Kn65. Donald E. Knuth. On the Translation of Languages from Left to Right

• LitD:Kn65.pdf, Kn65↗ ⊥
• Donald E. Knuth
• 1965
• On the Translation of Languages from Left to Right
• INFORMATION AND CONTROL 8, 607-639 (1965)
• my python implementation

LR(k) Grammatik, Parsing und grundlegenden Eigenschaften

LR(k)

• L = Left to right parsing
• R = Reverse, Rightmost Derivation (Auflösung von Rules, Tree Linearisierung: immer rechtestes NonTerminal auflösen)
• k Lookahead Terminals for unique parsing

i Intro and Defs

• "intermediates" I (Rules): A, B, C
• "terminals" T: a, b, c
• S denotes the "principal intermediate character"
• ε empty string
• production is a relation A → Θ mit Θ Strings über I ∪ T
• Grammar G: {A_p → X_p1 ... X_pnp | rulenumber p ∈ [1,s], n_p >= 0 }
• φ → ψ direkte Ableitung: ein Intermediat aus φ ersetzen durch eine Produktion
• φ ⇒ ψ transitiver Abschluss von →
• φ ⇛ ψ transitiver+reflexiver Abschluss von → i.e. φ → ψ or φ = ψ
• language einer Grammatik: {α String über T | S ⇒ α}
• sentential form: any string α for which S ⇒ α
• derivation tree or parse diagram
  • Linearisierungen indem ein Intermediate nach dem anderen expandiert wird, Reihenfolge ist irrelevant, also z.B. leftmost oder rightmost auswählen
• "handle of a tree" to be the leftmost set of adjacent leaves forming a complete branch
• "pruning off" handles (d.h. leaves der handle entfernen) ==> Schritt für Schritt zurück nach S <==> rightmost derivation in revers
• (n, p) is "handle of α = X₁ ... X_n ... X_t (String over T ∪ I)" iff ∃ derivation tree von α mit der handle X_r+1 ... X_n for production p
• "k-sentential form" is a sentential form followed by k ⊣ with ⊣ ∉ T ∪ I
• a grammar is "LR(k)" iff for any k-sentiential
  • iff any handle is always uniquely determined by the string to its left and the k terminal characters to its right. Formally: iff ∀
  • α = X₁ ... X_n ... X_n+k Y₁ ... Y_u is a k-sentential form without Intermediates from X_n+1 to Y_u
  • β = X₁ ... X_n ... X_n+k Z₁ ... Z_v is a k-sentential form without Intermediates from X_n+1 to Z_v
  • (n, p) handle of α and (n', p') handle of β
  • then n = n' and p = p'
  • this means we do all possible reductions at n, and then step one forward - no backtracking needed
  • remarks
    • handle implies, no reductions left of n, however, we have to consider all possible trees
    • the pushdown automata the infinit number of possible prefixes (for n unlimited) to a finite number of combinations

ii. ANALYSIS OF LR(k) GRAMMARS

method 1: reduce to regular Grammar

define Grammar G' from G by

• Terminals: T ∪ I ∪ {⊣}
• Intermediates: [A, α] with α ∈ (T ∪ ⊣)^k and [p]
• H_k(σ) = {α | ∃ β: σ ⇒ αβ}
• rules
  • [A_p, α] → X_p1 ... X_p(j-1) [X_pj, β] with j <= n_p and β = H_k(X_p(j+1) ... X_pnp α)
  • [A_p, α] → X_p1 ... X_pnp α [p]

now, the following to are equivalent

• [S, ⊣^k] ⇒ X₁ ... X_n ... X_n+k [p] in G'
• there exists a k-sentential form of G: X₁ ... X_n ... X_n+k Y₁ ... Y_u with handle (n, p) and X_n+1 ... Y_u not intermediates

easy to see for Knuth, I don't see how to prove the left-most property of handle (may be is not necessary because it follows from the second property below?)

thus

• G is LR(k) if and only if
• [S, ⊣^k] ⇒ θ [p] and [S, ⊣^k] ⇒ θ φ [q] implies φ = ε and p = q in G'

but G' is regular and well known methods exist to check this

method 2: LR(k) pushdown parser

careful for emtpy productions A → ε! modify H_k(σ) to omit all derivations when an intermediate as the initial character is replaced by ε.

state = set of [p, j, α] meaning we have parsed β X_p1 ... X_pj and there is a sentential form β A_p α ...

stack: S₀X₁S₁X₂S₂ ... X_nS_n | Y₁ ... Y_k ω: left from the | are the already parsed characters and states, right follows the lookahead und the rest of the input

algorithm on stack as above:

• step1 compute recursively closure S' of S' = S_n ∪ {[q, 0, β] | ∃ [p,j; α] in S', X_p(j+1) = q and β ∈ H_k(X_p(j+2) ... X_pnp α)} : add all productions that could newly start
• step2 compute
  • Z = {β | ∃ [p, j; alpha;] in S', j < n_p and β ∈ H_k(X_p(j+1) ... X_pnp α)} : in rule p, but not at end
  • Z_p = {α | ∃ [p, n_p; α]} : at end of rule p
  • if these sets are not disjoint, error: grammar is not LR(k)
  • if lookahead Y₁ ... Y_k in Z then shift Y₁ onto stack ==> S₀X₁S₁X₂S₂ ... X_nS_nY₁ | Y₂...
  • elif lookahead in Y_p then reduce stack by rule p ==> S₀X₁S₁X₂S₂ ... X_n-npS_n-npA_p | Y₁ ...
  • else syntax error
• step3 after renaming the stack has now the form S₀X₁S₁X₂S₂ ... X_nS_nX_n+1 | Y₁ ... Y_k ω
  • compute S' from S_n as in step1 (add all productions that could newly start)
  • compute S_n+1 = {[p, j+1, α] | [p, j; a] ∈ S' and X_pj = X_n+1}: new state after X_n+1
  • if S_n+1 = {[0, 1, ⊢_k]} and lookahead = ⊢_k then parsing succesfully finished
  • else push S_n+1 on the stack and go to step1