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Rask sortering

Det er en algoritme av Divide & Conquer-typen.

Dele opp: Omorganiser elementene og del arrays i to underarrayer og et element i mellom, søk etter at hvert element i venstre underarray er mindre enn eller lik gjennomsnittselementet og hvert element i høyre underarray er større enn midtelementet.

Erobre: Sorter to undermatriser rekursivt.

Kombinere: Kombiner den allerede sorterte matrisen.

Algoritme:

 QUICKSORT (array A, int m, int n) 1 if (n > m) 2 then 3 i ← a random index from [m,n] 4 swap A [i] with A[m] 5 o ← PARTITION (A, m, n) 6 QUICKSORT (A, m, o - 1) 7 QUICKSORT (A, o + 1, n) 

Partisjonsalgoritme:

Partisjonsalgoritmen omorganiserer underarrayene på et sted.

 PARTITION (array A, int m, int n) 1 x &#x2190; A[m] 2 o &#x2190; m 3 for p &#x2190; m + 1 to n 4 do if (A[p] <x) 1 5 6 7 8 then o &larr; + swap a[o] with a[p] a[m] return < pre> <p> <strong>Figure: shows the execution trace partition algorithm</strong> </p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort.webp" alt="DAA Quick sort"> <h3>Example of Quick Sort: </h3> <pre> 44 33 11 55 77 90 40 60 99 22 88 </pre> <p>Let <strong>44</strong> be the <strong>Pivot</strong> element and scanning done from right to left</p> <p>Comparing <strong>44</strong> to the right-side elements, and if right-side elements are <strong>smaller</strong> than <strong>44</strong> , then swap it. As <strong>22</strong> is smaller than <strong>44</strong> so swap them.</p> <pre> <strong>22</strong> 33 11 55 77 90 40 60 99 <strong>44</strong> 88 </pre> <p>Now comparing <strong>44</strong> to the left side element and the element must be <strong>greater</strong> than 44 then swap them. As <strong>55</strong> are greater than <strong>44</strong> so swap them.</p> <pre> 22 33 11 <strong>44</strong> 77 90 40 60 99 <strong>55</strong> 88 </pre> <p>Recursively, repeating steps 1 &amp; steps 2 until we get two lists one left from pivot element <strong>44</strong> &amp; one right from pivot element.</p> <pre> 22 33 11 <strong>40</strong> 77 90 <strong>44</strong> 60 99 55 88 </pre> <p> <strong>Swap with 77:</strong> </p> <pre> 22 33 11 40 <strong>44</strong> 90 <strong>77</strong> 60 99 55 88 </pre> <p>Now, the element on the right side and left side are greater than and smaller than <strong>44</strong> respectively.</p> <p> <strong>Now we get two sorted lists:</strong> </p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-2.webp" alt="DAA Quick sort"> <p>And these sublists are sorted under the same process as above done.</p> <p>These two sorted sublists side by side.</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-3.webp" alt="DAA Quick sort"> <br> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-4.webp" alt="DAA Quick sort"> <h3>Merging Sublists:</h3> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-5.webp" alt="DAA Quick sort"> <p> <strong> SORTED LISTS</strong> </p> <p> <strong>Worst Case Analysis:</strong> It is the case when items are already in sorted form and we try to sort them again. This will takes lots of time and space.</p> <h3>Equation:</h3> <pre> T (n) =T(1)+T(n-1)+n </pre> <p> <strong>T (1)</strong> is time taken by pivot element.</p> <p> <strong>T (n-1)</strong> is time taken by remaining element except for pivot element.</p> <p> <strong>N:</strong> the number of comparisons required to identify the exact position of itself (every element)</p> <p>If we compare first element pivot with other, then there will be 5 comparisons.</p> <p>It means there will be n comparisons if there are n items.</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-6.webp" alt="DAA Quick sort"> <h3>Relational Formula for Worst Case:</h3> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-7.webp" alt="DAA Quick sort"> <h3>Note: for making T (n-4) as T (1) we will put (n-1) in place of &apos;4&apos; and if <br> We put (n-1) in place of 4 then we have to put (n-2) in place of 3 and (n-3) <br> In place of 2 and so on. <p>T(n)=(n-1) T(1) + T(n-(n-1))+(n-(n-2))+(n-(n-3))+(n-(n-4))+n <br> T (n) = (n-1) T (1) + T (1) + 2 + 3 + 4+............n <br> T (n) = (n-1) T (1) +T (1) +2+3+4+...........