Induction and
RecursionPart 2
1

Covering a Board with L-shaped
Trominoes
Prove that for any integer
if one corner square is removed
from a
checkerboard, the remaining squares can be
completely covered by L-shaped trominoes.
࠵?
≥
1
2
࠵?
×
2
࠵?

Proof by induction on n
Basis case:
Show that a
checkerboard minus a corner square can be
covered by a
trominoe.
Inductive Hypothesis: Assume that a
checkerboard minus a
corner can be completely covered by l-shaped trominoes.
2
1
×
2
1
2
࠵?
×
2
࠵?

Inductive Step: Show that a
checkerboard minus a corner
can be completely covered by trominoes. Consider an
checkerboard minus a corner.
2
࠵?
+1
×
2
࠵?
+1
2
࠵?
+1
×
2
࠵?
+1

Inductive Step: Show that a
checkerboard minus a corner
can be completely covered by trominoes. Consider an
checkerboard minus a corner.
Divide into four quadrants…use I.H. on each quadrant….
2
࠵?
+1
×
2
࠵?
+1
2
࠵?
+1
×
2
࠵?
+1

Inductive Step: Show that a
checkerboard minus a corner
can be completely covered by trominoes. Consider an
checkerboard minus a corner.
use
I.H. 4 times, different corners
2
࠵?
+1
×
2
࠵?
+1
2
࠵?
+1
×
2
࠵?
+1

Inductive Step: Show that a
checkerboard minus a corner
can be completely covered by trominoes. Consider an
checkerboard minus a corner.
4 times
plus
one
2
࠵?
+1
×
2
࠵?
+1
2
࠵?
+1
×
2
࠵?
+1

Combinations
Define
as the number of sets of size
that can be chosen
from a set of size .
Like the sum of the first
numbers, the argument seems better
than the proof. Let’s recall the argument for a formula for
.
(
n
k
)
k
n
n
(
n
k
)

Combinations - the argument
If we are looking to pick a subset of
elements from a set of size
, we have
choices to write down a first element,
choices
for the second element, and so on down to
choices for
the last element, giving us
permutations.
But, we are talking combinations. (i.e., choosing a set
of committee members, not ways of arranging of them around a
table). Any set of
members can be written in how many different
ways?
k
n
n
n
−
1
n
−
(
k
−
1)
n
⋅
(
n
−
1)
⋅
(
n
−
2)
⋅
…
⋅
(
n
−
(
k
−
1))
k

Combinations - the argument
If we are looking to pick a subset of
elements from a set of size
, we have
choices to write down a first element,
choices
for the second element, and so on down to
choices for
the last element, giving us
permutations.
But, we are talking combinations. (i.e., choosing a set
of committee members, not ways of arranging of them around a
table). Any set of
members can be written in how many different
ways?
k
n
n
n
−
1
n
−
(
k
−
1)
n
⋅
(
n
−
1)
⋅
(
n
−
2)
⋅
…
⋅
(
n
−
(
k
−
1))
k
We can write any of
members in the first position, any of the
remaining
in the second position, and so on, giving us
k
k
−
1
k
⋅
(
k
−
1)
⋅
(
k
−
2)
⋅
…
⋅
3
⋅
2
⋅
1

Combinations - the argument
If we are looking to pick a subset of
elements from a set of size
, we have
choices to write down a first element,
choices
for the second element, and so on down to
choices for
the last element, giving us
permutations.
But, we are talking combinations. (i.e., choosing a set

#### You've reached the end of your free preview.

Want to read all 72 pages?

- Spring '14