We develop a computationally-intensive web app in Dart.
For the purposes of making sure our computations perform well across all platforms and browsers, we run benchmarks using the "test 0.12.15+3" package. However, we found out that in some cases, the difference in executing the same code differs by orders of magnitude. Below is the code that demonstrates the problem (sorting arrays using custom comparators).
I wonder if there is an explanation for the difference in time between the code being executed in chrome, and as test with choice of chrome as platform (pub run test -p chrome test/perf_test.dart) ?
Is benchmarking via the test package a viable option, or should we look elsewhere?
import 'package:quiver/iterables.dart' as it;
void main(){
Stopwatch sw = new Stopwatch();
int len = 1000000;//0000
num v = 0;
List ln0 = new List();
for(int i = 0; i < len; i++)
ln0.add(v++);
//T1
sw..reset()..start();
ln0.sort((x,y) => x - y);
sw..stop();
print('Num Comparator take, mcs: ${sw.elapsedMicroseconds}');
//T2
List ln1= it.range(len).toList();
sw..reset()..start();
ln1.sort((x,y) => x - y);
print('Suddenly, Num Comparator take, mcs: ${sw.elapsedMicroseconds}');
//T3
List li2 = it.range(len).map((numm)=>numm.toInt()).toList();
sw..reset()..start();
li2.sort((x,y)=>x - y);
sw..stop();
print('Int Comparator take, mcs: ${sw.elapsedMicroseconds}');
List<String> items = new List(len);
for(int i =0, len = items.length; i<len; i++){
List tl = ['a','b','c','d','e','f','g','h','i','j','']..shuffle();
items[i] = tl.join('');
}
Function _comparer = (String s1, String s2) {
if (s1 == null && s2 == null) return 0;
if (s1 == null) return 1;
if (s2 == null) return -1;
if (s1 == '' && s2 == '') return 0;
if (s1 == '') return 1;
if (s2 == '') return -1;
return s1.compareTo(s2);
};
//T4
List ls1 = new List.from(items);
sw..reset()..start();
ls1.sort((s1, s2) => s1.compareTo(s2));
sw..stop();
print('Standart String comparator take, mcs: ${sw.elapsedMicroseconds}');
//T5
List ls2 = new List.from(items);
sw..reset()..start();
ls2.sort(_comparer);
sw..stop();
print('String comparator(miss null,empty) take, mcs: ${sw.elapsedMicroseconds}');
}
test is not supposed to be used for benchmarks. Use insted https://pub.dartlang.org/packages/benchmark_harness. It helps to cope with benchmark related requirements like warmup phase and number of runs to get more realistic results.
Related
I am trying to prove a property in Dafny, which makes use of powers.
Concretely, this one: forall x,y in Reals : 2xy <= x^2+y^2. I implemented this idea in the following lemma:
lemma product2_lessEqual_powProduct (x:real, y:real)
requires 0.0<x<=1.0 && 0.0<y<=1.0
ensures 2.0*x*y <= (x*x)+(y*y)
{}
Which is verified with no problem (I guess some automatic induction is performed below).
However, I would like to use an own power function in order to make power(x,2) instead of x*x. Thus, I took a power function from https://github.com/bor0/dafny-tutorial/blob/master/pow.dfy, which is as follows:
function method power(A:int, N:nat):int
{
if (N==0) then 1 else A * power(A,N-1)
}
method pow(A:int, N:int) returns (x:int)
requires N >= 0
ensures x == power(A, N)
{
x := 1;
var i := N;
while i != 0
invariant x == power(A, (N-i))
{
x := x * A;
i := i - 1;
}
}
However, since I am using real values for the basis of the exponential, I modified it a bit so that it works for exponentials:
function method power(A:real, N:nat):real
{
if (N==0) then 1.0 else A * power(A,N-1)
}
method pow(A:real, N:int) returns (x:real)
requires N >= 0
ensures x == power(A, N)
{
x := 1.0;
var i := N;
while i != 0
invariant x == power(A, (N-i))
{
x := x * A;
i := i - 1;
}
}
Thus, I wanted to test it with the previous lemma:
lemma product2_lessEqual_powProduct (x:real, y:real)
requires 0.0<x<=1.0 && 0.0<y<=1.0
ensures 2.0*x*y <= power(x,2)+power(y,2)
{}
Surprisingly, it tells me the typical A postcondition might not hold on this return path.Verifier.
