MATLAB® supports an important exception, called reductions,
to the rule that loop iterations must be independent. A reduction
variable accumulates a value that depends on all the iterations
together, but is independent of the iteration order. MATLAB allows
reduction variables in parfor
-loops.
Reduction variables appear on both side of an assignment statement,
such as any of the following, where expr
is a MATLAB expression.
X = X + expr | X = expr + X |
X = X - expr | See Associativity in Reduction Assignments in Further Considerations with Reduction Variables |
X = X .* expr | X = expr .* X |
X = X * expr | X = expr * X |
X = X & expr | X = expr & X |
X = X | expr | X = expr | X |
X = [X, expr] | X = [expr, X] |
X = [X; expr] | X = [expr; X] |
X = min(X, expr) | X = min(expr, X) |
X = max(X, expr) | X = max(expr, X) |
X = union(X, expr) | X = union(expr, X) |
X = intersect(X, expr) | X = intersect(expr, X) |
Each of the allowed statements listed in this table is referred to as a reduction assignment, and, by definition, a reduction variable can appear only in assignments of this type.
The following example shows a typical usage of a reduction variable X
:
X = ...; % Do some initialization of X parfor i = 1:n X = X + d(i); end
This loop is equivalent to the following, where each d(i)
is
calculated by a different iteration:
X = X + d(1) + ... + d(n)
If the loop were a regular for
-loop, the
variable X
in each iteration would get its value
either before entering the loop or from the previous iteration of
the loop. However, this concept does not apply to parfor
-loops:
In a parfor
-loop, the value of X
is
never transmitted from client to workers or from worker to worker.
Rather, additions of d(i)
are done in each worker,
with i
ranging over the subset of 1:n
being
performed on that worker. The results are then transmitted back to
the client, which adds the workers' partial sums into X
.
Thus, workers do some of the additions, and the client does the rest.
The following requirements further define the reduction assignments associated with a given variable.
Required (static): For any reduction variable, the same reduction function or operation must be used in all reduction assignments for that variable. |
The parfor
-loop on the left is not valid
because the reduction assignment uses +
in one
instance, and [,]
in another. The parfor
-loop
on the right is valid:
Invalid | Valid |
---|---|
parfor i = 1:n if testLevel(k) A = A + i; else A = [A, 4+i]; end % loop body continued end | parfor i = 1:n if testLevel(k) A = A + i; else A = A + i + 5*k; end % loop body continued end |
Required (static): If the
reduction assignment uses * or [,] ,
then in every reduction assignment for X , X must
be consistently specified as the first argument or consistently specified
as the second. |
The parfor
-loop on the left below is not
valid because the order of items in the concatenation is not consistent
throughout the loop. The parfor
-loop on the right
is valid:
Invalid | Valid |
---|---|
parfor i = 1:n if testLevel(k) A = [A, 4+i]; else A = [r(i), A]; end % loop body continued end | parfor i = 1:n if testLevel(k) A = [A, 4+i]; else A = [A, r(i)]; end % loop body continued end |
This section provides more detail about reduction assignments, associativity, commutativity, and overloading of reduction functions.
Reduction Assignments. In addition to the specific forms of reduction assignment listed in the table in Reduction Variables, the only other (and more general) form of a reduction assignment is
X = f(X, expr) | X = f(expr, X) |
Required (static): f can
be a function or a variable. If it is a variable, it must not be affected
by the parfor body (in other words, it is a broadcast
variable). |
If f
is a variable, then for all practical
purposes its value at run time is a function handle. However, this
is not strictly required; as long as the right-hand side can be evaluated,
the resulting value is stored in X
.
The parfor
-loop below on the left will
not execute correctly because the statement f = @times
causes f
to
be classified as a temporary variable and therefore is cleared at
the beginning of each iteration. The parfor
on
the right is correct, because it does not assign f
inside
the loop:
Invalid | Valid |
---|---|
f = @(x,k)x * k; parfor i = 1:n a = f(a,i); % loop body continued f = @times; % Affects f end | f = @(x,k)x * k; parfor i = 1:n a = f(a,i); % loop body continued end |
Note that the operators &&
and ||
are
not listed in the table in Reduction Variables. Except for &&
and ||
,
all the matrix operations of MATLAB have a corresponding function f
,
such that u op v
is equivalent to f(u,v)
.
For &&
and ||
, such
a function cannot be written because u&&v
and u||v
might
or might not evaluate v
, but f(u,v)
always evaluates v
before
calling f
. This is why &&
and ||
are
excluded from the table of allowed reduction assignments for a parfor
-loop.
Every reduction assignment has an associated function f
.
The properties of f
that ensure deterministic behavior
of a parfor statement are discussed in the following sections.
Associativity in Reduction Assignments. Concerning the
function f
as used in the definition of a reduction
variable, the following practice is recommended, but does not generate
an error if not adhered to. Therefore, it is up to you to ensure that
your code meets this recommendation.
