CS100J Lecture 5

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

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.

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

4
Sequential Refinement

• The refinement is structured so that solving P1
  through Pn in sequence solves P.
• / Solve problem P /
• / Solve subproblem P1 /
• . . .
• / Solve subproblem P2 /
• . . .
• .
• .
• .
• / Solve subproblem Pn /
• . . .

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. /
• . . .

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. /
• . . .

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 xs 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

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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 Xk1,,Xn /
• . . .
• / Reverse the order of X1,,Xn /
• . . .

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 ?
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 /
• . . .

11
Case Analysis, cont.

• Example 1
• / Let x be the absolute value of y. /
• if ( y lt 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.

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.

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).

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.

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

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