# Error Correcting Codes _I_

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Volume 2. Number5                                                                                                                                                                                1996
Nay-Dec.

~ BiJ~ Ii aq II fiJ (-)
25th               United     States              of       America
Mathematical                                                                                                                                           ~ S m
Part I (9am-noon,May 2, 1996)
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num~s n sm n (n = 2, 4, 6, ...,180) IS                                       ~P).ljf- 0 ~:(£3!.1if~~Mir{j\l:i-                                                     AF     Area of MCP
cot 1 .~m.~~~fiiJJEJJ.'                                                                              !i1JQ~~~~                                                     FB     Area of ABCP

Problem 2. For any nonempty set S of fiiJ~WJ. 0
real numbers,let a(S) denotethe sum of                                                                                                                                     ~       * ~O
~ .l:.jzI!;'=-~.1:\ 1JI.!J;f§ ~ ~ CJM
the elementsof S. Given a set A of n .:fJtfrlj:\$'r.~~:PJ~tt'g-flM¥[fij":m:                                                                                ftm
positive integers, considerthe collection ~tt'g'l1.W~~ :
-.                                                                                                                 AF BD CE =1             0
of all distinct sumsa(S) as S rangesover                                                                                                                             FB DC EA
the nonempty subsetsof A. Prove tbat
this collection of sumscan bepartitioned                                                                                                                        ~
~ ~ JJ. ~ * iflJA "@"][ (Giovanni
into n classesso that in eachclass, the                                                                                                                   Ceva)1£+-t;-tlt*C~a~fI'i1' ffrPJf&
ratio of the largest sum to the smallest                                                                                                                                          ~
A fi Z ~ r"@"][ (Ceva) JJ. j 0 "@"
sumdoesnot exceed     2.                           B             c                                                                                        ][~JJ."CJm~lPJlJ~m~)! , f.§.-.t~
Problem 3. Let ABC be a ttiangle.                                                                                     (II-)                               m~):5~:tiP.ifIJm T m"III~OOtRffi
Prove that there is a line 1 (in the plane                                                                                                                ~fI'i1:=f:.~' f~ff**mm    0
of triangle ABq          such that the                                      !il-~moo':='~~:;FJ/;}#~AD'
intersection of the interior            of                                  ~irJ~~~~      ' ~!l:t                                                         g§][~:@.~~\$Q~~{t?~Vi?m
triangle ABC and the interior of its                                                                                                                      m~\$Q,tt.~re~\$Q~~*.ff
Area of MBD                    BD
reflectionA' B' C' in I h~ areamore than                                                                                       =-0                        ~°tt.g§][~:@.*~'~~~\$Q
2/3 the area of triangle ABC.                                                                    Areaof MDC                          DC                   ~:
(continuedonpage 4)                                                                       A
~.=.~ ~ABC~g§][~                         [~
(Cevians)
~
Jf:. fIt-
F        ;~..                                         AF BD CE
=1                  (*)
/..-   .::.t:
:~                                          FB DC EA
B              D             C                    RU!l:t=-g§"K~#1!i            0
(1].::.)
ffl
1l:t~~:@.al;Jm\$~ I!i ~ DJff§: ~ ~
111=, iii ~ (J{j ~ AD ' BE~D :@.:
W;Jlt             ~                 B~=-~][.f!:'f'-~#I5'                                                      fH
CRf§ ~ ~ fii]-~                              p 0 ~IJffll:.))I!; #.iI         ;tt~~*g§][.f!£\~#15       0 ~irlJDJ
.:=. ~ (J{j ':tl ijJ ~
][.f!CG, ~U
AreaofMBD -- BD                                   AreaofMBD
Area of MDC  DC                                   Area of MDC                        ~.~.~=1                      (**)
GR DC EA
mffl\$}.~'~ :
A
;E=-=- , ~U-=-= a-c
a c      a c
b          d               b      d    b-d
E
CJlf
BD                Area of MBP                                     B         D
DC                Area of MCP
CI)-=-)
IRJ~
CE Area of MCP
EA            Area of MAP                                               (continuedonpage 2)
Mathematical Excalibur. Vol.2. No.               Nav-Dec. 96                                                                          Pa~e 2

