Algorithm: Current day of week is always #2 in order sum up: I want to create an algorithm (or two) where the outcome is a list of numbers (1-7) where #2 is today, #1 is yesterday and #3.. is tomorrow and thereafter. 
For example;

If today is Wednesday (3)
  1. Monday = 7
  2. Tuesday = 1
  3. Wednesday = 2
  4. Thursday = 3
  5. Friday = 4
and if today is Friday (5)
   - Monday = 5
   - Tuesday = 6
   - Wednesday = 7
   - Thursday = 1
   - Friday = 2

Limitation:
The only numbers you can calculate with are  


*

*The number of the week (i.e. Tuesday = 2, Friday = 5)  

*'Today', as a variable (i.e. if Today = Wednesday, Today = 3)


i.e.

NewNumber = 7 - (today +  WeekDay)  

Above probably doesn't makes sense, but for demonstration of variables only.
I'd agree with two algorithms, one for every day from today until end of week, and one for the past days (i.e. if today =  Wednesday, #1 applies to wed-sun and #2 applies to mon-tue). 
What would be the correct algorithm for this matter?

Let these numbers represent the days (absolutely, by name):
$$\begin{array}{|c|c|}
\hline \text{weekday} & \text{#}\\
\hline\text{Monday} & 1 \\
\text{Tuesday} & 2 \\
\text{Wednesday} & 3 \\
\text{Thursday} & 4 \\
\text{Friday} & 5 \\
\text{Saturday} & 6 \\
\text{Sunday} & 7 \\\hline
\end{array}$$
And let the current day be denoted $T$ (from that table).
Then the algorithm you're looking for is $\boxed{f(d) = 1+(4+T+d)\%7}.$
For example, if today is Tuesday, $T = 2$. Say I wanted what day it was yesterday: then $f(1) = 1+(4+2+1)\%7 = 1+0 = 1$, which corresponds to Monday, the correct answer.
