Algoritmos e inducci´on Luis Sierra InstitutodeComputaci´on FacultaddeIngenier´ıa UniversidaddelaRepu´blica Febrero del 2012 Pr´acticas compartidas para la ensen˜anza de inform´atica: el aula y el trabajo Universidad Nacional de Quilmes, Argentina LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 1/40 Outline 1 Inducci´on como justificaci´on 2 Inducci´on como descubrimiento 3 Sumar 4 Tipos de datos 5 Programaci´on 6 Visualizaci´on 7 Conclusiones LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 2/40 Inducci´oncomojustificaci´on n n+1 LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 3/40 Inducci´oncomojustificaci´on Gauss n (cid:88) n(n+1) i = 2 i=0 LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 4/40 We verify that the formula holds true for the smallest possible value If we know that the formula holds true for the numbers < N, then it also holds true for the number N. From that we conclude that the formula holds true for all the numbers. Inducci´oncomojustificaci´on Induccio´n como justificaci´on n (cid:88) n(n+1) i = 2 i=0 Proof by Mathematical Induction follows this pattern: We want to verify that a formula, algebraic expression, holds true for all the values of the parameter, a whole number. LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 5/40 If we know that the formula holds true for the numbers < N, then it also holds true for the number N. From that we conclude that the formula holds true for all the numbers. Inducci´oncomojustificaci´on Induccio´n como justificaci´on n (cid:88) n(n+1) i = 2 i=0 Proof by Mathematical Induction follows this pattern: We want to verify that a formula, algebraic expression, holds true for all the values of the parameter, a whole number. We verify that the formula holds true for the smallest possible value LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 5/40 From that we conclude that the formula holds true for all the numbers. Inducci´oncomojustificaci´on Induccio´n como justificaci´on n (cid:88) n(n+1) i = 2 i=0 Proof by Mathematical Induction follows this pattern: We want to verify that a formula, algebraic expression, holds true for all the values of the parameter, a whole number. We verify that the formula holds true for the smallest possible value If we know that the formula holds true for the numbers < N, then it also holds true for the number N. LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 5/40 Inducci´oncomojustificaci´on Induccio´n como justificaci´on n (cid:88) n(n+1) i = 2 i=0 Proof by Mathematical Induction follows this pattern: We want to verify that a formula, algebraic expression, holds true for all the values of the parameter, a whole number. We verify that the formula holds true for the smallest possible value If we know that the formula holds true for the numbers < N, then it also holds true for the number N. From that we conclude that the formula holds true for all the numbers. LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 5/40 (cid:80)n+1i = (n+1)(n+2) i=0 2 ⇐= (cid:80)0 = 0(0+1) i=0 2 n+1+(cid:80)n i = (n+1)(n+2) ⇐= i=0 2 ⇐= (*) 0 = 0. n(n+1) (n+1)(n+2) n+1+ = 2 2 ⇐= 2(n+1)+n(n+1) = (n+1)(n+2). Inducci´oncomojustificaci´on Induccio´n como justificaci´on (∀n∈N::P.n) n (cid:88) n(n+1) P.n := i = 2 i=0 Paso: (∀n ∈ N : P.n : P.(n+1)) Base: P.0 LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 6/40 (cid:80)n+1i = (n+1)(n+2) i=0 2 ⇐= n+1+(cid:80)n i = (n+1)(n+2) i=0 2 ⇐= (*) n(n+1) (n+1)(n+2) n+1+ = 2 2 ⇐= 2(n+1)+n(n+1) = (n+1)(n+2). Inducci´oncomojustificaci´on Induccio´n como justificaci´on (∀n∈N::P.n) n (cid:88) n(n+1) P.n := i = 2 i=0 Paso: (∀n ∈ N : P.n : P.(n+1)) Base: P.0 (cid:80)0 = 0(0+1) i=0 2 ⇐= 0 = 0. LuisSierra (InCo-Uruguay) Algoritmoseinduccio´n Quilmes2012 6/40
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