CS211 Spring 2004 Lecture 06 Loops and their invariants. Software engineering reason for using loop invariants

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1 CS211 Spring 2004 Lecture 06 Loops and teir invariants Reading material: Tese notes. Weiss: Noting on invariants. ProgramLive: Capter 7 and 8 O! Tou ast damnale iteration and art, indeed, ale to corrupt a saint. Saespeare, Henry IV, Pt I, 1 ii Use not vain repetition, as te eaten do. Mattew V, 48 Your if is te only peacemaer; muc virtue if if. Saespeare, As You Lie It. /** Toen Scan.getToen() is first toen of a sentence for E. Parse it, giving error messages if tere are mistaes. After te parse, Scan.getToen() sould e te symol following te parsed E. Rule: E::= T { <+ -> T */ pulic static void parsee() { Toen toen= Scan.getToen(); int ind= toen.getkind(); wile (ind == Toen.PLUS ind == Toen.MINUS) { Expression rigt= parset(); exp= new BinOp(exp, toen, rigt); toen= Scan.getToen(); ind= toen.getkind(); return exp; To understand any single part of te loop, you first ave to understand te wole loop and its initialization! Tat is BAD BAD BAD!!! 1 2 Rule: E::= T { <+ -> T */ pulic static void parsee() { Toen toen= Scan.getToen(); int ind= toen.getkind(); // exp is te oject for te expression parsed so far // toen is next toen to process and ind is its ind wile (ind == Toen.PLUS ind == Toen.MINUS) { Expression rigt= parset(); exp= new BinOp(exp, toen, rigt); toen= Scan.getToen(); ind= toen.getkind(); invariant is true efore and after eac iteration // { R: exp is te oject for te expression parsed so far // toen is te next toen to process, and its not part of tis E return exp; 3 Rule: E::= T { <+ -> T */ pulic static void parsee() { Toen toen= Scan.getToen(); int ind= toen.getkind(); // exp is te oject for te expression parsed so far // toen is next toen to process and ind is its ind Loop question 1. Does te initialization mae te invariant true? 4 Rule: E::= T { <+ -> T */ pulic static void parsee() { Rule: E::= T { <+ -> T */ pulic static void parsee() { // exp is te oject for te expression parsed so far // toen is next toen to process and ind is its ind wile (ind == Toen.PLUS ind == Toen.MINUS) { // { R: exp is te oject for te expression parsed so far // toen is te next toen to process, and its not part of tis E Loopy question 2. Wen does loop stop? Can we tell tat result R is true ased on te invariant and te fact tat te loop condition is false? 5 // exp is te oject for te expression parsed so far // toen is next toen to process and ind is its ind wile ( ) { Expression rigt= parset(); Loopy question 3. Does eac iteration mae progress toward termination? 6

2 Rule: E::= T { <+ -> T */ pulic static void parsee() { // exp is te oject for te expression parsed so far // toen is next toen to process and ind is its ind wile (ind == Toen.PLUS ind == Toen.MINUS) { Expression rigt= parset(); exp= new BinOp(exp, toen, rigt); toen= Scan.getToen(); ind= toen.getkind(); Loopy question 4. Does eac iteration eep te invariant true? 7 initialization; // { invariant P wile ( B) { S // { R initialization; // { P wile ( B ) { // { P and B S // { P // {P and! B // { R Te four loopy questions 1. Does te initialization mae P true? 2. From P and falsity of B, can we conclude tat R is true? 3. Does eac iteration mae progress toward termination? 4. Does eac iteration eep P true? 8 Aout array segments [..-1]? as - elements +3 [..+3-1] as +3 - = 3 elements +2 [..+2-1] as +2 - = 2 elements +1 [..+1-1] as +1 - = 1 elements [..+0-1] as +0 - = 0 elements [..-1] is te empty segment starting at [] In te notation [i..j], we always require i! j+1 9 Linear searc: Find first occurrence of v in [..-1] Better spec.: Store an integer in i to trutify: postcondition: (0) v is not in [..i-1] Precondition (1) Eiter i= or v = [i]? i Postcondition v is not ere? and (i = or v = [i]) Oviously, start te searc from [], move forward. Wat is te invariant? Wat is te formula for te numer of values in [..i-1]? i - 10 Linear searc: Find first occurrence of v in [..-1] Binary searc: find v in sorted segment [..-1] i= ; i // v is not ere? wile (i!= && v!= [i]) { i= i+1;? Store in j to trutify: sorted j If v is in, ten [j] is its rigtmost occurrence. If v is not in, ten v elongs rigt after [j]. See Weiss and ProgramLive Section i v is not ere? and (i = or v = [i]) Wat if [..-1] is empty? Tat is, wat if =? Set j to -1, ecause ten: [..j] is [..-1] (empty) and [j+1..-1] is [..-1] (empty) 11 12

