Who can write the bigger number?
José Soto
September 18, 2008
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Disclaimer...
✔ Different sources in the web. I can provide them if you want.
✔ A big chunk of this comes from the Essay “Who can name the bigger number?” of
Scott Aaronson.
✔ You can talk with him in CSAIL.
✔ Probably not in his line of research.
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Let’s play a game.
✔ Every player is given a piece of paper.
✔ He/she must write the biggest number he/she can think of.
✔ The winner is the one that writes the biggest number.
5
3
4
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Possible outcomes
A billion
999999999999999999999999999999999999999999999
111111111111111111111111111111111111111111111111111111111111111111111111111111111
‘the sum of the numbers of my opponents plus one’
‘the biggest number whose description fits in this piece of paper’
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Problem of definition.
New rules:
✔ Using standard math notation, English words, or both. Name, using at most 100
symbols, a single whole number.
✔ Precise enough so that the number can be determined, by consulting only your
piece of paper and, if necessary, the published literature.
✔ A reasonable referee should be able to prove that the number is uniquely
determined and to compare between the answers.
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What can we do with English alone? (numbers in
‘words’)
Standard names
✔ Using the American and modern British “short scale” (billion = 109 ):
million
billion
trillion
quadrillion
quintillion
sextillion
septillion
octillion
nonillion
decillion
106
109
1012
1015
1018
1021
1024
1027
1030
1033
undecillion
duodecillion
tredecillion
quattuordecillion
quindecillion
sexdecillion
septendecillion
octodecillion
novemdecillion
vigintillion
1036
1039
1042
1045
1048
1051
1054
1057
1060
1063
n-th zillion
= 103(n+1) .
✔ Most dictionaries also include centillion. (10303 ).
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What can we do with English alone? (numbers in
‘words’)
Nonstandard names:
✔ (J. Conway and A. Wechsler) extended this system so that every integer has a
name in English:
✘ Combine latin prefixes for the first 1000 numbers and the suffix ‘illion’ to
express the n-th zillion.
✘ Use the prefix nill to express the zeroth ’zillion’.
✘ Group in blocks of thousands and concatenate the prefixes.
✘ e.g. billinillitrillion: would be the ‘two million, zero thousand and third’-zillion.
This is 103·(2.000.003+1) .
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What can we do with English alone? (numbers in
‘words’)
Nonstandard names:
✔ (J. Conway and A. Wechsler) extended this system so that every integer has a
name in English:
✘ Combine latin prefixes for the first 1000 numbers and the suffix ‘illion’ to
express the n-th zillion.
✘ Use the prefix nill to express the zeroth ’zillion’.
✘ Group in blocks of thousands and concatenate the prefixes.
✘ e.g. billinillitrillion: would be the ‘two million, zero thousand and third’-zillion.
This is 103·(2.000.003+1) .
✔ [Kasner and Newman 1940]
Googol = 10100 = 10 · 103·(32+1) = ten duotrigintillions. (c.f. 1090 and 10120 )
✔ Googolplex = 10googol .
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Googolplex
A Googolplex in the previous system would be written as:
Ten trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli
trestrigintatrecentilli trestrigintatrecentilli trestrigintatrecentilli duotringintatrecentillion.
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Using simple arithmetic.
✔ By only allowing sum and multiplication, we can’t do much right?
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Using simple arithmetic.
✔ By only allowing sum and multiplication, we can’t do much right?
✔ By allowing exponentiation, we get much bigger numbers:
99
(Try to express 9
9
without exponentiation.)
✔ Essentially any upper bound ever used in any mathematical proof can be written
concisely using just iterated exponentiation.
✔ First example of huge numbers used in a Mathematical proof:
Skewes numbers.
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Skewes Numbers
✔ Let π(x) : # primes ≤ x.
li(x) =
Z
0
x
dt
ln t
✔ Prime number theorem implies π(x) ∼ li(x).
✔ For all ‘small’ values of x, π(x) ≤ li(x). Want to know the smallest natural
number for which π(x) > li(x).
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Skewes Numbers
✔ Let π(x) : # primes ≤ x.
li(x) =
Z
0
x
dt
ln t
✔ Prime number theorem implies π(x) ∼ li(x).
