Theoretical Computer Science Cheat Sheet Definitions Series ... - TUG

38.
40.
42.
44.
46.
48.
Theoretical Computer Science Cheat Sheet
Identities Cont.
[ ] n + 1
= ∑ [ ]( n k
n∑
[ k
n∑
] [ ]
= n
m + 1 k m)
m]
n−k 1 k x
n∑
⟨ ⟩( )
n x + k
= n!
, 39. =
,
k![
m
x − n k 2n
{
k
k=0
k=0
k=0
n
=
m}
∑ ( ){ }
[ n k + 1
n
(−1) n−k , 41. =
k m + 1
m]
∑ [ ]( )
n + 1 k
(−1) m−k ,
k + 1 m
k
k
{ }
m + n + 1
m∑
{ }
[ ]
n + k
m + n + 1
m∑
[ ] n + k
= k , 43.
= k(n + k) ,
m
k
m
k
(
k=0
k=0
n
=
m)
∑ { }[ ]
( n + 1 k n
(−1) m−k , 45. (n − m)! =
k + 1 m
m)
∑ [ ]{ }
n + 1 k
(−1) m−k , for n ≥ m,
k + 1 m
{ }
k
k
n
= ∑ ( )( )[ ] [ ]
m − n m + n m + k
n
, 47. = ∑ ( )( ){ }
m − n m + n m + k
,
n − m m + k n + k k
n − m m + k n + k k
{ }(
k
)
k
n l + m
= ∑ { }{ }( [ ]( )
k n − k n n l + m
, 49.
=
l + m l
l m k)
∑ [ ][ ]( k n − k n
.
l + m l
l m k)
k
k
Trees
Every tree with n
vertices has n − 1
edges.
Kraft inequality:
If the depths
of the leaves of
a binary tree are
d 1 , . . . , d n :
n∑
2 −d i
≤ 1,
i=1
and equality holds
only if every internal
node has 2
sons.
Master method:
T (n) = aT (n/b) + f(n), a ≥ 1, b > 1
If ∃ϵ > 0 such that f(n) = O(n log b a−ϵ )
then
T (n) = Θ(n log b a ).
If f(n) = Θ(n log b a ) then
T (n) = Θ(n log b a log 2 n).
If ∃ϵ > 0 such that f(n) = Ω(n log b a+ϵ ),
and ∃c < 1 such that af(n/b) ≤ cf(n)
for large n, then
T (n) = Θ(f(n)).
Substitution (example): Consider the
following recurrence
T i+1 = 2 2i · T 2 i , T 1 = 2.
Note that T i is always a power of two.
Let t i = log 2 T i . Then we have
t i+1 = 2 i + 2t i , t 1 = 1.
Let u i = t i /2 i . Dividing both sides of
the previous equation by 2 i+1 we get
t i+1 2i
=
2i+1 2 i+1 + t i
2 i .
Substituting we find
u i+1 = 1 2 + u i, u 1 = 1 2 ,
which is simply u i = i/2. So we find
that T i has the closed form T i = 2 i2i−1 .
Summing factors (example): Consider
the following recurrence
T (n) = 3T (n/2) + n, T (1) = 1.
Rewrite so that all terms involving T
are on the left side
T (n) − 3T (n/2) = n.
Now expand the recurrence, and choose
a factor which makes the left side “telescope”
Recurrences
1 ( T (n) − 3T (n/2) = n )
3 ( T (n/2) − 3T (n/4) = n/2 )
. . .
3 log n−1( 2
T (2) − 3T (1) = 2 )
Let m = log 2 n. Summing the left side
we get T (n) − 3 m T (1) = T (n) − 3 m =
T (n) − n k where k = log 2 3 ≈ 1.58496.
Summing the right side we get
m−1
∑
m−1
n ∑ (
2 i 3i = n 3
) i
2 .
i=0
i=0
Let c = 3 2
. Then we have
m−1
∑
( c
n c i m )
− 1
= n
c − 1
i=0
= 2n(c log 2 n − 1)
= 2n(c (k−1) log c n − 1)
= 2n k − 2n,
and so T (n) = 3n k − 2n. Full history recurrences
can often be changed to limited
history ones (example): Consider
∑i−1
T i = 1 + T j , T 0 = 1.
Note that
j=0
T i+1 = 1 +
Subtracting we find
T i+1 − T i = 1 +
= T i .
i∑
T j .
j=0
i∑ ∑i−1
T j − 1 −
j=0
And so T i+1 = 2T i = 2 i+1 .
j=0
T j
Generating functions:
1. Multiply both sides of the equation
by x i .
2. Sum both sides over all i for
which the equation is valid.
3. Choose a generating function
G(x). Usually G(x) = ∑ ∞
i=0 xi g i .
3. Rewrite the equation in terms of
the generating function G(x).
4. Solve for G(x).
5. The coefficient of x i in G(x) is g i .
Example:
g i+1 = 2g i + 1, g 0 = 0.
Multiply and sum:
∑
i≥0
g i+1 x i = ∑ i≥0
2g i x i + ∑ i≥0
x i .
We choose G(x) = ∑ i≥0 xi g i . Rewrite
in terms of G(x):
G(x) − g 0
= 2G(x) + ∑ x i .
x
i≥0
Simplify:
G(x)
= 2G(x) + 1
x
1 − x .
Solve for G(x):
x
G(x) =
(1 − x)(1 − 2x) .
Expand this ( using partial fractions:
2
G(x) = x
1 − 2x − 1 )
1 − x
⎛
⎞
= x ⎝2 ∑ 2 i x i − ∑ x i ⎠
i≥0 i≥0
So g i = 2 i − 1.
= ∑ i≥0(2 i+1 − 1)x i+1 .