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CSE 599 Lecture 7: Information Theory, Thermodynamics and Reversible ComputingPowerPoint Presentation

CSE 599 Lecture 7: Information Theory, Thermodynamics and Reversible Computing

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CSE 599 Lecture 7: Information Theory, Thermodynamics and Reversible Computing

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CSE 599 Lecture 7: Information Theory, Thermodynamics and Reversible Computing

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- What have we done so far?
- Theoretical computer science: Abstract models of computing
- Turing machines, computability, time and space complexity

- Theoretical computer science: Abstract models of computing
- Physical Instantiations
- Digital Computing
- Silicon switches manipulate binary variables with near-zero error

- DNA computing
- Massive parallelism and biochemical properties of organic molecules allow fast solutions to hard search problems

- Neural Computing
- Distributed networks of neurons compute fast, parallel, adaptive, and fault-tolerant solutions to hard pattern recognition and motor control problems

- Digital Computing

- Information theory and Kolmogorov Complexity
- What is information?
- Definition based on probability theory
- Error-correcting codes and compression
- An algorithmic definition of information (Kolmogorov complexity)

- Thermodynamics
- The physics of computation
- Relation to information theory
- Energy requirements for computing

- Reversible Computing
- Computing without energy consumption?
- Biological example
- Reversibe logic gates Quantum computing (next week!)

- 3 principal results:
- Shannon’s source-coding theorem
- The main theorem of information content
- A measure of the number of bits needed to specify the expected outcome of an experiment

- Shannon’s noisy-channel coding theorem
- Describes how much information we can transmit over a channel
- A strict bound on information transfer

- Kolmogorov complexity
- Measures the algorithmic information content of a string
- An uncomputable function

- Shannon’s source-coding theorem

- First try at a definition…
- Suppose you have stored n different bookmarks on your web browser.
- What is the minimum number of bits you need to store these as binary numbers?
- Let I be the minimum number of bits needed. Then,
2I n I log2 n

- So, the “information” contained in your collection of n bookmarks is I0 = log2 n

- Consider a set of alternatives: X = {a1, a2, a3, …aK}
- When the outcome is a3, we say x = a3

- I0(X) is the amount of information needed to specify the outcome of X
- I0(X) = log2½X½
- We will assume base 2 from now on (unless stated otherwise)
- Units are bits (binary digits)

- I0(X) = log2½X½
- Relationship between bits and binary digits
- B = {0, 1}
- X = BM = set of all binary strings of length M
- I0(X) = log½BM½= log½2M½=M bits

- Appeal to your intuition…
- Which of these two messages contains more “information”?
“Dog bites man”

or

“Man bites dog”

- Appeal to your intuition…
- Which of these two messages contains more “information”?
“Dog bites man”

or

“Man bites dog”

- Same number of bits to represent each message!
- But, it seems like the second message contains a lot more information than the first. Why?

- Surprising events (unexpected messages) contain more information than ordinary or expected events
- “Dog bites man” occurs much more frequently than “Man bites dog

- Messages about less frequent events carry more information
- So, information about an event varies inversely with the probability of that event
- But, we also want information to be additive
- If message xy contains sub-parts x and y, we want:
- I(xy) = I(x) + I(y)

- If message xy contains sub-parts x and y, we want:
- Use the logarithm function: log(xy) = log(x) + log(y)

- Define the information contained in a message x in terms of log of the inverse probability of that message:
- I(x) = log(1/P(x)) = - log P(x)

- First defined rigorously and studied by Shannon (1948)
- “A mathematical theory of communication” – electronic handout (PDF file) on class website.

