Post on 02-Feb-2020
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Distribution of Missing Sums in Sumsets
Oleg Lazarev, Princeton UniversityAdvisor: Steven Miller, Williams College
SMALL Research Program 2011, Williams College
2011 Young Mathematicians ConferenceThe Ohio State University, August 21, 2011
1
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Background
Let A ⊆ N ∪ {0}.
DefinitionSumset: A + A = {x + y : x , y ∈ A}Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}
Example: if A = {1,2,5}, then
A + A = {2,3,4,6,7,10}
Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.
Key Question: What is the structure of A + A?
2
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Background
Let A ⊆ N ∪ {0}.
DefinitionSumset: A + A = {x + y : x , y ∈ A}
Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}
Example: if A = {1,2,5}, then
A + A = {2,3,4,6,7,10}
Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.
Key Question: What is the structure of A + A?
3
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Background
Let A ⊆ N ∪ {0}.
DefinitionSumset: A + A = {x + y : x , y ∈ A}Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}
Example: if A = {1,2,5}, then
A + A = {2,3,4,6,7,10}
Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.
Key Question: What is the structure of A + A?
4
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Background
Let A ⊆ N ∪ {0}.
DefinitionSumset: A + A = {x + y : x , y ∈ A}Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}
Example: if A = {1,2,5}, then
A + A = {2,3,4,6,7,10}
Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.
Key Question: What is the structure of A + A?
5
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Background
Let A ⊆ N ∪ {0}.
DefinitionSumset: A + A = {x + y : x , y ∈ A}Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}
Example: if A = {1,2,5}, then
A + A = {2,3,4,6,7,10}
Why study sumsets?
Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.
Key Question: What is the structure of A + A?
6
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Background
Let A ⊆ N ∪ {0}.
DefinitionSumset: A + A = {x + y : x , y ∈ A}Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}
Example: if A = {1,2,5}, then
A + A = {2,3,4,6,7,10}
Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.
Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.
Key Question: What is the structure of A + A?
7
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Background
Let A ⊆ N ∪ {0}.
DefinitionSumset: A + A = {x + y : x , y ∈ A}Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}
Example: if A = {1,2,5}, then
A + A = {2,3,4,6,7,10}
Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.
Key Question: What is the structure of A + A?
8
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Background
Let A ⊆ N ∪ {0}.
DefinitionSumset: A + A = {x + y : x , y ∈ A}Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}
Example: if A = {1,2,5}, then
A + A = {2,3,4,6,7,10}
Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.
Key Question: What is the structure of A + A?9
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets
Consider finite A ⊆ [0,n− 1] chosen randomly with uniformdistribution from all subsets of [0,n − 1].
Question: What is the structure of A + A for such A? whatthe distribution of |A + A| for such A?
Theorem: Martin, O’Bryant
E|A + A| = 2n − 1− 10 + O((3/4)n/2)
Theorem: ZhaoFor each fixed k , P(A ⊆ [0,n − 1] : |A + A| = 2n − 1− k) has alimit as n→∞
Note: Both theorem can be more naturally stated in terms ofmissing sums (independent of n).
10
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets
Consider finite A ⊆ [0,n− 1] chosen randomly with uniformdistribution from all subsets of [0,n − 1].Question: What is the structure of A + A for such A? whatthe distribution of |A + A| for such A?
Theorem: Martin, O’Bryant
E|A + A| = 2n − 1− 10 + O((3/4)n/2)
Theorem: ZhaoFor each fixed k , P(A ⊆ [0,n − 1] : |A + A| = 2n − 1− k) has alimit as n→∞
Note: Both theorem can be more naturally stated in terms ofmissing sums (independent of n).
11
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets
Consider finite A ⊆ [0,n− 1] chosen randomly with uniformdistribution from all subsets of [0,n − 1].Question: What is the structure of A + A for such A? whatthe distribution of |A + A| for such A?
Theorem: Martin, O’Bryant
E|A + A| = 2n − 1− 10 + O((3/4)n/2)
Theorem: ZhaoFor each fixed k , P(A ⊆ [0,n − 1] : |A + A| = 2n − 1− k) has alimit as n→∞
Note: Both theorem can be more naturally stated in terms ofmissing sums (independent of n).
12
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets
Consider finite A ⊆ [0,n− 1] chosen randomly with uniformdistribution from all subsets of [0,n − 1].Question: What is the structure of A + A for such A? whatthe distribution of |A + A| for such A?
Theorem: Martin, O’Bryant
E|A + A| = 2n − 1− 10 + O((3/4)n/2)
Theorem: ZhaoFor each fixed k , P(A ⊆ [0,n − 1] : |A + A| = 2n − 1− k) has alimit as n→∞
Note: Both theorem can be more naturally stated in terms ofmissing sums (independent of n).
13
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets
Consider finite A ⊆ [0,n− 1] chosen randomly with uniformdistribution from all subsets of [0,n − 1].Question: What is the structure of A + A for such A? whatthe distribution of |A + A| for such A?
Theorem: Martin, O’Bryant
E|A + A| = 2n − 1− 10 + O((3/4)n/2)
Theorem: ZhaoFor each fixed k , P(A ⊆ [0,n − 1] : |A + A| = 2n − 1− k) has alimit as n→∞
Note: Both theorem can be more naturally stated in terms ofmissing sums (independent of n).
14
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets, Continued
Why is the expectation so high? E |A + A| ∼ 2n − 11.
