CORRECTNESS ISSUES AND LOOP INVARIANTS

Transcription

1 The next everal lecture 2 Study algorithm for earching and orting array. Invetigate their complexity how much time and pace they take Formalize the notion of average-cae and wort-cae complexity CORRECTNESS ISSUES AND LOOP INVARIANTS Lecture 8 CS2110 Spring 2015 We want you to know thee algorithm Not y memorizing code ut y Being ale to develop the algorithm from their pecification and, when neceary, a mall idea We give you ome guideline and intruction on how to develop an algorithm from it pecification. Deal mainly with developing loop. 3 Many (mot) of you could ue intruction on developing algorithm, keeping thing imple String[] dummy =.plit(""); // turn into tring array int len =.length()-1; // length of tring String a = ; // will e revere of for (int = len; > -1; --){ a= a.dummy[]; if (.equal(a)) return true; ele return fale; return.equal() Thi umitted code for ody of ipalindrome didn t work ecaue plit wan t ued properly and it wan t deugged Why calculate the revere of 4 Some principle and trategie for development Don t introduce a variale without a good reaon. Put local variale a cloe to their firt ue a poile. Structure expreion to make them readale. Make the tructure of the program reflect the tructure of the data. Never have lot of yntax error. Interpere coding and teting: code a little, tet a little. Write the cla invariant while putting in field declaration. Write a method pec efore writing the method ody. Ue aert tatement to check method precondition a along a it doen t complicate program too much and doen t change the time-complexity of the method. 5 Show development of ipalindrome /** Return true iff i a palindrome */ pulic tatic oolean ipalindrome(string ) Our intruction aid to viit each char of only once! 6 ipalindrome: Set ipal to i a palindrome (forget aout return for now. Store value in ipal. Think of checking equality of outer char, then char inide them, then char inide them, etc. 0.length() ac ca Key idea: Generalize thi to a picture that i true efore/after each iteration 1

2 7 ipalindrome: Set ipal to i a palindrome (forget aout return for now. Store value in ipal. Generalize to a picture that i true efore/after each iteration 0.length() ac ca 0 h k.length() Thee ection are each other revere 8 ipalindrome: Set ipal to i a palindrome int h= 0; Initialization to make picture true int k=.length() 1; // [0..h-1] i the revere of [k+1..] Stop when reult i known Continue when it not while ( h < k &&.charat(h) ==.charat(k) ) { h= h+1; k= k-1; ipal= h >= k; 0 h k.length() Make progre toward termination AND keep picture true Thee ection are each other revere 9 ipalindrome /** Return true iff i a palindrome */ pulic tatic oolean ipal(string ) { int h= 0; int k=.length() 1; // invariant: [0..h-1] i revere of [k+1..] while (h < k) { if (.charat(h)!=.charat(k)) return fale; h= h+1; k= k-1; return true; h 0 k.length() Thee ection are each other revere Loop invariant invariant ecaue it true efore/ after each loop iteration 10 Engineering principle Break a project up into part, making them a independent a poile. When the part are contructed, put them together. Each part can e undertood y itelf, without mentioning the other. Reaon for introducing loop invariant Invariant: i true efore and after each iteration 11 Given c >= 0, tore ^c in x z= 1; x= ; y= c; if (y i even) { ele { {z = ^c Algorithm to compute ^c. Can t undertand any piece of it without undertanding all. In fact, only way to get a handle on it i to execute it on ome tet cae. Need to undertand initialization without looking at any other code. Need to undertand condition y!= 0 without looking at loop ody Etc. 12 initialization; // invariant P while (B) {S Upon termination, we know P true, B fale init {P true B S fale {P and! B invariant mean unchanging. Loop invariant: an aertion a true-fale tatement that i true efore and after each iteration of the loop every time B i to e evaluated. Help u undertand each part of loop without looking at all other part. 2

