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iragen [17]
3 years ago
8

A TV game show Jeopardy! contestant who uses a strategy to seek out the hidden Daily Double questions, which are usually the thr

ee highest-paying and most difficult questions in the categories instead of the more common approach of selecting a single category and selecting questions gradually increasing in the degree of difficulty, has been highly successful. What is this an application of?
Computers and Technology
1 answer:
jek_recluse [69]3 years ago
8 0

Answer:

Game theory.

Explanation:

In the TV game show Jeopardy! Game theory allows contestant to use a strategy to seek out the hidden Daily Double questions, instead of the more common approach of selecting a single category and selecting questions gradually increasing in the degree of difficulty.

Game theory is the process of using mathematical model in the study of strategic interaction between two or more players in a game or situation having laid down rules and outcomes.

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b goes with 1 c goes with 3 and a goes with 2

Explanation:

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Recall the CFG G4 that we gave in Example 2.4. For convenience, let’s rename its variables with single letters as follows. E → E
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Answer:

Check the attached image

Explanation:

the diagram for explaning all the solutions is in the attaged images below.

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Def build_dictionary(string):
MissTica

Answer:

Follows are the code to this question:

def build_word_counter(string):#defining a method build_word_counter that accepts string as a parameter  

   d = dict()#defining d variable as a dictionary

   for x in string:#defining for loop that use string

       x = x.lower()#Use lower method  

       if not d.get(x):#defining if block that holds value in d variable

           d[x] = 1#use dictionary to hold value in 1

       else:#defining else block

           d[x] += 1#use dictionary to increment value by 1

   return d#return dictionary

print(build_word_counter(["Like","like","LIKE"]))#use print method to call method

def build_letter_distribution(string):#defining a method build_letter_distribution that accepts string variable

   d = dict()#defining d variable as a dictionary

   for i in string:#defining for loop to count string value

       for j in i:#defining for loop to hold string as number

           if not d.get(j):#defining if block that doesn't get value in dictionary  

               d[j] = 1#hold value in dictionary  

           else:#defining else block

               d[j] += 1#use dictionary to increment the dictionary value

   return d#return dictionary

print(build_letter_distribution(["Like","test","abcdEFG"]))#use print method to return its value

Output:

please find the attached file.

Explanation:

In the first program a method, "build_word_counter" is defined that accepts a string variable in its parameter, and inside the method, a dictionary variable "d" is defined that uses the for loop to hold string value and convert all value into lower case.

In the next step, a conditional statement is used that holds value in the dictionary and count its word value.

In the second program a method, "build_letter_distribution" is defined that holds string variable in its parameter, and inside the method, a dictionary variable "d" is defined that uses the two for loop and in if the block, it holds value in the dictionary and returns its value, and use print method to print its return its value.

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