ECS 120 Lecture Notes - Lecture 3: Parse Tree, Regular Language, Regular Expression

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aquamarineserval950

31 May 2018

School

UC-Davis

Department

Engineering: Computer Science

Course

ECS 120

Professor

Doty

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Week 4

Lecture 10

review:

1. Last Time: operations on regular languages

2. Given M & M’ is L(M) = L(M’)

a. Minimize both of the machines

b. Make a DFSM that does the following

i. A-B U B-A

ii. If there is a path?

iii. Is it empty?

c. This method guarantees that we can determine if L(M) & L(M’) are the same.

New information:

1. Regular expressions are often converted into NFA’s then DFA’s

a. We do not have to do this conversion by hand but this is a quiz question

New topic Context Free Grammars

Proof: there are some languages that are not regular?

Regular languages means that a DFSM or NDFSM can be used to accept the language.

There are regular languages. But not all languages are regular.

L= ��where n > 0

Find a FSM that accepts this language = this is impossible

1. Some strings in the language

a. 000111, 01, 001

b. Not in language

i. 10, 110, 0, 1, 101010101, 011, 001,

Proof that L= ��where n > 0 is not regular

1. Suppose that L is a regular language then some DFSM called M will be able to accept

the language L.

2. Let p be the number of states in M.

a. We are saying that M is a real FSM machine that can do this, therefore it has

some number of states inside of it, we do not know how many states but just

assume that there is some number p of them

3. Pick n to be greater than p

a. N is the number that 0 & 1 is raised to

b. Theres always a number greater than p.

4. Imagine the FSM working on a string �� where n>p.

5. n>p means that there has to be some loop going on somewhere. If there wasnt a loop

we could never accept �� where n>p.

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a. So weve established there is a loop, well if there is a loop how do we know that

the loop cant be taken multiple times? We dont know this

6. ��= xyz

a. Note in the example we are assuming p = 12

b. X = 0000

c. y= 0

d. z= 0111111

i. Note this stuff is subjective of where to divide, but proffesor and textbook

divide in the same way.

7. What would happen if n = P+1 so a total of 13.

a. Well if the NDFSM M that we said exsist does exist it must accept this language.

b. So y must be the loop. Meaning if we are a zero in y, we can keep taking zeros

and staying in y.

c. The issue with this is that if there are more zeros then ones’ are machine will still

accept.

8. So if NDFSM M exists it must accept 00011, but this is not apart of our language.

Therefore there is a contradiction. This contradiction leads us to believe that no such

machine can exists.

Definition:

Regular language: any language that a FSM can be built for.

Pigeon hole/ pumping up proof template.

Critical elements of the proof:

1. Suppose L is regular

2. Let p be the number of states in M

3. Pick n>p.

4. Longer strings will have to revisit some states

5. Pump up, meaning add more 0’s into the y

6. Pump down, take out zeros

a. Why pump down or up, it forces the FSM to accept when it shouldnt be accepting

and this contradiction is the reason we know that the language is not a regular

language.

Lecture 11

Review:

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Document Summary

Lecture 10 review: last time: operations on regular languages, given m & m" is l(m) = l(m", minimize both of the machines, make a dfsm that does the following. Is it empty: this method guarantees that we can determine if l(m) & l(m") are the same. New information: regular expressions are often converted into nfa"s then dfa"s, we do not have to do this conversion by hand but this is a quiz question. Regular languages means that a dfsm or ndfsm can be used to accept the language. Find a fsm that accepts this language = this is impossible. L= (cid:882)(cid:883) where n > 0: some strings in the language, 000111, 01, 001, not in language. This contradiction leads us to believe that no such machine can exists. Regular language: any language that a fsm can be built for. Lecture 11 style of proof a pumping lemma.

Related Questions

1.Write a function encipher(s, n) that takes as inputs an arbitrary string s and a non-negative integer n between 0 and 25, and that returns a new string in which the letters in s have been "rotated" by n characters forward in the alphabet, wrapping around as needed. For example:

>>> encipher('hello', 1) result: 'ifmmp'

>>> encipher('hello', 2) result: 'jgnnq'

>>> encipher('hello', 4) result: 'lipps'

Upper-case letters should be "rotated" to upper-case letters, even if you need to wrap around. For example:

>>> encipher('XYZ', 3)

result: 'ABC'

Lower-case letters should be "rotated" to lower-case letters:

>>> encipher('xyz', 3)

result: 'abc'

Non-alphabetic characters should be left unchanged:

>>> encipher('#caesar!', 2)

result: '#ecguct!'

Hints/reminders:

You can use the built-in functions ord and chr convert from single-character strings to integers and back:

>>> ord('a')

result: 97

>>> chr(97)

result: 'a'

You can use the following test to determine if a character is between 'a' and 'z' in the alphabet:

if 'a' <= c <= 'z':

A similar test will work for upper-case letters.

