Desmitificando la programación dinámica

Cómo construir y codificar algoritmos de programación dinámica

Tal vez hayas oído hablar de ello al prepararte para codificar entrevistas. Tal vez lo hayas superado en un curso de algoritmos. Tal vez esté tratando de aprender a codificar por su cuenta y en algún momento le dijeron que es importante comprender la programación dinámica. El uso de la programación dinámica (DP) para escribir algoritmos es tan esencial como se teme.

¿Y quién puede culpar a quienes se apartan de él? La programación dinámica parece intimidante porque está mal enseñada. Muchos tutoriales se centran en el resultado, explicando el algoritmo, en lugar del proceso, encontrando el algoritmo. Esto fomenta la memorización, no la comprensión.

Durante mi clase de algoritmos este año, armé mi propio proceso para resolver problemas que requieren programación dinámica. Partes provienen de mi profesor de algoritmos (¡a quien se le debe mucho crédito!), Y partes de mi propia disección de algoritmos de programación dinámica.

Pero antes de compartir mi proceso, comencemos con lo básico. De todos modos, ¿qué es la programación dinámica?

Definición de programación dinámica

La programación dinámica equivale a dividir un problema de optimización en subproblemas más simples y almacenar la solución a cada subproblema de modo que cada subproblema solo se resuelva una vez.

Para ser honesto, esta definición puede no tener mucho sentido hasta que vea un ejemplo de un subproblema. Está bien, aparecerá en la siguiente sección.

Lo que espero transmitir es que DP es una técnica útil para problemas de optimización, aquellos problemas que buscan la solución máxima o mínima dadas ciertas restricciones, porque analiza todos los subproblemas posibles y nunca recalcula la solución a ningún subproblema. Esto garantiza corrección y eficiencia, lo que no podemos decir de la mayoría de técnicas utilizadas para resolver o aproximar algoritmos. Esto solo hace que DP sea especial.

En las dos secciones siguientes, voy a explicar lo que es un sub-problema es, y luego motivar por qué el almacenamiento de soluciones - una técnica conocida como memoization - cuestiones de programación dinámica.

Subproblemas sobre subproblemas sobre subproblemas

Los subproblemas son versiones más pequeñas del problema original. De hecho, los subproblemas a menudo parecen una versión modificada del problema original. Si se formulan correctamente, los subproblemas se complementan entre sí para obtener la solución al problema original.

Para darle una mejor idea de cómo funciona esto, busquemos el subproblema en un ejemplo de problema de programación dinámica.

Imagina que estás en la década de 1950 trabajando en una computadora IBM-650. Ya sabes lo que esto significa: ¡tarjetas perforadas! Su trabajo es para hombre o mujer, el IBM-650 por un día. Te dan un número natural n tarjetas perforadas para ejecutar. Cada tarjeta perforada i debe ejecutarse a una hora de inicio predeterminada s_i y dejar de ejecutarse en una hora de finalización predeterminada f_i . Solo se puede ejecutar una tarjeta perforada en el IBM-650 a la vez. Cada tarjeta perforada también tiene un valor asociado v_i basado en la importancia que tiene para su empresa.

Problema : Como persona a cargo del IBM-650, debe determinar el programa óptimo de tarjetas perforadas que maximice el valor total de todas las tarjetas perforadas ejecutadas.

Debido a que analizaré este ejemplo con gran detalle a lo largo de este artículo, solo te burlaré de su subproblema por ahora:

Sub-problema : El horario máximo valor para tarjetas perforadas i través de n tal que las tarjetas perforadas están ordenados por hora de inicio.

Observe cómo el subproblema descompone el problema original en componentes que forman la solución. Con el subproblema, puede encontrar el valor máximo programado para tarjetas perforadas n-1 a n , y luego para tarjetas perforadas n-2 a n , y así sucesivamente. Al encontrar las soluciones para cada subproblema, puede abordar el problema original en sí: el programa de valor máximo para las tarjetas perforadas 1 a n . Dado que el subproblema se parece al problema original, se pueden utilizar subproblemas para resolver el problema original.