Main characteristic of typical A + A: middle is full.Many ways to write middle elements as sums
50 100 150 200
5
10
15
20
25
Figure: Comparison of predicted and observed number ofrepresentations of possible elements of the sumset, Miller
Key fact: if k < n, then P(k 6∈ A + A) ∼(3
4
)k/2
15
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets, Continued
Why is the expectation so high? E |A + A| ∼ 2n − 11.Main characteristic of typical A + A: middle is full.
Many ways to write middle elements as sums
50 100 150 200
5
10
15
20
25
Figure: Comparison of predicted and observed number ofrepresentations of possible elements of the sumset, Miller
Key fact: if k < n, then P(k 6∈ A + A) ∼(3
4
)k/2
16
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets, Continued
Why is the expectation so high? E |A + A| ∼ 2n − 11.Main characteristic of typical A + A: middle is full.Many ways to write middle elements as sums
50 100 150 200
5
10
15
20
25
Figure: Comparison of predicted and observed number ofrepresentations of possible elements of the sumset, Miller
Key fact: if k < n, then P(k 6∈ A + A) ∼(3
4
)k/2
17
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Structure of Random Sets, Continued
Why is the expectation so high? E |A + A| ∼ 2n − 11.Main characteristic of typical A + A: middle is full.Many ways to write middle elements as sums
50 100 150 200
5
10
15
20
25
Figure: Comparison of predicted and observed number ofrepresentations of possible elements of the sumset, Miller
Key fact: if k < n, then P(k 6∈ A + A) ∼(3
4
)k/2
18
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
New Results
Theorem: Bounds on the distribution (Lazarev, 2011)
0.70k � P(A + A has k missing sums)� 0.80k
From numerical data, conjecture that
P(A + A has k missing sums) ∼ 0.78k
and Log(P(A + A has k missing sums)) is eventually linear:
Figure: Log P(k missing sums) for 5 · 108 trial A of size 120
19
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
New Results
Theorem: Bounds on the distribution (Lazarev, 2011)
0.70k � P(A + A has k missing sums)� 0.80k
From numerical data, conjecture that
P(A + A has k missing sums) ∼ 0.78k
and Log(P(A + A has k missing sums)) is eventually linear:
Figure: Log P(k missing sums) for 5 · 108 trial A of size 120
20
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
New Results
Theorem: Bounds on the distribution (Lazarev, 2011)
0.70k � P(A + A has k missing sums)� 0.80k
From numerical data, conjecture that
P(A + A has k missing sums) ∼ 0.78k
and Log(P(A + A has k missing sums)) is eventually linear:
Figure: Log P(k missing sums) for 5 · 108 trial A of size 120
21
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
New Results
Theorem: Bounds on the distribution (Lazarev, 2011)
0.70k � P(A + A has k missing sums)� 0.80k
From numerical data, conjecture that
P(A + A has k missing sums) ∼ 0.78k
and Log(P(A + A has k missing sums)) is eventually linear:
Figure: Log P(k missing sums) for 5 · 108 trial A of size 120
22
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
More Results
Theorem: Variance (Lazarev, 2011)Exact series for the variance (too long to fit on page...);also Var ∼ 35.98.
Theorem: Asymptotic formula for distribution ofconsecutive missing sums (Lazarev, 2011)
(12
)(k+i)/2
� P(A+A is missing [k , k+i])�(
12
)(k+i)/2
(1+εi)k
Key Obstacle: Dependent random variablesKey Approach: Graph theory! (and Fibonacci Numbers?)
23
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
More Results
Theorem: Variance (Lazarev, 2011)Exact series for the variance (too long to fit on page...);also Var ∼ 35.98.
Theorem: Asymptotic formula for distribution ofconsecutive missing sums (Lazarev, 2011)
(12
)(k+i)/2
� P(A+A is missing [k , k+i])�(
12
)(k+i)/2
(1+εi)k
Key Obstacle: Dependent random variablesKey Approach: Graph theory! (and Fibonacci Numbers?)
24
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
More Results
Theorem: Variance (Lazarev, 2011)Exact series for the variance (too long to fit on page...);also Var ∼ 35.98.
Theorem: Asymptotic formula for distribution ofconsecutive missing sums (Lazarev, 2011)
(12
)(k+i)/2
� P(A+A is missing [k , k+i])�(
12
)(k+i)/2
(1+εi)k
Key Obstacle: Dependent random variables
Key Approach: Graph theory! (and Fibonacci Numbers?)
25
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
More Results
Theorem: Variance (Lazarev, 2011)Exact series for the variance (too long to fit on page...);also Var ∼ 35.98.
Theorem: Asymptotic formula for distribution ofconsecutive missing sums (Lazarev, 2011)
(12
)(k+i)/2
� P(A+A is missing [k , k+i])�(
12
)(k+i)/2
(1+εi)k
Key Obstacle: Dependent random variablesKey Approach: Graph theory!
(and Fibonacci Numbers?)
26
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
More Results
Theorem: Variance (Lazarev, 2011)Exact series for the variance (too long to fit on page...);also Var ∼ 35.98.
Theorem: Asymptotic formula for distribution ofconsecutive missing sums (Lazarev, 2011)
(12
)(k+i)/2
� P(A+A is missing [k , k+i])�(
12
)(k+i)/2
(1+εi)k
Key Obstacle: Dependent random variablesKey Approach: Graph theory! (and Fibonacci Numbers?)
27
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bounds on the Distribution
28
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Upper Bound
Weaker Upper bound: P(A + A has k missing sums) < 0.93k
Proof sketch:
Recall P(k 6∈ A + A) =(3
4
)k/2
If k elements are missing, then missing one at least k/2from the edges
P(A + A has k missing sums) < P(k/2 6∈ A + A) <(34
)k/4 ∼ 0.93k
Note: Since the expectation is around 10 and expansion decay,all the higher moments are bounded as n→∞.
