Demistificējoša dinamiskā programmēšana

Kā izveidot un kodēt dinamiskos programmēšanas algoritmus

Varbūt esat par to dzirdējis, gatavojoties interviju kodēšanai. Varbūt jūs esat to cīnījies algoritmu kursā. Varbūt jūs mēģināt patstāvīgi uzzināt, kā kodēt, un kaut kur pa ceļam jums teica, ka ir svarīgi saprast dinamisko programmēšanu. Dinamiskās programmēšanas (DP) izmantošana algoritmu rakstīšanai ir tikpat būtiska, kā no tā baidās.

Un kas var vainot tos, kas no tā saraujas? Dinamiskā programmēšana šķiet biedējoša, jo tā ir slikti mācīta. Daudzas konsultācijas koncentrējas uz rezultātu - izskaidro algoritmu, nevis procesu - uz algoritma atrašanu . Tas veicina iegaumēšanu, nevis izpratni.

Šī gada algoritmu nodarbības laikā es saliku savu procesu problēmu risināšanai, kurām nepieciešama dinamiska programmēšana. Daļa no tā nāk no mana algoritmu profesora (kuram pienākas liels nopelns!), Un daļas no manis paša dinamiskajiem programmēšanas algoritmiem.

Bet pirms es dalīšos ar savu procesu, sāksim ar pamatiem. Kas vispār ir dinamiska programmēšana?

Definēta dinamiskā programmēšana

Dinamiskā programmēšana nozīmē optimizācijas problēmas sadalīšanu vienkāršākās apakšproblēmās un katras apakšproblēmas risinājuma glabāšanu tā, lai katra apakšproblēma tiktu atrisināta tikai vienu reizi.

Ja godīgi, šai definīcijai var nebūt pilnīgas jēgas, kamēr nav redzams apakšproblēmas piemērs. Tas ir labi, tas nāk klajā nākamajā sadaļā.

Es ceru pateikt, ka DP ir noderīgs paņēmiens optimizācijas problēmu risināšanai, tām problēmām, kuras meklē maksimālu vai minimālu risinājumu, ņemot vērā noteiktus ierobežojumus, jo tas izskata visas iespējamās apakšproblēmas un nekad nepārrēķina nevienas apakšproblēmas risinājumu. Tas garantē pareizību un efektivitāti, ko mēs nevaram teikt par lielāko daļu algoritmu risināšanai vai tuvināšanai izmantoto paņēmienu. Tas vien padara DP īpašu.

Nākamajās divās sadaļās es paskaidrošu, kas ir apakšproblēma , un pēc tam motivēju, kāpēc dinamiskā programmēšanā ir svarīgi saglabāt risinājumus - tehniku, kas pazīstama kā piezīme .

Apakšproblēmas par apakšproblēmām

Apakšproblēmas ir mazākas sākotnējās problēmas versijas. Faktiski apakšproblēmas bieži izskatās kā sākotnējās problēmas pārveidota versija. Ja apakšproblēmas ir pareizi formulētas, tās balstās viena uz otru, lai iegūtu sākotnējās problēmas risinājumu.

Lai gūtu labāku priekšstatu par to, kā tas darbojas, atradīsim apakšproblēmu dinamiskās programmēšanas problēmas piemērā.

Izlikieties, ka strādājat pagājušā gadsimta piecdesmitajos gados un strādājat ar datoru IBM-650. Jūs zināt, ko tas nozīmē - perfokartes! Jūsu darbs ir vīrietis vai sieviete - IBM-650 uz vienu dienu. Jums tiek dots dabiskais skaitlis n perfokartes, kuras palaist. Katra perfokarte i ir jāpalaiž noteiktā sākuma laikā s_i un jāpārtrauc darboties noteiktā finiša laikā f_i . Vienlaikus ar IBM-650 var darboties tikai viena perfokarte. Katrai perfokartei ir arī saistīta vērtība v_i, pamatojoties uz to, cik svarīgi tas ir jūsu uzņēmumam.

Problēma : kā personai, kas atbild par IBM-650, jums jānosaka optimālais perfokartu grafiks, kas maksimāli palielina visu palaisto perfokartu kopējo vērtību.

Tā kā šajā rakstā es ļoti detalizēti izskatīšu šo piemēru, pagaidām es jūs tikai ķircināšu ar tās apakšproblēmu:

Apakšproblēma : maksimālais vērtību grafiks perfokartēm no i līdz n tā, ka perfokartes tiek sakārtotas pēc sākuma laika.

