Lecture 9: SOS Lower Bound for Knapsack Lecture Outline Part I: - - PowerPoint PPT Presentation

Lecture 9: SOS Lower Bound for Knapsack

Lecture Outline

• Part I: Knapsack Eqations and Pseudo-

expectation Values

• Part II: Johnson Scheme
• Part III: Proving PSDness
• Part IV: Further Work

Part I: Knapsack Eqations and Pseudo-expectation Values

• Knapsack problem: Given weights 𝑥1, … , 𝑥𝑜

and a knapsack with total capacity 𝐷, what is the maximum weight that can be carried?

• In other words, defining 𝑥𝐽 = σ𝑗∈𝐽 𝑥𝑗 for each

subset 𝐽 ⊆ [1, 𝑜], what is max{𝑥𝐽: 𝐽 ⊆ 1, 𝑜 , 𝑥𝐽 ≤ 𝐷}?

• Here we’ll consider the simple case where

𝑥𝑗 = 1 for all 𝑗 and 𝐷 ∈ [0, 𝑜] is not an integer.

• Answer is 𝐷 , but can SOS prove it?

Knapsack Problem

• Want 𝑦𝑗 = 1 if 𝑗 ∈ 𝐽 and 𝑦𝑗 = 0 otherwise.
• Knapsack equations:

1. ∀𝑗, 𝑦𝑗

2 = 𝑦𝑗

2. σ𝑗=1

𝑜

𝑦𝑗 = 𝑙

• Here we take 𝑙 ∈ [0, 𝑜] to be a non-integer.
• Equations are infeasible because σ𝑗=1

𝑜

𝑦𝑗 ∈ ℤ

Knapsack Equations

• Theorem[Gri01]: SOS needs degree at least

2min{𝑙, 𝑜 − 𝑙} to refute these equations

• We’ll follow the presentation of [MPW15] and

show a lower bound of min{𝑙, 𝑜 − 𝑙}

• Note: This presentation was already in the

retracted paper [MW13]

SOS Lower Bound for Knapsack

• Recall: To prove an SOS lower bound, we

generally do the following:

• 1. Come up with pseudo-expectation values ෨

𝐹 which

• bey the required linear equations
• 2. Show that the moment matrix 𝑁 is PSD
• Here we’ll use symmetry for part 1 and some

combinatorics for part 2.

Review: SOS Lower Bound Strategy

• Define 𝑦𝐽 = ς𝑗∈𝐽 𝑦𝑗
• ∀𝐽, (σ𝑘=1

𝑜

𝑦𝑘)𝑦𝐽 = σ𝑘∈𝐽 𝑦𝑘𝑦𝐽 + σ𝑘∉𝐽 𝑦𝑘𝑦𝐽 = 𝑙𝑦𝐽

• If ෨

𝐹[𝑦𝐽] only depends on |𝐽|, ∀𝐽, 𝑘 ∉ 𝐽, |𝐽| ෨ 𝐹[𝑦𝐽] + 𝑜 − |𝐽| ෨ 𝐹[𝑦𝐽∪{𝑘}] = 𝑙 ෨ 𝐹[𝑦𝐽] ∀𝐽, 𝑘 ∉ 𝐽, ෨ 𝐹 𝑦𝐽∪{𝑘} = 𝑙 − |𝐽| 𝑜 − |𝐽| ෨ 𝐹[𝑦𝐽]

• Thus, ෨

𝐹[𝑦𝐽] = 𝑙 𝑙−1 …(𝑙−|𝐽|+1)

𝑜 𝑜−1 …(𝑜−|𝐽|+1) =

𝑙 |𝐽| 𝑜 |𝐽|

Pseudo-expectation Values

• ෨

𝐹[𝑦𝐽] =

𝑙 |𝐽| 𝑜 |𝐽|

• Could have predicted this as follows: If we had

a set 𝐵 of 1s of size 𝑙, then of the

𝑜 |𝐽|

possible sets of size |𝐽|,

𝑙 |𝐽| of them will be

contained in 𝐵.

• Bayesian view: ෨

𝐹[𝑦𝐽] is the expected value of 𝑦𝐽 given what we can compute (in SOS).

• Here it is a true expectation if 𝑙 ∈ ℤ

Viewing ෨ 𝐹 as an Expectation

• Recall from last lecture: If we have

constraints 𝑦𝑗

2 = 𝑦𝑗 or 𝑦𝑗 2 = 1, it is sufficient

to consider ෨ 𝐹[𝑕2] for multilinear 𝑕.

