Analyzing Algorithms
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- Annice Young
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1 Analyzing Algorithms
2 Insertion Sort (a) (d) (b) (e) (c) (f)
3 Insertion Sort
4 Insertion Sort (a) (b) (c) (d) (e) (f) INSERTION-SORT.A/ 1 for j D to A:length key D AŒj 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 4 i D j 1 5 while i>0and AŒi > key 6 AŒi C 1 D AŒi 7 i D i 1 8 AŒi C 1 D key Loop invariants and the correctness of insertion sort
5 Insertion Sort (a) (b) (c) (d) (e) (f) INSERTION-SORT.A/ 1 for j D to A:length key D AŒj 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 4 i D j 1 5 while i>0and AŒi > key 6 AŒi C 1 D AŒi 7 i D i 1 8 AŒi C 1 D key Loop invariants and the correctness of insertion sort
6 Correctness of Insertion Sort Is this algorithm correct?
7 Correctness of Insertion Sort Is this algorithm correct? Of course yes, we studied it in 31.
8 Correctness of Insertion Sort Is this algorithm correct? Of course yes, we studied it in 31. Why is correct? Who is able to prove this result?
9 Proof By Induction Claim:S(n) is true for all n >= k Basis: Show formula is true when n = k Inductive hypothesis: Assume formula is true for an arbitrary n Step: Show that formula is then true for n+1 `
10 Induction Example: Gaussian Closed Form Prove n = n(n+1) /
11 Induction Example: Gaussian Closed Form Prove n = n(n+1) / Basis: If n = 0, then 0 = 0(0+1) /
12 Induction Example: Gaussian Closed Form Prove n = n(n+1) / Basis: If n = 0, then 0 = 0(0+1) / Inductive hypothesis:
13 Induction Example: Gaussian Closed Form Prove n = n(n+1) / Basis: If n = 0, then 0 = 0(0+1) / Inductive hypothesis: Assume n = n(n+1) / Step (show true for n+1):
14 Induction Example: Gaussian Closed Form Prove n = n(n+1) / Basis: If n = 0, then 0 = 0(0+1) / Inductive hypothesis: Assume n = n(n+1) / Step (show true for n+1): n + n+1 = ( n) + (n+1) = n(n+1)/ + n+1 = [n(n+1) + (n+1)]/ = (n+1)(n+)/ = (n+1)(n+1 + 1) /
15 Induction Example: Geometric Closed Form Prove a 0 + a a n = (a n+1-1)/(a - 1) for all a 1 Basis: show that a 0 = (a 0+1-1)/(a - 1) a 0 = 1 = (a 1-1)/(a - 1) Inductive hypothesis: Assume a 0 + a a n = (a n+1-1)/(a - 1) Step (show true for n+1): a 0 + a a n+1 = a 0 + a a n + a n+1 = (a n+1-1)/(a - 1) + a n+1 = (a n+1+1-1)/(a - 1)
16 Insertion Sort and Loop Invariant INSERTION-SORT.A/ 1 for j D to A:length key D AŒj 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 4 i D j 1 5 while i>0and AŒi > key 6 AŒi C 1 D AŒi 7 i D i 1 8 AŒi C 1 D key Loop invariants and the correctness of insertion sort
17 Insertion Sort and Loop Invariant INSERTION-SORT.A/ 1 for j D to A:length key D AŒj 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 4 i D j 1 5 while i>0and AŒi > key 6 AŒi C 1 D AŒi 7 i D i 1 8 AŒi C 1 D key Loop invariants and the correctness of insertion sort At the start of each iteration of the for loop of lines 1 8, the subarray AŒ1 : : j 1 consists of the elements originally in AŒ1 : : j 1,butinsorted order. We use loop invariants to help us understand why an algorithm is correct. W must show three things about a loop invariant:
18 Insertion Sort and Loop Invariant (In the book) At the start of each iteration of the for loop of lines 1 8, the subarray AŒ1 : : j 1 consists of the elements originally in AŒ1 : : j 1,butinsorted order. We use loop invariants to help us understand why an algorithm is correct. W must Initialization: show three things It is true about priora toloop the first invariant: iteration of the loop. Maintenance: If it is true before an iteration of the loop, it remains true before the next iteration. Termination: When the loop terminates, the invariant gives us a useful property that helps show that the algorithm is correct. When the first two properties hold, the loop invariant is true prior to every iteration
19 Insertion Sort and Loop Invariant (In the book) At the start of each iteration of the for loop of lines 1 8, the subarray AŒ1 : : j 1 consists of the elements originally in AŒ1 : : j 1,butinsorted order. We Basis: use loop invariants to help us understand why an algorithm is correct. W must show three things about a loop invariant: Initialization: We start by showing that the loop invariant holds before the first loop iteration, when j D. 1 The subarray AŒ1 : : j 1, therefore,consists of just the single element AŒ1, whichisinfacttheoriginalelementinaœ1. Moreover, this subarray is sorted (trivially, of course), which shows that the loop invariant holds prior to the first iteration of the loop. Maintenance: Next, we tackle the second property: showing that each iteration
