#ATTENDANCE QUIZ FOR LECTURE 23 of Dr. Z.'s Math454(02) Rutgers University

# Please Edit this .txt page with Answers

#Email  ShaloshBEkhad@gmail.com 
#Subject: p23
#with an attachment called
#p23FirstLast.txt
#(e.g. p23DoronZeilberger.txt)

#Right after finishing watching the lecture but no later than Dec. 4, 2020, 8:00pm


THE NUMBER OF ATTENDANCE QUESTIONS WERE:3

PLEASE LIST ALL THE QUESTIONS FOLLOWED, AFTER EACH, BY THE ANSWER

#1. Use Lagrange inversion formula to find an explict express for the coeff. of x^n
#in the power series of u{x) that satisfies the functional eq. 
#i) u(x) = x*(1+u(x))^2
#ii) u(x) = x*(1+u(x))^3
#iii) for any k, pos. integer, u(x) = x*(1+u(x))^k

#i) u(x) = x*(1+u(x))^2 = x*PHI(u(x)) -> to convert this into PHI(z), PHI(z) should have formula (1+z)^2
LIF((1 + z)^2, z, 10);
         [1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796]
ifactor(%);
[                                      2                          
[1, (2), (5), (2) (7), (2) (3) (7), (2)  (3) (11), (3) (11) (13), 

                                            2               ]
  (2) (5) (11) (13), (2) (11) (13) (17), (2)  (13) (17) (19)]

#this is actually catalan number from previous lectures(A000108)

#the formula for this seq is then, binom(2n,n)/n+1 = (2*n)!/(n!*(n + 1)!)

#ii)By same methodology,

LIF((1 + z)^3, z, 10);
    [1, 3, 12, 55, 273, 1428, 7752, 43263, 246675, 1430715]

#which is equivalent to binom(3n,n)/(2n+1)

iii)so, from the formula from i) and ii) we now now that

for k where (1+z)^k as PHI(z),

the explicit formula is binom(kn,n)/((k-1)n+1)

########################################################################################

#2.
#L:=[13,7,15,10,13,16,6,2,11,12,4,6,5,11,16,1]
#
#f(1)=13,f(2)=7,...
#(i)draw this directed graph
#
# 1  2  3 4  5  6  7 8  9 10 11 12 13  14  15   16
#[13,7,15,10,13,16,6,2,11,12,4,  6, 5, 11, 16,  1]
#
#    14
#     |
#     +
# 9->11->4->10->12
#                |
#                +  
#      8-> 2->7->6->16->1->13<->5
#                   +   
#                   | 
#                3->15
#
#(ii)find the Doubly-rotted tree that is outputted by the joyal bijection
#there is only one cycle (13 5)
#
#f(5) = 13, f(13) = 5
#sorting,
#5 15
#13 5 <- our main stem for D.R tree.
#
#
#  13* -> 5**
#  |
#  1 - 16 - 6 - 7 - 2 - 8
#      |    |
#     15    12 - 10 -4 -11- 9
#      |                 |
#      3                14
#
#this is very much same as Joyal(L). SO its correct

##########################################################################################

#3. Finish the Pruffer from lecture

#Starting from the leftoff part
#
#
#
#                     2 - 13
#                     |
# 9 -> 8 -> 7 -> 6 -> 3 -> 10 -> 11
#                |
#                12
#
#a4=8 code so far [2,4,11,8]
#
#                     2 - 13
#                     |
#      8 -> 7 -> 6 -> 3 -> 10 -> 11
#                |
#                12
#
#a5=7 code so far [2,4,11,8,7]
#
#                     2 - 13
#                     |
#           7 -> 6 -> 3 -> 10 -> 11
#                |
#                12
#
#a6=6 code so far [2,4,11,8,7,6]
#
#                     2 - 13
#                     |
#                6 -> 3 -> 10 -> 11
#                |
#                12
#
#a7=10 code so far [2,4,11,8,7,6,10]
#
#                     2 - 13
#                     |
#                6 -> 3 -> 10
#                |
#                12
#
#a8=3 code so far [2,4,11,8,7,6,10,3]
#
#                     2 - 13
#                     |
#                6 -> 3
#                |
#                12
#
#a9=6 code so far [2,4,11,8,7,6,10,3,6] 
#but from here its one way starting from 12 because 13 is the biggest here. so following codes will be 
# [2,4,11,8,7,6,10,3,6,3,2] -> length 11
#correct.