- Our previous definition is a special case:
- Suppose you had n equally likely items (e.g. bookmarks)
- For any item x, P(x) = 1/n
- I(x) = log(1/P(x)) = log n
- Same as before (minimum number of bits needed to store n items)

- Kolmogorov, 1933
- P(a) >= 0where a is an event
- P(l) = 1where l is the certain event
- P(a + b) = P(a) + P(b)where a and b are mutually exclusive

- Kolmogorov (axiomatic) definition is computable
- Probability theory forms the basis for information theory
- Classical definition based on event frequencies (Bernoulli) is uncomputable:

- Joint probability of two events a and b: P(ab)
- Independence
- Events a and b are independent if P(ab) = P(a)P(b)

- Conditional probability: P(a|b) = probability that event a happens given that b has happened
- P(a|b) = P(ab)/P(b)
- P(b|a) = P(ba)/P(a) = P(ab)/P(a)

- We just proved Bayes’ Theorem:
- P(a)is called the a priori probability of a
- P(a½b) is called the a posteriori probability of a

1.Information is defined in the context of a set of alternatives. The amount of information quantifies the number of bits needed to specify an outcome from the alternatives

2.The amount of information is independent of the semantics (only depends on probability)

3.Information is always positive

4.Information is measured on a logarithmic scale

- Probabilities are multiplicative, but information is additive

- Message y contains duplicates: y = xx
- Message x has probability P(x)
- What is the information content of y?
- Is I(y) = 2 I(x)?

- Message y contains duplicates: y = xx
- Message x has probability P(x)
- What is the information content of y?
- Is I(y) = 2 I(x)?

- I(y) = log(1/P(xx)) = log[1/(P(x|x)P(x))]
= log(1/P(x|x)) + log(1/P(x))

= 0 + log(1/P(x))

= I(x)

- Duplicates convey no additional information!

- The average self-information or entropy of an ensemble X= {a1, a2, a3, …aK}
- E expected (or average) value

- 0 <= H(X) <= I0(X)
- Equals I0(X) = log½X½ if all the ak’s are equally probable
- Equals 0 if only one ak is possible

- Consider the case where k = 2
- X = {a1, a2}
- P(a1) = ; P(a2) = 1–

- Entropy is a measure of randomness of the source producing the events
- Example 1 : Coin toss: Heads or tails with equal probability
- H = -(½ log ½ + ½ log ½) = -(½ (-1) + ½ (-1)) = 1 bit per coin toss

- Example 2 : P(heads) = ¾ and P(tails) = ¼
- H = -(¾ log ¾ + ¼ log ¼) = 0.811 bits per coin toss
- As things get less random, entropy decreases
- Redundancy and regularity increases

- If we have N different symbols, we can encode them in log(N) bits. Example: English - 26 letters 5 bits
- So, over many, many messages, the average cost/symbol is still 5 bits.
- But, letters occur with very different probabilities! “A” and “E” much more common than “X” and “Q”. The log(N) estimate assumes equal probabilities.
- Question: Can we encode symbols based on probabilities so that the average cost/symbol is minimized?

- Also called the fundamental theorem. In words:
- You can compress N independent, identically distributed (i.i.d.) random variables, each with entropy H, down to NH bits with negligible loss of information (as N)
- If you compress them into fewer than NH bits you will dramatically lose information

- The theorem:
- Let X be an ensemble with H(X) = H bits. Let Hd(X) be the entropy of an encoding of X with allowable probability of error d
- Given any > 0 and 0 < < 1, there exists a positive integer No such that, for N > No,

- What do the two inequalities tell us?
- The number of bits that we need to specify outcomes x with vanishingly small error probability does not exceed H +
- If we accept a vanishingly small error, the number of bits we need to specify x drops to N(H + )
- The number of bits that we need to specify outcomes x with large allowable error probability is at least H –

- Question: How do we compress the outcomes XN?
- With vanishingly small probability of error
- How do we assign the elements of X such that the number of bits we need to encode XN drops to N(H + )

- Symbol coding: Given x = a3 a2 a7 … a5
- Generate codeword (x) = 01 1010 00
- Want Io((x)) ~ H(X)

- Well-known coding examples
- Zip, gzip, compress, etc.
- The performance of these algorithms is, in general, poor when compared to the Shannon limit

- A code is a function : X B+
- B = {0, 1}
- B+ the set of finite strings over B
- B+ = {0, 1, 00, 01, 10, 11, 000, 001, …}

- (x) = (x1) (x2) (x3) … (xN)

- A code is uniquely decodable (UD) iff
- : X+ B+ is one-to-one

- A code is instantaneous iff
- No codeword is the prefix of another
- (x1) is not a prefix of (x2)

- Given X = {a1, a2, …aK}, with associated probabilities P(ak)
- Given a code with codeword lengths n1, n2, …nk
- The expected code length

