1. Compute the values: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9 2.What is the pattern? 8.1 Examples Defined by Recurrence EquationsExample 8.1.1: DerangementsImagine a party where couples arrive together, but at the end of the evening,every person leaves with a new partner. For each n E P, let D„ denote the number ofdifferent ways n couples can be "deranged" – that is, rearranged in couples so noone is paired with the person they arrived with.// one couple cannot be deranged./3 one and only one way to derange two couples.// If the couples arrive as Aa, Bb, and Cc,// then A would be paired with b or c.//Then D1 = 0D2D3 = 2If A is paired with b, C must be paired with a (not c),and then B with c.////If A is paired withC,B must be paired with a (not b),//and then C with b.// How big are D4, D5, and D10? How can we calculate them? Is there a formula?Let's develop a strategy for counting derangements when n >= 4. Suppose then women are A1, A2, A3,Woman A1 may be "re-paired" with any of the n –let's say she's paired with az where 2 <= k <= n. Now let's consider az's originalpartner, woman Ag: she might take a, or she might refuse a1 and take someone else.If A1 is paired with ar and Ak is paired with a1, there are nderange, and that may be done in exactly D,–2If Aj is paired with ak but Ak refuses a1, we could pretend that Ak and a1together, so now, there are nexactly Dn-1 different ways.For each of the n –An, and each A; arrives with man a;.1 men az or az or . . . or an;2 couples left todifferentways.arrived1 couples left to derange which may be done in1 men that A1 might choose, there are {Dn-2 + Dn-1}different ways to complete the derangement. Therefore, when n >= 4,D, = (n – 1){Dn-2+Dn-1}.(8.1.2)// This also applies when n =// Equation 8.1.2 is an example of a second-order recurrence equation,// where the generic entry is expressed in terms the previous two entries.3. Using Eq. 8.1.2 together with the values of D, and D2, we can evaluate D, forany value of n:// in principleD3D4D5D6D7D8D9D10(3 – 1){D2+D1} = 2{1+0}(4 – 1){D3 +D2} = 3{2+1}(5 – 1){D4+D3} = 4{9+2}(6 – 1){D5+D4} = 5{44+9}(7 – 1){D6+D5} = 6{265 +44}(8 – 1){D7+D6} = 7{1854+265}(9 – 1){D8+D7} = 8{14833+1854}(10 – 1){D9+Ds} = 9{133496 +14833}2442651 85414 833133 4961 334 961// It's strange that 1 334 961 = 10 × (133 496) + 1. Or is it?// Is there a (convenient and compact) formula for D, that we can use to calculate// its values?The sequence on P defined by S,= A × n! where A is any real number satisfiesthe recurrence equation (8.1.2). If n >= 3 then(n – 1){S„-2+Sn-1} = (n – 1){A(n – 2)! +A(n – 1)!}(n – 1)A(n – 2)!{1+[n – 1]}А(п — 1) (п — 2)!{n}= A x n!= Sn.%3|||// But will this "formula" apply when n =// Does there exist a real number A such that D,// No, because// and1 or n =2?A(n!) when n = 1 or n = 2?if 0 = D1 = A(1!), then A must equal 0,if 1 =D2 = A(2!), then A must equal ½.We can however use this formula to prove that D, is O(n!) by provingTheorem 8.1.1: For all n >= 2, (1/3)n! <= Dn <= (1/2)n!Consider the table of valuesn(1/3)n!Dn(1/2)n!11/31/222/311 = 2/236/3 = 223 = 6/2424/3 = 89.12 = 24/2120/3 = 404460= 120/26.720/3 = 240265360 = 720/2|| || || ||| || ||||

Given:- Compute the values: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9

Answered: 1. Compute the values: D2-2D1, D3-3D2,…

Engineering

Computer Science

1. Compute the values: D2-2D1, D3-3D2, D4-4D3, D5-5D4, …, D10-10D9 2.What is the pattern?

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

1. Compute the values: D₂-2D₁, D₃-3D₂, D₄-4D₃, D₅-5D₄, …, D₁₀-10D₉

2.What is the pattern?

