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[SPAM] [obm-l] Re: [obm-l] segunda fase - nível universitário 2007

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Subject: [SPAM] [obm-l] Re: [obm-l] segunda fase - nível universitário 2007
From: <rodrigocientista@xxxxxxxxxxxx>
Date: Sat, 8 Dec 2007 22:50:31 -0300
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Ol=C3=A1 Marcelo,

voc=C3=AA leu a demonstra=C3=A7=C3=A3o abaixo?gostaria de saber se ela =
cont=C3=A9m algum erro
abra=C3=A7os
  ----- Original Message -----=20
  From: Marcelo Salhab Brogliato=20
  To: obm-l@xxxxxxxxxxxxxx=20
  Sent: Sunday, December 09, 2007 1:50 AM
  Subject: Re: [obm-l] segunda fase - n=C3=ADvel universit=C3=A1rio 2007


  eita... desculpe! tava pensando e sem querer apertei um atalho e =
enviou... hehe ;)

  abra=C3=A7os,
  Salhab



  On Dec 9, 2007 2:50 AM, Marcelo Salhab Brogliato < msbrogli@xxxxxxxxx> =
wrote:

    Ol=C3=A1 Rodrigo,


    Dado um inteiro positivo n, mostre que existe um inteiro positivo N =
com a seguinte propriedade: se A =C3=A9 um subconjunto de {1,2,...,N}  =
com pelo menos N/2 elementos, ent=C3=A3o existe um inteiro positivo =
m<=3D N - n   tal que  |A interse=C3=A7=C3=A3o com {m+1, m+2,..., =
m+k}|>=3Dk/2 para todo k =3D 1, 2, =E2=80=A6, n.

    =20





    |AUB| =3D |A| + |B| - |AinterB|

    |A inter {m+1, m+2, ..., m+k}| =3D |A| + |{m+1, m+2, ..., m+k}| - |A =
uniao {m+1, m+2, ..., m+k}|=20
    |A inter {m+1, m+2, ..., m+k}| =3D |A| + k - |A uniao {m+1, m+2, =
..., m+k}|=20








    On Dec 6, 2007 6:19 PM, Rodrigo Cientista =
<rodrigocientista@xxxxxxxxxxxx > wrote:

      PROBLEMA 2:
      Dado um inteiro positivo n, mostre que existe um inteiro positivo =
N com a seguinte propriedade: se A =C3=A9 um subconjunto de {1,2,...,N}  =
com pelo menos N/2 elementos, ent=C3=A3o existe um inteiro positivo =
m<=3D N - n   tal que  |A interse=C3=A7=C3=A3o com {m+1, m+2,..., =
m+k}|>=3Dk/2=20

      para todo k =3D 1, 2, =E2=80=A6, n.

      =
*************************************************************************=
*************************************************
      (gostaria de coment=C3=A1rios sobre esta demonstra=C3=A7=C3=A3o, =
falhas, se conhecem alguma demonstra=C3=A7=C3=A3o pra esse problema, =
pois ainda n=C3=A3o tem o gabarito)=20

      suponha existir x > N - n tal que  |A interse=C3=A7=C3=A3o com =
{x+1, x+2,..., x+k}|>=3Dk/2

      como x + n > N, pelo menos um elemento de  {x+1, x+2,..., x+k} =
ser=C3=A1 maior que qualquer elemento de A; escolhendo-se um n =3D 1, a =
afirma=C3=A7=C3=A3o acima =C3=A9 falsa=20

      assim, se  |A interse=C3=A7=C3=A3o com {m+1, m+2,..., m+k}|>=3Dk/2 =
=3D=3D> existe m <=3D N - n

      chamemos S =3D {m+1, m+2,..., m+k}

      m + n <=3D N =3D=3D> m + k <=3D N para todo k =3D 1, 2, =E2=80=A6, =
n =3D=3D>

       =3D=3D> S =C3=A9 subconjunto de {1,2,...,N}, ou =C3=A9 o =
pr=C3=B3prio conjunto {1,2,...,N} na hip=C3=B3tese em que  N =3D n=20

      quando N =3D n =C3=A9 trivial que |A interse=C3=A7=C3=A3o com =
{m+1, m+2,..., m+k}|>=3Dk/2 (=3D k/2 na verdade)

      suponha N > n =3D=3D> N/2 > n/2 =3D=3D> |{1,2,...,N}| > |S| =
=3D=3D> |A| > |S|/2 =3D n/2

      como S est=C3=A1 contido em {1,2,...,N} =3D=3D> =C3=A9 sempre =
poss=C3=ADvel tomar-se um subconjunto A de {1,2,...,N} tal que S/2 =
esteja contido em A=20


