#`OK to post homework
#`Jiecheng Guo, 2020.09.14, Assignment 1`
#first question
diff(x^3,x);
3*x^2
sqrt(x+3);
(x+3)^(1/2)
p := (x+2)*(x+3);
p := (x+2)*(x+3)
expand(p);
x^2+5*x+6
#Question2
 	
F(6) = {[1,1,1,1,1,1], [2,1,1,1,1], [1,2,1,1,1], [1,1,2,1,1], [1,1,1,2,1], [1,1,1,1,2], [2,2,1,1], [2,1,2,1], [2,1,1,2], [1,2,2,1], [1,2,1,2], [1,1,2,2], [2,2,2]}

[[1, 1, 1, 1, 1, 1], [1, 1, 1, 1, 2], [1, 1, 2, 2], [2, 2, 2]] = {[2, 2, 2], [1, 1, 2, 2], [1, 2, 1, 2], [1, 2, 2, 1], [2, 1, 1, 2], [2, 1, 2, 1], [2, 2, 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]}

#question3
#1).L->[op(L),a]is a command that copy the original sequence L and add the new element a
#2).Since it is impossible to use positive steps to get negative coordinate 
#3).Because it is the original and there is only one way to get there, which is staying there, so it should be null
#question4
F:= proc(n) local W1,W2,w1,w2:
#input the number
option remember:
if n < 0 then
#if the imput is negative
	RETURN ({}):
fi:
#if the imput is 0
if n = 0 then
	RETURN ({[]}):
fi:
if n = 1 then
#if the imput is only one step
	RETURN ({[1]}):
fi:
#normal
W1 := F(n-1):
W2 := F(n-2):

{seq([op(w1),1],w1 in W1), seq([op(w2),2],w2 in W2)}:

end:
F(6)
{[2, 2, 2], [1, 1, 2, 2], [1, 2, 1, 2], [1, 2, 2, 1], [2, 1, 1, 2], [2, 1, 2, 1], [2, 2, 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]}
#question5
NuF:= proc(n) option remember:


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

if n <= 1 then 
	RETURN (1):
fi:

NuF(n-1) + NuF(n-2):

end:




NuF(1000);