Efficient way to multiply a large set of small numbers - algorithm

Question

This question was asked in an interview.
You have an array of small integers. You have to multiply all of them. You need not worry about overflow you have ample support for that. What can you do to speed up the multiplication on your machine?
Would multiple additions be better in this case?
I suggested multiplying using a divide and conquer approach but the interviewer was not impressed. What could be the best possible solution for this?

Answer 1

Here are some thoughts:
Divide-and-Conquer with Multithreading: Split the input apart into n different blocks of size b and recursively multiply all the numbers in each block together. Then, recursively multiply all n / b blocks back together. If you have multiple cores and can run parts of this in parallel, you could save a lot of time overall.
Word-Level Parallelism: Let's suppose that your numbers are all bounded from above by some number U, which happens to be a power of two. Now, suppose that you want to multiply together a, b, c, and d. Start off by computing (4U2a + b) × (4U2c + d) = 16U4ac + 4U2ad + 4U2bc + bd. Now, notice that this expression mod U2 is just bd. (Since bd < U2, we don't need to worry about the mod U2 step messing it up). This means that if we compute this product and take it mod U2, we get back bd. Since U2 is a power of two, this can be done with a bitmask.
Next, notice that
4U2ad + 4U2bc + bd < 4U4 + 4U4 + U2 < 9U4 < 16U4
This means that if we divide the entire expression by 16U4 and round down, we will end up getting back just ad. This division can be done with a bitshift, since 16U4 is a power of two.
Consequently, with one multiplication, you can get back the values of both ac and bd by applying a subsequent bitshift and bitmask. Once you have ac and bd, you can directly multiply them together to get back the value of abcd. Assuming that bitmasks and bitshifts are faster than multiplies, this reduces the number of multiplications necessary by 33% (two instead of three here).
Hope this helps!

Answer 2

Your divide and conquer suggestion was a good start. It just needed more explanation to impress.
With fast multiplication algorithms used to multiply large numbers (big-ints), it is much more efficient to multiply similar sized multiplicands than a series of mismatched sizes.
Here's an example in Clojure
; Define a vector of 100K random integers between 2 and 100 inclusive
(def xs (vec (repeatedly 100000 #(+ 2 (rand-int 99)))))
; Naive multiplication accumulating linearly through the array
(time (def a1 (apply *' xs)))
"Elapsed time: 7116.960557 msecs"
; Simple Divide and conquer algorithm
(defn d-c [v]
(let [m (quot (count v) 2)]
(if (< m 3)
(reduce *' v)
(*' (d-c (subvec v 0 m)) (d-c (subvec v m))))))
; Takes less than 1/10th the time.
(time (def a2 (d-c xs)))
"Elapsed time: 600.934268 msecs"
(= a1 a2) ;=> true (same result)
Note that this improvement does not rely on a set limit for the size of the integers in the array (100 chosen arbitrarily and to demonstrate the next algorithm), but only that they be similar in size. This is a very simple divide an conquer. As the numbers get larger and more expensive to multiply, it would make sense to invest more time in iteratively grouping them by similar size. Here I am relying on random distribution and chance that the sizes will stay similar, but it is still going to be significantly better than the naive approach even for the worst case.
As suggested by Evgeny Kluev in the comments, for a large number of small integers, there is going to be a lot of duplication, so efficient exponentiation is also better than naive multiplication. This depends a lot more on the relative parameters than the divide and conquer, that is the numbers must be sufficiently small relative to the count for enough duplicates to accumulate to bother, but certainly performs well with these parameters (100K numbers in the range 2-100).
; Hopefully an efficient implementation
(defn pow [x n] (.pow (biginteger x) ^Integer n))
; Perform pow on duplications based on frequencies
(defn exp-reduce [v] (reduce *' (map (partial apply pow) (frequencies v))))
(time (def a3 (exp-reduce xs)))
"Elapsed time: 650.211789 msecs"
Note the very simple divide and conquer performed just a wee better in this trial, but would be even relatively better if fewer duplicates were expected.
Of course we can also combine the two:
(defn exp-d-c [v] (d-c (mapv (partial apply pow) (frequencies v))))
(time (def a4 (exp-d-c xs)))
"Elapsed time: 494.394641 msecs"
(= a1 a2 a3 a4) ;=> true (all equal)
Note there are better ways to combine these two since the result of the exponentiation step is going to result in various sizes of multiplicands. The value of added complexity to do so depends on the expected number of distinct numbers in the input. In this case, there are very few distinct numbers so it wouldn't pay to add much complexity.
Note also that both of these are easily parallelized if multiple cores are available.

