1 / 16

CS100J Lecture 5

CS100J Lecture 5. Previous Lecture Programming Concepts Rules of thumb learn and use patterns inspiration from hand-working problem boundary conditions validation Pattern for processing input values up to (but not including) a stopping signal Example Processing exam grades

marthacarpenter
Télécharger la présentation

CS100J Lecture 5

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS100J Lecture 5 • Previous Lecture • Programming Concepts • Rules of thumb • learn and use patterns • inspiration from hand-working problem • boundary conditions • validation • Pattern for processing input values up to (but not including) a stopping signal • Example • Processing exam grades • Java Constructs • Casts and Rounding • Reading, Lewis & Loftus, Section 3.9 • This Lecture • Programming Concepts • Programming by stepwise refinement • a pattern • sequential refinement • case analysis • iterative refinement • Use of comments as higher-level statements Lecture 5

  2. Programming By Stepwise Refinement • An “algorithm” for you to follow when writing a program that solves problem P: if ( P is simple enough to code immediately ) Write the code that solves P; else { Refine P into subproblems; Write the code that solves each subproblem; } • Stepwise refinement is an example of • a divide-and-conquer algorithm, because it breaks problems down into simpler parts, • a recursive algorithm, because you use the same “algorithm” to write the code for any given subproblem. • The refinement of P into subproblems must include a description of how the code segments solving the subproblems combine to form code that solves P. • You can write the code segments that solve the subproblems in any order. Lecture 5

  3. Ways to Refine P into Subproblems • A program pattern • Do whatever n times • Process input values up until (but not including) a stopping value. • A sequential refinement • A case analysis • An iterative refinement Lecture 5

  4. Sequential Refinement • The refinement is structured so that solving P1 through Pnin sequence solves P. /* Solve problem P */ { /* Solve subproblem P1*/ . . . /* Solve subproblem P2*/ . . . . . . /* Solve subproblem Pn*/ . . . } Lecture 5

  5. Sequential Refinement, cont. • Example 1: /* Drive from Ithaca to NYC. */ { /* Drive from Ithaca to Binghamton */ . . . /* Drive from Binghamton to NYC */ . . . } • Example 2: /* Drive from Ithaca to NYC. */ { /* Drive from Ithaca to Albany. */ . . . /* Drive from Albany to NYC. */ . . . } Lecture 5

  6. Sequential Refinement, cont. • Example 1, continued /* Drive from Ithaca to NYC. */ { /* Drive from Ithaca to Binghamton. */ . . . /* Drive from Binghamton to NYC. */ /* Drive from Binghamton to Stroudsburg. */ . . . /* Drive from Stroudsburg to NYC. */ . . . } Lecture 5

  7. Sequential Refinement, cont. • Example 3 Let X1,…,Xn be an ordered sequence of variables. Let k be an integer between 1 and n. /* Rotate the values in X1,…,Xn left k places, where values shifted off the left end reenter at the right. */ E.g., suppose k = 3, n = 7, and the x’s contain letters. Before: x1 x2 x3 x4 x5 x6 x7 a b c d e f g After: x1 x2 x3 x4 x5 x6 x7 d e f g a b c Lecture 5

  8. Sequential Refinement, cont. /* Rotate the values in X1,…,Xn left k places, where values shifted off the left end reenter at the right. */ { /* Reverse the order of X1,…,Xk */ . . . /* Reverse the order of Xk+1,…,Xn */ . . . /* Reverse the order of X1,…,Xn */ . . . } Lecture 5

  9. Sequential Refinement, cont. • Example 3, continued E.g., suppose k = 3 and n = 7 x1 x2 x3 x4 x5 x6 x7 a b c d e f g c b a d e f g c b a g f e d d e f g a b c time Lecture 5

  10. Case Analysis • The refinement is structured so that solving one of the subproblems P1,…,Pn solves P. Which Pi is solved is determined during execution by testing conditions. /* Solve problem P */ if (P is an instance of case 1) /* Solve subproblem 1 */ . . . else if (P is an instance of case 2) /* Solve subproblem 2 */ . . . else if (P is an instance of case 3) /* Solve subproblem 3 */ . . . . . . else /* P is an instance of case n */ /* Solve subproblem n */ . . . Lecture 5

  11. Case Analysis, cont. • Example 1 /* Let x be the absolute value of y. */ if ( y < 0 ) /* Let x be -y. */ . . . else /* Let x be y. */ . . . • Example 2 /* Let x be the absolute value of y. */ x = Math.abs(y); I.e., sometimes case analysis is counter-productive and there is a uniform way to solve the problem. Lecture 5

  12. Iterative Refinement • The refinement of P is structured so that repeated solution of subproblem P’ eventually solves P. /* Solve problem P */ while (P has not yet been solved) /* Solve subproblem P’ */ . . . • Question: How can repeatedly doing P’ solve P ? • Answer: P’ must be parameterized in terms of some state. Each execution of P’ must change the state so that progress is made, i.e., with each iteration, we move to a state that is “closer” to a solution for P. • The notion of “distance” must be well-founded, i.e., it must converge to 0 in a finite number of steps that get “closer”. • Different notions of “distance” lead to different programs. Lecture 5

  13. Iterative Refinement, cont. • Example: Running a maze /* There are n2 rooms arranged in an n-by-n grid. Some adjacent rooms have connecting doors. No doors lead outside. You are in the upper-leftmost room facing left. A sequence of doors leads to the lower-rightmost room. Get there. */ • Rule of thumb: Work some test cases by hand. Inspiration: Keep your left hand on the wall and keep walking (the left-hand rule). Lecture 5

  14. Iterative Refinement, cont. /* There are n2 rooms arranged in an n-by-n grid. Some adjacent rooms have connecting doors. No doors lead outside. You are in the upper-leftmost room facing left. A sequence of doors leads to the lower-rightmost room. Get there. */ while ( you are not in lower rightmost room ) /* Get “closer” to lower rightmost room. */ . . . • Different notions of “closer” lead to different versions of /* Get “closer” to lower rightmost room. */ . . . Notion A: wall length away, using the “left-hand rule”. Notion B: # of rooms away, using the “left-hand rule”. Lecture 5

  15. Iterative Refinement, cont. • Refinement A (using wall-length distance) /* There are n2 rooms arranged in an n-by-n grid. Some adjacent rooms have connecting doors. No doors lead outside. You are in the upper-leftmost room facing left. A sequence of doors leads to the lower-rightmost room. Get there. */ while ( you are not in lower rightmost room ) /* Get at least one wall “closer” to the lower rightmost room. */ if (you are facing a door){ Go through the door; Turn 90 degrees counter-clockwise; } else Turn 90 degrees clockwise; Lecture 5

  16. Iterative Refinement, cont. • Refinement B (using #rooms distance) /* There are n2 rooms arranged in an n-by-n grid. Some adjacent rooms have connecting doors. No doors lead outside. You are in the upper-leftmost room facing left. A sequence of doors leads to the lower-rightmost room. Get there. */ while ( you are not in lower rightmost room ) { /* Get at least one room “closer” to the lower rightmost room. */ while (you are not facing a door) Turn 90 degrees clockwise; Go through door; Turn 90 degrees counter-clockwise; } Lecture 5

More Related