Coding%20Theory,%20Card%20Tricks%20and%20Hat%20Problems

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Coding Theory, Card Tricks and Hat Problems
Ho Chi Minh City, September 30, 2007

• Michel Waldschmidt
• Université P. et M. Curie - Paris VI
• Centre International de Mathématiques Pures et
  Appliquées - CIMPA

http//www.math.jussieu.fr/miw/coursHCMUNS2007.ht
ml
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Ho Chi Minh City
September 30, 2007
Coding Theory, Card Tricks and Hat Problems
Starting with card tricks, we show how
mathematical tools are used to detect and to
correct errors occuring in the transmission of
data. These so-called "error-detecting codes"
and "error-correcting codes" enable
identification and correction of the errors
caused by noise or other impairments during
transmission from the transmitter to the
receiver. They are used in compact disks to
correct errors caused by scratches, in satellite
broadcasting, in digital money transfers, in
telephone connexions, they are useful for
improving the reliability of data storage media
as well as to correct errors cause when a hard
drive fails. The National Aeronautics and Space
Administration (NASA) has used many different
error-correcting codes for deep-space
telecommunications and orbital missions.
http//www.math.jussieu.fr/miw/
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Ho Chi Minh City
September 30, 2007
Coding Theory, Card Tricks and Hat Problems
Most of the theory arises from earlier
developments of mathematics which were far
removed from any concrete application. One of the
main tools is the theory of finite fields, which
was invented by Galois in the XIXth century, for
solving polynomial equations by means of
radicals. The first error-correcting code
happened to occur in a sport newspaper in Finland
in 1930. The mathematical theory of information
was created half a century ago by Claude Shannon.
The mathematics behind these technical devices
are being developped in a number of scientific
centers all around the world, including in
Vietnam and in France.
http//www.math.jussieu.fr/miw/
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I know which card you selected

• Among a collection of playing cards, you select
  one without telling me which one it is.
• I ask you some questions where you answer yes or
  no.
• Then I am able to tell you which card you
  selected.

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2 cards

• You select one of these two cards
• I ask you one question and you answer yes or no.
• I am able to tell you which card you selected.

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2 cards one question suffices

• Question is-it this one?

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4 cards
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First question is-it one of these two?
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Second question is-it one of these two ?
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4 cards 2 questions suffice
Y Y
Y N
N Y
N N
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8 Cards
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First question is-it one of these?
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Second question is-it one of these?
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Third question is-it one of these?
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8 Cards 3 questions
YYY
YYN
YNY
YNN
NYY
NYN
NNY
NNN
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Yes / No

• 0 / 1
• Yin / Yang - -
• True / False
• White / Black
• / -
• Heads / Tails (tossing a coin)

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3 questions, 8 solutions
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
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16 Cards

• If you select one card among a set of 16, I
  shall know which one it is, once you answer my 4
  questions by yes or no.

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(No Transcript)
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Label the 16 cards
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In binary expansion
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Ask the questions so that the answers are
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The 4 questions

• Is the first digit 0 ?
• Is the second digit 0 ?
• Is the third digit 0 ?
• Is the fourth digit 0 ?

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More difficult

• One answer may be wrong!

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One answer may be wrong

• Consider the same problem, but you are allowed to
  give (at most) one wrong answer.
• How many questions are required so that I am able
  to know whether your answers are right or not?
  And if they are right, to know the card you
  selected?

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Detecting one mistake

• If I ask one more question, I shall be able to
  detect if there is one of your answers which is
  not compatible with the others.
• And if you made no mistake, I shall tell you
  which is the card you selected.

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Detecting one mistake with 2 cards

• With two cards I just repeat twice the same
  question.
• If both your answers are the same, you did not
  lie and I know which card you selected
• If your answers are not the same, I know that one
  is right and one is wrong (but I dont know which
  one is correct!).

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4 cards
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First question is-it one of these two?
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Second question is-it one of these two?
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Third question is-it one of these two?
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4 cards 3 questions
Y Y Y
Y N N
N Y N
N N Y
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4 cards 3 questions
0 0 0
0 1 1
1 0 1
1 1 0
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Correct triple of answers
Wrong triple of answers
One change in a correct triple of answers yields
a wrong triple of answers
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Boolean addition

• even even even
• even odd odd
• odd even odd
• odd odd even

• 0 0 0
• 0 1 1
• 1 0 1
• 1 1 0

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Parity bit

• Use one more bit which is the Boolean sum of the
  previous ones.
• Now for a correct answer the sum of the bits
  should be 0.
• If there is exactly one error, the parity bit
  will detect it the sum of the bits will be 1
  instead of 0.