- No instantaneous, UD code can achieve a smaller than a Huffman code

- Feynman example: Encoding an alphabet
- Code is instantaneous and UD: 00100001101010 = ANOTHER

- Code achieves close to Shannon limit
- H(X) = 2.06 bits; = 2.13 bits

1

- I(X;Y) is the average mutual information between X and Y
- Definition: Channel capacity
- The information capacity of a channel is: C = max[I(X;Y)]

- The channel may add noise
- Corrupting our symbols

Input

Output

channel

x

y

I(X;Y) what we know

about X given Y

H(X) entropy

of input ensemble X

Problem: A binary source sends equiprobable messages in a time T, using the alphabet {0, 1} with a symbol rate R. As a result of noise, a “0” may be mistaken for a “1”, and a “1” for a “0”, both with probability q. What is the channel capacity C?

X

Y

Channel is discrete

and memoryless

Assume no noise (no errors)

T is the time to send the string, R is the rate

The number of possible message strings is 2RT

The maximum entropy of the source is Ho = log(2RT ) bits

The source rate is (1/T) Ho = R bits per second

The entropy of the noise (per transmitted bit) is

Hn = qlog[1/q] + (1–q)log[1/(1–q)]

The channel capacity C (bits/sec) = R – RHn = R(1 – Hn)

C is always less than R (a fixed fraction of R)!

We must add code bits to correct the received message

- We want to send a message string of length M
- We add codebits to M, thereby increasing its length to Mc
- How are M, Mc, and q related?

- M = Mc(1 – Hn)
- Intuitively, from our example
- Also see pgs. 106 – 110 of Feynman
- Note: this is an asymptotic limit
- May require a hugeMc

- The Theorem:
- There is a nonnegative channel capacity C associated with each discrete memoryless channel with the following property: For any symbol rate R < C, and any error rate > 0, there is a protocol that achieves a rate >= R and a probability of error <=

- In words:
- If the entropy of our symbol stream is equal to or less than the channel capacity, then there exists a coding technique that enables transmission over the channel with arbitrarily small error
- Can transmit information at a rate H(X) <= C

- Shannon’s theorem tells us the asymptotically maximum rate
- It does not tell us the code that we must use to obtain this rate
- Achieving a high rate may require a prohibitively long code

- Error-correcting codes allow us to detect and correct errors in symbol streams
- Used in all signal communications (digital phones, etc)
- Used in quantum computing to ameliorate effects of decoherence

- Many techniques and algorithms
- Block codes
- Hamming codes
- BCH codes
- Reed-Solomon codes
- Turbo codes

- An example: Construct a code that corrects a single error
- We add m check bits to our message
- Can encode at most (2m – 1) error positions

- Errors can occur in the message bits and/or in the check bits
- If n is the length of the original message then 2m – 1 >= (n + m)

- Examples:
- If n = 11, m = 4: 24 – 1= 15 >= (n + m) = 15
- If n = 1013, m = 10: 210 – 1= 1023 >= (n + m) = 1023

- We add m check bits to our message

- Example: An 11/15 SEC Hamming code
- Idea: Calculate parity over subsets of input bits
- Four subsets: Four parity bits

- Check bit x stores parity of input bit positions whose binary representation holds a “1” in position x:
- Check bit c1: Bits 1,3,5,7,9,11,13,15
- Check bit c2: Bits 2,3,6,7,10,11,14,15
- Check bit c3: Bits 4,5,6,7,12,13,14,15
- Check bit c4: Bits 8,9,10,11,12,13,14,15

- Idea: Calculate parity over subsets of input bits
- The parity-check bits are called a syndrome
- The syndrome tells us the location of the error

Position in message

binarydecimal

0001 1

0010 2

0011 3

0100 4

0101 5

0110 6

0111 7

1000 8

1001 9

1010 10

1011 11

1100 12

1101 13

1110 14

1111 15

- The check bits specify the error location
- Suppose check bits turn out to be as follows:
- Check c1 = 1 (Bits 1,3,5,7,9,11,13,15)
- Error is in one of bits 1,3,5,7,9,11,13,15

- Check c2 = 1 (Bits 2,3,6,7,10,11,14,15)
- Error is in one of bits 3,7,11,15

- Check c3 = 0 (Bits 4,5,6,7,12,13,14,15)
- Error is in one of bits 3,11

- Check c4 = 0 (Bits 8,9,10,11,12,13,14,15)
- So error is in bit 3!!

