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Firdavs [7]
3 years ago
9

Lab Goal : This lab was designed to demonstrate the similarities and differences in a for loop and a while loop.Lab Description

: Write a for loop that accomplishes the same goal as a while loop and write a while loop that accomplishes the same goal as a for loop. Finally, write a loop your way: write either a for loop or a while loop that produces the appropriate output.Sample Output :***** While Loop String Cleaner ****I am Sam I am with the letter a removed by a while loop is I m Sm I m***** For Loop String Cleaner ****I am Sam I am with the letter a removed by a for loop is I m Sm I m***** For Loop Common Divisor ****The for loop determined the common divisors of 528 and 60 are12 6 4 3 2***** While Loop Common Divisor ****The while loop determined the common divisors of 528 and 60 are12 6 4 3 2***** My Total Loop My Way ****The total of even numbers from 1 to 1000 using a for loop is 250500The total of even numbers from 1 to 1000 using a while loop is 250500*Only one of the two output statements are required. For extra credit, do both
Computers and Technology
1 answer:
Mandarinka [93]3 years ago
8 0

Answer:

See Explanation

Explanation:

Required:

Use for and while loop for the same program

<u>(1) String Cleaner</u>

#For Loop

name = "I am Sam"

result = ""  

<em>for i in range(0, len(name)):  </em>

<em>    if name[i]!= 'a':  </em>

<em>        result = result + name[i] </em>

print(result)

#While Loop

name = "I am Sam"

result = ""  

<em>i = 0 </em>

<em>while i < len(name): </em>

<em>    if name[i]!= 'a':  </em>

<em>        result = result + name[i] </em>

<em>    i+=1 </em>

print(result)

<u>(2): Common Divisor</u>

#For Loop

num1 = 528

num2 = 60

div = num2

if num1 > num2:

   div = num1

<em>for i in range(2,div):</em>

<em>    if num1%i == 0 and num2%i==0:</em>

<em>        print(i,end = " ")</em>

print()

#While Loop

num1 = 528

num2 = 60

div = num2

if num1 > num2:

   div = num1

i = 2

<em>while i <div:</em>

<em>    if num1%i == 0 and num2%i==0:</em>

<em>        print(i,end = " ")</em>

   i+=1

The iterates statements show the difference in the usage of both loops.

For the for loop, the syntax is:

<em>for [iterating-variable] in range(begin,end-1)</em>

<em>-------</em>

<em>---</em>

<em>--</em>

<em />

For the while loop, the syntax is:

<em>while(condition)</em>

<em>-------</em>

<em>---</em>

<em>--</em>

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Answer:

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3 years ago
Create a string called alphabet containing 'abcdefghijklmnopqrstuvwxyz', then perform the following separate slice operations to
spin [16.1K]

Answer:

There is no short answer.

Explanation:

First let's create the string:

  • alphabetString = "abcdefghijklmnopqrstuvwxyz";

The first half of the string using slice method can be written as:

  • alphabetString.slice(0, 13);

The first half of the string using only the ending index can be written as:

  • alphabetString.slice(-13);

When we put - at the start of the index number, the counting begins at the last element with -1 and goes backwards.

The second half of the string can be written as:

  • alphabetString.slice(13,26);

The second half of the string using only the starting index can be written as:

  • alphabetString.slice(13);

To get the every second letter in the string, we need a for loop:

  • for( let x = 0; x < alphabetString.length(); x = x + 2){

                 alphabetString.slice(x);

}

To get the entire string in reverse, we can use the reverse method that is built-in:

  • alphabetString.reverse();

To get the every third letter of the string, we can again use a for loop:

  • for( let x = -1; x = -27; x = x - 3){

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I hope this answer helps.

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Given the following word addresses: 3, 180, 43, 2,191, 88, 190, 14, 181, 44, 186, 253
professor190 [17]

Answer:

A. index bits = 2, tag bits = 2, block offset bits = 1, it is a miss.

B. index bits = 2, tag bits = 1, block offset bits = 0, it is a hit

C. the miss rate is 0

Explanation:

a. number of blocks = 12

number of blocks per set = 3

number of set = number of blocks / number of blocks per set = 12/3 = 4

word size = 24

block size = 2

the block offset = log_{2} block size

   = log_{2} 2 = 1

the index bits = log_{2} number of set = log_{2} 4 = 2

the tag bits = (log_{2} word size) - offset - index = (log_{2} 24) -2 - 1 = 5 -2 - 1 = 2

b. word size = 8

block size = 1

the block offset = log_{2} block size

   = log_{2} 1 = 0

the index bits = log_{2} number of set = log_{2} 4 = 2

the tag bits = (log_{2} word size) - offset - index = (log_{2} 8) -2 - 1 = 3 - 0- 2= 1

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