Dynamic programming.
Dynamic programming is a powerful algorithmic design paradigm.
The key idea is to save state to avoid recomputation:
break a large computational problem up into smaller
subproblems, store the answers to those smaller subproblems, and,
eventually, use the stored answers to solve the original problem.
This avoids recomputing the same quantity over and over again, and
a potential exponential blowup in the running time. We now
consider two examples.
 Binomial coefficients.
The binomial
coefficient C(n, k)
is the number of ways of choosing a subset of k elements from
a set of n elements. It arises in probability and statistics,
e.g., the probability of flipping a biased coin (with probability p
of heads) n times and getting exactly k heads is
C(n, k) p^{k} (1p)^{k}.
One formula for computing binomial coefficients is
C(n, k) = n! / (k! (nk)!).
This formula is not so amenable to direct computation because
the intermediate results may overflow, even if the final answer does not.
For example C(100, 15) = 253338471349988640 fits in a 64bit long,
but the binary representation of 100! is 525 bits long.
Pascal's identity expresses C(n, k) in terms of
smaller binomial coefficients:
To understand why Pascal's identity is valid, consider some arbitrary element x.
To choose k of n elements, we either select element x (in which case
we still need to choose k1 of the remaining n1 elements), or we don't
select element x (in which case we still need to choose k of the remaining
n1 elements).
The naive implementation below fails spectacularly for medium n or k,
not because of overflow, but rather because the same subproblems are
solved repeatedly.
// DO NOT RUN THIS CODE FOR LARGE INPUTS public static long binomial(int n, int k) { if (k == 0) return 1; if (n == 0) return 0; return binomial(n1, k) + binomial(n1, k1); }

This makes the
algorithm take exponential time (the number of recursive calls
is around C(n, k)). To compensate, we must avoid recomputing
the same quantities over and over. One way to do this is to
store the results of all of the resulting subproblems in an nbyk array.
Before checking whether to compute C(n, k), we first consult
the table to see if it has already been computed. This ensures
that we compute C(n, k) at most once for each choice of n and k.
This version of dynamic programming is referred to as topdown.
Instead of using a recursive procedure, we could fill up the
entries in the nbyk array. We must organize the computation so
that all entries that we need are filled in prior to using
them in subsequent computations.
Bottomup dynamic programming fills in the
2D array starting with the values that are easiest to compute.
We begin with the base cases: C(n, 0) = 1 for all n ≥ 0,
and C(0, k) = 0 for all k ≥ 1.
Then we fill in all the values when n = 1, then n = 2, and
so forth.
This ensures that everything we need is precomputed before
we ever access it.
Program Binomial.java takes two
command line integer N and K and prints out C(N, K)
using a combination of Pascal's identity and bottomup
dynamic programming.
for (int k = 1; k <= K; k++) binomial[0][k] = 0; for (int n = 0; n <= N; n++) binomial[n][0] = 1;
for (int n = 1; n <= N; n++) for (int k = 1; k <= K; k++) binomial[n][k] = binomial[n1][k1] + binomial[n1][k];

To compute C(6, 4), the program computes the following table of
values from lefttoright (k = 1 to K) and then toptobottom (n = 1 to N).
n\k 0 1 2 3 4  0 1 0 0 0 0 1 1 1 0 0 0 2 1 2 1 0 0 3 1 3 3 1 0 4 1 4 6 4 1 5 1 5 10 10 5 6 1 6 15 20 15

 Longest common subsequence.
Now we consider a more sophisticated application of dynamic programming
to a central problem arising in computational biology and other domains.
Given two strings s and t, we wish to determine how similar they are.
Some examples include: comparing two DNA sequences for homology (similarity),
two English words for spelling, two Java files for repeated code.
It also arises in molecular biology, gas chromatography, and bird
song analysis.
One simple strategy is
to find the length of the longest common subsequence (LCS).
If we delete some characters from s and some characters from t,
and the resulting two strings are equal, we call the resulting
string a common subsequence. The LCS problem is to find
a common subsequence that is as long as possible. For example
the LCS of ggcaccacg and acggcggatacg is
ggcaacg.
ggcaccacg acggcggatacg

