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RideAnS [48]
2 years ago
9

(a) Show how to use (the Boolean formula satisfiability program) satisfiable num (and substitute) to find a satisfying assignmen

t that minimizes the number of variables set to TRUE. Write the pseudo-code. HINT: eurtottesebtsumselbairavynamwohenimretedtsrif. HINT for hint: sdrawkcabtidaer. (b) How fast is your algorithm? Justify.
Computers and Technology
1 answer:
tia_tia [17]2 years ago
6 0

Answer:

Check the explanation

Explanation:

(a)

# We need to try all options(TRUE/FALSE) of all the Variables

# to find the correct arrangement.

permutations <- permutate(0,1,n) [Find all permutions O(2^n)]

for permutation in permutations:

for X in variables:

if(permutation[i] == 1)

substitute(H,X,true)

else

substitute(H,X,false)

if(satisfiable(H)) return permutation

increment i

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array of String objects, words, has been properly declared and initialized. Each element of words contains a String consisting o
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Answer:

for(String s:words)

   if(s.endsWith("ing"))

 System.out.println(s);

Explanation:

Create an enhanced for loop that iterates through the words array

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If you find one that ends with "ing", print the element

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2 years ago
python Consider this data sequence: "3 11 5 5 5 2 4 6 6 7 3 -8". Any value that is the same as the immediately preceding value i
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int firstNumber,secondNumber = -1, duplicates = 0;

do {

cin >> firstNumber;

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secondNumber = firstNumber;

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2 years ago
Problem 1a. Write a function named hasFinalLetter that takes two parameters 1. strList, a list of non-empty strings 2. letters,
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Answer:

The answer is the programming in Python language has strings and characters that has to be declared in the method.

Explanation:

#method

def hasFinalLetter(strList,letters):

output = #output list

#for every string in the strList

for string in strList:

#findout the length of each string in strList

length = len(string)

#endLetter is last letter in each string

endLetter = string[length-1]

#for each letter in the letters list

for letter in letters:

#compare with endLetter

#if we found any such string

#add it to output list

if(letter == endLetter):

output.append(string)

#return the output list

return output

#TestCase 1 that will lead to return empty list

strList1 = ["user","expert","login","compile","Execute","stock"]

letters1 = ["a","b","y"]

print hasFinalLetter(strList1,letters1)

#TestCse2

strList2 = ["user","expert","login","compile","Execute","stock"]

letters2 = ["g","t","y"]

print hasFinalLetter(strList2,letters2)

#TestCase3

strList3 = ["user","expert","login","compile","Execute","stock"]

letters3 = ["k","e","n","t"]

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2 years ago
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