Answer 3

If many of the small integers occur multiple times, you could start by counting every unique integer. If c(n) is the number of occurrences of integer n, the product can be computed as
P = 2 ^ c(2) * 3 ^ c(3) * 4 ^ c(4) * ...
For the exponentiation steps, you can use exponentiation by squaring which can reduce the number of multiplications considerably.

Answer 4

If the count of numbers really is large compared to the range, then we have seen two asymptotic solutions presented to reduce the complexity considerably. One was based on successive squaring to compute c^k in O(log k) time for each number c, giving O(C mean(log k)) time if the largest number is C and k gives the exponent for each number between 1 and C. The mean(log k) term is maximized if every number appears an equal number of times, so if you have N numbers then the complexity becomes O(C log(N/C)), which is very weakly dependent on N and essentially just O(C) where C specifies the range of numbers.
The other approach we saw was sorting numbers by the number of times they appear, and keeping track of the product of leading numbers (starting with all numbers) and raising this to a power so that the least frequent number is removed from the array, and then updating the exponents on the remaining element in the array and repeating. If all numbers occur the same number of times K, then this gives O(C + log K) which is an improvement over O(C log K). However, say the kth number appears 2^k times. Then this will still give O(C^2 + C log(N/C)) time which is technically worse than the previous method O(C log(N/C)) if C > log(N/C). Thus, if you don't have good information on how evenly distributed the occurrences of each number are, you should go with the first approach, just take the appropriate power of each distinct number that appears in the product by using successive squaring, and take the product of the results. Total time O(C log (N/C)) if there are C distinct numbers and N total numbers.

Answer 5

To answer this question we need to interpret in some way the assumption from OP: need not worry about overflow. In larger part of this answer it is interpreted as "ignore overflows". But I start with several thoughts about other interpretation ("use multiprecision arithmetic"). In this case process of multiplication may be approximately split into 3 stages:
Multiply together small sets of small numbers to get a large set of not-so-small numbers. Some of the ideas from second part of this answer may be used here.
Multiply together these numbers to get a set of large numbers. Either trivial (quadratic time) algorithm or Toom–Cook/Karatsuba (sub-quadratic time) methods may be used.
Multiply together large numbers. Either Fürer's or Schönhage–Strassen algorithm may be used. Which gives O(N polylog N) time complexity for the whole process.
Binary exponentiation may give some (not very significant) speed improvement, because most (if not every) of complex multiplication algorithms mentioned here do squaring faster than multiplication of two unequal numbers. Also we could factorize every "small" number and use binary exponentiation only for prime factors. For evenly distributed "small" numbers this will decrease number of exponentiations by factor log(number_of_values) and slightly improve balance of squarings/multiplications.
Divide and conquer is OK when numbers are evenly distributed. Otherwise (for example when input array is sorted or when binary exponentiation is used) we could do better by placing all multiplicands into priority queue, ordered (may be approximately ordered) by number length. Then we could multiply two shortest numbers and place the result back to the queue (this process is very similar to Huffman encoding). There is no need to use this queue for squaring. Also we should not use it while numbers are not long enough.
More information on this could be found in the answer by A. Webb.
If overflows may be ignored we could multiply the numbers with linear-time or better algorithms.
Sub-linear time algorithm is possible if input array is sorted or input data is presented as set of tuples {value, number of occurrences}. In latter case we could perform binary exponentiation of each value and multiply results together. Time complexity is O(C log(N/C)), where C is number of different values in the array. (See also this answer).
If input array is sorted, we could use binary search to find positions where value changes. This allows to determine how many times each value occurs in the array. Then we could perform binary exponentiation of each value and multiply results together. Time complexity is O(C log N). We could do better by using one-sided binary search here. In this case time complexity is O(C log(N/C)).
But if input array is not sorted, we have to inspect each element, so O(N) time complexity is the best we can do. Still we could use parallelism (multithreading, SIMD, word-level parallelism) to obtain some speed improvement. Here I compare several such approaches.
To compare these approaches I've chosen very small (3-bit) values, which are pretty tightly packed (one value per 8-bit integer). And implemented them in low-level language (C++11) to get easier access to bit manipulation, specific CPU instructions, and SIMD.
Here are all the algorithms:
accumulate from standard library.
Trivial implementation with 4 accumulators.
Word-level parallelism for multiplication, as described in the answer by templatetypedef. With 64-bit word size this approach allows up to 10-bit values (with only 3 multiplications instead of each 4) or it may be applied twice (and I did it in the tests) with up to 5-bit values (requiring only 5 multiplications out of each 8).
Table lookup. In the tests 7 multiplications out of each 8 are substituted by single table lookup. With values larger than in these tests, number of substituted multiplications decreases, slowing down the algorithm. Values larger than 11-12 bits make this approach useless.
Binary exponentiation (see more details below). Values larger than 4 bits make this approach useless.
SIMD (AVX2). This implementation can use up to 8-bit values.
Here are sources for all tests on Ideone. Note that SIMD test requires AVX2 instruction set from Intel. Table lookup test requires BMI2 instruction set. Other tests do not depend on any particular hardware (I hope). I run these tests on 64-bit Linux, compiled with gcc 4.8.1, optimization level -O2.
Here are some more details for binary exponentiation test:
for (size_t i = 0; i < size / 8; i += 2)
{
auto compr = (qwords[i] << 4) | qwords[i + 1];
constexpr uint64_t lsb = 0x1111111111111111;
if ((compr & lsb) != lsb) // if there is at least one even value
{
auto b = reinterpret_cast<uint8_t*>(qwords + i);
acc1 *= accumulate(b, b + 16, acc1, multiplies<unsigned>{});
if (!acc1)
break;
}
else
{
const auto b2 = compr & 0x2222222222222222;
const auto b4 = compr & 0x4444444444444444;
const auto b24 = b4 & (b2 * 2);
const unsigned c7 = __builtin_popcountll(b24);
acc3 += __builtin_popcountll(b2) - c7;
acc5 += __builtin_popcountll(b4) - c7;
acc7 += c7;
}
}
const auto prod4 = acc1 * bexp<3>(acc3) * bexp<5>(acc5) * bexp<7>(acc7);
This code packs values even more densely than in the input array: two values per byte. Low-order bit of each value is handled differently: since we could stop after 32 zero bits is found here (with result "zero"), this case cannot alter performance very much, so it is handled by simplest (library) algorithm.
Out of 4 remaining values, "1" is not interesting, so we need to count only occurrences of "3", "5" , and "7" with bitwise manipulations and "population count" intrinsic.
Here are the results:
source size: 4 Mb: 400 Mb:
1. accumulate: 0.835392 ns 0.849199 ns
2. accum * 4: 0.290373 ns 0.286915 ns
3. 2 mul in 1: 0.178556 ns 0.182606 ns
4. mult table: 0.130707 ns 0.176102 ns
5. binary exp: 0.128484 ns 0.119241 ns
6. AVX2: 0.0607049 ns 0.0683234 ns
Here we can see that accumulate library algorithm is pretty slow: for some reason gcc could not unroll the loop and use 4 independent accumulators.
It is not too difficult to do this optimization "by hand". The result is not particularly fast. But if we allocate 4 threads for this task, CPU would approximately match memory bandwidth (2 channels, DDR3-1600).
Word-level parallelism for multiplications is almost twice as fast. (We'll need only 3 threads to match memory bandwidth).
Table lookup is even faster. But its performance degrades when input array cannot fit in L3 cache. (We'll need 3 threads to match memory bandwidth).
Binary exponentiation has approximately the same speed. But with larger inputs this performance does not degrade, it even slightly improves because exponentiation itself uses less time compared to value counting. (We'll need only 2 threads to match memory bandwidth).
As expected, SIMD is the fastest. Its performance slightly degrades when input array cannot fit in L3 cache. Which means we are close to memory bandwidth with single thread.

