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<html>
<!-- THIS FILE WAS GENERATED BY A SCRIPT: DO NOT EDIT IT! -->
<head>
<link href="style.css" rel="stylesheet" type="text/css"/>
<title>
Design and Analysis of Algorithms: Dynamic Programming
</title>
</head>
<body>
<div id="header">
<div id="logo">
<img src="graphics/Julia.png">
</div>
<div id="user-tools">
<a href="index.html">Home</a>
<a href="about.html">About</a>
<a href="feedback.html">Feedback</a>
</div>
</div>
<h1>
Design and Analysis of Algorithms: Dynamic Programming
</h1>
<div style="text-align:center">
<p>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/0/03/Shortest_path_optimal_substructure.svg/250px-Shortest_path_optimal_substructure.svg.png">
</p>
</div>
<details>
<summary class="sum1">
What is dynamic programming?
</summary>
<p>
Memoization!
<br>
<br>
The name is largely a marketing construct. Here is the
inventor of the term, Richard Bellman, on how it came about:
<br>
<br>
"I spent the Fall quarter (of 1950) at RAND.
My first task was to find a name
for multistage decision processes.
An interesting question is, Where did the
name, dynamic programming, come from?
The 1950s were not good years for
mathematical research.
We had a very interesting gentleman in Washington named Wilson.
He was Secretary of Defense,
and he actually had a pathological fear
and hatred of the word research.
I'm not using the term lightly; I'm using it precisely.
His face would suffuse, he would turn red,
and he would get violent
if people used the term research in his presence.
You can imagine how he felt,
then, about the term mathematical.
The RAND Corporation was employed by the Air
Force, and the Air Force had Wilson as its boss, essentially.
Hence, I felt I had to do something to shield Wilson
and the Air Force from the fact that I was
really doing mathematics inside the RAND Corporation.
What title, what name,
could I choose? In the first place I was
interested in planning, in decision making, in thinking.
But planning, is not a good word for various reasons.
I decided therefore to use the word "programming".
I wanted to get across the
idea that this was dynamic,
this was multistage, this was time-varying.
I thought, let's kill two birds with one stone.
Let's take a word that has an
absolutely precise meaning, namely dynamic,
in the classical physical sense.
It also has a very interesting property as an adjective,
and that it's impossible
to use the word dynamic in a pejorative sense.
Try thinking of some combination
that will possibly give it a pejorative meaning.
It's impossible.
Thus, I thought dynamic programming was a good name.
It was something not even a
Congressman could object to.
So I used it as an umbrella for my activities."
<br>
(Source:
https://en.wikipedia.org/wiki/Dynamic_programming#History)
<br>
<br>
Note that Bellman's claim that "dynamic" can be use
pejoratively is surely false: most people would not favor
"dynamic ethnic cleansing"!
<br>
<br>
</p>
<details>
<summary class="sum2">
<b>Algorithms that use dynamic programming</b>:
</summary>
<ul>
<li>Recurrent solutions to lattice models for protein-DNA
binding
</li>
<li>Backward induction as a solution method for
finite-horizon discrete-time dynamic optimization
problems
</li>
<li>Method of undetermined coefficients can be used to
solve the Bellman equation in infinite-horizon,
discrete-time, discounted, time-invariant dynamic
optimization problems
</li>
<li>Many string algorithms including longest common
subsequence, longest increasing subsequence, longest
common substring, Levenshtein distance (edit distance)
</li>
<li>Many algorithmic problems on graphs can be solved
efficiently for graphs of bounded treewidth or bounded
clique-width by using dynamic programming on a tree
decomposition of the graph.
