Parallel Algorithms: Difference between revisions

Revision as of 19:46, 27 February 2020

Parallel Algorithms notes from CMSC751 with Uzi Vishkin.
This class is based on his book Thinking in Parallel Some Basic Data-Parallel Algorithms and Techniques

XMT Language

XMTC is a single-program multiple-data (SPMD) extension of C.

Spawn creates threads
Threads expire at Join
$ represents the number of the thread
PS Ri Rj is an atomic prefix sum
- Stores Ri + Rj in Ri
- Stores the original value of Ri in Rj

int x = 0;

// Spawn n threads
spawn(0, n-1) {
  int e = 1;
  if (A[$] != 0) {
    // Sets e=x and increments x
    ps(e,x);
  }
}

Models

PRAM

Parallel Random-Access Machine/Model
You're given n synchronous processors each with local memory and access to a shared memory.
Each processor can write to shared memory, read to shared memory, or do computation in local memory.
You tell each processor what to do at each time step.

Types of PRAM

exclusive-read exclusive-write (EREW)
concurrent-read exclusive-write (CREW)
concurrent-read concurrent-write (CRCW)
- Arbitrary CRCW - an arbitrary processor writing succeeds
- Priority CRCW - the lowest numbered processor writing succeeds
- Common CRCW - writing succeeds only if all processors write the same value

Drawbacks

Does not reveal how the algorithm will run on PRAMs with different number of proc
Fully specifying allocation requires an unnecessary level of detail

Work Depth

You provide a sequence of instructions. At each time step, you specify the number of parallel operations.

WD-presentation Sufficiency Theorem

Given an algorithm in WD mode that takes $\displaystyle x=x(n)$ operations and $\displaystyle d=d(n)$ time, the algorithm can be implemented in any p-processor PRAM with $\displaystyle O(x/p + d)$ time.

Notes

Other resources call this Brent's theorem
$\displaystyle x$ is the work and $\displaystyle d$ is the depth

Speedup

A parallel algorithm is work-optimal if W(n) grows asymptotically the same as T(n).
A work-optimal parallel algorithm is work-time-optimal if T(n) cannot be improved by another work-optimal algorithm.
If T(n) is best known and W(n) matches it, this is called linear-speedup

NC Theory

Nick's Class Theory

Good serial algorithms: Poly time
Good parallel algorithm: poly-log $\displaystyle O(\log ^c n)$ time, poly processors

Technique: Balanced Binary Trees

Example Applications:

Prefix sum
Largest element
Nearest-one element
Compaction

Inputs

Array A[1..n] of elements
Associative binary operation, denoted * (e.g. addition, multiplication, min, max)

Outputs

Array B containing B(i) = A(1) * ... * A(i)

Algorithm

$\displaystyle O(n)$ work and $\displaystyle O(\log n)$ time

for i, 1 <= i <= n pardo
  B(0, i) = A(i)
  // Summation step up the tree
  for h=1 to log n
    B(h, i) = B(h-1, 2i-1) * B(h-1, 2i)
  // Down the tree
  for h=log n to 1
    if i even, i <= n/2^h
      C(h, i) = C(h+1, i/2)
    if i == 1
      C(h, i) = B(h, i)
    if i odd, 3 <= i <= n/2^h
      C(h, i) = C(h+1, i/2) * B(h, i)
return C(0, :)

Technique: Merge Sorting Cluster

Technique: Partitioning

Merging: Given two sorted arrays, A and B, create a sorted array C consisting of elements in A and B
Ranking: Given an element x, find Rank(x, A) = i such that $\displaystyle A(i) \leq x \leq A(i+1)$

Equivalence of Merging and Ranking

Given an algorithm for merging, we know A(i) belongs in C(j) so Rank(A(i), B) = j-i
Given an algorithm for ranking, we get A(i) belongs in C(j) where j = Rank(A(i), B) + i

Naive algorithm

Apply binary search element wise on concurrent read system. O(log n) time, O(nlog n) work

for 1 <= i <= n pardo
  Rank(A(i), B)
  Rank(B(i), A)

Partitioning paradigm

Input size: n

Partition the input into p jobs of size approx. n/p
Do small jobs concurrently using a separate (possibly serial) algo for each.

