set Given S, assume each s S has a weight |s|. A GF of S is F ( z) = S an = {s S: s = n} , F ( z) = an z n S NOTE: Since S = a n , S = FS (1) .
n n
s S
z
s
. Letting
Ex: In how many ways can we give money using 1 coin? (z + z 5 + z 10 + z 25 ) [z 5 ]( z + z 5 + z 10 + z 25 ) =1 [z 6 ]( z + z 5 + z 10 + z 25 ) = 0 Ex: In how many ways can we give money using 2 coins (order doesnt matter)? (z + z 5 + z 10 + z 25 ) (z + z 5 + z 10 + z 25 ) No! pWhy not? [z 6 ]( z + z 5 + z 10 + z 25 )( z + z 5 + z 10 + z 25 ) = 2 Ex: Need to separate 2 kinds of data: # of ways & resource consumed. Thus, put 1 datum in coefficient & 1 in exponent. (Polya) In how many ways can we make change for 50 using an arbitrary number of coins? Order doesnt matter. In how many ways can we leave pennies for k? 1 P = 0 + p + pp + ppp + ... =1 + z + z 2 + = z6 P =1 1-z If we can spend pennies & nickels? 1 N = P + nP + nnP + ... = ( z 0 + z 5 + z 10 + )P = P z6 N = 2 1 - z5 If we can spend pennies, nickels, dimes & quarters? 1 D = N + dN + ddN + ... = ( z 0 + z 10 + z 20 + ) N = N z6 D = 2 1 - z 10 1 Q = D + qD + qqD + ... (z 0 + z 25 + z 50 + )D = z6 Q = 2 25 D 1-z 1 1 1 1 Answer is [z 50 ] . Q is generating function of the # of ways to 1 - z 1 - z 5 1 - z 10 1 - z 25 make change for n. pWhat if we have 3 pennies, 2 nickels & a dime? 1+ z + z 2 + z 3 1+ z 5 + z 10 1+ z 10 1 1 pWhat if we have green & blue pennies? P = 1-z1-z 1 pWhy not P( z ) = ? 1 - 2z p Suppose we have n distinguishable objects labeled x1,,xn. In how many ways can we 0 take a subset of k of these objects? For the jth object, z + x j z = 1 + x j z . Ignoring k,
[ ] [ ] [ ] [ ]
(
)(
)(
)
were interested in
0 n j 1 j
(1 + x z ) , GF for ways to choose a subset from the objects. [z ] (1 + x z ) =1 [z ] (1 + x z ) = x + + x If we want the number of ways to
n j 1 j n j 1 j 1 n n j 1 k n k
(
) (
)
choose the objects,
(1+ z) = (1+ z) . [z ](1+ z)
n
n
n = n = . Final answer is 1
[z ](1+ z) = [z ] j z
j
n
j
n = k
LN5-1
Ex: Assume a set of dice is really fair if probabilities of each event {2,,12} are equal (uniform distribution). 1 k 1 2 z2 z 2 1 - z 11 P( z ) = z = ( z + + z 12 ) = (1 + + z 10 ) = 11 11 1 - z 11 12 k 2 11 6 The 2 dice have PGFs R( z ) = r1 z + + r6 z and S( z ) = s1 z + + s6 z 6 . P( z ) = R( z )S( z ) = (r1 z + + r6 z 6 )( s1 z + + s6 z 6 ) = z 2 ( r1 + + r6 z 5 )( s1 + + s6 z 5 ) . 1 1 - z 11 = (r1 + + r6 z 5 )( s1 + + s6 z 5 ) Leftside has 5 pairs of complex conjugate 11 1 - z roots e , 1k5. Rightside has 2 real roots. Hence, impossible. Using generating functions to solve recurrences: -Create a power series in z (assume tn=0 for n<0) G(z)=t0+t1z +t2z2 + ... -Convert the power series to closed form -Manipulate the closed form so it's a sum of simple functions with known expansions -Convert back to power series & match terms 5tn1 6tn2 if n > 1 Ex: tn = 7 if n = 1 1 if n = 0 As a general recurrence, tn=5tn-1 - 6tn-2 This is fine for n>1 and n<0. tn =5tn-1 - 6tn-2 +[n=0] Fine except for n=1. tn =5tn-1- 6tn-2 + [n=0] +2[n=1]. - For any n, tn zn=5tn-1zn - 6tn-2zn + [n=0]zn +2[n=1]zn
