Afmystificering af dynamisk programmering

Sådan konstrueres og kodes dynamiske programmeringsalgoritmer

Måske har du hørt om det i forberedelsen til kodningssamtaler. Måske har du kæmpet igennem det i et algoritmekursus. Måske forsøger du at lære at kode på egen hånd og fik at vide et eller andet sted undervejs, at det er vigtigt at forstå dynamisk programmering. Brug af dynamisk programmering (DP) til at skrive algoritmer er lige så vigtigt som det frygtes.

Og hvem kan bebrejde dem, der krymper væk fra det? Dynamisk programmering virker skræmmende, fordi den er dårligt lært. Mange tutorials fokuserer på resultatet - forklarer algoritmen i stedet for processen - finder algoritmen. Dette tilskynder til memorisering, ikke forståelse.

I løbet af min algoritmeklasse i år opsamlede jeg min egen proces til løsning af problemer, der kræver dynamisk programmering. Dele af det kommer fra min algoritmeprofessor (som jeg har stor kredit!) Og dele fra min egen dissektion af dynamiske programmeringsalgoritmer.

Men inden jeg deler min proces, lad os starte med det grundlæggende. Hvad er alligevel dynamisk programmering?

Dynamisk programmering defineret

Dynamisk programmering svarer til at nedbryde et optimeringsproblem i enklere underproblemer og lagre løsningen på hvert underproblem, så hvert underproblem kun løses en gang.

For at være ærlig giver denne definition muligvis ikke total mening, før du ser et eksempel på et underproblem. Det er okay, det kommer op i næste afsnit.

Hvad jeg håber at formidle er, at DP er en nyttig teknik til optimeringsproblemer, de problemer, der søger den maksimale eller minimale løsning givet visse begrænsninger, fordi den ser igennem alle mulige underproblemer og aldrig omregner løsningen til noget underproblem. Dette garanterer korrekthed og effektivitet, hvilket vi ikke kan sige om de fleste teknikker, der bruges til at løse eller tilnærme algoritmer. Dette alene gør DP speciel.

I de næste to afsnit vil jeg forklare, hvad et underproblem er, og derefter motivere, hvorfor lagring af løsninger - en teknik kendt som memoization - betyder noget i dynamisk programmering.

Underproblemer på Underproblemer på Underproblemer

Underproblemer er mindre versioner af det oprindelige problem. Faktisk ligner underproblemer ofte en omformuleret version af det oprindelige problem. Hvis de formuleres korrekt, bygger underproblemer på hinanden for at få løsningen på det oprindelige problem.

For at give dig en bedre idé om, hvordan dette fungerer, lad os finde underproblemet i et eksempel på et dynamisk programmeringsproblem.

Lad som om du er tilbage i 1950'erne og arbejder på en IBM-650 computer. Du ved hvad det betyder - stempelkort! Dit job er at mand eller kvinde IBM-650 i en dag. Du får et naturligt antal n hulkort til at køre. Hvert stempelkort i skal køres til en forudbestemt starttid s_i og stoppe med at køre ved en forudbestemt sluttid f_i . Kun ét hulkort kan køre på IBM-650 på én gang. Hvert hulkort har også en tilknyttet værdi v_i baseret på hvor vigtigt det er for din virksomhed.

Problem : Som den person, der er ansvarlig for IBM-650, skal du bestemme den optimale tidsplan for hulkort, der maksimerer den samlede værdi af alle kørte stempelkort.

Fordi jeg gennemgår dette eksempel i detaljer i hele denne artikel, driller jeg dig kun med dets underproblem indtil videre:

Underproblem : Den maksimale værdiskema for hulkort i gennem n, således at hulkortene sorteres efter starttidspunktet.

Læg mærke til, hvordan underproblemet nedbryder det originale problem i komponenter, der bygger op på løsningen. Med delproblemet kan du finde den maksimale tidsplan for hulkort n-1 til n og derefter for hulkort n-2 til n osv. Ved at finde løsninger til hvert enkelt underproblem kan du derefter tackle det originale problem i sig selv: den maksimale værdiskema for hulkort 1 til n . Da delproblemet ligner det oprindelige problem, kan underproblemer bruges til at løse det oprindelige problem.

Når du har løst hvert underproblem, skal du huske eller gemme det i dynamisk programmering. Lad os finde ud af hvorfor i det følgende afsnit.

Motiverende memoisering med Fibonacci-numre

Når du fik besked på at implementere en algoritme, der beregner Fibonacci-værdien for et givet nummer, hvad ville du gøre? De fleste mennesker, jeg kender, vælger en rekursiv algoritme, der ser sådan ud i Python:

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

Denne algoritme opfylder sit formål, men til en enorm pris. Lad os for eksempel se på, hvad denne algoritme skal beregne for at løse for n = 5 (forkortet 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)

Træet ovenfor repræsenterer hver beregning, der skal foretages for at finde Fibonacci-værdien for n = 5. Bemærk, hvordan delproblemet for n = 2 løses tre gange. For et relativt lille eksempel (n = 5) er det meget gentagen og spildt beregning!

Hvad hvis vi i stedet for at beregne Fibonacci-værdien for n = 2 tre gange, oprettede vi en algoritme, der beregner den en gang, gemmer dens værdi og får adgang til den lagrede Fibonacci-værdi for hver efterfølgende forekomst af n = 2? Det er præcis, hvad memoization gør.

Med dette i tankerne har jeg skrevet en dynamisk programmeringsløsning til Fibonacci-værdiproblemet:

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]

Læg mærke til, hvordan løsningen på returværdien kommer fra memoization array memo [], som iterativt udfyldes af for loop. Med "iterativt" mener jeg, at memo [2] beregnes og gemmes før memo [3], memo [4],… og memo [ n ]. Fordi memo [] er udfyldt i denne rækkefølge, kan løsningen for hvert underproblem (n = 3) løses af løsningerne på de foregående underproblemer (n = 2 og n = 1), fordi disse værdier allerede var gemt i memo [] på et tidligere tidspunkt.

Memoization betyder ingen genberegning, hvilket giver en mere effektiv algoritme. Memoization sikrer således, at dynamisk programmering er effektiv, men det vælger det rigtige delproblem, der garanterer, at et dynamisk program gennemgår alle muligheder for at finde det bedste.

Nu hvor vi har adresseret memoisering og underproblemer, er det tid til at lære den dynamiske programmeringsproces. Spænd ind.

Min dynamiske programmeringsproces

Trin 1: Identificer underproblemet i ord.

Alt for ofte vil programmører henvende sig til at skrive kode, før de tænker kritisk på det aktuelle problem. Ikke godt. En strategi til at fyre din hjerne op, før du rører ved tastaturet, er at bruge ord, engelsk eller på anden måde, til at beskrive det delproblem, du har identificeret inden for det oprindelige problem.

Hvis du løser et problem, der kræver dynamisk programmering, skal du tage et stykke papir og tænke på de oplysninger, du har brug for for at løse dette problem. Skriv underproblemet med dette i tankerne.

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!

Mange tak til Steven Bennett, Claire Durand og Prithaj Nath for korrekturlæsning af dette indlæg. Tak til professor Hartline for at få mig så begejstret for dynamisk programmering, at jeg skrev om det udførligt.

Nyd hvad du læser? Spred kærligheden ved at lide og dele dette stykke. Har du tanker eller spørgsmål? Kontakt mig på Twitter eller i kommentarerne nedenfor.