Recommended: To get deterministic
behavior of parfor -loops, the reduction function f must
be associative. |
To be associative, the function f
must satisfy
the following for all a
, b
,
and c
:
f(a,f(b,c)) = f(f(a,b),c)
The classification rules for variables, including reduction
variables, are purely syntactic. They cannot determine whether the f
you
have supplied is truly associative or not. Associativity is assumed,
but if you violate this, different executions of the loop might result
in different answers.
Note:
While the addition of mathematical real numbers is associative,
addition of floating-point numbers is only approximately associative,
and different executions of this |
For example, the statement on the left yields 1, while the statement
on the right returns 1 + eps
:
(1 + eps/2) + eps/2 1 + (eps/2 + eps/2)
With the exception of the minus operator (-
),
all the special cases listed in the table in Reduction Variables have a corresponding (perhaps approximately)
associative function. MATLAB calculates the assignment X = X - expr
by
using X = X + (-expr)
. (So, technically, the function for calculating this
reduction assignment is plus
, not minus
.)
However, the assignment X = expr - X
cannot be written using an associative function, which
explains its exclusion from the table.
Commutativity in Reduction Assignments. Some associative
functions, including +
, .*
, min
, and max
, intersect
, and union
,
are also commutative. That is, they satisfy the following for all a
and b
:
f(a,b) = f(b,a)
Examples of noncommutative functions are *
(because
matrix multiplication is not commutative for matrices in which both
dimensions have size greater than one), [,]
, and [;]
.
Noncommutativity is the reason that consistency in the order of arguments
to these functions is required. As a practical matter, a more efficient
algorithm is possible when a function is commutative as well as associative,
and parfor
is optimized to exploit commutativity.
Recommended: Except in the
cases of * , [,] , and [;] ,
the function f of a reduction assignment should
be commutative. If f is not commutative, different
executions of the loop might result in different answers. |
Violating the restriction on commutativity in a function used for reduction, could result in unexpected behavior, even if it does not generate an error.
Unless f
is a known noncommutative built-in,
it is assumed to be commutative. There is currently no way to specify
a user-defined, noncommutative function in parfor
.
Overloading in Reduction Assignments. Most associative
functions f
have an identity element e
,
so that for any a
, the following holds true:
f(e,a) = a = f(a,e)
Examples of identity elements for some functions are listed in this table.
Function | Identity Element |
---|---|
+ | 0 |
* and .* | 1 |
[,] and [;] | [] |
& | true |
| | false |
MATLAB uses the identity elements of reduction functions when it knows them. So, in addition to associativity and commutativity, you should also keep identity elements in mind when overloading these functions.
Recommended: An overload of
+ , * , .* , [,] ,
or [;] should be associative if it is used in a
reduction assignment in a parfor . The overload
must treat the respective identity element given above (all with class double )
as an identity element. |
Recommended: An overload of + , .* , union ,
or intersect should be commutative. |
There is no way to specify the identity element for a function.
In these cases, the behavior of parfor
is a little
less efficient than it is for functions with a known identity element,
but the results are correct.
Similarly, because of the special treatment of X =
X - expr
, the following is recommended.
Recommended: An overload of
the minus operator (- ) should obey the mathematical
law that X - ( is equivalent to (X - . |
Suppose each iteration of a loop performs some calculation, and you are interested in finding which iteration of a loop produces the maximum value. This is a reduction exercise that makes an accumulation across multiple iterations of a loop. Your reduction function must compare iteration results, until finally the maximum value can be determined after all iterations are compared.
First consider the reduction function itself. To compare an iteration's result against another's, the function requires as input the current iteration's result and the known maximum result from other iterations so far. Each of the two inputs is a vector containing an iteration's result data and iteration number.
function mc = comparemax(A, B) % Custom reduction function for 2-element vector input if A(1) >= B(1) % Compare the two input data values mc = A; % Return the vector with the larger result else mc = B; end
Inside the loop, each iteration calls the reduction function
(comparemax
), passing in a pair of 2-element vectors:
The accumulated maximum and its iteration index (this
is the reduction variable, cummax
)
The iteration's own calculation value and index
If the data value of the current iteration is greater than the
maximum in cummmax
, the function returns a vector
of the new value and its iteration number. Otherwise, the function
returns the existing maximum and its iteration number.
The code for the loop looks like the following, with each iteration
calling the reduction function comparemax
to compare
its own data [dat i]
to that already accumulated
in cummax
.
% First element of cummax is maximum data value % Second element of cummax is where (iteration) maximum occurs cummax = [0 0]; % Initialize reduction variable parfor ii = 1:100 dat = rand(); % Simulate some actual computation cummax = comparemax(cummax, [dat ii]); end disp(cummax);