           Abra sua conta no Yahoo! Mail, o =C3=BAnico sem limite de =
espa=C3=A7o para armazenamento!
      http://br.mail.yahoo.com/

      =
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=20
      Instru=EF=BF=BD=C3=B5es para entrar na lista, sair da lista e usar =
a lista em
      http://www.mat.puc-rio.br/~obmlistas/obm-l.html
      =
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=20





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<DIV><FONT face=3DArial size=3D2>Ol=C3=A1 Marcelo,</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>voc=C3=AA leu&nbsp;a =
demonstra=C3=A7=C3=A3o abaixo?gostaria de=20
saber se ela cont=C3=A9m algum erro</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>abra=C3=A7os</FONT></DIV>
<BLOCKQUOTE=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
  <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
  <DIV=20
  style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
  <A title=3Dmsbrogli@xxxxxxxxx =
href=3D"mailto:msbrogli@xxxxxxxxx";>Marcelo Salhab=20
  Brogliato</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Dobm-l@xxxxxxxxxxxxxx=20
  href=3D"mailto:obm-l@xxxxxxxxxxxxxx";>obm-l@xxxxxxxxxxxxxx</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Sunday, December 09, 2007 =
1:50=20
  AM</DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> Re: [obm-l] segunda =
fase - n=C3=ADvel=20
  universit=C3=A1rio 2007</DIV>
  <DIV><BR></DIV>eita... desculpe! tava pensando e sem querer apertei um =
atalho=20
  e enviou... hehe ;)<BR><BR>abra=C3=A7os,<BR>Salhab<BR><BR><BR>
  <DIV class=3Dgmail_quote>On Dec 9, 2007 2:50 AM, Marcelo Salhab =
Brogliato &lt;<A=20
  href=3D"mailto:msbrogli@xxxxxxxxx";> msbrogli@xxxxxxxxx</A>&gt; =
wrote:<BR>
  <BLOCKQUOTE class=3Dgmail_quote=20
  style=3D"PADDING-LEFT: 1ex; MARGIN: 0pt 0pt 0pt 0.8ex; BORDER-LEFT: =
rgb(204,204,204) 1px solid">Ol=C3=A1=20
    Rodrigo,
    <DIV class=3DIh2E3d><BR><BR>Dado um inteiro positivo n, mostre que =
existe um=20
    inteiro positivo N com a seguinte propriedade: se A =C3=A9 um =
subconjunto de=20
    {1,2,...,N} &nbsp;com pelo menos N/2 elementos, ent=C3=A3o existe um =
inteiro=20
    positivo m&lt;=3D N - n &nbsp; tal que &nbsp;|A interse=C3=A7=C3=A3o =
com {m+1, m+2,...,=20
    m+k}|&gt;=3Dk/2 para todo k =3D 1, 2, =E2=80=A6,=20
    n.<BR><BR>&nbsp;<BR><BR><BR><BR><BR></DIV>|AUB| =3D |A| + |B| -=20
    |AinterB|<BR><BR>|A inter {m+1, m+2, ..., m+k}| =3D |A| + |{m+1, =
m+2, ...,=20
    m+k}| - |A uniao {m+1, m+2, ..., m+k}| <BR>|A inter {m+1, m+2, ..., =
m+k}| =3D=20
    |A| + k - |A uniao {m+1, m+2, ..., m+k}|=20
    <DIV>
    <DIV></DIV>
    <DIV class=3DWj3C7c><BR><BR><BR><BR><BR><BR><BR>
    <DIV class=3Dgmail_quote>On Dec 6, 2007 6:19 PM, Rodrigo Cientista =
&lt;<A=20
    href=3D"mailto:rodrigocientista@xxxxxxxxxxxx"=20
    target=3D_blank>rodrigocientista@xxxxxxxxxxxx </A>&gt; wrote:<BR>
    <BLOCKQUOTE class=3Dgmail_quote=20
    style=3D"PADDING-LEFT: 1ex; MARGIN: 0pt 0pt 0pt 0.8ex; BORDER-LEFT: =
rgb(204,204,204) 1px solid">PROBLEMA=20
      2:<BR>Dado um inteiro positivo n, mostre que existe um inteiro =
positivo N=20
      com a seguinte propriedade: se A =C3=A9 um subconjunto de =
{1,2,...,N} &nbsp;com=20
      pelo menos N/2 elementos, ent=C3=A3o existe um inteiro positivo =
m&lt;=3D N - n=20
      &nbsp; tal que &nbsp;|A interse=C3=A7=C3=A3o com {m+1, m+2,..., =
m+k}|&gt;=3Dk/2=20
      <BR><BR>para todo k =3D 1, 2, =E2=80=A6,=20
      =
n.<BR><BR>***************************************************************=
***********************************************************<BR>(gostaria =

