v2_n5 Error Correcting Codes (Part I).pdf

8/10/2019 v2_n5 Error Correcting Codes (Part I).pdf

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Volume 2. Number5

Nay-Dec. 996

Olympiad Corner

~ BiJ~ Ii aq II fiJ (-)

25th United States of America

Mathematicallympiad: ~ S m

Part I (9am-noon,May 2, 1996)

Pr bl IPr th th fth :W:~~~~:fj::;or\"m.m~fiiJ ' ~:5J,~

0 em ..o:e at e average 0 .e ~ffij:1fif~#;~~fiiJ'l1.:W~f&~~~

num~s n sm n (n = 2, 4, 6, ...,180) IS ~P).ljf- 0 ~:(3 .1if~~Mir{j\l:i-

cot 1 .~m.~~~fiiJJEJJ.' i1JQ~~~~

Problem 2. For any nonempty set S of fiiJ~WJ. 0

real numbers, et a(S) denote he sum of

the elementsof S. Given a set A of n

positive integers, consider he collection -.

of all distinct sumsa(S) as S rangesover

the nonempty subsetsof A. Prove tbat

this collection of sumscan bepartitioned

into n classesso that in each class, the

ratio of the largest sum to the smallest

sumdoesnot exceed . B

Problem 3. Let ABC be a ttiangle.

Prove that there s a line 1 (in the plane

of triangle ABq such that the

intersection of the interior of

triangle ABC and the interior of its

reflectionA' B' C' in I h~ areamore than

2/3 the area of triangle ABC.

&

AF

FB

Area of MCP

Area of ABCP

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ftm

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AF BD CE

=1 0

FB DC EA

~ ~ JJ.~ ~ * iflJA "@"][ (Giovanni

Ceva)1+-t;-tlt*C~a~fI'i1 ffrPJf&

A fi Z ~ r"@"][ (Ceva) JJ. 0 "@"

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c

(II-)

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Area of MBD BD

=-0

DCreaof MDC

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~:

~.=.~ ~ABC~g][~ (Cevians)~

rn~:.Jj;iJ;~~~]AD" BE~CF~

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(continuedon page 4)

A

F

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=1

FB DC EA

RU l:t=-g"K~#1 i 0

(*)

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B D C

(1].::.)

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CRf ~ ~ fii] ~ p 0 ~IJ fll:.))I ; #.iI

.:=. ~ (J{j~ ':tl ijJ ~

AreaofMBD BD AreaofMBD

--

Area of MDC DC Area of MDC

mffl$}.~'~ :

a c a c

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Area of MBP

Area of MCP

B

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CI)-=-)

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EA

Area of MCP

Area of MAP



2/4

Mathematical Excalibur. Vol.2. No.

Nav-Dec. 96

Pa~e 2

~gffl~~~~fiiJ (-)

(continued from page 1)

Error Correcting Codes (Part I)

Tsz-MeiKo

l:t~ (*):fa (**)"DJ~

AF AG

-=-

Supposeone would like to transmit a

message, say "HELLO.. .", from one

computer o another. One possibleway

is to use a table to encode he message

into binary digits. Then the receiver

would be able to decode he message

with a similar table. One such table is

the American Standard Code for

Information Interchange ASCII) shown

in Figure 1. The letter H would be

encodedas 1001000, he letter E would

be encoded s 1000101, tc. ~igure 2).

A 1000001 S 1010011 1100001 1110011

1000010 1010100 1100010 1110100

1000011 10101011100011 1110101

1000100 1010110 1100100 1110110

1000101 1010111 11001011110111

1000110 1011000 11qO110 l1l1000

1000111 1011001 1100111 l1l1001

1001000 1011010 1101000 l1l1010

1001001 '0110000 1101001 0101110

1001010 0110001 11101010 0101100

1001011 0110010 Ic 1101011 0ll1111

1001100 0110011 1 1101100 0101001

M 1001101 0110100 '

~1101101 l1l1011

N 1001110 0110101 1101110 010l1l1

J 100l1l1, 0110110 110l1l1 0100110

'" 1010000

I 011011111110000 I

I 0101011

1010001 0111000 c:r1110001 0101101

1~10010 0111001 r 1110010 0l1l101

FB GB

~Jl:tFWG~ABJ::~IffJ-1 i 0

:gg-][:iE~fII\]~:iE~~{PJ.~~ .f.:$

.L~.t87E-~Wfll\]E-i'lm~'~...:

E-I=P *1 (medians) # I i' E-~

(altitudes) I i' E- $j- ~ *1 (angle-

bisectors) I i 0 ~ E- i'l1 i $j- 7Jl ;:g t

m'L' (centroid)' ~ 'L' (orthocentre)~D

rf'\JWIiI'L'(incentre) 0

-~.#l im -.w-ar PJ.~H~~i1 ?

~-gg-][::iE:JJ.J'iJ~::iE:JJ.tf.m'rRJ~frj

~~o **-arjfrJffl#jf=-~~

~OO~~'f*' ~m (::I:/L\re~..$t

JjX,pjij~~~2:1J'iJ.~)m-~: 6' 1.

Ja~=-~~

ij, --I-.

E,'...,."