</li>
<li>The Cocke-Younger-Kasami (CYK) algorithm which
determines whether and how a given string can be
generated by a given context-free grammar
</li>
<li>Knuth's word wrapping algorithm that minimizes
raggedness when word wrapping text
</li>
<li>The use of transposition tables and refutation tables
in computer chess
</li>
<li>The Viterbi algorithm (used for hidden Markov models)</li>
<li>The Earley algorithm (a type of chart parser)</li>
<li>The Needleman-Wunsch algorithm and other algorithms
used in bioinformatics, including sequence alignment,
structural alignment, RNA structure prediction
</li>
<li>Floyd's all-pairs shortest path algorithm</li>
<li>Optimizing the order for chain matrix multiplication</li>
<li>Pseudo-polynomial time algorithms for the subset sum,
knapsack and partition problems
</li>
<li>The dynamic time warping algorithm for computing the
global distance between two time series
</li>
<li>The Selinger (a.k.a. System R) algorithm for relational
database query optimization
</li>
<li>De Boor algorithm for evaluating B-spline curves</li>
<li>Duckworth-Lewis method for resolving the problem when
games of cricket are interrupted
</li>
<li>The value iteration method for solving Markov decision
processes
</li>
<li>Some graphic image edge following selection methods
such as the "magnet" selection tool in Photoshop
</li>
<li>Some methods for solving interval scheduling problems</li>
<li>Some methods for solving the travelling salesman
problem, either exactly (in exponential time) or
approximately (e.g. via the bitonic tour)
</li>
<li>Recursive least squares method</li>
<li>Beat tracking in music information retrieval</li>
<li>Adaptive-critic training strategy for artificial neural
networks
</li>
<li>Stereo algorithms for solving the correspondence
problem used in stereo vision
</li>
<li>Seam carving (content-aware image resizing)</li>
<li>The Bellman-Ford algorithm for finding the shortest
distance in a graph
</li>
<li>Some approximate solution methods for the linear search
problem
</li>
<li>Kadane's algorithm for the maximum subarray problem</li>
</ul>
</details>
<details>
<summary class="sum2">
Dynamic programming video
</summary>
<div id="python_anywhere_dynprog"
class="python-console">
<iframe width="560" height="315"
src="https://www.youtube.com/embed/W2ote4jCuYw"
frameborder="0" allowfullscreen></iframe>
</div>
</details>
</details>
<details>
<summary class="sum1">
Rod cutting
</summary>
<p>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/8/8b/Wood_from_victoria_mountain_ash.jpg/250px-Wood_from_victoria_mountain_ash.jpg">
<br>
<br>
Nothing special here about steel rods: the algorithm applies to
any good that can be sub-divided, but only in multiples of some
unit, like lumber, or meat, or cloth.
</p>
<details>
<summary class="sum2">
Recursive top-down implmentation
</summary>
<p>
Keeps calculating the same cuts again and again, much like
naive, recursive Fibonacci.
<br>
<br>
Running time is exponential in n. Why?
<br>
Our textbook gives us the equation:
<br>
<br>
<img src="graphics/RecRodCutEqn.gif">
<br>
<br>
This is equivalent to:
<br>
T(n) = 1 + T(n - 1) + T(n - 2) + ... T(1)
<br>
For n = 1, there are 2<sup>0</sup> ways to solve the
problem.
<br>
For n = 2, there are 2<sup>1</sup> ways to solve the
problem.
<br>
For n = 3, there are 2<sup>2</sup> ways to solve the
problem.
<br>
Each additional foot of rod gives us 2 * (previous number
of ways of solving problem), since we have all the previous
solutions, either with a cut of one foot for the new
extension, or without a cut there. (Similar to why each row
of Pascal's triangle gives us the next power of two.)
<br>
<br>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/f/f6/Pascal%27s_triangle_5.svg/250px-Pascal%27s_triangle_5.svg.png">
<br>
<br>
So, we have the series:
<br>
2<sup>n - 1</sup> + 2<sup>n - 2</sup> + 2<sup>n -
3</sup>... + 2<sup>0</sup> + 1
<br>
And this equals 2<sup>n</sup>. Why?