En la programación dinámica, después de resolver cada subproblema, debe memorizarlo o almacenarlo. Averigüemos por qué en la siguiente sección.

Motivar la memorización con números de Fibonacci

Cuando se le diga que implemente un algoritmo que calcula el valor de Fibonacci para cualquier número dado, ¿qué haría? La mayoría de las personas que conozco optarían por un algoritmo recursivo que se parece a esto en Python:

def fibonacciVal(n): if n == 0: return 0 elif n == 1: return 1 else: return fibonacciVal(n-1) + fibonacciVal(n-2)

Este algoritmo cumple su propósito, pero a un costo enorme . Por ejemplo, veamos lo que debe calcular este algoritmo para resolver n = 5 (abreviado como F (5)):

F(5) / \ / \ / \ F(4) F(3) / \ / \ F(3) F(2) F(2) F(1) / \ / \ / \ F(2) F(1) F(1) F(0) F(1) F(0) / \ F(1) F(0)

El árbol de arriba representa cada cálculo que se debe hacer para encontrar el valor de Fibonacci para n = 5. Observe cómo el subproblema para n = 2 se resuelve tres veces. Para un ejemplo relativamente pequeño (n = 5), ¡eso es mucho cálculo repetido y desperdiciado!

¿Qué pasa si, en lugar de calcular el valor de Fibonacci para n = 2 tres veces, creamos un algoritmo que lo calcula una vez, almacena su valor y accede al valor de Fibonacci almacenado para cada aparición posterior de n = 2? Eso es exactamente lo que hace la memorización.

Con esto en mente, escribí una solución de programación dinámica para el problema del valor de Fibonacci:

def fibonacciVal(n): memo = [0] * (n+1) memo[0], memo[1] = 0, 1 for i in range(2, n+1): memo[i] = memo[i-1] + memo[i-2] return memo[n]

Observe cómo la solución del valor de retorno proviene de la matriz de memorización memo [], que se completa iterativamente con el bucle for. Por “iterativamente”, me refiero a que el memo [2] se calcula y almacena antes del memo [3], el memo [4],… y el memo [ n ]. Debido a que memo [] se llena en este orden, la solución para cada subproblema (n = 3) se puede resolver con las soluciones de sus subproblemas anteriores (n = 2 yn = 1) porque estos valores ya estaban almacenados en memo [] en un momento anterior.

La memorización significa que no hay recálculo, lo que lo convierte en un algoritmo más eficiente. Así, la memorización asegura que la programación dinámica sea eficiente, pero es elegir el subproblema correcto lo que garantiza que un programa dinámico pase por todas las posibilidades para encontrar el mejor.

Ahora que hemos abordado la memorización y los subproblemas, es hora de aprender el proceso de programación dinámica. Abróchese el cinturón.

Mi proceso de programación dinámica

Paso 1: Identifica el subproblema en palabras.

Con demasiada frecuencia, los programadores recurrirán a la escritura de código antes de pensar críticamente sobre el problema en cuestión. No está bien. Una estrategia para encender su cerebro antes de tocar el teclado es usar palabras, en inglés o de otro tipo, para describir el subproblema que ha identificado dentro del problema original.

Si está resolviendo un problema que requiere programación dinámica, tome un papel y piense en la información que necesita para resolver este problema. Escriba el subproblema teniendo esto en cuenta.

For example, in the punchcard problem, I stated that the sub-problem can be written as “the maximum value schedule for punchcards i through n such that the punchcards are sorted by start time.” I found this sub-problem by realizing that, in order to determine the maximum value schedule for punchcards 1 through n such that the punchcards are sorted by start time, I would need to find the answer to the following sub-problems:

  • The maximum value schedule for punchcards n-1 through n such that the punchcards are sorted by start time
  • The maximum value schedule for punchcards n-2 through n such that the punchcards are sorted by start time
  • The maximum value schedule for punchcards n-3 through n such that the punchcards are sorted by start time
  • (Et cetera)
  • The maximum value schedule for punchcards 2 through n such that the punchcards are sorted by start time

If you can identify a sub-problem that builds upon previous sub-problems to solve the problem at hand, then you’re on the right track.