Now we describe a systematic method for computing the LCS of
two strings x and y using dynamic programming.
Let M and N
be the lengths of x and y, respectively.
We use the notation x[i..M] to denote the suffix of
x starting at position i, and y[j..N]
to denote the suffix of y starting at position j.
If x and y begin with the same letter, then we should
include that first letter in the LCS. Now our problem reduces
to finding the LCS of the two remaining substrings x[1..M]
and y[1..N].
On the other hand, if the two strings start with different
letters, both characters cannot be part of a common subsequence,
so we must remove one or the other.
In either case, the problem reduces to finding the LCS of
two strings, at least one of which is strictly shorter.
If we let opt[i][j] denote the length of the LCS
of x[i..M] and y[j..N], then the following recurrence
expresses it in terms of the length of LCSs of shorter suffixes.
opt[i][j] = 0 if i = M or j = N = opt[i+1][j+1] + 1 if x_{i} = y_{j} = max(opt[i][j+1], opt[i+1][j]) otherwise

Program LCS.java is a bottomup
translation of this recurrence.
We maintain a two dimensional array opt[i][j] that
is the length of the LCS for the two strings x[i..M]
and y[j..N].
For the input strings ggcaccacg and acggcggatacg,
the program computes the following table by filling in values
from righttoleft (j = N1 to 0) and bottomtotop (i = M1 to 0).
0 1 2 3 4 5 6 7 8 9 10 11 12 x\y a c g g c g g a t a c g  0 g 7 7 7 6 6 6 5 4 3 3 2 1 0 1 g 6 6 6 6 5 5 5 4 3 3 2 1 0 2 c 6 5 5 5 5 4 4 4 3 3 2 1 0 3 a 6 5 4 4 4 4 4 4 3 3 2 1 0 4 c 5 5 4 4 4 3 3 3 3 3 2 1 0 5 c 4 4 4 4 4 3 3 3 3 3 2 1 0 6 a 3 3 3 3 3 3 3 3 3 3 2 1 0 7 c 2 2 2 2 2 2 2 2 2 2 2 1 0 8 g 1 1 1 1 1 1 1 1 1 1 1 1 0 9 0 0 0 0 0 0 0 0 0 0 0 0 0

One final challenge is to recover the optimal solution
itself, not just its value.
The key idea is to retrace the steps of the dynamic programming
algorithm backwards, rediscovering the path of choices (highlighted in
red in the table above) from opt[0][0] to
opt[M][N].
To determine the choice that led to opt[i][j],
we consider the three possibilities:

x[i] matches y[j]. In this case,
we must have opt[i][j] = opt[i+1][j+1] + 1, and the
next character in the LCS is x[i].

The LCS does not contain x[i]. In this case,
opt[i][j] = opt[i+1][j].

The LCS does not contain y[j]. In this case,
opt[i][j] = opt[i][j+1].
The algorithm takes time and space proportional to MN.
Space usage.
Usually with dynamic programming you run out of space before time.
However, sometimes it is possible to avoid using an MbyN
array and getting by with just one or two arrays of length M and N.
For example, it is not hard to modify
Binomial.java to do exactly
this. (See Exercise 1.)
Similarly, for the longest common subsequence problem, it is
easy to avoid the 2D array if you only need the length of
LCS. Finding the alignment itself in linear space is substantially
more challenging (but possible using Hirschberg's divideandconquer
algorithm).
Dynamic programming history.
Bellman. LCS by Robinson in 1938.
CockeYoungerKasami (CYK) algorithm for parsing context free grammars,
FloydWarshall for allpairs shortest path,
BellmanFord for arbitrage detection (negative cost cycles),
longest common subsequence for diff, edit distance for
global sequence alignment, bitonic TSP.
Knapsack problem, subset sum, partitioning.
Application = multiprocessor scheduling, minimizing VLSI circuit size.
Root finding.
Goal: given function f(x), find x* such that f(x*) = 0. Nonlinear
equations can have any number of solutions.
x^{2} + y^{2} = 1 no real solutions e^{x} + 17 = 0 one real solution x^{2} 4x + 3 = 0 has two solutions (1, 3) sin(x) = 0 has infinitely many solutions

Unconstrained optimization.
Goal: given function f(x), find x* such that f(x) is
maximized or minimized. If f(x) is differentiable, then we
are looking for an x* such that f'(x*) = 0. However, this
may lead to local minima, maxima, or saddle points.
Bisection method.
Goal: given function f(x), find x* such that f(x*) = 0.
Assume you know interval [a, b] such that f(a) < 0 and f(b) > 0.
Newton's method.
Quadratic approximation. Fast convergence if close enough to answer.
The update formulas below are for finding the root of f(x) and f'(x).
root finding: x_{k+1} = x_{k}  f'(x_{k})^{1} f(x_{k}) optimization: x_{k+1} = x_{k}  f''(x_{k})^{1} f'(x_{k})

