The formula above is only valid, for the breakup of a jet system into a hadron plus a remainder-system, when the remainder mass is large. If the fragmentation algorithm were to be used all the way from the end to the one, the mass of the last hadron to be formed at the end would be completely constrained by global energy and momentum conservation, and could not be on its mass shell. In theory it is known how to take such effects into account [Edé00], but the resulting formulae are wholly unsuitable for Monte Carlo implementation.

The practical solution to this problem is to carry out the fragmentation both from the and the end, such that for each new step in the fragmentation process, a random choice is made as to from what side the step is to be taken. If the step is on the side, then is interpreted as fraction of the remaining of the system, while is interpreted as fraction for a step from the end. At some point, when the remaining mass of the system has dropped below a given value, it is decided that the next breakup will produce two final hadrons, rather than a hadron and a remainder-system. Since the momenta of two hadrons are to be selected, rather than that of one only, there are enough degrees of freedom to have both total energy and total momentum completely conserved.

The mass at which the normal fragmentation process is stopped and the final two hadrons formed is not actually a free parameter of the model: it is given by the requirement that the string everywhere looks the same, i.e. that the rapidity spacing of the final two hadrons, internally and with respect to surrounding hadrons, is the same as elsewhere in the fragmentation process. The stopping mass, for a given setup of fragmentation parameters, has therefore been determined in separate runs. If the fragmentation parameters are changed, some retuning should be done but, in practice, reasonable changes can be made without any special arrangements.

Consider a fragmentation process which has already split off a number
of hadrons from the and
sides, leaving behind a
a
remainder system. When this system breaks by the
production of a
pair, it is decided to make this pair
the final one, and produce the last two hadrons
and
, if

(249) |

(250) |

This procedure does not work all that well for heavy flavours, since it
does not fully take into account the harder fragmentation function
encountered. Therefore, in addition to the check above, one further
test is performed for charm and heavier flavours, as follows. If the
check above allows more particle production,
a heavy hadron
is formed, leaving a
remainder
. The range of allowed values, i.e. the
fraction of remaining that may be taken by the
hadron, is constrained away from 0 and 1 by the
mass and minimal mass of the
system.
The limits of the physical range is obtained when the
system only consists of one single particle, which then
has a well-determined
transverse mass
. From the value obtained with the
infinite-energy fragmentation function formulae, a rescaled value
between these limits is given by

(251) |

(252) |

The consequence of the procedure above is that, the more the infinite energy fragmentation function is peaked close to , the more likely it is that only two particles are produced. The procedure above has been constructed so that the two-particle fraction can be calculated directly from the shape of and the (approximate) mass spectrum, but it is not unique. For the symmetric Lund fragmentation function, a number of alternatives tried all give essentially the same result, whereas other fragmentation functions may be more sensitive to details.

Assume now that two final hadrons have been picked. If the transverse mass of the remainder-system is smaller than the sum of transverse masses of the final two hadrons, the whole fragmentation chain is rejected, and started over from the and endpoints. This does not introduce any significant bias, since the decision to reject a fragmentation chain only depends on what happens in the very last step, specifically that the next-to-last step took away too much energy, and not on what happened in the steps before that.

If, on the other hand, the remainder-mass is large enough, there
are two kinematically allowed solutions for the final two hadrons:
the two mirror images in the rest frame of the remainder-system. Also
the choice between these two solutions is given by the consistency
requirements, and can be derived from studies of infinite-energy jets.
The probability for the reverse ordering, i.e. where the rapidity and
the flavour orderings disagree, is given by the area law as

When baryon production is included, some particular problems arise
(also see section ). First consider
situations. In the naïve iterative scheme, away from the middle
of the event, one already has a quark and is to choose a matching
diquark flavour or the other way around. In either case the choice
of the new flavour can be done taking into account the number of
**SU(6)** states available for the quark-diquark combination.
For a case where the final
breakup is an
antidiquark-diquark one, the weights for forming
and
enter at the same time, however. We do not know how to
handle this problem; what is done is to use weights as usual for the
baryon to select , but then consider
as given (or the other way around with equal
probability). If
turns out to be
an antidiquark-diquark combination, the whole fragmentation chain is
rejected, since we do not know how to form corresponding hadrons.
A similar problem arises, and is solved in the same spirit, for a
configuration in which the (or ) was chosen
as third-last particle. When only two particles remain to be
generated, it is obviously too late to consider having a
configuration. This is as it should, however, as can be found by
looking at all possible ways a hadron of given rank can be a baryon.

While some practical compromises have to be accepted in the joining procedure, the fact that the joining takes place in different parts of the string in different events means that, in the end, essentially no visible effects remain.