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#### Degree

Doctor of Philosophy

#### Program

Electrical and Computer Engineering

#### Supervisor

Arash Reyhani-Masoleh

#### Abstract

The utilization of global communications network for supporting new electronic applications is growing. Many applications provided over the global communications network involve exchange of security-sensitive information between different entities. Often, communicating entities are located at different locations around the globe. This demands deployment of certain mechanisms for providing secure communications channels between these entities. For this purpose, cryptographic algorithms are used by many of today's electronic applications to maintain security. Cryptographic algorithms provide set of primitives for achieving different security goals such as: confidentiality, data integrity, authenticity, and non-repudiation. In general, two main categories of cryptographic algorithms can be used to accomplish any of these security goals, namely, asymmetric key algorithms and symmetric key algorithms. The security of asymmetric key algorithms is based on the hardness of the underlying computational problems, which usually require large overhead of space and time complexities. On the other hand, the security of symmetric key algorithms is based on non-linear transformations and permutations, which provide efficient implementations compared to the asymmetric key ones. Therefore, it is common to use asymmetric key algorithms for key exchange, while symmetric key counterparts are deployed in securing the communications sessions. This thesis focuses on finding efficient hardware implementations for symmetric key cryptosystems targeting mobile communications and resource constrained applications.

First, efficient lightweight hardware implementations of two members of the Welch-Gong (WG) family of stream ciphers, the WG$\left(29,11\right)$ and WG-$16$, are considered for the mobile communications domain. Optimizations in the WG$\left(29,11\right)$ stream cipher are considered when the $GF\left(2^{29}\right)$ elements are represented in either the Optimal normal basis type-II (ONB-II) or the Polynomial basis (PB). For WG-$16$, optimizations are considered only for PB representations of the $GF\left(2^{16}\right)$ elements. In this regard, optimizations for both ciphers are accomplished mainly at the arithmetic level through reducing the number of field multipliers, based on novel trace properties. In addition, other optimization techniques such as serialization and pipelining, are also considered.

After this, the thesis explores efficient hardware implementations for digit-level multiplication over binary extension fields $GF\left(2^{m}\right)$. Efficient digit-level $GF\left(2^{m}\right)$ multiplications are advantageous for ultra-lightweight implementations, not only in symmetric key algorithms, but also in asymmetric key algorithms. The thesis introduces new architectures for digit-level $GF\left(2^{m}\right)$ multipliers considering the Gaussian normal basis (GNB) and PB representations of the field elements. The new digit-level $GF\left(2^{m}\right)$ single multipliers do not require loading of the two input field elements in advance to computations. This feature results in high throughput fast multiplication in resource constrained applications with limited capacity of input data-paths. The new digit-level $GF\left(2^{m}\right)$ single multipliers are considered for both the GNB and PB. In addition, for the GNB representation, new architectures for digit-level $GF\left(2^{m}\right)$ hybrid-double and hybrid-triple multipliers are introduced. The new digit-level $GF\left(2^{m}\right)$ hybrid-double and hybrid-triple GNB multipliers, respectively, accomplish the multiplication of three and four field elements using the latency required for multiplying two field elements. Furthermore, a new hardware architecture for the eight-ary exponentiation scheme is proposed by utilizing the new digit-level $GF\left(2^{m}\right)$ hybrid-triple GNB multipliers.

COinS


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# DES key expansion: why are these two descriptions different?

I would like someone to explain me how to generate DES subkeys, because in the documentation that I saw, sometimes they say:

"We rotate the initial key according to the number of iterations given by this sequence: 1, 2, 4, 6, 8, 10, 12, 14, 15, 17, 19, 21, 23, 25, 27, 28."

and sometimes they say:

"The two halves are each rotated to the left by an amount depending on the iteration stage, in the following order: 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1."

Do you have any idea why these descriptions are different?

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## migrated from stackoverflow.comJan 2 '13 at 16:11

This question came from our site for professional and enthusiast programmers.

Why are you using DES? We're not in the 70s anymore. – Ilmari Karonen Dec 28 '12 at 18:50
only, in the school, the teacher – simonTifo Dec 28 '12 at 19:22

I'm not going to look up the DES key schedule such, but the connection between your two sequences

$$( a_i ) = ( 1, 2, 4, 6, 8, 10, 12, 14, 15, 17, 19, 21, 23, 25, 27, 28 )$$

and

$$( b_i ) = ( 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1 )$$

is obvious: $b_i = a_i - a_{i - 1}$ (with $a_0$ taken to be 0).

That is, the second sequence gives the difference between successive elements of the first sequence.

