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<!DOCTYPE html>
<html>
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<head>
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content="width=device-width, initial-scale=1">
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<title>
Design and Analysis of Algorithms: Divide and Conquer
</title>
</head>
<body>
<header id="header">
<a id="logo" href="index.html">
<img src="graphics/Julia.png" alt="">
<span class="brand-text">Design & Analysis of Algorithms</span>
</a>
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<a href="about.html">About</a>
<a href="feedback.html">Feedback</a>
</nav>
</header>
<h1>
Design and Analysis of Algorithms: Divide and Conquer
</h1>
<div style="text-align:center">
<p>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/4/45/Sierpinski_triangle.svg/250px-Sierpinski_triangle.svg.png">
</p>
</div>
<details>
<summary class="sum1">
Introduction
</summary>
<p>
What is this?
</p>
<ul>
<li><b>Divide</b> the problem into a number of sub problems that are
smaller instances of the same problem.
</li>
<li><b>Conquer</b> the sub problems by solving them recursively. If
the subproblem sizes are small enough, however, just solve
the subproblems in a straightforward manner.
</li>
<li><b>Combine</b> the solutions for the sub problems into the
solution for the original problem.
</li>
</ul>
<figure>
<iframe width="560" height="315"
src="https://www.youtube.com/embed/6SUmp_Cn-SU"
frameborder="0" allowfullscreen></iframe>
<figcaption>
Introduction to divide-and-conquer
</figcaption>
</figure>
<details>
<summary class="sum2">
Recurrences
</summary>
<p>
Merge sort recurrence:
<br>
<br>
<img src="graphics/RecEq1.gif">
<br>
<br>
We "solve" these by finding a closed-form equation that
describes the recurrence but without recursion.
<br>
<b>Solution</b>: T(n) = Θ(n lg n)
<br>
<br>
<b>Methods</b>:
<br>
</p>
<ul>
<li><b>Substitution method</b>: Guess a solution and then use
induction to prove it.
</li>
<li><b>Recursive-tree method</b>: Convert the recurrence into a
tree whose nodes represent costs incurred at each level.
</li>
<li><b>Master method</b>:
<br>
Solves recurrences of the form:
<br>
T(n) = aT(n / b) + f(n)
<br>
where a ≥ 1, b > 1.
</li>
</ul>
<p>
<br>
<b>Technicalities</b>
<br>
We often omit floors, ceilings, and boundary conditions. For
instance, if n is odd, we may say n / 2 anyway.
</p>
</details>
</details>
<details>
<summary class="sum1">
4.1 The maximum-subarray problem
</summary>
<p>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/2/25/Maximum_Subarray_Visualization.svg/220px-Maximum_Subarray_Visualization.svg.png">
<br>
<br>
Only makes since in an array with both negative and positive
values: otherwise the answer is either the whole array or the
maximum member.
<br>
<br>
</p>
<details>
<summary class="sum2">
Brute-force solution
</summary>
<p>
Try every combination of two elements!
<br>
A n choose 2 problem, so order of Ω(n<sup>2</sup>).
<br>
n choose 2 will be about 1/2 n<sup>2</sup>, since it equals
n(n - 1) / 2. So we can establish a lower bound by setting c =
1/3, for instance, and n choose 2 will always be bounded from
below by c*n<sup>2</sup>.
</p>
</details>
<details>
<summary class="sum2">
A transformation
</summary>
<p>
We look at the problem differently: let's find the nonempty,
contiguous subarray of our array whose values have the largest sum.
We call this the <b>maximum subarray</b>.
</p>
</details>
<details>
<summary class="sum2">
A solution using divide-and-conquer
</summary>
<p>
To solve this problem, we divide an array A into three subarrays,
and ask what is the maximum subarray in each:
</p>
<ol>
<li>From A[low] to A[midpoint - 1].</li>
<li>Crossing the mid-point.</li>
<li>From A[midpoint + 1] to A[high]</li>
</ol>
<p>
Problems 1 and 3 are simply this same problem on a smaller array!