Example: Total work O(n), total time O(log n)

// Partitioning
for 1 <= i <= p pardo
  b(i) = Rank((n/p)(i-1)+1, B)
  a(i) = Rank((n/p)(i-1)+1, A)
// Total slice size is <= 2n/p. Total 2p slices.
// E.g. slice size is 2 * log(n) if p=n/log(n).
// Work
for 1 <= i <= p pardo
  k = ceil(b(i)/(n/p)) * (n/p)
  merge slice A[(n/p)(i-1)+1:min(a(i), end_a] and B[b(i):end_b]

Notes

We do not need to compute A_end or B_end. Just continue merging until we hit some index where (i % x)=0 in A or B.

Techinque: Divide and Conquer

Merge Sort

Input: Array A[1..n]
Complexity:
Time: $\displaystyle T(n) \leq T(n/2) + \alpha \log n$
Work: $\displaystyle T(n) \leq 2 * T(n/2) + \beta n$
Time $\displaystyle O(\log^2 n)$. Work: $\displaystyle O(n \log n)$

MergeSort(A, B):
  if n == 1
    return B(1) = A(1)
  call in parallel:
    MergeSort(A[1:n/2], C[1:n/2])
    MergeSort(A[n/2+1:n], C[n/2+1:n])
  Merge C[1:n/2] and C[n/2:n] using O(log n) algorithm

Technique: Informal Work-Depth (IWD) and Accelerating Cascades

Technique: Accelerating Cascades

Consider: for problem of size n, there are two parallel algorithms.

Algo A: $\displaystyle W_1(n)$ work, $\displaystyle T_1(n)$ time
Algo B: Work $\displaystyle W_2(n) \gt W_1(n)$. Time $\displaystyle T_2(n) \lt T_1(n)$. Faster but less efficient.

Assume Algo A is a "reducing algorithm"
We start with Algo A until the size is below some threshold. Then we switch to Algo B.

Problem: Selection

Algorithm 1

Partition elements into rows of $\displaystyle \log n$ size
For each row, find the median within the row
Find the median of medians (MoM) in $\displaystyle O(n)$
Put all rows with median <= MoM above and all rows with median >= Mom below
Now $\displaystyle m/4$ elements are smaller and $\displaystyle m/4$ are larger
- Known as the reducing lemma

This algorithm solves the selection problem in $\displaystyle O(\log^2 n)$ time and $\displaystyle O(n)$ work.

Accelerating Cascades

What we have:
- Algorithm 1 has $\displaystyle O(\log n)$ iterations.

Each iteration reduces a size m instance in $\displaystyle O(\log m)$ time and $\displaystyle O(m)$ work to an instance of size $\displaystyle \leq 3m/4$

- Algorithm 2 runs in $\displaystyle O(\log n)$ time and $\displaystyle O(n \log n)$ work.
Step 1: Use algo 1 to reduce from n to $\displaystyle n / \log n$
Step 2: Apply algorithm 2

Informal Work-Depth (IWD)

At each time unit there is a set containing a number of instructions to be performed concurrently.

Integer Sorting

There is a theorem that sorting with only comparisons is worst case at least $\displaystyle O(n\log n)$

Input: Array A[1..n], integers are range [0..r-1]
Sorting: rank from smallest to largest
Assume n is divisible by r ($\displaystyle r=\sqrt{n}$)