<br>
<img src="graphics/TwoToTheN.png">
<br>
<br>
<b>Example</b>: 2<sup>4</sup> = 2<sup>3</sup> +
2<sup>2</sup> + 2<sup>1</sup> + 2<sup>0</sup> + 1
<br>
Or, 16 = 8 + 4 + 2 + 1 + 1
</p>
</details>
<details>
<summary class="sum2">
Using dynamic programming for optimal rod-cutting
</summary>
<p>
Much like we did with the naive,
<a
href="https://github.com/gcallah/algorithms/blob/master/Python/DivideAndConquer/fibonacci.py">
recursive Fibonacci</a>,
we can "memoize" the recursive rod-cutting algorithm and
achieve huge time savings.
<br>
<br>
That is an efficient top-down approach. But we can also do
a bottom-up approach, which will have the same run-time
order but may be slightly faster due to fewer function
calls. (The algorithm uses an additional loop instead of
recursion to do its work.)
</p>
</details>
<details>
<summary class="sum2">
Subproblem graphs
</summary>
<p>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/0/06/Fibonacci_dynamic_programming.svg/108px-Fibonacci_dynamic_programming.svg.png">
<br>
<br>
The above is the Fibonacci sub-problem graph for fib(5). As
you can see, F<sub>5</sub> must solve F<sub>4</sub> and
F<sub>3</sub>. But F<sub>4</sub> must <i>also</i> solve
F<sub>3</sub>. It also must solve F<sub>2</sub>, which
F<sub>3</sub> must solve as well. And so on.
<br>
<br>
This is the sort of graph we want to see if dynamic
programming is going to be a good approach: a recursive
solution involves repeatedly solving the same problems.
<br>
<br>
This is quite different than, say, a parser, where the code
sub-problems are very unlikely to be the same chunks of
code again and again, unless we are parsing the code of a
very bad programmer who doesn't understand functions!
</p>
</details>
<details>
<summary class="sum2">
Reconstructing a solution
</summary>
<p>
In this section, we see how to <i>record</i> the solution
we arrived at, rather than simply return the optimal
revenue possible. The owner of Serling Enterprises will
surely be much more pleased with this code than the earlier
versions.
</p>
</details>
<details>
<summary class="sum2">
Run the Python code
</summary>
<p>
In the console below, type or paste:
<br/>
<code>
!git clone https://gist.github.com/80d2a774f08f686f675f8a9254570da0.git
<br/>
cd 80d2a774f08f686f675f8a9254570da0
<br/>
from dynamic_programming import *
<br/>
</code>
</p>
<div class="python-console">
<iframe style="width: 640; height: 480;"
name="embedded_python_anywhere"
src="https://www.pythonanywhere.com/embedded3/" scrolling="yes">
</iframe>
<figcaption>
Python console
</figcaption>
</div>
<p>
Now let's run our ext_bottom_up_cut_rod() code.
(Link to full source code below.)
Type or paste:
<code>
<br/>
p4
<br/>
(revs, cuts, max_rev) = ext_bottom_up_cut_rod(p4, 4)
</code>
</p>
<p>
You can explore more, by designing your own price
arrays! Just type in:
<br/>
<code>
my_name = [x, y, z...]
</code>
<br/>
where 'my_name' is whatever name you want to give your
price array, and x, y, z, etc. are the prices for a cut
of length 1, 2, 3, etc.
</p>
</details>
<details>
<summary class="sum2">
A video on rod cutting
</summary>
<div style="text-align:center">
<iframe width="560" height="315"
src="https://www.youtube.com/embed/IRwVmTmN6go"
frameborder="0" allowfullscreen></iframe>
</div>
</details>
</details>
<details>
<summary class="sum1">
Matrix-chain multiplication
</summary>
<p>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/a/a8/Catalan-Hexagons-example.svg/400px-Catalan-Hexagons-example.svg.png">
<br>
<br>
There are many ways to parenthesize a series of matrix
multiplications. For instance, if we are parenthesizing
A<sub>1</sub> * A<sub>2</sub> * A<sub>3</sub> * A<sub>4</sub>,
we could parenthesize this in the following ways:
<br>
<br>
(A<sub>1</sub> (A<sub>2</sub> (A<sub>3</sub> A<sub>4</sub>)))
<br>
(A<sub>1</sub> ((A<sub>2</sub> A<sub>3</sub>) A<sub>4</sub>))
<br>
((A<sub>1</sub> A<sub>2</sub>) (A<sub>3</sub> A<sub>4</sub>))
<br>
((A<sub>1</sub> (A<sub>2</sub> A<sub>3</sub>)) A<sub>4</sub>)
<br>
(((A<sub>1</sub> A<sub>2</sub>) A<sub>3</sub>) A<sub>4</sub>)
<br>
<br>
Which way we choose to do so can make a huge difference in
run-time!
