Summary

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2.5 Summary

The case of m×n checkerboard, where m ≥ 6, can be obtained by rotation of the rectangles. For example,6×5 checkerboard can be considered to 5×6 checkerboard, so it has a (2,2)-monochromatic-rectangle.

We can convert the above theorems to graphic problems. We have the following proposition.

• Ifn>6, every 2-coloring of K_3,ncontains a monochromaticK_2,2.

• Ifn>6, every 2-coloring of K_4,ncontains a monochromaticK_2,2.

• Ifn>4, every 2-coloring of K_5,ncontains a monochromaticK_2,2.

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Example 2.7. Every 2-coloring ofK_3,7exists a monochromaticK_2,2subgraph.

Figure 2.10: The subgraph induced by {X₁, X₃, Y₂, Y₇} is a monochromatic copy ofK_2,2.

Example 2.8. There is a 2-coloring ofK_3,6no a monochromaticK_2,2subgraph.

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Example 2.9. Every 2-coloring ofK_4,7exists a monochromaticK_2,2subgraph.

Figure 2.11: The subgraph induced by {X₁, X₃, Y₂, Y₇} is a monochromatic copy ofK_2,2.

Example 2.10. There is a 2-coloring ofK_4,6no a monochromaticK_2,2subgraph.

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Example 2.11. Every 2-coloring ofK_5,5exists a monochromaticK_2,2subgraph.

Figure 2.12: The subgraph induced by {X₄, X₅, Y₂, Y₅} is a monochromatic copy ofK_2,2.

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3.1 The Case of 2 × n Checkerboard

If there aret (2,1)-monochromatic-rectangles of the same color, then the checkerboard has a (2,t)-monochromatic-rectangle. Otherwise, there is no (2,t)-monochromatic-rectangle.

3.2 The Case of 3 × n Checkerboard

Theorem 3.1. Ifn > (6t−6), wheret≥2, then in every 2-colored 3×n checkerboard. There is a (2,t)-monochromatic-rectangle.

Proof. Ifn= (6t−⁶) +1=6t−5 (Only prove that every coloring of two colors 3× (^6t− 5)checkerboard, there is a (2,t)-monochromatic-rectangle.) By pigeonhole principle, there are at least ⌈^6t₂⁻⁵⌉ = 3t−2 columns that have at least two of same color grids. Without loss of generality, let the color be black. Then d_i ≥ 2 i = 1, 2, . . . ,(3t−2) , where d_i is the number of black grids of thei^thcolumn of the checkerboard. Assume two colored3× (6t−5) checkerboard has a coloring such that there is no (2,t)-monochromatic-rectangles, then any t columns don’t contain the same black (2,1)-monochromatic-rectangles, each column contains C₂^di distinct black monochromatic-rectangles, and the total number of distinct black

(2,1)-‧

3×n checkerboard. There is a (2,t)-monochromatic-rectangle. □

3.3 The Case of 4 × n Checkerboard

Theorem 3.2. Ifn> (6t−6), wheres≥2, then in every 2-colored 4×n checkerboard. There is a (2,t)-monochromatic-rectangle.

Proof. By Theorem 3.1, in every 2-colored 3× (6t−6) checkerboard, there is a (2,t)-monochromatic-rectangle. Therefore, in every 2-colored4× (^6t−⁶)checkerboard, there is a (2,t)-monochromatic-rectangle. □

3.4 The Case of 5 × n Checkerboard

Theorem 3.3. Ifn > (5t−6), wheret≥2, then in every 2-colored 5×n checkerboard. There is a (2,t)-monochromatic-rectangle, where t is even. AndC₂^d1+C^d₂²+· · · +C₂^dn > (2t−3)·C₂⁵ whered_iis the number of black grids of thei^thcolumn of the checkerboard.

Proof. Supposet = 2k−2, where k is integer greater than two, we use induction on k, If

− − × ( −

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checkerboard, there is a(2k−2, 2)-monochromatic-rectangle.

Basis step: Whenk =2, by the Lemma 2.6 we have in every 2-coloring of 5×5 checkerboard, there is a (2,2)-monochromatic-rectangle. Suppose k = s is true, s ≥ 2 and s ∈ N for all 2-coloring of 5× (10s−15) checkerboard, there is a (2, 2s−2)-monochromatic-rectangle.

By pigeonhole principle, there are at least ⌈^5×(10s₂⁻¹⁵⁾⌉ = ^25s−37 grids of the same color.