Algorithm

Partition A into n/r subarrays $\displaystyle B_1,...,B_{n/r}$
- Sort each subarray separately using serial bucket sort (counting sort)
- Compute number(v,s) # of elements of value v in $\displaystyle B_s$ for $\displaystyle 0\leq v \leq r-1$ and $\displaystyle 1 \leq s \leq n/r$
- Compute serial(i) = # of elements in the block $\displaystyle B_s$ such that A(j)=A(i) and $\displaystyle j \lt i $ $\displaystyle 1 \leq j \neq n$
Run prefix sum on number(v,1),number(v,2),...,number(v,n/r) into ps(v,1), ps(v,2),..., ps(v,n/r)
Compute prefix sums of cardinality(0),.., cardinality(r-1) into global_ps(0),...,global_ps(r-1)
The rank of element $\displaystyle i$ is $\displaystyle 1+serial(i)+ps(v,s-1)+global_ps(v-1)$

Complexity

Step 1:$\displaystyle T=O(r),\;W=O(r)$ per subarray.
- Total: $\displaystyle T=O(r),\; W=O(n)$
Step 2: r computations each $\displaystyle T=O(\log(n/r)),\; W=O(n/r)$
- Total $\displaystyle T=O(\log n),\; W=O(n)$
Step 3: $\displaystyle T=O(\log r),\; W=O(r)$
Step 4: $\displaystyle T=O(1)\; W=O(n)$
Total: $\displaystyle T=O(r + \log n),\; W=O(n)$

Notes

Running time is not poly-log

Theorems

The integer sorting algorithm runs in $\displaystyle O(r+\log n)$ time and $\displaystyle O(n)$ work
The integer sorting algorithm can be applied to run in time $\displaystyle O(k(r^{1/k}+\log n))$ and work $\displaystyle O(kn)$

Radix sort using the basic integer sort (BIS) algorithm.
If your range is 0 to n and and your radix is $\displaystyle \sqrt{n}$ then you will need $\displaystyle log_{\sqrt{n}}(r) = 2$ rounds.

2-3 trees; Technique: Pipelining

Dictionary: Search, insert, delete
Problem: How to parallelize to handle batches of queries

2-3 tree

A 2-3 tree is a rooted tree which has the properties:

Each node has 2-3 ordered children
For any internal node, every directed path to a leaf is the same length

Types of queries:

search(a)
insert(a)
delete(a)

Serial Algorithms

Deletion: discard(a)
- Delete connection between a and parent of a
- If parent has 1 child
  - If parent's sibling has 2 children, move child to sibling and discard(parent)
  - If parent's sibling has 3 children, take one child from sibling

Parallel Algorithms

Assume concurrent read model

Insert: Suppose we want to insert sorted elements $\displaystyle c_1,...,c_k$ into a tree with n elements.
- Insert element $\displaystyle c_{k/2}$.
- Then insert in parallel $\displaystyle (c_1,...,c_{k/2-1})$ and $\displaystyle (c_{k/2+1},...,c_k)$
- Time is $\displaystyle O(\log k \log n)$ using $\displaystyle k$ processors.
Deletion: Suppose we want to delete sorted elements $\displaystyle c_1,...,c_k$
- Backwards Insertion
- for t=0 to $\displaystyle \log k$
  - if $\displaystyle i \equiv 2^t (\operatorname{mod}2^{t+1})$
    - discard(c_i)
- Time is $\displaystyle O(\log n \log k)$ without pipelining
- With pipelining, complexity is $\displaystyle O(\log n + \log k)$

Pipelining

There are $\displaystyle \log k$ waves of the absorb procedure.
Apply pipelining to make the time $\displaystyle O(\log n + \log k)$

Maximum Finding

Given an array A=A(1),...,A(n), find the largest element in the array

Constant time, $\displaystyle O(n^2)$ Work

Compare every pair of elements in A[1...n]

for i=1 to n pardo
  B(i) = 0
for 1 <= i,j <= n pardo
  if A(i) <= A(j) and i < j
    B(i) = 1
  else
    B(j) = 1
for 1 <= i <= n pardo
  if B(i) == 0
    A(i) is the max

$\displaystyle O(\log \log n)$ time and $\displaystyle O(n\log \log n)$ work algorithm

Split A into $\displaystyle \sqrt{n}$ subarrays
Find the max of each subarray recursively
Find the max of all subarrays in $\displaystyle O(n)$ using the constant time algo

