COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
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272 Donald E. Knuth and Peter B. Bendix Word problems in universal algebras 273<br />
we present here a method for solving certain word problems which are<br />
general enough to be of wide interest.<br />
The principal restriction is that we require all of the relations to be<br />
comparable in the sense of $ 2: we require that<br />
3. w (? (3.4)<br />
for each relation in R. In such a case we say R is a set of reductions. It<br />
follows from Theorem 2 that<br />
a + x’ implies a =- a’. (3.5)<br />
4. The completeness theorem. Let R be a set of reductions. We say a<br />
word a is irreducible with respect to R if there is no a’ such that a --) a’.<br />
It is not difficult to design an algorithm which determines whether or not<br />
a given word is irreducible with respect to R. If R = {(II, ol), . . . , (I,, em)},<br />
we must verify that no subword of a has the form of Al, or 22,. . . , or A,,,.<br />
If a is reducible with respect to R, the algorithm just outlined can be<br />
extended so that it finds some a’ for which a + a’. Now the same procedure<br />
can be applied to x’, and if it is reducible we can find a further word a”,<br />
andsoon.Wehavea+a’-a”-+...; so by (3.5) and the corollary to Theorem<br />
2, this process eventually terminates.<br />
Thus, there is an algorithm which, given any word a and any set of reductions<br />
R, finds an irreducible word a0 such that a = ao, with respect to R.<br />
We have therefore shown that each word is equivalent to at least one<br />
irreducible word. It would be very pleasant if we could also show that<br />
each word is equivalent to at most one irreducible word; for then the algorithm<br />
above solves the word problem! Take any two words a and p,<br />
and use the given algorithm to find irreducible a0 and PO. If a s /3, then<br />
a0 = DO, so by hypothesis a0 must be equal to PO. If a + /?, then a0 f PO,<br />
so a0 must be unequal to PO. In effect, a0 and /&, are canonical representatives<br />
of the equivalence classes.<br />
This pleasant state of affairs is of course not true for every set of reductions<br />
R, but we will see that it is true for surprisingly many sets and therefore<br />
it is an important property worthy of a special name. Let us say R<br />
is a complete set of reductions if no two distinct irreducible words are equivalent,<br />
with respect to R. We will show in the next section that there is an<br />
algorithm to determine whether or not a given set of reductions is complete.<br />
First we need to characterize the completeness condition in a more<br />
useful way.<br />
Let “+*” denote the reflexive transitive completion of “+“, so that<br />
a +*p means that there are words ao, al,. . . , a,, for some n 3 0 such<br />
that a = a,,, xj+aj+l for On, and a,=lQ.<br />
THEOREM 4. A set of reductions R is complete if and only if the following<br />
“lattice condition” is satisfied:<br />
Jfx-x’anda-a” there exists a wordy such that a’ + *y and a” - * y.<br />
Proof. If x + a’ and a -+ XI’, we can find irreducible words aA and aA’<br />
such that a’-+* x; and a’+* 31;‘. Since aA E a;‘, we may take y = ah = aA’<br />
if R is complete.<br />
Conversely let us assume that the lattice condition holds; we will show<br />
that R is complete. First, we show that if a -+* a0 and a -+* aA, wherea,-,<br />
and aA are irreducible, we must have a0 = a;. For if not, the set of all<br />
x which violate this property has no infinite decreasing sequence so there<br />
must be a “smallest” a (with respect to the Z- relation) such that a -* ao,<br />
SC+ *aA f ao, where both a0 and aA are irreducible. Clearly a is not itself<br />
irreducible, since otherwise a0 = x = a;. So we must have a + x0, a $ aA,<br />
and there must be elements al, a; such that a + al +* ao, a - a; +* a:.<br />
By the lattice condition there is a word y such that al +* y and xi -+ * y.<br />
Furthermore there is an irreducible word y. such that y +* ~0. Now by<br />
(3.5), a > al, so (by the way we chose a) we must have a0 = ~0. Similarly<br />
the fact that a r Z; implies that ah = yo. This contradicts the assumption<br />
that a0 $ ah.<br />
NOW to show that R is complete, we will prove the following fact:<br />
Vu G t!$a+*ao,andB-* /IO, where a0 and /?o are irreducible, then a0 = PO.<br />
Let the derivation of the relation a G /I be a = og+q++. . . ++o,, = B, where<br />
“u5’ denotes “+” or “+ “. If n = 0, we have a = ,B, hence a0 = PO by<br />
the proof in the preceding paragraph. If n = 1, we have either a -+ /? or<br />
B * a, and again the result holds by the preceding paragraph. Finally<br />
if n =- 1, let ol+* o;, where 0; is irreducible. By induction on n, we<br />
have 0; = jo, and also 01 = ao. Therefore R and the proof are both<br />
complete.<br />
5. The superposition process. Our immediate goal, in view of Theorem 4,<br />
is to design an algorithm which is capable of testing whether or not the<br />
“lattice condition” is satisfied for all words.<br />
En terms of the definitions already given, the hypothesis that a - a’ and<br />
a --f a” has the following detailed meaning: There are subwords p1 and<br />
pz of a, so that 3: has the form<br />
a = q&y1 = y2;32y)2.<br />
(5.1)<br />
There are also relations (]%I, gl), (1.2, 02) in R, and words 81, . . ., e,,<br />
0’1, . . .: o,, such that<br />
and<br />
p1 = Wl, . . .> 8,; Rl), p2 = S(Ol, . . ., 0,; 3.2) (5.2)<br />
a’ = q1S(6J1, . . . . 8,; pl)yl, a" = ~~S(o.1, . . . . 0,; ~2)w2. (5.3)<br />
The lattice condition will hold if we can find a word y such that a’ +* y<br />
and a” -.* y.<br />
Several possibilities arise, depending on the relative positions of t%<br />
and p2 in (5.1). If b1 and ,19~ are disjoint (have no common symbols), then