✔ For all ‘small’ values of x, π(x) ≤ li(x). Want to know the smallest natural
number for which π(x) > li(x).
✔ Assuming Riemann Hypothesis. Skewes (1933) proved an upper bound of
79
ee
e
1010
≈ 10
34
for this number.
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Skewes Numbers
✔ Let π(x) : # primes ≤ x.
li(x) =
Z
0
x
dt
ln t
✔ Prime number theorem implies π(x) ∼ li(x).
✔ For all ‘small’ values of x, π(x) ≤ li(x). Want to know the smallest natural
number for which π(x) > li(x).
✔ Assuming Riemann Hypothesis. Skewes (1933) proved an upper bound of
79
ee
e
1010
34
≈ 10
for this number.
1010
3
✔ Without Riemann Hypothesis. Skewes (1955) proved an upper bound of 1010
for this number.
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Skewes Numbers
✔ Let π(x) : # primes ≤ x.
li(x) =
Z
0
x
dt
ln t
✔ Prime number theorem implies π(x) ∼ li(x).
✔ For all ‘small’ values of x, π(x) ≤ li(x). Want to know the smallest natural
number for which π(x) > li(x).
✔ Assuming Riemann Hypothesis. Skewes (1933) proved an upper bound of
79
ee
e
1010
34
≈ 10
for this number.
1010
3
✔ Without Riemann Hypothesis. Skewes (1955) proved an upper bound of 1010
for this number.
✔ Bays and Hudson (1999) proved that there are (many) crosses around
1.39822 · 10316 and conjectured that the first cross is not before 10167 .
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Beyond exponentiation.
✔ A product is a repeated sum. An exponentiation is just a repeated product. Want
to guess what’s next?
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Beyond exponentiation.
✔ A product is a repeated sum. An exponentiation is just a repeated product. Want
to guess what’s next?
✔ Tetration:
b
···a
aa
a= a
| {z }
b copies of a
✔ These numbers are HUGE:
6
66
6 = 66
66
>> Skewes numbers.
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Beyond exponentiation.
✔ A product is a repeated sum. An exponentiation is just a repeated product. Want
to guess what’s next?
✔ Tetration:
b
···a
aa
a= a
| {z }
b copies of a
✔ These numbers are HUGE:
6
66
6 = 66
66
>> Skewes numbers.
✔ More familiar, its inverse log∗a :
(log∗a n is the number of times you have to take loga from n to obtain something
smaller than 1).
✔ This function grows really slow.
✔ Fürer (2007) algorithm to multiply 2 n-bits integers runs in n log n · 2O(log
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∗
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n)
.
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Beyond exponentiation: Knuth’s arrow notation
✔ a ↑ b = ab . (exponentiation).
a ↑↑ b = b a (tetration).
a ···a
✔ a ↑↑↑ b = a ↑↑ a ↑↑ a ↑↑ · · · ↑↑ a = | {z a a} (pentation).
|
{z
}
b copies of a
b copies of a
✔ In general
a ↑n b = a ↑n−1 a ↑n−1 a ↑n−1 · · · ↑n−1 a .
|
{z
}
b copies of a
✔ Humongous numbers: 3 ↑↑↑ 3 = 3 ↑↑ (3 ↑↑ 3) = 3 ↑↑ (327 ) =
= A tower of 327 = 7.625.597.484.987 numbers 3.
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Primitive recursive functions.
✔ The functions that we have defined so far are all primitive recursive functions.
✔ These are the functions which can be defined using only the succesor function,
nested conditional (if-then-else) statements and fixed iteration (for) loops.
✔ These function tend to correspond closely with our intuition of what a computable
function must be.
✔ There are computable functions that are not primitive recursive.
✔ Simple example of non-primitive recursive function: The Ackermann function.
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The Ackermann function
Ackermann (1928) gave an example of a primitive (computable) function which is not
primitive recursive. His example was later simplified into the following function:


if m = 0
n + 1
A(m, n) = A(m − 1, 1)
if m > 0 and n = 0


A(m − 1, A(m, n − 1)) if m > 0 and n > 0.