Can anyone explain why this happens? Why is it verifying with primitive operations of Dafny, but not when I define them functions? And how could I prove this lemma now?
Even though second parameter of power is concrete and small, Dafny is not doing enough unrolling to prove desired fact. Adding {:fuel 2} to power makes proof go through. You can read more about fuel here https://dafny.org/dafny/DafnyRef/DafnyRef.html#sec-fuel
function method {:fuel 2} power(A:real, N:nat):real
{
if (N==0) then 1.0 else A * power(A,N-1)
}
method pow(A:real, N:int) returns (x:real)
requires N >= 0
ensures x == power(A, N)
{
x := 1.0;
var i := N;
while i != 0
invariant x == power(A, (N-i))
{
x := x * A;
i := i - 1;
}
}
lemma product2_lessEqual_powProduct (x:real, y:real)
requires 0.0<x<=1.0 && 0.0<y<=1.0
ensures 2.0*x*y <= power(x,2)+power(y,2)
{}
It's surprising until you realize that there is a mathematical theory for A*A, but power(A, 2) requires two unfolding of power to have a meaning.
If you want your function to work seamlessly with the theory and prove your last lemma, you can give it precise postconditions:
function method power(A:real, N:nat): (result: real)
ensures N == 1 ==> result == A
ensures N == 2 ==> result == A*A
{
if (N==0) then 1.0 else A * power(A,N-1)
}
I tested it, your second lemma verifies.
So the Fibonacci number for log (N) — without matrices.
Ni // i-th Fibonacci number
= Ni-1 + Ni-2 // by definition
= (Ni-2 + Ni-3) + Ni-2 // unwrap Ni-1
= 2*Ni-2 + Ni-3 // reduce the equation
= 2*(Ni-3 + Ni-4) + Ni-3 //unwrap Ni-2
// And so on
= 3*Ni-3 + 2*Ni-4
= 5*Ni-4 + 3*Ni-5
= 8*Ni-5 + 5*Ni-6
= Nk*Ni-k + Nk-1*Ni-k-1
Now we write a recursive function, where at each step we take k~=I/2.
static long N(long i)
{
if (i < 2) return 1;
long k=i/2;
return N(k) * N(i - k) + N(k - 1) * N(i - k - 1);
}
Where is the fault?
You get a recursion formula for the effort: T(n) = 4T(n/2) + O(1). (disregarding the fact that the numbers get bigger, so the O(1) does not even hold). It's clear from this that T(n) is not in O(log(n)). Instead one gets by the master theorem T(n) is in O(n^2).
Btw, this is even slower than the trivial algorithm to calculate all Fibonacci numbers up to n.
The four N calls inside the function each have an argument of around i/2. So the length of the stack of N calls in total is roughly equal to log2N, but because each call generates four more, the bottom 'layer' of calls has 4^log2N = O(n2) Thus, the fault is that N calls itself four times. With only two calls, as in the conventional iterative method, it would be O(n). I don't know of any way to do this with only one call, which could be O(log n).
An O(n) version based on this formula would be:
static long N(long i) {
if (i<2) {
return 1;
}
long k = i/2;
long val1;
long val2;
val1 = N(k-1);
val2 = N(k);
if (i%2==0) {
return val2*val2+val1*val1;
}
return val2*(val2+val1)+val1*val2;
}
which makes 2 N calls per function, making it O(n).
public class fibonacci {
public static int count=0;
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
System.out.println("value of i ="+ i);
int result = fun(i);
System.out.println("final result is " +result);
}
public static int fun(int i) {
count++;
System.out.println("fun is called and count is "+count);
if(i < 2) {
System.out.println("function returned");
return 1;
}
int k = i/2;
int part1 = fun(k);
int part2 = fun(i-k);
int part3 = fun(k-1);
int part4 = fun(i-k-1);
return ((part1*part2) + (part3*part4)); /*RESULT WILL BE SAME FOR BOTH METHODS*/
//return ((fun(k)*fun(i-k))+(fun(k-1)*fun(i-k-1)));
}
}
I tried to code to problem defined by you in java. What i observed is that complexity of above code is not completely O(N^2) but less than that.But as per conventions and standards the worst case complexity is O(N^2) including some other factors like computation(division,multiplication) and comparison time analysis.
The output of above code gives me information about how many times the function
fun(int i) computes and is being called.
OUTPUT
So including the time taken for comparison and division, multiplication operations, the worst case time complexity is O(N^2) not O(LogN).