~gffl~~~~fiiJ                   (-)
(continued from page 1)
Error Correcting Codes (Part I)
.l:t~ (*):fa (**)"DJ~                                                                         Tsz-MeiKo
AF        AG
-=-                       Supposeone would like to transmit a
FB        GB                                                                        message.For an evenparity code,a 0 or
message, say "HELLO.. .", from one                                         so
1 is appended that the total number of
~Jl:tFWG~ABJ::~IffJ-1!i          0
computerto another. One possibleway                         1's is an evennumber. The letters Hand
is to use a table to encodethe message                      E would be representedby 10010000
:gg-][:iE~fII\]~:iE~~{PJ.~~ .f.§:\$         into binary digits. Then the receiver                      and 10001011 respectively. With an
.L~.t87E-~Wfll\]E-i'lm~'~...:              would be able to decode the message                        even parity code, the receiver can detect
E-I=P *1 (medians) # I!i'    E-~    with a similar table. One such table is                    one transmission error, but unable to
#
(altitudes) I!i' E- \$j- ~ *1 (angle- the American Standard Code for                                    correctit. For example,if 10010000(for
~                    tl
bisectors) I!i 0 ~ E- i'l1!i \$j- 7J ;:g t£ Information Interchange(ASCII) shown                       the letter H) is receivedas 10010100, the
m'L' (centroid)' ~ 'L' (orthocentre)~Din Figure 1. The letter H would be                              receiverknows that there is at least one
rf'\JWIiI'L'(incentre)    0                encodedas 1001000,the letter E would                       error during transmission since the
as          etc.
be encoded 1000101, ~igure 2).                                                     has
=-~.#l!im           -.w-ar PJ.~H~~i1!?                                                                i.e., the total number of l' s is an odd
A 1000001                    S 1010011     1100001    1110011
~-gg-][::iE:JJ.£J'iJ~::iE:JJ.tf.m'rRJ~frj
1000010                       1010100    1100010   1110100
number.
m~~o **£-arjfrJffl#jf=-~~                              1000011        10101011100011       1110101
~OO~~'f*'~m (::I:/L\re~..\$t                            1000100
1000101
1010110
1010111
1100100   1110110
11001011110111
m-~:!6'!1.
JjX,pjij~~~2:1£J'iJ.~)                                 1000110        1011000    11qO110   l1l1000
ft"o                                                   1000111        1011001    1100111   l1l1001
1001000        1011010    1101000   l1l1010
1001001      '0110000     1101001   0101110       ij,'--I-.
mJa~
=-~#1!i1l:t'~'1fijJ;fIJ =-~~                           1001010        0110001 11101010     0101100       E,'...,."
3Ift~=-~~"'I;~~~.m'{tl.                                1001011
1001100
0110010 Ic 1101011
0110011 1 1101100
0ll1111
0101001             Figure4. Even parity code
f3J~rI1J~irJ~a~~o                                    M 1001101               ~
0110100 ' 1101101    l1l1011
N 1001110         0110101    1101110    010l1l1
Is there an encodingmethod so that the
=-~~(l{j5j-~~::fif-@M.ffijm~                         J 100l1l1,
'" 1010000
0110110    110l1l1
I 011011111110000      I
0100110
receiver would be able to correct
~.~~:                                                 1010001         0111000 c:r1110001
I 0101011
0101101
transmissionerrors? Figure 5 showsone
A                            1~10010        0111001 r 1110010    0l1l101
such method by arranging the bit
Figure 1. ASCII code         sequence    (e.g., 1001) into a rectangular
block and add parity bits to both rows
and columns. For the example shown,
1001 would be encodedas 10011111(by
B          D               c                                                   fIrst appendingthe row parities and then
(/I)~)                                                                  the columnparities). If there is an error
during transmission, say at position 2,
the receiver can similarly arrange the
Figure2. Two computers   talking   received sequence 11011111 into a
rectangularblock and detectthat there is
The receiver will be able to decodethe an error in row 1 and column2.
messagecorrectly if there is no error
~1I:t.                                             during transmission. However, if there
BD
-=-0        AB                     decode the message incorrectly. For                                      ~??)