+n+1-1</p> <p>[Adding 1 and subtracting 1 for making AP series]</p> <p>T (n) = (n-1) T (1) +T (1) +1+2+3+4+........ + n-1 <br> T (n) = (n-1) T (1) +T (1) + <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-8.webp" alt="DAA Quick sort">-1</p> <p> <strong>Stopping Condition: T (1) =0</strong> </p> <p>Because at last there is only one element left and no comparison is required.</p> <p>T (n) = (n-1) (0) +0+<img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-8.webp" alt="DAA Quick sort">-1</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-9.webp" alt="DAA Quick sort"> <p> <strong>Worst Case Complexity of Quick Sort is T (n) =O (n<sup>2</sup>)</strong> </p> <h3>Randomized Quick Sort [Average Case]:</h3> <p>Generally, we assume the first element of the list as the pivot element. In an average Case, the number of chances to get a pivot element is equal to the number of items.</p> <pre> Let total time taken =T (n) For eg: In a given list p 1, p 2, p 3, p 4............pn If p 1 is the pivot list then we have 2 lists. I.e. T (0) and T (n-1) If p2 is the pivot list then we have 2 lists. I.e. T (1) and T (n-2) p 1, p 2, p 3, p 4............pn If p3 is the pivot list then we have 2 lists. I.e. T (2) and T (n-3) p 1, p 2, p 3, p 4............p n </pre> <p>So in general if we take the <strong>Kth</strong> element to be the pivot element.</p> <p> <strong>Then,</strong> </p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-10.webp" alt="DAA Quick sort"> <p>Pivot element will do n comparison and we are doing average case so,</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-11.webp" alt="DAA Quick sort"> <p> <strong>So Relational Formula for Randomized Quick Sort is:</strong> </p> <pre> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-12.webp" alt="DAA Quick sort"> = n+1 +<img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-13.webp" alt="DAA Quick sort">(T(0)+T(1)+T(2)+...T(n-1)+T(n-2)+T(n-3)+...T(0)) <br> = n+1 +<img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-13.webp" alt="DAA Quick sort">x2 (T(0)+T(1)+T(2)+...T(n-2)+T(n-1)) </pre> <pre> n T (n) = n (n+1) +2 (T(0)+T(1)+T(2)+...T(n-1)........eq 1 </pre> <p>Put n=n-1 in eq 1</p> <pre> (n -1) T (n-1) = (n-1) n+2 (T(0)+T(1)+T(2)+...T(n-2)......eq2 </pre> <p>From eq1 and eq 2</p> <p>n T (n) - (n-1) T (n-1)= n(n+1)-n(n-1)+2 (T(0)+T(1)+T(2)+?T(n-2)+T(n-1))-2(T(0)+T(1)+T(2)+...T(n-2)) <br> n T(n)- (n-1) T(n-1)= n[n+1-n+1]+2T(n-1) <br> n T(n)=[2+(n-1)]T(n-1)+2n <br> n T(n)= n+1 T(n-1)+2n</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-14.webp" alt="DAA Quick sort"> <p>Put n=n-1 in eq 3</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-15.webp" alt="DAA Quick sort"> <p>Put 4 eq in 3 eq</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-16.webp" alt="DAA Quick sort"> <p>Put n=n-2 in eq 3</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-17.webp" alt="DAA Quick sort"> <p>Put 6 eq in 5 eq</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-18.webp" alt="DAA Quick sort"> <p>Put n=n-3 in eq 3</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-19.webp" alt="DAA Quick sort"> <p>Put 8 eq in 7 eq</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-20.webp" alt="DAA Quick sort"> <br> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-21.webp" alt="DAA Quick sort"> <p>From 3eq, 5eq, 7eq, 9 eq we get</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-22.webp" alt="DAA Quick sort"> <br> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-23.webp" alt="DAA Quick sort"> <p>From 10 eq</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-24.webp" alt="DAA Quick sort"> <p>Multiply and divide the last term by 2</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-25.webp" alt="DAA Quick sort"> <p> <strong>Is the average case complexity of quick sort for sorting n elements.</strong> </p> <p> <strong>3. Quick Sort [Best Case]:</strong> In any sorting, best case is the only case in which we don&apos;t make any comparison between elements that is only done when we have only one element to sort.</p> <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-26.webp" alt="DAA Quick sort"> <hr></h3></x)>