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8 Cards
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4 questions for 8 cards
Use the 3 previous questions plus the parity bit
question
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First question is-it one of these?
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Second question is-it one of these?
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Third question is-it one of these?
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Fourth question is-it one of these?
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16 cards, at most one wrong answer 5 questions
to detect the mistake
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Ask the 5 questions so that the answers are
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Correcting one mistake

• Again I ask you questions where your answer is
  yes or no, again you are allowed to give at most
  one wrong answer, but now I want to be able to
  know which card you selected - and also to tell
  you whether and when you lied.

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With 2 cards

• I repeat the same question three times.
• The most frequent answer is the right one vote
  with the majority.

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With 4 cards
I repeat my two questions three times each,
which makes 6 questions
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With 8 Cards
I repeat 3 times my 3 questions, which makes 9
questions
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With 16 cards, this process requires 3?412
questions
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16 cards, 7 questions

• Recall if you select one card among 16, only 3
  questions are required if all answers are correct.

• With 4 questions, we detect one mistake.

• We shall see that 7 questions allow to recover
  the exact result if there is at most one wrong
  answer.

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The Hat Problem
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The Hat Problem

• Three people are in a room, each has a hat on his
  head, the color of which is black or white. Hat
  colors are chosen randomly. Everybody sees the
  color of the hat on everyones head, but not on
  their own. People do not communicate with each
  other.
• Everyone gets to guess (by writing on a piece of
  paper) the color of their hat. They may write
  Black/White/Abstain.

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• The people in the room win together or lose
  together.
• The team wins if at least one of the three people
  did not abstain, and everyone who did not abstain
  guessed the color of their hat correctly.
• How will this team decide a good strategy with a
  high probability of winning?

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• Simple strategy they agree that two of them
  abstain and the other guesses randomly.
• Probability of winning 1/2.
• Is it possible to do better?

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• Hint
• Improve the odds by using the available
  information everybody sees the color of the hat
  on everyones head but himself.

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• Better strategy if a member sees two different
  colors, he abstains. If he sees the same color
  twice, he guesses that his hat has the other
  color.

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• The two people with white hats see one white
  hat and one black hat, so they abstain.

The one with a black hat sees two white hats,
so he writes black.
They win!
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• The two people with black hats see one white
  hat and one black hat, so they abstain.

The one with a white hat sees two black hats,
so he writes white.
They win!
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Everybody sees two white hats, and therefore
writes black on the paper.

• They lose

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Everybody sees two black hats, and therefore
writes white on the paper.

• They lose

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• Winning

two white or two black
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• Losing

three white or three black
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• The team wins exactly when the three hats do not
  have all the same color, that is in 6 cases out
  of a total of 8
• Probability of winning 3/4.

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• Now if the hats are no more distributed randomly
  and if the master of the game knows the previous
  strategy, he may select often the three hats all
  of the same color.

• Is there another strategy giving the same
  probability 3/4, for randomly distributed colors?

Answer YES!
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• The team selects one distribution of hats and
  bets that it will not be this one,

neither the opposite distribution where black
is replaced by white and white by black
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• If the distribution of the colors is one of
  these, they will lose.

• If the distribution of the colors is one of the 6
  other possibilities, exactly one of the three
  people will be able to guess correctly the color
  of his hat, and the two others will abstain, so
  the team will win.

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Hat problem with 7 people
For 7 people in the room in place of 3, which is
the best strategy and its probability of
winning?
Answer the best strategy gives a probability
of winning of 7/8
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Winning at the lottery
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Tails and Ends

• Toss a coin 3 consecutive times
• There are 8 possible sequences of results
• (0,0,0), (0,0,1), (0,1,0), (0,1,1),
• (1,0,0), (1,0,1), (1,1,0), (1,1,1).

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If you bet (0,1,0), you have

• All three correct results for (0,1,0).
• Exactly two correct results if the sequence is
  either (0,1,1), (0,0,0) or (1,1,0),
• Exactly one correct result if the sequence is
  either (0,0,1), (1,1,1) or (1,0,0),
• No correct result at all for (1,0,1).