Figure4. Even parity code

Is there an encodingmethod so that the

receiver would be able to correct

transmissionerrors? Figure 5 showsone

such method by arranging the bit

sequence e.g., 1001) nto a rectangular

block and add parity bits to both rows

and columns. For the example shown,

1001 would be encodedas 10011111 by

fIrst appending he row parities and then

the column parities). If there s an error

during transmission, say at position 2,

the receiver can similarly arrange the

received sequence 11011111 into a

rectangularblock and detect hat there s

an error in row 1 and column2.

A

Figure 1. ASCII code

B

c

(/I)~)

Figure2. Two computers alking

The receiver will be able to decode he

message correctly if there is no error

during transmission. However, f there

are transmissionerrors, the receivermay

decode the message incorrectly. For

example, the letter H (1001000) would

be receivedas J (1001010) f there s an

error at position6.

~??)

D

AB

-=-0

DC AC

9J- lj ;f,m I i.:;l- ~ '/1 ijJ m 11}] J:t

'I)(I I ,/

I I

,Ix

*1

I 1

I I

(f~.)

Figure 5. A code that can correct1 error.

Figure 3. Error at position6.

One possible way to detect ransmission

errors is to add redundant bits, i.e.,

append extra bits to the original

The abovemethod can be used o correct

one error but rather costly. For every

four bits, one would need to transmit an

extra four redundant bits. Is there a

better way to do the encoding? n 1950,

Hamming ound an ngeniousmethod o


message.For an evenparity code,a 0 or

1 is appended o that the total number of

1's is an evennumber. The letters Hand

E would be representedby 10010000

and 10001011 respectively. With an

even parity code, he receiver can detect

one transmission error, but unable to

correct t. For example, f 10010000 for

the letter H) is receivedas 10010100, he

receiverknows that there is at least one

error during transmission since the

receivedbit sequence as an odd parity,

i.e., the total number of l' s is an odd

number.


3/4

Mathematical Excalibur, Vol. 2, No.

Nov-Dec. 96

P~e3

Problem Corner

We welcome readers to submit solutions

to the problems posed below for

publication consideration. Solutions

should be preceded by the solver's name,

address, school affiliation and grade

level. Please send submissions to Dr.

Kin-Yin li, Dept of Mathematics, Hong

Kong University of Science and

Technology, Clear Water Bay, Kowloon.

The deadline for submitting solutions is

Ian 31, 1997.

Problem 46. For what integer a does

X2 x +a divide Xl3 + x + 90?

Problem 47. If x, y, z are real numbers

such that X2+ y2 + Z2= 2, then show that

x + y + z ~ xyz + 2.

Problem 48. Squares ABDE and BCFG

are drawn outside of triangle ABC.

Prove that triangle ABC is isosceles if

DG is parallel to AC.

Problem 49. Let UI, U2, U3, ...be a

sequence of integers such that UI = 29,

U2= 45 and Un+2 Un+? -Un for n = 1, 2,

3, Show that 1996 divides infinitely

many terms of this sequence. (Source:

1986 Canadian Mathematical Olympiad

with modification)

Problem 42. What are the possible

valuesof oJ 2+ x + 1 -.J;2 ~ as x

ranges ver all realnumbers?

Solution: William CHEUNG Pot-man

(STFALeung KauKui College,Form 6).

Problem 50. Four integers are marked

on a circle. On each step we

simultaneously eplace each number by

the difference between his number and

next number on the circle in a given

direction (that is, the numbersa, b, c, d

are replaced by a -b, b -c, c -d, d -

a). Is it possibleafter 1996 such steps o

have numbers a, b, c, d such that the

numbers Ibc -adl, lac -bdl, lab-

cd I are primes? (Source: unused

problem n the 1996 MO.)

Problem 44. For an acute riangle ABC,

let H be the foot of the perpendicular

from A to BC. Let M, N be the feet of

the perpendiculars rom H to AB, AC,

respectively. Define LA to be the line

through A perpendicular to MN and

similarly define LB and Lc. Show that

LA, LB and Lc pass through a common

point O. (This was an unusedproblem

proposedby Iceland n a past MO.)

Let A=(x,O),B=(-t,~), c=(t,~).

The expression.J 2 + x + 1 -j;2=-;1

is just AB -AC. As x ranges over all

real numbers, A moves along the real

axis and he triangle inequality yields

-1 =-BC


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Mathematical Excalibur. Vol. 2, No. 5, Nov-Dec.96

PMe4

point P such hat LPAB = 10, LPBA =

20, LPCA = 30, LPAC = 40. Prove

that triangle ABC is isosceles.

Problem Corner

(continued rom page 3)

LA passes hrough the circumcenter O.

Similarly, LB and Lc will pass

through O.

:

~

, , Co

6

)(

Problem 6. Determine (with proof)

whether there is a subset X of the

integers with the following property: for

any integer n there is exactly one

solution of a + 2b = n with a, b E X.

~'""DC>A~'~"" -a

'\~~

t> ~i~

a+b+c

Adding 3 such inequalities, we get the

desired nequality. In fact, equality can

occur f and only if a = b = c = I.

-

..

.

Other commended olvers: POON Wing

Chi (La Salle College, Form 7) and YU

Chon Ling (HKU).

~~, /jl}.~t'

.i ??

~

-~""c:>C>

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v2_n5 Error Correcting Codes (Part I).pdf