Step 2: Write out the sub-problem as a recurring mathematical decision.

Once you’ve identified a sub-problem in words, it’s time to write it out mathematically. Why? Well, the mathematical recurrence, or repeated decision, that you find will eventually be what you put into your code. Besides, writing out the sub-problem mathematically vets your sub-problem in words from Step 1. If it is difficult to encode your sub-problem from Step 1 in math, then it may be the wrong sub-problem!

There are two questions that I ask myself every time I try to find a recurrence:

  • What decision do I make at every step?
  • If my algorithm is at step i, what information would it need to decide what to do in step i+1? (And sometimes: If my algorithm is at step i, what information did it need to decide what to do in step i-1?)

Let’s return to the punchcard problem and ask these questions.

What decision do I make at every step? Assume that the punchcards are sorted by start time, as mentioned previously. For each punchcard that is compatible with the schedule so far (its start time is after the finish time of the punchcard that is currently running), the algorithm must choose between two options: to run, or not to run the punchcard.

If my algorithm is at stepi, what information would it need to decide what to do in stepi+1? To decide between the two options, the algorithm needs to know the next compatible punchcard in the order. The next compatible punchcard for a given punchcard p is the punchcard q such that s_q (the predetermined start time for punchcard q) happens after f_p (the predetermined finish time for punchcard p) and the difference between s_q and f_p is minimized. Abandoning mathematician-speak, the next compatible punchcard is the one with the earliest start time after the current punchcard finishes running.

If my algorithm is at stepi, what information did it need to decide what to do in stepi-1? The algorithm needs to know about future decisions: the ones made for punchcards i through n in order to decide to run or not to run punchcard i-1.

Now that we’ve answered these questions, perhaps you’ve started to form a recurring mathematical decision in your mind. If not, that’s also okay, it becomes easier to write recurrences as you get exposed to more dynamic programming problems.

Without further ado, here’s our recurrence:

OPT(i) = max(v_i + OPT(next[i]), OPT(i+1))

This mathematical recurrence requires some explaining, especially for those who haven’t written one before. I use OPT(i) to represent the maximum value schedule for punchcards i through n such that the punchcards are sorted by start time. Sounds familiar, right? OPT(•) is our sub-problem from Step 1.

In order to determine the value of OPT(i), we consider two options, and we want to take the maximum of these options in order to meet our goal: the maximum value schedule for all punchcards. Once we choose the option that gives the maximum result at step i, we memoize its value as OPT(i).

The two options — to run or not to run punchcard i — are represented mathematically as follows:

v_i + OPT(next[i])

This clause represents the decision to run punchcard i. It adds the value gained from running punchcard i to OPT(next[i]), where next[i] represents the next compatible punchcard following punchcard i. OPT(next[i]) gives the maximum value schedule for punchcards next[i] through n such that the punchcards are sorted by start time. Adding these two values together produces maximum value schedule for punchcards i through n such that the punchcards are sorted by start time if punchcard i is run.

OPT(i+1)

Conversely, this clause represents the decision to not run punchcard i. If punchcard i is not run, its value is not gained. OPT(i+1) gives the maximum value schedule for punchcards i+1 through n such that the punchcards are sorted by start time. So, OPT(i+1) gives the maximum value schedule for punchcards i through n such that the punchcards are sorted by start time if punchcard i is not run.

In this way, the decision made at each step of the punchcard problems is encoded mathematically to reflect the sub-problem in Step 1.

Step 3: Solve the original problem using Steps 1 and 2.

In Step 1, we wrote down the sub-problem for the punchcard problem in words. In Step 2, we wrote down a recurring mathematical decision that corresponds to these sub-problems. How can we solve the original problem with this information?

OPT(1)

It’s that simple. Since the sub-problem we found in Step 1 is the maximum value schedule for punchcards i through n such that the punchcards are sorted by start time, we can write out the solution to the original problem as the maximum value schedule for punchcards 1 through n such that the punchcards are sorted by start time. Since Steps 1 and 2 go hand in hand, the original problem can also be written as OPT(1).

