Tutorial 4 GEM2507 Physical Question from Everyday Life Information and Nano

Tutorial 4GEM2507

Physical Question from Everyday Life

Information and Nano

1. Write the number 2500 in binaryThe 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

Conversion between base 2 and 10

Information Theory

What is the number 100 in binary?

100 2050 20

This reads as we have the divisor 50 and the remainder 0 when 100 is divided by 2.

25 2112 206 203 211 210 Stop until you get 0!

Rea

d in

this

di

rect

ion!

The number 100 in binary is 1100100.

Taken from Kenneth Hong’s slide on Life of Complex system

Conversion between base 2 and 10

Information Theory

What is the number 1100100 in decimal?

1 1 0 0 1 0 0

26 25 24 23 22 21 20

Right to left!

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

= 100Taken from Kenneth Hong’s slide on Life of Complex system

Number of bits required

Information Theory

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


Optimal coding

Information Theory

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.


Optimal coding

Information Theory

Number of bits required:

For symbol “1”: -log2(1/8) = 3 bitsFor symbol “2”: -log2(1/8) = 3 bitsFor symbol “3”: -log2(1/4) = 2 bitsFor 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.


Optimal coding

Information Theory

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


Optimal coding

Information Theory

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!


Optimal coding

Information Theory

In general, the pattern of the optimal codes would be as follow.

01011011101111011111011111101111111

Let’s look at the case of 8 symbols:10100100010000100000100000010000000

OR

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.


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

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:

What is Turing Machine?

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 MachineWhat 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

0Head

State

Tape

Rules are defined in this format: (current state, current symbol, new state, new symbol, left/right)

S

Start with the symbol S on the tape and in state zero.Then apply the rules until the machine stops.

> = Right, - = write or do nothing

7. Design a Turing machine that can write the word “Life” its tape

Turing Machine

0Head

State

Tape


S

R = Right, - = write or do nothing

Let us go through this algorithm step by step

In the beginning (this is also called the initial condition) we have:

Step 0 - Read Apply rule 1.

Rules:

1) 0,’S’, 1, ‘L’, R

2) 1, ‘-’, 2, I, R

3) 2, ‘-’, 3, ‘F’, R

4) 3, ‘-’, 4, ‘E’,R

5) 4,’ –’, -,’ –’, -

Turing Machine

1Head

State

Tape


L



Step 1 - Move Right

Rules:

1) 0,’S’, 1, ‘L’, R

2) 1, ‘-’, 2, I, R

3) 2, ‘-’, 3, ‘F’, R

4) 3, ‘-’, 4, ‘E’,R

5) 4,’ –’, -,’ –’, -

Turing Machine

1Head

State

Tape


L



Step 2 - Read Apply rule 2

Rules:

1) 0,’S’, 1, ‘L’, R

2) 1, ‘-’, 2, I, R

3) 2, ‘-’, 3, ‘F’, R

4) 3, ‘-’, 4, ‘E’,R

5) 4,’ –’, -,’ –’, -

Turing Machine

2Head

State

Tape


L



Step 3 - Move Right

I

Rules:

1) 0,’S’, 1, ‘L’, R

2) 1, ‘-’, 2, I, R

3) 2, ‘-’, 3, ‘F’, R

4) 3, ‘-’, 4, ‘E’,R

5) 4,’ –’, -,’ –’, -

Turing Machine

2Head

State

Tape


L




I

Rules:

1) 0,’S’, 1, ‘L’, R

2) 1, ‘-’, 2, I, R

3) 2, ‘-’, 3, ‘F’, R

4) 3, ‘-’, 4, ‘E’,R

5) 4,’ –’, -,’ –’, -

Turing Machine

3Head

State

Tape


L



Step 5-Move Right

I F

Rules:

1) 0,’S’, 1, ‘L’, R

2) 1, ‘-’, 2, I, R

3) 2, ‘-’, 3, ‘F’, R

4) 3, ‘-’, 4, ‘E’,R

5) 4,’ –’, -,’ –’, -

Turing Machine

4Head

State

Tape


L




I F E

Rules:

1) 0,’S’, 1, ‘L’, R

2) 1, ‘-’, 2, I, R

3) 2, ‘-’, 3, ‘F’, R

4) 3, ‘-’, 4, ‘E’,R

5) 4,’ –’, -,’ –’, -

Turing Machine

4Head

State

Tape


L




I F E

The head will stay there without doing anything at this position after applying Rule 5

Rules:

1) 0,’S’, 1, ‘L’, R

2) 1, ‘-’, 2, I, R

3) 2, ‘-’, 3, ‘F’, R

4) 3, ‘-’, 4, ‘E’,R

5) 4,’ –’, -,’ –’, -

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

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

20/64 = 0.3125

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

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Tutorial 4 GEM2507 Physical Question from Everyday Life Information and Nano