29
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Upper Bound
Weaker Upper bound: P(A + A has k missing sums) < 0.93k
Proof sketch:Recall P(k 6∈ A + A) =
(34
)k/2
If k elements are missing, then missing one at least k/2from the edges
P(A + A has k missing sums) < P(k/2 6∈ A + A) <(34
)k/4 ∼ 0.93k
Note: Since the expectation is around 10 and expansion decay,all the higher moments are bounded as n→∞.
30
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Upper Bound
Weaker Upper bound: P(A + A has k missing sums) < 0.93k
Proof sketch:Recall P(k 6∈ A + A) =
(34
)k/2
If k elements are missing, then missing one at least k/2from the edges
P(A + A has k missing sums) < P(k/2 6∈ A + A) <(34
)k/4 ∼ 0.93k
Note: Since the expectation is around 10 and expansion decay,all the higher moments are bounded as n→∞.
31
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Upper Bound
Weaker Upper bound: P(A + A has k missing sums) < 0.93k
Proof sketch:Recall P(k 6∈ A + A) =
(34
)k/2
If k elements are missing, then missing one at least k/2from the edges
P(A + A has k missing sums) < P(k/2 6∈ A + A) <(34
)k/4 ∼ 0.93k
Note: Since the expectation is around 10 and expansion decay,all the higher moments are bounded as n→∞.
32
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Upper Bound
Weaker Upper bound: P(A + A has k missing sums) < 0.93k
Proof sketch:Recall P(k 6∈ A + A) =
(34
)k/2
If k elements are missing, then missing one at least k/2from the edges
P(A + A has k missing sums) < P(k/2 6∈ A + A) <(34
)k/4 ∼ 0.93k
Note: Since the expectation is around 10 and expansion decay,all the higher moments are bounded as n→∞.
33
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Upper Bound
Weaker Upper bound: P(A + A has k missing sums) < 0.93k
Proof sketch:Recall P(k 6∈ A + A) =
(34
)k/2
If k elements are missing, then missing one at least k/2from the edges
P(A + A has k missing sums) < P(k/2 6∈ A + A) <(34
)k/4 ∼ 0.93k
Note: Since the expectation is around 10 and expansion decay,all the higher moments are bounded as n→∞.
34
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Lower Bound
Lower bound: P(A + A has k missing sums) > 0.01 · 0.70k
Proof sketch: Construction.
Let the first k/2 be missing from AFor the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n
P(A + A has k missing sums) > 0.01(1
2
)k/2 ∼ 0.01 · 0.70k
35
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Lower Bound
Lower bound: P(A + A has k missing sums) > 0.01 · 0.70k
Proof sketch: Construction.Let the first k/2 be missing from A
For the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n
P(A + A has k missing sums) > 0.01(1
2
)k/2 ∼ 0.01 · 0.70k
36
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Lower Bound
Lower bound: P(A + A has k missing sums) > 0.01 · 0.70k
Proof sketch: Construction.Let the first k/2 be missing from AFor the rest of elements, pick any set that fills in
Martin/O’Bryant: P( fills in) > 0.01 independent of n
P(A + A has k missing sums) > 0.01(1
2
)k/2 ∼ 0.01 · 0.70k
37
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Lower Bound
Lower bound: P(A + A has k missing sums) > 0.01 · 0.70k
Proof sketch: Construction.Let the first k/2 be missing from AFor the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n
P(A + A has k missing sums) > 0.01(1
2
)k/2 ∼ 0.01 · 0.70k
38
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Lower Bound
Lower bound: P(A + A has k missing sums) > 0.01 · 0.70k
Proof sketch: Construction.Let the first k/2 be missing from AFor the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n
P(A + A has k missing sums) > 0.01(1
2
)k/2 ∼ 0.01 · 0.70k
39
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Lower Bound
Lower bound: P(A + A has k missing sums) > 0.01 · 0.70k
Proof sketch: Construction.Let the first k/2 be missing from AFor the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n
P(A + A has k missing sums) > 0.01(1
2
)k/2 ∼ 0.01 · 0.70k
40
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Bound on Distribution: Lower Bound
Lower bound: P(A + A has k missing sums) > 0.01 · 0.70k
Proof sketch: Construction.Let the first k/2 be missing from AFor the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n
P(A + A has k missing sums) > 0.01(1
2
)k/2 ∼ 0.01 · 0.70k41
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Variance
42
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Variance
Can study Var|A + A| instead.Recall that
Var|A + A| = E(|A + A|2)− (E|A + A|)2
Therefore just need E(|A + A|2):
E(|A + A|2) =12n
∑A⊆[0,n−1]
|A + A|2
=12n
∑A⊆[0,n−1]
∑i,j∈A+A
1
=12n
∑0≤i,j≤2n−2
∑A:i,j∈A+A
1
=∑
0≤i,j≤2n−2
P(A : i and j ∈ A + A)
43
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Variance
Can study Var|A + A| instead.Recall that
Var|A + A| = E(|A + A|2)− (E|A + A|)2
Therefore just need E(|A + A|2):
E(|A + A|2) =12n
∑A⊆[0,n−1]
|A + A|2
=12n
∑A⊆[0,n−1]
∑i,j∈A+A
1
=12n
∑0≤i,j≤2n−2
∑A:i,j∈A+A
1
=∑
0≤i,j≤2n−2
P(A : i and j ∈ A + A)
44
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Problem: Dependent Random Variables
It is sufficient to study P(A : i and j 6∈ A + A):
P(A : i and j ∈ A + A) = 1− P(A : i 6∈ A + A)− P(A : j 6∈ A + A)+P(A : i and j 6∈ A + A).