3 Binary searc: find v in sorted segment [..-1] Binary searc: find v in sorted segment [..-1] j j j i j i initialize: j= -1; i= ; j= -1; i= ; wile (? ) { Wen can it stop? Wen sould it continue? Binary searc: find v in sorted segment [..-1] Binary searc: find v in sorted segment [..-1] j j j= -1; i= ; wile ( j +1 < i ) { j i How does it mae progress toward termination? 15 j= -1; i= ; wile ( j +1 < i ) { int e= (j+i) / 2; // { j < e < i if ([e] <= v) j= e; else i= e; j i 1. Finds rigtmost value, if in. 2. Finds position were value elongs, if not in. 3. Wors on an empty array. 16 Buggy Binary Searc from a new text on programming /**Searc (sorted) for target using a inary searc. If found, store its position in p, oterwise store -1 in p */ int start= 0; // start of searc region int end=.lengt; // end of searc region int p= -1; // position of target // wile tere is still someting list left to searc and te element as not een // found wile (start <= end && position == -1) { int middle = (start+end)/2; if ( target < [middle]) end= middle - 1; else if (target > [middle]) {start= middle + 1; else position= middle; fails if.lengt = 1, [0] = 0, and target = 1 17 Buggy Binary Searc from a 2002 text on programming /**Searc (sorted) for target using a inary searc. If found, store its position in p, oterwise store -1 in p */ pulic static int execute(int[], int target) { int lo= 0; int i=.lengt; wile (true) { int centre = (i+lo)/2; if (centre == lo) { // only 2 items to test so it is centre, centre+1, or not in if ([centre] == target) return centre; else ([centre+1] == target) return center+1; else return -1; if ([centre] < target) lo= centre; else if (target < [centre]) i= centre; else return centre; suscript out of range if.lengt = 2, [0] = 10 [1] = 20 target = 30 18

4 Buggy Binary Searc from a 2002 text on programming Te mistaes are logical errors, not simple typos. Te comments are generally not elpful. Variales names give no elp at all, in fact are misleading int start= 0; // start of searc region int end=.lengt; // end of searc region int lo= 0; int i=.lengt; Neiter text, lie most texts, give asolutely no elp on ow to develop suc algoritms. You eiter can do it or you can t and you can see tat te autors can t. Dutc national flag: contains red, wite, and lue alls. precondition:? wat invariant to use? Weiss (page 169) says tat inary searc is surprisingly tricy to code. Tat s ecause e does not follow a metodology for developing loops. You can coose your style. Program lie tese texts, getting not only difficulty to understand and complicated code, or learn a simple way of developing loops using invariants and definitions of variales. 19 postcondition: reds wites lues 20 Dutc national flag: contains red, wite, and lue alls. Dutc national flag: contains red, wite, and lue alls. precondition:? precondition:? reds wites lues? 0 w n reds wites lues? How to mae progress in te repetend? (ody of te loop, te ting to e repeated). Cases: [] is lue: increment [] is wite: swap [], [], increment and [] is red: swap [], []; swap [], [w]; increment,, and w postcondition: reds wites lues postcondition: reds wites lues Dutc national flag: contains red, wite, and lue alls. Dutc national flag: contains red, wite, and lue alls. precondition:? precondition:? 0 w n invariant : reds wites? lues w= 0; = 0; = n-1; 0 w n // reds wites? lues Cases: [] is lue: Swap [], []; decrement [] is wite: increment [] is red: swap [], [w]; increment and w postcondition: reds wites lues wile ( <= ) { if ([] is lue) {swap [], []; = -1; else ([] is wite) = +1; else {Swap [], [w]; = +1; w= w+1; postcondition: reds wites lues 23 24

5 Partition algoritm of quicsort (wic we will see later) Plive, Sec. 8.5 (loo at CD), Weiss, page 298 Partition algoritm of quicsort (wic we will see later) precondition: x? precondition: x? (x is not a program variales; it is used just to sow wat is initially in []) Permute te values of [..] and store a value in j to trutify: j postcondition: <= x x >= x j R1: x <= x >= x initial: [..] = { possile final [..] = { possile final [..] = { [j] 25 Swap [] and [j] j postcondition: <= x x >= x 26 Partition algoritm of quicsort (wic we will see later) Partition algoritm of quicsort (wic we will see later) precondition: x? precondition: x? i j x <= x? >= x j R1: x <= x >= x i= +1; j= ; i j x <= x? >= x wile (i <= j) { if ([i] <= x) i= i+1; else {Swap [i], [j]; j= j-1; j R1: x <= x >= x 27 28