Ievērojiet, kā apakšproblēma sadala sākotnējo problēmu komponentos, kas veido risinājumu. Izmantojot apakšproblēmu, jūs varat atrast maksimālo vērtību grafiku perfokartēm n-1 līdz n un pēc tam perfokartēm n-2 līdz n utt. Atrodot katras atsevišķās apakšproblēmas risinājumus, jūs varat risināt pašu sākotnējo problēmu: maksimālo vērtību grafiku perfokartēm no 1. līdz n . Tā kā apakšproblēma izskatās kā sākotnējā problēma, sākotnējās problēmas risināšanai var izmantot apakšproblēmas.

Dinamiskā programmēšanā pēc katras apakšproblēmas atrisināšanas jums tā ir jāpiezīmē vai jāglabā. Noskaidrosim, kāpēc nākamajā sadaļā.

Motivējoša atmiņa ar Fibonači numuriem

Kad jums būtu jāievieš algoritms, kas aprēķina Fibonači vērtību jebkuram skaitlim, ko jūs darītu? Lielākā daļa cilvēku, kurus es zinu, izvēlētos rekursīvu algoritmu, kas Python izskatās apmēram šādi:

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

Šis algoritms sasniedz savu mērķi, bet par milzīgām izmaksām. Piemēram, apskatīsim, kas šim algoritmam jāaprēķina, lai atrisinātu n = 5 (saīsināti kā 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)

Iepriekš redzamais koks apzīmē katru aprēķinu, kas jāveic, lai atrastu Fibonači vērtību n = 5. Ievērojiet, kā trīs reizes tiek atrisināta apakšu problēma n = 2 . Salīdzinoši nelielam piemēram (n = 5) tas ir daudz atkārtotu un izšķērdētu aprēķinu!

Ko darīt, ja tā vietā, lai trīs reizes aprēķinātu Fibonacci vērtību n = 2, mēs izveidotu algoritmu, kas to aprēķina vienu reizi, saglabā tā vērtību un piekļūst glabātajai Fibonacci vērtībai par katru nākamo n = 2 gadījumu? Tas ir tieši tas, ko memoization dara.

Paturot to prātā, es esmu uzrakstījis dinamisku programmēšanas risinājumu Fibonači vērtības problēmai:

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]

Ievērojiet, kā atgriešanās vērtības risinājums rodas no piezīmju masīva piezīmes [], kuru iteratīvi aizpilda for cilpa. Ar “iteratīvi” es domāju, ka piezīme [2] tiek aprēķināta un saglabāta pirms piezīmes [3], piezīmes [4],… un piezīmes [ n ]. Tā kā piezīme [] ir aizpildīta šādā secībā, katras apakšproblēmas (n = 3) risinājumu var atrisināt ar iepriekšējo apakšproblēmu (n = 2 un n = 1) risinājumiem, jo ​​šīs vērtības jau tika saglabātas piezīme [] agrāk.

Memoizēšana nozīmē, ka nav jāveic atkārtota aprēķināšana, kas padara efektīvāku algoritmu. Tādējādi atmiņa nodrošina, ka dinamiska programmēšana ir efektīva, taču pareizās apakšproblēmas izvēle garantē, ka dinamiska programma iziet visas iespējas, lai atrastu labāko.

Tagad, kad mēs esam pievērsušies piezīmēm un apakšproblēmām, ir pienācis laiks apgūt dinamisko programmēšanas procesu. Piesprādzēties.

Mans dinamiskais programmēšanas process

1. darbība: Apakšproblēmas noteikšana vārdos.

Pārāk bieži programmētāji pievērsīsies koda rakstīšanai, pirms kritiski domās par problēmu. Nav labi. Viena stratēģija smadzeņu aktivizēšanai, pirms pieskaraties tastatūrai, ir angļu vai citu vārdu lietošana, lai aprakstītu sākotnējo problēmu identificēto apakšproblēmu.

Ja jūs risināt problēmu, kurai nepieciešama dinamiska programmēšana, paņemiet papīru un padomājiet par informāciju, kas jums nepieciešama šīs problēmas risināšanai. Paturot to prātā, izrakstiet apakšproblēmu.

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!

Liels paldies Stīvenam Benettam, Klērai Durandai un Prithajam Natham par šī amata korektūru. Paldies profesoram Hārtlinam, ka viņš mani tik ļoti sajūsmināja par dinamisko programmēšanu, ka es par to ilgi rakstīju.

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