Recursion and Fixed-Point Theory

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Recursion and Fixed-Point Theory
Programming Language Principles Lecture 14

• Prepared by
• Manuel E. Bermúdez, Ph.D.
• Associate Professor
• University of Florida

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Recursion and Fixed-Point Theory

• We know that the meaning of
• let xP in Q
• is the value of
• (?x.Q)P
• So, what's the meaning of (the de-sugarized
  version of)
• let rec f n P in Q ?

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Consider Factorial

• let rec f n n eq 0 ? 1 nf(n-1) in f 3
• Without the 'rec', we'd have
• (? f.f 3) ? n.n eq 0 ? 1 nf(n-1)
• Note the last 'f' is free.
• The 'rec' makes the last 'f' bound to the ? n.
  ... expression.

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With 'rec', somehow

• (? f.f 3) ? n.n eq 0 ? 1
  nf(n-1)
• gt? (? n.n eq 0 ? 1 nf(n-1)) 3
• gt? 3 eq 0 ? 1 3f(3-1)
• gt? 3f(2)
• gtmagic 3(? n.n eq 0 ? 1 nf(n-1)) 2
• gt? 3(2 eq 0 ? 1 2f(2-1))
• gt? 32f(1)
• gtmagic 32(? n.n eq 0 ? 1 nf(n-1)) 1
• gt? 32(1 eq 0 ? 1 1f(1-1))
• gt? 321f(0)
• gtmagic 321(? n.n eq 0 ? 1 nf(n-1)) 0
• gt? 321(0 eq 0 ? 1 0f(0-1))
• gt? 3211
• gt? 6

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To dispel the magic, need some math

• Let F be a function. A value w is called a
  "fixed-point" of F if F w w.
• Example
• f ? x.x 2 - 6 has two fixed points,
• 3 and -2, because
• (? x.x 2 - 6) 3 gt? 3 2 - 6 gt? 3, and
• (? x.x 2 - 6) -2 gt? -2 2 - 6 gt? -2
• Only two fixed-points
• solutions to x2-x-6 0.

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Fixed-points can be functions

• T ? f. ? ().(f nil)2-6 has two fixed
  points,
• ? ().3 and ? ().-2, because
• (? f. ? ().(f nil)2-6) (? ().3)
• gt? ? ().((? ().3) nil)2-6
• gt? ? ().32-6
• gt? ? ().3
• and
• (? f. ? ().(f nil)2-6) (? ().-2)
• gt? ? ().((? ().-2) nil)2-6
• gt? ? ().-22-6
• gt? ? ().-2

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Let's Dispel the Magic

• let rec f n P in Q
• gt let rec f ? n.P in Q (fcn_form)
• lt let rec f (? f. ? n.P) f in Q
  (abstraction)
• lt let rec f F f in Q where F ? f. ? n.P
  (abstraction).
• Now there are no free occurrences of f in F, and
  only one free occurrence of f overall.

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Dispelling magic, (contd)

• But, let rec f F f in Q means we want f to be
  a fixed point of F !
• So, re-phrase it as
• let f A_fixed_point_of F in Q.
• The "A-fixed_point_of" operator is called a
  "fixed point finder". We call it "Y, or Ystar,
  or Y.

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Dispelling magic, (contd)

• So, we have
• let f Y F in Q
• gt (? f.Q) (Y F)
• (? f.Q) (Y (? f. ? n.P))
• Note No free occurrences of f !
• So, explaining the purpose of "rec" is the same
  as finding a fixed point of F.

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How do we find a suitable Y ?

• WE DON'T NEED TO !
• We only need to use some of its characteristics
• f Y F
• F f f
• Substituting 1) in 2), we obtain
• F (Y F) Y F
• This is the fixed point identity.

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Let's extend some earlier definitions

• Definition
• Axiom ? (rho) Y F gt? F (Y F).
• (Extended) Definition
• An applicative expression M is not in normal form
  if it is of the form Y N.
• (Extended) Definition
• A reduction sequence is in normal order if at
  every step we reduce the left-most Y or ?.

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Let's Do the Factorial
?

• This time, no magic. Just science.
• F ? f. ? n.n eq 0 ? 1 nf(n-1), so
• (? f.f 3) (Y F)
• gt? Y F 3
• gt? F (Y F) 3
• (? f. ? n.n eq 0 ? 1 nf(n-1)) (Y F) 3
• gt? (? n.n eq 0 ? 1 n Y F (n-1)) 3
• gt? ,? 3 Y F (2)
• gt? 3 F (Y F) 2

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Let's Do the Factorial (contd)

• 3 (? f. ? n.n eq 0 ? 1 nf(n-1)) (Y
  F) 2
• gt? 3 (? n.n eq 0 ? 1 n Y F (n-1)) 2
• gt?,? 32 Y F (1)
• gt? 32 F (Y F) 1
• 32 (? f. ? n.n eq 0 ? 1 nf(n-1)) (Y
  F) 1
• gt? 32 (? n.n eq 0 ? 1 n Y F (n-1)) 1
• gt?,? 321 Y F (0)
• gt? 321 F (Y F) 0
• 321 (? f. ? n.n eq 0 ? 1 nf(n-1))
  (Y F) 0
• gt? 321 (? n.n eq 0 ? 1 n Y F (n-1)) 0
• gt?,? 3211
• gt? 6

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Note

• Normal order is required. In PL order the
  evaluation would not terminate.
• Lemma YF is in fact the factorial function.
• Proof (by induction, see notes).

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Recursion via lambda calculus.

• The following value of Y will work
• Y ? F.(? f.f f) (? g. F (g g))
• Lemma Y satisfies the fixed-point identity.
• Proof see notes.

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Recursion in the CSE Machine

• Alternatives
• Use Y ? F.(? f.f f) (? g.F(g g)).
• Pretty tedious. Won't work, because it requires
  normal order.
• Delay evaluation of arguments in Y above, (as in
  Cond, earlier).
• Could use
• Z ? F.(? f.f f) (? h.F(? x.h h x)).
• EXTREMELY tedious.
• Our choice a variation of the fixed-point
  identity.

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Subtree transformation for rec

• Build AST for factorial, and standardize it.
• RPAL subtree transformation rule for rec.
  (remember we skipped it ?)
• Example factorial (see diagram)

• rec gt
• / \
• X gamma
• / \ / \
• X E Ystar lambda
• / \
• X E

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Control Structures and CSE Machine Evaluation

• See diagram.
• CSE Rule 12
• Applying Y to F
• Results in in (Y F). We represent it using a
  single symbol ?
• CSE Rule 13
• Applying Y to F.
• Results in F (Y F)

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Recursion and Fixed-Point Theory
Programming Language Principles Lecture 14

• Prepared by
• Manuel E. Bermúdez, Ph.D.
• Associate Professor
• University of Florida