-=-9J-flj ;f,m I!i.:;l-
.J:\~ilio
#
DC         AC

~ '/1 ijJ m 11}]JJ:t
~

(f~.)
example, the letter H (1001000)would
be receivedas J (1001010)if there is an
error at position6.                           0 °1 1
I    * I   I
0 I I ,/

I I
'I)(

,Ix

Figure 5. A code that can correct1 error.

The abovemethod can be usedto correct
one error but rather costly. For every
Figure3. Error at position6.     four bits, one would need to transmit an
extra four redundant bits. Is there a
One possibleway to detect transmission better way to do the encoding?In 1950,
errors is to add redundant bits, i.e., Hammingfound an ingeniousmethodto

~
append extra bits to the original                         (continuedon page 4)
Mathematical Excalibur, Vol. 2, No.        Nov-Dec. 96                                                           P~e3

Supposex, y are nonnegative integers Solution: CHAN Ming ChiD (La Salle
Problem Corner                   suchthat (xy -7)2 = r + y2. Then (xy - College, Form 6), CHAN Wing Sum
We welcome readers to submit solutions    6)2+ 13 = (x + y)2 by algebra. So       (HKUST), CHENG Wing Kin (S.K.H.
to the problems posed below for            13 = [(x+y) + (xy-6)][(x+y) -(xy-6)].  Lam Woo Memorial SecondarySchool,
publication    consideration.   Solutions                                         Form 5), CHEUNG Cheuk Lun
should be preceded by the solver's name, Since 13 is prime, the factors on the (S.T.F.A. Leung Kau Kui College, Form
address, school affiliation and grade right side can only be :tl or :t13. There 6), William       CHEUNG Pok-man
level.   Please send submissions to Dr.   are four possibilities yielding (x,y) = (S.T.F.A. Leung Kau Kui College, Form
Kin-Yin li, Dept of Mathematics, Hong (0,7), (7,0), (3,4), (4,3).                 6), Yves CHEUNG Yui Ho (S.T.F.A.
Kong University       of Science and                                              Leung Kau Kui College, Form 5), CHUI
Technology, Clear Water Bay, Kowloon. Other commended       solvers: CHAN Ming Yuk Man (Queen Elizabeth School,
The deadline for submitting solutions is Chin (La Salle College. Form 6), Form 7), FUNG Tak K wan (La Salle
Ian 31, 1997.                             CHENG Wing Kin (S.K.H. Lam Woo College, Form 7), LEUNG Wing Lon
Memorial SecondarySchool, Form 5), (STFA Leung Kau Kui College, Form
Problem 46. For what integer a does William CHEUNG Pok-man (S.T.F.A. 6), LIU Wai Kwong (Poi Tak
X2-x +a divide Xl3 + x + 90?              Leung Kau Kui College, Form 6), Yves Canossian College), Henry NG Ka
CHEUNG Yui Ho(S.T.F.A. Leung Kau Man (STFA Leung Kau Kui College,
Problem 47. If x, y, z are real numbers Kui College, Form 5), CHING Wai Form 6), Gary NG Ka Wing (STFA
such that X2+ y2 + Z2= 2, then show that Hung (S.T.F.A. Leung Kau Kui College, Leung Kau Kui College, Form 4),
x + y + z ~ xyz + 2.                      Form 5), CHUI Yuk Man (Queen POON Wing Chi (La Salle College,
Elizabeth School, Form 7), LIU Wai Form 7), TSANG Sai Wing (Valtorta
Problem 48. Squares ABDE and BCFG         Kwong (pui Tak Canossian College), College, Form 6), YU Chon Ling
are drawn outside of triangle ABC.        POON Wing Chi (La Salle College, (HKU), YUEN Chu Ming (Kiangsu-
Prove that triangle ABC is isosceles if Form 7), TING Kwong Chi & David ChekiangCollege (Shatin), Form 6) and
DG is parallel to AC.                     GIGGS (SKH Lam Woo Memorial YUNG Fai (CUHK).
SecondarySchool, Form 5), YU Chon
Problem 49. Let UI, U2, U3, ...be a Ling (HKU) and YUNG Fai (CUHK).               There are C;O = 1140 3-elementsubsets
