Демистификација динамичког програмирања

Како конструисати и кодирати алгоритме динамичког програмирања

Можда сте чули за то током припрема за кодирање интервјуа. Можда сте се мучили кроз то на курсу алгоритама. Можда покушавате да научите како да кодирате сами, а негде успут су вам рекли да је важно да разумете динамичко програмирање. Коришћење динамичког програмирања (ДП) за писање алгоритама је онолико битно колико се и плаши.

И ко може кривити оне који се суздржавају од тога? Динамичко програмирање делује застрашујуће јер је лоше научено. Многи водичи се фокусирају на исход - објашњавање алгоритма, уместо на процес - проналажење алгоритма. Ово подстиче меморисање, а не разумевање.

Током предавања из алгоритма ове године, саставио сам сопствени процес за решавање проблема који захтевају динамичко програмирање. Делови тога потичу од мог професора алгоритма (коме дугујем велику част!), А делови мог сопственог сецирања алгоритама динамичког програмирања.

Али пре него што поделим свој процес, почнимо са основама. Шта је уопште динамичко програмирање?

Дефинисано динамичко програмирање

Динамичко програмирање своди се на то да се проблем оптимизације рашчлањује на једноставније потпроблеме и чува решење за сваки под-проблем тако да се сваки под-проблем решава само једном.

Да будем искрен, ова дефиниција можда неће имати потпуни смисао док не видите пример под-проблема. У реду је, то се појављује у следећем одељку.

Оно што се надам да кажем је да је ДП корисна техника за проблеме оптимизације, оне проблеме који траже максимално или минимално решење с обзиром на одређена ограничења, јер прегледава све могуће подпроблеме и никада не прерачунава решење за било који подпроблем. Ово гарантује исправност и ефикасност, што не можемо рећи за већину техника које се користе за решавање или приближавање алгоритама. Само ово ДП чини посебним.

У следећа два одељка објаснићу шта је под-проблем , а затим ћу мотивисати зашто је складиштење решења - техника позната као мемоизација - важно у динамичком програмирању.

Под-проблеми на Под-проблеми на Под-проблеми

Подпроблеми су мање верзије оригиналног проблема. У ствари, под-проблеми често изгледају као преформулисана верзија оригиналног проблема. Ако су правилно формулисани, под-проблеми се надограђују једни на друге како би се добило решење изворног проблема.

Да бисмо вам пружили бољу представу о томе како ово функционише, пронађимо под-проблем у примеру проблема са динамичким програмирањем.

Претварајте се да сте се вратили у педесете и радили на рачунару ИБМ-650. Знате шта ово значи - боксери! Ваш посао је мушкарцу или жени ИБМ-650 на један дан. Добили сте природни број н пунцхцардс за покретање. Свака пунцхцард и мора се покренути у неко унапред одређено време почетка с_и и зауставити извођење у неко унапред одређено време завршетка ф_и . На ИБМ-650 одједном може радити само једна бушотина. Свака пунцхцард такође има придружену вредност в_и на основу тога колико је важна за вашу компанију.

Проблем : Као особа одговорна за ИБМ-650, морате одредити оптималан распоред пунцхцардс-а који максимизира укупну вредност свих покренутих пунцхцардс-а.

Будући да ћу детаљно проћи кроз овај пример у овом чланку, засад ћу вас задиркивати само његовим под-проблемом:

Подзадатак : Распоред максималних вредности за пунцхцардс од и до н такав да су пунцхцардс сортирани према времену почетка.

Приметите како под-проблем рашчлањује оригинални проблем на компоненте које граде решење. Уз под-проблем можете пронаћи распоред максималних вредности за бушилице од н-1 до н , а затим за бушилице од н-2 до н и тако даље. Проналажењем решења за сваки поједини потпроблем, тада се можете позабавити самим оригиналним проблемом: распоредом максималних вредности за бушилице од 1 до н . Будући да под-проблем изгледа као оригинални проблем, под-проблеми се могу користити за решавање оригиналног проблема.

У динамичком програмирању, након што решите сваки под-проблем, морате га меморисати или сачувати. Откријмо зашто у следећем одељку.

Мотивисање мемоизације Фибоначијевим бројевима

Шта бисте урадили када вам се каже да примените алгоритам који израчунава Фибоначијеву вредност за било који дати број? Већина људи које познајем одлучили би се за рекурзивни алгоритам који у Питхону изгледа отприлике овако:

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

Овај алгоритам остварује своју сврху, али уз огромну цену. На пример, погледајмо шта овај алгоритам мора израчунати да би решио за н = 5 (скраћено Ф (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)

Стабло горе представља свако израчунавање које се мора извршити да би се пронашла Фибоначијева вредност за н = 5. Обратите пажњу на то како се под-проблем за н = 2 решава три пута. За релативно мали пример (н = 5), то је много поновљених и узалудних прорачуна!

Шта ако смо, уместо да три пута израчунамо Фибоначијеву вредност за н = 2, креирали алгоритам који је израчунава једном, чува њену вредност и приступа сачуваној Фибоначијевој вредности за сваку следећу појаву н = 2? То је управо оно што мемоизатион ради.

Имајући ово на уму, написао сам решење за динамичко програмирање проблема Фибонаццијеве вредности:

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]

Приметите како решење повратне вредности долази из мемоизацијског низа мемо [], који итеративно попуњава петља фор. Под „итеративно“ мислим да се белешка [2] израчунава и чува пре белешки [3], белешке [4],… и белешке [ н ]. Будући да се мемо [] попуњава овим редоследом, решење за сваки под-проблем (н = 3) може се решити решењима његових претходних под-проблема (н = 2 и н = 1), јер су ове вредности већ биле ускладиштене у белешка [] у раније време.

Мемоизација не значи поновно израчунавање, што чини ефикаснији алгоритам. Дакле, мемоизација осигурава ефикасност динамичког програмирања, али одабир правог под-проблема који гарантује да динамички програм пролази кроз све могућности како би пронашао најбољи.

Сад кад смо се позабавили памћењем и под-проблемима, време је да научимо процес динамичког програмирања. Закопчајте се.

Мој процес динамичког програмирања

Корак 1: Идентификујте под-проблем речима.

Програмери се пречесто окрећу писању кода пре него што критички размисле о проблему који је у питању. Није добро. Једна од стратегија за паљење мозга пре него што додирнете тастатуру је коришћење речи, енглеске или неке друге, за описивање потпроблема који сте идентификовали у оригиналном проблему.

Ако решавате проблем који захтева динамичко програмирање, узмите папир и размислите о информацијама које су вам потребне за решавање овог проблема. Имајте на уму под-проблем.

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!

Велико хвала Стевену Беннетту, Цлаире Дуранд и Притхај Натх-у на лектури овог поста. Хвала професору Хартлинеу што сте ме толико узбудили због динамичког програмирања да сам о њему опширно писао.

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