Without loss of generality, let the color be black. So, we have ^10s∑⁻¹⁵

i=₁

d_i = 25s−^{37, and} C₂^d1+C₂^d2+· · · +^C₂^d10s−15 > (2s−³)·^C5₂

Induction step: When k = s+1, n = 10(s+1)−15 = 10s−5, by pigeonhole principle, there are at least ⌈^5×(10s₂ ⁻⁵⁾⌉ = 25s−12 grids of the same color in 5× (10s−5) checker-board. Without loss of generality, let the color be black. We have ^10s∑⁻⁵

i=₁

d_i = 25s−12. By induction hypothesis, ^10s∑⁻⁵

i=₁₁

d_i = 25s−37, so ¹⁰∑

i=₁

d_i = 25 and C₂^di is the number of black (2,1)-monochromatic-rectangles in thei^thcolumn, i = 1, 2,· · · ,(10s−5). Assume two col-ored5× (10s−5)checkerboard has a coloring such that there is no (2, 2s) -monochromatic-rectangles, then any2s columns don’t contain the same black (2,1)-monochromatic-rectangles, each column containsC^d₂ⁱ distinct black (2,1)-monochromatic-rectangles, and the total number of distinct black (2,1)-monochromatic-rectangles is not more than(2s−1)·C⁵₂. So we have

C₂^d1 +C^d₂² +· · · +C₂^d10s−5 ≤ (2s−1)·C⁵₂ (3.4.1)

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By the definition of combination we get

d₁(d₁−¹)

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(d²₂+d²₃+· · · +d²₁₀)·9≥ (d₂+d₃+· · · +d₁₀)²

(63−^d2₁)·9≥ (25−^d1)²

10d²₁−^50d1+58≤⁰

1.83 ≤d₁ ≤3.17

andd₁is integer, sod₁ must be2 or 3. Similarly, d₂, d₃,· · · , d₁₀ must be2 or 3 But d₁+d₂+

· · · +d₁₀ =25, so d₁, d₂,· · · , d₁₀ consists of five2^′s and five 3^′s Therefore, d²₁+d²₂+· · · + d²₁₀ =65̸=63, we reach a contradiction.

Case2 : d²₁+d²₂+· · · +d²₁₀ =64

d²₂+d²₃+· · · +^d2₁₀ =64−^d2₁

and

d₂+d₃+· · · +d₁₀ =25−d₁

By Cauchy–Schwarz inequality

(d²₂+d²₃+· · · +d²₁₀)·9≥ (d₂+d₃+· · · +d₁₀)²

(64−^d2₁)·9≥ (25−^d1)²

10d²₁−^50d1+49≤⁰

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So for all 2-colored5× (^10s−⁵)checkerboard, there is a (2,2s)-monochromatic rectangle.

By induction,∀k ≥2 and k ∈N, in every 2-colored 5× (5t−6)checkerboard, there is a (2, t)-monochromatic rectangle, wheret =2k−2.

Lemma 3.4. In every 2-colored5×11 checkerboard, there is a (2,3)-monochromatic-rectangle.

Proof. By pigeonhole principle, there are at least ⌈^5×₂¹¹⌉ = 28 grids of the same color.

Without loss of generality, let the color be black. Then we have ¹¹∑

i=₁

d_i ≥ 28, where d_i is the number of black grids of the i^th column of the checkerboard. Assume two colored 5×¹¹ checkerboard has a coloring such that there is no monochromatic-rectangles, then any three columns don’t contain the same black (2,1)-monochromatic-rectangles, each column contains C₂^di distinct black monochromatic-rectangles, and the total number of distinct black (2,1)-monochromatic-rectangles is not more than2·C⁵₂. So we have

C₂^d1 +C₂^d2+· · · +^C₂^d11 ≤²×^C₂^{5 (3.4.2)}

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So we have

t²+45t+36≤0

Butt ∈N, the last inequality is contradiction. Therefore, every 2-colored 5×11 checkerboard yields a (2,3)-monochromatic-rectangle. □

Figure 3.1: There is a 2-colored 5×10 checkerboard containing no a (2,3)-monochromatic-rectangle.

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whered_iis the number of black grids of thei^thcolumn of the checkerboard.

Proof. Suppose t = 2k−1, where k is integer greater than two, we use induction on k, If n = (10k−¹⁰) +1=10k−9 (Only prove that every 2-colored 5× (^10k−⁹)checkerboard, there is a(2, 2k−1)-monochromatic-rectangle.

Basis step: When k = 2, by the Lemma 3.4 we have in every 2-colored 5×11 checker-board, there is a (2,3)-monochromatic-rectangle. Suppose k = s is true, s ≥ 2 and s ∈ N for all 2-colored5× (10s−9)checkerboard, there is a(2, 2s−1)-monochromatic-rectangle.

By pigeonhole principle, there are at least ⌈^5×(10s₂ ⁻⁹⁾⌉ = 25s−22 grids of the same color.

Without loss of generality, let the color be black. So, we have ^10t∑⁻⁹

i=₁

d_i = 25s−22, and C₂^d1+C₂^d2+· · · +^C₂^d10s−9 > (2s−2)·^C₂⁵

Induction step: When k = s+1, n = 10(s+1)−⁹ = 10s+1, by pigeonhole principle, there are at least⌈^5×(10s₂ ⁺¹⁾⌉ = ^25s+3 grids of the same color in 5× (^10s+1)checkerboard.

