Tutorial 4
GEM2507
Physical Question from Everyday Life
Information and Nano
1. Write the number 2500 in binary
The decimal number 2500 is 100111000100
2. How many bits does one need to encode a position in a string of 200000
symbols?
The smallest power of 2 greater than 200,000 is 218 = 262,144. Hence the minimal
number needed to encode a position in a string of 200,000 symbols is 18
3. Suppose you are given a five-sided biased die with a probability of 1/8 to
obtain a 1 or 2, and probability of ¼ to obtain a 3, 4, or 5. What is an
optimal code for transmitting the throws of this die?
Dice
Probability
Length of code
Example of
code
1
1/8
-log(1/8) =3
111
2
1/8
-log(1/8) =3
110
3
1/4
-log(1/4) =3
10
4
1/4
-log(1/4) =3
01
5
1/4
-log(1/4) =3
00
= Average number of bits per
throw
Information Theory
Conversion between base 2 and 10
Read in this direction!
What is the number 100 in binary?
0
0
1
0
0
1
1
This reads as we have the
divisor 50 and the
remainder 0 when 100 is
divided by 2.
100
2
50
2
25
2
12
2
6
2
3
2
1
2
Stop until you get 0!
0
The number 100 in binary is 1100100.
Taken from Kenneth Hong’s slide on Life of Complex system
Information Theory
Conversion between base 2 and 10
What is the number 1100100 in decimal?
1
1
0
0
1
0
Right to left!
0
26
25
24
23
22
20
21
0 x 20 + 0 x 21 + 1 x 22 + 0 x 23 + 0 x 24 + 1 x 25 + 1 x 26
= 0 x 1 + 0 x 2 + 1 x 4 + 0 x 8 + 0 x 16 + 1 x 32 + 1 x 64
= 0 + 0 + 4 + 0 + 0 + 32 + 64
= 100
Taken from Kenneth Hong’s slide on Life of Complex system
Information Theory
Number of bits required
How many bits does one need to encode any position in a
string of 800 symbols?
The position of the symbols goes like 1, 2, 3, 4, 5, ….. 800.
To determine the number of bits needed to encode any
position, we only need to determine the number of bits
required to encode the highest position. The highest
position in this case is 800.
We just need to convert the number 800 in binary and count
the bits required!
Answer: 10 bits
Taken from Kenneth Hong’s slide on Life of Complex system
Information Theory
Optimal coding
Suppose you are given a four-sided biased dice with
probability of 1/8 to obtain a “1” or a “2”, a probability of 1/4
to obtain a “3” and a probability of 1/2 to obtain a “4”. What
is the optimal code for transmitting the throws of this dice?
To solve this problem, we have to first determine the
number of bits required to encode each of the symbols
respectively according to the formula –log2P.
Next, based on these, design the coding scheme wisely
without ambiguity. These will be the optimal code to
transmit this information.
Taken from Kenneth Hong’s slide on Life of Complex system
Information Theory
Optimal coding
Number of bits required:
For symbol “1”: -log2(1/8) = 3 bits
For symbol “2”: -log2(1/8) = 3 bits
For symbol “3”: -log2(1/4) = 2 bits
For symbol “4”: -log2(1/2) = 1 bit
Based on these, we know that the optimal codes for the
symbol “1”, “2”, “3” and “4” are 3 bits, 3 bits, 2 bits and
1 bit respectively.
Taken from Kenneth Hong’s slide on Life of Complex system
Information Theory
Optimal coding
To design the optimal coding scheme, we always start from the
simplest one, i.e. the one requires the minimum number of bits.
We could choose the following coding scheme:
“4” = 0, “3” = 10, “2” = 110, “1” = 111
Of course, the coding scheme is not unique. We could also
choose:
“4” = 1, “3” = 01, “2” = 001, “1” = 000
Taken from Kenneth Hong’s slide on Life of Complex system
Information Theory
Optimal coding
However, the following scheme is not correct:
“4” = 0, “3” = 11, “2” = 110, “1” = 111
Let’s say we have the string 110. Should be decode it to be the
symbol “2” or “34”?
Therefore, the design of the optimal code must be careful so
that no ambiguity would be caused!
Taken from Kenneth Hong’s slide on Life of Complex system
Information Theory
Optimal coding
In general, the pattern of the optimal codes would be as follow.