Example: P(A : 3 and 7 6∈ A + A)Conditions:
i = 3 : 0 or 3 6∈ A j = 7 : 0 or 7 6∈ Aand 1 or 2 6∈ A and 1 or 6 6∈ A
and 2 or 5 6∈ Aand 3 or 4 6∈ A.
Since there are common integers in both lists, the events3 6∈ A + A and 7 6∈ A + A are dependent.
45
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Problem: Dependent Random Variables
It is sufficient to study P(A : i and j 6∈ A + A):
P(A : i and j ∈ A + A) = 1− P(A : i 6∈ A + A)− P(A : j 6∈ A + A)+P(A : i and j 6∈ A + A).
Example: P(A : 3 and 7 6∈ A + A)
Conditions:
i = 3 : 0 or 3 6∈ A j = 7 : 0 or 7 6∈ Aand 1 or 2 6∈ A and 1 or 6 6∈ A
and 2 or 5 6∈ Aand 3 or 4 6∈ A.
Since there are common integers in both lists, the events3 6∈ A + A and 7 6∈ A + A are dependent.
46
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Problem: Dependent Random Variables
It is sufficient to study P(A : i and j 6∈ A + A):
P(A : i and j ∈ A + A) = 1− P(A : i 6∈ A + A)− P(A : j 6∈ A + A)+P(A : i and j 6∈ A + A).
Example: P(A : 3 and 7 6∈ A + A)Conditions:
i = 3 : 0 or 3 6∈ A j = 7 : 0 or 7 6∈ Aand 1 or 2 6∈ A and 1 or 6 6∈ A
and 2 or 5 6∈ Aand 3 or 4 6∈ A.
Since there are common integers in both lists, the events3 6∈ A + A and 7 6∈ A + A are dependent.
47
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Problem: Dependent Random Variables
It is sufficient to study P(A : i and j 6∈ A + A):
P(A : i and j ∈ A + A) = 1− P(A : i 6∈ A + A)− P(A : j 6∈ A + A)+P(A : i and j 6∈ A + A).
Example: P(A : 3 and 7 6∈ A + A)Conditions:
i = 3 : 0 or 3 6∈ A j = 7 : 0 or 7 6∈ Aand 1 or 2 6∈ A and 1 or 6 6∈ A
and 2 or 5 6∈ Aand 3 or 4 6∈ A.
Since there are common integers in both lists, the events3 6∈ A + A and 7 6∈ A + A are dependent.
48
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Solution: Use Graphs!
Transform the conditions into a graph!
For each integers in [0,7], add a vertex with that integer.Then connect two vertices if add up to 3 or 7.
Example i = 3, j = 7:
01
2
34
5
6
7
7 0 3 4
6 1 2 5
One-to-one correspondence between conditions/edges(and integers/vertices).
49
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Solution: Use Graphs!
Transform the conditions into a graph!For each integers in [0,7], add a vertex with that integer.
Then connect two vertices if add up to 3 or 7.
Example i = 3, j = 7:
01
2
34
5
6
7
7 0 3 4
6 1 2 5
One-to-one correspondence between conditions/edges(and integers/vertices).
50
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Solution: Use Graphs!
Transform the conditions into a graph!For each integers in [0,7], add a vertex with that integer.Then connect two vertices if add up to 3 or 7.
Example i = 3, j = 7:
01
2
34
5
6
7
7 0 3 4
6 1 2 5
One-to-one correspondence between conditions/edges(and integers/vertices).
51
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Solution: Use Graphs!
Transform the conditions into a graph!For each integers in [0,7], add a vertex with that integer.Then connect two vertices if add up to 3 or 7.
Example i = 3, j = 7:
01
2
34
5
6
7
7 0 3 4
6 1 2 5
One-to-one correspondence between conditions/edges(and integers/vertices).
52
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Solution: Use Graphs!
Transform the conditions into a graph!For each integers in [0,7], add a vertex with that integer.Then connect two vertices if add up to 3 or 7.
Example i = 3, j = 7:
01
2
34
5
6
7
7 0 3 4
6 1 2 5
One-to-one correspondence between conditions/edges(and integers/vertices).
53
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Solution: Use Graphs!
Transform the conditions into a graph!For each integers in [0,7], add a vertex with that integer.Then connect two vertices if add up to 3 or 7.
Example i = 3, j = 7:
01
2
34
5
6
7
7 0 3 4
6 1 2 5
One-to-one correspondence between conditions/edges(and integers/vertices).
54
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Solution: Use Graphs!
Transform the conditions into a graph!For each integers in [0,7], add a vertex with that integer.Then connect two vertices if add up to 3 or 7.
Example i = 3, j = 7:
01
2
34
5
6
7
7 0 3 4
6 1 2 5
One-to-one correspondence between conditions/edges(and integers/vertices).
55
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Interpretation of Graphs
Transformed into:
7 0 3 4 6 1 2 5
Need to pick integers so that each condition is satisfied.Therefore, need to pick vertices so that each edge has avertex chosen.So need to pick a vertex cover!Each integer is equally likely so can ignore vertex labels!
56
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Interpretation of Graphs
Transformed into:
7 0 3 4 6 1 2 5
Need to pick integers so that each condition is satisfied.
Therefore, need to pick vertices so that each edge has avertex chosen.So need to pick a vertex cover!Each integer is equally likely so can ignore vertex labels!