sequence of integers such that UI = 29,                                       of X. For a 3-element subsetwhose 3
=
U2= 45 and Un+2 Un+? -Un for n = 1, 2,      Problem 42. What are the possible numbershave product not divisible by 4,
3,     Show that 1996 divides infinitely                                      the numbersare either all odd (there are
X2
valuesof oJ + x + 1 -.J;2 ~ as x
or
cjO = 120 such subsets) two odd and
many terms of this sequence. (Source:            overall realnumbers?
ranges
with modification)                          Solution: William CHEUNG Pot-man      divisible by 4 (therc are C~Ox5=225
Form 6).
(STFALeungKauKui College,             such subsets). So the answer to the
Problem 50. Four integers are marked                                              problemis 1140 -120 -225 = 795.
on a circle.         On each step we         Let A=(x,O),B=(-t,~),    c=(t,~).
simultaneouslyreplace each number by                                            Problem 44. For an acutetriangle ABC,
the difference betweenthis number and                      X2
The expression.J + x + 1 -j;2=-;1
let H be the foot of the perpendicular
next number on the circle in a given    is just AB -AC. As x ranges over all from A to BC. Let M, N be the feet of
direction (that is, the numbersa, b, c, d
real numbers, A moves along the real the perpendicularsfrom H to AB, AC,
are replacedby a -b, b -c, c -d, d -    axis and the triangle inequality yields respectively. Define LA to be the line
a). Is it possibleafter 1996 suchstepsto                                        through A perpendicular to MN and
have numbers a, b, c, d such that the        -1 =-BC<AB-,.AC<        BC= 1.
similarly define LB and Lc. Show that
numbers Ibc -adl,       lac -bdl, lab-  All numbers on the intergal (-1,1) are LA, LB and Lc pass through a common
cd I are primes?        (Source: unused possible.                               point O. (This was an unusedproblem
problem in the 1996 IMO.)                                                       proposedby Iceland in a past IMO.)
Othercommended    solvers: CHAN Ming
Chiu (La Salle College, Form 6), Solution: William CHEUNG Pok-man
*****************               CHENG Wing Kin (S.K.H. Lam Woo (STFALeungKauKui College,               F0ffil6).
Memorial SecondarySchool, FOrm 5),
Solutions                LIU Wai Kwong (Poi Tak Canossian Let LA interSect the circumcircle of
*****************               College), POON Wing Chi (La Salle MBC at A and E. Since LAMH = 900 =
College, FOrm 7), YU Chon Ling LANH, A, M, H, N are concyclii::. So
Problem 41. Find all nonnegative (HKU) and YUNG Fai (CUHK).                     LMAH = LMNH = 90° -LANM                  =
integers x, y satisfying (xy -7Y = x2-+                                         LNAE = LCBE. Now LABE = LCBE
2-
y.                                      Problem 43.       How many 3-elemen't + LABC= LMAH + LABC = 90°. So
AE is a dianleter of the circumcircle and
Solution: Gary NG Ka Wing (STFA subse~of the setX= {I, 2, 3, ..., 20} are
there such that the product of the 3
Leung Kau Kui College,Form 4).
numbersin the subset diVisible by 4?
is                                  (continuedon page 4)
Mathematical Excalibur. Vol. 2, No. 5, Nov-Dec.96                                                                                               PMe4

Problem Corner                                      point P such that LPAB = 10°, LPBA =
(continuedfrom page 3)                              20°, LPCA = 30°, LPAC = 40°. Prove
LA passesthrough the circumcenter O.
Similarly, LB and Lc will pass
through O.
that triangle ABC is isosceles.

