no file is needed to show the setup You are a young engineer

no file is needed to show the setup

You are a young engineer preparing a construction project in a remote location. Much of the construction is outdoors, yet little weather data exists for the site. Luckily, you have been provided some raw data, but, it is merely a data file containing four rows of data, each associated with the month of each column. Given this data file, create a temperature plot for your project manager\'s presentation. MATLAB is an extremely effective and easy plotting tool, so you decide to use MATLAB. Given the data file weather.dat Average Monthly High Temperature (degrees Celsius) Average Monthly Low Temperature (degrees Celsius) Months (1-12, January - December) Average Monthly Rainfall (centimeters) Make a Plot, Temperatures (average high and low) vs month of year, to visually capture this data. Your manager will of course want the data in English units. When you load the data file weather.dat, make sure you save your new matrix weather_lastname remember, when you load weather.dat, MATLAB will automatically create a matrix and save it to a variable with the same name (i.e. \"weather\"), you need to rename this matrix \"weather_lastname\". Save your MATLAB file Problem_Set_8_4Jastname.

Solution

Let L be a daily language. Then there exists a relentless ‘c’ specified for each string w in L

|w| c

We can break w into 3 strings, w = xyz, specified

|y| > 0
|xy| c
For all k zero, the string xykz is additionally in L.
Applications of Pumping Lemma
Pumping Lemma is to be applied to point out that sure languages don\'t seem to be regular. It ought to ne\'er be wont to show a language is regular.

If L is regular, it satisfies Pumping Lemma.

If L is non-regular, it doesn\'t satisfy Pumping Lemma.

Method to prove that a language L isn\'t regular
At first, we\'ve got to assume that L is regular.

So, the pumping lemma ought to hold for L.

Use the pumping lemma to get a contradiction

Select w specified |w| c

Select y specified |y| one

Select x specified |xy| c

Assign the remaining string to z.

Select k specified the ensuing string isn\'t in L.

Hence L isn\'t regular.

Problem
Prove that L = i 0 isn\'t regular.

Solution

At first, we tend to assume that L is regular and n is that the range of states.

Let w = anbn. so |w| = 2n n.

By pumping lemma, let w = xyz, wherever |xy| n.

Let x = ap, y = aq, and z = arbn, wherever p + letter of the alphabet + r = n.p 0, q 0, r 0. so |y| zero

Let k = 2. Then xy2z = apa2qarbn.

Number of as = (p + 2q + r) = (p + letter of the alphabet + r) + letter of the alphabet = n + letter of the alphabet

Hence, xy2z = an+q bn. Since letter of the alphabet zero, xy2z isn\'t of the shape anbn.

Thus, xy2z isn\'t in L. thence L isn\'t regular.

Complement of a DFA
If (Q, , , q0, F) be a DFA that accepts a language L, then the complement of the DFA are often obtained by swapping its acceptive states with its non-accepting states and the other way around.

Let L be a daily language. Then there exists a relentless ‘c’ specified for each string w in L

|w| c

We can break w into 3 strings, w = xyz, specified

|y| > 0
|xy| c
For all k zero, the string xykz is additionally in L.
Applications of Pumping Lemma
Pumping Lemma is to be applied to point out that sure languages don\'t seem to be regular. It ought to ne\'er be wont to show a language is regular.

If L is regular, it satisfies Pumping Lemma.

If L is non-regular, it doesn\'t satisfy Pumping Lemma.

Method to prove that a language L isn\'t regular
At first, we\'ve got to assume that L is regular.

So, the pumping lemma ought to hold for L.

Use the pumping lemma to get a contradiction

Select w specified |w| c

Select y specified |y| one

Select x specified |xy| c

Assign the remaining string to z.

Select k specified the ensuing string isn\'t in L.

Hence L isn\'t regular.

Problem
Prove that L = i 0 isn\'t regular.

Solution

At first, we tend to assume that L is regular and n is that the range of states.

Let w = anbn. so |w| = 2n n.

By pumping lemma, let w = xyz, wherever |xy| n.

Let x = ap, y = aq, and z = arbn, wherever p + letter of the alphabet + r = n.p 0, q 0, r 0. so |y| zero

Let k = 2. Then xy2z = apa2qarbn.

Number of as = (p + 2q + r) = (p + letter of the alphabet + r) + letter of the alphabet = n + letter of the alphabet

Hence, xy2z = an+q bn. Since letter of the alphabet zero, xy2z isn\'t of the shape anbn.

Thus, xy2z isn\'t in L. thence L isn\'t regular.

Complement of a DFA
If (Q, , , q0, F) be a DFA that accepts a language L, then the complement of the DFA are often obtained by swapping its acceptive states with its non-accepting states and the other way around.Let L be a daily language. Then there exists a relentless ‘c’ specified for each string w in L

|w| c

We can break w into 3 strings, w = xyz, specified

|y| > 0
|xy| c
For all k zero, the string xykz is additionally in L.
Applications of Pumping Lemma
Pumping Lemma is to be applied to point out that sure languages don\'t seem to be regular. It ought to ne\'er be wont to show a language is regular.

If L is regular, it satisfies Pumping Lemma.

If L is non-regular, it doesn\'t satisfy Pumping Lemma.

Method to prove that a language L isn\'t regular
At first, we\'ve got to assume that L is regular.

So, the pumping lemma ought to hold for L.

Use the pumping lemma to get a contradiction

Select w specified |w| c

Select y specified |y| one

Select x specified |xy| c

Assign the remaining string to z.

Select k specified the ensuing string isn\'t in L.

Hence L isn\'t regular.

Problem
Prove that L = i 0 isn\'t regular.

Solution

At first, we tend to assume that L is regular and n is that the range of states.

Let w = anbn. so |w| = 2n n.

By pumping lemma, let w = xyz, wherever |xy| n.

Let x = ap, y = aq, and z = arbn, wherever p + letter of the alphabet + r = n.p 0, q 0, r 0. so |y| zero

Let k = 2. Then xy2z = apa2qarbn.

Number of as = (p + 2q + r) = (p + letter of the alphabet + r) + letter of the alphabet = n + letter of the alphabet

Hence, xy2z = an+q bn. Since letter of the alphabet zero, xy2z isn\'t of the shape anbn.

Thus, xy2z isn\'t in L. thence L isn\'t regular.

Complement of a DFA
If (Q, , , q0, F) be a DFA that accepts a language L, then the complement of the DFA are often obtained by swapping its acceptive states with its non-accepting states and the other way around.