probs) over some ODD number of choices!
> The universality of the odd-parity of the optimal solution is truly
> surprising. I posted this as a problem to sci.math many years ago,
> and got several correct explanations.
> Perhaps you might like to tackle it yourselves?
Let M be the k-by-k game matrix with dominated (or "impossible")
strategies deleted and s its saddle-point solution vector. The well-known
game theorem states that M.s is constant, and in fact M.s = 0 since
the game is fair. Thus M is singular since s is nonzero. Bill Taylor
asks us to demonstrate that k is odd. It is sufficient to show that
with even k, M of the required form must have non-zero determinant.
M has zeros along the diagonal and +1 or -1 everywhere else. Its
determinant has k! terms but most of them are zero. There will be
A_k non-zero terms, each the product of only +1 and -1, and thus equal
to +1 or -1. If A_k is odd the determinant will be non-zero.
The best I can do for A_k is the recurrence
A_1 = 0
A_2 = 1
A_3 = 2
A_k = (k-1) {(k-1)A_(k-2) + (k-2)A_(k-3)}
and A_k is seen to be even when k is odd and vice versa. QED.
This proof seems "neat," but is still dissatisfying: is there a more intuitive
explanation?
Regards, James
> -------------------------------------------------------------------------------
> Equilibrium - the state of a system after everything
> interesting has already happened.
> -------------------------------------------------------------------------------
"In the long run, ... you're dead." --- John M. Keynes
For my real e-mail address, go to
http://freepages.genealogy.rootsweb.com/~jamesdow
Sent via Deja.com http://www.deja.com/
Before you buy.
Bill Taylor
>> scissors-paper-rock...
>> ...if one increases the number of candidates to 5 or more,
>>
>> The optimal strategy will always be some randomized mixture (though
>> not necessarily with equal probs) over some ODD number of choices!
> with even k, M of the required form must have non-zero determinant.
>
> M has zeros along the diagonal and +1 or -1 everywhere else. Its
> determinant has k! terms but most of them are zero. There will be
> A_k non-zero terms, each the product of only +1 and -1, and thus equal
> to +1 or -1. If A_k is odd the determinant will be non-zero.
> ... ...
> and A_k is seen to be even when k is odd and vice versa. QED.
Excellent! This is a simpler proof than those I received before. Thanks.
> The best I can do for A_k is the recurrence
> A_1 = 0
> A_2 = 1
> A_3 = 2
> A_k = (k-1) {(k-1)A_(k-2) + (k-2)A_(k-3)}
Not quite sure where you got that last line from. I figured it by this
double recursion:
A_(n+1) = #terms in matrix(n+1) = n B_n
where B_n is a matrix order nxn with zeros in a diagonal just off the main diag
and #terms in B_(n+1) = A_n + n B_n
which easily gives A_(n+1) = n(A_n + A_(n-1))
from which one can see A_n is between (n-1)! and n!
As you say, it is trivial to prove the parity alternates; though in fact
the A_n series 0,1,2,9,44,265... *does* have at least a semi-closed form;
it is subfactorial(n), the number of no-fixed-point permutations of 1 to n.
A_n = n! ( 1/0! - 1/1! + 1/2! - 1/3! + ... +- 1/n! )
> This proof seems "neat," but is still dissatisfying: is there a more intuitive
> explanation?
Agreed on both counts. Your proof is very neat. But it *would* be nice
to get a "game-theoretical" solution, rather than a merely algebraic one!
Thanks again.
-------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
-------------------------------------------------------------------------------
Its not the winning, its the taking apart.
-------------------------------------------------------------------------------
James Dow Allen
> I recently posted a puzzle based on the game "Rock, Paper, Scissors". Can
> you come up with other good examples of three object sets, A, B, and C, in
> which A beats B, B beats C, and C beats A?
> How about a set of four or more objects?
Fire overcomes metal (melting)
Metal overcomes wood (chopping)
Wood overcomes earth (rooting)
Earth overcomes water (absorbing)
Water overcomes fire (quenching)
The Chinese imposed a second cycle on the same five elements.
Metal carries water
Water feeds wood
Wood fuels fire
Fire forms earth
Earth contains metal
James D. Allen