Problem 2 can be solved by finding the maximum subarrays in
low-to-mid and in mid+1-to-high.
<br>
<br>
The recurrence is the same as for merge sort.
</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/25ffc0600a866535adef05c5d8eca34a.git
<br/>
cd 25ffc0600a866535adef05c5d8eca34a
<br/>
from find_max_subarray import *
<br/>
A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
<br/>
</code>
</p>
<div class="python-console">
<iframe style="width: 640px; height: 480px;"
name="embedded_python_anywhere"
src="https://www.pythonanywhere.com/embedded3/" scrolling="yes">
</iframe>
<figcaption>
Python console
</figcaption>
</div>
<p>
To run the example from the textbook, type:
<br/>
<code>
A
<br/>
find_max_subarray(A, 0, 15)
</code>
</p>
<p>
Now you can experiment with the algorithm by typing
in your own array (my_array = [x, y, z])
and running find_max_array(my_array).
</p>
</details>
<details>
<summary class="sum2">
Video on maximum subarray problem
</summary>
<figure>
<iframe width="560" height="315"
src="https://www.youtube.com/embed/OexQs_cYgAQ"
frameborder="0" allowfullscreen></iframe>
<figcaption>
Maximum sub-array video
</figcaption>
</figure>
</details>
<details>
<summary class="sum2">
Quiz
</summary>
<ol>
<li>
The problem with the brute force max-subarray solution is
</li>
<ol type="a" class="nested">
<li>
<input type="radio" name="q1" value="a">
it is too complicated
</li>
<li>
<input type="radio" name="q1" value="b">
the force used might break the code
</li>
<li>
<input type="radio" name="q1" value="c">
it is too slow
</li>
</ol>
<li>
The three possible arrays containing the maximum sequence are
</li>
<ol type="a" class="nested">
<li>
<input type="radio" name="q2" value="a">
An array of ints, an array of doubles, and an array of strings
</li>
<li>
<input type="radio" name="q2" value="b">
The array from A[0] to the mid-point; an array that crosses the midpoint, and the array from A[midpoint + 1] to the end
</li>
<li>
<input type="radio" name="q2" value="c">
The whole array, the null array, and the middle array
</li>
</ol>
<li>
What does FIND-MAXIMUM-SUBARRAY return when all elements of A are negative?
</li>
<ol type="a" class="nested">
<li>
<input type="radio" name="q3" value="a">
Largest Positive Number in A
</li>
<li>
<input type="radio" name="q3" value="b">
Largest Negative Number in A
</li>
<li>
<input type="radio" name="q3" value="c">
First index of A
</li>
<li>
<input type="radio" name="q3" value="d">
Lase index of A
</li>
</ol>
</ol>
<details>
<summary class="sum3">
Answers
</summary>
<p>
1. c; 2. b; 3. b;
</p>
</details>
</details>
</details>
<details>
<summary class="sum1">
4.2 Strassen's algorithm for matrix multiplication
</summary>
<details>
<summary class="sum2">
Recursive Square-Matrix Multiply
</summary>
<p>
We divide each of our initial matrices into four
sub-matrices, and multiply them. Which we do by dividing
each of them into four...
<br>
<br>
In the base case when each matrix has only one member, we
just multiply them and return the result.
<br>
<br>
So what is our recurrence? Each step except the base case
multiplies eight matrices of size n / 2. So they
contribute 8T(n / 2) to running time. There are also four
matrix additions of matrices containing n<sup>2</sup> / 4
entries -- squared because n specifies an n x n matrix. So
this contributes Θ(n<sup>2</sup>) time.