- Check c1 = 1 (Bits 1,3,5,7,9,11,13,15)

- Example: Encode 10111011011
- Code position:15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
- Code symbol: 1 0 1 1 1 0 1 c4 1 0 1 c3 1 c2 c1
- Codeword:1 0 1 1 1 0 1 1 1 0 1 1 1 0 1
- Notice that we can generate the code bits on the fly!

- What if we receive 101100111011101?
- c4 = 1 101100111011101
- c3 = 0 101100111011101
- c2 = 1 101100111011101
- c1 = 1 101100111011101
- The error is in location 1011 = 1110

- Computers represent information as stored symbols
- Not probabilistic n the Shannon sense)
- Can we quantify information from an algorithmic standpoint?

- Kolmogorov complexity K(s) of a finite binary string s is the single, natural number representing the minimum length (in bits) of a program p that generates s when run on a Universal Turing machine U
- K(s) is the algorithmic information content of s
- Quantifies the “algorithmic randomness” of the string

- K(s) is an uncomputable function
- Similar argument to the halting problem
- How do we know when we have the shortest program?

- Similar argument to the halting problem

- Randomness of a string defined by shortest algorithm that can print it out.
- Suppose you were given the binary string x:
“11111111111111….11111111111111111111111” (1000 1’s)

- Instead of 1000 bits, you can compress this string to a few tens of bits, representing the length |P| of the program:
- For I = 1 to 1000
- Print “1”

- For I = 1 to 1000
- So, K(x) <= |P|
- Possible project topic: Quantum Kolmogorov complexity?

5-minute break…

Next: Thermodynamics and Reversible Computing

- Physics imposes fundamental limitations on computing
- Computers are physical machines
- Computers manipulate physical quantities
- Physical quantities represent information

- The limitations are both technological and theoretical
- Physical limitations on what we can build
- Example: Silicon-technology scaling

- Major limiting factor in the future: Power Consumption
- Theoretical limitations of energy consumed during computation
- Thermodynamics and computation

- Physical limitations on what we can build

- How much energy must we use to carry out a computation?
- The theoretical, minimum energy

- Is there a minimum energy for a certain rate of computation?
- A relationship between computing speed and energy consumption

- What is the link between energy and information?
- Between information–entropy and thermodynamic–entropy

- Is there a physical definition for information content?
- The information content of a message in physical units

- Computation has no inherent thermodynamic cost
- A reversible computation, that proceeds at an infinitesimal rate, consumes no energy

- Destroying information requires kTln2 joules per bit
- Information-theoretic bits (not binary digits)

- Driving a computation forward requires kTln(r) joules per step
- r is the rate of going forward rather than backward

- First law: Conservation of energy
- (heat put into system) + (work done on system) = increase in energy of a system
- DQ + DW = DU
- Total energy of the universe is constant

- Second law: It is not possible to have heat flow from a colder region to a hotter region i.e. DQ/T >= 0
- Change in Entropy DS = DQ/T
- Equality holds only for reversible processes
- The entropy of the universe is always increasing

- A basic heat engine: Q2 = Q1 – W
- T1 and T2 are temperatures
- T1 > T2

- Reversible heat engines are those that have:
- No friction
- Infinitesimal heat gradients

- The Carnot cycle: Motivation was steam engine
- Reversible
- Pumps heat DQ from T1 to T2
- Does work W = DQ

- No engine that takes heat Q1 at T1 and delivers heat Q2 at T2 can do more work than a reversible engine
- W = Q1 – Q2 = Q1(T1 – T2) / T1

- Heat will not, by itself, flow from a cold object to a hot object

- If we add heat Q reversibly to a system at fixed temperature T, the increase in entropy of the system is S = Q/T

- S is a measure of degrees of freedom
- The probability of a configuration
- The probability of a point in phase space

- In a reversible system, the total entropy is constant
- In an irreversible system, the total entropy always increases

- The probability of a configuration

- Assume a gas containing N atoms
- Occupies a volume V1
- Ideal gas: No attraction or repulsion between particles

- Now shrink the volume
- Isothermally (at constant temperature, immerse in a bath)
- Reversibly, with no friction

- How much work does this require?