We recommend writing a helper function rot(c, n) that rotates a single character c forward by n spots in the alphabet. We have given you a template for this helper function in ps3pr3.py that checks to ensure that c is a single-character string. We wrote rot13(c) in lecture; rot(c, n) will be very close to rot13(c)! You can test your rot(c, n) as follows:

>>> rot('a', 1)

result: 'b'

>>> rot('y', 2)

result: 'a'

>>> rot('A', 3)

result: 'D'

>>> rot('Y', 3)

result: 'B'

>>> rot('!', 4)

result: '!'

Once you have rot(c, n), you can write a recursive encipher function.

Once you think you have everything working, here are three more examples to try:

>>> encipher('xyza', 1)

result: 'yzab'

>>> encipher('Z A', 2)

result: 'B C'

>>> encipher('Caesar cipher? I prefer Caesar salad.', 25)

result: 'Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.'

2.Write a function decipher(s) that takes as input an arbitrary string s that has already been enciphered by having its characters "rotated" by some amount (possibly 0). decipher should return, to the best of its ability, the original English string, which will be some rotation (possibly 0) of the input string s. For example:

>>> decipher('Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.')

result: 'Caesar cipher? I prefer Caesar salad.'

Here are two more examples:

>>> decipher('Hu lkbjhapvu pz doha ylthpuz hmaly dl mvynla lclyfaopun dl ohcl slhyulk.')

result: 'An education is what remains after we forget everything we have learned.'

>>> decipher('python')

result: 'eniwdc'

------------------------------------

def rot(c, n):
""" your docstring goes here """
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

def letter_prob(c):
""" if c is the space character (' ') or an alphabetic character,
returns c's monogram probability (for English);
returns 1.0 for any other character.
adapted from:

"""
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

if c == ' ': return 0.1904
if c == 'e' or c == 'E': return 0.1017
if c == 't' or c == 'T': return 0.0737
if c == 'a' or c == 'A': return 0.0661
if c == 'o' or c == 'O': return 0.0610
if c == 'i' or c == 'I': return 0.0562
if c == 'n' or c == 'N': return 0.0557
if c == 'h' or c == 'H': return 0.0542
if c == 's' or c == 'S': return 0.0508
if c == 'r' or c == 'R': return 0.0458
if c == 'd' or c == 'D': return 0.0369
if c == 'l' or c == 'L': return 0.0325
if c == 'u' or c == 'U': return 0.0228
if c == 'm' or c == 'M': return 0.0205
if c == 'c' or c == 'C': return 0.0192
if c == 'w' or c == 'W': return 0.0190
if c == 'f' or c == 'F': return 0.0175
if c == 'y' or c == 'Y': return 0.0165
if c == 'g' or c == 'G': return 0.0161
if c == 'p' or c == 'P': return 0.0131
if c == 'b' or c == 'B': return 0.0115
if c == 'v' or c == 'V': return 0.0088
if c == 'k' or c == 'K': return 0.0066
if c == 'x' or c == 'X': return 0.0014
if c == 'j' or c == 'J': return 0.0008
if c == 'q' or c == 'Q': return 0.0008
if c == 'z' or c == 'Z': return 0.0005
return 1.0

