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Jun 18, 2020, 11:03:33 AM6/18/20

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I stumbled on a domain of mathematics which is

about as simple as imaginable. It might seem like graph

theory but is much too simple even to be called that.

It might be a good toy domain to kindle youngsters' interest.

Definition: A Flock is a finite non-null set of chickens.

(Think of 'chicken' as an abstract object; never mind

whether the following apply to real chickens or their pecking.)

Property of 'Peck': For any chickens x, y in a Flock,

exactly one of the following is true:

(a) x Pecks y

(b) y Pecks x

(c) x = y

Remark: Pecking need not be transitive. If x Pecks y

and y Pecks z, it may be the case that z Pecks x.

Definition: K is an Emperor of Flock F if and only if

K∈F and

K Pecks every chicken in the Flock except K himself.

Definition: K is a King of Flock F if and only if

K∈F and

for every z∈F who Pecks K, there is some

y∈F with K Pecks y and y Pecks z.

Theorem 1: Every Flock has at least one King.

Theorem 2: A Flock has an Emperor if and only if it has exactly one King.

Theorem 3: ---- [Fill in the blank -- a "secret lemma"!]

Theorem 4: No Flock has exactly two Kings.

Theorems 1 and 2 are easy. The challenge is to prove Theorem 4.

What makes this interesting is that there is a Theorem 3 which is

easier to prove than Theorem 4, and for which Theorem 4 is an easy

corollary. Can you (or a youngster) guess the secret Theorem 3

which assists this proof?

(Or -- also a fun challenge -- prove Theorem 4 without discovering

and applying the "secret lemma.")

James D. Allen

about as simple as imaginable. It might seem like graph

theory but is much too simple even to be called that.

It might be a good toy domain to kindle youngsters' interest.

Definition: A Flock is a finite non-null set of chickens.

(Think of 'chicken' as an abstract object; never mind

whether the following apply to real chickens or their pecking.)

Property of 'Peck': For any chickens x, y in a Flock,

exactly one of the following is true:

(a) x Pecks y

(b) y Pecks x

(c) x = y

Remark: Pecking need not be transitive. If x Pecks y

and y Pecks z, it may be the case that z Pecks x.

Definition: K is an Emperor of Flock F if and only if

K∈F and

K Pecks every chicken in the Flock except K himself.

Definition: K is a King of Flock F if and only if

K∈F and

for every z∈F who Pecks K, there is some

y∈F with K Pecks y and y Pecks z.

Theorem 1: Every Flock has at least one King.

Theorem 2: A Flock has an Emperor if and only if it has exactly one King.

Theorem 3: ---- [Fill in the blank -- a "secret lemma"!]

Theorem 4: No Flock has exactly two Kings.

Theorems 1 and 2 are easy. The challenge is to prove Theorem 4.

What makes this interesting is that there is a Theorem 3 which is

easier to prove than Theorem 4, and for which Theorem 4 is an easy

corollary. Can you (or a youngster) guess the secret Theorem 3

which assists this proof?

(Or -- also a fun challenge -- prove Theorem 4 without discovering

and applying the "secret lemma.")

James D. Allen

Nov 22, 2020, 7:19:16 AM11/22/20

to

https://sites.lafayette.edu/liebnerj/files/2014/10/LVAIC2013.pdf

--

"All things extant in this world,

Gods of Heaven, gods of Earth,

Let everything be as it should be;

Thus shall it be!"

- Magical chant from "Magical Shopping Arcade Abenobashi"

"Drizzle, Drazzle, Drozzle, Drome,

Time for this one to come home!"

- Mr. Wizard from "Tooter Turtle"

Jan 14, 2021, 4:48:29 PM1/14/21

to

on the 'Net and elsewhere. Your cite asks for a proof of Theorem 1.

This is easy by induction: Just show that if K is a King of Flock F, then

either K or T will be a King of (F union {T}).

More interesting is Theorem 4. What makes this "fun" is that, as I wrote,

> > The challenge is to prove Theorem 4.

> > What makes this interesting is that there is a Theorem 3 which is much
> > easier to prove than Theorem 4, but for which Theorem 4 is an easy

> > corollary. Can you guess the secret Theorem 3

> > which assists this proof?

jamesdowallen at Gmail

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