Step 4: Determine the dimensions of the memoization array and the direction in which it should be filled.

Did you find Step 3 deceptively simple? It sure seems that way. You may be thinking, how can OPT(1) be the solution to our dynamic program if it relies on OPT(2), OPT(next[1]), and so on?

You’re correct to notice that OPT(1) relies on the solution to OPT(2). This follows directly from Step 2:

OPT(1) = max(v_1 + OPT(next[1]), OPT(2))

But this is not a crushing issue. Think back to Fibonacci memoization example. To find the Fibonacci value for n = 5, the algorithm relies on the fact that the Fibonacci values for n = 4, n = 3, n = 2, n = 1, and n = 0 were already memoized. If we fill in our memoization table in the correct order, the reliance of OPT(1) on other sub-problems is no big deal.

How can we identify the correct direction to fill the memoization table? In the punchcard problem, since we know OPT(1) relies on the solutions to OPT(2) and OPT(next[1]), and that punchcards 2 and next[1] have start times after punchcard 1 due to sorting, we can infer that we need to fill our memoization table from OPT(n) to OPT(1).

How do we determine the dimensions of this memoization array? Here’s a trick: the dimensions of the array are equal to the number and size of the variables on which OPT(•) relies. In the punchcard problem, we have OPT(i), which means that OPT(•) only relies on variable i, which represents the punchcard number. This suggest that our memoization array will be one-dimensional and that its size will be n since there are n total punchcards.

If we know that n = 5, then our memoization array might look like this:

memo = [OPT(1), OPT(2), OPT(3), OPT(4), OPT(5)]

However, because many programming languages start indexing arrays at 0, it may be more convenient to create this memoization array so that its indices align with punchcard numbers:

memo = [0, OPT(1), OPT(2), OPT(3), OPT(4), OPT(5)]

Step 5: Code it!

To code our dynamic program, we put together Steps 2–4. The only new piece of information that you’ll need to write a dynamic program is a base case, which you can find as you tinker with your algorithm.

A dynamic program for the punchcard problem will look something like this:

def punchcardSchedule(n, values, next): # Initialize memoization array - Step 4 memo = [0] * (n+1) # Set base case memo[n] = values[n] # Build memoization table from n to 1 - Step 2 for i in range(n-1, 0, -1): memo[i] = max(v_i + memo[next[i]], memo[i+1]) # Return solution to original problem OPT(1) - Step 3 return memo[1]

Congrats on writing your first dynamic program! Now that you’ve wet your feet, I’ll walk you through a different type of dynamic program.

Paradox of Choice: Multiple Options DP Example

Although the previous dynamic programming example had a two-option decision — to run or not to run a punchcard — some problems require that multiple options be considered before a decision can be made at each step.

Time for a new example.

Pretend you’re selling the friendship bracelets to n customers, and the value of that product increases monotonically. This means that the product has prices {p_1, …, p_n} such that p_i ≤ p_j if customer j comes after customer i. These n customers have values {v_1, …, v_n}. A given customer i will buy a friendship bracelet at price p_i if and only if p_iv_i; otherwise the revenue obtained from that customer is 0. Assume prices are natural numbers.

Problem: You must find the set of prices that ensure you the maximum possible revenue from selling your friendship bracelets.

Take a second to think about how you might address this problem before looking at my solutions to Steps 1 and 2.

Step 1: Identify the sub-problem in words.

Sub-problem: The maximum revenue obtained from customers i through n such that the price for customer i-1 was set at q.

I found this sub-problem by realizing that to determine the maximum revenue for customers 1 through n, I would need to find the answer to the following sub-problems:

  • The maximum revenue obtained from customers n-1 through n such that the price for customer n-2 was set at q.
  • The maximum revenue obtained from customers n-2 through n such that the price for customer n-3 was set at q.
  • (Et cetera)

Notice that I introduced a second variable q into the sub-problem. I did this because, in order to solve each sub-problem, I need to know the price I set for the customer before that sub-problem. Variable q ensures the monotonic nature of the set of prices, and variable i keeps track of the current customer.

