MERGESORT BBM ALGORITHMS DEPT. OF COMPUTER ENGINEERING. Mergesort

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1 BBM ALGORITHMS DEPT. OF COMPUTER ENGINEERING Mergesort Basc plan. Dvde array nto two halves. Recursvely sort each half. Merge two halves. MERGESORT nput sort left half sort rght half merge results M E R G E S O R T E X A M P L E E E G M O R R S T E X A M P L E E E G M O R R S A E E L M P T X A E E E E G L M M O P R R S T X Mergesort overvew Acnowledgement: The course sldes are adapted from the sldes prepared by R. Sedgewc and K. Wayne of Prnceton Unversty. 2 Abstract n-place merge lo md md+1 h sorted sorted 3 4

2 Abstract n-place merge Abstract n-place merge lo md md+1 h copy to auxlary array 5 6 Abstract n-place merge Abstract n-place merge EA E G M R A C E R T compare mnmum n each subarray compare mnmum n each subarray E E G M R A C E R T 7 8

3 Abstract n-place merge Abstract n-place merge A E G M R A C E R T A EC G M R A C E R T compare mnmum n each subarray compare mnmum n each subarray 9 10 Abstract n-place merge Abstract n-place merge A C G M R A C E R T A C GE M R A C E R T compare mnmum n each subarray compare mnmum n each subarray 11 12

4 Abstract n-place merge Abstract n-place merge A C E M R A C E R T A C E ME R A C E R T compare mnmum n each subarray compare mnmum n each subarray Abstract n-place merge Abstract n-place merge A C E E R A C E R T A C E E RE A C E R T compare mnmum n each subarray compare mnmum n each subarray E E G M R A C E R T 15 16

5 Abstract n-place merge Abstract n-place merge A C E E E A C E R T A C E E E AG C E R T compare mnmum n each subarray compare mnmum n each subarray Abstract n-place merge Abstract n-place merge A C E E E G C E R T A C E E E G MC E R T compare mnmum n each subarray compare mnmum n each subarray 19 20

6 Abstract n-place merge Abstract n-place merge A C E E E G M E R T A C E E E G M ER R T compare mnmum n each subarray compare mnmum n each subarray Abstract n-place merge Abstract n-place merge A C E E E G M R R T A C E E E G M R R T one subarray exhausted, tae from other one subarray exhausted, tae from other 23 24

7 Abstract n-place merge Abstract n-place merge A C E E E G M R R T A C E E E G M R R T one subarray exhausted, tae from other one subarray exhausted, tae from other Abstract n-place merge Abstract n-place merge lo h A C E E E G M R R T A C E E E G M R R T both subarrays exhausted, done sorted 27 28

8 Mergng Q. How to combne two sorted subarrays nto a sorted whole. A. Use an auxlary array. Mergng: Java mplementaton prvate statc vod merge(comparable[] a, Comparable[] aux, nt lo, nt md, nt h) assert ssorted(a, lo, md); // precondton: a[lo..md] sorted assert ssorted(a, md+1, h); // precondton: a[md+1..h] sorted nput copy merged result A A C 0 7 E E G M R C E R T 2 A C E 1 7 E E G M R E R T 3 A C E E 2 7 E G M R E R T 4 A C E E E 2 8 G M R E R T 5 A C E E E G 3 8 G M R R T 6 A C E E E G M 4 8 M R R T 7 A C E E E G M R 5 8 R R T 8 A C E E E G M R R 5 9 R T 9 A C E E E G M R R T 6 10 T A C E E E G M R R T for (nt = lo; <= h; ++) aux[] = a[]; nt = lo, = md+1; for (nt = lo; <= h; ++) f ( > md) a[] = aux[++]; else f ( > h) a[] = aux[++]; else f (less(aux[], aux[])) a[] = aux[++]; else a[] = aux[++]; assert ssorted(a, lo, h); lo // postcondton: a[lo..h] sorted md A G L O R H I M S T h copy merge Abstract n-place merge trace A G H I L M Mergesort: Java mplementaton Mergesort: trace publc class Merge prvate statc vod merge(comparable[] a, Comparable[] aux, nt lo, nt md, nt h) /* as before */ prvate statc vod sort(comparable[] a, Comparable[] aux, nt lo, nt h) f (h <= lo) return; nt md = lo + (h - lo) / 2; sort (a, aux, lo, md); sort (a, aux, md+1, h); merge(a, aux, lo, md, h); publc statc vod sort(comparable[] a) aux = new Comparable[a.length]; sort(a, aux, 0, a.length - 1); lo h merge(a, 0, 0, 1) merge(a, 2, 2, 3) merge(a, 0, 1, 3) merge(a, 4, 4, 5) merge(a, 6, 6, 7) merge(a, 4, 5, 7) merge(a, 0, 3, 7) merge(a, 8, 8, 9) merge(a, 10, 10, 11) merge(a, 8, 9, 11) merge(a, 12, 12, 13) merge(a, 14, 14, 15) merge(a, 12, 13, 15) merge(a, 8, 11, 15) merge(a, 0, 7, 15) M E R G E S O R T E X A M P L E E M R G E S O R T E X A M P L E E M G R E S O R T E X A M P L E E G M R E S O R T E X A M P L E E G M R E S O R T E X A M P L E E G M R E S O R T E X A M P L E E G M R E O R S T E X A M P L E E E G M O R R S T E X A M P L E E E G M O R R S E T X A M P L E E E G M O R R S E T A X M P L E E E G M O R R S A E T X M P L E E E G M O R R S A E T X M P L E E E G M O R R S A E T X M P E L E E G M O R R S A E T X E L M P E E G M O R R S A E E L M P T X A E E E E G L M M O P R R S T X Trace of merge results for top-down mergesort lo md h result after recursve call

