4.2 Generalising the counting condition is
So, if we have |xav|yw1|z1aw2|z2a . . . wn|zna such that
Ψ(av) =
n
X
i=1
ki·Ψ(w1. . . awi)
with all rationalki ≥0, then the word is unsolvable since all the words w1. . . awi increase the token count on all critical places while av with the same relative occurrence of transitions decreases the token count.
The sequence |xabca|ycaaab|za is unsolvable. We cannot obtain the same rela-tive Parikh vector on both sides of y, i.e. Ψ(abca) =k1 ·Ψ(c) +k2·Ψ(ca) +k3 · Ψ(caa)+k4·Ψ(caaab)cannot be solved. There is a different argument, though: the transition a must have a negative effect on all critical places, since a fires directly before y. On the other hand, the effect of abca is also negative while the effect of caaab is positive, meaning that Ψ(caaab)−Ψ(abca) = Ψ(a) also must have a positive effect on the critical places, a contradiction.
Let us now examine the unsolvable sequence |xaccbc|yccab|za. Since accbc has a negative effect on the critical places and ccab has a positive effect, we conclude fromΨ(accbc)−Ψ(ccab) = Ψ(c) that chas a negative effect. On the other hand, cc enablesa after y, showing that cmust have a positive effect, a contradiction.
Finally, the sequence|xaccab|ycbaa|zais also unsolvable. This case is a bit more complicated. Due to Ψ(accab)−Ψ(cbaa) = Ψ(c) we know that c has a negative effect on the critical places. But since Ψ(cba)−Ψ(ab) = Ψ(c), where cba has a positive effect andab(in front of y) has a negative effect,c itself must also have a positive effect, a contradiction.
With these types of argument, all minimal unsolvable words over {a, b, c} up to at least length 10 can be shown unsolvable. The general argument (that also holds in the first examples) is that if we find a transition [sequence] which must have a positive and a negative effect for the same critical place(s), then we have found a separation failure and the word is unsolvable. This sounds a lot like a region approach.
Note that the two-letter case is a special case of our general argument of ‘a positive and a negative effect’ for unsolvable words. In a word |xav|ybw|za with v, w∈ {a, b}∗, b must have a positive effect, since a is disabled at y but not at z, and it cannot enable itself – sob must do it. If#a(av)·#b(bw)≤#b(av)·#a(bw)
– i.e. the condition for unsolvability of words over two letters – we can find factors j, k >0 such that j·#a(av) =k·#a(bw) and j ·#b(av)≥k·#b(bw), giving a a negative effect on the critical place(s).
Let us now think about the following Question Q: If a positive and a negative effect on the same place does not follow from any logical deduction, is the word then solvable? This could be possible. Any quantitative (not sign-based) relation on token effects might be achievable by multiplying all arc weights and token counts by some high number and then doing small changes somewhere in the net.
If the answer to Question Q is “yes”, then we can also ‘easily’ detect whether a word is minimal unsolvable, and we also know why minimal unsolvable words must start and end with the same letter and why no separation failure against the other letters occurs anywhere in such words.
If a minimal unsolvable word starting and ending with awould contain a sepa-ration failure against another letter (sayb), then the contradictory argument could not use the whole word, i.e. we could at least cut off a prefix up to (but excluding) the first b and a postfix after the last b and still have an unsolvable word.
Example 22. Consider the wordabccabbcabaa. This word has, according to APT, two separation failures, one between the two c’s, one between the two b’s. So, for
|xabc|yc|z1abbc|z2ab|z3a|z4a,
in Ψ(z1 −y) = Ψ(c), transition c has a positive effect on the critical places at y while inΨ(z4−y)−3·Ψ(y−x) =−Ψ(c)it has a negative effect on the same places.
Since the latter calculation uses the whole word (fromx to the finala which defines z4), cutting off any letters at the start or end would invalidate this argument. It also seems to be the only way to obtain the necessary contradiction.
If we find another separation failure at which we do not need the whole word for our contradiction, we could cut off letters at the start or end and the reduced word would remain unsolvable. So it would not be minimal. Let us check the other separation failure:
|x1abcc|x2ab|ybc|z1ab|z2a|z3a.
