| [This is a *stunning* result in recreational math. My understanding
is that it was essentially cracked by J.H.Conway on a flight from the UK to
the US. I copy without permission a draft article that was put together by
Conway for the Cambridge University recreational math journal, 'Eureka'.
I've omitted most of the key table, and one proof. If anyone is really
desperate, I believe that it's fairly easy to generate the rest of the
table using what's given and the Splitting Theorem.]
THE WIERD AND WONDERFUL CHEMISTRY OF AUDIOACTIVE DECAY
J.H.Conway
INTRODUCTION
Suppose we start with a list of numbers (i.e. positive integers) [or
any symbols at all] say
5 5 5 5 5
We might describe this in words in the usual way as 'five fives', and
write down
5 5
as the *derived* list. This we describe as 'two fives', and so it yields the
next derived list
2 5
which is 'one two, one five', giving
1 2 1 5
namely 'one one, one two, one one, one five', or
1 1 1 2 1 1 1 5
and so on. What happens when an arbitrary list of positive integers is
repeatedly derived like this?
I note that more usually, one is given a sequence such as
55555 ; 55 ; 25 ; 1215 ; 11121115 ;
and asked to guess the generating rule or the next term. The history of such
problems is described elsewhere in this issue of 'Eureka'. [I don't have
a copy of this issue, I would guess that it's about 1984 or so.]
The numbers in our lists are usually single-digit ones, so we'll call
them digits, and usually try to cram them together as we have just done. But
occasionally, we want to indicate the way the numbers in the later lists are
derived, and we can do this neatly by inserting commas recalling the commas
and quotes in our verbal descriptions, thus:
5 5 5 5 5
,5 5,
,2 5,
,1 2,1 5,
,1 1,1 2,1 1,1 5,
... .
The insertion of these commas into a list or portion thereof is
called *parsing*.
We'll often denote repetitions by indices in the usual way
exponentiation, so that the derivation rule is
A B C D
a b c d ... -> AaBbCcDd...
When we do this, it is always understood that the repetitions are
collected maximally, so that here we must have
a <> b, b <> c, c <> d, ... .
Since what we write down is often only a *chunk* of the entire list
(that is a consecutive subsequence of its terms), we often use the square
brackets '[' or ']' to indicate that the apparent left or right end really
is the end. We also introduce the formal digits:
0 to give alternative ways of indicating the ends (see below)
X for an arbitrary digit, possibly 0, and
n' for any digit (maybe 0) other than n.
Thus
0 A B C X A B C A B C
X a b c or 0 a b c means the same as [a b c
A B C 0 A B C X A B C
a b c X or a b c 0 means the same as a b c ]
A B C 0' A B C
a b c X means a b c followed by at least one more digit
and A B C 0'
a b c 2' means this digit is not 2.
I'm afraid that this heap of conventions makes it quite hard to check
the proofs, since they cover more cases than [one] naively expects. To
separate these cases would make this article very long and tedious, and the
reader who really wants to check all the details is warned that he should
first spend some time practising the derivation process. He should note
that when we write
L -> L* -> L**
we mean just that every list of type L derives to one of type L*, every list
of type L* derives to one of type L**, and so on. So when in our proof of
the Ending Theorem we have
n n' 1
n ] ----> n ] -> n ]
(n<>2)
the fact that the left arrow is asserted only when n<>2 does not excuse us
from checking the right arrow for n = 2. (but we needn't check either of them
for n = 1, since n > 1 was explicitly enforced in the prior discussion).
By applying the drivation process n times to a list L, we obtain
what we call its nth descendant, L_n. The list itself is counted among its
descendants, as the 0th.
Sometimes a list factors as the product LR of two lists L,R whose
descendants never interfere with each other, in the sense that (LR)_n = L_nR_n
for all n. In this case, we say that LR *splits* as L.R (dots is lists will
always have this meaning). It is plain that this happens just when (L or R is
empty or) the last digit of L_n always differs from the first one of R_n.
Can you find a simple criterion for this? (When you give up, you'll find the
answer in our Splitting Theorem).
Obviously we call a list with no non-trivial splittings an *atom*, or
*element*. Then every list is a split product, or *compound* of a certain
number of elements, which we call the elements it *involves*. There are
infinitely many distinct elements, but most of them only arise from specially
chosen starting lists. However, there are some very interesting elements
that are involved in the descendants of *every* list except the boring ones
[] and [22].
