Iteration

1
Iteration
2
Outline

• Prerequisites
• Basic Recursion
• Processing Lists
• Objectives
• Iteration Concepts
• (do )
• (let )
• (set! )
• Example Iterative Fibonacci

3
Background

• Thus far, we have achieved repetitive processing
  by way of recursion
• Determine a terminating condition
• Call the function repetitively
• Step towards the terminating condition
• Example summing a list of numbers
• (define (sum lst)
• (if (empty? lst)
• 0
• ( (first lst) (sum (rest lst)))))

4
Iteration

• Used extensively in procedural and
  object-oriented languages
• Also has three components
• Determine a terminating condition
• Use an index to manage the repetition
• Step towards the terminating condition
• Example summing numbers from 1 to n
• sum number -gt number
• total the numbers from i to n
• (define (sum n)
• (do ((i 1 ( i 1))
• (sum 0 ( sum i)))
• ((gt i n) sum)))

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General Form of (do )

• (do ((ltvariable1gt ltinit1gt ltstep1gt)
• ...)
• (lttestgt ltexpressiongt ...)
• ltcommandgt ...)
• (do ) specifies
• a set of variables to be bound,
• how they are to be initialized at the start, and
• how they are to be updated on each iteration.
• When a termination condition is met, the loop
  exits after evaluating the ltexpressiongts.
• (do ) expressions are evaluated as follows
• The ltinitgt expressions are evaluated (in some
  order),
• the results of the ltinitgt expressions are stored
  in the bindings of the ltvariablegts, and then
• the iteration phase begins.

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General Form of (do ) continued

• (do ((ltvariable1gt ltinit1gt ltstep1gt)
• ...)
• (lttestgt ltexpressiongt ...)
• ltcommandgt ...)
• Each iteration begins by evaluating lttestgt
• if the result is false, the ltcommandgt expressions
  are evaluated in order for effect,
• the ltstepgt expressions are evaluated in some
  unspecified order,
• the ltvariablegts are bound to the results of the
  ltstepgts, and
• the next iteration begins.

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General Form of (do ) continued

• (do ((ltvariable1gt ltinit1gt ltstep1gt)
• ...)
• (lttestgt ltexpressiongt ...)
• ltcommandgt ...)
• If lttestgt evaluates to a true value, then
• the ltexpressiongts are evaluated from left to
  right and
• the value(s) of the last ltexpressiongt is(are)
  returned.
• If no ltexpressiongts are present, then the value
  of the do expression is unspecified.
• The region of the binding of a ltvariablegt
  consists of the entire do expression except for
  the ltinitgts. It is an error for a ltvariablegt to
  appear more than once in the list of do
  variables.
• A ltstepgt may be omitted, in which case the effect
  is the same as if (ltvariablegt ltinitgt ltvariablegt)
  had been written instead of (ltvariablegt ltinitgt).

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iteration.scm
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Code Used

• sum number -gt number
• total the numbers from i to n
• (define (sum n)
• (do ((i 1 ( i 1))
• (sum 0 ( sum i)))
• ((gt i n) sum)
• no body required
• ))
• Tests
• (sum 3) -gt 6
• (sum 6) -gt 21
• (sum 100) -gt 5050

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Questions?
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The use of (let )

• Pure functional programming does not permit
  persistence (remembering stuff) in any form
• This is nice, but there are times when you don't
  want to redo an expensive computation
• Consider this fragment from the fractal exercise
  earlier
• (let ((pc (new-location pa pb center radius)))
• (and (chord pa pc center radius
  color)
• (chord pc pb center radius
  color))))
• In this case, it took some effort to compute the
  new location referred to as pc.

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The use of (let )

• Without (let ), we would have had to do
• (and (chord pa
• (new-location pa pb center radius)
• center
• radius
• color)
• (chord (new-location pa pb center radius)
• pb
• center
• radius
• color))
• In summary, (let ) allows us to remember
  results within a function and re-use them as
  necessary

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The use of (set! )

• Just as (let ) allows us to save values we do
  not want to recompute,
• (set! ) allows us to replace values in
  iterative calculations
• Example
• (define (test from to)
• (let ((x from))
• (begin
• (display "x was ") (display x) (newline)
• (set! x to)
• (display "x is now ") (display x))))
• Note test with (1 2), ('fred 'joe), (list
  cons)

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Iterative Solution to Fibonacci

• Now that we have some ability to save values,
  let's re-think Fibonacci.
• (fib n) is computed as
• (fib (n-1) (fib (n-2))
• Suppose we did this iteratively starting at 2,
  saving the previous values of fib as we went

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Iterative Fibonacci
N is 8
I is 2
Fib-2 1
Fib-1 1
Value 1
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
16
Iterative Fibonacci
N is 8
I is 3
Fib-2 1
Fib-1 1
Value 1
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
17
Iterative Fibonacci
N is 8
I is 3
Fib-2 1
Fib-1 1
Value 1
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
18
Iterative Fibonacci
N is 8
I is 3
Fib-2 1
Fib-1 1
Value 1
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
19
Iterative Fibonacci
N is 8
I is 3
Fib-2 1
Fib-1 1
Value 2
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
20
Iterative Fibonacci
N is 8
I is 3
Fib-2 1
Fib-1 1
Value 2
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
21
Iterative Fibonacci
N is 8
I is 4
Fib-2 1
Fib-1 2
Value 3
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
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Iterative Fibonacci
N is 8
I is 5
Fib-2 2
Fib-1 3
Value 5
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
23
Iterative Fibonacci
N is 8
I is 6
Fib-2 3
Fib-1 5
Value 8
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
24
Iterative Fibonacci
N is 8
I is 7
Fib-2 5
Fib-1 8
Value 13
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return Value
25
Iterative Fibonacci
N is 8
I is 8
Fib-2 8
Fib-1 13
Value 21
no
I gt N?
Increase I Move data up Value fib-2 fib-1
yes
Return 21
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fib.scm
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Code Used

• fib-iter number -gt number
• compute the fibonacci number
• (define (fib-iter n)
• (let ((nm2 1)
• (nm1 1))
• (do ((i 2 ( i 1))
• (value 1 ( nm2 nm1)))
• ((gt i n) value)
• (begin
• (set! nm2 nm1)
• (set! nm1 value)))))

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Questions?
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Summary

• You should now know
• (do )
• (let )
• (set! )

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