Maximum amount of work to spend on the computation

The environment variable `MAXEFFORT`

determines
the amount of effort allowed for heuristical parts of a computation,
measured in "working units". The default value is `MAXEFFORT`

=
1000000.

Possible values: Non-negative floating-point number; or `infinity`

.

`MAXEFFORT`

determines the maximum number of
"working units" that may be spent on internal heuristics.

One working unit roughly corresponds to 1000 evaluation steps done by an average kernel function.

Whatever `MAXEFFORT`

is set to, every MuPAD^{®} function
returns a correct though possibly unsimplified result; in particular,
some functions may return unevaluated. `MAXEFFORT`

determines
the amount of additional time spent on obtaining a better or more
simplified result; a value of `infinity`

means that all built-in heuristics
are really tried, a value of `0`

means that all heuristics
that might take considerable effort are left out.

A function whose result is uniquely specified has no way to
react to `MAXEFFORT`

.

Other functions carry out, in any event, all computations necessary
to obtain some correct result; `MAXEFFORT`

only determines
the time available for improving that result. In case of functions
that may return unevaluated immediately (e.g., `solve`

or `int`

), or may return their input immediately
(as, e.g., `simplify`

),
or may answer a question by `UNKNOWN`

immediately (as, e.g., `is`

), all of their time
consumption is counted to be spent on heuristics (purely heuristic
functions).

Purely heuristic functions will usually return immediately if
their input is quite complicated in relation to the effort allowed.
This is also true if the user has provided that input on the interactive
level. In order to pose a difficult problem where a longer running
time is acceptable, `MAXEFFORT`

should be increased.

A simplification achieved by heuristic methods may speed up
the deterministic parts, such that a small value of `MAXEFFORT`

does
not necessarily decrease the total computing time.

The user may employ `MAXEFFORT`

in his own
functions as follows: any function may use the amount of effort given
by `MAXEFFORT`

partly for own overhead, and distribute
the rest on the functions it calls. To do this, the caller has to `save`

the variable `MAXEFFORT`

and
set it to whatever it wants to make available to the called function.
Depending on whether the call is necessary to obtain a correct result
at all and whether the called function is a heuristic one, there are
the following cases to handle. If the call is necessary and the called
function is deterministic, `MAXEFFORT`

has no influence.
If the call is not absolutely necessary and as far as the called function
is deterministic, the caller has to subtract the necessary amount
as own overhead from `MAXEFFORT`

if enough is available;
otherwise, such call must not take place. As far as the called function
works heuristically (for whatever reason it was called), it has to
limit its efforts to the amount given by `MAXEFFORT`

.

In no event may the value of `MAXEFFORT`

on
entering a procedure be different from the value on leaving it, even
not in case of an error. `save`

must
be used to ensure this.

No function may distribute and/or use more than the amount it
has been given by its caller. The own overhead should be estimated;
if it is supposedly small, `MAXEFFORT`

may be ignored.

In order to avoid casual, not reproducible effects, e.g., by
other programs running on the same computer, `MAXEFFORT`

should
not be used in connection with time measurement using `time`

or `rtime`

. For example,
the running time saved in one recursive call according to time measurement
must not be supplied to another recursive call.

The decomposition of an integer into prime factors is unique;
hence the result of `ifactor`

is
uniquely determined, such that `ifactor`

does not react to `MAXEFFORT`

:

MAXEFFORT:= 0: ifactor(2^10 + 1)

The function solve may return unevaluated. Hence it will do so if there is no effort left to spend on the computation:

MAXEFFORT:= 0: solve(ln(x) + x = 3, x)

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