• Reason: For every polynomial 𝑕, there is a

multilinear polynomial 𝑕′ with deg 𝑕′ ≤ deg(𝑕) such that ෨ 𝐹 𝑕′2 = ෨ 𝐹[𝑕2].

• Thus, it is sufficient to consider the restriction
• f 𝑁 to multilinear indices.

Reduction to Multilinear Indices

• Lemma: If we also have the constraint

σ𝑗=1

𝑜

𝑦𝑗 = 𝑙, for every polynomial 𝑕 of degree at most

𝑒 2, there is a homogeneous, multilinear

polynomial 𝑕′ of degree exactly

𝑒 2 such that

෨ 𝐹 𝑕′2 = ෨ 𝐹[𝑕2].

• Proof idea: Use the following reductions:

1. ∀𝑗, 𝑦𝑗

2𝑔 = 𝑦𝑗𝑔

2. ∀𝐽 ⊆ 1, 𝑜 : 𝐽 <

𝑒 2, 𝑦𝐽 = σ𝑗∉𝐽

𝑜

𝑦𝐽∪{𝑗} 𝑙−|𝐽|

. To see this, note that σ𝑗=1

𝑜

𝑦𝑗 𝑦𝐽 = 𝑙𝑦𝐽 = 𝐽 𝑦𝐽 + σ𝑗∉𝐽

𝑜 𝑦𝐽∪{𝑗}

Reduction to Degree

𝑒 2 Indices

• Corollary: To prove that 𝑁 ≽ 0, it is sufficient to

prove that the submatrix of 𝑁 with multilinear entries of degree exactly

𝑒 2 is PSD.

Reduction to Degree

𝑒 2 Indices

Part II: Johnson Scheme

Johnson Scheme

• Algebra of matrices 𝑁 such that:
• 1. The rows and columns of 𝑁 are indexed by

subsets of [1, 𝑜] of size 𝑠 for some 𝑠. 2. 𝑁𝐽𝐾 only depends on |𝐽 ∩ 𝐾|

• Equivalently, the Johnson Scheme is the algebra
• f matrices which are invariant under

permutations of [1, 𝑜].

• Claim: The matrices 𝑁 in the Johnson scheme

are all symmetric and commute with each other

Johnson Scheme Claim Proof

• Claim: For all 𝐵, 𝐶 in the Johnson scheme, 𝐵𝑈 =

𝐵, 𝐵𝐶 is in the Johnson scheme as well, and 𝐵𝐶 = 𝐶𝐵

• Proof: For the first part, ∀𝐽, 𝐾, 𝐵𝐽𝐾 = 𝐵𝐾𝐽 because

𝐽 ∩ 𝐾 = |𝐾 ∩ 𝐽|. For the second part, 𝐵𝐶𝐽𝐿 = σ𝐾∈ 𝑜

𝑠 𝐵𝐽𝐾𝐶𝐾𝐿. Now observe that for any

permutation 𝜏 of [1, 𝑜], 𝐵𝐶𝐽𝐿 = σ𝐾∈ 𝑜

𝑠 𝐵𝐽𝐾𝐶𝐾𝐿 =

σ𝐾∈ 𝑜

𝑠 𝐵𝜏(𝐽)𝐾𝐶𝐾𝜏(𝐿) = 𝐵𝐶𝜏 𝐽 𝜏(𝐿)

• For the third part, 𝐵𝐶 = 𝐵𝐶 𝑈 = 𝐶𝑈𝐵𝑈 = 𝐶𝐵

Johnson Scheme Picture for 𝑠 = 1

1 2 3 4 5 6 1 2 3 4 5 6

𝐽 ∩ 𝐾 = 0 𝐽 ∩ 𝐾 = 1

Johnson Scheme Picture for 𝑠 = 2

12 13 14 15 16 23 24 25 26 34 35 36 45 46 56 12 13 14 15 16 23 24 25 26 34 35 36 45 46 56

𝐽 ∩ 𝐾 = 1 𝐽 ∩ 𝐾 = 2 𝐽 ∩ 𝐾 = 0

• Natural basis for Johnson Scheme: Define

𝐸𝑏 ∈ ℝ

𝑜 𝑠 × 𝑜 𝑠 to have entries 𝐸𝑏 𝐽𝐾 = 1 if

𝐽 ∩ 𝐾 = 𝑏 and 𝐸𝑗 𝐽𝐾 = 0 if 𝐽 ∩ 𝐾 ≠ 𝑏.