9 Mergesort: anmaton Mergesort: anmaton 50 random tems 50 reverse-sorted tems algorthm poston algorthm poston n order current subarray n order current subarray not n order not n order Mergesort: emprcal analyss Runnng tme estmates: Laptop executes 10 8 compares/second. Supercomputer executes compares/second. nserton sort (N 2 ) mergesort (N log N) computer thousand mllon bllon thousand mllon bllon Mergesort: number of compares and array accesses Proposton. Mergesort uses at most N lg N compares and 6 N lg N array accesses to sort any array of sze N. Pf setch. The number of compares C (N) and array accesses A (N) to mergesort an array of sze N satsfy the recurrences: C (N) C ( N / 2 ) + C ( N / 2 ) + N for N > 1, wth C (1) = 0. home nstant 2.8 hours 317 years nstant 1 second 18 mn super nstant 1 second 1 wee nstant nstant nstant left half rght half merge A (N) A ( N / 2 ) + A ( N / 2 ) + 6 N for N > 1, wth A (1) = 0. We solve the recurrence when N s a power of 2. Bottom lne. Good algorthms are better than supercomputers. D (N) = 2 D (N / 2) + N, for N > 1, wth D (1) =

10 Mergng: Java mplementaton prvate statc vod merge(comparable[] a, Comparable[] aux, nt lo, nt md, nt h) assert ssorted(a, lo, md); // precondton: a[lo..md] sorted assert ssorted(a, md+1, h); // precondton: a[md+1..h] sorted for (nt = lo; <= h; ++) aux[] = a[]; nt = lo, = md+1; for (nt = lo; <= h; ++) f ( > md) a[] = aux[++]; else f ( > h) a[] = aux[++]; else f (less(aux[], aux[])) a[] = aux[++]; else a[] = aux[++]; 2N assert ssorted(a, lo, h); 2N // postcondton: a[lo..h] sorted copy merge 2N Dvde-and-conquer recurrence: proof by pcture lg N Proposton. If D (N) satsfes D (N) = 2 D (N / 2) + N for N > 1, wth D (1) = 0, then D (N) = N lg N. Pf 1. [assumng N s a power of 2] D (N / 2) D (N / 4) D (N / 4) D (N) D (N / 2 ) D (N / 4) D (N / 2) D (N / 4) N 2 (N/2)... 2 (N/2 ) = N = N 4 (N/4) = N... = N D (2) D (2) D (2) D (2) D (2) D (2) D (2) D (2) N/2 (2) = N N lg N Dvde-and-conquer recurrence: proof by expanson Proposton. If D (N) satsfes D (N) = 2 D (N / 2) + N for N > 1, wth D (1) = 0, then D (N) = N lg N. Pf 2. [assumng N s a power of 2] D (N) = 2 D (N/2) + N D (N) / N = 2 D (N/2) / N + 1 = D (N/2) / (N/2) + 1 = D (N/4) / (N/4) = D (N/8) / (N/8) = D (N/N) / (N/N) = lg N gven dvde both sdes by N algebra apply to frst term apply to frst term agan stop applyng, D(1) = 0 Dvde-and-conquer recurrence: proof by nducton Proposton. If D (N) satsfes D (N) = 2 D (N / 2) + N for N > 1, wth D (1) = 0, then D (N) = N lg N. Pf 3. [assumng N s a power of 2] Base case: N = 1. Inductve hypothess: D (N) = N lg N. Goal: show that D (2N) = (2N) lg (2N). D (2N) = 2 D (N) + 2N = 2 N lg N + 2N = 2 N (lg (2N) 1) + 2N = 2 N lg (2N) gven nductve hypothess algebra QED 39 40