Can you guess how many of them there are? (Hint: we have given them
the names
Hydrogen, Helium, Lithium, ..., Uranium !)
It's also true, but ASTONISHINGLY hard to prove!) that *every* list
eventually decays into a compound of these common elements, together perhaps
with a few others (namely Neptunium and Plutonium, which each have various
isotopes). Moreover, all lists except the two boring ones ultimately
increase in length exponentially at the same constant rate. (This rate is
roughly 1.36538482: it can be precisely defined as the largest root of a
certain algebraic equation of degree 71.) Also, the relative abundance of
the elements settle down to fixed numbers (zero for Neptunium and Plutonium).
THus of every million atoms, on average almost exactly 117329 will be of
Hydrogen, the commonest element, while rather fewer than 7 will be of Arsenic,
the rarest one.
You should get to know the common elements, as enumerated in our
Periodic Table. The abundance (in atoms per million) is given first,
followed by the atomic number and symbol as in ordinary chemistry. The actual
list of numbers defining the element is the numerical part of the remainder
of the entry, which if read in full gives the derivate of the element of next
higher atomic number, split into atoms. Thus for example the last line of
the periodic table tells us that Hydrogen (H) is our name for the number-list
22, and that the next higher element (He) derives to the compound
Hf.Pa.H.Ca.Li
which we might call
"Hafnium-Proactinium-Hydrogen-Calcium-Lithide" !
Not everything is in the Periodic Table! For instance, the single
digit list [1] isn't. But watch: 1
11
21
1211
111221
312211
13112221
11132.13211 = Hf.Sn
After a few moves it has become Hafnium Stannide! This is an instance
of our "Universal Theorem".
abundance: n E_n E_n inside the derivate of E_n+1
================================================================
32.7267131 92 U 3
4731.13472 91 Pa 13
3465.05591 90 Th 1113
89 Ac 3113
88 Ra 132113
87 Fr 1113122113
86 Rn 311311222113
85 At Ho.1322113
84 Po 1113222113
83 Bi 3113322113
82 Pb Pm.123222113
81 Tl 111213322113
80 Hg 31121123222113
79 Au 132112211213322113
78 Pt 111312212221121123222113
(Good grief, I'm not typing all this lot in. Work it out for yourselves.)
1274.99013 2 He 13112221133211322112211213322112
117328.997 1 H Hf.Pa.22.Ca.Li
================================================================
THE THEORY
Start with some easy theorems that restrict the possible lists after
the first few moves. Any chunk of a list that has lasted at least n moves
will be called an n-day-old list.
THE ONE-DAY THEOREM. Chunks of type
4+ 3 3
,a x,b x, x x y
don't happen in day old lists.
a b
Proof: The first possibility comes from x x , which however should have been
a+b
written x , in the previous day's list. The other two, however parsed,
imply cases of the first.
THE TWO-DAY THEOREM. No digit 4 or more can be born on the second or a later
day. Also, a chunk 3X3 (in particular 333) can't appear in a 2-day-old list.
4+
Proof: The first possibility comes from a chunk x , while the second, which
3 3
we now know we must parse ,3 x,3 y, can only come from a chunk x y , of the
previous day's list.
When tracking particular lists we'll use these facts without explicit
mention.
THE STARTING THEOREM. Let R be a chunk of any 2-day-old list, considered as
a list in its own right. The its descendents ultimately cycle in one of
the four ways:
[]
1 1 3 1 3'
[1 X -> [1 -> [3 X
2
[2 ]
2 1 1 2 3 2 1 3'
[2 1 X -> [2 1 -> [2 3 X
If R is not already *in* such a cycle, it has descendents that begin
with three distinct digits.
2
Proof: If R is non-empty and doesn't start with 2 , then it
*either* starts with 1 and is of one of the types:
1 (0or1) 1 2 3 2 2 (1 or 1') 3
[1 X or [1 (2 or 2 or 3 ) or [1 X or [1
*or* starts with 2, and is one of the types
1 (2 or 2') 3
[2 X or [2
*or* starts with 3, and is one of the types:
1 (3 or 3') 2 (3 or 3')
[3 X or [3 X
1
*or* starts with some n > 3 and has form [n .