- From mechanics
- work = force × distance
- force = pressure × (area of piston)
- volume change = (area of piston) × distance
- Solving:

- From gas theory
- The idea gas law:
- N number of molecules
- k Boltzmann’s constant (in joules/Kelvin)

- Solving:

- W is negative because we are doing work on the gas:V2 < V1
- W would be positive if the gas did work for us

- Where did the work go?
- Isothermal compression
- The temperature is constant (same before and after)

- First law: The work went into heating the bath
- Second law: We decreased the entropy of the gas
and increased the entropy of the bath

- Isothermal compression

- The total energy of the gas, U, remains unchanged
- Same number of particles
- Same temperature

- The “free energy” Fe, and the entropy S both change
- Both are related to the number of states (degrees of freedom)
- Fe = U – TS

- Both are related to the number of states (degrees of freedom)
- For our experiment, change in free energy is equal to the work done on the gas and U remains unchanged

DFe is the (negative) heat siphoned off into the bath

- Imagine that our gas contains only one molecule
- Take statistical averages of same molecule over time rather than over a population of particles
- Halve the volume
- Fe increases by +kTln2
- S decreases by kln2
- But U is constant

- What’s going on?
- Our knowledge of the possible locations of the particle has changed!
- Fewer places that themolecule can be in, now that volume has been halved
- The entropy, a measure of the uncertainty of a configuration, has decreased

- Take the probability of a gas configuration to be P
- Then S ~ klnP
- Random configurations (molecules moving haphazardly) have large P and large S
- Ordered configurations (all molecules moving in one direction) have small P and small S

- Then S ~ klnP
- The less we know about a gas…
- the more states it could be in
- and the greater the entropy

- A clear analogy with information theory

- Analysis is from Bennett: Tape cells with particles coding 0 (left side) or 1 (right side)
- If we know the message on a tape
- Then randomizing the tape can do useful work
- Increasing the tape’s entropy

- Then randomizing the tape can do useful work

What is the fuel value of the tape

(i.e. what is the fuel value of our knowledge)?

- The procedure
- Tape cell comes in with known particle location
- Orient a piston depending on whether cell is a 0 or a 1
- Particle pushes piston outward
- Increasing the entropy by kln2
- Providing free energy of kTln2 joules per bit

- Tape cell goes out with randomized particle location

- Define fuel value of tape = (N – I)kTln2
- N is the number of tape cells
- I is information (Shannon)

- Examples
- Random tape (I = N) has no fuel value
- Known tape (I = 0) has maximum fuel value

- Define the information in the tape to be the amount of free energy required to reset the tape
- The energy required to compress each bit to a known state
- Only the “surprise” bits cost us energy
- Doesn’t take any energy to reset known bits

- Cost to erase the tape: IkTln2 joules

For known bits, just move the partition (without changing the volume)

- A reversible computation, that proceeds at an infinitesimal rate, destroying no information, consumes no energy
- Regardless of the complexity of the computation
- The only cost is in resetting the machine at the end
- Erasing information costs energy

- Reversible computers are like heat engines
- If we run a reversible heat engine at an infinitesimal pace, it consumes no energy other than the work that it does

- We want our computations to run in finite time
- We need to drive the computation forward
- Dissipates energy (kinetic, thermal, etc.)

- We need to drive the computation forward
- Assume we are driving the computation forward at a rate r
- The computation is r times as likely to go forward as go backward

- What is the minimum energy per computational step?