<br>
<br>
So our recurrence is:
<br>
<br>
<img src="graphics/RecEq6.gif">
<br>
<br>
The master method will show us that the solution to this
recurrence is:
<br>
T(n) = Θ(n<sup>3</sup>)
</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/87e1f86c634c1538062041ca153bc466.git
<br/>
cd 87e1f86c634c1538062041ca153bc466
<br/>
from divide_conquer_matrix import *
<br/>
A = [[1, 3], [7, 5]]
<br/>
B = [[6, 8], [4, 2]]
<br/>
</code>
</p>
<div class="python-console">
<iframe style="width: 640px; height: 480px;"
name="embedded_python_anywhere"
src="https://www.pythonanywhere.com/embedded3/" scrolling="yes">
</iframe>
<figcaption>
Python console
</figcaption>
</div>
<p>
To run the example from the textbook, type:
<br/>
<code>
A,B
<br/>
square_matrix_multiply(A, B)
<br/>
square_matrix_multiply_recursive(A, B)
</code>
</p>
<p>
Now you can experiment with the algorithm by typing
in your own Matrix (my_matrix = [x, y, z])
and running square_matrix_multiply(a_matrix, b_matrix) or
square_matrix_multiply_recursive(a_matrix, b_matrix).
</p>
</details>
<details>
<summary class="sum2">
Strassen's Algorithm
</summary>
<p>
<img
src="https://upload.wikimedia.org/wikipedia/commons/thumb/2/2e/Strassen_algorithm.svg/800px-Strassen_algorithm.svg.png">
<br>
<br>
By adding ten additions, we can cut the divide portion of
our algorithm down to seven multiplications instead of
eight.
<br>
<br>
Let's try one!
<br>
<br>
Here is the method:
<br>
<br>
For two matrices:
<br>
<br>
<img src="graphics/RecEq8.gif">
<br>
<br>
Define:
<br>
<br>
P<sub>1</sub> = A(F - H)
<br>
P<sub>2</sub> = H(A + B)
<br>
P<sub>3</sub> = E(C + D)
<br>
P<sub>4</sub> = D(G - E)
<br>
P<sub>5</sub> = (A + D) * (E + H)
<br>
P<sub>6</sub> = (B - D) * (G + H)
<br>
P<sub>7</sub> = (A - C) * (E + F)
<br>
<br>
Then:
<br>
<br>
<img src="graphics/RecEq9.gif">
<br>
<br>
So let's try this example:
<br>
<br>
<img src="graphics/RecEq7.gif">
<br>
<br>
<b>Important Lesson</b>
<br>
<br>
There are often serious trade-offs between set-up time and
aymptotic run-time. One must carefully consider how large
one's inputs are likely to be before opting for a complex
algorithm like Strassen's. On modern hardware optimized for
matrix multiplication, matrix sizes often need to be in the
thousands before Strassen's algorithm yields significant
gains.
</p>
</details>
</details>
<details>
<summary class="sum1">
4.3 The substitution method for solving recurrences
</summary>
<p>
<b>Note:</b> I am presenting these three methods in my
notes in textbook order. But in lectures, I present the
substitution method last, because we can best make sense of
our "guess" for a solution if we understand the other two
methods first. I suggest students tackle recursion-tree,
then master method, and <i>then</i> substitution.
</p>
<details>
<summary class="sum2">
Towers of Hanoi
</summary>
<p>
Shared web material <a
href="https://gcallah.github.io/DiscreteMathematics/advcounting.html#Hanoi">located here</a>.
</p>
</details>
<details>
<summary class="sum2">
Another substitution-method problem
</summary>
<p>
Let's look at the recurrence:
<br>
T(n) = T(n/3) + T(2n/3) + <i>n</i>
<br>
T(0) = 1
<br>
T(1) = 1
<br>
<br>
This does not have the form the master method
requires. And if we sketch a recursion tree, not every
node is the same on a level, so it is different from
what we usually deal with there. But if we <i>do</i>
diagram a recursion tree, we will see that the work
looks constant at each level, like a master theorem
case 2. And since the function part of the equation is
f(n) = n, let's "guess":
<br>
T(n) ≤ cn log <i>n</i>
<br>
<br>
But that is just our hunch: we have to prove it!
<br>
<br>
Let's look at base cases for our inductive proof.
Use floors for divisions! Is:
</p>
<ul>
<li>T(0) ≤ c(0 * log 0) --> nonsense! (There is no log 0.)