Newton's method only reliable if started "close enough" to solution.
Bad example (Smale): f(x) = x^3  2*x + 2. If you start in
the interval [0.1, 0.1] , Newton's method reaches a stable
2cycle. If started to the left of the negative real root, it
will converge.
To handle general differentiable or
twice differentiable functions of one variable,
we might declare an interface
public interface Function { public double eval(double x); public double deriv(double x); }

Program Newton.java runs Newton's method
on a differentiable function to compute points x* where f(x*) = 0 and
f'(x*) = 0.
The probability of finding an electron in the 4s excited state of hydrogen
ar radius r is given by:
f(x) = (1  3x/4 + x^{2}/8  x^{3}/192)^{2} e^{x/2},
where x is the radius in units of the Bohr radius (0.529173E8 cm).
Program BohrRadius.java contains the
formula for f(x), f'(x), and f''(x).
By starting Newton's method at 0, 4, 5, and 13, and 22, we obtain
all three roots and all five local minima and maxima.
Newton's method in higher dimensions. [probably omit or leave as an exercise]
Use to solve system of nonlinear equations.
In general, there are no good methods for solving a nonlinear system of equations
x_{k+1} = x_{k}  J(x_{k})^{1} f(x_{k})

where J is the Jacobian matrix of partial derivatives.
In practice, we don't explicitly compute the inverse.
Instead of computing y = J^{1}f, we solve the
linear system of equations Jy = f.
To illustrate the method, suppose we want to find a solution (x, y) to the following
system of two nonlinear equations.
x^{3}  3xy^{2}  1 = 0 3x^{2}y  y^{3} = 0

In this example, the Jacobian is given by
J = [ 3x^{2}  3y^{2} 6xy ] [ 6x 3x^{2}  3y^{2} ]

If we start Newton's method at the point (0.6, 0.6), we quickly obtain
one of the roots (1/2, sqrt(3)/2) up to machine accuracy.
The other roots are (1/2, sqrt(3)/2) and (1, 0).
Program TestEquations.java uses
the interface Equations.java and
EquationSolver.java to
solve the system of equations. We use the Jama matrix library to
do the matrix computations.
Optimization.
Use same method to optimize a function of several variables.
Good methods exist if multivariate function is sufficiently smooth.
x_{k+1} = x_{k}  H(x_{k})^{1} g(x_{k})

Need gradient g(x) = ∇f(x) and Hessian H(x) = ∇^{2}f(x).
Method finds an x* where g(x*) = 0,
but this could be a maxima, minima, or saddle point.
If Hessian is positive definite (all eigenvalues are positive)
then it is a minima; if all eigenvalues are negative, then it's
a maxima; otherwise it's a saddle point.
Also, 2nd derivatives change slowly, so it may not be necessary to
recalculate the Hessian (or its LU decomposition) at each step.
In practice, it is expensive to compute the Hessian exactly, so other
so called quasiNewton methods are preferred, including the
BroydenFletcherGoldfarbShanno (BFGS) update rule.
Linear programming.
Create matrix interface.
Generalizes twoperson zerosum games, many problems in combinatorial
optimization, ....
run AMPL from the web.
Programming = planning. Give some history. Decision problem not known
to be in P for along time.
In 1979, Khachian resolved the question in the affirmative
and made headlines in the New York Times with a geometric divideandconquer
algorithm known as the ellipsoid algorithm. It requires
O(N^{4}L) bit operations where N is the number of variables
and L is the number of bits in the input. Although this was a landmark
in optimization, it did not immediately lead to a practical algorithm.
In 1984, Karmarkar proposed a projective scaling algorithm that takes
O(N^{3.5}L) time. It opened up the door for efficient implementations
because by typically performing much better than its worst case guarantee.
Various interior point methods were proposed in the 1990s, and the
best known complexity bound is O(N^{3} L). More importantly, these
algorithm are practical and competitive with the simplex method. They
also extend to handle even more general problems.
Simplex method.
Linear programming solvers.
In 1947, George Dantzig proposed the simplex algorithm for linear programming.
One of greatest and most successful algorithms of all time.
Linear programming, but not industrial strength.
Program LPDemo.java illustrates how
to use it.
The classes MPSReader and MPSWriter
can parse input files and write output files in the standard
MPS format.
Test LP data files in
MPS format.
More applications.
ORObjects also has graph coloring, traveling salesman problem,
vehicle routing, shortest path.