- Computation is a transition between states
- State transitions have an associated energy diagram
- Assume forward state E2 has a lower energy than backward state E1
- “A” is the activation energy for a state transition

- Thermal fluctuations cause the computer to move between states
- Whenever the energy exceeds “A”

- State transitions have an associated energy diagram

We also used this model in neural networks (e.g. Hopfield networks)

- The probability of a transition between states differing in positive energy DE is proportional to exp(–DE/kT)
- Our state transitions have unequal probabilities
- The energy required for a forward step is (A – E1)
- The energy required for a backward step is (A – E2)

- The (reaction) rate r depends only on the energy difference between successive states
- The bigger (E1 – E2), the more likely the state transitions, and the faster the computation

- Energy expended per step = E1 – E2= kTlnr

- We can drive a computation even if the forward and backward states have the same energy
- As long as there are more forward states than backward states

- The computation proceeds by diffusion
- More likely to move into a state with greater availability
- Thermodynamic entropy drives the computation

- Protein synthesis is an example...
- of (nearly) reversible computation
- of the copy computation
- of a computation driven forward by thermodynamic entropy

- Protein synthesis is a 2-stage process
- 1. DNA forms mRNA
- 2. mRNA forms a protein

- We will consider step 1

- DNA comprises a double-stranded helix
- Each strand comprises alternating phosphate and sugar groups
- One of four bases attaches to each sugar
- Adenine (A)
- Thymine (T)
- Cytosine (C)
- Guanine (G)

- (base + sugar + phosphate) group is called a nucleotide

- DNA provides a template for protein synthesis
- The sequence of nucleotides forms a code

- RNA polymerase attaches itself to a DNA strand
- Moves along, building an mRNA strand one base at a time

- RNA polymerase catalyzes the copying reaction
- Within the nucleus there is DNA, RNA polymerase, and triphosphates (nucleotides with 2 extra phosphates), plus other stuff
- The triphosphates are
- adenosine triphosphate (ATP)
- cytosine triphosphate (CTP)
- guanine triphosphate (GTP)
- uracil triphosphate (UTP)

- The mRNA strand is complementary to the DNA
- The matching pairs are
DNARNA

A U

T A

C G

G C

- The matching pairs are
- As each nucleotide is added, two phosphates are released
- Bound as a pyrophosphate

- Catalysts influence the rate of a biochemical reaction
- But not the direction

- Chemical reactions are reversible
- RNA polymerase can unmake an mRNA strand
- Just as easily as it can make one
- Grab a pyrophosphate, attach to a base, and release

- RNA polymerase can unmake an mRNA strand
- The direction of the reaction depends on the relative concentrations of the pyrophosphates and triphosphates
- More triphosphates than pyrophosphates: Make RNA
- More pyrophosphates than triphosphates: Unmake RNA

- The relative concentrations of pyrophosphate and triphosphate define the number of states available
- Cells hydrolyze pyrophosphate to keep the reactions going forward

- How much energy does a cell use to drive this reaction?
- Energy = kTlnr = (S2 – S1)T ~ 100kT/bit

- Cells create protein engines (mRNA) for 100kT/bit
- 0.03µm transistors consume 100kT per switching event
- Think of representational efficiency
- What does each system get for 100kT?

- Digital logic uses an impoverished representation
- 104switching events to perform an 8-bit multiply
- Semiconductor scaling doesn’t improve the representation

- We pay a huge thermodynamic cost to use discrete math

- 104switching events to perform an 8-bit multiply

- Two reversible gates: controlled not (CN) and controlled controlled not (CCN).

A B C A’ B’ C’0 0 0 0 0 0

0 0 1 0 0 10 1 0 0 1 0

0 1 1 0 1 1

1 0 0 1 0 0

1 0 1 1 0 11 1 0 1 1 1

1 1 1 1 1 0

ABA’ B’000 0010 1101 1

111 0

CCN is complete: we can form any Boolean

function using only CCN gates: e.g. AND if C = 0

- Reversible Logic Gates and Quantum Computing
- Quantum versions of CN and CCN gates
- Quantum superposition of states allows exponential speedup

- Shor’s fast algorithm for factoring and breaking the RSA cryptosystem
- Grover’s database search algorithm
- Physical substrates for quantum computing

- Guest Lecturer: Dan Simon, Microsoft Research
- Introductory lecture on quantum computing and Shor’s algorithm

- Discussion and review afterwards
- Homework # 4 due: submit code and results electronically by Thursday (let us know if you have problems meeting the deadline)
- Sign up for project and presentation times
- Feel free to contact instructor and TA if you want to discuss your project
- Have a great weekend!