Scrabble is a word game in which words are constructed from letter tiles, each letter tile containing a point value. The value of a word is the sum of each tile's points added to any points provided by the word's placement on the game board. Write a program using the given dictionary of letters and point values that takes a word as input and outputs the base total value of the word (before being put onto a board). Ex: If the input is: PYTHON the output is: 14

Complete question:

Scrabble is a word game in which words are constructed from letter tiles, each letter tile containing a point value. The value of a word is the sum of each tile's points added to any points provided by the word's placement on the game board. Write a program using the given dictionary of letters and point values that takes a word as input and outputs the base total value of the word (before being put onto a board). Ex:  If the input is:  PYTHON

the output is: 14

part of the code:

tile_dict = { 'A': 1, 'B': 3, 'C': 3, 'D': 2, 'E': 1, 'F': 4, 'G': 2, 'H': 4, 'I': 1, 'J': 8,  'K': 5, 'L': 1, 'M': 3, 'N': 1, 'O': 1, 'P': 3, 'Q': 10, 'R': 1, 'S': 1, 'T': 1,  'U': 1, 'V': 4, 'W': 4, 'X': 8, 'Y': 4, 'Z': 10 }

Complete the program as thus:

word = input("Word: ").upper()

points = 0

for i in range(len(word)):

   for key, value in tile_dict.items():

       if key == word[i]:

           points+=value

           break

print("Points: "+str(points))

Explanation:

This gets input from the user in capital letters

word = input("Word: ").upper()

This initializes the number of points to 0

points = 0

This iterates through the letters of the input word

for i in range(len(word)):

For every letter, this iterates through the dictionary

   for key, value in tile_dict.items():

This locates each letters

       if key == word[i]:

This adds the point

           points+=value

The inner loop is exited

           break

This prints the total points

print("Points: "+str(points))

answer; i believe the correct answer is layers;

answer; always; never;