57
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Interpretation of Graphs
Transformed into:
7 0 3 4 6 1 2 5
Need to pick integers so that each condition is satisfied.Therefore, need to pick vertices so that each edge has avertex chosen.
So need to pick a vertex cover!Each integer is equally likely so can ignore vertex labels!
58
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Interpretation of Graphs
Transformed into:
7 0 3 4 6 1 2 5
Need to pick integers so that each condition is satisfied.Therefore, need to pick vertices so that each edge has avertex chosen.So need to pick a vertex cover!
Each integer is equally likely so can ignore vertex labels!
59
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Interpretation of Graphs
Transformed into:
7 0 3 4 6 1 2 5
Need to pick integers so that each condition is satisfied.Therefore, need to pick vertices so that each edge has avertex chosen.So need to pick a vertex cover!Each integer is equally likely so can ignore vertex labels!
60
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Vertex Covers
Have:
7 0 3 4 6 1 2 5
Example:7,0,4 and 6,2 form a vertex cover⇐⇒If 7,0,4,6,2 6∈ A, then 3,7 6∈ A + A
Lemma(Lazarev, 2011)
P(i , j 6∈ A + A) = P(pick a vertex cover for graph)
61
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Vertex Covers
Have:
7 0 3 4 6 1 2 5
Example:7,0,4 and 6,2 form a vertex cover
⇐⇒If 7,0,4,6,2 6∈ A, then 3,7 6∈ A + A
Lemma(Lazarev, 2011)
P(i , j 6∈ A + A) = P(pick a vertex cover for graph)
62
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Vertex Covers
Have:
7 0 3 4 6 1 2 5
Example:7,0,4 and 6,2 form a vertex cover⇐⇒If 7,0,4,6,2 6∈ A, then 3,7 6∈ A + A
Lemma(Lazarev, 2011)
P(i , j 6∈ A + A) = P(pick a vertex cover for graph)
63
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Vertex Covers
Have:
7 0 3 4 6 1 2 5
Example:7,0,4 and 6,2 form a vertex cover⇐⇒If 7,0,4,6,2 6∈ A, then 3,7 6∈ A + A
Lemma(Lazarev, 2011)
P(i , j 6∈ A + A) = P(pick a vertex cover for graph)
64
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Number of Vertex Covers
Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.
Case 1: If the first vertex is chosen:x ? ? ? ?
Need an vertex cover for the rest of the graph: g(n − 1).Case 2: If the first vertex is not chosen:
o x ? ? ?
Need an vertex cover for the rest of the graph: g(n − 2).Fibonacci recursive relationship!
g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2
65
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Number of Vertex Covers
Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.
Case 1: If the first vertex is chosen:
x ? ? ? ?
Need an vertex cover for the rest of the graph: g(n − 1).Case 2: If the first vertex is not chosen:
o x ? ? ?
Need an vertex cover for the rest of the graph: g(n − 2).Fibonacci recursive relationship!
g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2
66
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Number of Vertex Covers
Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.
Case 1: If the first vertex is chosen:x ? ? ? ?
Need an vertex cover for the rest of the graph: g(n − 1).Case 2: If the first vertex is not chosen:
o x ? ? ?
Need an vertex cover for the rest of the graph: g(n − 2).Fibonacci recursive relationship!
g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2
67
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Number of Vertex Covers
Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.
Case 1: If the first vertex is chosen:x ? ? ? ?
Need an vertex cover for the rest of the graph: g(n − 1).
Case 2: If the first vertex is not chosen:
o x ? ? ?
Need an vertex cover for the rest of the graph: g(n − 2).Fibonacci recursive relationship!
g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2
68
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Number of Vertex Covers
Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.
Case 1: If the first vertex is chosen:x ? ? ? ?
Need an vertex cover for the rest of the graph: g(n − 1).Case 2: If the first vertex is not chosen:
o x ? ? ?
Need an vertex cover for the rest of the graph: g(n − 2).Fibonacci recursive relationship!
g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2
69
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Number of Vertex Covers
Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.
Case 1: If the first vertex is chosen:x ? ? ? ?
Need an vertex cover for the rest of the graph: g(n − 1).Case 2: If the first vertex is not chosen:
o x ? ? ?
Need an vertex cover for the rest of the graph: g(n − 2).
Fibonacci recursive relationship!
g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2
70
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Number of Vertex Covers
Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.
Case 1: If the first vertex is chosen:x ? ? ? ?
Need an vertex cover for the rest of the graph: g(n − 1).Case 2: If the first vertex is not chosen:
o x ? ? ?
Need an vertex cover for the rest of the graph: g(n − 2).Fibonacci recursive relationship!
g(n) = g(n − 1) + g(n − 2)
=⇒ g(n) = Fn+2
71
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Number of Vertex Covers
Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.
Case 1: If the first vertex is chosen:x ? ? ? ?
Need an vertex cover for the rest of the graph: g(n − 1).Case 2: If the first vertex is not chosen:
o x ? ? ?
Need an vertex cover for the rest of the graph: g(n − 2).Fibonacci recursive relationship!
g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2
72
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
General i , j
In particular
P(3 and 7 6∈ A + A) =128 F4+2F4+2 =
14
since there were two graphs each of length 4.
In general, graphs will have many components withdifferent lengths.For odd i < j < n:
P(A : i and j 6∈ A + A)
=1
2j+1 F12
((j−i)
⌈i+1j−i
⌉−(i+1)
)2⌈
i+1j−i
⌉+2
× F12
(j+1−(j−i)
⌈i+1j−i
⌉)2⌈
i+1j−i
⌉+4
Need to find the formulas for when i , j are even....And to get variance, need to sum up over all i < j < 2n....