✔ It is equivalent to A(m, n) = (2 ↑m−2 (n + 3)) − 3.
A(0)
A(1)
A(2)
✔
A(3)
=
=
=
=
1
2
2
2
+
·
↑
4 −3
5 −3
6 −3
A(4) = 2 ↑↑ 7 −3
=
=
=
=
A(n) = A(n, n).
3
7
61
7
2
22
2 − 3 = 22
22
−3
✔ In fact it can be proven that it grows faster than any primitive recursive function.
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Ackermann inverse
Perhaps more useful than the Ackermann function is its inverse:
✔ We can define an inverse α(x) = max{n : A(n) ≤ x}, thus, for any reasonable
number α(x) < 5.
✔ This function grows extremely slow and appears naturally in many contexts.
✔ Example: The fastest known Minimum Spanning Tree Algorithm [Chazelle 2000]
runs in ‘almost’ linear time O(mα(n)) where m is the number of edges and n is
the number of vertices of the graph. It is an open problem if it can be done in
O(m) time.
✔ Example: Davenport-Schinzel sequences.
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Aside: Graham Number
✔ Problem: Consider an n-dimensional hypercube, and connect each pair of vertices
to obtain a complete graph on 2n vertices. Is it possible to color each of the edges
of this graph using red and black such that no monochromatic K4 lies on a two
dimensional plane?
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Aside: Graham Number
✔ Problem: Consider an n-dimensional hypercube, and connect each pair of vertices
to obtain a complete graph on 2n vertices. What is the minimum n such that it is
impossible to color each of the edges of this graph using red and black such that
no monochromatic K4 lies on a two dimensional plane? Denote it by N .
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Aside: Graham Number
✔ Problem: Consider an n-dimensional hypercube, and connect each pair of vertices
to obtain a complete graph on 2n vertices. What is the minimum n such that it is
impossible to color each of the edges of this graph using red and black such that
no monochromatic K4 lies on a two dimensional plane? Denote it by N .
✔ N ≥ 6. Graham (1971) gave an upper bound for N involving a function similar to
Ackermann’s.
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Aside: Graham Number
✔ Problem: Consider an n-dimensional hypercube, and connect each pair of vertices
to obtain a complete graph on 2n vertices. What is the minimum n such that it is
impossible to color each of the edges of this graph using red and black such that
no monochromatic K4 lies on a two dimensional plane? Denote it by N .
✔ N ≥ 6. Graham (1971) gave an upper bound for N involving a function similar to
Ackermann’s.
✔ Let g1 = 3 ↑↑↑↑ 3,
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Aside: Graham Number
✔ Problem: Consider an n-dimensional hypercube, and connect each pair of vertices
to obtain a complete graph on 2n vertices. What is the minimum n such that it is
impossible to color each of the edges of this graph using red and black such that
no monochromatic K4 lies on a two dimensional plane? Denote it by N .
✔ N ≥ 6. Graham (1971) gave an upper bound for N involving a function similar to
Ackermann’s.
✔ Let g1 = 3 ↑↑↑↑ 3, g2 = 3 ↑g1 3.
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Aside: Graham Number
✔ Problem: Consider an n-dimensional hypercube, and connect each pair of vertices
to obtain a complete graph on 2n vertices. What is the minimum n such that it is
impossible to color each of the edges of this graph using red and black such that
no monochromatic K4 lies on a two dimensional plane? Denote it by N .
✔ N ≥ 6. Graham (1971) gave an upper bound for N involving a function similar to
Ackermann’s.
✔ Let g1 = 3 ↑↑↑↑ 3, g2 = 3 ↑g1 3. And in general gn = 3 ↑gn−1 3.
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Aside: Graham Number
✔ Problem: Consider an n-dimensional hypercube, and connect each pair of vertices
to obtain a complete graph on 2n vertices. What is the minimum n such that it is
impossible to color each of the edges of this graph using red and black such that
no monochromatic K4 lies on a two dimensional plane? Denote it by N .
✔ N ≥ 6. Graham (1971) gave an upper bound for N involving a function similar to
Ackermann’s.
