Demistificiranje dinamičkog programiranja

Kako konstruirati i kodirati algoritme dinamičkog programiranja

Možda ste o tome čuli pripremajući se za kodiranje intervjua. Možda ste se mučili kroz tečaj algoritma. Možda pokušavate naučiti kako sami kodirati, a negdje usput su vam rekli da je važno razumjeti dinamičko programiranje. Korištenje dinamičkog programiranja (DP) za pisanje algoritama jednako je bitno koliko se i boji.

A tko može kriviti one koji se klone toga? Dinamično programiranje djeluje zastrašujuće jer je loše poučeno. Mnogi se vodiči usredotočuju na ishod - objašnjavanje algoritma, umjesto na postupak - pronalaženje algoritma. To potiče pamćenje, a ne razumijevanje.

Tijekom predavanja iz algoritma ove godine sastavio sam vlastiti proces rješavanja problema koji zahtijevaju dinamičko programiranje. Dijelovi toga dolaze od mog profesora algoritma (kojemu pripada velika zasluga!), A dijelovi od mog vlastitog seciranja algoritama dinamičkog programiranja.

No, prije nego što podijelim svoj postupak, krenimo s osnovama. Što je uopće dinamičko programiranje?

Definirano dinamičko programiranje

Dinamičko programiranje svodi se na to da se problem optimizacije raščlanjuje na jednostavnije pod-probleme i pohranjuje rješenje za svaki pod-problem tako da se svaki pod-problem rješava samo jednom.

Da budem iskren, ova definicija možda neće imati potpuni smisao dok ne vidite primjer potproblema. To je u redu, to se pojavljuje u sljedećem odjeljku.

Nadam se da ću reći da je DP korisna tehnika za probleme optimizacije, one probleme koji traže maksimalno ili minimalno rješenje s obzirom na određena ograničenja, jer promatra sve moguće podprobleme i nikada ne izračunava rješenje bilo kojeg pod problema. To jamči ispravnost i učinkovitost, što ne možemo reći za većinu tehnika korištenih za rješavanje ili približavanje algoritama. Samo ovo DP čini posebnim.

U sljedeća dva odjeljka objasnit ću što je pod-problem , a zatim motivirati zašto je pohranjivanje rješenja - tehnike poznate kao memoizacija - važno u dinamičkom programiranju.

Pod-problemi na Pod-problemi na Pod-problemi

Podproblemi su manje verzije izvornog problema. Zapravo, pod-problemi često izgledaju poput preoblikovane verzije izvornog problema. Ako su pravilno formulirani, potproblemi se nadograđuju jedni na druge kako bi se dobilo rješenje izvornog problema.

Da bismo vam pružili bolju ideju o tome kako to funkcionira, pronađimo potproblem u primjeru problema s dinamičkim programiranjem.

Pravite se da ste se vratili u pedesete godine prošlog stoljeća i radili na računalu IBM-650. Znate što ovo znači - bušilice! Vaš posao je muškarcu ili ženi IBM-650 na jedan dan. Dobit ćete prirodni broj n punchcards za pokretanje. Svaka punchcard i mora se pokrenuti u neko unaprijed određeno vrijeme početka s_i i zaustaviti izvođenje u neko unaprijed određeno vrijeme završetka f_i . Na IBM-650 odjednom može raditi samo jedna bušotina. Svaka punchcard također ima pridruženu vrijednost v_i ovisno o tome koliko je važna za vašu tvrtku.

Problem : Kao osoba odgovorna za IBM-650, morate odrediti optimalan raspored punchcards-a koji maksimizira ukupnu vrijednost svih pokrenutih punchcards-a.

Budući da ću detaljno proći kroz ovaj primjer u ovom članku, zasad ću vas zadirkivati ​​samo njegovim problemom:

Podzadatak : Raspored maksimalnih vrijednosti za punchcards od i do n takav da se punchcards sortiraju prema vremenu početka.

Primijetite kako pod-problem raščlanjuje izvorni problem na komponente koje grade rješenje. S pod-problemom možete pronaći raspored maksimalnih vrijednosti za punchcards n-1 do n , a zatim za punchcards n-2 do n , i tako dalje. Pronalaženjem rješenja za svaki pojedini pod-problem, tada možete riješiti sam izvorni problem: raspored maksimalnih vrijednosti za punchcards od 1 do n . Budući da pod-problem izgleda kao izvorni problem, pod-problemi se mogu koristiti za rješavanje izvornog problema.

U dinamičkom programiranju, nakon što riješite svaki pod-problem, morate ga zapamtiti ili pohraniti. Otkrijmo zašto u sljedećem odjeljku.

Motivirajuće memoiranje Fibonaccijevim brojevima

Što biste učinili kada vam se kaže da implementira algoritam koji izračunava Fibonaccijevu vrijednost za bilo koji zadani broj? Većina ljudi koje poznajem odlučili bi se za rekurzivni algoritam koji u Pythonu izgleda otprilike ovako:

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

Ovaj algoritam ostvaruje svoju svrhu, ali uz ogromnu cijenu. Na primjer, pogledajmo što ovaj algoritam mora izračunati da bi riješio za n = 5 (skraćeno 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)

Stablo gore predstavlja svaki proračun koji se mora izvršiti da bi se pronašla Fibonaccijeva vrijednost za n = 5. Primijetite kako se tri puta rješava pod-problem za n = 2 . Za relativno mali primjer (n = 5), to je puno ponovljenih i uzaludnih proračuna!

Što ako smo, umjesto da tri puta izračunamo Fibonaccijevu vrijednost za n = 2, stvorili algoritam koji je izračunava jednom, pohranjuje njezinu vrijednost i pristupa pohranjenoj Fibonaccijevoj vrijednosti za svaku sljedeću pojavu n = 2? To je upravo ono što memoization radi.

Imajući ovo na umu, napisao sam rješenje za dinamičko programiranje problema Fibonaccijeve vrijednosti:

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]

Primijetite kako rješenje povratne vrijednosti dolazi iz memoizacijskog niza memo [], koji iterativno popunjava petlja for. Pod "iterativno" mislim da se bilješka [2] izračunava i pohranjuje prije bilješke [3], bilješke [4],… i bilješke [ n ]. Budući da se memo [] popunjava ovim redoslijedom, rješenje za svaki potproblem (n = 3) može se riješiti rješenjima njegovih prethodnih potproblema (n = 2 i n = 1) jer su te vrijednosti već bile pohranjene u bilješka [] u ranije vrijeme.

Memoizacija ne znači ponovno izračunavanje, što čini učinkovitiji algoritam. Dakle, memoizacija osigurava učinkovitost dinamičkog programiranja, ali odabir pravog pod-problema koji jamči da dinamički program prolazi kroz sve mogućnosti kako bi pronašao najbolji.

Sad kad smo se pozabavili pamćenjem i pod-problemima, vrijeme je da naučimo proces dinamičkog programiranja. Zakopčajte se.

Moj proces dinamičkog programiranja

Korak 1: Identificirajte potproblem riječima.

Previše često programeri će se okrenuti pisanju koda prije nego što kritički razmisle o problemu koji je u pitanju. Nije dobro. Jedna od strategija za paljenje mozga prije nego što dodirnete tipkovnicu je korištenje riječi, engleske ili neke druge, za opisivanje potproblema koji ste prepoznali u izvornom problemu.

Ako rješavate problem koji zahtijeva dinamičko programiranje, uzmite papir i razmislite o informacijama koje su vam potrebne za rješavanje ovog problema. Imajte na umu pod-problem.

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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