Answer 6

I have one solution. Let us discuss it with other solutions.
The key part of question is how to reduce times of multiply. And integers are small but set is big.
My solution:
use an small array to record how many times each number appears.
Remove number 1 from array. You don't need to count it.
Find the number which appears least times n. Then multiply all numbers and get result K. Then count K^n.
Remove this number (For instance, you can switch it with the last number of array and reduce size of array for 1). So next time you won't consider this number any more. At same time, the appearance times of other numbers need to be reduced with the times of removed number.
Once again get the number which appears least times. Do same thing as step 2.
Repeatedly do step 2-4 and complete counting.
Let me use an example to show how many multiply we need to do: Assume
we have 5 numbers [1, 2, 3, 4, 5]. Number 1 appears 100 times, number
2 appears 150 times, number 3 appears 200 times, number 4 appears 300
times, and number 5 appears 400 times.
method 1: multiply it directly or use divide and conquer
we need 100+150+200+300+400-1 = 1149 multiply to get result.
method 2: we do (1^100)(2^150)(3^200)(4^300)(5^400)
(100-1)+(150-1)+(200-1)+(300-1)+(400-1)+4 = 1149.[same as method 1]
Cause n^m will do m-1 multiply in deed. Plus you need time to go through all numbers, though this time is short.
method in this post:
First, you need time to go through all numbers. It can be discarded compare to time of multiply.
The real counting you are making is:
((2*3*4*5)^150)*((3*4*5)^50)*((4*5)^100)*(5^100)
Then you need to do multiply 3+149+2+49+1+99+99+3 = 405 times