Transcript Richard

Thought Experiment
• Take a piece of paper one thousandth of an inch thick
• Now fold it in half and then in half again
• Do this 50 times (I know that this is not practicable)
• How thick do you think it will be ?
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Each time we fold the paper, it doubles in thickness
So after the first fold it is 2x as thick
After the second fold it is 2x2 ie 4 times as thick
After 50 folds it is 2x2x2 .......... 2x2 times as thick
• That is 250
• Divide by a thousand to get the thickness in inches
• Divide by 63,360 to get the thickness in miles
• And we get ......
250 = 1125899906842624
Dividing by a thousand = 1125899906842 inches
Dividing by 63360 = 17,769,885 miles
Carol recently gave a talk on the metal
Bismuth.
One of its interesting facts was it's half-life
This is currently calculated at :1.9 x 1019 years
If we assume that there are 64
generations between now and the start of
the Christian Era
Then you will have approximately
1.8 x 1019 ancestors
There is a story about an Indian temple which contains a
large room with three posts in it surrounded by 64 golden
disks. Brahmin priests, acting out the command of an
ancient prophecy, have been moving these disks, in
accordance with the immutable rules of the Brahma,
since that time. The puzzle is therefore also known as the
Tower of Brahma puzzle. According to the legend, when
the last move of the puzzle will be completed, the world
will end
if the priests were able to move disks at a
rate of one per second, using the smallest
number of moves, it would take them 264−1
seconds or roughly 585 billion years or
18,446,744,073,709,551,615 turns to finish
ie 1.8 x 1019 moves
The world will certainly have ended !!!
Just a little larger are the
4.3 x 1019
different starting positions for a
Rubik's Cube
Any ideas of the smallest
number of moves required to
solve the cube from any
starting position ?
(Known as God's Number)
Any cube can be solved in 22 moves at
most, but there is strong evidence that
only 20 moves are required.
Hence God's Number is either 20 or 22
Consider this 200 digit number
914743972814745128948036774162014302835642105
034332853395613272769334542296093046464719250
945181147710162588965929074414263498975565041
455709602039255036791052451991423388060824942
54050610000000000000
You now have 70 seconds to extract its 13th root
assuming that you wish to claim the world record
The answer is
2,407,899,893,032,210
Compare this to :Age of the universe
13.798×109 years
or 4.354×1017 seconds
Number of elementary particles in the
observable universe
between 1080 and1097
But these numbers are mere
beginners
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Googol
Googolplex
Skewe's number
Gijswijt's sequence
Ramsey Numbers
Graham's Number
Googol
Originally named by Edward Kasner
a Googol = 10100
ie
1000000000000000000000000000000000
0000000000000000000000000000000000
0000000000000000000000000000000
(The search engine Google copied the named but
spelled it incorrectly)
Any idea what's special about this
number ?
57885161
2
-1
this is approximately 1010000000
It's the largest prime number
currently know
It was discovered in 2013 and is some
17.5 million digits long
It is also the 48th Mersenne Prime
ie of the type 2n-1
It's the largest known number that cannot be
expressed in terms of smaller numbers
Googolplex
100
10
10
googol
or
10
There isn't enough room or time to write out
this number
If you can write 2 digits/second, you would need 1.5
x 1092 years, which is some 1082 times greater than
the age of the universe
Skewes' Number
• Gauss generated a formula to
give the number of Primes up
to any given number.
• It is known that it generally
over-estimates the true
number.
• Stanley Skewes showed that
Gauss's formula would underestimate when we got above
the number that bears his
name
Which is
1
100
34
10
This result, however depends upon the Riemann Hypothesis being true
If the Riemann Hypothesis is ever shown to be false, then Skewes
number increases to
1
100
10
963
Gijswijt's sequence
1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3,
2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2,
3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 1, 1, 2, 1, 1, 2, 2,
2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 2,
2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2,
2, 3, 3, 2, 2, 2, 3, 2, 1.............
where does 4, 5, etc appear ?
The first time 3 appears is in the ninth
position.
4 appears for the first time in the 221st
position.
The first 5 occurs in position
10100000000000000000000000
23
10
ie 10
The other numbers will eventually appear,
although with no sense of urgency
Ramsey Numbers
You are going to have a party.
How many people do you need to invite to
be sure that at least 3 people will be
mutual aquaintances, or at least 3 people
will be mutual strangers ?
However, if we need to know the minimum
number of guests such that either 5 people
are mutually aquainted or mutually
strangers,
there are some 2903 or 6.7 x 10271 cases to
try out using brute computer force alone !
We do know that between 43 and 49 guests
would be required
Graham's Number
If we now consider
Ramsey's numbers but
venture into
multi-dimensional space
the numbers get very
large.
So large, that we need
a whole new way to
represent them
Once again, say we have some points, but
now they are the corners of an n
dimensional hypercube. They are still all
connected by blue and red lines. For any 4
points, there are 6 lines connecting them.
Can we find 4 points that all lie on one
plane,
and the 6 lines connecting them are all the
same colour?
Graham's Number gives an upper limit of the
points needed for this to be certain
Knuth's up-arrow notation
Invented by Donald Knuth in 1976
It uses a series of up arrows and looks like :-
g64=3↑↑↑↑ g63 times ↑↑↑↑3
If each digit could occupy a Planck length 10-35m,
then you could get
1035 x 1035 x 1035 digits ( ie 10105) in a cubic meter,
there would still not be enough space in the
universe to write it down
We do know the last 500 digits of his
number - the last digit is a 7
Also the lower limit is 13