Without loss of generality, let the color be black. We have^10s

+₁ color-ing such that there is no(2, 2s+1)-monochromatic-rectangles, then any2s+1 columns don’t contain the same black (2,1)-monochromatic-rectangles, each column containsC₂^didistinct black rectangles, and the total number of distinct black (2,1)-monochromatic-rectangles is not more than2s·^C5₂. So we have

C^d₂¹ +C₂^d2+· · · +C₂^d10s+1 ≤2s·C⁵₂ (3.4.4)

By induction hypothesis

C₂^d11 +C^d₂¹² +· · · +C₂^d10s+1 > (2s−2)·C₂⁵

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By the definition of combination we get

d₁(d₁−1)

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d₂+d₃+· · · +d₁₀ =25−d₁

By Cauchy–Schwarz inequality

(d²₂+d²₃+· · · +^d2₁₀)·⁹≥ (^d2+d₃+· · · +^d10)²

(63−^d2₁)·⁹≥ (²⁵−^d1)²

10d²₁−50d₁+58≤0

1.83 ≤d₁ ≤3.17

andd₁is integer, sod₁ must be2 or 3. Similarly, d₂, d₃,· · · , d₁₀ must be2 or 3 But d₁+d₂+

· · · +d₁₀ =25, so d₁, d₂,· · · , d₁₀ consists of five2^′s and five 3^′s Therefore, d²₁+d²₂+· · · + d²₁₀ =65̸=63, we reach a contradiction. Case2 : d²₁+d²₂+· · · +^d2₁₀ =64

d²₂+d²₃+· · · +^d2₁₀ =64−^d2₁

and

d₂+d₃+· · · +d₁₀ =25−d₁

By Cauchy–Schwarz inequality

(d²₂+d²₃+· · · +^d2₁₀)·9≥ (^d2+d₃+· · · +^d10)²

(64−^d2₁)·⁹≥ (²⁵−^d1)²

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10d²₁−50d₁+49≤0

1.34 ≤^d1 ≤3.66

andd₁is integer, so d₁must be2 or 3. Similarly, d₂, d₃,· · · , d₁₀ must be2 or 3. But d₁+ d₂+· · · +^d10 =25, so d₁, d₂,· · · ^{, d}10 consists of five2^′s and five 3^′s. Therefore, d²₁+d²₂+

· · · +^d2₁₀ =65 ̸=64, we reach a contradiction.

So for all 2-colored5× (10s−5)checkerboard, there is a(2, 2s+1)-monochromatic rect-angle.

By induction,∀k ≥2 and k ∈N, in every 2-colored 5× (5t−5)checkerboard, there is a (2, t)-monochromatic rectangle,wheret=2k−1. □

3.5 Summary

We can convert the above theorems to graphic problems. We have the following proposition.

Lets≥²

• Ifn>6(t−1), every 2-coloring of K_3,nexists a monochromaticK_2,tsubgraph.

• Ifn>6(t−1), every 2-coloring of K_4,nexists a monochromaticK_2,tsubgraph.

• Ifn> (10t−16), every 2-coloring ofK_5,nexists a monochromaticK_2,(_2t−2) subgraph.

• Ifn> (10t−10), every 2-coloring ofK_5,nexists a monochromaticK_2,(2t−1) subgraph.

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Chapter 4

(3,2)-Monochromatic-rectangles in a Checkerboard

4.1 The Case of 3 × n Checkerboard

If there are two columns of all grids are of the same color, then the checkerboard has a (3,2)-monochromatic-rectangle. Otherwise, there is no (3,2)-monochromatic-rectangle.

4.2 The Case of 4 × n Checkerboard

If every column of a 4×n checkerboard has two black grids and two white grids, then it doesn’t have s a (3,2)monochromatic-rectangle.Therefore, for every two color 4×n checker-board exists a coloring such that has no (3,2)-monochromatic-rectangles in the4×n checker-board.

4.3 The Case of 5 × n Checkerboard

Lemma 4.1. In every 2-colored5×n checkerboard, n = 21 is the smallest number such that there exists a (3,2)-monochromatic-rectangle.

5×20 checkerboard has no

(3,2)-‧

monochromatic-rectangles. In a column, there are at mostC₃⁵distinct black (3,1)-monochromatic-rectangles. So we can distribute theC₃⁵distinct black (3,1)-monochromatic-rectangles and the C₃⁵distinct white (3,1)-monochromatic-rectangles to the 20 columns, then the two colored5×20 checkerboards have no (3,2)-monochromatic-rectangles.

By pigeonhole principle, there are at least ⌈²¹₂⌉ = 11 columns with at least three grids of the same color. Without loss of generality, let the color be black. Then d_i ≥ ^{3 i} = 1, 2, . . . , 11 , whered_iis the number of black grids of thei^thcolumn of the checkerboard. Assume 2-colored 5×21 checkerboard has a coloring such that there is no (3,2)monochromatic-rectangles, then any two columns don’t contain the same black (3,1)-monochromatic-rectangles, each column contains C₃^di distinct black (3,1)-monochromatic-rectangles, and the total number of distinct black (3,1)-monochromatic-rectangles is not more thanC⁵₃. So we have

C₃^d1+C₃^d2 +· · · +C^d₃¹¹ ≤C₃⁵

11≤10 iwe reach a contradiction in the last inequality. So, If n>21, then in every 2-coloring of5×n checkerboard. There is a (3,2)monochromatic-rectangle. □

在文檔中 The Existence of (s,t)-Monochromatic-rectangles in a 2-colored Checkerboard (頁 17-33)