73
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
General i , j
In particular
P(3 and 7 6∈ A + A) =128 F4+2F4+2 =
14
since there were two graphs each of length 4.
In general, graphs will have many components withdifferent lengths.
For odd i < j < n:
P(A : i and j 6∈ A + A)
=1
2j+1 F12
((j−i)
⌈i+1j−i
⌉−(i+1)
)2⌈
i+1j−i
⌉+2
× F12
(j+1−(j−i)
⌈i+1j−i
⌉)2⌈
i+1j−i
⌉+4
Need to find the formulas for when i , j are even....And to get variance, need to sum up over all i < j < 2n....
74
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
General i , j
In particular
P(3 and 7 6∈ A + A) =128 F4+2F4+2 =
14
since there were two graphs each of length 4.
In general, graphs will have many components withdifferent lengths.For odd i < j < n:
P(A : i and j 6∈ A + A)
=1
2j+1 F12
((j−i)
⌈i+1j−i
⌉−(i+1)
)2⌈
i+1j−i
⌉+2
× F12
(j+1−(j−i)
⌈i+1j−i
⌉)2⌈
i+1j−i
⌉+4
Need to find the formulas for when i , j are even....And to get variance, need to sum up over all i < j < 2n....
75
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
General i , j
In particular
P(3 and 7 6∈ A + A) =128 F4+2F4+2 =
14
since there were two graphs each of length 4.
In general, graphs will have many components withdifferent lengths.For odd i < j < n:
P(A : i and j 6∈ A + A)
=1
2j+1 F12
((j−i)
⌈i+1j−i
⌉−(i+1)
)2⌈
i+1j−i
⌉+2
× F12
(j+1−(j−i)
⌈i+1j−i
⌉)2⌈
i+1j−i
⌉+4
Need to find the formulas for when i , j are even....
And to get variance, need to sum up over all i < j < 2n....
76
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
General i , j
In particular
P(3 and 7 6∈ A + A) =128 F4+2F4+2 =
14
since there were two graphs each of length 4.
In general, graphs will have many components withdifferent lengths.For odd i < j < n:
P(A : i and j 6∈ A + A)
=1
2j+1 F12
((j−i)
⌈i+1j−i
⌉−(i+1)
)2⌈
i+1j−i
⌉+2
× F12
(j+1−(j−i)
⌈i+1j−i
⌉)2⌈
i+1j−i
⌉+4
Need to find the formulas for when i , j are even....And to get variance, need to sum up over all i < j < 2n....
77
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Variance Formula
Var|A + A| = −40 + 4∑
i<j<n
P(i, j 6∈ A + A)
= −40 + O(cn)
+4∑
i,j odd
1
2j+1F
12
((j−i)
⌈i+1j−i
⌉−(i+1)
)2⌈
i+1j−i
⌉+2
F12
(j+1−(j−i)
⌈i+1j−i
⌉)2⌈
i+1j−i
⌉+4
+4∑
i even,j odd
1
2j+1F
2⌈
i/2+1j−i
⌉+1
F12
((j−i−1)
⌈i+1j−i
⌉−(i+1)+2
⌈i/2+1
j−i
⌉−1)
2⌈
i+1j−i
⌉+2
F12
(j+1−(j−i−1)
⌈i+1j−i
⌉−2⌈
i/2+1j−i
⌉)2⌈
i+1j−i
⌉+4
+4∑
i odd,j even
1
2j+1F
2⌈
j/2+1j−i
⌉F12
((j−i−1)
⌈i+1j−i
⌉−(i+1)+2
⌈j/2+1
j−i
⌉−2)
2⌈
i+1j−i
⌉+2
F12
(j+2−(j−i−1)
⌈i+1j−i
⌉+2⌈
j/2+1j−i
⌉)2⌈
i+1j−i
⌉+4
+4∑
i,j even
1
2j+1F
2⌈
i/2+1j−i
⌉+1
F2⌈ j+2
j−i
⌉ ×
F12
((j−i−2)
⌈i+1j−i
⌉−(i+1)+2
⌈i/2+1
j−i
⌉+2⌈
j/2+1j−i
⌉−3)
2⌈
i+1j−i
⌉+2
F12
(j+2−(j−i−2)
⌈i+1j−i +2
⌈i/2+1
j−i
⌉+2⌈
j/2+1j−i
⌉⌉)2⌈
i+1j−i
⌉+4
So clearlyVar |A + A| ∼ 35.98
78
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Variance Formula
Var|A + A| = −40 + 4∑
i<j<n
P(i, j 6∈ A + A)
= −40 + O(cn)
+4∑
i,j odd
1
2j+1F
12
((j−i)
⌈i+1j−i
⌉−(i+1)
)2⌈
i+1j−i
⌉+2
F12
(j+1−(j−i)
⌈i+1j−i
⌉)2⌈
i+1j−i
⌉+4
+4∑
i even,j odd
1
2j+1F
2⌈
i/2+1j−i
⌉+1
F12
((j−i−1)
⌈i+1j−i
⌉−(i+1)+2
⌈i/2+1
j−i
⌉−1)
2⌈
i+1j−i
⌉+2
F12
(j+1−(j−i−1)
⌈i+1j−i
⌉−2⌈
i/2+1j−i
⌉)2⌈
i+1j−i
⌉+4
+4∑
i odd,j even
1
2j+1F
2⌈
j/2+1j−i
⌉F12
((j−i−1)
⌈i+1j−i
⌉−(i+1)+2
⌈j/2+1
j−i
⌉−2)
2⌈
i+1j−i
⌉+2
F12
(j+2−(j−i−1)
⌈i+1j−i
⌉+2⌈
j/2+1j−i
⌉)2⌈
i+1j−i
⌉+4
+4∑
i,j even
1
2j+1F
2⌈
i/2+1j−i
⌉+1
F2⌈ j+2
j−i
⌉ ×
F12
((j−i−2)
⌈i+1j−i
⌉−(i+1)+2
⌈i/2+1
j−i
⌉+2⌈
j/2+1j−i
⌉−3)
2⌈
i+1j−i
⌉+2
F12
(j+2−(j−i−2)
⌈i+1j−i +2
⌈i/2+1
j−i
⌉+2⌈
j/2+1j−i
⌉⌉)2⌈
i+1j−i
⌉+4
So clearlyVar |A + A| ∼ 35.98
79
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums
80
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Improved Upper Bound
Use P(i , j 6∈ A + A) to improve the upper bound forP(A + A has k missing sums )
From previous section:
P(i , j 6∈ A + A) ∼(
φ2
2 · 51/4
)j (51/4
φ
)i
where φ is the golden ratioIf A + A is missing at k elements, then missing at least twoelements greater than k :
P(A+A has k missing sums ) <
(φ2
2 · 51/4
)k (51/4
φ
)k
∼ 0.80k
General idea: The better we know P(a1, · · · ,aj 6∈ A + A),the better we know the P(A + A has k missing sums )
81
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Improved Upper Bound
Use P(i , j 6∈ A + A) to improve the upper bound forP(A + A has k missing sums )