8.1 Examples Defined by Recurrence Equations
Example 8.1.1: Derangements
Imagine a party where couples arrive together, but at the end of the evening,
every person leaves with a new partner. For each n E P, let D„ denote the number of
different ways n couples can be "deranged" – that is, rearranged in couples so no
one is paired with the person they arrived with.
// one couple cannot be deranged.
/3 one and only one way to derange two couples.
// If the couples arrive as Aa, Bb, and Cc,
// then A would be paired with b or c.
//
Then D1 = 0
D2
D3 = 2
If A is paired with b, C must be paired with a (not c),
and then B with c.
//
//
If A is paired with
C,
B must be paired with a (not b),
//
and then C with b.
// How big are D4, D5, and D10? How can we calculate them? Is there a formula?
Let's develop a strategy for counting derangements when n >= 4. Suppose the
n women are A1, A2, A3,
Woman A1 may be "re-paired" with any of the n –
let's say she's paired with az where 2 <= k <= n. Now let's consider az's original
partner, woman Ag: she might take a, or she might refuse a1 and take someone else.
If A1 is paired with ar and Ak is paired with a1, there are n
derange, and that may be done in exactly D,–2
If Aj is paired with ak but Ak refuses a1, we could pretend that Ak and a1
together, so now, there are n
exactly Dn-1 different ways.
For each of the n –
An, and each A; arrives with man a;.
1 men az or az or . . . or an;
2 couples left to
different
ways.
arrived
1 couples left to derange which may be done in
1 men that A1 might choose, there are {Dn-2 + Dn-1}
different ways to complete the derangement. Therefore, when n >= 4,
D, = (n – 1){Dn-2+Dn-1}.
(8.1.2)
// This also applies when n =
// Equation 8.1.2 is an example of a second-order recurrence equation,
// where the generic entry is expressed in terms the previous two entries.
3.

Using Eq. 8.1.2 together with the values of D, and D2, we can evaluate D, for
any value of n:
// in principle
D3
D4
D5
D6
D7
D8
D9
D10
(3 – 1){D2+D1} = 2{1+0}
(4 – 1){D3 +D2} = 3{2+1}
(5 – 1){D4+D3} = 4{9+2}
(6 – 1){D5+D4} = 5{44+9}
(7 – 1){D6+D5} = 6{265 +44}
(8 – 1){D7+D6} = 7{1854+265}
(9 – 1){D8+D7} = 8{14833+1854}
(10 – 1){D9+Ds} = 9{133496 +14833}
2
44
265
1 854
14 833
133 496
1 334 961
// It's strange that 1 334 961 = 10 × (133 496) + 1. Or is it?
// Is there a (convenient and compact) formula for D, that we can use to calculate
// its values?
The sequence on P defined by S,= A × n! where A is any real number satisfies
the recurrence equation (8.1.2). If n >= 3 then
(n – 1){S„-2+Sn-1} = (n – 1){A(n – 2)! +A(n – 1)!}
(n – 1)A(n – 2)!{1+[n – 1]}
А(п — 1) (п — 2)!{n}
= A x n!
= Sn.
%3|
||
// But will this "formula" apply when n =
// Does there exist a real number A such that D,
// No, because
// and
1 or n =
2?
A(n!) when n = 1 or n = 2?
if 0 = D1 = A(1!), then A must equal 0,
if 1 =
D2 = A(2!), then A must equal ½.
We can however use this formula to prove that D, is O(n!) by proving
Theorem 8.1.1: For all n >= 2, (1/3)n! <= Dn <= (1/2)n!
Consider the table of values
n
(1/3)n!
Dn
(1/2)n!
1
1/3
1/2
2
2/3
1
1 = 2/2
3
6/3 = 2
2
3 = 6/2
4
24/3 = 8
9.
12 = 24/2
120/3 = 40
44
60
= 120/2
6.
720/3 = 240
265
360 = 720/2
|| || || ||| || ||||

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