✔ Let g1 = 3 ↑↑↑↑ 3, g2 = 3 ↑g1 3. And in general gn = 3 ↑gn−1 3.
✔ Graham (1971) proved that 6 ≤ N ≤ g64 .
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Aside: Graham Number
✔ Problem: Consider an n-dimensional hypercube, and connect each pair of vertices
to obtain a complete graph on 2n vertices. What is the minimum n such that it is
impossible to color each of the edges of this graph using red and black such that
no monochromatic K4 lies on a two dimensional plane? Denote it by N .
✔ N ≥ 6. Graham (1971) gave an upper bound for N involving a function similar to
Ackermann’s.
✔ Let g1 = 3 ↑↑↑↑ 3, g2 = 3 ↑g1 3. And in general gn = 3 ↑gn−1 3.
✔ Graham (1971) proved that 6 ≤ N ≤ g64 .
✔ No one has improved the upper bound, but Exoo(2003) proved that N ≥ 11.
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In popular culture: What is xkcd?
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What’s next.
We could define more and more notation to create bigger and bigger numbers. Just to
have a taste of what has been done.
Conway’s Chained arrow notation: a → b → c → . . . → n. (Not binary!)
✔ It is such powerful that with only 4 parameters can clobber Knuth arrow’s notation.
(m → n → p = m ↑p n).
✔ For larger chains, if there is a 1, truncate anything afterwards, if not, look at two
last elements.
X → 1 → Y = X.
X → p + 1 → q + 1 = X → (X → p → q + 1) → q.
= X → (X → (X → p − 1 → q + 1) → q) → q.
= X → (X → (. . . (X → (X) → q) . . . ) → q) → q.
✔ Compact!
3 → 3 → 64 → 2 < g64 < 3 → 3 → 65 → 2.
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Can we do better?
Of course:
✔ Bower’s array notation [a, b, . . . , n]. It is such powerful that [n, n, n, n, n] is already
bigger than n → n → · · · → n (n copies of n).
✔ Exploding arrays.
✔ Multidimensional array constructions. (arrays of multiarrays, etc...).
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Can we do better?
Of course:
✔ Bower’s array notation [a, b, . . . , n]. It is such powerful that [n, n, n, n, n] is already
bigger than n → n → · · · → n (n copies of n).
✔ Exploding arrays.
✔ Multidimensional array constructions. (arrays of multiarrays, etc...).
✔ However this is just reusing and creating more and more notation to make
everything more and more compact... Can we do better?
✔ What if we allow ourselves to use non-recursive (non-computable) functions?
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What can computers do
✔ General model of computation: Turing Machine.
✔ Informal description: A machine that on an infinite tape on which symbols can be
written, which it can read or write one at a time using a tape head. The machine
also has a finite number of internal states, including a starting state and a halting
(accepting) state.
✔ When the tape head reads a symbol on the tape, it checks its internal state and,
following a transition function it will
✘ Rewrite the symbol in the tape.
✘ Change to another state.
✘ Move left or right.
✔ A function f :
N → N is computable, if there exists a TM that, given a tape with
n encoded on it, halts leaving f (n) in the tape.
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Computable functions and Busy Beavers
✔ ChurchTuring thesis: If an algorithm (procedure that terminates) exists then there
exists an equivalent Turing Machine for it.
✔ We can classify the T.M. according to the number of internal states. Hence we can
enumerate them.
✔ Radó [1962] asked the following question: Given a T.M. with n states and tape
initially empty (every cell is 0). Run the machine. If it halts, what is the maximum
numbers of steps it do?
✔ Define that number to be BB(n). We can prove that
this function grows faster than any computable function.
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Is BB(n) useful?
Only in theory...
✔ If we knew, say, BB(1000), we could give an algorithm to prove several open
problems in Math.
✔ Goldbach’s conjecture: Every even integer greater than 2 can be written as the
sum of two primes.
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Is BB(n) useful?
Only in theory...
✔ If we knew, say, BB(1000), we could give an algorithm to prove several open
problems in Math.
✔ Goldbach’s conjecture: Every even integer greater than 2 can be written as the
sum of two primes.
