BuK/IIm20/Vorlesung 11 Merkelt

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== Folie 6 ==
$mult(0, n) = 0$

$mult(S(m), n) = add(mult(m, n), n)$


$pot(0, n) = 1$

$pot(S(m), n) = mult(pot(m, n), n)$

== Folie 16 ==
$S^2_*(q,m) = S(S(...S(m))...)$


Nach Schema:

$S^2_*(0,m) = f(m)$

$S^2_*(S(n),m) = g(n,h(n,m),m)$


$f(m) = U^1_1(m) = m$

$g(a,b,c) = S(U^3_2(a,b,c))$

$h(n,m) = S^2_*(n,m)$


finale $\mu$-rekursive Vorschrift:

$S^2_*(0,m) = U^1_1(m) = m$

$S^2_*(S(n),m) = S(U^3_2(n,S^2_*(n,m),m))$

== Folie 32 ==

Nur die Tabellen benutzen:

{| class="wikitable"
|'''nr'''
|0
|1
|2
|3
|4
|5
|6
|7
|8
|9
|10
|-
|'''0'''
|1
|2
|3
|4
|5
|6
|7
|8
|9
|10
|11
|-
|'''1'''
|2
|3
|4
|5
|6
|7
|8
|9
|10
|11
|
|-
|'''2'''
|3
|5
|7
|9
|11
|
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|-
|'''3'''
|5
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|'''4'''
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|'''5'''
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|}

Mit Schema weiter verfolgen:

{| class="wikitable"
|'''nr'''
|0
|1
|2
|3
|4
|5
|6
|7
|8
|9
|10
|-
|'''0'''
|1
|2
|3
|4
|5
|6
|7
|8
|9
|10
|11
|-
|'''1'''
|2
|3
|4
|5
|6
|7
|8
|9
|10
|11
|''12''
|-
|'''2'''
|3
|5
|7
|9
|11
|''13''
|''15''
|''17''
|''19''
|''21''
|''23''
|-
|'''3'''
|5
|''13''
|''29''
|''61''
|''125''
|''253''
|''509''
|''1021''
|''2045''
|''4093''
|''8189''
|-
|'''4'''
|''13''
|
|
|
|
|
|
|
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|-
|'''5'''
|
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|}

Bildungsvorschriften:

(0) $m + 1$

(1) $m + 2$

(2) $2m + 3$

(3) $2^{m+3} - 3$

== Folie 35 ==

== Folie 39 ==
[[Bild:Nmerkelt_11_39.png]]

== Folie 41 ==

=== Fakultät ===

[[Bild:Nmerkelt_2021-06-19_15-08-48.png]]


=== Abstandsfunktion ===

[[Bild:Nmerkelt_2021-06-19_15-23-16.png]]

== Folie 45 (1) ==
[[Bild:Nmerkelt_2021-06-19_16-24-08.png]]

[[Bild:Nmerkelt_2021-06-19_16-26-02.png]]

== Folie 45 (2) ==

$F_1(4,6) = F_0(F_1(4,5),F_1(4,5)+5+1) = F_0(185,191) = 376$

$F_1(4,5) = F_0(F_1(4,4),F_1(4,4)+4+1) = F_0(90,95) = 185$

$F_1(4,4) = F_0(F_1(4,3),F_1(4,3)+3+1) = F_0(43,47) = 90$

$F_1(4,3) = F_0(F_1(4,2),F_1(4,2)+2+1) = F_0(20,23) = 43$

$F_1(4,2) = F_0(F_1(4,1),F_1(4,1)+1+1) = F_0(9,11) = 20$

$F_1(4,1) = F_0(F_1(4,0),F_1(4,0)+0+1) = F_0(4,5) = 9$

$F_1(4,0) = 4$


$F_2(2,1) = F_1(F_2(2,0),F_2(2,0)+0+1) = F_1(2,3) = 27$

$F_2(2,0) = 2$

$F_1(2,3) = F_0(F_1(2,2),F_1(2,2)+2+1) = F_0(12,15) = 27$

$F_1(2,2) = F_0(F_1(2,1),F_1(2,1)+1+1) = F_0(5,7) = 12$

$F_1(2,1) = F_0(F_1(2,0),F_1(2,0)+0+1) = F_0(2,3) = 5$

$F_1(2,0) = 2$