• Easy to express matrices in this basis, but not

so easy to show PSDness

Basis for Johnson Scheme

• Want a convenient basis of PSD matrices.
• Building block: Define 𝑤𝐵 so that 𝑤𝐵 𝐽 = 1 if

A ⊆ 𝐽 and 0 otherwise

• PSD basis for Johnson Scheme: Define 𝑄

𝑏 ∈

ℝ

𝑜 𝑠 × 𝑜 𝑠 to be 𝑄

𝑏 = σ𝐵⊆ 1,𝑜 : 𝐵 =𝑏 𝑤𝐵𝑤𝐵 𝑈

• 𝑄

𝑏 has entries 𝑄 𝑏 𝐽𝐾 = |𝐽∩𝐾| 𝑏

because 𝑤𝐵𝑤𝐵

𝑈 = 1 if and only if 𝐵 ⊆ 𝐽 ∩ 𝐾 and there

are |𝐽∩𝐾|

𝑏

such 𝐵 ⊆ [1, 𝑜] of size 𝑏.

PSD Basis for Johnson Scheme

Basis for 𝑠 = 1

1 2 3 4 5 6 1 2 3 4 5 6

𝐸0 𝐽𝐾 = 1 𝐸0 𝐽𝐾 = 0

Basis for 𝑠 = 1

1 2 3 4 5 6 1 2 3 4 5 6

𝐸1 𝐽𝐾 = 0 𝐸1 𝐽𝐾 = 1

PSD Basis for 𝑠 = 1

1 2 3 4 5 6 1 2 3 4 5 6

𝑄0 𝐽𝐾 = = 1 𝑄0 𝐽𝐾 = 1 = 1

PSD Basis for 𝑠 = 1

1 2 3 4 5 6 1 2 3 4 5 6

𝑄

1 𝐽𝐾 =

1 = 0 𝑄

1 𝐽𝐾 =

1 1 = 1

PSD Basis for 𝑠 = 2

12 13 14 15 16 23 24 25 26 34 35 36 45 46 56 12 13 14 15 16 23 24 25 26 34 35 36 45 46 56

𝑄0 𝐽𝐾 = 2 = 1 𝑄0 𝐽𝐾 = 1 = 1 𝑄0 𝐽𝐾 = = 1

PSD Basis for 𝑠 = 2

12 13 14 15 16 23 24 25 26 34 35 36 45 46 56 12 13 14 15 16 23 24 25 26 34 35 36 45 46 56

𝑄

1 𝐽𝐾 =

2 1 = 2 𝑄

1 𝐽𝐾 =

1 1 = 1 𝑄

1 𝐽𝐾 =

1 = 0

PSD Basis for 𝑠 = 2

12 13 14 15 16 23 24 25 26 34 35 36 45 46 56 12 13 14 15 16 23 24 25 26 34 35 36 45 46 56

𝑄2 𝐽𝐾 = 2 2 = 1 𝑄2 𝐽𝐾 = 1 2 = 0 𝑄2 𝐽𝐾 = 2 = 0

• Basis for Johnson Scheme: 𝐸𝑏 𝐽𝐾 = 𝜀𝑏|𝐽∩𝐾|
• PSD Basis for Johnson Scheme : 𝑄

𝑏 𝐽𝐾 = |𝐽∩𝐾| 𝑏

• Want to shift between bases.
• Lemma:

1. 𝑄

𝑏 = σ𝑐=𝑏 𝑠 𝑐 𝑏 𝐸𝑐

2. 𝐸𝑏 = σ𝑐=𝑏

𝑠

−1 𝑐−𝑏 𝑐

𝑏 𝑄𝑐

• First part is trivial, second part follows from a

bit of combinatorics.