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Among 8 possibilities

• each bet
• is winning in exactly 1 case
• has exactly two correct results in 3 cases
• has exactly one correct result in 3 cases
• has no correct result at all in only 1 case

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• Goal To be sure of having at least 2 correct
  results
• Clearly, one bet is not sufficient
• Are two bets sufficient?
• Recall that there are 8 possible results, and
  that each bet has at least 2 correct results in 4
  cases.

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Answer YES, two bets
suffice!

• For instance bet
• (0,0,0) and (1,1,1)
• Whatever the result is, one of the two digits
• 0 and 1
• occurs more than once.
• Hence one (and only one) of the two bets
• has at least two correct results.

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• The 8 sequences of three digits
• 0 and 1
• split into two groups
• 4 having two or three 0s
• 000, 001, 010, 100
• and
• 4 having two or three 1s
• 111, 110, 101, 011

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Other solutions

• Select any two bets with all three different
  digits, say
• 1 0 1 and 0 1 0
• The result either is one of these, or else has
  two common digits with one of these and just one
  common digit with the other.

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• Select 101 and 010
• The 8 sequences of three digits
• 0 and 1
• split into two groups
• 4 having two or three digits which
  are the same as 101
• 101, 100, 111, 001
• and
• 4 having two or three digits which are the
  same as 010
• 010, 011, 000, 110

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Toss a coin 7 consecutive times

• There are 27 128 possible sequences of results.
• How many bets are required in such a way that you
  are sure one at least of them has at most one
  wrong answer?

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Tossing a coin 7 times

• Each bet has all correct answers once every 128
  cases.
• It has just one wrong answer 7 times either the
  first, second, seventh guess is wrong.
• So it has at most one wrong answer 8 times among
  128 possibilities.

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Tossing a coin 7 times

• Now 128 8 ? 16. Indeed 128 is 27, and 8 is 23.
  27 23 ? 24
• Therefore you cannot achieve your goal with less
  than 16 bets.
• Coding theory tells you how to select your 16
  bets, exactly one of them will have at most one
  wrong answer.

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SPORT TOTO

• A match between two players (or teams) may give
  three possible results either player 1 wins, or
  player 2 wins, or else there is a draw.
• There is a lottery, and a winning ticket needs to
  have at least three correct bets. How many
  tickets should one buy to be sure to win?

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4 matches, 3 correct forecasts

• For 4 matches, there are 34 81 possibilities.
• A bet on 4 matches is a sequence of 4 symbols
  1, 2, x. Each such ticket has exactly 3 correct
  answers 8 times. For instance the ticket
  (1, 2, x, 1) has 3 correct answers for the
  results
• (2, 2, x, 1), (1, 1, x, 1), (1, 2, 1, 1), (1, 2,
  x, 2)
• (x, 2, x, 1), (1, x, x, 1), (1, 2, 2, 1), (1, 2,
  x, x)

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4 matches, 3 correct forecasts

• Each ticket has all correct answers once and 3
  correct answers 8 times
• Hence it has at least 3 correct answers in 9
  cases.
• Altogether there are 81 9 ? 9 possible results.
• So at least 9 tickets are required to be sure to
  win.
• How to select such 9 tickets?

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9 tickets (in columns)

• x x x 1 1 1 2 2
  2
• x 1 2 x 1 2 x
  1 2
• x 1 2 1 2 x 2 x
  1
• x 1 2 2 x 1 1 2
  x

This is an error correcting code on the
alphabet 1,2,x with rate 1/2
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Error correcting codes
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Coding Theory

• Coding theory is the branch of mathematics
  concerned with transmitting data across noisy
  channels and recovering the message. Coding
  theory is about making messages easy to read
  don't confuse it with cryptography which is the
  art of making messages hard to read!

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Claude Shannon

• In 1948, Claude Shannon, working at Bell
  Laboratories in the USA, inaugurated the whole
  subject of coding theory by showing that it was
  possible to encode messages in such a way that
  the number of extra bits transmitted was as small
  as possible. Unfortunately his proof did not give
  any explicit recipes for these optimal codes.

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Richard Hamming

• Around the same time, Richard Hamming, also at
  Bell Labs, was using machines with lamps and
  relays having an error detecting code. The digits
  from 1 to 9 were send on ramps of 5 lamps with
  two on and three out. There were very frequent
  errors which were easy to detect and then one had
  to restart the process.