11 Mergesort analyss: memory Proposton. Mergesort uses extra space proportonal to N. Pf. The array needs to be of sze N for the last merge. Mergesort: practcal mprovements Use nserton sort for small subarrays. Mergesort has too much overhead for tny subarrays. Cutoff to nserton sort for 7 tems. two sorted subarrays A C D G H I M N U V B E F J O P Q R S T A B C D E F G H I J M N O P Q R S T U V merged result Def. A sortng algorthm s n-place f t uses c log N extra memory. Ex. Inserton sort, selecton sort, shellsort. Challenge for the bored. In-place merge. [Kronrod, 1969] prvate statc vod sort(comparable[] a, Comparable[] aux, nt lo, nt h) f (h <= lo + CUTOFF - 1) Inserton.sort(a, lo, h); nt md = lo + (h - lo) / 2; sort (a, aux, lo, md); sort (a, aux, md+1, h); merge(a, aux, lo, md, h); Mergesort: practcal mprovements Stop f already sorted. Is bggest tem n frst half smallest tem n second half? Helps for partally-ordered arrays. Mergesort: practcal mprovements Elmnate the copy to the auxlary array. Save tme (but not space) by swtchng the role of the nput and auxlary array n each recursve call. A B C D E F G H I J M N O P Q R S T U V A B C D E F G H I J M N O P Q R S T U V prvate statc vod sort(comparable[] a, Comparable[] aux, nt lo, nt h) f (h <= lo) return; nt md = lo + (h - lo) / 2; sort (a, aux, lo, md); sort (a, aux, md+1, h); f (!less(a[md+1], a[md])) return; merge(a, aux, lo, md, h); prvate statc vod merge(comparable[] a, Comparable[] aux, nt lo, nt md, nt h) nt = lo, = md+1; for (nt = lo; <= h; ++) f ( > md) aux[] = a[++]; else f ( > h) aux[] = a[++]; else f (less(a[], a[])) aux[] = a[++]; merge from to else aux[] = a[++]; prvate statc vod sort(comparable[] a, Comparable[] aux, nt lo, nt h) f (h <= lo) return; nt md = lo + (h - lo) / 2; sort (aux, a, lo, md); sort (aux, a, md+1, h); merge(aux, a, lo, md, h); 43 swtch roles of and 44

12 Mergesort: vsualzaton Bottom-up mergesort frst subarray second subarray frst merge frst half sorted Basc plan. Pass through array, mergng subarrays of sze 1. Repeat for subarrays of sze 2, 4, 8, 16,... sz = 1 merge(a, 0, 0, 1) merge(a, 2, 2, 3) merge(a, 4, 4, 5) merge(a, 6, 6, 7) merge(a, 8, 8, 9) merge(a, 10, 10, 11) merge(a, 12, 12, 13) merge(a, 14, 14, 15) sz = 2 merge(a, 0, 1, 3) merge(a, 4, 5, 7) merge(a, 8, 9, 11) merge(a, 12, 13, 15) sz = 4 merge(a, 0, 3, 7) merge(a, 8, 11, 15) sz = 8 merge(a, 0, 7, 15) a[] M E R G E S O R T E X A M P L E E M R G E S O R T E X A M P L E E M G R E S O R T E X A M P L E E M G R E S O R T E X A M P L E E M G R E S O R T E X A M P L E E M G R E S O R E T X A M P L E E M G R E S O R E T A X M P L E E M G R E S O R E T A X M P L E E M G R E S O R E T A X M P E L E G M R E S O R E T A X M P E L E G M R E O R S E T A X M P E L E G M R E O R S A E T X M P E L E G M R E O R S A E T X E L M P E E G M O R R S A E T X E L M P E E G M O R R S A E E L M P T X A E E E E G L M M O P R R S T X Trace of merge results for bottom-up mergesort second half sorted result 45 Bottom lne. No recurson needed! 46 Bottom-up mergesort: Java mplementaton Bottom-up mergesort: vsual trace publc class MergeBU prvate statc Comparable[] aux; prvate statc vod merge(comparable[] a, nt lo, nt md, nt h) /* as before */ publc statc vod sort(comparable[] a) nt N = a.length; aux = new Comparable[N]; for (nt sz = 1; sz < N; sz = sz+sz) for (nt lo = 0; lo < N-sz; lo += sz+sz) merge(a, lo, lo+sz-1, Math.mn(lo+sz+sz-1, N-1)); Bottom lne. Concse ndustral-strength code, f you have the space

