Fibonacci Numbers and Dynamic Programming

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Questions and Answers

Los mexicanos no son todos ______.

mariachis

Los ______ no son todos futbolistas.

brasileños

Los chilenos no ______ todos mineros.

son

No todos los ______ bailan tango.

<p>argentinos</p>
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No todos los uruguayos beben ______.

<p>mate</p>
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No todos los Colombianos beben ______.

<p>café</p>
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No todos los ______ comen paella.

<p>españoles</p>
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No todos los árabes son ______.

<p>musulmanes</p>
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No todos los estadounidenses les gustan ______.

<p>hamburguesas</p>
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No todos los irlandeses ______ bromistas.

<p>son</p>
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Los artistas no siempre ______ comprendidos en su tiempo.

<p>son</p>
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No todos los libros ______ iguales en calidad.

<p>son</p>
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No todos los sueños ______ alcanzables, pero vale la pena intentarlo.

<p>son</p>
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Los inviernos no ______ siempre fríos en todas partes del mundo.

<p>son</p>
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No todos los gatos ______ cariñosos, algunos son más independientes.

<p>son</p>
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Las películas de terror no ______ para todos, algunas personas las encuentran demasiado aterradoras.

<p>son</p>
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No todos los consejos ______ útiles, depende de la situación.

<p>son</p>
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Los veranos no ______ siempre soleados, también puede haber días de lluvia.

<p>son</p>
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No todos los héroes ______ valientes, algunos tienen miedo pero actúan a pesar de ello.

<p>son</p>
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Los postres no ______ siempre dulces, hay algunos que son agridulces o salados.

<p>son</p>
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Flashcards

Los mexicanos no son todos ____?

Not all Mexicans are mariachi players.

Los ____ no son todos futbolistas?

Not all Argentinians are soccer players.

Los chilenos no ____ todos mineros?

Not all Chileans are miners.

No todos los ____ bailan tango?

Not all Brazilians dance the tango.

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No todos los uruguayos beben ____?

Not all Uruguayans drink coffee.

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No todos los Colombianos beben ____?

Not all Colombians drink mate.

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No todos los ____ comen paella?

Not all Spanish people eat paella.

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No todos los árabes son ____?

Not all Arabs are Muslims.

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No todos los estadounidenses les gustan ____?

Not all Americans like hamburgers.

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No todos los irlandeses ____ bromistas?

Not all Irish people are jokers.

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Study Notes

Fibonacci Numbers

  • Fibonacci numbers defined recursively: F(n) = 0 if n = 0, F(n) = 1 if n = 1, F(n) = F(n-1) + F(n-2) if n > 1.
  • A naive recursive algorithm to compute F(n) has exponential running time.
  • The exponential running time is because the algorithm recomputes the same Fibonacci numbers multiple times.
  • Memoization stores computed Fibonacci numbers to avoid recomputation, reducing running time to O(n).
  • Dynamic programming computes Fibonacci numbers in a bottom-up fashion, achieving O(n) running time.
  • Space complexity of dynamic programming can be reduced to O(1) by storing only the last two Fibonacci numbers.

Dynamic Programming

  • Dynamic programming is a technique for solving problems with overlapping subproblems by storing solutions to subproblems in a table, avoiding recomputation.
  • Conditions for dynamic programming applicability: optimal substructure and overlapping subproblems.
  • Optimal substructure: an optimal solution contains optimal sub-solutions.
  • Overlapping subproblems: the same subproblems are solved repeatedly.

Weighted Interval Scheduling

  • This is an example problem of Dynamic Programming
  • Input involves n intervals, each with a start time (s_i), a finish time (f_i), and a weight (w_i).
  • The Goal is to find a set of non-overlapping intervals with maximum total weight.
  • Intervals are sorted by finish time (f_1 ≤ f_2 ≤ ... ≤ f_n).
  • p(j) is the largest index i < j where interval i is compatible with interval j (f_i ≤ s_j); if no such interval exists, p(j) = 0.
  • Let OPT(j) be the value of an optimal solution for intervals 1 to j.
  • Case 1: If interval j is in an optimal solution, the solution includes interval j and an optimal solution for intervals 1 to p(j).
  • Case 2: If interval j is not in an optimal solution, the solution is an optimal solution for intervals 1 to j-1.
  • Recurrence relation: OPT(j) = 0 if j = 0, else OPT(j) = max(w_j + OPT(p(j)), OPT(j-1)) if j > 0.

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