It is therefore visible in
1
2 3 2 3' [n --------+
1 3 [3 X ------+ 1 [3 X ------+ | 3
[1 2 ----- + | [1 ]-------+ | | [2 ------+
| | | | | |
v v v v v v
1 3 1 2 2 1 1 2 1 2 2 1' 1 2' 1 1 3 1 3'
[3 X -> [1 3 -> [1 X -> [2 X -> [1 2 -> [1 X -> [2 X -> [1 X -> [1 -> [3 X
^ v
| |
+-------<-----+
which establishes the desired results for it.
2
This proves the Theorem excpet for lists of type [2 R*, all of whose
decendents also start with 22. this happens only if no descendent of R*
starts with a 2, and so we can complete the proof by applying the results we've
just found to R*.
THE SPLITTING THEOREM. A 2-day-old list LR splits as L.R just if one of L and
R is empty or L and R are of the types shown in one of:
L | R
----------------+-------------------------------
n] | [m (n >= 4, m =< 3)
| 1 1 3 1 3' 1
2] | [1 X or [1 or [3 X or [n (n >= 4)
| 2 1 1 2 3 2 1 3' 2
2'] | [2 1 X or [2 1 or [2 3 X or [2 ]
Proof: This follows immediately from the splitting theorem, applied to R, and
the obvious fact that the last digit of L is constant.
THE ENDING THEOREM. The end of a list ultimately cycles in one of the ways:
2.Tc] -> 2.Mo]
^ |
| v
2.Zr] <- 2.Nb]
2.Pu] <-> 2.Np]
2.Li] <-> 2.He]
2.W] <-> 2.Ta]
2'.22] -> self
Proof: I don't want to type it in.
We are now ready for our first major result:
THE CHEMICAL THEOREM.
(a) the descendants of any of the 92 elements in our Periodic Table are
compounds of those elements.
(b) all sufficiently late descendants of any of these elements other than
Hydrogen involve *all* of the 92 elements.
(c) The descendents of *any* list other than [] or [22] ultimately involve
*all* of those 92 elements simultaneously.
(d) These 92 elements are precisely the common elements as defined in the
introduction.
Proof: (a) follows instantly from the form of the Periodic table.
(b) It also follows that if the element E_n of atomic number n appears
at some time t, then all of the elements of any line m < n will appear at the
later time t+n-m. In particular.
E_n at t => Hf & Li at t+n-1 (if n >= 2)
Hf & Li at t => Hf & Li at t+2 & t+71
Hf at t => Sr & U at t+72-38
U at t => E_n at t+92-n
From these we successively deduce that if any of these 92 elements other than
Hydrogen is involved at some time t_0, hafnium and Lithium will simultaneously
be involved at *some* strictly later time =< t_0 + 100, and then both these
will exist at *all* times >= t_0 + 200, Uranium at at times >= t_0 + 300, and
*every* one of these 92 elements at all times >= t_0 + 400.
(c) If L is not of the form L*22], this now follows from the
observation that Calcium (number-list 12) is a descendant of L, since it appears
in both the bottom lines of Figure 1. Otherwise we can replace L by L*, which
does not end in a 2.
(d) follows instantly from (a),(b),(c) & the definition of common
elements.
Now e'll call an arbitrary list *common* just if it's a compound of
common atoms.
THE ARITHMETICAL THEOREM
(a) The lengths of all common lists other than boring old [] and [22]
increase exponentially at the same rate � > 1.
(b) The relative abundances of the elements in such lists tend to
certain fixed values, all strictly positive.
Note: Since each common element has at least 1 and at most 42 digits, we can
afford to measure the length by either digits or atoms - we prefer to use
atoms. The numerical value of � is 1.36538482; the abundances are tabulated
in the Periodic Table.
Proof: Let v be the 92-component vector whose (i)-entry is the number of
atoms of atomic number i in some such list. Then at each derivation
step, v is multiplied by the matrix whose (i,j)-entry is the numver of times
E_j is involved in the derivate of E_i. Now our Chemical Theorem shows that
some power of M has strictly positive (i,j) entries for all i = 1 (the
(1,j)-entry will be 0 for j <> 1, 1 for j = 1, since every descendant of a
single atom of Hydrogen is another such).
