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

1
CSE 599 Lecture 7 Information Theory,
Thermodynamics and Reversible Computing

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

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Overview of Todays Lecture

• 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!)

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Information and Algorithmic Complexity

• 3 principal results
• Shannons 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
• Shannons 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

4
What is information?

• 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

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Deterministic information I0

• 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)
• 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

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Is this definition satisfactory?

• Appeal to your intuition
• Which of these two messages contains more
  information?
• Dog bites man
• or
• Man bites dog

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Is this definition satisfactory?

• 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?

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Enter probability theory

• 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)
• Use the logarithm function log(xy) log(x)
  log(y)

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New Definition of Information

• 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)

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Review Axioms of probability theory

• Kolmogorov, 1933
• P(a) gt 0 where a is an event
• P(l) 1 where 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

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Review Results from probability theory

• 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(ab) probability
  that event a happens given that b has happened
• P(ab) P(ab)/P(b)
• P(ba) 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

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Summary Postulates of information theory

• 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

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In-Class Example

• 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)?

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In-Class Example

• 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)) log1/(P(xx)P(x))
• log(1/P(xx)) log(1/P(x))
• 0 log(1/P(x))
• I(x)
• Duplicates convey no additional information!

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Definition Entropy

• The average self-information or entropy of an
  ensemble X a1, a2, a3, aK
• E ? expected (or average) value

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Properties of Entropy

• 0 lt H(X) lt I0(X)
• Equals I0(X) log½X½ if all the aks 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 ?

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Examples

• 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

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Question

• 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?

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Shannons noiseless source-coding theorem

• 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 ? gt 0 and 0 lt ? lt 1, there exists a
  positive integer No such that, for N gt No,

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Comments on the theorem

• 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 ?

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Source coding (data compression)

• 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

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Source-coding definitions

• 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)

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Huffman coding

• 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

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Constructing 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
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Information channels
Input
Output
channel
x
y
I(XY) ? what we know about X given Y
H(X) ? entropy of input ensemble X

• I(XY) is the average mutual information between
  X and Y
• Definition Channel capacity
• The information capacity of a channel is C
  maxI(XY)
• The channel may add noise
• Corrupting our symbols

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Example Channel capacity

• 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
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Example Channel capacity (cont)

• 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 qlog1/q (1q)log1/(1q)
• 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

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How many code bits must we add?

• 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 huge Mc

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Shannons Channel-Coding Theorem

• The Theorem
• There is a nonnegative channel capacity C
  associated with each discrete memoryless channel
  with the following property For any symbol rate
  R lt C, and any error rate ? gt 0, there is a
  protocol that achieves a rate gt R and a
  probability of error lt ?
• 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) lt C
• Shannons 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

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Error-correction codes

• 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

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Hamming 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 gt (n m)
• Examples
• If n 11, m 4 24 1 15 gt (n
  m) 15
• If n 1013, m 10 210 1 1023 gt (n
  m) 1023

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Hamming codes (cont.)

• 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
• The parity-check bits are called a syndrome
• The syndrome tells us the location of the error

Position in message binary decimal 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
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Hamming codes (cont)

• 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!!

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Hamming codes (cont.)

• 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

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Kolmogorov Complexity (Algorithmic Information)

• 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?

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Kolmogorov Complexity Example

• Randomness of a string defined by shortest
  algorithm that can print it out.
• Suppose you were given the binary string x
• 11111111111111.11111111111111111111111 (1000
  1s)
• 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
• So, K(x) lt P
• Possible project topic Quantum Kolmogorov
  complexity?

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5-minute break

• Next Thermodynamics and Reversible Computing

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Thermodynamics and the Physics of Computation

• 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

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Principal Questions of Interest

• 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 informationentropy and
  thermodynamicentropy
• Is there a physical definition for information
  content?
• The information content of a message in physical
  units

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Main Results

• 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

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Basic thermodynamics

• 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
  gt 0
• Change in Entropy DS DQ/T
• Equality holds only for reversible processes
• The entropy of the universe is always increasing

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Heat engines

• A basic heat engine Q2 Q1 W
• T1 and T2 are temperatures
• T1 gt 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

43
Heat engines (cont.)
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The Second Law

• 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

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Thermodynamic entropy

• 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

46
Thermodynamic versus Information Entropy

• 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?

47
Compressing the gas

• 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 ? Boltzmanns constant (in joules/Kelvin)
• Solving

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A few notes

• W is negative because we are doing work on the
  gasV2 lt 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

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Free energy and entropy

• 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
• 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
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Special Case N 1

• 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

• Whats 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

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Thermodynamic entropy revisited

• 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
• 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

52
The fuel value of knowledge

• 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 tapes entropy

What is the fuel value of the tape (i.e. what is
the fuel value of our knowledge)?
53
Bennetts idea

• 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

54
The energy value of knowledge

• 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

55
Feynmans tape-erasing machine

• 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
• Doesnt 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)
56
Reversible Computing

• 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

57
Energy cost versus speed

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

58
Energy-driven computation

• 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

We also used this model in neural networks (e.g.
Hopfield networks)
59
State transitions

• 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)

60
Driving computation by energy differences

• 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

61
Driving computation by state availability

• 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

62
Rate-Driven Reversible Computing A Biological
Example

• 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

63
DNA

• 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

64
RNA polymerase

• 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)

65
mRNA

• The mRNA strand is complementary to the DNA
• The matching pairs are
• DNA RNA
• A U
• T A
• C G
• G C
• As each nucleotide is added, two phosphates are
  released
• Bound as a pyrophosphate

66
The process
67
RNA polymerase is a catalyst

• 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
• 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

68
DNA, entropy, and states

• 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

69
Efficiency of a representation

• 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
• 104 switching events to perform an 8-bit multiply
• Semiconductor scaling doesnt improve the
  representation
• We pay a huge thermodynamic cost to use discrete
  math

70
Example 2 Computing using Reversible Logic Gates

• Two reversible gates controlled not (CN) and
  controlled controlled not (CCN).

A B C A B C0 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
A B A B0 0 0 00 1 0 11 0 1
1 1 1 1 0
CCN is complete we can form any Boolean
function using only CCN gates e.g. AND if C 0
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Next Week Quantum Computing

• Reversible Logic Gates and Quantum Computing
• Quantum versions of CN and CCN gates
• Quantum superposition of states allows
  exponential speedup
• Shors fast algorithm for factoring and breaking
  the RSA cryptosystem
• Grovers database search algorithm
• Physical substrates for quantum computing

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Next Week

• Guest Lecturer Dan Simon, Microsoft Research
• Introductory lecture on quantum computing and
  Shors 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!