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# Name:Treasa Bency Biju Jose
# Date: 09-13-2020
# Assignment #1

I. 
diff(x^3, x);
                                 2
                              3 x 

5!;
                              120

567^(1/3);
                               (1/3)
                            567     

Matrix(2, 2, [[2, 3], [6, 5]]);
                             [2  3]
                             [6  5]

II.
# F(6): {[6], [1, 5], [2, 4], [3, 3], [4, 2], [5, 1], 
# [1, 1, 4], [1, 2, 3], [1, 3, 2], [1, 4, 1], [2, 1, 3], [2, 2, 2], [2, 3, 1], [3, 1, 2], [3, 2, 1], [4, 1, 1], 
# [1, 1, 1, 3], [1, 1, 2, 2], [1, 1, 3, 1], [1, 2, 1, 2], [1, 2, 2, 1], [1, 3, 1, 1], [2, 1, 1, 2], [2, 1, 2, 1], [2, 2, 1, 1], [3, 1, 1, 1], 
# [1, 1, 1, 1, 2], [1, 1, 1, 2, 1], [1, 1, 2, 1, 1], [1, 2, 1, 1, 1], [2, 1, 1, 1, 1], 
# [1, 1, 1, 1, 1, 1]}

III.
L1 := [Mercury, Venus, Earth];
                 L1 := [Mercury, Venus, Earth]

L2 := [op(L1), Mars];
              L2 := [Mercury, Venus, Earth, Mars]


# The reason we use an empty set {} when m or n are negative is because, when they are negative there's 
# no place to walk and hence the set is negative.
# We declare set {[]} when m and n are 0 because its an empty walk which is also considered a walk. 


IV.
F:=proc(n) local S,S1,i,s:
option remember:

if n<0 then
 RETURN({}):
fi:

if n=0 then
 RETURN({[]}):
fi:

S:={}:

for i from 1 to n do
 S1:=F(n-i):
 S:=S union {seq( [op(s),i], s in S1)}:
od:

S:
end:

F(6);
  {[6], [1, 5], [2, 4], [3, 3], [4, 2], [5, 1], [1, 1, 4], 

    [1, 2, 3], [1, 3, 2], [1, 4, 1], [2, 1, 3], [2, 2, 2], 

    [2, 3, 1], [3, 1, 2], [3, 2, 1], [4, 1, 1], [1, 1, 1, 3], 

    [1, 1, 2, 2], [1, 1, 3, 1], [1, 2, 1, 2], [1, 2, 2, 1], 

    [1, 3, 1, 1], [2, 1, 1, 2], [2, 1, 2, 1], [2, 2, 1, 1], 

    [3, 1, 1, 1], [1, 1, 1, 1, 2], [1, 1, 1, 2, 1], 

    [1, 1, 2, 1, 1], [1, 2, 1, 1, 1], [2, 1, 1, 1, 1], 

    [1, 1, 1, 1, 1, 1]}




V.
NuF := n -> nops(F(n));
  NuF := proc (n) options operator, arrow; nops(F(n)) end proc

NuF(6);
                               32

nops(F(6));
                               32

NuF(1000) and nops(F(1000)) are really long codes and take long time to evaluate