</p>
<p>
Why is this different than rod cutting? Think about this
for a moment, and see if you can determine why the problems
are not the same.
</p>
<details>
<summary class="sum4"><b>The reason</b></summary>
<p>
In rod cutting, a cut of 4-2-2 is the <i>same cut</i> as a
cut of 2-2-4, and the same as a cut of 2-4-2.
<br>
That is not at all the case for matrix parenthesization.
</p>
</details>
<details>
<summary class="sum2">
Counting the number of parenthesizations
</summary>
<p>
The number of solutions is exponential in <i>n</i>, thus
brute-force is a bad technique for solving this
problem.
</p>
</details>
<details>
<summary class="sum2">
Applying dynamic programming
</summary>
<details>
<summary class="sum3">
Step 1: The structure of an optimal parenthesization
</summary>
<p>
For any place at level <i>n</i> where we place
parentheses, we must have optimal parentheses
at level <i>n</i> + 1.
Otherwise, we could substitute
in the optimal <i>n</i> + 1
level parentheses, and level n would be better!
<br/>
Cut-and-paste proof.
</p>
</details>
<details>
<summary class="sum3">
Step 2: A recursive solution
</summary>
<p>
If we know the optimal place to split A<sub>1</sub>...
A<sub>n</sub> (call it k), then the optimal solution is
that split, plus the optimal solution for A<sub>1</sub>...
A<sub>k</sub> and the optimal solution for
A<sub>k+1</sub>... A<sub>n</sub>. Since we don't know k, we
try each possible k in turn, compute the optimal
sub-problem for each such split, and see which pair of
optimal sub-problems yields the optimal (minimum, in this
case) total.
<br>
<br>
"For example, if we have four matrices ABCD, we compute the
cost required to find each of (A)(BCD), (AB)(CD), and
(ABC)(D), making recursive calls to find the minimum cost
to compute ABC, AB, CD, and BCD. We then choose the best
one."
(https://en.wikipedia.org/wiki/Matrix_chain_multiplication)
<br>
<br>
An easy way to understand this:
<br>
Let's say we need to get from class at NYU Tandon to a
ballgame at Yankee Stadium in the Bronx as fast as
possible. If we choose Grand Central Station as the optimal
high-level split, we must also choose the optimal ways to
get from NYU to Grand Central, and from Grand Central to
Yankee Stadium. It won't do to choose Grand Central, and
then walk from NYU to Grand Central, and CitiBike from
Grand Central to Yankee Stadium: there are faster ways to
do each sub-problem!
</p>
</details>
<details>
<summary class="sum3">
Step 3: Computing the optimal costs
</summary>
<p>
CLRS does not offer a recursive version here (they do later
in the chapter); they go
straight to the bottom-up approach of storing each
lowest-level result in a table, avoiding recomputation, and
then combine those lower-level results into higher-level
ones. The indexing here is very tricky and hard to follow
in one's head, but it is worth trying to trace out what is
going on by following the code. I have as usual included
some print statements to help.
</p>
</details>
<details>
<summary class="sum3">
Step 4: Constructing an optimal solution
</summary>
<p>
Finally, we use the results computed in step 3 to actually
provide the optimal solution, by actually determing where
the parentheses go.