Edit: OK, I did look it up, see link above. Basically, the first sequence gives the number of bits by which the halves of the original key need to be rotated to obtain the $i$-th subkey, whereas the second sequence gives the number of bits by which the halves of the previous subkey need to be rotated to obtain the next one. That's all there is to it.

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thank you :), you are right – simonTifo Dec 28 '12 at 20:31

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# Linearizing Stefan-Boltzmann equation

1. Dec 12, 2007

### J_I_F

1. The problem statement, all variables and given/known data
I want to linearize both the Stefan-Boltzmann equation and the adiabatic relation to draw a relationship between δL/L(0) and δR/R(0).

2. Relevant equations
Stefan-Boltzmann equation in the form of L=4πσ(R^2)(T^4)

3. The attempt at a solution
I have L(0) + δL = 4πσ((R(0) + δR)^2)*((T(0) + δT)^4) for the S-B part, but don't know how to get it into this desired form:

δL/L(0) = (2δR/R(0)) + (4δT/T(0))

Last edited: Dec 12, 2007
2. Dec 13, 2007

### dwintz02

So you have:

$$L_0+\delta L=4\pi\sigma(R_0+\delta R)^2(T_0+\delta T)^4$$

You will have to use a Taylor Expansion to expand the terms that can be considered very small. For example:

$$(a+x)^4=a^4(1+\frac{x}{a})^4 \approx a^4(1+\frac{4x}{a})$$

This only works when $$\frac{x}{a}$$ is very small compared to 1, as is the case in your example. If you're still having trouble, show your work and we can go from there.

Last edited: Dec 13, 2007
3. Dec 13, 2007

### J_I_F

That helped a bunch, thanks.


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### Home > CCA2 > Chapter 8 > Lesson 8.2.3 > Problem8-105

8-105.

Make a sketch of a graph $p(x)$ so that $p(x) = 0$ would have the indicated number and type of solutions.

1. $5$ real solutions

Your graph should have five $x$-intercepts.

1. $3$ real and $2$ complex

Draw a graph with three $x$-intercepts that when shifted vertically would have five $x$-intercepts.

1. $4$ complex

Draw a graph with no $x$-intercepts. But, when it is shifted down it should have four $x$-intercepts.

1. $4$ complex and $2$ real

1. For parts (a) through (d), what is the lowest degree each function could have?

The degree needs to be at least as high as the number of solutions.


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# Math Help - cdf of a transformation of two random variables

1. ## cdf of a transformation of two random variables

X and Y have the following joint pdf.

$f(x,y)=\frac{1}{4}$ for $-1, zero elsewhere.

Find the cdf of $Z=X+Y$

2. Originally Posted by akolman
X and Y have the following joint pdf.

$f(x,y)=\frac{1}{4}$ for $-1, zero elsewhere.

Find the cdf of $Z=X+Y$

$F(z) = \Pr(Z < z) = \Pr(X + Y < z) = \Pr(Y < -X + z)$.

A diagram is essential to see why there are two cases to consider and where the integral terminals come from in each case.

Case 1: $z \leq 0$ (that is, $-2 < z \leq 0$)

$F(z) = \int_{x = -1}^{x = z + 1} \int_{y = -1}^{y = -x + z} \frac{1}{4} \, dy \, dx = \, ....$

Case 2: $z > 0$ (that is, $0 < z \leq 2$)

$F(z) = \int_{x = -1}^{x = z - 1} \int_{y = -1}^{y = 1} \frac{1}{4} \, dy \, dx + \int_{x = z-1}^{x = 1} \int_{y = -1}^{y = -x + z} \frac{1}{4} \, dy \, dx = \, ....$

As I've said, a diagram is essential to see where these results have come from. The calculations are left for you to do.


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## Geometry: Common Core (15th Edition)

$m \angle C = 43.6^{\circ}$
In this exercise, we are given the measure of an angle and two sides. We can use the law of sines to find the measure of another angle in the triangle. The law of sines states the following: $\frac{sin A}{a}$ = $\frac{sin C}{c}$, where $A$ and $C$ are the measures of two angles in the triangle and $a$ and $c$ are the measures of the sides opposite those two angles, respectively. The side opposite to $\angle A$ is $\overline{BC}$, and the side opposite to $\angle C$ is $\overline{AB}$. We want to find $m \angle C$. Let's plug in what we know into the formula: $\frac{sin 80}{10}$ = $\frac{sin m \angle C}{7}$ Multiply each side by $7$: $7\frac{sin 80}{10}$ = sin $m \angle C$ Evaluate the expression on the left: sin $m \angle C \approx 0.6894$ Take $sin^{-1}$ to solve for $C$: $m \angle C = 43.6^{\circ}$