olivemink636

9. Write a method with the following header public static void replace (String oldFile, string newfile, char oldChar, char newChar) throws IOException This method opens the file whose name is given by the String parameter oldFile (the input file) for reading, and opens the file whose name is given by the string parameter newlile (the output file) for writing. It then reads every lime in the input file. For each line, it creates a new line identical to the original line, creept for the fact that all occurrences of char oldchar in the original String have been replaced by occurrences of char newchar in the new line. When the new line has been properly created, it write it to the output file. Once all the lines in the input file have been read and a corresponding line has been written to the output file, the method clases both files. Additional notes: • You MAY assume that none of the parameters this method accepts are null; in other words, your method does not have to handle the case where one or more of the parameters are null. You MAY assume that all the 10-related classes have already heen imported properly . Finally, because the replace() method is specified to throw an exception object of type TORxception, you MAY assume that I/O related problems will occur while reading from the input file or writing to the output file. In other words, your method DOES NOT have to check that no such problems occur, and DOES NOT have to manage Texceptions using try-catch blocks. WRITE YOUR replace() METHOD IN THE SPACE BELOW: YOUR replace() METHOD CONTINUED: USE THIS PAGE IF YOU NEED ADDITIONAL SPACE. CLEARLY INDICATE WHICH QUESTION(S) YOU ARE ANSWERING HERE. USE THIS PAGE IF YOU NEED MORE ADDITIONAL SPACE. CLEARLY INDICATE WHICH QUES- TION(S) YOU ARE ANSWERING HERE. • String (package java.lang) Methods: - public String chal wale Al t ri g od med in the chance au - public int length(): R e ach of the string - public char chattin R echar w heefd index public chart to Chachay: Consthis String to a new c r y - public boolean equals Object anbject): Ce this string to and yect - public boolean equalsigmoce Case String anotherstring]: Cam p ing sider this String to anotherString - public int compareToString anotherString: Compare this String to anotherstring graphically returns a negative ef this string before anotherstring, apithe value of this Stringatering distries publie iw paneelgoro (String another Camp i ng Garden ces, string to anoth i ng li hically returns te value of this thing care anotherstring, apie as if this string e r anotherstring, Strings - public int indexin ch Rather within this string of the och ch-lil does not - public int index fint.ch, int from index): Returns the index within this string of the first Roe of character ch, starting the search at position from Index retums - 1 il ch does not in this String - public int indexor (String ste): Returns the index within this string of the first currence of ringt-d eur - public int indexOf(stringate, int from index): Resures the index within the string of the fare currence of shariaparting police froiden liste des not count i ng - public String subat ring it begintndax) Rema new string which is a webstring of this string compend of the chesting at position begintndex (inclusive) - puble string subat ring intaginindex intendindex: Returns a new sering that waring of this string, composed of the characters starting at position beginIndex (inclusive, and ending at - public String replace(char oldChar, char newchar): Returns a new string resulting from replacing a Tences of olchar in this string with new Char. puble string tolon Case(): Returns a new string consisting of all the characters in this sting - publle ting toppe : Roum ingiting of all the c on in this thing uverted to perce - publie ting tet): Returns a copy of this line with leading and wailing whitespace • Scanner (package java.util) Methods: - publie danner File source) thrown jatorio Foundation sa Scanner the fun the specified file - public scanner InputStream source): Cits Scanner that process and from the input stream - puble Scannerstag source): Corsa ww Scanner that produces wales w ed from the specified String -puble vold close(): Cwes this Scanner -puble boolean asext(): Return this can het in its input puble boolean asext.Double(): Return to the taken in this season al ante pasadala w ith Doble() ed publie boolean hat Into: Reutte le inse gule ed wat w eet Int() had - public boolean hati : Return true if this w ee - puble boolean has longt) Return true if the r in this cannerm ane das along the next long) with puble ting at the t e rm from the canner puble double Smerthea ter puble intention there the l ot. puble w ith the need them - puble Intolog m u long • Printstream ( kape Java.io) Methods: a - Pri le rlle) throw alo.Pilo t Pelat ihan the p rile - Printing the throw r alo. Pet Stichw e d the wind - le void lose them - le void phone boolean ve - void prichar public void print chart 6): the way of char public void print double d o uble bed, public void print (int ) Fins int value - public void printet a n net public void printing : Strings - public void printin (1 Tm the line by writing the line para ring - public void printinhoolna ) Prints hoolean when thinorminates the public void printin (cthar c h ar wheelthening the line public void print in char i ta way of cars and then the line public void print in double d in double valued and the terminates the line public void printin(int 1) Pints int value and then into the line public void printin ( jat ova t terminates the line pulle void print intring): Tits Strings then to the line Note that the printwriter class defines the same methods and constructors (except for the fact that the structors are called Print.writer instead of PrintStream) • Math (package java.lang) Methods: - please le podle double bear the value of a t the of - puble balle blanche a Returns they u se of dele with pe r than realto - pietate double random Ruma doble 0.0 wi th .. puble stat la doble sin double a: Rethe puble statie doble con double a Return ther l e, where a ininin we angle, wherea in in e -puble statie doble tan double a) R he i ngold , whereas o m public statie double toeges double agrad): C a ngrad v ideo puble salle de anable and we a r e - - estatie delle au t olewa idle de ahholooble • Character (packape java.lang) Methods: - public static boolean ibigit charch: D i chtechisadil public static int digit(charch, int radix: Return the meal of cho ch in the di radix-lich des representa digit - public statie char forbigit chat digit, ut radix) Rems the character of digit in the disa - public statie boolean isletter charch): Demines if character chalet publle statie boolean islowerCase(char ch): Determines if character ch is a character - public static boolean ipparate(char eh): Determines ir ductes chinuppercase cura nd - publicatie boliwhip (arch Termine if character his white a intola - public static char tolowrane(char chevert hech to low - public static char tolpperCase(char ch): Covers Chococh topp • Array list (package java.util) Methods: - public ArrayList(): Create a new em ArrayList which contains clements of type - public int (1 : R m the number of clans in this list - public boolean isimpty(): Rens true if this in contains no elements - public boolean contains (object o): Return true if this lintutins el compr a re performed using the equals() method on - public int IndexOf(obj ): Returns the index of the first currence of elemento in this list, or lir this listes contain this element companies are performed wing the equals() method - publiek galimt in ): Returns the elementat position index in this list - publieket fint inde element): Replace the element at the position index in this list with the - puble honlanad a ): Appends the specifiedalent to the end of this list. - public void (int in aw e t the specified element the price index in this - publiek veint index): Remove the de s ition index in the list - publie hoolean remove(object a s the fir e of the president of this if it is p r epared wing the equals() methodno - public void clear(): Remes all of the elements from this • Vile (package java.io) Methods: - puble File(String pathname: Creates a File representing the file at the given pathane

ECS 120 Lecture Notes - Lecture 3: Parse Tree, Regular Language, Regular Expression

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