13 Bottom-up mergesort: vsual trace Bottom-up mergesort: vsual trace Complexty of sortng Decson tree (for 3 dstnct tems a, b, and c) Computatonal complexty. Framewor to study effcency of algorthms for solvng a partcular problem X. Model of computaton. Allowable operatons. Cost model. Operaton count(s). Upper bound. Cost guarantee provded by some algorthm for X. Lower bound. Proven lmt on cost guarantee of all algorthms for X. Optmal algorthm. Algorthm wth best possble cost guarantee for X. yes b < c a < b yes no code between compares (e.g., sequence of exchanges) no yes a < c no heght of tree = worst-case number of Example: sortng. Model of computaton: decson tree. Cost model: # compares. Upper bound: ~ N lg N from mergesort. Lower bound:? Optmal algorthm:? lower bound ~ upper bound can access nformaton only through compares (e.g., Java Comparable framewor) a b c yes a c b a < c no c a b b a c yes b c a b < c no c b a 51 (at least) one leaf for each possble orderng 52

14 Compare-based lower bound for sortng Proposton. Any compare-based sortng algorthm must use at least lg ( N! ) ~ N lg N compares n the worst-case. Pf. Assume array conssts of N dstnct values a1 through an. Worst case dctated by heght h of decson tree. Bnary tree of heght h has at most 2 h leaves. N! dfferent orderngs at least N! leaves. Compare-based lower bound for sortng Proposton. Any compare-based sortng algorthm must use at least lg ( N! ) ~ N lg N compares n the worst-case. Pf. Assume array conssts of N dstnct values a1 through an. Worst case dctated by heght h of decson tree. Bnary tree of heght h has at most 2 h leaves. N! dfferent orderngs at least N! leaves. h 2 h # leaves N! h lg ( N! ) ~ N lg N Strlng's formula at least N! leaves no more than 2 h leaves Complexty of sortng Model of computaton. Allowable operatons. Cost model. Operaton count(s). Upper bound. Cost guarantee provded by some algorthm for X. Lower bound. Proven lmt on cost guarantee of all algorthms for X. Optmal algorthm. Algorthm wth best possble cost guarantee for X. Example: sortng. Model of computaton: decson tree. Cost model: # compares. Upper bound: ~ N lg N from mergesort. Lower bound: ~ N lg N. Optmal algorthm = mergesort. Frst goal of algorthm desgn: optmal algorthms. Complexty results n context Other operatons? Mergesort s optmal wth respect to number of compares (e.g., but not wth respect to number of array accesses). Space? Mergesort s not optmal wth respect to space usage. Inserton sort, selecton sort, and shellsort are space-optmal. Challenge. Fnd an algorthm that s both tme- and space-optmal. [stay tuned] Lessons. Use theory as a gude. Ex. Don't try to desgn sortng algorthm that guarantees ½ N lg N compares

