本文通过四个步骤深入浅出地介绍了动态规划的核心思想，包括成为上帝视角、决策问题、选择变量和最优决策。通过具体例子展示了如何将动态规划应用到实际问题中，如连续段最大和、任务调度等，旨在帮助读者理解和掌握动态规划的解决方法。
Firstly
, let me put forth my own thought process for solving DP problems (since its short), and then refer you to other sources.
NOTE: All DPs can be (re)formulated as recursion. The extra effo rt you put in in finding out what is the underlying recursion will go a long way in helping you in future DP problems.
STEP1: Imagine you are GOD. Or as such, you are a third-person overseer of the problem.
STEP2: As God, you need to decide wha t choice to make. Ask a decision question .
STEP3: In order to make an informed choice, you need to ask "what variables would help me make my informed choice?". This is an important step and you may have to ask "but this is not enough info, so what more do I need" a few times.
STEP4: Make the choice that gives you your best result.
In the above, the variables alluded to in Step3 are what is generally called the "state" of your DP. The decision in Step2 is thought of as "from my current state, what all states does it depend upon?"
Trust me: I've solved loads of TC problems on DP just using the above methodology. It took me about max 2 months to imbibe this methodology, so unlike everyone's "Keep practicing" advice (which I would liken to Brownian motion ), I'm suggesting the above 'intuition'.
Btw, if you can identify me by my advi ce, good for y ou. The point of going anonymous is mainly so that people don't blindly upvote when they see " X added an answer to Algorithms : ..."
Now for some examples:
Q. Given an array,find the largest sumalong a contiguous segment of it.
Simple enough, and you probably know the answer, but lets see what the 4 steps amount to.
Step1: I am an overseer (just a psychological step)
Step2: I go through the array, and ask, " does the largest sum begin at this point? "
Step3: I need to know what is the largest sum that begins at the next point, in order to decide if it can be extended or not.
Step4: I either extend it to the next point, or I cut it off here: f(i) = max(arr[i], arr[i] + f(i+1)).
The above particular example was demonstrated to me by a friend and prior to his demonstration (of how easy DP is), I was completely baffled by DPs myself.
Q. I have a set of N jobs, and two machines A and B. Job i takes A[i] time on machine A, and B[i] time on machine B. What is the minimum amount of time I need to finish all the jobs?
Step1 : I am the overseer (this problem lends itself more naturally to being "God").
Step2: I go through jobs from 1 to N, and have to decide on which machine to schedule job i .
Step3 : If I schedule the job on machine A , then I then have a load of A[i] + best way to schedule the remaining jobs. If its on B, then I have a load of B[i] + best way to schedule the remaining jobs.
But wait! "best way to schedule remaining jobs" doesn't account for the fact that the i'th job will potentially interfere. Thus, I need to infact also pass on the "total load" on each machine in order to make my informed choice.
Therefore, I need to modify my question to: " Given that the current load on A isa, and on B isb, and I have to schedule jobs i to N, what is the best way to do so? "
Step4: f(a, b, i) = { min(a, b) : i == N+1, else min(f(a + A[i], b, i+1), f(a, b + B[i], i+1): i <= N}. Final answer is in f(0, 0, 1).
Now, enough of examples (answer is becoming too long). If you have any DP question you'd like me to work my magic on, post it as a comment :)
NOTE: All DPs can be (re)formulated as recursion. The extra effo rt you put in in finding out what is the underlying recursion will go a long way in helping you in future DP problems.
STEP1: Imagine you are GOD. Or as such, you are a third-person overseer of the problem.
STEP2: As God, you need to decide wha t choice to make. Ask a decision question .
STEP3: In order to make an informed choice, you need to ask "what variables would help me make my informed choice?". This is an important step and you may have to ask "but this is not enough info, so what more do I need" a few times.
STEP4: Make the choice that gives you your best result.
In the above, the variables alluded to in Step3 are what is generally called the "state" of your DP. The decision in Step2 is thought of as "from my current state, what all states does it depend upon?"
Trust me: I've solved loads of TC problems on DP just using the above methodology. It took me about max 2 months to imbibe this methodology, so unlike everyone's "Keep practicing" advice (which I would liken to Brownian motion ), I'm suggesting the above 'intuition'.
Btw, if you can identify me by my advi ce, good for y ou. The point of going anonymous is mainly so that people don't blindly upvote when they see " X added an answer to Algorithms : ..."
Now for some examples:
Q. Given an array,find the largest sumalong a contiguous segment of it.
Simple enough, and you probably know the answer, but lets see what the 4 steps amount to.
Step1: I am an overseer (just a psychological step)
Step2: I go through the array, and ask, " does the largest sum begin at this point? "
Step3: I need to know what is the largest sum that begins at the next point, in order to decide if it can be extended or not.
Step4: I either extend it to the next point, or I cut it off here: f(i) = max(arr[i], arr[i] + f(i+1)).
The above particular example was demonstrated to me by a friend and prior to his demonstration (of how easy DP is), I was completely baffled by DPs myself.
Q. I have a set of N jobs, and two machines A and B. Job i takes A[i] time on machine A, and B[i] time on machine B. What is the minimum amount of time I need to finish all the jobs?
Step1 : I am the overseer (this problem lends itself more naturally to being "God").
Step2: I go through jobs from 1 to N, and have to decide on which machine to schedule job i .
Step3 : If I schedule the job on machine A , then I then have a load of A[i] + best way to schedule the remaining jobs. If its on B, then I have a load of B[i] + best way to schedule the remaining jobs.
But wait! "best way to schedule remaining jobs" doesn't account for the fact that the i'th job will potentially interfere. Thus, I need to infact also pass on the "total load" on each machine in order to make my informed choice.
Therefore, I need to modify my question to: " Given that the current load on A isa, and on B isb, and I have to schedule jobs i to N, what is the best way to do so? "
Step4: f(a, b, i) = { min(a, b) : i == N+1, else min(f(a + A[i], b, i+1), f(a, b + B[i], i+1): i <= N}. Final answer is in f(0, 0, 1).
Now, enough of examples (answer is becoming too long). If you have any DP question you'd like me to work my magic on, post it as a comment :)
Please have a look at Mimino's answers on DP:Dynamic Programming: What are systematic ways to prepare for dynamic programming?
https://www.quora.com/Dynamic-Programming/How-can-one-start-solving-Dynamic-Programming-problems
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