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The first correcting codes

• For his researches, Hamming was allowed to have
  the machine working during the week-end only, and
  they were on the automatic mode. At each error
  the machine stopped until the next monday
  morning.
• "If it can detect the error," complained
  Hamming, "why can't it correct it! "

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The origin of Hammings code

• He decided to find a device so that the machine
  would not only detect the errors but also correct
  them.
• In 1950, he published details of his work on
  explicit error-correcting codes with information
  transmission rates more efficient than simple
  repetition.
• His first attempt produced a code in which four
  data bits were followed by three check bits which
  allowed not only the detection but the correction
  of a single error.

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Codes and Geometry

• 1949 Marcel Golay (specialist of radars)
  produced two remarkably efficient codes.
• Eruptions on Io (Jupiters volcanic moon)
• 1963 John Leech uses Golays ideas for sphere
  packing in dimension 24 - classification of
  finite simple groups
• 1971 no other perfect code than the two found by
  Golay.

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Error Correcting Codes Data Transmission

• Telephone
• CD or DVD
• Image transmission
• Sending information through the Internet
• Radio control of satellites

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Applications of error correcting codes

• Transmitions by satellites
• Compact discs
• Cellular phones

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• Olympus Mons on Mars Planet
• Image from Mariner 2 in 1971.

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• Information was sent to the earth using an
  error correcting code which corrected 7 bits
  among 32.
• In each group of 32 bits, 26 are control bits
  and the 6 others contain the information.

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• Between 1969 and 1973 the NASA Mariner probes
  used a powerful Reed--Muller code capable of
  correcting 7 errors out of 32 bits transmitted,
  consisting now of 6 data bits and 26 check bits!
  Over 16,000 bits per second were relayed back to
  Earth.

The North polar cap of Mars, taken by Mariner 9
in 1972.
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Voyager 1 and 2 (1977)

• Journey Cape Canaveral, Jupiter, Saturn, Uranus,
  Neptune.
• Sent information by means of a binary code which
  corrected 3 errors on words of length 24.

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Mariner spacecraft 9 (1979)

• Sent black and white photographs of Mars
• Grid of 600 by 600, each pixel being assigned one
  of 64 brightness levels
• Reed-Muller code with 64 words of 32 letters,
  minimal distance 16, correcting 7 errors, rate
  3/16

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Voyager (1979-81)

• Color photos of Jupiter and Saturn
• Golay code with 4096212 words of 24 letters,
  minimal distance 8, corrects 3 errors, rate 1/2.
• 1998 lost of control of Soho satellite recovered
  thanks to double correction by turbo code.

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NASA's Pathfinder mission on Mars

• The power of the radio transmitters on these
  craft is only a few watts, yet this information
  is reliably transmitted across hundreds of
  millions of miles without being completely
  swamped by noise.

Sojourner rover and Mars Pathfinder lander
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Listening to a CD

• On a CD as well as on a computer, each sound is
  coded by a sequence of 0s and 1s, grouped in
  octets
• Further octets are added which detect and correct
  small mistakes.
• In a CD, two codes join forces and manage to
  handle situations with vast number of errors.

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Coding the sound on a CD
On CDs the signal in encoded digitally. To guard
against scratches, cracks and similar damage,
two "interleaved" codes which can correct up to
4,000 consecutive errors (about 2.5 mm of track)
are used.

• Using a finite field with 256 elements, it is
  possible to correct 2 errors in each word of 32
  octets with 4 control octets for 28 information
  octets.

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A CD of high quality may have more than 500
000 errors!

• After processing of the signal in the CD player,
  these errors do not lead to any disturbing noise.
• Without error-correcting codes, there would be no
  CD.

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1 second of radio signal 1 411 200 bits

• The mathematical theory of error correcting codes
  provides more reliability and at the same time
  decreases the cost. It is used also for data
  transmission via the internet or satellites

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Codes and Mathematics

• Algebra
• (discrete mathematics finite fields, linear
  algebra,)
• Geometry
• Probability and statistics

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Finite fields and coding theory

• Solving algebraic equations with
  radicals Finite fields theory
  Evariste Galois
  (1811-1832)
• Construction of regular polygons with rule and
  compass
• Group theory

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Coding Theory
transmission
Source Coded Text Coded Text Receiver
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Error correcting codes
Transmission with noise
Source Coded Text Coded Text Receiver
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Principle of coding theory

• Only certain words are allowed (code
  dictionary of valid words).
• The  useful  letters carry the information,
  the other ones (control bits) allow detecting or
  correcting errors.