Shifting Between Bases

• Lemma:

1. 𝑄

𝑏 = σ𝑐=𝑏 𝑠 𝑐 𝑏 𝐸𝑐

2. 𝐸𝑏 = σ𝑐=𝑏

𝑠

−1 𝑐−𝑏 𝑐

𝑏 𝑄𝑐

• Proof of the second part: Observe that

σ𝑐=𝑏

𝑠

−1 𝑐−𝑏 𝑐

𝑏 𝑄𝑐 = σ𝑏′=𝑏 𝑠

σ𝑐=𝑏

𝑏′

−1 𝑐−𝑏 𝑐

𝑏 𝐸𝑐

• Must show that for all 𝑏′ ≥ 𝑏,

σ𝑐=𝑏

𝑏′

−1 𝑐−𝑏

𝑏′ 𝑐 𝑐 𝑏 = 𝜀𝑏′𝑏

• In-class exercise: Prove this

Shifting Between Bases Proof

• Need to show: σ𝑐=𝑏

𝑏′

−1 𝑐−𝑏

𝑏′ 𝑐 𝑐 𝑏 = 𝜀𝑏′𝑏

• Answer: Observe that

𝑏′ 𝑐 𝑐 𝑏 = 𝑏′!𝑐! 𝑐!(𝑏′−𝑐)!𝑏!(𝑐−𝑏)! = 𝑏′! 𝑏! 𝑏′−𝑏 ! 𝑏′−𝑏 ! 𝑏′−𝑐 ! 𝑐−𝑏 !

• Our expression is equal to

𝑏′! 𝑏! 𝑏′−𝑏 ! σ𝑘=0 𝑛

−1 𝑘

𝑛 𝑘

where 𝑛 = 𝑏′ − 𝑏

• Now note that σ𝑘=0

𝑛

−1 𝑘

𝑛 𝑘

= 1 + −1

𝑛,

which equals 1 if 𝑛 = 0 and 0 if 𝑛 > 0.

Shifting Between Bases Proof

Part III: Proving PSDness

• Recall that ෨

𝐹[𝑦𝐽] =

𝑙 |𝐽| 𝑜 |𝐽|

• 𝑁𝐽𝐾 =

𝑙 |𝐽∪𝐾| 𝑜 |𝐽∪𝐾|

• Thus, 𝑁 = σ𝑏=0

𝑠

𝑙 2𝑠−𝑏 𝑜 2𝑠−𝑏 𝐸𝑏

Decomposition of 𝑁

• To prove 𝑁 ≽ 0, it is sufficient to express 𝑁

as a non-negative linear combination of the matrices 𝑄

𝑏.

PSD Decomposition

Example: Decomposition for 𝑠 = 1

1 2 3 4 5 6 1 2 3 4 5 6

𝑁𝐽𝐾 = 𝑙 𝑙 − 1 𝑜(𝑜 − 1) 𝑁𝐽𝐾 = 𝑙 𝑜

• 𝑁 =

𝑙 𝑜 𝐸1 + 𝑙 𝑙−1 𝑜(𝑜−1) 𝐸0 = 𝑙 𝑜 𝑄 1 + 𝑙 𝑙−1 𝑜 𝑜−1 (𝑄0 − 𝑄 1)

• 𝑁 =

𝑙 𝑜 − 𝑙 𝑙−1 𝑜(𝑜−1) 𝑄 1 + 𝑙 𝑙−1 𝑜(𝑜−1) 𝑄0 = 𝑙(𝑜−𝑙) 𝑜(𝑜−1) 𝑄 1 + 𝑙 𝑙−1 𝑜(𝑜−1) 𝑄0

• Claim: 𝑁 = σ𝑏=0

𝑠

𝑙 2𝑠 𝑜 2𝑠 ⋅ 𝑜−𝑙 𝑏 𝑙−2𝑠+𝑏 𝑏

𝑄

𝑏

• For the proof, see the appendix
• Corollary: 𝑁 ≽ 0 if 𝑙 ≥ 2𝑠 and 𝑜 − 𝑙 ≥ 𝑠

(where 𝑒 = 2𝑠)

PSD Decomposition

• {𝑄

𝑏} is a nice basis to work with because it is

relatively easy to go between {𝐸𝑏} and {𝑄

𝑏}.

• However, in some sense, it’s not the right

basis to use.

• Want a basis {𝑄

𝑏 ′} such that all symmetric PSD

matrices are a non-negative linear combination of the {𝑄

𝑏 ′}.

• With the right basis, can get a higher degree

lower bound.

Improving Degree Lower Bound

• Let 𝐾 be the all ones matrix.
• For the case 𝑒 = 2, 𝑠 = 1, 𝑄0 = 𝐾 and 𝑄

1 = 𝐽𝑒

• Better basis: 𝑄0

′ = 𝐾, 𝑄 1 ′ = 𝑜−1 𝑜 𝐽𝑒 − 1 𝑜 𝐾

Example

Part IV: Further Work

• Can we take advantage of symmetry in the

problem more generally?

• Yes!