Let � be an eigenvalue of M with the largest possible modulus, and v_0
a corresponding eigenvector. Then the non-zero entries of (v_0)M^n are
proportional to �^n, while the non-zero entries in the successive images of
all other vectors have sizes that increase at *at most* this rate. Since
the 92 coordinate vectors (which we'll call H, He, ..., U in the obvious way)
spand the space, at least one of them must increase at rate �. On the other
hand, our Chemical Theorem shows that the descendants of *each* of He, Li, ...
U increase as fast as *any* of them, and this is at some rate > 1, while H
is a fixed vector (rate 1). These remarks establish our Theorem.
We have essentially proved the Frobenius-Perron Theorem, that the
dominant eigenvalue of a matrix with positive entries is positive and occurs
just once, but I don't want to frighten you with these long names.
The Transuranic Elements
For eah number n > 1, we define two particular atoms :-
an isotope of Plutonium (Pu) : 31221132221222112112322211n
an isotope of Neptunium (Np) : 1311222113321132211221121332211n
For n = 2 these are Lithium (Li) and Helium (He); for n = 3 they are Tungsten
(W) and Tantalum (Ta), while for n >=4 they are the transuranic elements.
We won't bother to specify the number n in our notation. We can enlarge our
92-dimensional vector space by adding any number of new pairs of co-ordinate
vectors Pi,Np corresponding to pairs of transuranic elements.
Our proof of the Ending Theorem shows that every digit 4 or more
ultimately ends up as the last digit in one of the appropriate pair of
Transuranic Elements, and 9see the bottom left of Figure 1) that we have the
decomposition
Pu -> Np -> Hf.Pa.H.Ca.Pu
Now Pu + Np is an eigenvector of eigenvalue �1 modulo the subspace
corresponding to the common elements, since Pu <-> Np modulo that space.
Because these eigenvalues are strictly less than � in modulus, the relative
abundances of the transuranic elements tend to 0.
So far, I can proudly say that this magnificent theory is essentially
all my own work. But the next theorem, the finest achievement so far in
Audioactive Chemistry, is the result of the combined labours of three
brilliant workers.
THE UNIVERSAL THEOREM *Any* list decays into a compound of common and
transuranic elements after a bounded number of derivation steps. As a
consequence, *every* list other than the two boring ones increases at the
magic rate �, and the relative abundances of the atoms in its descendants
approach the values we have already described.
Proof: for this Universal Theorem would fill the rest of Eureka! Richard
Parker and I found the first proof over a period of about a month of very
intensive work (or rather, play!). We first produced a very subtle and
complicated argument which (almost) reduced the problem to tracking a few
hundred cases, and then handled these on dozens of sheets of paper (now lost).
Mike Guy later found a much simpler proof that used tracking and backtracking
in roughly equal proportions. Guys proof still took lots of pages, but at
least it yielded the critical number (26) of steps. Can *you* find a proof
in only a few pages? Please! [I have no idea whether this resistable
offer is still open.]
THE DEGREE OF �.
Plainly � is an algebraic number of degree at most 92. However, we
can reduce this bound to 71 by exhibiting a 21-dimensional invariant subspace
on which all the eigenvalues of M are 0 or �1. To see this, define the
vectors
v_1 = H, v_2 = He-Ta, v_3 = Li-W, ..., v_20 = Ca-Pa, but v21 = Sr-U,
or in atomic number notation
v_1 = E_1, v_2 = E_2-E_73, v_3 = E_3-E_74, ..., v_20 = E_20-E_91, but
v_21 = E_38-E_92,
and observe that
v_21 -> v_0 -> v_19 -> v_18 -> ... -> v_3 <-> v_2, v_1 -> v_1
An alternative basis for this space consists of the eigenvectors
v_1, v_3+v_2, v_3-v_2, v_21
of eigenvalues
1, 1, -1, 0,
together with the following Jordan block of size 0 and eigenvalue 0:-
v_20-v_18 -> v_19-v_17 -> ... -> v_5-v_3 -> v4-v_2 -> 0
This shows that M is one of those "infinitely rare" matrices that
can't be diagonalized. Don't expect to follow these remarks unless you've
understood more of linaer algebra than I fear most of your colleagues have!
[Conway in "lecturer" mode. He's actually one of the better teachers I
suffered.]
I haven't been able to compute any non-trivial explicit polynomial
satisfied by �. Can you do this? Nor do I know if the actual degree
of � is 71. Can you find out? [Again, this was written five years ago, and
I don't know if this is still open.]
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