{VERSION 3 0 "IBM INTEL NT" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "He ading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }} {SECT 0 {SECT 1 {PARA 3 "" 0 "" {TEXT -1 35 "Soluciones del examen 2B \+ de Maple. " }}{PARA 3 "" 0 "" {TEXT -1 37 "Matem\341tica Discreta. Cur so 2003/2004." }}{PARA 0 "" 0 "" {TEXT 264 67 "Ingenier\355a T\351cnic a en Inform\341tica de Sistemas (tarde) y de Gesti\363n." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 13 "Ejercicio 1: " } {TEXT -1 86 "Encuentra la soluci\363n, si existe, al siguiente sistema de ecuaciones en congruencias: " }}{PARA 0 "" 0 "" {TEXT -1 18 "x = 6 (mod 1117)" }}{PARA 0 "" 0 "" {TEXT -1 14 "x= 5 (mod 541)" }}{PARA 0 "" 0 "" {TEXT -1 15 "x= 6 (mod 5585)" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 112 "Es f\341cil ver que 5585 es m\372lti plo de 1117, as\355, basta resolver el sistema formado por las \372lti mas dos ecuaciones" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "with(n umtheory): chrem([5,6],[541,5585]);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 263 13 "Ejercicio 2: " }{TEXT -1 159 "Determinar el n\372mero de palabras distintas que pueden const ruirse utilizando todas las letras de la palabra ACTUALIZACION, tantas veces como aparecen en ella. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "with(combinat): numbperm([a,c,t,u,a,l,i,z,a,c,i,o,n]);" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 12 "Ejercicio 3:" }{TEXT -1 25 " D etermina si la matriz " }{TEXT 259 3 "B2 " }{TEXT -1 75 "(ver fichero \+ 2B.mws) puede ser la matriz de adyacencias de un grafo simple." }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6734 "B2:=matrix([[0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, \+ 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0 , 1, 0, 1, 0, 1, 0], [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, \+ 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0 , 1, 0, 1, 0, 1, 0, 1], [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, \+ 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0 , 1, 0, 1, 0, 1, 0, 1, 0], [1, 0, 1, 0, 1, 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0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, \+ 1, 0, 1, 0], [1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, \+ 1, 0, 1, 0, 1], [0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, \+ 1, 0, 1, 0, 1, 0]]):" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "a:=0:\nfo r i from 1 to rowdim(B2) do\na:=a+B2[i,i] od:\na;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "equal(B2,transpose(B2));" }}}{PARA 0 "" 0 "" {TEXT -1 148 "La matriz tiene s\363lo ceros en la diagonal y es s im\351trica. Siendo todos sus coeficientes ceros o unos, puede ser la \+ matriz de adyacencias de un grafo" }{TEXT 266 1 " " }{TEXT -1 7 "simpl e." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 12 "Ej ercicio 4:" }{TEXT -1 33 " Determina si el grafo que tiene " }{TEXT 262 2 "B2" }{TEXT -1 41 " como matriz de adyacencias es euleriano." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "for i from 1 to rowdim(B2) do\ns[i]:=sum(B2[i,j],j=1..rowdim(B2)) \+ od:\nseq(s[k],k=1..rowdim(B2));" }}}{PARA 0 "" 0 "" {TEXT -1 47 "No lo es, ya que tiene v\351rtices de grado impar." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "1" 0 }{VIEWOPTS 1 1 0 1 1 1803 }