✔ It is very easy to construct a Turing Machine M with not so many states (say at
most 1000) that looks for the first counterexample.
✔ If M halts that means that the conjecture is false, if it does not halt, then it is
true.
✔ The following algorithm finds the veracity of Goldbach’s conjecture: Run M for
BB(1000) steps,
✘ If it halts then the conjecture is false
✘ If it does not halts after BB(1000) steps then the conjecture is true.
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So what’s next?
✔ Composing the BB function with some constructible function is cheating. We need
a new paradigm to beat the BB function.
✔ What about higher levels of BB functions? If we use Turing Machines with access
to an oracle that solves the Halting problem we can define a second level of BB
function. (BB2 (n)?)
✔ Can define higher levels: Arithmetic hierarchy (Σ0k , Π0k for i ∈ N).
✔ Essentially any statements about natural numbers that involve a finite sequence of
∀ and ∃, where the quantifiers are over natural numbers could be ‘solved’ in some
level of this hierarchy: If we knew the value of the appropriate BBk (n) we could
run this imaginary non constructive super Turing Machine to decide its true value.
(For instance the twin prime conjecture could be ‘solved’ in the second level).
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Warning: We should stop.
✔ We could continue creating functions that grow faster and faster using stronger
and stronger paradigms. It’s not that hard to imagine ways to go outside the
Arithmetic Hierarchy. (Second order hierarchy, Analytical Hierarchy, etc.)
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Warning: We should stop.
✔ We could continue creating functions that grow faster and faster using stronger
and stronger paradigms. It’s not that hard to imagine ways to go outside the
Arithmetic Hierarchy. (Second order hierarchy, Analytical Hierarchy, etc.)
✔ However, the game we were trying to play will end in a huge draw since no referee
could decide the winner.
✔ Just comparing BB(1000) with, say, A(g64 , g64 ) is already an impossible task
(I’d bet that the first one is bigger, but I don’t think anyone will be ever able to
prove it).
✔ It’s even worse, comparing numbers created using these definition is a task that
most reasonable logic systems can’t decide: Logic systems themselves can be
modeled using these super machines.
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Sooooo...
What would you write?
THANKS.
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Example: DS sequences
go back
Davenport-Schinzel sequences Given n, s positive integers, a DS(n, s) sequence is a
sequence of integers u1 , . . . , um ∈ {1, . . . , n}
1. For all i < m, ui 6= ui+1 .
2. For any a 6= b there are no subsequences of length s + 2 alternating a and b
e.g. (1,2,10,1,50,2,50,1) is a is DS(50, 4) but not DS(50, 3) sequence.
Let λs (n) be the maximum size of a DS(s, n) sequence.
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Example: DS sequences
go back
Davenport-Schinzel sequences Given n, s positive integers, a DS(n, s) sequence is a
sequence of integers u1 , . . . , um ∈ {1, . . . , n}
1. For all i < m, ui 6= ui+1 .
2. For any a 6= b there are no subsequences of length s + 2 alternating a and b
e.g. (1,2,10,1,50,2,50,1) is a is DS(50, 4) but not DS(50, 3) sequence.
Let λs (n) be the maximum size of a DS(s, n) sequence.
Known bounds [Agarwal, Sharirb, Shor 87]
λ1
λ2
λ3
λ4
λs
n
2n − 1
Θ(nα(n))
Θ(n · 2α(n) )
Θ(n · 2poly(α(n),s) )
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BB(n) grows fast
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✔ Intuitive proof: Suppose that F is a computable function such that
F (n) > BB(n), then we can create another machine that, on an input (n, m)
decides if the n-th TM machine halts on input m (simply let the machine count
the number of states k of the n-th machine and simulate it for F (k) steps, if it has
not halted yet, then it never will).
✔ But this second machine can not exist! It would solve the Halting Problem.
✔ So BB(n) grows faster than ANYTHING defined so far...
✔ Actually BB(1) = 1, BB(2) = 6, BB(3) = 21, BB(4) = 107 and no upper bound
is known for BB(5). (the best lower bound is like 47 millions).
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