Using Symmetry

• Proposition: Whenever there are valid pseudo-

expectation values, there are valid pseudo- expectation values which are symmetric.

• Proof: Let 𝑇 be the group of symmetries of the
• problem. If we have pseudo-expectation values

෨ 𝐹, then for any 𝜏 ∈ 𝑇, ෪ 𝐹′ f = ෩ E[𝜏 𝑔 ] is also

• valid. Since the conditions for pseudo-

expectation values are convex, ෫ 𝐹𝑏𝑤𝑕 𝑔 =

෨ 𝐹 σ𝜏∈𝑇 𝜏 𝑔 |𝑇|

is valid as well and is symmetric.

Using Symmetry

• Gatermann and Parrilo [GP04] show how

symmetry can be used to drastically reduce the search space for finding pseudo- expectation values.

• Recently, Raymond, Saunderson, Singh, and

Thomas [RSST16] showed that if the problem is symmetric, it can be solved with a semidefinite program whose size is independent of 𝑜.

Using Symmetry

• One way to give intuition for the lower bound:

SOS “thinks” that we are choosing 𝑙 elements

• ut of 𝑜 and takes the corresponding pseudo-

expectation values.

• SOS is very bad at determining functions must

be integers and needs degree ≥ 𝑙 to detect a problem.

Obtaining Lower Bounds Directly

• Is there a way to say that this intuition is good

enough to obtain a lower bound without going through the combinatorics?

• Unless I’m mistaken, yes (this is work in

progress).

Obtaining Lower Bounds Directly

References

• [GP04] K. Gatermann and P. Parrilo. Symmetry groups, semidefinite programs, and

sums of squares. J. Pure Appl. Algebra, 192(1-3):95–128, 2004.

• [Gri01] D. Grigoriev. Complexity of Positivstellensatz proofs for the knapsack.

Computational Complexity 10(2):139–154, 2001

• [MW13] R. Meka and A. Wigderson Association schemes, non-commutative

polynomial concentration, and sum-of-squares lower bounds for planted clique. https://arxiv.org/abs/1307.7615v1

• [MPW15] R. Meka, A Potechin, A. Wigderson, Sum-of-squares lower bounds for

planted clique. STOC p.87–96, 2015

• [RSST16] A. Raymond, J. Saunderson, M. Singh, R. Thomas. Symmetric sums of

squares over k-subset hypercubes. https://arxiv.org/abs/1606.05639, 2016

Appendix: PSD Decomposition Calculations

Picture for 𝑠 = 2

12 13 14 15 16 23 24 25 26 34 35 36 45 46 56 12 13 14 15 16 23 24 25 26 34 35 36 45 46 56

𝑁𝐽𝐾 =

𝑙 3 𝑜 3

𝑁𝐽𝐾 =

𝑙 2 𝑜 2

𝑁𝐽𝐾 =

𝑙 4 𝑜 4

Decomposition for 𝑠 = 2

• 𝑁 =

𝑙 2 𝑜 2 𝐸2 + 𝑙 3 𝑜 3 𝐸1 + 𝑙 4 𝑜 4 𝐸0

• 𝑁 =

𝑙 2 𝑜 2 𝑄2 + 𝑙 3 𝑜 3 (𝑄

1−2𝑄2) +

𝑙 4 𝑜 4 (𝑄0−𝑄

1 + 𝑄2)

• 𝑁 =

𝑙 2 𝑜 2 − 2 𝑙 3 𝑜 3 + 𝑙 4 𝑜 4

𝑄2 +

𝑙 3 𝑜 3 − 2 𝑙 4 𝑜 4

𝑄

1 +

𝑙 4 𝑜 4 𝑄0

• 𝑙

4 𝑜 4 =

𝑙(𝑙−1)(𝑙−2)(𝑙−3) 𝑜(𝑜−1)(𝑜−2)(𝑜−3)

• 𝑙

3 𝑜 3 − 𝑙 4 𝑜 4

=

𝑙(𝑙−1)(𝑙−2)( 𝑜−3 − 𝑙−3 ) 𝑜(𝑜−1)(𝑜−2)(𝑜−3)

=

𝑙(𝑙−1)(𝑙−2)(𝑜−𝑙) 𝑜(𝑜−1)(𝑜−2)(𝑜−3)

Decomposition for 𝑠 = 2

• Claim:

𝑙 2 𝑜 2 − 2 𝑙 3 𝑜 3 + 𝑙 4 𝑜 4

=

𝑙(𝑙−1) 𝑜−𝑙 𝑜−𝑙−1 𝑜(𝑜−1)(𝑜−2)(𝑜−3)

• Proof: Consider

𝑜(𝑜−1)(𝑜−2)(𝑜−3) 𝑙(𝑙−1)

𝑙 2 𝑜 2 − 2 𝑙 3 𝑜 3 + 𝑙 4 𝑜 4

. This equals 𝑜 − 2 𝑜 − 3 − 2 𝑙 − 2 𝑜 − 3 + 𝑙 − 2 𝑙 − 3 which equals 𝑜 − 2 − 𝑙 − 2 𝑜 − 3 − (𝑙 − 2)(𝑜 − 3 − (𝑙 − 3)) = 𝑜 − 𝑙 𝑜 − 3 − 𝑙 − 2 = (𝑜 − 𝑙)(𝑜 − 𝑙 − 1)

General Pattern

• 𝑁 = 𝑙(𝑙−1) 𝑜−𝑙

𝑜−𝑙−1 𝑜(𝑜−1)(𝑜−2)(𝑜−3) 𝑄2 + 𝑙(𝑙−1)(𝑙−2)(𝑜−𝑙) 𝑜(𝑜−1)(𝑜−2)(𝑜−3) 𝑄 1 + 𝑙(𝑙−1)(𝑙−2)(𝑙−3) 𝑜 𝑜−1 𝑜−2 𝑜−3 𝑄0

• Can you see the pattern?
• General Pattern: 𝑁 =

𝑙 2𝑠 𝑜 2𝑠

σ𝑏=0

𝑠

𝑜−𝑙 𝑏 𝑙−2𝑠+𝑏 𝑏

𝑄

𝑏

General Pattern Proof

• Claim: 𝑁 =

𝑙 2𝑠 𝑜 2𝑠

σ𝑏=0

𝑠

𝑜−𝑙 𝑏 𝑙−2𝑠+𝑏 𝑏

𝑄

𝑏

• This gives 𝑁 =

𝑙 2𝑠 𝑜 2𝑠

σ𝑏=0

𝑠

σ𝑐=𝑏

𝑠

𝑜−𝑙 𝑏 𝑙−2𝑠+𝑏 𝑏

𝑐 𝑏 𝐸𝑐

• 𝑁 =

𝑙 2𝑠 𝑜 2𝑠

σ𝑐=0

𝑠

σ𝑏=0

𝑐

𝑜−𝑙 𝑏 𝑙−2𝑠+𝑏 𝑏

𝑐 𝑏 𝐸𝑐

• Need to show: σ𝑏=0

𝑐

𝑜−𝑙 𝑏 𝑙−2𝑠+𝑏 𝑏

𝑐 𝑏 =

𝑜−2𝑠+𝑐 𝑐 𝑙−2𝑠+𝑐 𝑐

General Pattern Proof

• Claim: σ𝑏=0

𝑐

𝑜−𝑙 𝑏 𝑙−2𝑠+𝑏 𝑏

𝑐 𝑏 =

𝑜−2𝑠+𝑐 𝑐 𝑙−2𝑠+𝑐 𝑐

• Proof: Note that

𝑙−2𝑠+𝑐 𝑐 𝑙−2𝑠+𝑏 𝑏

=

𝑙−2𝑠+𝑐 𝑐−𝑏 𝑐 𝑏

, so this is equivalent to the following: σ𝑏=0

𝑐 𝑜−𝑙 𝑏 𝑙−2𝑠+𝑐 𝑐−𝑏

=

𝑜−2𝑠+𝑐 𝑐

General Pattern Proof

• Claim:σ𝑏=0

𝑐 𝑜−𝑙 𝑏 𝑙−2𝑠+𝑐 𝑐−𝑏

=

𝑜−2𝑠+𝑐 𝑐

• Proof: One way to choose 𝑐 elements out of

[1, 𝑜 − 2𝑠 + 𝑐] elements is to first choose the number 𝑏 of elements which will be in 1, 𝑜 − 𝑙 . We then choose 𝑏 elements from 1, 𝑜 − 𝑙 and choose the remaining 𝑐 − 𝑏 elements from 𝑜 − 𝑙 + 1, 𝑜 − 2𝑠 + 𝑐 , which gives

𝑜−𝑙 𝑏 𝑙−2𝑠+𝑐 𝑐−𝑏

choices for each 𝑏 ∈ [0, 𝑐].