Complexity

$\displaystyle T(n) \leq T(\sqrt{n}) + c_1$
$\displaystyle W(n) \leq \sqrt{n}W(\sqrt{n}) + c_2 n$
$\displaystyle T(n) = O(\log \log n)$, $\displaystyle W(n) = O(n\log \log n)$

$\displaystyle O(\log \log n)$ time and $\displaystyle O(n)$ time

Step 1: Partition into blocks of size $\displaystyle \log \log n$. Then we have $\displaystyle n/ \log \log n$ blocks.
- Apply serial linear time algorithm to find maximum of each block
- $\displaystyle O(\log \log n)$ time and $\displaystyle O(n)$ work.
Step 2: Apply doubly log algorithm to $\displaystyle n / \log \log n$ maxima.
- $\displaystyle O(\log \log n)$ time and $\displaystyle O(n)$ work.

Random Sampling

Maximum finding in $\displaystyle O(1)$ time and $\displaystyle O(n)$ work with very high probability

Step 1: Using aux array B of size $\displaystyle b^{7/8}$. Independently fill with random elements from A
Step 2: Find the max $\displaystyle m$ in array B in $\displaystyle O(1)$ time and $\displaystyle O(n)$ work
- Last 3 pulses of the recursive doubly-log time algorithm:
- Pulse 1: B is partitioned into $\displaystyle n^{3/4}$ blocks of size $\displaystyle n^{1/8}$ each. Find the max of each block.
- Pulse 2: $\displaystyle n^{3/4}$ maxima are partitioned into $\displaystyle n^{1/2}$ blocks of size $\displaystyle n^{1/4}$ each.
- Pulse 2: Find the max m of $\displaystyle n^{1/2}$ maxima in $\displaystyle O(1)$ time and $\displaystyle O(n)$ work.
Step 3: While there is an element larger than m, throw the new element into an array of size $\displaystyle n^{7/8}$.

Compute the maximum of the new array.

Complexity

Step 1 takes $\displaystyle O(1)$ time and $\displaystyle O(n^{7/8})$ work
Step 2 takes $\displaystyle O(1)$ time and $\displaystyle O(n)$ work
Each time Step 3 takes $\displaystyle O(1)$ time and $\displaystyle O(n)$ work
With high probability, we only need one iteration (see theorem below) so the time is $\displaystyle O(1)$ and work is $\displaystyle O(n^{7/8})$

Theorem 8.2

The algorithm find the maximum among $\displaystyle n$ elements. With high probability it runs in $\displaystyle O(1)$ time and $\displaystyle O(n)$ work. The probability of not finishing in the above time is $\displaystyle O(1/n^c)$.

Proof

See page 58 of the classnotes.

List Ranking Cluster

Techniques: Euler tours, pointer jumping, randomized and deterministic symmetry breaking

Resources

Ignore

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Parallel Algorithms: Difference between revisions

Revision as of 19:46, 27 February 2020

XMT Language

Models

PRAM

Types of PRAM

Drawbacks

Work Depth

WD-presentation Sufficiency Theorem

Speedup

NC Theory

Technique: Balanced Binary Trees

Technique: Merge Sorting Cluster

Technique: Partitioning

Techinque: Divide and Conquer

Technique: Informal Work-Depth (IWD) and Accelerating Cascades

Technique: Accelerating Cascades

Problem: Selection

Informal Work-Depth (IWD)

Integer Sorting

Algorithm

Complexity

Theorems

2-3 trees; Technique: Pipelining

2-3 tree

Serial Algorithms

Parallel Algorithms

Pipelining

Maximum Finding

Constant time, \(\displaystyle O(n^2)\) Work

\(\displaystyle O(\log \log n)\) time and \(\displaystyle O(n\log \log n)\) work algorithm

\(\displaystyle O(\log \log n)\) time and \(\displaystyle O(n)\) time

Random Sampling

Theorem 8.2

List Ranking Cluster

Resources

Ignore