Let’s look at the case of 8 symbols:
0
10
110
1110
OR
11110
111110
1111110
1111111
1
01
001
0001
00001
000001
0000001
0000000
Of course, these patterns depend on our distribution of the
probabilities: 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128, 1/128.
Taken from Kenneth Hong’s slide on Life of Complex system
4. Given the fact that the four types of nucleotides in mRNA only need to
code for 20 amino acids and furthermore a stop sign, is it possible to design a
code that is shorter than the standard genetic code? If so, how could this be
done?
There are several possibilities
1. There could be 6 different nucleotides. Using 6 different nucleotides, we just need two
nucleotides to code for 20 amino acids since 62 = 36 > (20+1)
2. Reduce the number of different amino acids to 15. This would still leave us with a
practically infinite number of different possible proteins. Then only two nucleotides are
needed to code for all the amino acids plus one stop codon since 42 = 16 = (15+1)
5. Can a personal computer be considered a universal Turing machine?
Yes. Indeed personal computer is the physical realization of universal turing machine.
Any algorithmic process can be simulated efficiently using Turing machines
6. Are there any amino acids that are effectively only coded for by two
nucleotids? (That is, the third nucleotide can be ignored)
Yes, From the genetic code on page 18.14, we can see that all the codons which have
cytosine (C) as the second nucleotide code for the same amino acid regardless of the
third nucleotide. Another example is valine which is coded for by GUU, GUC, GUA and
GUA
Turing Machine
What is Turing Machine?
A Turing machine is a very simple machine, but, logically
speaking, has all the power of any digital computer. It may
be described as follows:
1. There is an infinite tape with squares on it.
2. Each square may contain a symbol from a finite alphabet but
there can be only finitely many non-blank square on the
tape.
3. There is a read-write head which is positioned somewhere
on the tape.
4. The machine is in one of a finite number of states.
5. There is a set of rules that tell the machine what to do given
a certain state and a certain symbol read from the tape.
Turing Machine
What is Turing Machine?
The rule is of the following form:
(current state, current symbol, new state, new symbol, left/right)
This rule means that if the Turing machine is now in current
state, and the symbol under the read/write head is current
symbol, change its internal state to new state, replace the
symbol on the tape at its current position by new symbol,
and move the read/write head one square in the given
direction (left or right).
If a Turing machine is in a condition for which it has no rule,
it halts.
Turing Machine
7. Design a Turing machine that can write the word “Life” its tape
Start with the symbol S on the tape and in state zero.
Then apply the rules until the machine stops.
S
State
Head
0
> = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
Turing Machine
Let us go through this algorithm step by step
In the beginning (this is also called the initial condition) we have:
Apply rule 1.
Step 0 - Read
Rules:
S
1) 0,’S’, 1, ‘L’, R
State
2) 1, ‘-’, 2, I, R
3) 2, ‘-’, 3, ‘F’, R
Head
0
4) 3, ‘-’, 4, ‘E’,R
5) 4,’ –’, -,’ –’, -
R = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
Turing Machine
Let us go through this algorithm step by step
Right
Step 1 - Move
L
Rules:
State
Head
1
1) 0,’S’, 1, ‘L’, R
2) 1, ‘-’, 2, I, R
3) 2, ‘-’, 3, ‘F’, R
4) 3, ‘-’, 4, ‘E’,R
5) 4,’ –’, -,’ –’, -
R = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
Turing Machine
Let us go through this algorithm step by step
Apply rule 2
Step 2 - Read
L
Rules:
State
Head
1
1) 0,’S’, 1, ‘L’, R
2) 1, ‘-’, 2, I, R
3) 2, ‘-’, 3, ‘F’, R
4) 3, ‘-’, 4, ‘E’,R
5) 4,’ –’, -,’ –’, -
> = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
Turing Machine
Let us go through this algorithm step by step
Step 3 - Move
Right
L I
Rules:
State
Head
2
1) 0,’S’, 1, ‘L’, R
2) 1, ‘-’, 2, I, R
3) 2, ‘-’, 3, ‘F’, R
4) 3, ‘-’, 4, ‘E’,R
5) 4,’ –’, -,’ –’, -
> = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
Turing Machine
Let us go through this algorithm step by step
Step 4 - Read
Apply rule 3
L I
Rules:
State
Head
2
1) 0,’S’, 1, ‘L’, R
2) 1, ‘-’, 2, I, R
3) 2, ‘-’, 3, ‘F’, R
4) 3, ‘-’, 4, ‘E’,R
5) 4,’ –’, -,’ –’, -
> = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
Turing Machine
Let us go through this algorithm step by step
Step 5-Move
Right
L I
F
Rules:
State
Head
3
1) 0,’S’, 1, ‘L’, R
2) 1, ‘-’, 2, I, R
3) 2, ‘-’, 3, ‘F’, R
4) 3, ‘-’, 4, ‘E’,R
5) 4,’ –’, -,’ –’, -
R = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
Turing Machine
Let us go through this algorithm step by step
Step 6 - Read
Apply rule 4
L I
F E
Rules:
State
Head
4
1) 0,’S’, 1, ‘L’, R
2) 1, ‘-’, 2, I, R
3) 2, ‘-’, 3, ‘F’, R
4) 3, ‘-’, 4, ‘E’,R
5) 4,’ –’, -,’ –’, -
R = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
Turing Machine
Let us go through this algorithm step by step
Step 6 - Read
Apply rule 5
L I
F E
Rules:
State
Head
4
1) 0,’S’, 1, ‘L’, R
2) 1, ‘-’, 2, I, R
3) 2, ‘-’, 3, ‘F’, R
4) 3, ‘-’, 4, ‘E’,R
5) 4,’ –’, -,’ –’, -
The head will stay there without doing anything at this position after applying Rule 5
R = Right, - = write or do nothing
Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)
8. What is the function of allolactose?
Allolactose binds to the lactose repressor. When doing so, expression of the genes
necessary for the lactose digestion is greatly increased
Allolactose molecule
9. If I flip a coin 6 times, how likely is it to obtain exactly 3 heads and 3 tails?
The probability of having k successes in n trials is given by
Where p is probability of success
So, the probability of obtaining 3 heads and 3 tails is
20/64 = 0.3125
10. What is the basic mechanism of an Atomic Force Microscope (AFM)?
In AFM, a tiny cantilever is brought very close to the surface of the sample
to be investigated. The cantilever then interacts with the surface due to
contact, van der Waals or electrostatic forces. Resolution of AFM is
fractions of a nanometer, more than 1000 times better than the optical
diffraction limit. Operating modes of AFM is divided into static (contact)
mode and dynamic (non-contact mode).
Static mode
•Tip deflection is used as a feedback signal.
•Force is kept constant
Dynamic mode
• Cantilever is externally oscillated at or close to its harmonics.
• The change in oscillation amplitude, phase and resonance frequency,
which are modified by tip-sample interactions forces, provides information
about the sample's characteristics
11. Name one way in which a quantum dot can be modified to change its color?
Example of colloidal semiconductor quantum dots:
Cadmium Selenide, Cadmium Sulfide, Indium Arsenide
and Indium Phosphate. Q-dots contain 100 to 100,000
atoms within the quantum dot volume, with a diameter of
10 to 50 atoms (2 - 10 nm). At 10 nm in diameter, nearly
3 million quantum dots could be lined up end to end and
fit within the width of a human thumb.
The smaller the quantum dot, the wider the band gap.
Changing the band gap, and thus the color of light a
quantum dot absorbs or emits, requires only adding or
subtracting atoms from the quantum dot.
Quantum dots of the same material but different
sizes (here, cadmium selenide in suspension)
have different band gaps and emit different
colors
12. Would it be possible to use quantum dots to generate
infrared radiation?
Yes. Q-dots that generate infrared light is bigger that Q-dots
that generate visible light because infrared light has longer
wavelength than visible light.
13. What happens to the energy gap of a quantum dot when its size is increased?
The energy gap decreases and hence the wavelength of
the light emitted is longer
14. Would it be possible to have a quantum dot of 0.01 nm size?
No, because the size of hydrogen atom is about 0.1 nm. Since the hydrogen atom is
the smallest atom of all, quantum dot needs to be substantially larger than 0.1 nm
15. Give a sketch with the rolling vector R(7,7). Is this a zigzag formation?
No. It is armchair formation as shown below
http://www.youtube.com/watch?v=i4Ax8sY2U4A&feature=related