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Detecting one error by sending twice the message

• Send twice each bit
• 2 code words among 422 possible words
• (1 useful letter among 2)

• Code words
• (two letters)
• 0 0
• and
• 1 1
• Rate 1/2

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• Principle of codes detecting one error
• Two distinct code words
• have at least two distinct letters

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Detecting one error with the parity bit

• Code words (three letters)
• 0 0 0
• 0 1 1
• 1 0 1
• 1 1 0
• Parity bit (x y z) with zxy.
• 4 code words (among 8 words with 3 letters),
• 2 useful letters (among 3).
• Rate 2/3

2
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Code Words Non Code Words

• 0 0 0 0 0 1
• 0 1 1 0 1 0
• 1 0 1 1 0 0
• 1 1 0 1 1 1
• Two distinct code words
• have at least two distinct letters.

2
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Check bit

• In the International Standard Book Number (ISBN)
  system used to identify books, the last of the
  ten-digit number is a check bit.
• The Chemical Abstracts Service (CAS) method of
  identifying chemical compounds, the United States
  Postal Service (USPS) use check digits.
• Modems, computer memory chips compute checksums.
• One or more check digits are commonly embedded in
  credit card numbers.

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Correcting one errorby repeating three times

• Code words
• (three letters)
• 0 0 0
• 1 1 1
• Rate 1/3

• Send each bit three times
• 2 code words
• among 8 possible ones
• (1 useful letter among 3)

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• Correct 0 0 1 as 0 0 0
• 0 1 0 as 0 0 0
• 1 0 0 as 0 0 0
• and
• 1 1 0 as 1 1 1
• 1 0 1 as 1 1 1
• 0 1 1 as 1 1 1

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• Principle of codes correcting one error
• Two distinct code words have at least three
  distinct letters

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Hamming Distance between two words

• number of places in which the two words
  
• differ
• Examples
• (0,0,1) and (0,0,0) have distance 1
• (1,0,1) and (1,1,0) have distance 2
• (0,0,1) and (1,1,0) have distance 3
• Richard W. Hamming (1915-1998)

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Hamming distance 1
120
Hammings unit sphere

• The unit sphere around a word includes the words
  at distance at most 1

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At most one error
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Words at distance at least 3
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Decoding
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Hamming Distance

• The Hamming distance between two words is the
  number of places where letters differ.
• A code detects n errors iff the Hamming distance
  between two distinct code words is at least 2n.
• It corrects n errors iff the Hamming distance
  between two distinct code words is at least
  2n1.

125
The code (0 0 0) (1 1 1)

• The set of words with three letters (eight
  elements) splits into two balls
• The centers are (0,0,0) and (1,1,1)
• Each of the two balls consists of elements at
  distance at most 1 from the center

126
Two or three 0
Two or three 1

• (0,0,1)
• (0,0,0) (0,1,0)
• (1,0,0)

• (1,1,0)
• (1,1,1) (1,0,1)
• (0,1,1)

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Another error correcting code on 3 letters

• Select two words with mutual distance 3
  two words with three distinct letters, say
  (0,0,1) and (1,1,0)
• For each of them, consider the ball of elements
  at distance at most 1

128

• (0,0,0)
• (0,0,1) (0,1,1)
• (1,0,1)

• (1,1,1)
• (1,1,0) (1,0,0)
• (0,1,0)

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Back to the Hat Problem

• Replace white by 0 and black by 1
• hence the distribution of colors becomes a
  word of three letters on the alphabet 0 , 1
• Consider the centers of the balls (0,0,0) and
  (1,1,1).
• The team bets that the distribution of colors is
  not one of the two centers.

130
Assume the distribution of hats does not
correspond to one of the centers (0, 0, 0) and
(1, 1, 1). Then

• One color occurs exactly twice (the word has both
  digits 0 and 1).
• Exactly one member of the team sees twice the
  same color this corresponds to 0 0 in case he
  sees two white hats, 1 1 in case he sees two
  black hats.
• Hence he knows the center of the ball (0 , 0 ,
  0) in the first case, (1, 1, 1) in the second
  case.
• He bets the missing digit does not yield the
  center.