3 13 Simple example to illutrate methodology Store um of 0..n in // { n >= 0 k= 1; = 0; // = um of 0..k-1 && = + k; { = um of 0..n Firt loopy quetion. Doe it tart right Doe initialization make invariant true Ye! = um of 0..k-1 = <utitute initialization> 0 = um of = <arithmetic> 0 = um of 0..0 We undertand initialization without looking at any other code 14 Simple example to illutrate methodology Store um of 0..n in // { n >= 0 k= 1; = 0; // = um of 0..k-1 && = + k; { = um of 0..n Second loopy quetion. Doe it top right Upon termination, i potcondition true Ye! inv &&! k <= n => <look at inv> inv && k = n+1 => <ue inv> = um of 0..n+1-1 We undertand that potcondition i true without looking at init or repetend Simple example to illutrate methodology Simple example to illutrate methodology 15 Store um of 0..n in // { n >= 0 k= 1; = 0; // = um of 0..k-1 && = + k; { = um of 0..n Third loopy quetion. Progre Doe the repetend make progre toward termination Ye! Each iteration increae k, and when it get larger than n, the loop terminate We undertand that there i no infinite looping without looking at init and focuing on ONE part of the repetend. 16 Store um of 0..n in // { n >= 0 k= 1; = 0; // = um of 0..k-1 && = + k; { = um of 0..n Fourth loopy quetion. Invariant maintained y each iteration I thi property true {inv && k <= n repetend {inv Ye! { = um of 0..k-1 = + k; { = um of 0..k k= k+1; { = um of 0..k-1 4 loopy quetion to enure loop correctne Note on range m..n 17 {precondition Q init; // invariant P while (B) { S {R Four loopy quetion: if anwered ye, algorithm i correct. Firt loopy quetion; Doe it tart right I {Q init {P true Second loopy quetion: Doe it top right Doe P &&! B imply R Third loopy quetion: Doe repetend make progre Will B eventually ecome fale Fourth loopy quetion: Doe repetend keep invariant true I {P &&! B S {P true 18 Range m..n contain n+1 m int: m, m+1,..., n (Think aout thi a Follower (n+1) minu Firt (m) ) 2..4 contain 2, 3, 4: that i = 3 value 2..3 contain 2, 3: that i = 2 value 2..2 contain 2: that i = 1 value 2..1 contain : that i = 0 value Convention: notation m..n implie that m <= n+1 Aume convention even if it i not mentioned! If m i 1 larger than n, the range ha 0 value array egment [m..n]: m n 3

4 Can t undertand thi example without invariant! For loopy quetion to reaon aout invariant 19 Given c >= 0, tore ^c in z z= 1; x= ; y= c; // invariant y >= 0 && // z*x^y = ^c if (y i even) { ele { {z = ^c Firt loopy quetion. Doe it tart right Doe initialization make invariant true Ye! z*x^y = <utitute initialization> 1*^c = <arithmetic> ^c We undertand initialization without looking at any other code 20 Given c >= 0, tore ^c in x z= 1; x= ; y= c; // invariant y >= 0 AND // z*x^y = ^c if (y i even) { ele { {z = ^c Second loopy quetion. Doe it top right When loop terminate, i z = ^c Ye! Take the invariant, which i true, and ue fact that y = 0: z*x^y = ^c = <y = 0> z*x^0 = ^c = <arithmetic> z = ^c We undertand loop condition without looking at any other code For loopy quetion to reaon aout invariant For loopy quetion to reaon aout invariant 21 Given c >= 0, tore ^c in x z= 1; x= ; y= c; // invariant y >= 0 AND // z*x^y = ^c if (y i even) { ele { {z = ^c Third loopy quetion. Doe repetend make progre toward termination Ye! We know that y > 0 when loop ody i executed. The loop ody decreae y. We undertand progre without looking at initialization 22 Given c >= 0, tore ^c in x z= 1; x= ; y= c; // invariant y >= 0 AND // z*x^y = ^c if (y i even) { ele { {z = ^c Fourth loopy quetion. Doe repetend keep invariant true Ye! Becaue of propertie: For y even, x^y = (x*x)^(y/2) z*x^y = z*x*x^(y-1) We undertand invariance without looking at initialization Develop inary earch for v in orted array Develop inary earch in orted array for v 23 pre: 0.length 24 pre: 0.length pot: Example: pre: 0 h.length <= v > v length If v i 4, 5, or 6, h i 5 If v i 7 or 8, h i 6 If v in, h i index of rightmot occurrence of v. If v not in, h i index efore where it elong. Store a value in h to make thi true: 0 h.length pot: <= v > v Get loop invariant y comining pre- and potcondition, adding variale t to mark the other oundary 0 h t.length <= v > v 4

