The `ROUND_ROBIN` algorithm can lead to overburdened endpoints, especially when weights
    are used. The `LEAST_REQUEST` algorithm, distributes more evenly across and is far less
    likely to overburden endpoints. A number of experiments (by both the Istio and
    Envoy teams) have shown that `LEAST_REQUEST` outperforms `ROUND_ROBIN` in virtually all

.md-footer-meta {
  padding-bottom: 2em;
}

.user-list {
  display: flex;
  flex-wrap: wrap;
  margin-bottom: 2rem;
}

.user-list-center {
  justify-content: space-evenly;
}

.user {
  margin: 1em;
  min-width: 7em;
}

.user .avatar-wrapper {
  width: 80px;
  height: 80px;
  margin: 10px auto;
  overflow: hidden;
  border-radius: 50%;
  position: relative;

}

// -----

// Check number of replicated inputs is evenly divisible by 'n'.
func.func @verifier_replicate_bad_operandSegmentSizes(%arg0: tensor<*xi32>) {
  "tf_device.replicate" (%arg0, %arg0, %arg0, %arg0) ({
// expected-error@-1 {{'tf_device.replicate' op expects number of replicated inputs (4) to be evenly divisible by 'n' (3)}}
  ^entry(%input0: tensor<*xi32>, %input1: tensor<*xi32>):
    tf_device.return

// NUMA nodes to allocate any 'remainder' CPUs from (in cases where the total
// number of CPUs to allocate cannot be evenly distributed across the chosen
// set of NUMA nodes). This "balance score" is calculated as the standard
// deviation of how many CPUs will be available on each NUMA node after all
// evenly distributed and remainder CPUs are allocated. The subset with the

    if (expectedBucketSize == 0) {
        // The elements in the list are so small that they can't even be divided into {expectedBucketNumber}.
        // For example, how do you split [0,0,0,0,0] into 3 buckets?
        // In this case, we simply put the elements into these buckets evenly.
        return list.chunked(list.size / expectedBucketNumber, smallElementAggregateFunction)
    }

     * guard instead.
     */
    public static <T> ProducerGuard<T> adaptive() {
        return new AdaptiveProducerGuard<T>();
    }

    /**
     * Creates a {@link ProducerGuard} which evenly spreads calls over a fixed number of locks.
     * This means that in some cases two different keys can block on the same lock. The benefit of
     * this strategy is that it uses only a fixed amount of memory. If your code depends on

        return mlir::emitError(
            location,
            llvm::formatv(
                "incorrect input sharding configuration received. "
                "{0}-th dimension of the input must be evenly divisible by {1}",
                split_dimension, num_split));
      }

      shape[split_dimension] = shape[split_dimension] / num_split;
      output_type =

	// any possible naming scheme will lead to ambiguity to some extent.
	// We picked "pcpu" because it the established docs hints at vCPU already.
	FullPhysicalCPUsOnly bool
	// Flag to evenly distribute CPUs across NUMA nodes in cases where more
	// than one NUMA node is required to satisfy the allocation.
	DistributeCPUsAcrossNUMA bool
	// Flag to ensure CPUs are considered aligned at socket boundary rather than

	for {
		if _, r = t.div(r, q, f); len(r) != 0 {
			break // f doesn't divide q evenly
		}
		tab = append(tab, f)
		f = nat(nil).sqr(f) // nat(nil) to ensure a new f for each table entry
	}

	// Factor q using the table entries, if any.
	// We start with the largest factor f = tab[len(tab)-1]
	// that evenly divides q. It does so at most once because
	// otherwise f·f would also divide q. That can't be true

        try {
          assertEquals(p + "/" + q, p, IntMath.divide(p, q, UNNECESSARY) * q);
          assertTrue(p + "/" + q + " not expected to divide evenly", dividesEvenly);
        } catch (ArithmeticException e) {
          assertFalse(p + "/" + q + " expected to divide evenly", dividesEvenly);
        }
      }
    }
  }

  public void testZeroDivIsAlwaysZero() {
    for (int q : NONZERO_INTEGER_CANDIDATES) {