Dünaamilise programmeerimise demüstifitseerimine

Kuidas koostada ja kodeerida dünaamilisi programmeerimisalgoritme

Võib-olla olete sellest intervjuude kodeerimisel valmistumisel kuulnud. Võib-olla olete algoritmide kursusel selle nimel vaeva näinud. Võib-olla proovite õppida ise kodeerima ja teile öeldi kusagil teel, et dünaamilise programmeerimise mõistmine on oluline. Dünaamilise programmeerimise (DP) kasutamine algoritmide kirjutamiseks on sama oluline kui kardetakse.

Ja kes saaks süüdistada neid, kes sellest eemale tõmbuvad? Dünaamiline programmeerimine tundub hirmutav, kuna seda on halvasti õpetatud. Paljud õpetused keskenduvad tulemusele - algoritmi selgitamisele protsessi asemel - algoritmi leidmisele . See soodustab meeldejätmist, mitte mõistmist.

Selle aasta algoritmide tunnis panin kokku omaenda protsessi dünaamilist programmeerimist nõudvate probleemide lahendamiseks. Osa sellest pärineb minu algoritmide professorilt (kellele kuulub suur au!) Ja osad minu enda dünaamiliste programmeerimisalgoritmide lahkamisest.

Aga enne kui jagan oma protsessi, alustame põhitõdedest. Mis on dünaamiline programmeerimine üldse?

Dünaamiline programmeerimine määratletud

Dünaamiline programmeerimine tähendab optimeerimisprobleemi jaotamist lihtsamateks alamprobleemideks ja iga alamprobleemi lahenduse salvestamist nii, et iga alamprobleem lahendatakse ainult üks kord.

Kui aus olla, ei pruugi see määratlus olla täiesti mõistlik enne, kui näete alamprobleemi näidet. See on okei, see tuleb järgmises osas.

Loodan öelda, et DP on kasulik meetod optimeerimisprobleemide jaoks - need probleemid, mis otsivad teatud piirangute korral maksimaalset või minimaalset lahendust, sest see vaatab läbi kõik võimalikud alamprobleemid ja ei arvesta kunagi ühegi alamprobleemi lahendust. See tagab õigsuse ja tõhususe, mida me ei saa öelda enamiku algoritmide lahendamiseks või ligikaudseks muutmiseks kasutatavate tehnikate kohta. Juba see muudab DP eriliseks.

Järgmises kahes jaotises selgitan, mis on alamprobleem , ja motiveerin siis, miks on dünaamilises programmeerimises oluline lahendite salvestamine - tehnika, mida nimetatakse memodeks .

Alamprobleemid alamprobleemide kohta

Alamprobleemid on algse probleemi väiksemad versioonid. Tegelikult näevad alamprobleemid sageli algse probleemi ümber sõnastatud versiooni. Õige sõnastuse korral toetuvad alamprobleemid üksteisele, et leida lahendus algsele probleemile.

Selle toimimise kohta parema ettekujutuse leidmiseks leiame dünaamilise programmeerimise probleemi alamprobleemi.

Teeskle, et töötate tagasi 1950. aastatel ja töötate IBM-650 arvutiga. Teate, mida see tähendab - perfokaardid! Teie ülesandeks on ühe päeva jooksul IBM-650 mees või naine. Teile antakse käivitamiseks loomulik number n perfokaarti. Iga perfokaart i tuleb käivitada mingil ettemääratud algusajal s_i ja lõpetada töötamine mõnel ettemääratud lõppajal f_i . IBM-650-l saab korraga töötada ainult üks perfokaart. Igal perfokaardil on ka seotud väärtus v_i, lähtudes sellest, kui oluline see teie ettevõttele on.

Probleem : IBM-650 eest vastutajana peate määrama perfokaartide optimaalse ajakava, mis maksimeerib kõigi käitatavate perfokaartide koguväärtuse.

Kuna ma vaatan selle näite kogu artiklis väga üksikasjalikult läbi, siis praegu kiusan teid ainult selle alamprobleemiga:

Sub-probleem : maksimaalne väärtus ajakava perfokaardid i kaudu n selline, et perfokaardid on järjestatud vastavalt algusaeg.

Pange tähele, kuidas alamprobleem jaotab algse probleemi komponentideks, mis lahenduse loovad. Alamprobleemi abil saate leida perfokaartide n-1 kuni n ja seejärel perfokaartide n-2 kuni n maksimaalse väärtuse ajakava . Leides lahendused igale üksikule alamprobleemile, saate seejärel tegeleda algse probleemiga: perfokaartide 1 kuni n maksimaalse väärtuse ajakava . Kuna alamprobleem näeb välja nagu algne probleem, saab algprobleemi lahendamiseks kasutada alamprobleeme.

