Solved by Expert Tutors
Solved by Expert Tutors
Question

# Linear algebra / matrices - need solutions to 3 problems from one

document and 3 problems from another, for a total of 6 solutions.

Hill Substitution Ciphers Text Reference: Section 4.1, p. 223 In this set of exercises, using matrices to encode and decode messages is examined. Perhaps the simplest way to encode a message is to simply replace each letter of the alphabet with another letter. This is the method used in the “Cryptograms” often found in puzzle books or newspapers, and is called a substitution cipher . Consider this cipher array for a substitution cipher: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Y C W O G R D B P I Z X K L N M T S E F H J A V U Q To encode a message like VECTOR SPACE, encode each letter in turn and get JGWFNSEMYWG Notice that the space between the words hasn’t been encoded; this character could be added (as could any other punctuation or symbol) to the cipher array and encoded also. To decode a message, the decipher array is needed, which for our substitution cipher is A B C D E F G H I J K L M N O P Q R S T U V W X Y Z W H B G S T E U J V M N P O D I Z F R Q Y X C L A K Thus the secret message FBGSNNEFGSWSNAEYFOYAL is easily revealed as THE ROOSTER CROWS AT DAWN Amajordrawbackofthesubstitutioncipheristhatitisveryeasyforapersonto“crack”thecode; that is, to determine the decipher array from an encoded message. If you have done a crytogram before, you know how this is done: the relative frequency of letters in English is known, as are the frequencies of certain groups of letters like TH or ST. See Reference 3 (p.16 and p.19) for sample tables. Common short words like THE and OF also help the code cracker. To make the code harder to crack, groups of letters can be encoded at the same time. For example, consider splitting the above message into units of three letters each, although any length of block would be allowable. The message THE ROOSTER CROWS AT DAWN is thus converted to THE ROO STE RCR OWS ATD AWN 1
If the number of letters is not a multiple of three, the Fnal set of letters can be padded with random letters; thus VECTOR SPACE could be split up as VEC TOR SPA CEX. Since linear algebra will come in handy for the encoding process, replace each letter by its position in the alphabet. The positions are numbered from 0 to 25 for reasons which will shortly become apparent. A B C D E ± G H I J K L M N O P Q R S T U V W X Y Z 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 The batches of three letters can be converted into batches of three numbers, which can be thought of as vectors .±orexamp le ,THEROOSTERCROWSATDAWNhasbecome T H E 19 7 4 R O O 17 14 14 S T E 18 19 4 R C R 17 2 17 O W S 14 22 18 A T D 0 19 3 A W N 0 22 13 In order to encode messages using the numbers instead of letters, an easy way of calculating using only the numbers 0 through 25 is needed. A way to do this is called modular arithmetic .Sums , differences, and products are calculated as normal, but if the result is larger than 25 or less than 0, it is replaced by the remainder left when the result is divided by 26. Thus only numbers from 0 to 25 result from this arithmetic. This form of arithmetic is denoted by stating that the calculations are to be done modulo 26 ,andbyplacingthesymbol (mod 26) after the calculation. Examples : 1+2=3(mod26) 13 × 2=0(mod26) ,since 13 × 2=26=26(1)+0 7+24=5(mod26) ,since 7+24=31=26(1)+5 10 15 = 21(mod 26) ,since 10 15 = 5=26( 1) + 21 Just as numbers can be added and multiplied, vectors may be added and scalar multiplication may be performed: Examples : 19 7 4 + 14 22 18 = 7 3 22 (mod 26) 10 × 14 22 18 = 10 12 24 (mod 26) 2
Show entire document
Equilibrium Temperature Distributions The purpose of this set of exercises is to discuss a physical situation in which solving a system of linear equations becomes necessary: that of determining the equilibrium temperature of a thin plate. Methods for solving these systems will be compared, and some will be shown to be more efFcient than others. Consider a thin square plate whose faces are insulated from heat. Suppose that the temperature along the four edges of the plate is known, and further suppose that those temperatures are held constant. After some time has passed, the temperature inside the plate will reach an equilibrium. ±inding this equilibrium temperature distibution at the points on the plate is desirable, given only the temperature data from the edges of the plate. Unfortunately, the exact determination of this temperature distribution is a difFcult problem. An appproximation to the exact distribution may be found by discretizing the problem ;tha tis ,byon lycons ider ingafewpo in tsonthep la teand approximating the temperature at those points. Apropertyfromthermodynamicshelpswiththediscretizationoftheproblem .Thepropertysays that if equilibrium has been acheived, then the temperature at a point is the average value of the temperature at surrounding points. The Mean Value Property :I fap la teha sreachedthe rma lequ i l ib r ium ,and P is a point on the plate, and C is a circle centered at P fully contained in the plate, then the temperature at P is the average value of the temperature function over C . P C ±igure 1: The Mean Value Property See ±igure 1 for a picture of this situation. In order to discretize the problem, place a grid over the plate and concentrate only on the points where the grid lines cross; the temperature at only those points will be considered. The grid is fashioned so that some grid points lie on the boundary of the 1
plate; assume that the temperature at these points equals the external temperature. At grid points inside the plate, assume the following version of the Mean Value Property. The Discretized Mean Value Property : If a plate has reached thermal equilibrium, and P is a grid point not on the boundary of the plate, then the temperature at P is the average of the temperatures at the four closest grid points to P . Example: Consider placing the following grid on the square plate; see Figure 2. There are four points inside the plate to consider; the temperatures at these points are labelled x 1 , x 2 , x 3 ,and x 4 . Assume that the exterior temperatures are as labelled in Figure 2. The Discretized Mean Value Property gives rise to the following four equations: 10 o 0 o 20 o 30 o x 1 x 2 x 3 x 4 Figure 2: Grid on a Square Plate x 1 = 40 + x 2 + x 3 4 ,x 2 = 30 + x 1 + x 4 4 ,x 3 = 30 + x 1 + x 4 4 ,x 4 = 20 + x 2 + x 3 4 which is equivalent to the following system of linear equations: 4 x 1 - x 2 - x 3 =4 0 - x 1 +4 x 2 - x 4 =3 0 - x 1 +4 x 3 - x 4 =3 0 - x 2 - x 3 +4 x 4 =2 0 What if the external temperatures change? A new system of equations would need to be solved, but it would be very similar to the one previously considered. To handle this dif±culty more easily, the system may be rewritten in the form 2
Show entire document

risus a

, ultrices ac magna. Fusce dui lectus, congue vel laoreet ac, dictum vitae odio. Donec aliquet. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nam lacinia pulvinar tortor nec facilisis. Pellentesque dapibus efficitur laoreet. Nam risus ante, dapibus a molestie consequat, ultrices ac magna. Fusce dui lectu

o

Subject: Math

### Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

• ### -

Study Documents

Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

Browse Documents