Question

WRITE THE ALGORITHM PSEUDOCODE FOR EACH OF THE FOLLOWING(NO CODE JUST PSEUDOCODE)

1.DETERMINE THE DAY OF THE WEEK FOR ANY GIVEN DATE WITHIN THE SAME WEEK

(EX. TODAY IS 03/24 TUE, SO 03/23 MUST BE MON)

2. DETERMINE THE DAY OF THE WEEK FOR ANY GIVEN DATE WITHIN THE SAME MONTH

(EX. TODAY IS 03/24 TUE, SO 03/15 MUST BE SUN)

3. DETERMINE THE DAY OF THE WEEK FOR ANY GIVEN DATE WITHIN THE SAME NON LEAP YEAR

(DO NOT COPY OR USE THE TOMOHIKO SAKAMOTO OR DOOMSDAY ALGORITHMS)

Answer 1

Hello, please find the below detailed explanation for the asked question. And in case of any doubt please don't forget to ask your queries in the comment section.

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Solution 1:

For this firstly, create an array of string for 7 days i.e.

string arr[] = {"SUN", "MON", "TUE", "WED", "THU", "FRI", "SAT"};

then calculate the sum,

sum = [array index of today + (asked date - today date)]

and then compute i = sum%7 and print the array arr[i].

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Example :

Today = 03/24 TUE

Asked = 03/23

Hence, sum = [2 + 23 - 24] = 1

i = 1%7 = 1

and arr[1] = "MON"

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Solution 2:

For this firstly, create an array of string for 7 days i.e.

string arr[] = {"SUN", "MON", "TUE", "WED", "THU", "FRI", "SAT"};

then calculate the sum,

if asked date > today date then, sum = [array index of today + 7 + (asked date - today date)%7]

else, sum = [array index of today + 7 - (today date - asked date)%7]

and then compute i = sum%7 and print the array arr[i].

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Example :

Today = 03/24 TUE

Asked = 03/15

Since, Asked date < today date

Hence, sum = [2 + 7 - (24 - 15)%7] = 7

i = 7%7 = 0

and arr[1] = "SUN"

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Solution 3:

For this firstly, create an array of string for 7 days i.e.

string arr[] = {"SUN", "MON", "TUE", "WED", "THU", "FRI", "SAT"};

Then, calculate the number of days between asked and today date. This will be equal to days.

now calculate days%7.

if(today date > asked date){

print arr[(array index of today + 7 - (days%7)) % 7]

}

else{

print arr[(array index of today + 7 + (days%7)) % 7]

}

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Hope, you will understand the explanation. In case of any query, please feel free to ask. And please don't forget to give a like to this explanation.