15 Complexty results n context (contnued) Lower bound may not hold f the algorthm has nformaton about: The ntal order of the nput. The dstrbuton of ey values. The representaton of the eys. Comparable nterface: revew Comparable nterface: sort usng a type's natural order. publc class Date mplements Comparable<Date> prvate fnal nt month, day, year; Partally-ordered arrays. Dependng on the ntal order of the nput, we may not need N lg N compares. nserton sort requres only N-1 compares f nput array s sorted Duplcate eys. Dependng on the nput dstrbuton of duplcates, we may not need N lg N compares. stay tuned for 3-way qucsort Dgtal propertes of eys. We can use dgt/character compares nstead of ey compares for numbers and strngs. stay tuned for radx sorts publc Date(nt m, nt d, nt y) month = m; day = d; year = y; publc nt compareto(date that) f (ths.year < that.year ) return -1; f (ths.year > that.year ) return +1; f (ths.month < that.month) return -1; f (ths.month > that.month) return +1; f (ths.day < that.day ) return -1; f (ths.day > that.day ) return +1; return 0; natural order Comparator nterface Comparator nterface: system sort Comparator nterface: sort usng an alternate order. publc nterface Comparator<Key> To use wth Java system sort: Create Comparator obect. Pass as second argument to Arrays.sort(). Requred property. Must be a total order. Ex. Sort strngs by: Natural order. Case nsenstve. Spansh. Brtsh phone boo.... nt compare(key v, Key w) compare eys v and w pre-1994 order for dgraphs ch and ll and rr Now s the tme s Now the tme café cafetero cuarto churro nube ñoño McKnley Macntosh 59 Strng[] a;... Arrays.sort(a);... uses natural order uses alternate order defned by Comparator<Strng> obect Arrays.sort(a, Strng.CASE_INSENSITIVE_ORDER);... Arrays.sort(a, Collator.getInstance(new Locale("es")));... Arrays.sort(a, new BrtshPhoneBooOrder());... Bottom lne. Decouples the defnton of the data type from the defnton of what t means to compare two obects of that type. 60

16 Comparator nterface: usng wth our sortng lbrares To support comparators n our sort mplementatons: Use Obect nstead of Comparable. Pass Comparator to sort() and less() and use t n less(). Comparator nterface: mplementng To mplement a comparator: Defne a (nested) class that mplements the Comparator nterface. Implement the compare() method. nserton sort usng a Comparator publc statc vod sort(obect[] a, Comparator comparator) nt N = a.length; for (nt = 0; < N; ++) for (nt = ; > 0 && less(comparator, a[], a[-1]); --) exch(a,, -1); prvate statc boolean less(comparator c, Obect v, Obect w) return c.compare(v, w) < 0; prvate statc vod exch(obect[] a, nt, nt ) Obect swap = a[]; a[] = a[]; a[] = swap; publc class Student publc statc fnal Comparator<Student> BY_NAME = new ByName(); publc statc fnal Comparator<Student> BY_SECTION = new BySecton(); prvate fnal Strng name; prvate fnal nt secton;... one Comparator for the class prvate statc class ByName mplements Comparator<Student> publc nt compare(student v, Student w) return v.name.compareto(w.name); prvate statc class BySecton mplements Comparator<Student> publc nt compare(student v, Student w) return v.secton - w.secton; ths technque wors here snce no danger of overflow Comparator nterface: mplementng To mplement a comparator: Defne a (nested) class that mplements the Comparator nterface. Implement the compare() method. Stablty A typcal applcaton. Frst, sort by name; then sort by secton. Selecton.sort(a, Student.BY_NAME); Selecton.sort(a, Student.BY_SECTION); Andrews 3 A Lttle Fura 1 A Brown Battle 4 C Whtman Rohde 2 A Forbes Chen 3 A Blar Chen 3 A Blar Arrays.sort(a, Student.BY_NAME); Arrays.sort(a, Student.BY_SECTION); Fox 3 A Dcnson Fox 3 A Dcnson Andrews 3 A Lttle Fura 1 A Brown Fura 1 A Brown Andrews 3 A Lttle Battle 4 C Whtman Chen 3 A Blar Fox 3 A Dcnson Rohde 2 A Forbes Andrews 3 A Lttle Chen 3 A Blar Gazs 4 B Brown Kanaga 3 B Brown Rohde 2 A Forbes Kanaga 3 B Brown Gazs 4 B Brown Battle 4 C Whtman Fura 1 A Brown Fox 3 A Dcnson Gazs 4 B Brown Kanaga 3 B Brown Rohde 2 A Forbes Kanaga 3 B Brown Battle 4 C Whtman Gazs 4 B Students n secton 3 no longer sorted by name. A stable sort preserves the relatve order of tems wth equal eys