20 Insertion Sort and Loop Invariant (In the book) At the start of each iteration of the for loop of lines 1 8, the subarray AŒ1 : : j 1 consists of the elements originally in AŒ1 : : j 1,butinsorted order. loop invariant holds prior to the first iteration of the loop. We Induction use loop Hypothesis invariants + Stepto help us understand why an algorithm is correct. W must show three things about a loop invariant: Maintenance: Next, we tackle the second property: showing that each iteration maintains the loop invariant. Informally, the body of the for loop works by moving AŒj 1, AŒj, AŒj 3, andsoonbyonepositiontotheright until it finds the proper position for AŒj (lines 4 7), at which point it inserts the value of AŒj (line 8). The subarray AŒ1 : : j then consists of the elements originally in AŒ1 : : j,butinsortedorder.incrementingj for the next iteration of the for loop then preserves the loop invariant. Amoreformaltreatmentofthesecondpropertywouldrequireustostateand
21 Insertion Sort and Loop Invariant (In the book) At the start of each iteration of the for loop of lines 1 8, the subarray AŒ1 : : j 1 consists of the elements originally in AŒ1 : : j 1,butinsorted order. We Use use loop invariants to help us understand why an algorithm is correct. W must show informal of the loop three analysis thingstoabout showathat loop theinvariant: second property holds for the outer loop. Termination: Finally, we examine what happens when the loop terminates. The condition causing the for loop to terminate is that j>a:length D n. Because each loop iteration increases j by 1, wemusthavej D n C 1 at that time. Substituting n C 1 for j in the wording of loop invariant, we have that the subarray AŒ1 : : n consists of the elements originally in AŒ1 : : n, butinsorted order. Observing that the subarray AŒ1 : : n is the entire array, we conclude that the entire array is sorted. Hence, the algorithm is correct. We shall use this method of loop invariants to show correctness later in this
22 Algorithm complexity What is the Algorithm complexity?
23 Analysis of Algorithms Analysis is performed with respect to a computational model We will usually use a generic uniprocessor random-access machine (RAM) All memory equally expensive to access No concurrent operations All reasonable instructions take unit time Except, of course, function calls Constant word size Unless we are explicitly manipulating bits
24 Algorithm complexity INSERTION-SORT.A/ cost times 1 for j D to A:length c 1 n key D AŒj c n 1 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 0 n 1 4 i D j 1 c 4 n P 1 n 5 while i>0and AŒi > key c 5 t j P n 6 AŒi C 1 D AŒi c.t 6 j 1/ P 7 i D i 1 c n.t 7 j 1/ 8 AŒi C 1 D key c 8 n 1 The running time of the algorithm is the sum of running times
25 Algorithm complexity INSERTION-SORT.A/ cost times 1 for j D to A:length c 1 n key D AŒj c n 1 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 0 n 1 4 i D j 1 c 4 n P 1 n 5 while i>0and AŒi > key c 5 t j P n 6 AŒi C 1 D AŒi c.t 6 j 1/ P 7 i D i 1 c n.t 7 j 1/ 8 AŒi C 1 D key c 8 n 1 P The running time of the algorithm is the sum of running times columns, obtaining C C C T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 t j C c 6.t j 1/ C c 7.t j 1/ C c 8.n 1/ : X or inputs of a given size, an algo
26 Best Case Running Time INSERTION-SORT.A/ cost times 1 for j D to A:length c 1 n key D AŒj c n 1 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 0 n 1 4 i D j 1 c 4 n P 1 n 5 while i>0and AŒi > key c 5 t j P n 6 AŒi C 1 D AŒi c.t 6 j 1/ P 7 i D i 1 c n.t 7 j 1/ 8 AŒi C 1 D key c 8 n 1 The running time of the algorithm is the sum of running times
27 Best Case Running Time INSERTION-SORT.A/ cost times 1 for j D to A:length c 1 n key D AŒj c n 1 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 0 n 1 4 i D j 1 c 4 n 1 5 while i>0andx P n AŒi > key c 5 t j P n 6 AŒi C 1 D AŒi c.t 6 j 1/ P 7 i D i 1 c n.t 7 j 1/ columns, obtaining 8 AŒi C 1 D key c 8 n 1 The running time of the algorithm is the sum of running times T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 D X T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5.n 1/ C c 8.n 1/ D.c 1 C c C c 4 C c 5 C c 8 /n.c C c 4 C c 5 C c 8 /: We can express this running time as C for constants and P t j C c 6.t j 1/ C c 7 The vector is already ordered C C C.t j 1/ C c 8.n 1/ : or inputs of a given size, an algo
