Fundamentals%20of%20Perfect%20Developer

1
Fundamentals of Perfect Developer

• A one-day hands-on tutorial
• Answers to Exercises

2
Suggested answers to Exercise 1

• const zeroToOneHundred seq of int 0..100
• property assert 42 in zeroToOneHundred, 101
  in zeroToOneHundred
• function divides(i, j int) boolpre j gt 0 i
  j 0
• const squaresOfPrimes seq of int for those
  i2..100 - forall j2..lti - divides(i,
  j) yield i 2
• function max(S set of int) intpre S.empty
  that xS - forall yS - y lt x

3
Suggested answers to Exercise 4

• A recursive solution to the first problem is
• function numLeadingSpaces(s string)
  natdecrease s ( s.empty s.head
  0, 1 numLeadingSpaces(s.tail)
• )
• A non-recursive solution to the first is
• function numLeadingSpaces(s string) nat that
  j0..s - (j s sj )
  (forall k0..ltj - sk )
• The following member functions may be useful
• take drop findFirst prepend

4
Suggested answers 4 (contd)

• function removeLeadingSpaces(s string)
  string s.drop(numLeadingSpaces(s))
• function firstWord(s string) string ( let n
  s.findFirst( ) n lt 0 s,
  s.take(n) )

5
Suggested answers 4 (contd)

• function splitIntoWords(s string) seq of
  stringdecrease s ( let stripped
  removeLeadingSpaces(s) stripped.empty
  seq of string, ( let w
  firstWord(stripped) splitIntoWords(stri
  pped.drop(w)) .prepend(w) )
  )

6
Suggested answers to Exercise 5

• function min2a(x, y int) int
• satisfy result lt x,
• result lt y,
• result x result y
• via
• if x gt y value y value x fi
• end
• function min2b(x, y int) int
• satisfy result lt x,
• result lt y,
• result x result y
• via
• value (x gt y y, x)
• end

7
Suggested answers 5 (contd)

• function findFirst1(s seq of int, x int) int
• satisfy 0 lt result lt s,
• result s sresult x,
• forall j0..ltresult - sj x
• via
• loop
• var i (nat in 0..s)! 0
• keep forall j0..lti' - sj x
• until i' s
• decrease s - i'
• if si x value i fi
• i! 1
• end
• value s
• end

8
Suggested answers 5 (contd)

• function findFirst1(s seq of int, x int) int
• satisfy 0 lt result lt s,
• result s sresult x,
• forall j0..ltresult - sj x
• via
• var i (nat in 0..s)! 0
• loop
• change i
• keep forall j0..lti' - sj x
• until i' s si' x
• decrease s - i'
• i! 1
• end
• value i
• end

9
Suggested answers 5 (contd)

• function numLeadingSpaces(s string) nat
• that j0..s
• - (j s sj )
• (forall k0..ltj - sk )
• via
• var i (nat in 0..s)! 0
• loop
• change i
• keep forall j0..lti'- sj
• until i' s si'
• decrease s - i'
• i! 1
• end
• value i
• end

10
Suggested answers 5 (contd)

• function splitIntoWords(s string) seq of string
• decrease s
• ( let stripped removeLeadingSpaces(s)
• stripped.empty
• seq of string,
• ( let w firstWord(stripped)
• assert w.empty
• splitIntoWords(stripped.drop(w)).prepen
  d(w)
• )
• )
• ...

11
Suggested answers 5 (contd)

• via
• var rslt seq of string ! seq of string
• loop
• var i (nat in 0..s)! 0
• change rslt
• keep splitIntoWords(s)
• rslt' splitIntoWords(s.drop(i'))
• until i' s
• decrease s - i'
• i! numLeadingSpaces(s.drop(i))
• if i lt s
• let w firstWord(s.drop(i))
• rslt! rslt.append(w), i! w
• fi
• end
• value rslt
• end

12
Suggested answers to Exercise 6
function longest(s seq of string) string
decrease s ( s.empty "",
( let temp
longest(s.front) s.last gt
temp s.last,
temp ) )
13
Suggested answers 6 (contd)
class ListOfStrings abstract var list seq
of string internal var long string
invariant long longest(list) interface
build post list! seq of string via
list! seq of string, long! "
end
14
Suggested answers 6 (contd)
schema !add(s string) post list!
list.append(s) via list!
list.append(s) if s gt long
long! s fi end function
longest string longest(list) via
value long end end