</p>
<p>
Here is the code from our textbook, implemented in
<a
href="https://github.com/gcallah/algorithms/blob/master/Python/DynamicProgramming/DynamicProgramming.ptml">
Python</a>,
runnning on the example where A<sub>1</sub> is 10 x 100,
A<sub>2</sub> is 100 x 5, and A<sub>3</sub> is 5 x 50:
</p>
<figure>
<img src="graphics/MatrixChainRun.png">
<figcaption>
The actual output of our Python code.
</figcaption>
</figure>
<p>
The structure of m:
</p>
<table>
<tr>
<td>
0
</td>
<td>
5000
</td>
<td>
7500
</td>
</tr>
<tr>
<td>
∞
</td>
<td>
0
</td>
<td>
25000
</td>
</tr>
<tr>
<td>
∞
</td>
<td>
∞
</td>
<td>
0
</td>
</tr>
</table>
</details>
<details>
<summary class="sum3">
Memoization
</summary>
<p>
We can memoize the recursive version and change its run
time from Ω(2<sup>n</sup>) to O(n<sup>3</sup>).
</p>
</details>
</details>
<details>
<summary class="sum2">
A video on matrix chains.
</summary>
<div style="text-align:center">
<iframe width="560" height="315"
src="https://www.youtube.com/embed/u6Y055g4mOE"
frameborder="0" allowfullscreen></iframe>
</div>
</details>
</details>
<details>
<summary class="sum1">
Elements of dynamic programming
</summary>
<details>
<summary class="sum2">
Optimal substructure
</summary>
<p>
A problem exhibits <i>optimal substructure</i> if an optimal
solution to the problem contains within it optimal
solutions to subproblems.
</p>
</details>
<details>
<summary class="sum2">
Overlapping subproblems
</summary>
<p>
The problem space must be "small," in that a recursive
algorithm visits the same sub-problems again and again,
rather than continually generating new subproblems. The
recursive Fibonacci is an excellent example of this!
</p>
</details>
<details>
<summary class="sum2">
Reconstructing an optimal solution
</summary>
<p>
Storing our choices in a table as we make them allows quick
and simple reconstruction of the optimal solution.
</p>
</details>
<details>
<summary class="sum2">
Memoization
</summary>
<p>
As mentioned above, recursion with memoization is often a viable
alternative to the bottom-up approach. Which to choose
depends on several factors, one of which being that a
recursive approach is often easier to understand.
If our algorithm is going to handle small data sets,
or not run very often, a recursive approach
with memoization may be the right answer.
</p>
</details>
</details>
<details>
<summary class="sum1">
Longest common subsequence
</summary>
<details>
<summary class="sum2">
Step 1: Characterizing a longest common subsequence
</summary>
<p>
'Let X be "XMJYAUZ" and Y be "MZJAWXU".
The longest common subsequence between X and Y is "MJAU".'
(https://en.wikipedia.org/wiki/Longest_common_subsequence_problem)
<br>
<br>
<img src="graphics/CommonSubsequence.png">
<br>
<br>
Brute force solution runs in exponential time: not so good!
<br>
<br>
But the problem has an optimal substructure:
<br>
<br>
X = gregorsamsa
<br>
Y = reginaldblack
<br>
LCS: regaa
<br>
Our match on the last 'a' is at position X<sub>11</sub> and
Y<sub>11</sub>. The previous result string ('rega') must have been
the LCS before X<sub>11</sub> and Y<sub>11</sub>:
otherwise, we could substitute in <i>that actual</i> LCS
for 'rega' and have a longer overall LCS.
</p>
</details>
<details>
<summary class="sum2">
Step 2: A recursive solution
</summary>
<p>
Caution: here some sub-problems are ruled out! If
X<sub>i</sub> and Y<sub>j</sub> are different, we consider
the sub-problems of finding the LCS for X<sub>i</sub> and
Y<sub>j - 1</sub> and for X<sub>i - 1</sub> and
Y<sub>j</sub>, but not for X<sub>i</sub> and Y<sub>j</sub>.
Why not? Well, if they aren't equal, they can't be the
endpoint of an LCS.