Problem 6. Determine                       (with proof)
whether there is a subset X of the
:
~
6
0 ,
)(
,     Co

integers with the following property: for
Other commended solvers: Calvin                                                                                            code
Figure6. Hamming
any integer n there is exactly one
CHEUNG Cheuk Lon (STFA Leung
solution of a + 2b = n with a, b E X.
Kau Kui College, FOrin 5), LIU Wai                                                               If there is one error during transmission,
Kwong (Pui Tak Canossian College),                                                               say 1001001 received as 1011001, the
POON Wing Chi (La Salle College,                                  ~'""DC><::5~--                 receiver can check the parities of the
Foml 7) and YU Chon Ling (HKU).                                                                  three circles to find that the error is in
Error Correcting Codes(part I) circles B and C but not in A. This (7,4)
Problem 45. Let a, b, c> 0 and abc=l                 (continued from page 2)                     Hammingcode(the notation (7,4) means
Showthat                                                                                         that every4 information bits are encoded
add the redundancy. To encode a four- as a 7 bit sequence)          can be generalized.
ab             bc              ca""            bit. sequence PIP1P3P4(say 1001), one For example, one may draw 4
+               +~                  l
as+bs+ab         bs+cs+bc        cs+as+ca           would first draw three intersecting          intersecting spheres in a three-
(This was an unused problem in                      circles A, B, C and put the information      dimensional space to obtain a (15,11)
bits Pl, Pl, PJ,P4 into the four overlapping Hamming code. Hamming has also
IMO96.)
regions AnB, AnC, BnC and AnBnC              proved that his coding method is
Solution: YUNG Fai (CUHK) ,                         (Figure 6). Then three parity bits Ps, P6, optimum for single error correction.
P1 are generated so that the total number
of l's in each circle is an even number.                                                (... to be continued)
Expanding (a3-b3)(a2 -b1 ~ 0, we get
For the example shown, 1001 would be
a5 + b5 ~ a2b2(a+b). So using this and
abc = 1, we get                        encoded as 1001001.                                         R~h                         ~ IJ'1-e-!&\
ab                    ab              C2
x- 2                                                                                               J:.tt3)
c.ti:;t:
as + b5 + ab                               C

()f')
a+b+c
Adding 3 such inequalities, we get the
desiredinequality. In fact, equality can
t>A~'~""j-a

'\~~
-~
occur if and only if a = b = c = I.                                                                               t> ~i~

Other commended   solvers: POON Wing
..
..
Chi (La Salle College, Form 7) and YU
Chon Ling (HKU).
~~, /jl}.~t'
.-" ~t yotl.~ A Xyz                          .i ??
Igo. 1~~lv1'~ ~A8c.';
--~""c:>C><:5"               ~                                                 ~
/r ". Y. z ~:fJ            f3c,
Y
L). )C z. 1m j 1£4t            £1/fe-c.-'I@~
(continuedfrom page 1)
A )( Y z. -iff) h :JtJ"~~ ~ FtBc.1~
~
ColA'"
Co;'"  c.~nt,.t. 0
Part n (lpm-4pm,May 2, 1996)
ABc.
~'~1'f1 G ~ 1/':", ~ £).
Problem 4. An n-term sequence X2,   (Xl'                                                                 ,go
..&iJ.-:1ji." -, r}f";" o. f-r.H
..., xn) in which each term is either 0 or                                                       {iiJ--1li,~~l,~A              o~::~H       = /=2.
1 is called a binary sequenceof lengthn.
Let anbe the numberof binary sequences                                                    o!.!      ! iJ.' 9~A'f ~ ..), ~~ .A X Yz.!j: I~~
of length n containing no three
50 Yo ,:tt       ISOD, l<~ fJ L). XYz. p~~
consecutive  terms equal to 0, 1, 0 in that
Cir-c.umC£l:Itr~ N , ~          GN : 0&          : GrH
order. Let bn be the number of binary                                                     --=                                   1 ~ 2..          ""
sequencesof length n that contain no
four consecutive  terms equal to 0, 0, 1, 1
or 1, 1, 0, 0 in that order. Prove that                                                           ~).. ~ "of/v'
N                                           .
'1'.'Z.
bn+l=2an for all positive integersn.                                                                  "!1~A. ~'f!1
,.V. ~
-:s:
~~
S. T    (HA. 1-18. 11c.
Problem 5. Triangle ABC has the                                                                           I N t..t: '04.~ AASc. ~'ft. ".."e
following property: there is an interior                                                                             point

,~
~

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