Dünaamilises programmeerimises peate pärast iga alamprobleemi lahendamist selle memodestama või salvestama. Uurime, miks järgmises jaotises.

Motiveeriv memo Fibonacci numbritega

Kui teil palutakse rakendada algoritmi, mis arvutab iga numbri Fibonacci väärtuse, mida te teete? Enamik inimesi, keda ma tean, valiks rekursiivse algoritmi, mis näeb Pythonis välja umbes selline:

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

See algoritm täidab oma eesmärki, kuid tohutute kuludega. Näiteks vaatame, mida see algoritm peab arvutama, et lahendada n = 5 (lühendatult 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)

Ülaltoodud puu tähistab kõiki arvutusi, mis tuleb teha, et leida n = 5 Fibonacci väärtus. Pange tähele, kuidas n = 2 alamprobleem lahendatakse kolm korda. Suhteliselt väikese näite (n = 5) puhul on see palju korduvat ja raisatud arvutust!

Mis oleks, kui selle asemel, et kolm korda arvutada Fibonacci väärtus n = 2, loome algoritmi, mis arvutab selle üks kord, salvestab selle väärtuse ja pääseb juurde iga järgneva n = 2 esinemise korral salvestatud Fibonacci väärtusele? See on täpselt , mida memoization teeb.

Seda silmas pidades olen kirjutanud dünaamilise programmeerimislahenduse Fibonacci väärtusprobleemile:

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]

Pange tähele, kuidas tagastusväärtuse lahendus tuleneb memodeerimise massiivi memost [], mille for silmus täidab iteratiivselt. „Korduvalt” all mõtlen, et memo [2] arvutatakse ja salvestatakse enne memot [3], memot [4],… ja memot [ n ]. Kuna memo [] täidetakse selles järjekorras, saab iga alamprobleemi (n = 3) lahenduse lahendada selle eelmiste alamprobleemide (n = 2 ja n = 1) lahendustega, kuna need väärtused olid juba salvestatud memo [] varasemal ajal.

Mälestamine ei tähenda ümberarvutamist, mis muudab algoritmi tõhusamaks. Seega tagab memodeerimine dünaamilise programmeerimise efektiivsuse, kuid just õige alamprobleemi valimine tagab, et dünaamiline programm läbib kõik võimalused parima leidmiseks.

Nüüd, kui oleme tegelenud memode ja alamprobleemidega, on aeg õppida dünaamilist programmeerimisprotsessi. Pannal sisse.

Minu dünaamiline programmeerimisprotsess

1. samm: tuvastage sõnadega alamprobleem.

Liiga sageli pöörduvad programmeerijad koodi kirjutamise poole, enne kui mõtlevad kriitiliselt käsitletavale probleemile. Pole hea. Enne klaviatuuri puudutamist on aju süttimise strateegiaks sõnade kasutamine inglise keeles või muul viisil, et kirjeldada algprobleemis tuvastatud alamprobleemi.

Kui lahendate probleemi, mis nõuab dünaamilist programmeerimist, haarake paber ja mõelge teabele, mida selle probleemi lahendamiseks vajate. Kirjutage seda silmas pidades välja alamprobleem.

Näiteks märkisin perfokaardiprobleemis, et alamprobleemi võib kirjutada kui "perfokaartide i kuni n maksimaalse väärtuse ajakava, nii et perfokaardid on sorditud algusaja järgi". Leidsin selle alamprobleemi, mõistes, et perfokaartide 1 kuni n maksimaalse väärtuste ajakava määramiseks nii, et perfokaardid oleksid sorditud algusaja järgi, peaksin leidma vastuse järgmistele alamprobleemidele:

  • Perfokaartide n-1 kuni n maksimaalne väärtusgraafik, nii et perfokaardid on sorditud algusaja järgi
  • Perfokaartide n-2 kuni n maksimaalne väärtusgraafik nii, et perfokaardid on sorditud algusaja järgi
  • 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!

Suur tänu Steven Bennettile, Claire Durandile ja Prithaj Nathile selle postituse korrektuuri eest. Aitäh professor Hartline'ile, kes mind dünaamilise programmeerimise vastu nii põnevile ajas, et kirjutasin sellest pikalt.

Nautige seda, mida loete? Levitage armastust, meeldides ja jagades seda tükki. Kas teil on mõtteid või küsimusi? Võtke minuga ühendust Twitteris või allpool toodud kommentaarides.