Logika Dan Algoritma, Pertemuan 8

7/30/2019 Logika Dan Algoritma, Pertemuan 8

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LOGICS AND ALGORITHMSby : Hendr a Sup rayog i, S.Kom ., ST.

The 8th Subject

Sorting


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WHAT IS IT?

Process to make a random

array sorted.

45 77 2 10 -7 29 12 72 80 53 41 22

-7 2 10 12 22 29 41 45 53 72 77 80


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WHY?

...to make the searching

process faster, since binary

search can be performed on asorted array only.

45 77 2 10 -7 29 12 72 80 53 41 22

failed

-8 14 24 29 31 45 52 54 60 72 76 81

success


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THE METHODS

how to perform?

the simplest one, the bubble sort.

merge sort. maximum successive sort.

quick sort.

etc.

BUBBLE SORT

MAXIMUM

SUCCESSIVE SORT


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BUBBLE SORT

45 77 2 10 -7 29

start from 0 to N-2, perform the following steps.

compare the current element to the next element.

if the current element is greater than the next, then swap them. repeat the entire process (N-1) times.

45 77 2 10 -7 29

45 2 77 10 -7 29

45 2 10 77 -7 29

45 2 10 -7 77 29

45 2 10 -7 29 77

1

pass

1st progress


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BUBBLE SORT (CONT'D)

45 2 10 -7 29 77

2 45 10 -7 29 77

2 10 45 -7 29 77

2 10 -7 45 29 77

2 10 -7 29 45 77

2 10 -7 29 45 77

1

pass 2nd progress


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2 10 -7 29 45 77

2 10 -7 29 45 77

2 -7 10 29 45 77

2 -7 10 29 45 77

2 -7 10 29 45 77

2 -7 10 29 45 77

1

pass 3rd progress


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2 -7 10 29 45 77

-7 2 10 29 45 77

-7 2 10 29 45 77

-7 2 10 29 45 77

-7 2 10 29 45 77

-7 2 10 29 45 77

1

pass 4th progress

OK, in this case we need only 4 passes to make the sorted one.

The next step, in this case, is uneccessary.


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2 -7 10 29 45 77

-7 2 10 29 45 77

-7 2 10 29 45 77

-7 2 10 29 45 77

-7 2 10 29 45 77

-7 2 10 29 45 77

1

pass 5th progress

In a worst case, this pass must be performed.

Hence, the case in which initial position of the lowest value in the most right element.


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BUBBLE SORT (PSEUDOCODE)

PROCEDURE SWAP(A, B)

TEMP AA

BB TEMP

END PROC

PROC BUBBLESORT(DATA)

FOR I 0 TO LENGTH(DATA) 2 DOFOR J 0 TO LENGTH(DATA) 2 DOIF DATA[J] > DATA[J + 1] THEN

SWAP(DATA[J], DATA[J + 1])

END IF

END FOR

END FOR

END PROC


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MAXIMUM SUCCESSIVE SORT

45 77 2 10 -7 29

give initial condition for X = N 1.

compare all elements from 0 to X.

find the greatest, swap it into element X. subtract X with 1.

repeat from 2nd step until X reaches 0.

45 29 2 10 -7 771pass

1st progress


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MAXIMUM SUCCESSIVE SORT (CONT'D)

45 29 2 10 -7 77

-7 29 2 10 45 771

pass

2nd progress


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-7 29 2 10 45 77

-7 10 2 29 45 771

pass

3rd progress


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-7 10 2 29 45 77

-7 2 10 29 45 771

pass

4th progress

OK, in this case we need only 4 passes to make the sorted one.

The next step, in this case, is uneccessary.


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-7 2 10 29 45 77

-7 2 10 29 45 771

pass

5th progress

In a worst case, this pass must be performed.

Hence, the case in which initial position of the lowest value in the most right element.


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MAXIMUM SUCCESSIVE SORT (PSEUDOCODE)

PROCEDURE SWAP(A, B)

TEMP AA

BB TEMP

END PROC

PROC MAXIMUMSUCCESSIVESORT(DATA)

FOR X HIGH(DATA) DOWNTO 1 DOMAXPOS 0FOR I 1 TO X DOIF DATA[I] > DATA[MAXPOS] THEN

MAXPOS IEND IF

END FOR

SWAP(DATA[MAXPOS], DATA[X])

END FOR

END PROC


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ROUTINE TASKS

you have array which contains 300 elements, and every element is a record

structured with name, address, reg num, score, and age.

fill them with random value, make sure no duplicate in reg num, and makesure the reg num is absolutely randomized (unsorted).

create a function (Sort_By_RegNum) to sort this array by reg num.


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WHATS NEXT?

final task, just prepare for it...

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Logika Dan Algoritma, Pertemuan 8