17 Stablty Q. Whch sorts are stable? A. Inserton sort and mergesort (but not selecton sort or shellsort). sorted by tme sorted by locaton (not stable) sorted by locaton (stable) Chcago 09:00:00 Phoenx 09:00:03 Houston 09:00:13 Chcago 09:00:59 Houston 09:01:10 Chcago 09:03:13 Seattle 09:10:11 Seattle 09:10:25 Phoenx 09:14:25 Chcago 09:19:32 Chcago 09:19:46 Chcago 09:21:05 Seattle 09:22:43 Seattle 09:22:54 Chcago 09:25:52 Chcago 09:35:21 Seattle 09:36:14 Phoenx 09:37:44 Chcago 09:25:52 Chcago 09:03:13 Chcago 09:21:05 Chcago 09:19:46 Chcago 09:19:32 Chcago 09:00:00 Chcago 09:35:21 Chcago 09:00:59 Houston 09:01:10 Houston 09:00:13 Phoenx 09:37:44 Phoenx 09:00:03 Phoenx 09:14:25 Seattle 09:10:25 Seattle 09:36:14 Seattle 09:22:43 Seattle 09:10:11 Seattle 09:22:54 no longer sorted by tme Chcago 09:00:00 Chcago 09:00:59 Chcago 09:03:13 Chcago 09:19:32 Chcago 09:19:46 Chcago 09:21:05 Chcago 09:25:52 Chcago 09:35:21 Houston 09:00:13 Houston 09:01:10 Phoenx 09:00:03 Phoenx 09:14:25 Phoenx 09:37:44 Seattle 09:10:11 Seattle 09:10:25 Seattle 09:22:43 Seattle 09:22:54 Seattle 09:36:14 stll sorted by tme Stablty: nserton sort Proposton. Inserton sort s stable. publc class Inserton publc statc vod sort(comparable[] a) nt N = a.length; for (nt = 0; < N; ++) for (nt = ; > 0 && less(a[], a[-1]); --) exch(a,, -1); B1 A1 A2 A3 B2 1 0 A1 B1 A2 A3 B2 2 1 A1 A2 B1 A3 B2 3 2 A1 A2 A3 B1 B2 4 4 A1 A2 A3 B1 B2 Note. Need to carefully chec code ("less than" vs "less than or equal to"). A1 A2 A3 B1 B2 Pf. Equal tems never move past each other Stablty: selecton sort Proposton. Selecton sort s not stable. Stablty: shellsort Proposton. Shellsort sort s not stable. publc class Selecton publc statc vod sort(comparable[] a) nt N = a.length; for (nt = 0; < N; ++) nt mn = ; for (nt = +1; < N; ++) f (less(a[], a[mn])) mn = ; exch(a,, mn); mn B1 B2 A 1 1 A B2 B1 2 2 A B2 B1 A B2 B1 publc class Shell publc statc vod sort(comparable[] a) nt N = a.length; nt h = 1; whle (h < N/3) h = 3*h + 1; whle (h >= 1) for (nt = h; < N; ++) for (nt = ; > h && less(a[], a[-h]); -= h) exch(a,, -h); h h = h/3; B1 B2 B3 B4 A1 4 A1 B2 B3 B4 B1 Pf by counterexample. Long-dstance exchange mght move an tem past some equal tem. Pf by counterexample. Long-dstance exchanges. 1 A1 B2 B3 B4 B1 A1 B2 B3 B4 B

18 Stablty: mergesort Proposton. Mergesort s stable. Stablty: mergesort Proposton. Merge operaton s stable. publc class Merge prvate statc Comparable[] aux; prvate statc vod merge(comparable[] a, nt lo, nt md, nt h) /* as before */ prvate statc vod merge(comparable[] a, nt lo, nt md, nt h) for (nt = lo; <= h; ++) aux[] = a[]; prvate statc vod sort(comparable[] a, nt lo, nt h) f (h <= lo) return; nt md = lo + (h - lo) / 2; sort(a, lo, md); sort(a, md+1, h); merge(a, lo, md, h); nt = lo, = md+1; for (nt = lo; <= h; ++) f ( > md) a[] = aux[++]; else f ( > h) a[] = aux[++]; else f (less(aux[], aux[])) a[] = aux[++]; else a[] = aux[++]; publc statc vod sort(comparable[] a) /* as before */ A1 A2 A3 B D A4 A5 C E F G Pf. Suffces to verfy that merge operaton s stable. Pf. Taes from left subarray f equal eys