Here, we can argue for a negative effect fora: 2·Ψ(z1−y)−Ψ(y−x1) = −2·Ψ(a),
and for c: Ψ(z3 − y) − Ψ(y − x1) = Ψ(c), as well as positive effects for b: 2·Ψ(z2−y)−Ψ(y−x1) = 2·Ψ(b), and for c: Ψ(z3 −y)−2·Ψ(y−x2) = Ψ(c). The contradiction again comes from transition c and for that argument we need again the whole word. There seems to be no shorter argument. Since there are no other separation failures, the word must be minimal unsolvable.
We never need to show that a transition sequence has a positive and a negative effect at the same time. If we have
X
i
ji·Ψ(zi −y)−X
i
ki·Ψ(y−xi) = Ψ(τ) and
X
i
ji0·Ψ(zi−y)−X
i
k0iΨ(y−xi) = −Ψ(τ) then
X
i
(ji+ji0)·Ψ(zi−y) = X
i
(ki+k0i)·Ψ(y−xi).
It is sufficient to sum up the same Parikh vector in both the (allowed) prefixes and postfixes at y to get a contradiction and thus a separation failure.
Revisitabca|ycaaabaand mark the start and end points of the subwords ending and starting at y, respectively:
|x1abc|x2a|yc|z1a|z2a|z3ab|z4a.
NowΨ(z4−y) = Ψ(y−x1) + Ψ(y−x2), which contradicts caaabhaving a positive and abca and a having a negative effect on the critical places at y.
We can show the following ring of conclusions:
Take some word u1u2. . . um|v1v2. . . vna over Σ, such that each subword ui and vj except v1 starts with an a and no other subwords do. Assume there is a separation failure againsta for the ESSP in front of v1.
Then, from the theory of regions, we know that the linear system
#a(u1. . . um) #b(u1. . . um) #c(u1. . . um) . . .
#a(u2. . . um) #b(u2. . . um) #c(u2. . . um) . . .
... ... ...
#a(um) #b(um) #c(um) . . .
−#a(v1) −#b(v1) −#c(v1) . . .
−#a(v1v2) −#b(v1v2) −#c(v1v2) . . .
... ... ...
−#a(v1. . . vn) −#b(v1. . . vn) −#c(v1. . . vn) . . .
Ea
Eb
Ec
...
<0
is unsolvable. The subwords starting with a and ending at the separation point must reduce the number of tokens on the place to be constructed for the ESSP, while the subwords starting at the separation point and ending before an a must increase this token count (expressed by the negative coefficients).
Then, by Gordan’s Theorem [Dan63], the following dual linear system must have a non-zero, non-negative solution:
#a(u1. . . um) . . . #a(um) −#a(v1) . . . −#a(v1. . . vn)
#b(u1. . . um) . . . #b(um) −#b(v1) . . . −#b(v1. . . vn)
#c(u1. . . um) . . . #c(um) −#c(v1) . . . −#c(v1. . . vn)
... ... ... ...
x1
...
xm y1
...
yn
= 0
We can rewrite this as
#a(u1. . . um) . . . #a(um)
#b(u1. . . um) . . . #b(um)
#c(u1. . . um) . . . #c(um)
... ...
x1
...
xm
=
#a(v1) . . . #a(v1. . . vn)
#b(v1) . . . #b(v1. . . vn)
#c(v1) . . . #c(v1. . . vn)
... ...
y1
...
yn
or, using Parikh vectors, even shorter as
Ψ(u1. . . um) . . . Ψ(um)
x1
...
xm
=
Ψ(v1) . . . Ψ(v1. . . vn)
y1
...
yn
. We can interpret this as the existence of a sequenceσ whose Parikh vector can be expressed as a linear combination of the ui’s (always ending at the separation point) as well as of thevj’s (starting from there), i.e.
Ψ(σ) =X
i
Ψ(ui. . . um)·xi and Ψ(σ) =X
j
Ψ(v1. . . vj)·yj.
This sequence can be arbitrarily long (consider aw|waa with w ∈(Σ\ {a})+) for which the only solution to the linear system isx1 =y2 and y1 = 0).
The effect of this sequence on the placepto be constructed for the ESSP must be negative (Ep(σ)<0)) and positive (Ep(σ)>0) at the same time. We conclude that there must be a separation failure for this ESSP.
This not only is a positive solution to Question Q, it also shows that a gen-eralised counting condition for words over three or more letters, if the one were found, would be derivable from the theory of regions. In other words, one cannot hope for a significant enhancement of the synthesis efficiency in comparison with the region based approach.