</p>
</details>
<details>
<summary class="sum2">
Step 3: Computing the length of an LCS
</summary>
<p>
The solution here proceeds much like the earlier ones: find
an LCS in a bottom-up fashion, using tables to store
intermediate results and information for reconstructing the
optimal solution.
</p>
</details>
<details>
<summary class="sum2">
Step 4: Constructing an LCS
</summary>
<details>
<summary class="sum3">
Improving the code
</summary>
<p>
We could eliminate a table here, reduce aymptotic
run-time a bit there. But is the code more confusing?
Do we lose an ability (reconstructing the solution) we
might actually need later?
<br>
<br>
<b>An important principle</b>: Don't optimize unless it is
needed!
</p>
</details>
</details>
<details>
<summary class="sum2">
Video on LCS
</summary>
<div style="text-align:center">
<iframe width="560" height="315"
src="https://www.youtube.com/embed/P-mMvhfJhu8"
frameborder="0" allowfullscreen></iframe>
</div>
</details>
</details>
<details>
<summary class="sum1">
Optimal binary search trees
</summary>
<p>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/0/06/AVLtreef.svg/251px-AVLtreef.svg.png">
</p>
<details>
<summary class="sum2">
Step 1: The structure of an optimal binary search tree
</summary>
<p>
If a binary search tree is optimally construted, then both
its left and right sub-trees must be optimally constructed.
The usual "cut-and-paste" argument applies.
</p>
</details>
<details>
<summary class="sum2">
Step 2: A recursive solution
</summary>
<p>
As usual, this is straightforward, but too slow.
</p>
</details>
<details>
<summary class="sum2">
Step 3: Computing the expected search cost
</summary>
<p>
Very much like the matrix-chain-order code. Working code coming
soon!
</p>
</details>
<details>
<summary class="sum2">
Optimal binary search tree video
</summary>
<div style="text-align:center">
<iframe width="560" height="315"
src="https://www.youtube.com/embed/YXwmt55nl7I"
frameborder="0" allowfullscreen></iframe>
</div>
</details>
</details>
<details>
<summary class="sum1">
Source Code
</summary>
<p>
<a href="https://github.com/gcallah/algorithms/tree/master/Java/DynamicProgramming">Java</a><br>
<a href="https://github.com/gcallah/algorithms/tree/master/Ruby/DynamicProgramming">Ruby</a><br>
<a href="https://github.com/gcallah/algorithms/tree/master/Python/DynamicProgramming">Python</a><br>
<a href="https://github.com/gcallah/algorithms/tree/master/Clojure/DynamicProgramming">Clojure</a><br>
</p>
</details>
<details>
<summary class="sum1">
For Further Study
</summary>
<ul>
<li>
<a
href="https://en.wikipedia.org/wiki/Matrix_chain_multiplication">
Matrix-chain multiplication
</a>
</li>
<li>
<a
href="http://www.geeksforgeeks.org/dynamic-programming-set-4-longest-common-subsequence/">
Longest common subsequence
</a>
</li>
</ul>
</details>
<details>
<summary class="sum1">
Homework
</summary>
<ol>
<li>Change memoized-rod-cut to return a list of cuts to
make, instead of the maximum possible revenue.
Pseudo-code or real code are both fine.
</li>
<li>For the following table, determine the cost and
structure of an optimal binary search tree:
<br>
<br>
<table>
<tr>
<th>
<i>i</i>
</th>
<th>
0
</th>
<th>
1
</th>
<th>
2
</th>
<th>
3
</th>
<th>
4
</th>
<th>
5
</th>
</tr>
<tr>
<th>
p<sub>i</sub>
</th>
<td>
</td>
<td>
.05
</td>
<td>
.05
</td>
<td>
.25
</td>
<td>
.05
</td>
<td>
.05
</td>
</tr>
<tr>
<th>
q<sub>i</sub>
</th>
<td>
.05
</td>
<td>
.15
</td>
<td>
.05
</td>
<td>
.05
</td>
<td>
.05
</td>
<td>
.20
</td>
</tr>
</table>
</li>
</ol>
</details>
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