5 How doe it tart (what make the invariant true) When doe it end (when doe invariant look like potcondition) 25 pre: 0.length 26 pot: 0 h.length <= v > v 0 h t.length <= v > v 0 h t.length <= v > v Make firt and lat partition empty: h= -1; t=.length; h= -1; t=.length; Stop when ection i empty. That i when h = t-1. Therefore, continue a long a h!= t-1. How doe ody make progre toward termination (cut in half) and keep invariant true How doe ody make progre toward termination (cut in half) and keep invariant true 27 h= -1; t=.length; int e= (h+t)/2; 0 h t.length <= v > v 0 h e t.length <= v > v Let e e index of middle value of Section. Maye we can et h or t to e, cutting ection in half 28 h= -1; t=.length; int e= (h+t)/2; if ([e] <= v) h= e; 0 h t.length <= v > v 0 h e t.length <= v > v 0 h e t.length <= v <= v > v If [e] <= v, then o i every value to it left, ince the array i orted. Therefore, h= e; keep the invariant true. How doe ody make progre toward termination (cut in half) and keep invariant true Develop inary earch in orted array for v 29 0 h t.length h= -1; t=.length; int e= (h+t)/2; if ([e] <= v) h= e; ele t= e; <= v > v 0 h e t.length <= v > v 0 h e t.length <= v > v > v If [e] > v, then o i every value to it right, ince the array i orted. Therefore, t= e; keep the invariant true length pre: Store a value in h to make thi true: 0 h.length pot: <= v > v DON T TRY TO MEMORIZE CODE! Intead, learn to derive the loop invariant from the preand pot-condition and then to develop the loop uing the pre- and pot-condition and the loop invariant. PRACTICE THIS ON KNOWN ALGORITHMS! 5

6 Proceing array from eg to end (or end to eg) Proceing array from eg to end (or end to eg) 31 Many loop proce element of an array (or a String, or any lit) in order: [0], [1], [2], If the potcondition i R: [0...length-1] ha een proceed Then in the eginning, nothing ha een proceed, i.e. [0..-1] ha een proceed After k iteration, k element have een proceed: P: [0..k-1] ha een proceed 0 k.length invariant P: proceed not proceed 32 Replace.length in potcondition Tak: Proce [0...length-1] y freh variale k to get invariant k= 0; [0..k-1] ha een proceed {inv P while ( k!=.length ) { or draw it a a picture Proce [k]; // maintain invariant // progre toward termination {R: [0...length-1] ha een proceed 0 k.length inv P: proceed not proceed 33 Proceing array from eg to end (or end to eg) 34 Count the numer of zero in. Start with lat program and refine it for thi tak Tak: Proce [0...length-1] k= 0; {inv P while ( k!=.length ) { Proce [k]; // maintain invariant // progre toward termination {R: [0...length-1] ha een proceed Mot loop that proce the element of an array in order will have thi loop invariant and will look like thi. Tak: Set to the numer of 0 in [0...length-1] k= 0; = 0; {inv P while ( k!=.length ) { Proce if ([k] == [k]; 0) = // + maintain 1; invariant // progre toward termination {R: = numer of 0 in [0...length-1] 0 k.length 0 k.length inv P: proceed not proceed inv P: = # 0 here not proceed Be careful. Invariant may require proceing element in revere order! Proce element from end to eginning Thi invariant force proceing from eginning to end 0 k.length inv P: proceed not proceed k=.length 1; // how doe it tart while (k >= 0) { // how doe it end Proce [k]; // how doe it maintain invariant Thi invariant force proceing from end to eginning 0 k.length k= k 1; // how doe it make progre {R: [0...length-1] i proceed inv P: not proceed proceed 0 k.length inv P: not proceed proceed 6

7 Proce element from end to eginning 37 k=.length 1; while (k >= 0) { Proce [k]; k= k 1; {R: [0...length-1] i proceed Head up! It i important that you can look at an invariant and decide whether element are proceed from eginning to end or end to eginning! For ome reaon, ome tudent have difficulty with thi. A quetion like thi could e on the prelim! 0 k.length inv P: not proceed proceed 7