Counting Strategies
Mohammad Reza Karimi
Department of Computer Engineering
Sharif University of Technology
[email protected]
March 15, 2014
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Addition, Multiplication
.
Example
.
Determine the number of squares with all their vertices belonging to
the
. 10 × 10 array of lattice points.
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M. R. Karimi (SUT)
Counting Strategies
March 2014
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Addition, Multiplication
.
Example
.
Determine the number of squares with all their vertices belonging to
the
. 10 × 10 array of lattice points.
Solution idea: each “rotated square” can be inscribed in a horizontal
square.
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Counting Strategies
March 2014
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Addition, Multiplication
.
Example
.
Determine the number of squares with all their vertices belonging to
the
. 10 × 10 array of lattice points.
Solution idea: each “rotated square” can be inscribed in a horizontal
square. Final answer:
9
∑
(10 − k)2 · k
k=1
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Counting Strategies
March 2014
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Addition, Multiplication
.
Example
.
Determine the number of squares with all their vertices belonging to
the
. 10 × 10 array of lattice points.
Solution idea: each “rotated square” can be inscribed in a horizontal
square. Final answer:
9
∑
(10 − k)2 · k
k=1
.
Example
.
There are n sticks of length 1, 2, . . . , n. How many incongruent
triangles
can be formed by using three of the given sticks?
.
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Counting Strategies
March 2014
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Addition, Multiplication
.
Example
.
Determine the number of squares with all their vertices belonging to
the
. 10 × 10 array of lattice points.
Solution idea: each “rotated square” can be inscribed in a horizontal
square. Final answer:
9
∑
(10 − k)2 · k
k=1
.
Example
.
There are n sticks of length 1, 2, . . . , n. How many incongruent
triangles
can be formed by using three of the given sticks?
.
Solution idea: Triangle inequality.
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Counting Strategies
March 2014
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Grouping
.
Example
.
Determine the number of seven-letter codes such that
(i) no letters are repeated in the code; and
(ii)
. letters a and b are not next to each other.
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Counting Strategies
March 2014
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Grouping
.
Example
.
Determine the number of seven-letter codes such that
(i) no letters are repeated in the code; and
(ii)
. letters a and b are not next to each other.
Solution idea: Group a and b into a pack. Find complement answer.
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Counting Strategies
March 2014
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Combinations
.
Example
.
Twenty-five of King Arthur’s knights are seated at their customary
round table. Three of them are chosen- all choices of three being
equally likely-and are sent off to slay a troublesome dragon. What is
the probability that at least two of the three were seated next to each
other?
.
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Counting Strategies
March 2014
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Combinations
.
Example
.
Twenty-five of King Arthur’s knights are seated at their customary
round table. Three of them are chosen- all choices of three being
equally likely-and are sent off to slay a troublesome dragon. What is
the probability that at least two of the three were seated next to each
other?
.
Answer 1: n + n(n − 4) = n(n − 3)
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Counting Strategies
March 2014
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Combinations
.
Example
.
Twenty-five of King Arthur’s knights are seated at their customary
round table. Three of them are chosen- all choices of three being
equally likely-and are sent off to slay a troublesome dragon. What is
the probability that at least two of the three were seated next to each
other?
.
Answer 1: n + n(n − 4) = n(n − 3)
Answer 2: n(n − 2) − n = n(n − 3)
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Counting Strategies
March 2014
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Combinations
.
Example
.
Twenty-five of King Arthur’s knights are seated at their customary
round table. Three of them are chosen- all choices of three being
equally likely-and are sent off to slay a troublesome dragon. What is
the probability that at least two of the three were seated next to each
other?
.
Answer 1: n + n(n − 4) = n(n − 3)
Answer 2: n(n − 2) − n = n(n − 3)
Answer 3: n(n − 3) !!!
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Counting Strategies
March 2014
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Combinations
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Example
.
In how many ways can one arrange 5 indistinguishable armchairs and 5
indistinguishable armless chairs around a circular table? Two
arrangements are considered the same if one can be obtained from the
other
by rotation.
.
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Counting Strategies
March 2014
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Combinations
.
Example
.
In how many ways can one arrange 5 indistinguishable armchairs and 5
indistinguishable armless chairs around a circular table? Two
arrangements are considered the same if one can be obtained from the
other
by rotation.
.
Problem translation: (1 ≤ k ≤ 9)
{r1 , . . . , r5 } ≡ {r1 + k, . . . , r5 + k} (mod 10)
⇒ 5k ≡ 0 (mod 10)
⇒ k ≡ {2, 4, 6, 8} (mod 10)
⇒ {r1 , . . . , r5 } = {1, 3, 5, 7, 9} or {2, 4, 6, 8, 10}
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Counting Strategies
March 2014
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Combinations
.
Example
.
In how many ways can one arrange 5 indistinguishable armchairs and 5
indistinguishable armless chairs around a circular table? Two
arrangements are considered the same if one can be obtained from the
other
by rotation.
.
Problem translation: (1 ≤ k ≤ 9)
{r1 , . . . , r5 } ≡ {r1 + k, . . . , r5 + k} (mod 10)
⇒ 5k ≡ 0 (mod 10)
⇒ k ≡ {2, 4, 6, 8} (mod 10)
⇒ {r1 , . . . , r5 } = {1, 3, 5, 7, 9} or {2, 4, 6, 8, 10}
Final answer:
(10)
5
M. R. Karimi (SUT)
−2 2
+
10
2
Counting Strategies
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Combinatorical Identities
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Theorem (Chu Shih-Chieh’s Identity)
.