<br>
</li>
<li>T(1) ≤ c(1 * log 1) --> false! (There is no such c.)
<br>
</li>
<li>T(2) ≤ c(2 * log 2) --> true!
<br>
T(2) = T(floor(2/3)) + T(floor(4/3)) + 2 = 1 + 1 + 2 ≤
c(2 * log 2) = c(2 * 1)
<br>
So if we set c = 2 (or greater) the inequality holds.
</li>
</ul>
<p>
How many base cases do we need to examine? We will see!
<br>
But we can prove it for any given "small" n > 2 by
setting:
<br>
c ≥ T(n) / n log <i>n</i>
</p>
<p>
<b>Recursion step:</b>
<br>
We assume that for k where:
<br>
2 ≤ k < <i>n</i>
<br>
the claim is true.
<br>
Now, we need to show that if for sub-problems smaller
than n the claim is true, then it is true for n.
</p>
</details>
</details>
<details>
<summary class="sum1">
4.4 The recursion-tree method for solving recurrences
</summary>
<p>
There are two ways to use this method:
</p>
<ol>
<li>As a way to generate a guess for the substitution
method.
</li>
<li>As a way to generate a rigorous answer by itself.
</li>
</ol>
<p>
Analyze the tree:
<br>
<br>
<img
src="graphics/NSquaredTree1.png">
<br>
<br>
Calculate the work at each level:
<br>
<br>
<img
src="graphics/NSquaredTree2.png">
<br>
<br>
This produces the geometric series:
<br>
<br>
<img src="graphics/RecEq4.gif">
<br>
<br>
If we set <i>a</i> = <i>n</i><sup>2</sup>
and <i>r</i> = 1/2, then we have the
general sum of a converging geometric series:
<br>
<br>
<img src="graphics/RecEq5.gif">
<br>
<br>
So the solution here is O(<i>n</i><sup>2</sup>). The amount of work at
each level is reduced by a power of two, and so is
just a constant factor times the root.
<br>
<br>
<b>Consider these three examples</b>:
</p>
<ol>
<li>
<i>T</i>(<i>n</i>) = 4<i>T</i>(<i>n</i>/2) + <i>cn</i>
</li>
<li>
<i>T</i>(<i>n</i>) = 2<i>T</i>(<i>n</i>/2) + <i>cn</i>
</li>
<li>
<i>T</i>(<i>n</i>) = 2<i>T</i>(<i>n</i>/2) + <i>cn</i><sup>2</sup>
<br>
(This is the recurrence in the diagram above.)
</li>
</ol>
<p>
(We assume <i>c</i> > 0.)
<br>
Let's break down these cases:
</p>
<table>
<tr>
<th colspan="4">
<b><i>T</i>(<i>n</i>) = 4<i>T</i>(<i>n</i>/2)
+ <i>cn</i><sup>1</sup></b>
</th>
</tr>
<tr>
<th>
Level
</th>
<th>
# Nodes
</th>
<th>
Work at
<br>
Node
</th>
<th>
Work at
<br>
Level
</th>
</tr>
<tr>
<td>
0
</td>
<td>
1
</td>
<td>
<i>n</i>
</td>
<td>
<i>n</i>
</td>
</tr>
<tr>
<td>
1
</td>
<td>
4
</td>
<td>
<i>n</i>/2
</td>
<td>
2<i>n</i>
</td>
</tr>
<tr>
<td>
2
</td>
<td>
16
</td>
<td>
<i>n</i>/4
</td>
<td>
4<i>n</i>
</td>
</tr>
<tr>
<td>
3
</td>
<td>
64
</td>
<td>
<i>n</i>/8
</td>
<td>
8<i>n</i>
</td>
</tr>
<tr>
<td>
<i>i</i>
</td>
<td>
4<sup><i>i</i></sup>
</td>
<td>
<i>n</i>/2<sup><i>i</i></sup>
</td>
<td>
2<sup><i>i</i></sup><i>n</i>
</td>
</tr>
<tr>
<td>
<i>h</i> =
<br>
log<sub>2</sub><i>n</i>
</td>
<td>
4<sup><i>h</i></sup>
</td>
<td>
<i>T</i>(1)
</td>
<td>
4<sup><i>h</i></sup><i>T</i>(1)
</td>
</tr>
</table>
<p>
The runtime then is:
<br>
<img src="graphics/4TRec.gif">
<br>
<i>h</i> = log<sub>2</sub><i>n</i>
<br>
so the first part equals:
<br>
4<sup>log<sub>2</sub><i>n</i></sup>
= <i>n</i><sup>log<sub>2</sub>4</sup>
<br>
We pull out the n from the sum and it is
an increasing geometric series
that evaluates to <i>n</i> − 1.