La 44 vær den Dreie element og skanning gjort fra høyre til venstre

Sammenligner 44 til høyreelementene, og hvis høyreelementene er det mindre enn 44 , så bytt den. Som 22 er mindre enn 44 så bytt dem.

 <strong>22</strong> 33 11 55 77 90 40 60 99 <strong>44</strong> 88 

Sammenligner nå 44 til venstre sideelement og elementet må være større enn 44 så bytt dem. Som 55 er større enn 44 så bytt dem.

 22 33 11 <strong>44</strong> 77 90 40 60 99 <strong>55</strong> 88 

Rekursivt, gjenta trinn 1 og trinn 2 til vi får to lister en igjen fra pivotelementet 44 & en rett fra pivotelementet.

 22 33 11 <strong>40</strong> 77 90 <strong>44</strong> 60 99 55 88 

Bytt med 77:

 22 33 11 40 <strong>44</strong> 90 <strong>77</strong> 60 99 55 88 

Nå er elementet på høyre side og venstre side større enn og mindre enn 44 hhv.

Nå får vi to sorterte lister:

DAA Rask sortering

Og disse underlistene er sortert under samme prosess som ovenfor gjort.

Disse to sorterte underlistene side om side.

DAA Rask sortering
DAA Rask sortering

Slå sammen underlister:

DAA Rask sortering

SORTERT LISTER

Worst Case-analyse: Det er tilfelle når varene allerede er i sortert form og vi prøver å sortere dem på nytt. Dette vil ta mye tid og plass.

Ligning:

 T (n) =T(1)+T(n-1)+n 

T (1) er tid tatt av pivotelementet.

T (n-1) er tid tatt av gjenværende element bortsett fra pivotelementet.

N: antall sammenligninger som kreves for å identifisere den nøyaktige plasseringen av seg selv (hvert element)

Hvis vi sammenligner første element pivot med andre, vil det være 5 sammenligninger.

Det betyr at det vil være n sammenligninger hvis det er n elementer.

DAA Rask sortering

Relasjonsformel for verste fall:

DAA Rask sortering

Merk: for å gjøre T (n-4) som T (1) vil vi sette (n-1) i stedet for '4' og hvis
Vi setter (n-1) i stedet for 4, så må vi sette (n-2) i stedet for 3 og (n-3)
I stedet for 2 og så videre.

T(n)=(n-1) T(1) + T(n-(n-1))+(n-(n-2))+(n-(n-3))+(n-( n-4))+n
T (n) = (n-1) T (1) + T (1) + 2 + 3 + 4+............n
T (n) = (n-1) T (1) +T (1) +2+3+4+...........+n+1-1

[Legg til 1 og trekke fra 1 for å lage AP-serier]

T (n) = (n-1) T (1) +T (1) +1+2+3+4+........ + n-1
T (n) = (n-1) T (1) +T (1) + DAA Rask sortering-1

Stopptilstand: T (1) =0

For endelig er det bare ett element igjen og ingen sammenligning er nødvendig.

T (n) = (n-1) (0) +0+ DAA Rask sortering-1

DAA Rask sortering

Worst Case-kompleksiteten til hurtigsortering er T (n) =O (n2)

Randomisert hurtigsortering [Gjennomsnittlig tilfelle]:

Generelt antar vi det første elementet i listen som pivotelementet. I et gjennomsnittlig tilfelle er antall sjanser for å få et pivotelement lik antall elementer.

 Let total time taken =T (n) For eg: In a given list p 1, p 2, p 3, p 4............pn If p 1 is the pivot list then we have 2 lists. I.e. T (0) and T (n-1) If p2 is the pivot list then we have 2 lists. I.e. T (1) and T (n-2) p 1, p 2, p 3, p 4............pn If p3 is the pivot list then we have 2 lists. I.e. T (2) and T (n-3) p 1, p 2, p 3, p 4............p n 

Så generelt hvis vi tar Kth element for å være pivotelementet.

Deretter,

DAA Rask sortering

Pivot-elementet vil gjøre en sammenligning, og vi gjør gjennomsnittlig kasus så,

DAA Rask sortering

Så relasjonsformelen for randomisert hurtigsortering er:

 <img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-12.webp" alt="DAA Quick sort"> = n+1 +<img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-13.webp" alt="DAA Quick sort">(T(0)+T(1)+T(2)+...T(n-1)+T(n-2)+T(n-3)+...T(0)) <br> = n+1 +<img src="//techcodeview.com/img/daa-tutorial/50/quick-sort-13.webp" alt="DAA Quick sort">x2 (T(0)+T(1)+T(2)+...T(n-2)+T(n-1)) 
 n T (n) = n (n+1) +2 (T(0)+T(1)+T(2)+...T(n-1)........eq 1 

Sett n=n-1 i lik 1

 (n -1) T (n-1) = (n-1) n+2 (T(0)+T(1)+T(2)+...T(n-2)......eq2 

Fra eq1 og eq 2

nT (n) - (n-1) T (n-1)= n(n+1)-n(n-1)+2 (T(0)+T(1)+T(2)+? T(n-2)+T(n-1))-2(T(0)+T(1)+T(2)+...T(n-2))
n T(n)- (n-1) T(n-1)= n[n+1-n+1]+2T(n-1)
n T(n)=[2+(n-1)]T(n-1)+2n
n T(n)= n+1 T(n-1)+2n

kat timpf vekt
DAA Rask sortering

Sett n=n-1 i lik 3

DAA Rask sortering

Sett 4 eq i 3 eq

DAA Rask sortering

Sett n=n-2 i lik 3

DAA Rask sortering

Sett 6 ekv i 5 ekv

DAA Rask sortering

Sett n=n-3 i lik 3

DAA Rask sortering

Sett 8 ekv i 7 ekv

DAA Rask sortering
DAA Rask sortering

Fra 3eq, 5eq, 7eq, 9 eq får vi

DAA Rask sortering

Fra 10 ekv

Multipliser og del siste ledd med 2

Er den gjennomsnittlige sakskompleksiteten av rask sortering for sortering av n elementer.

3. Hurtigsortering [Best case]: I enhver sortering er best case det eneste tilfellet der vi ikke gjør noen sammenligning mellom elementer som bare gjøres når vi bare har ett element å sortere.