From previous section:
P(i , j 6∈ A + A) ∼(
φ2
2 · 51/4
)j (51/4
φ
)i
where φ is the golden ratio
If A + A is missing at k elements, then missing at least twoelements greater than k :
P(A+A has k missing sums ) <
(φ2
2 · 51/4
)k (51/4
φ
)k
∼ 0.80k
General idea: The better we know P(a1, · · · ,aj 6∈ A + A),the better we know the P(A + A has k missing sums )
82
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Improved Upper Bound
Use P(i , j 6∈ A + A) to improve the upper bound forP(A + A has k missing sums )
From previous section:
P(i , j 6∈ A + A) ∼(
φ2
2 · 51/4
)j (51/4
φ
)i
where φ is the golden ratioIf A + A is missing at k elements, then missing at least twoelements greater than k :
P(A+A has k missing sums ) <
(φ2
2 · 51/4
)k (51/4
φ
)k
∼ 0.80k
General idea: The better we know P(a1, · · · ,aj 6∈ A + A),the better we know the P(A + A has k missing sums )
83
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Improved Upper Bound
Use P(i , j 6∈ A + A) to improve the upper bound forP(A + A has k missing sums )
From previous section:
P(i , j 6∈ A + A) ∼(
φ2
2 · 51/4
)j (51/4
φ
)i
where φ is the golden ratioIf A + A is missing at k elements, then missing at least twoelements greater than k :
P(A+A has k missing sums ) <
(φ2
2 · 51/4
)k (51/4
φ
)k
∼ 0.80k
General idea: The better we know P(a1, · · · ,aj 6∈ A + A),the better we know the P(A + A has k missing sums )
84
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
Will study the particular case of P(a1, · · · ,aj 6∈ A + A) ofconsecutive missing sums: P(k , k + 1, · · · , k + i 6∈ A + A)
Example: P(16,17,18,19,20 6∈ A + A)
Start with original graph and remove some conditions (edges):
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
=⇒ Transforms to:
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
85
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
Will study the particular case of P(a1, · · · ,aj 6∈ A + A) ofconsecutive missing sums: P(k , k + 1, · · · , k + i 6∈ A + A)Example: P(16,17,18,19,20 6∈ A + A)
Start with original graph and remove some conditions (edges):
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
=⇒ Transforms to:
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
86
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
Will study the particular case of P(a1, · · · ,aj 6∈ A + A) ofconsecutive missing sums: P(k , k + 1, · · · , k + i 6∈ A + A)Example: P(16,17,18,19,20 6∈ A + A)
Start with original graph and remove some conditions (edges):
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
=⇒ Transforms to:
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
87
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
Will study the particular case of P(a1, · · · ,aj 6∈ A + A) ofconsecutive missing sums: P(k , k + 1, · · · , k + i 6∈ A + A)Example: P(16,17,18,19,20 6∈ A + A)
Start with original graph and remove some conditions (edges):
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
=⇒ Transforms to:
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
88
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
Will study the particular case of P(a1, · · · ,aj 6∈ A + A) ofconsecutive missing sums: P(k , k + 1, · · · , k + i 6∈ A + A)Example: P(16,17,18,19,20 6∈ A + A)
Start with original graph and remove some conditions (edges):
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
=⇒ Transforms to:
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
89
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
Will study the particular case of P(a1, · · · ,aj 6∈ A + A) ofconsecutive missing sums: P(k , k + 1, · · · , k + i 6∈ A + A)Example: P(16,17,18,19,20 6∈ A + A)
Start with original graph and remove some conditions (edges):
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
=⇒ Transforms to:
0 1 2 3 4 5 6 7 8
20 19 18 17 16 15 14 13 12 11 10 9
90
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
So get around 20/6 complete bipartite graphs like:
0 1 2
18 17 16
For a vertex cover, need one side to have all verticeschosen; occurs with probability ≤ 1
8 + 18 = 1
4
By independence, P(16,17,18,19,20) ≤(1
4
)20/6
In generalP(k , k + 1, k + 2, k + 3, k + 4) ≤
(14
)(k+4)/6 ∼ 0.79k+4
which is a slight improvement!