Step 2: Write out the sub-problem as a recurring mathematical decision.

There are two questions that I ask myself every time I try to find a recurrence:

  • What decision do I make at every step?
  • If my algorithm is at step i, what information would it need to decide what to do in step i+1? (And sometimes: If my algorithm is at step i, what information would it need to decide what to do in step i-1?)

Let’s return to the friendship bracelet problem and ask these questions.

What decision do I make at every step? I decide at which price to sell my friendship bracelet to the current customer. Since prices must be natural numbers, I know that I should set my price for customer i in the range from q — the price set for customer i-1 — to v_i — the maximum price at which customer i will buy a friendship bracelet.

If my algorithm is at stepi, what information would it need to decide what to do in stepi+1? My algorithm needs to know the price set for customer i and the value of customer i+1 in order to decide at what natural number to set the price for customer i+1.

With this knowledge, I can mathematically write out the recurrence:

OPT(i,q) = max~([Revenue(v_i, a) + OPT(i+1, a)])
such that max~ finds the maximum over all a in the range q ≤ a ≤ v_i

Once again, this mathematical recurrence requires some explaining. Since the price for customer i-1 is q, for customer i, the price a either stays at integer q or it changes to be some integer between q+1 and v_i. To find the total revenue, we add the revenue from customer i to the maximum revenue obtained from customers i+1 through n such that the price for customer i was set at a.

In other words, to maximize the total revenue, the algorithm must find the optimal price for customer i by checking all possible prices between q and v_i. If v_iq, then the price a must remain at q.

What about the other steps?

Working through Steps 1 and 2 is the most difficult part of dynamic programming. As an exercise, I suggest you work through Steps 3, 4, and 5 on your own to check your understanding.

Runtime Analysis of Dynamic Programs

Now for the fun part of writing algorithms: runtime analysis. I’ll be using big-O notation throughout this discussion . If you’re not yet familiar with big-O, I suggest you read up on it here.

Generally, a dynamic program’s runtime is composed of the following features:

  • Pre-processing
  • How many times the for loop runs
  • How much time it takes the recurrence to run in one for loop iteration
  • Post-processing

Overall, runtime takes the following form:

Pre-processing + Loop * Recurrence + Post-processing

Let’s perform a runtime analysis of the punchcard problem to get familiar with big-O for dynamic programs. Here is the punchcard problem dynamic program:

def punchcardSchedule(n, values, next): # Initialize memoization array - Step 4 memo = [0] * (n+1) # Set base case memo[n] = values[n] # Build memoization table from n to 1 - Step 2 for i in range(n-1, 0, -1): memo[i] = max(v_i + memo[next[i]], memo[i+1]) # Return solution to original problem OPT(1) - Step 3 return memo[1]

Let’s break down its runtime:

  • Pre-processing: Here, this means building the the memoization array. O(n).
  • How many times the for loop runs: O(n).
  • How much time it takes the recurrence to run in one for loop iteration: The recurrence takes constant time to run because it makes a decision between two options in each iteration. O(1).
  • Post-processing: None here! O(1).

The overall runtime of the punchcard problem dynamic program is O(n) O(n) * O(1) + O(1), or, in simplified form, O(n).

You Did It!

Well, that’s it — you’re one step closer to becoming a dynamic programming wizard!

One final piece of wisdom: keep practicing dynamic programming. No matter how frustrating these algorithms may seem, repeatedly writing dynamic programs will make the sub-problems and recurrences come to you more naturally. Here’s a crowdsourced list of classic dynamic programming problems for you to try.

So get out there and take your interviews, classes, and life (of course) with your newfound dynamic programming knowledge!

Many thanks to Steven Bennett, Claire Durand, and Prithaj Nath for proofreading this post. Thank you to Professor Hartline for getting me so excited about dynamic programming that I wrote about it at length.

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