28 Best Case Running Time INSERTION-SORT.A/ cost times 1 for j D to A:length c 1 n key D AŒj c n 1 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 0 n 1 4 i D j 1 c 4 n 1 5 while i>0andx P n AŒi > key c 5 t j P n 6 AŒi C 1 D AŒi c.t 6 j 1/ P 7 i D i 1 c n.t 7 j 1/ columns, obtaining 8 AŒi C 1 D key c 8 n 1 The running time of the algorithm is the sum of running times T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 D X T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5.n 1/ C c 8.n 1/ D.c 1 C c C c 4 C c 5 C c 8 /n.c C c 4 C c 5 C c 8 /: We can express this running time as C for constants and P t j C c 6.t j 1/ C c 7 The vector is already ordered C C C.t j 1/ C c 8.n 1/ : or inputs of a given size, an algo
29 Best Case Running Time INSERTION-SORT.A/ cost times 1 for j D to A:length c 1 n key D AŒj c n 1 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 0 n 1 4 i D j 1 c 4 n 1 5 while i>0andx P n AŒi > key c 5 t j P n 6 AŒi C 1 D AŒi c.t 6 j 1/ P 7 i D i 1 c n.t 7 j 1/ columns, obtaining 8 AŒi C 1 D key c 8 n 1 The running time of the algorithm is the sum of running times T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 D X T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5.n 1/ C c 8.n 1/ D.c 1 C c C c 4 C c 5 C c 8 /n.c C c 4 C c 5 C c 8 /: We can express this running time as C for constants and P t j C c 6.t j 1/ C c 7 The vector is already ordered C C C.t j 1/ C c 8.n 1/ : or inputs of a given size, an algo an C b ear func rder th
30 Worst Case Running Time INSERTION-SORT.A/ cost times 1 for j D to A:length c 1 n key D AŒj c n 1 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 0 n 1 4 i D j 1 c 4 n P 1 n 5 while i>0and AŒi > key c 5 t j P n 6 AŒi C 1 D AŒi c.t 6 j 1/ P 7 i D i 1 c n.t 7 j 1/ 8 AŒi C 1 D key c 8 n 1 The running time of the algorithm is the sum of running times
31 Worst Case Running Time INSERTION-SORT.A/ cost times 1 for j D to A:length c 1 n key D AŒj c n 1 3 // Insert AŒj into the sorted sequence AŒ1 : : j 1. 0 n 1 4 i D j 1 c 4 n P 1 n 5 while i>0and AŒi > key c 5 t j P n 6 AŒi C 1 D AŒi c.t 6 j 1/ P 7 i D i 1 c n.t 7 j 1/ columns, obtaining 8 AŒi C 1 D key c 8 n 1 The running time of the algorithm is the sum of running times T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 X??? P t j C c 6.t j 1/ C c 7 The vector is in reverse sorted order C C C.t j 1/ C c 8.n 1/ : or inputs of a given size, an algo
32 P Worst Case Running Time columns, obtaining T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 t j C c 6 j D n.n C 1/ and.j 1/ D X 1 n.n 1/ (see Appendix A for a re.t j 1/ C c 7 C C C.t j 1/ C c 8.n 1/ : or inputs of a given size, an algo The vector is in reverse sorted order
33 P Worst Case Running Time columns, obtaining T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 t j C c X 6 j D n.n C 1/ and.j 1/ D X 1 n.n 1/ (see Appendix A for a re.t j 1/ C c 7 C C C.t j 1/ C c 8.n 1/ : The vector is in reverse sorted order the worst case, the running time of INSERTION-SORT is n.n C 1/ T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 1 n.n 1/ n.n 1/ C c 6 C c 7 C c 8.n 1/ c c c D C C C C C C c c c C or inputs of a given size, an algo
34 P Worst Case Running Time columns, obtaining T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 t j C c X 6 j D n.n C 1/ and.j 1/ D X 1 n.n 1/ (see Appendix A for a re C.t j 1/ C c 7 C C C.t j 1/ C c 8.n 1/ : or inputs of a given size, an algo The vector is in reverse sorted order the c5 worst case, the running time of INSERTION-SORT is C c 6 C c 7 n C c 1 C c C c 4 C c 5 c 6 c 7 C c 8 n.c C c 4 C c 5 C c 8 /: for constants n.n C 1/ T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 1 n.n 1/ n.n 1/ C c 6 C c 7 C c 8.n 1/ c c c D C C C C C C c c c C C C C D D C e can express this worst-case running time as C c5 C c 6 C c 7 n C c 1 C c C c 4 C c 5 c 6 c 7.c C c 4 C c 5 C c 8 /: C c 8 n e can express this worst-case running time as for constants
35 P Worst Case Running Time columns, obtaining T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 t j C c X 6 X n.n C 1/ j D 1 and n.n 1/.j 1/ D (see Appendix A for a re an C bn C c c ;itisthusaq.t j 1/ C c 7 C C C.t j 1/ C c 8.n 1/ : or inputs of a given size, an algo The vector is in reverse sorted order the worst case, the running time of INSERTION-SORT is n.n C 1/ T.n/ D c 1 n C c.n 1/ C c 4.n 1/ C c 5 1 n.n 1/ n.n 1/ C c 6 C c 7 C c 8.n 1/ c c c D C C C C C C c c c C C C C c5 D C c 6 C c 7 n C c 1 C c C c 4 C c 5 c 6 c 7 C c 8 n.c C c 4 C c 5 C c 8 /: e can express this worst-case running time as for constants
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