n_choose_k.cc

Go to the documentation of this file.

// Copyright 2010-2024 Google LLC
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
//     http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
 
#include "ortools/algorithms/n_choose_k.h"
 
#include <cmath>
#include <cstdint>
#include <limits>
#include <vector>
 
#include "absl/log/check.h"
#include "absl/numeric/int128.h"
#include "absl/status/status.h"
#include "absl/status/statusor.h"
#include "absl/strings/str_format.h"
#include "absl/time/clock.h"
#include "absl/time/time.h"
#include "ortools/algorithms/binary_search.h"
#include "ortools/base/logging.h"
#include "ortools/base/mathutil.h"
 
namespace operations_research {
namespace {
// This is the actual computation. It's in O(k).
template <typename Int>
Int InternalChoose(Int n, Int k) {
  DCHECK_LE(k, n - k);
  // We compute n * (n-1) * ... * (n-k+1) / k! in the best possible order to
  // guarantee exact results, while trying to avoid overflows. It's not perfect:
  // we finish with a division by k, which means that me may overflow even if
  // the result doesn't (by a factor of up to k).
  Int result = 1;
  Int i = 0;
  while (i < k) {  // We always have k < n/2.
    result *= n--;
    result /= ++i;  // The product of i consecutive numbers is divisible by i.
  }
  return result;
}
 
constexpr int64_t kint64max = std::numeric_limits<int64_t>::max();
 
// This function precomputes the maximum N such that (N choose K) doesn't
// overflow, for all K.
// When `overflows_intermediate_computation` is true, "overflow" means
// "some overflow happens inside InternalChoose<int64_t>()", and when it's false
// it simply means "the result doesn't fit in an int64_t".
// This is only used in contexts where K ≤ N-K, which implies N ≥ 2K, thus we
// can stop when (2K Choose K) overflows, because at and beyond such K,
// (N Choose K) will always overflow. In practice that happens for K=31 or 34
// depending on `overflows_intermediate_computation`.
std::vector<int64_t> LastNThatDoesNotOverflowForAllK(
    bool overflows_intermediate_computation) {
  absl::Time start_time = absl::Now();
  // Given the algorithm used in InternalChoose(), it's not hard to
  // find out when (N choose K) overflows an int64_t during its internal
  // computation: that's when (N choose K) > kint64max / k.
 
  // For K ≤ 2, we hardcode the values of the maximum N.
  std::vector<int64_t> result = {
      kint64max,  // K=0
      kint64max,  // K=1
      // The binary search done below uses MathUtil::LogCombinations, which only
      // works on int32_t, and that's problematic for the max N we get for K=2.
      overflows_intermediate_computation
          ?  // Max N such that N*(N-1) < 2^63. N*(N-1) ≈ (N-0.5)².
          static_cast<int64_t>(0.5 + std::pow(2.0, 63.0 / 2))
          : 1l << 32,  // Max N such that N*(N-1) < 2^64.
  };
  // We find the last N with binary search, for all K. We stop growing K
  // when (2*K Choose K) overflows.
  for (int64_t k = 3;; ++k) {
    const double max_log_comb = overflows_intermediate_computation
                                    ? 63 * std::log(2) - std::log(k)
                                    : 63 * std::log(2);
    result.push_back(BinarySearch<int64_t>(
        /*x_true*/ k, /*x_false=*/(1l << 23) - 1, [k, max_log_comb](int64_t n) {
          return MathUtil::LogCombinations(n, k) <= max_log_comb;
        }));
    if (result.back() < 2 * k) {
      result.pop_back();
      break;
    }
  }
  DCHECK_EQ(result.size(),
            overflows_intermediate_computation
                ? 31    // 60 Choose 30 < 2^63/30 but 62 Choose 31 > 2^63/31.
                : 34);  // 66 Choose 33 < 2^63 but 68 Choose 34 > 2^63.
  VLOG(1) << "LastNThatDoesNotOverflowForAllK(): " << absl::Now() - start_time;
  return result;
}
 
bool NChooseKIntermediateComputationOverflowsInt64(int64_t n, int64_t k) {
  DCHECK_LE(k, n - k);
  static const auto* const result =
      new std::vector<int64_t>(LastNThatDoesNotOverflowForAllK(
          /*overflows_intermediate_computation=*/true));
  return k < result->size() ? n > (*result)[k] : true;
}
 
bool NChooseKResultOverflowsInt64(int64_t n, int64_t k) {
  DCHECK_LE(k, n - k);
  static const auto* const result =
      new std::vector<int64_t>(LastNThatDoesNotOverflowForAllK(
          /*overflows_intermediate_computation=*/false));
  return k < result->size() ? n > (*result)[k] : true;
}
}  // namespace
 
absl::StatusOr<int64_t> NChooseK(int64_t n, int64_t k) {
  if (n < 0) {
    return absl::InvalidArgumentError(absl::StrFormat("n is negative (%d)", n));
  }
  if (k < 0) {
    return absl::InvalidArgumentError(absl::StrFormat("k is negative (%d)", k));
  }
  if (k > n) {
    return absl::InvalidArgumentError(
        absl::StrFormat("k=%d is greater than n=%d", k, n));
  }
  // NOTE(user): If performance ever matters, we could simply precompute and
  // store all (N choose K) that don't overflow, there aren't that many of them:
  // only a few tens of thousands, after removing simple cases like k ≤ 5.
  if (k > n / 2) k = n - k;
  if (!NChooseKIntermediateComputationOverflowsInt64(n, k)) {
    return InternalChoose<int64_t>(n, k);
  }
  if (NChooseKResultOverflowsInt64(n, k)) {
    return absl::InvalidArgumentError(
        absl::StrFormat("(%d choose %d) overflows int64", n, k));
  }
  return static_cast<int64_t>(InternalChoose<absl::int128>(n, k));
}
 
}  // namespace operations_research