131

• The two others see two different colors, hence
  they do not know the center of the ball. They
  abstain.
• Therefore the team wins when the distribution of
  colors does not correspond to one of the centers
  of the balls.
• This is why the team wins in 6 cases.

132

• Now if the word corresponding to the distribution
  of the hats is one of the centers, all members of
  the team bet the wrong answer!
• They lose in 2 cases.

133
Another strategy

• Select two words with mutual distance 3
  two words with three distinct letters, say
  (0,0,1) and (1,1,0)
• For each of them, consider the ball of elements
  at distance at most 1

134

• The team bets that the distribution of colors is
  not one of the two centers (0,0,1), (1,1,0).
• A word not in the center has exactly one letter
  distinct from the center of its ball, and two
  letters different from the other center.

135
Assume the word corresponding to the distribution
of the hats is not a center. Then

• Exactly one member of the team knows the center
  of the ball. He bets the distribution does not
  correspond to the center.
• The others do not know the center of the ball.
  They abstain.
• Hence the team wins.

136
The binary code of Hamming (1950)

• It is a linear code (the sum of two code words
  is a code word) and the 16 balls of radius 1 with
  centers in the code words cover all the space of
  the 128 binary words of length 7
• (each word has 7 neighbors (71)?16 256).

137
Hamming code

• Words of 7 letters
• Code words (1624 among 12827)
• (a b c d e f g)
• with
• eabd
• facd
• gabc
• Rate 4/7

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Hammings code
139
How to compute e , f , g from a , b , c , d.
eabd
d
a
b
facd
c
gabc
140
16 code words of 7 letters

• 0 0 0 0 0 0 0
• 0 0 0 1 1 1 0
• 0 0 1 0 0 1 1
• 0 0 1 1 1 0 1
• 0 1 0 0 1 0 1
• 0 1 0 1 0 1 1
• 0 1 1 0 1 1 0
• 0 1 1 1 0 0 0

• 1 0 0 0 1 1 1
• 1 0 0 1 0 0 1
• 1 0 1 0 1 0 0
• 1 0 1 1 0 1 0
• 1 1 0 0 0 1 0
• 1 1 0 1 1 0 0
• 1 1 1 0 0 0 1
• 1 1 1 1 1 1 1

Two distinct code words have at least three
distinct letters
141
The Hat Problem with 7 people

• The team bets that the distribution of the hats
  does not correspond to the 16 elements of the
  Hamming code
• Loses in 16 cases (they all fail)
• Wins in 128-16112 cases (one bets correctly, the
  6 others abstain)
• Probability of winning 112/1287/8

142
Tossing a coin 7 times

• There are 128 possible results
• Each bet is a word of 7 letters on the alphabet
  0, 1
• How many bets do you need if you want to
  guarantee at least 6 correct results?

143

• Each of the 16 code words has 7 neighbors (at
  distance 1), hence the ball of which it is the
  center has 8 elements.
• Each of the 128 words is in exactly one of these
  balls.

144

• Make 16 bets corresponding to the 16 code words
  then, whatever the final result is, exactly one
  of your bets will have at least 6 correct
  answers.
• Notice that 16/1281/8. This is the probability
  of losing for the hat problem with 7 people using
  the Hamming code.

145
7 questions to find the selected card among 16,
with one possible wrong answer

• Replace the cards by labels from 0 to 15 and
  write the binary expansions of these
• 0000, 0001, 0010, 0011
• 0100, 0101, 0110, 0111
• 1000, 1001, 1010, 1011
• 1100, 1101, 1110, 1111

146
7 questions to find the selected number in
0,1,2,,15 with one possible wrong answer

• Is the first binary digit 1?
• Is the second binary digit 1?
• Is the third binary digit 1?
• Is the fourth binary digit 1?
• Is the number in 1,2,4,7,9,10,12,15?
• Is the number in 1,2,5,6,8,11,12,15?
• Is the number in 1,3,4,6,8,10,13,15?

147
Sphere Packing

• While Shannon and Hamming were working on
  information transmission in the States, John
  Leech invented similar codes while working on
  Group Theory at Cambridge. This research included
  work on the sphere packing problem and culminated
  in the remarkable, 24-dimensional Leech lattice,
  the study of which was a key element in the
  programme to understand and classify finite
  symmetry groups.

148
Sphere packing
The kissing number is 12
149
Sphere Packing

• Kepler Problem maximal density of a packing of
  identical sphères
•   p / Ö 18 0.740 480 49
• Conjectured in 1611.
• Proved in 1999 by Thomas Hales.
• Connections with crystallography.

