MERGESORT BBM ALGORITHMS DEPT. OF COMPUTER ENGINEERING ERKUT ERDEM. Mergesort. Feb. 27, 2014

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1 ergesort BB ALGOITHS Basc plan. Dvde array nto two halves. ecursvely sort each half. erge two halves. DPT. OF COPUT NGINING KUT D GSOT nput G S O T X A P L sort left half G O S T X A P L sort rght half G O S A L P T X merge results A G L O P S T X ergesort overvew Feb. 27, 2014 Acnowledgement: The course sldes are adapted from the sldes prepared by. Sedgewc and K. Wayne of Prnceton Unversty. 2 Abstract n-place merge Abstract n-place merge lo G sorted md md+1 A h C lo T G md md+1 A h C T sorted copy to auxlary array 3 4

2 Abstract n-place merge Abstract n-place merge compare mnmum n each subarray G A C T 5 6 Abstract n-place merge Abstract n-place merge A A G A C T compare mnmum n each subarray compare mnmum n each subarray 7 8

3 Abstract n-place merge Abstract n-place merge A C G A C T A C G A C T compare mnmum n each subarray compare mnmum n each subarray 9 10 Abstract n-place merge Abstract n-place merge A C G A C T A C A C T compare mnmum n each subarray compare mnmum n each subarray 11 12

4 Abstract n-place merge Abstract n-place merge A C A C T A C A C T compare mnmum n each subarray compare mnmum n each subarray G A C T Abstract n-place merge Abstract n-place merge A C A C T A C A C T compare mnmum n each subarray compare mnmum n each subarray 15 16

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

6 Abstract n-place merge Abstract n-place merge A C G T A C G T compare mnmum n each subarray one subarray exhausted, tae from other Abstract n-place merge Abstract n-place merge A C G T A C G 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 G T A C G T one subarray exhausted, tae from other both subarrays exhausted, done Abstract n-place merge ergng Q. How to combne two sorted subarrays nto a sorted whole. A. Use an auxlary array. lo h A C G T sorted nput copy merged result A A C 0 7 G C T 2 A C 1 7 G T 3 A C 2 7 G T 4 A C 2 8 G T 5 A C G 3 8 G T 6 A C G 4 8 T 7 A C G 5 8 T 8 A C G 5 9 T 9 A C G T 6 10 T A C G T Abstract n-place merge trace 27 28

8 ergng: Java mplementaton ergesort: 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; ++) copy 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 H I S T h merge publc class erge 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); A G H I L 29 lo md h ergesort: trace ergesort: anmaton 50 random tems 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) G S O T X A P L G S O T X A P L G S O T X A P L G S O T X A P L G S O T X A P L G S O T X A P L G O S T X A P L G O S T X A P L G O S T X A P L G O S T A X P L G O S A T X P L G O S A T X P L G O S A T X P L G O S A T X L P G O S A L P T X A G L O P S T X Trace of merge results for top-down mergesort result after recursve call algorthm poston n order current subarray not n order 31 32

9 ergesort: anmaton ergesort: emprcal analyss 50 reverse-sorted tems algorthm poston n order current subarray not n order unnng tme estmates: Laptop executes 10 8 compares/second. Supercomputer executes compares/second. nserton sort (N mergesort (N log N) computer thousand mllon bllon thousand mllon bllon home nstant 2.8 hours 317 years nstant 1 second 18 mn super nstant 1 second 1 wee nstant nstant nstant Bottom lne. Good algorthms are better than supercomputers ergesort: number of compares and array accesses Proposton. ergesort 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: Dvde-and-conquer recurrence: proof by pcture 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) N = N C (N) C ( N / 2 ) + C ( N / 2 ) + N for N > 1, wth C (1) = 0. D (N / 2) D (N / 2) 2 (N/2) = N 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. lg N D (N / 4) D (N / 4) D (N / 4) D (N / 2 ) D (N / 4) 4 (N/4) = N... 2 (N/2 ) = N... D (N) = 2 D (N / 2) + N, for N > 1, wth D (1) = 0. D (2) D (2) D (2) D (2) D (2) D (2) D (2) D (2) N/2 (2) = N N lg N 35 36

10 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] gven D (N) = 2 D (N/2) + N D (N) / N = 2 D (N/2) / N + 1 dvde both sdes by N = D (N/2) / (N/2) + 1 algebra = D (N/4) / (N/4) apply to frst term = D (N/8) / (N/8) 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 gven nductve hypothess... = D (N/N) / (N/N) apply to frst term agan = 2 N (lg (2N) 1) + 2N = 2 N lg (2N) algebra = lg N stop applyng, D(1) = 0 QD ergesort analyss: memory Proposton. ergesort uses extra space proportonal to N. Pf. The array needs to be of sze N for the last merge. A C D G H I N U V two sorted subarrays A B C D F G H I J N O P Q S T U V merged result Def. A sortng algorthm s n-place f t uses c log N extra memory. x. Inserton sort, selecton sort, shellsort. B F J O P Q S T Challenge for the bored. In-place merge. [Kronrod, 1969] ergesort: practcal mprovements Use nserton sort for small subarrays. ergesort has too much overhead for tny subarrays. Cutoff to nserton sort for 7 tems. 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); 39 40

11 ergesort: practcal mprovements Stop f already sorted. Is bggest tem n frst half smallest tem n second half? Helps for partally-ordered arrays. A B C D F G H I J N O P Q S T U V A B C D F G H I J N O P Q 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); ergesort: practcal mprovements lmnate 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. 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[++]; merge from to else f (less(a[], a[])) aux[] = a[++]; 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); 41 swtch roles of and 42 ergesort: vsualzaton Bottom-up mergesort frst subarray second subarray frst merge frst half sorted second half sorted result 43 Basc plan. Pass through array, mergng subarrays of sze 1. epeat 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) Trace of merge results for bottom-up mergesort Bottom lne. No recurson needed a[] G S O T X A P L G S O T X A P L G S O T X A P L G S O T X A P L G S O T X A P L G S O T X A P L G S O T A X P L G S O T A X P L G S O T A X P L G S O T A X P L G O S T A X P L G O S A T X P L G O S A T X L P G O S A T X L P G O S A L P T X A G L O P S T X 44

12 Bottom-up mergesort: Java mplementaton Bottom-up mergesort: vsual trace publc class ergebu 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, ath.mn(lo+sz+sz-1, N-1)); Bottom lne. Concse ndustral-strength code, f you have the space