(a) For all r, n ∈ N, n ≥ r we have
( ) (
)
( ) (
)
r
r+1
n
n+1
+
+ ··· +
=
r
r
r
r+1
(a) For all r, k ∈ N we have
( ) (
)
(
) (
)
r
r+1
r+k
r+k+1
+
+ ··· +
=
0
1
k
k
.
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Counting Strategies
March 2014
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Bijections
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Example
.
Let n be an integer n ≥ 2, and define the sequence S = (1, 2, . . . , n). A
sub-sequence of S is called arithmetic if it has at least two terms and it
is an arithmetic progression. An arithmetic sub-sequence is called
maximal if this progression cannot be lengthened by the inclusion of
another element of S. Determine the number of maximal arithmetic
.sub-sequences.
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Counting Strategies
March 2014
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Bijections
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Example
.
Let n be an integer n ≥ 2, and define the sequence S = (1, 2, . . . , n). A
sub-sequence of S is called arithmetic if it has at least two terms and it
is an arithmetic progression. An arithmetic sub-sequence is called
maximal if this progression cannot be lengthened by the inclusion of
another element of S. Determine the number of maximal arithmetic
.sub-sequences.
Solution outline: Each arithmetic progression can be distinguished by a
pair (a, d), where a is the first element of this progression, and d is
difference between two consecutive elements.
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Counting Strategies
March 2014
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Bijections
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Example
.
Let n be an integer n ≥ 2, and define the sequence S = (1, 2, . . . , n). A
sub-sequence of S is called arithmetic if it has at least two terms and it
is an arithmetic progression. An arithmetic sub-sequence is called
maximal if this progression cannot be lengthened by the inclusion of
another element of S. Determine the number of maximal arithmetic
.sub-sequences.
Solution outline: Each arithmetic progression can be distinguished by a
pair (a, d), where a is the first element of this progression, and d is
difference between two consecutive elements.
1 ≤ a ≤ ⌊ n2 ⌋ and a ≤ d ≤ n − a. So the answer is
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Counting Strategies
March 2014
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Bijections
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Example
.
Let n be an integer n ≥ 2, and define the sequence S = (1, 2, . . . , n). A
sub-sequence of S is called arithmetic if it has at least two terms and it
is an arithmetic progression. An arithmetic sub-sequence is called
maximal if this progression cannot be lengthened by the inclusion of
another element of S. Determine the number of maximal arithmetic
.sub-sequences.
Solution outline: Each arithmetic progression can be distinguished by a
pair (a, d), where a is the first element of this progression, and d is
difference between two consecutive elements.
2
1 ≤ a ≤ ⌊ n2 ⌋ and a ≤ d ≤ n − a. So the answer is ⌊ n4 ⌋
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Counting Strategies
March 2014
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Bijections
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Theorem
.
Combinatorial Equation Number of integer solutions to the equation
x1 + x2 + · · · + xk = n,
equals to
.
xi ≥ 0
(
)
n+k−1
k−1
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Counting Strategies
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Bijections
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Example
.
In how many ways we can make a “suitable” parentheses configuration
with
n ‘(’s and n ‘)’s? AKA Catalan Number
.
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Counting Strategies
March 2014
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Bijections
.
Example
.
In how many ways we can make a “suitable” parentheses configuration
with
n ‘(’s and n ‘)’s? AKA Catalan Number
.
Solution idea: problem ⇔ Path from (0, 0) to (2n, 0) with moves to NE
and SE, not going below y = 0 line.
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Counting Strategies
March 2014
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Bijections
.
Example
.
In how many ways we can make a “suitable” parentheses configuration
with
n ‘(’s and n ‘)’s? AKA Catalan Number
.
Solution idea: problem ⇔ Path from (0, 0) to (2n, 0) with moves to NE
and SE, not going below y = 0 line.
Make each “bad” path to a “general” path (in a (n − 1) × (n + 1) grid)
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Counting Strategies
March 2014
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Bijections
.
Example
.
In how many ways we can make a “suitable” parentheses configuration
with
n ‘(’s and n ‘)’s? AKA Catalan Number
.
Solution idea: problem ⇔ Path from (0, 0) to (2n, 0) with moves to NE
and SE, not going below y = 0 line.
Make each “bad”
to )a “general”
(2n)path
( 2n
(2n) path (in a (n − 1) × (n + 1) grid)
1
Final answer: n − n−1 = n+1 n
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Counting Strategies
March 2014
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Recursion
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Example
.
Give a recursive representation of the previous problem, then prove the
recursion
formula by induction (you already know the answer!)
.
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Counting Strategies
March 2014
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Recursion
.
Example
.
Give a recursive representation of the previous problem, then prove the
recursion
formula by induction (you already know the answer!)
.
Solution: Let Cn be the answer. We have (the first intersection with
y = 0 is at 2k)
n
∑
Cn =
Ck Cn−k
k=1
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Counting Strategies
March 2014
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Recursion
.
Example
.
Give a recursive representation of the previous problem, then prove the
recursion
formula by induction (you already know the answer!)
.
Solution: Let Cn be the answer. We have (the first intersection with
y = 0 is at 2k)
n
∑
Cn =
Ck Cn−k
k=1
Induction, too hard to solve!
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Counting Strategies
March 2014
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Happy New Year!!
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Counting Strategies
March 2014
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