150
Current trends

• In the past two years the goal of finding
  explicit codes which reach the limits predicted
  by Shannon's original work has been achieved. The
  constructions require techniques from a
  surprisingly wide range of pure mathematics
  linear algebra, the theory of fields and
  algebraic geometry all play a vital role. Not
  only has coding theory helped to solve problems
  of vital importance in the world outside
  mathematics, it has enriched other branches of
  mathematics, with new problems as well as new
  solutions.

151
Research going on in France

• INRIA Rocquencourt, Projet Codes
  http//www-rocq.inria.fr/codes/
• Marseille, Institut de Mathématiques de Luminy
  iml.univ-mrs.fr
• Limoges http//www.xlim.fr
• Toulon Groupe de Recherche en Informatique et
  Mathématiques GRIM http//grim.univ-tln.fr

152
Research going on in France

• Bordeaux Institut de Mathématiques de Bordeaux
  A2X Théorie des nombres et Algorithmique
  Arithmétique
• Equipe Codes et Réseaux
• http//www.math.u-bordeaux1.fr/maths/
  spip.php?article9

153
Research going on in France

• Toulouse Groupe de Recherche en Informatique et
  Mathématiques du Mirail (GRIMM)
• http//www.univ-tlse2.fr/LB028/0/
• fiche___laboratoire/RH02Equipes
• ENST (Télécom) Bretagne
• http//departements.enst-bretagne.fr/sc/recherche
  /turbo/

154
GDR IM ''Informatique Mathématique''

• Groupe de travail C2 Cryptographie et Codage
• http//www.gdr-im.fr/
• http//www-rocq.inria.fr/codes/Claude.Carlet/c
  omp.html

155
Directions of research

• Theoretical questions of existence of specific
  codes
• connection with cryptography
• lattices and combinatoric designs
• algebraic geometry over finite fields
• equations over finite fields

156
A last card trick

• The best card trick
• Michael Kleber, Mathematical Intelligencer 24
  (2002)

157
A card trick

• Among 52 cards, select 5 of them, do not show
  them to me, but give them to my assistant.
• After looking at these 5 cards, my assistant
  gives me four of them, one at a time.
• There is one card left which is known to you and
  to my assistant.
• I am able to tell you which one it is.

158
Which information did I receive?

• I received 4 cards, one at a time. With my
  assistant we agreed beforehand with an ordering.
• I receive these 4 cards in one of 24 possible
  orders.
• There are 4 choices for the first card, once the
  first card is selected there are 3 choices for
  the second card, and then 2 choices for the third
  one. And finally no choice for the last one.
• 24 4 ? 3 ? 2 ? 1

159
24 possible orders for 4 cards

• I receive these 4 cards in one of the 24
  following orders
• 1234, 1243, 1324, 1342, 1423, 1432,
• 2134, 2143, 2314, 2341, 2413, 2431,
• 3124, 3142, 3214, 3241, 3412, 3421,
• 4123, 4132, 4213, 4231, 4312, 4321.
• So the information I receive can be converted
  into a number from 1 to 24

160
But there are 52 cards!

• Therefore this idea is not sufficient.
• There are four colors
• Spade, Hart, Diamond, Club

My assistant received 5 cards.
161
Pigeonhole or Box Principle

• If there are more pigeons than holes, one at
  least of the holes hosts at least two pigeons.
• If there are more holes than pigeons, one at
  least of the holes is empty.

My assistant received 5 cards, there are 4
colours So one at least of the colors occurs
twice.
We agree that the first card I receive will tell
me the colour of the hidden card.
162
Information I receive from the next 3 cards

• Once I receive the first card, I know the color
  of the hidden card. I need to find out which one
  it is among the 12 others of the same color.
• Next I receive 3 cards in one of the 6 possible
  orders. I convert this information into a number
  from 1 to 6.
• My assistant had the choice between two cards for
  the first I received.

163
Count from 1 to 6
164
Coding Theory, Card Tricks and Hat Problems
Ho Chi Minh City, September 30, 2007

• Michel Waldschmidt
• Université P. et M. Curie - Paris VI
• Centre International de Mathématiques Pures et
  Appliquées - CIMPA

http//www.math.jussieu.fr/miw/coursHCMUNS2007.ht
ml