So the closed form for the recurrence is:
<br>
<i>n</i><sup>2</sup><i>T</i>(1) + <i>n</i>(<i>n</i> - 1)
<br>
<br>
The very last level dominates, as it already has
O(<i>n</i><sup>2</sup>) complexity.
</p>
<hr>
<table>
<tr>
<th colspan="4">
<b><i>T</i>(<i>n</i>) = 2<i>T</i>(<i>n</i>/2)
+ <i>cn</i><sup>1</sup></b>
</th>
</tr>
<tr>
<th>
Level
</th>
<th>
# Nodes
</th>
<th>
Equ. for
<br>
Node
</th>
<th>
Work
</th>
</tr>
<tr>
<td>
0
</td>
<td>
1
</td>
<td>
<i>n</i>
</td>
<td>
<i>cn</i>
</td>
</tr>
<tr>
<td>
1
</td>
<td>
2
</td>
<td>
<i>n</i>/2
</td>
<td>
<i>cn</i>
</td>
</tr>
<tr>
<td>
2
</td>
<td>
4
</td>
<td>
<i>n</i>/4
</td>
<td>
<i>cn</i>
</td>
</tr>
<tr>
<td>
3
</td>
<td>
8
</td>
<td>
<i>n</i>/8
</td>
<td>
<i>cn</i>
</td>
</tr>
<tr>
<td>
<i>i</i>
</td>
<td>
2<sup><i>i</i></sup>
</td>
<td>
<i>n</i>/2<sup><i>i</i></sup>
</td>
<td>
<i>cn</i>
</td>
</tr>
<tr>
<td>
<i>h</i> =
<br>
log<sub>2</sub> <i>n</i>
</td>
<td>
2<sup><i>h</i></sup>
</td>
<td>
<i>T</i>(1)
</td>
<td>
2<sup><i>h</i></sup><i>T</i>(1)
</td>
</tr>
</table>
<p>
And so we get:
<br>
<img src="graphics/EqCaseRec.gif">
<br>
2<sup><i>h</i></sup> = <i>n</i>.
<br>
The sum happens log <i>n</i> times, so we have <i>cn</i> * log <i>n</i>.
<br>
<br>
All levels contribute equally.
</p>
<hr>
<table>
<tr>
<th colspan="4">
<b><i>T</i>(<i>n</i>) = 2<i>T</i>(<i>n</i>/2)
+ <i>cn</i><sup>2</sup></b>
</th>
</tr>
<tr>
<th>
Level
</th>
<th>
# Nodes
</th>
<th>
Equ. for
<br>
Node
</th>
<th>
Work
</th>
</tr>
<tr>
<td>
0
</td>
<td>
1
</td>
<td>
n<sup>2</sup>
</td>
<td>
n<sup>2</sup>
</td>
</tr>
<tr>
<td>
1
</td>
<td>
2
</td>
<td>
(<i>n</i>/2)<sup>2</sup>
</td>
<td>
<i>n</i><sup>2</sup>/2
</td>
</tr>
<tr>
<td>
2
</td>
<td>
4
</td>
<td>
(<i>n</i>/4)<sup>2</sup>
</td>
<td>
<i>n</i><sup>2</sup>/4
</td>
</tr>
<tr>
<td>
3
</td>
<td>
8
</td>
<td>