      de coment=C3=A1rios sobre esta demonstra=C3=A7=C3=A3o, falhas, se =
conhecem alguma=20
      demonstra=C3=A7=C3=A3o pra esse problema, pois ainda n=C3=A3o tem =
o gabarito)=20
      <BR><BR>suponha existir x &gt; N - n tal que &nbsp;|A =
interse=C3=A7=C3=A3o com {x+1,=20
      x+2,..., x+k}|&gt;=3Dk/2<BR><BR>como x + n &gt; N, pelo menos um =
elemento de=20
      &nbsp;{x+1, x+2,..., x+k} ser=C3=A1 maior que qualquer elemento de =
A;=20
      escolhendo-se um n =3D 1, a afirma=C3=A7=C3=A3o acima =C3=A9 falsa =
<BR><BR>assim, se=20
      &nbsp;|A interse=C3=A7=C3=A3o com {m+1, m+2,..., m+k}|&gt;=3Dk/2 =
=3D=3D&gt; existe m &lt;=3D=20
      N - n<BR><BR>chamemos S =3D {m+1, m+2,..., m+k}<BR><BR>m + n =
&lt;=3D N =3D=3D&gt;=20
      m + k &lt;=3D N para todo k =3D 1, 2, =E2=80=A6, n =
=3D=3D&gt;<BR><BR>&nbsp;=3D=3D&gt; S =C3=A9=20
      subconjunto de {1,2,...,N}, ou =C3=A9 o pr=C3=B3prio conjunto =
{1,2,...,N} na=20
      hip=C3=B3tese em que &nbsp;N =3D n <BR><BR>quando N =3D n =C3=A9 =
trivial que |A=20
      interse=C3=A7=C3=A3o com {m+1, m+2,..., m+k}|&gt;=3Dk/2 (=3D k/2 =
na=20
      verdade)<BR><BR>suponha N &gt; n =3D=3D&gt; N/2 &gt; n/2 =
=3D=3D&gt; |{1,2,...,N}|=20
      &gt; |S| =3D=3D&gt; |A| &gt; |S|/2 =3D n/2<BR><BR>como S est=C3=A1 =
contido em=20
      {1,2,...,N} =3D=3D&gt; =C3=A9 sempre poss=C3=ADvel tomar-se um =
subconjunto A de=20
      {1,2,...,N} tal que S/2 esteja contido em A <BR><BR><BR>&nbsp; =
&nbsp;=20
      &nbsp;Abra sua conta no Yahoo! Mail, o =C3=BAnico sem limite de =
espa=C3=A7o para=20
      armazenamento!<BR><A href=3D"http://br.mail.yahoo.com/"=20
      =
target=3D_blank>http://br.mail.yahoo.com/</A><BR><BR>=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
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=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=20
      <BR>Instru=EF=BF=BD=C3=B5es para entrar na lista, sair da lista e =
usar a lista em<BR><A=20
      href=3D"http://www.mat.puc-rio.br/%7Eobmlistas/obm-l.html"=20
      =
target=3D_blank>http://www.mat.puc-rio.br/~obmlistas/obm-l.html</A><BR>=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
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References:
- [SPAM] [obm-l] segunda fase - nível universitário 2007
  - From: Rodrigo Cientista
- Re: [obm-l] segunda fase - nível universitário 2007
  - From: Marcelo Salhab Brogliato
- Re: [obm-l] segunda fase - nível universitário 2007
  - From: Marcelo Salhab Brogliato

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