2 ipk 11
tn z n = 5tn1z n 6tn2 z n + 2[n = 1]z n + [n = 0]z n
n n n n n n
Letting G(z) = tn z n yields
n n
G(z ) = 5z tn -1 z n -1 - 6z 2 tn -2 z n -2 + 2z +1 = 5zG(z )- 6z 2 G(z )+ 2z +1 2z + 1 1 5z + 6z 2 2z + 1 a b -Noting that this is close to a closed form + after solving for 2 = 1 3z 1 2z 1 5z + 6z a, b, by a +b =1,-2az -3bz =2z , we solve for a =5, 4 5 b =-4 and we rewrite G(z) = 1 3z 1 2z 1 - Noting that = 1 + qz + q 2 z 2 +...our solution is tn = 5*3n - 4*2n 1 qz In doing step 3, we use the method partial of fractions which says (in weakened form) ck, j 1 that we can find ck,j to solve = m j (ak z + bk ) k 1kn 1 jmk (ak z + bk ) - Regrouping yields the closed form G(z) =
1kn
LN5-2
Some common generating functions sequence G.F. 1 1,1,1,1,... 1 z 1 1,-1,1,-1,... 1+ z 1 1,0,1,0,... 1 z2 1 1,2,3,4,5,... (1 z)2 1 1,k ,k 2,k 3,... 1 kz k k 1,k, , ... (1+z )k 2 3 0,1,1/2,1/3,1/4,... 1 ln 1 z
tn 1 (-1)n [2|n] n+1 kn (1 + z)k =
0 j
j z j 1k j
k
1 [n 1] n 1 1/0!,1/1!,1/2!,1/3!,... ez n! (Flajolet) Admissible constructions THEOREM : If FA(z) and GB(z) are GFs for the sets A and B, then if A & B are disjoint, then FA ( z) + G B ( z) is the GF for A B, FA ( z)G B (z) is the GF for A B {( a,b) | a A b B} , 1 is the GF for all sequences of elements of A. 1- FA ( z) PROOF: Let FA ( z) = an z n and G B( z) = gn z n . Then FA ( z) + G B ( z) = a n z + bn z = ( an + bn ) z n =
n n n n n n k n -k n
FA ( z)G B (z) = a k bn -k z z
n k
= ak z k bn -k z n -k =
n k
s A B
z
s
.
a,b )
a =k b =n -k
( a,b ) A B
z (
.
1 2 3 = 1+ FA ( z) + FA ( z) + FA (z) + ... counts the classes U A k . 1- FA ( z) k 0 EX: How many possible k letter words in English? k=1 F1(z)=26z to get all 1 1 sequences F( z) = = = 26n z n . 1- F1 ( z) 1- 26z n 0 EX: A composition of an integer n is a sequence a1,,ak such that k1, ai1, 1ik, and ai = n (order doesnt matter). Let fn be the number of compositions of n. f0=0, f1=1,
k i 1
f2=2. We want F(z), a GF for the fn. Try the SUBROBLEM s: z k=1 F1 ( z) = 1- z
LN5-3
z k=2 F2 ( z) = . 1- z Arbitrary but fixed k, Fk ( z) = z . 1- z 1 1 z -1 = -1 = z 1 - F1 ( z) 1 - 2z 11-z
k
2
By the third case of the above theorem, F( z ) =
0,if n = 0 . yielding fn = n -1 2 ,if n > 0 EX: How many full (every node has degree 0 or 2) binary trees of n internal nodes (n+1 leaves) are there? B(z)= bn z n =1+ z + 2z2 + 5z3 + 14z4 + 42z5 + ...
n 2 B( z) = zB( z) B( z) +1 zB ( z) - B( z) +1 = 0. Solving for B(z): B(z) =
1 1- 4z 2z
Which root? B(0)=1negative root so B(z) = express
1- 1- 4z . To extract bn, we need to 2z
1- 1- 4z as a power series. 2z 1 / 2 (-4 z) n - 1 / 2(-4z) n 1- 1/ 2 1- (1- 4z) n n n 0 = = n 1 . 2z ...