91
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
So get around 20/6 complete bipartite graphs like:
0 1 2
18 17 16
For a vertex cover, need one side to have all verticeschosen; occurs with probability ≤ 1
8 + 18 = 1
4
By independence, P(16,17,18,19,20) ≤(1
4
)20/6
In generalP(k , k + 1, k + 2, k + 3, k + 4) ≤
(14
)(k+4)/6 ∼ 0.79k+4
which is a slight improvement!
92
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
So get around 20/6 complete bipartite graphs like:
0 1 2
18 17 16
For a vertex cover, need one side to have all verticeschosen; occurs with probability ≤ 1
8 + 18 = 1
4
By independence, P(16,17,18,19,20) ≤(1
4
)20/6
In generalP(k , k + 1, k + 2, k + 3, k + 4) ≤
(14
)(k+4)/6 ∼ 0.79k+4
which is a slight improvement!
93
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums in A+A
So get around 20/6 complete bipartite graphs like:
0 1 2
18 17 16
For a vertex cover, need one side to have all verticeschosen; occurs with probability ≤ 1
8 + 18 = 1
4
By independence, P(16,17,18,19,20) ≤(1
4
)20/6
In generalP(k , k + 1, k + 2, k + 3, k + 4) ≤
(14
)(k+4)/6 ∼ 0.79k+4
which is a slight improvement!
94
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums
But most general is:P(k , k + 1, · · · , k + i 6∈ A + A) ≤
(12
)(k+i)/2(1 + εi)
k
However the trivial lower bound is:(12
)(k+i)/2 ≤ P(k , k + 1, · · · , k + i 6∈ A + A)Why interesting? Bounds almost match!Essentially the only way to miss a block of i consecutivesums is to miss all elements before the block as well.
95
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums
But most general is:P(k , k + 1, · · · , k + i 6∈ A + A) ≤
(12
)(k+i)/2(1 + εi)
k
However the trivial lower bound is:(12
)(k+i)/2 ≤ P(k , k + 1, · · · , k + i 6∈ A + A)
Why interesting? Bounds almost match!Essentially the only way to miss a block of i consecutivesums is to miss all elements before the block as well.
96
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums
But most general is:P(k , k + 1, · · · , k + i 6∈ A + A) ≤
(12
)(k+i)/2(1 + εi)
k
However the trivial lower bound is:(12
)(k+i)/2 ≤ P(k , k + 1, · · · , k + i 6∈ A + A)Why interesting? Bounds almost match!
Essentially the only way to miss a block of i consecutivesums is to miss all elements before the block as well.
97
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Consecutive Missing Sums
But most general is:P(k , k + 1, · · · , k + i 6∈ A + A) ≤
(12
)(k+i)/2(1 + εi)
k
However the trivial lower bound is:(12
)(k+i)/2 ≤ P(k , k + 1, · · · , k + i 6∈ A + A)Why interesting? Bounds almost match!Essentially the only way to miss a block of i consecutivesums is to miss all elements before the block as well.
98
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Summary
We found
Exponential bounds for distribution of number of missingsumsSeries for the variance in terms of Fibonacci numbersAsymptotic formula for consecutive missing sums
by developing graph theoretic framework for studyingdependent random variables.
99
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Summary
We foundExponential bounds for distribution of number of missingsums
Series for the variance in terms of Fibonacci numbersAsymptotic formula for consecutive missing sums
by developing graph theoretic framework for studyingdependent random variables.
100
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Summary
We foundExponential bounds for distribution of number of missingsumsSeries for the variance in terms of Fibonacci numbers
Asymptotic formula for consecutive missing sumsby developing graph theoretic framework for studyingdependent random variables.
101
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Summary
We foundExponential bounds for distribution of number of missingsumsSeries for the variance in terms of Fibonacci numbersAsymptotic formula for consecutive missing sums
by developing graph theoretic framework for studyingdependent random variables.
102
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Summary
We foundExponential bounds for distribution of number of missingsumsSeries for the variance in terms of Fibonacci numbersAsymptotic formula for consecutive missing sums
by developing graph theoretic framework for studyingdependent random variables.
103
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Future Research
And currently investigating/other research ideas:
Is distribution of missing sums approximately exponential?Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?Higher moments: third moment involves P(i , j , k) whichhas more complicated graphsDistribution of missing differences inA− A = {x − y : x , y ∈ A}
104
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Future Research
And currently investigating/other research ideas:Is distribution of missing sums approximately exponential?
Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?Higher moments: third moment involves P(i , j , k) whichhas more complicated graphsDistribution of missing differences inA− A = {x − y : x , y ∈ A}
105
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Future Research
And currently investigating/other research ideas:Is distribution of missing sums approximately exponential?Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?
Higher moments: third moment involves P(i , j , k) whichhas more complicated graphsDistribution of missing differences inA− A = {x − y : x , y ∈ A}
106
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Future Research
And currently investigating/other research ideas:Is distribution of missing sums approximately exponential?Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?Higher moments: third moment involves P(i , j , k) whichhas more complicated graphs
Distribution of missing differences inA− A = {x − y : x , y ∈ A}
107
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Future Research
And currently investigating/other research ideas:Is distribution of missing sums approximately exponential?Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?Higher moments: third moment involves P(i , j , k) whichhas more complicated graphsDistribution of missing differences inA− A = {x − y : x , y ∈ A}
108
Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research
Thank you!
109