Ok if we use Analysis of the recursive Fibonacci program technique.Then we end up getting a simple equation
T(N) = 4* T(N/2) + O(1)
where O(1) is some constant time.
So let's apply Master's method on this equation.
According to Master's method
T(n) = aT(n/b) + f(n) where a >= 1 and b > 1
There are following three cases:
If f(n) = Θ(nc) where c < Logba then T(n) = Θ(nLogba)
If f(n) = Θ(nc) where c = Logba then T(n) = Θ(ncLog n)
If f(n) = Θ(nc) where c > Logba then T(n) = Θ(f(n))
And in our equation a=4 , b=2 & c=0.
As case 1 c < logba => 0 < 2 (which is log base 2 and equals to 2) is satisfied
hence T(n) = O(n^2).
For more information about how master's algorithm works please visit: Analysis of Algorithms
Your idea is correct, and it will perform in O(log n) provided you don't compute the same formula
over and over again. The whole point of having N(k) * N(i-k) is to have (k = i - k) so you only have to compute one instead of two. But if you only call recursively, you are performing the computation twice.
What you need is called memoization. That is, store every value that you already have computed, and
if it comes up again, then you get it in O(1).
Here's an example
const int MAX = 10000;
// memoization array
int f[MAX] = {0};
// Return nth fibonacci number using memoization
int fib(int n) {
// Base case
if (n == 0)
return 0;
if (n == 1 || n == 2)
return (f[n] = 1);
// If fib(n) is already computed
if (f[n]) return f[n];
// (n & 1) is 1 iff n is odd
int k = n/2;
// Applying your formula
f[n] = fib(k) * fib(n - k) + fib(k - 1) * fib(n - k - 1);
return f[n];
}
When working on a basic linear search I encountered an error with my Valid() predicate. It seems to only work when I uncomment the additional ensures statements on the constructor and the data method. That is, when I am very explicit about the contents.
I'm also having trouble with the postcondition of my search when the item isn't found.
Any suggestions on how to resolve these?
class Search{
ghost var Contents: set<int>;
var a : array<int>;
predicate Valid()
reads this, a;
{
a != null &&
a.Length > 0 &&
Contents == set x | 0 <= x < a.Length :: a[x]
}
constructor ()
ensures a != null;
ensures a.Length == 4;
//ensures a[0] == 0;
ensures Valid();
{
a := new int[4](_ => 0);
Contents := {0};
new;
}
method data()
modifies this, a;
requires Valid();
requires a != null;
requires a.Length == 4;
ensures a != null;
ensures a.Length == 4;
// ensures a[0] == 0;
// ensures a[1] == 1;
// ensures a[2] == 2;
// ensures a[3] == 3;
ensures Valid();
{
a[0] := 0;
a[1] := 1;
a[2] := 2;
a[3] := 3;
Contents := {0, 1, 2, 3};
}
method search(e: int) returns (r: bool)
modifies this, a;
requires Valid();
ensures Valid();
ensures r == (e in Contents)
ensures r == exists i: int :: 0 <= i < a.Length && a[i] == e
{
var length := a.Length - 1;
while (length >= 0)
decreases length;
{
var removed := a[length];
if (e == removed)
{
return true;
}
length := length - 1;
}
return false;
}
}
method Main()
{
var s := new Search();
s.data();
}
There are several orthogonal issues going on here.
First, you have noticed that Dafny is reluctant to reason about the part of Valid that describes Contents. This is a common problem when reasoning about sets in Dafny. Essentially, the only way Dafny will ever "notice" that something is a member of the set set x | 0 <= x < a.Length :: a[x] is if it already has the expression a[x] lying around somewhere. Your solution of including extra postconditions works because it mentions a lot of expressions of the form a[x]. Another solution is to include those facts as assertions instead of postconditions:
// in data()
assert a[0] == 0;
assert a[1] == 1;
assert a[2] == 2;
assert a[3] == 3;
Second, Dafny cannot show your search procedure satisfies its postcondition. You need a loop invariant to keep track of the progress of the search. See the guide for more information about how to design loop invariants.
Third, Dafny reports a problem with your Main about modifies clauses. You can fix this by adding a postcondition fresh(a) to the constructor. The problem here is that the data method claims to modify a, but Dafny can't tell if a is visible from the caller.
In trying to verify a generic FIFO queue backed by an array I ran into a confusing error. The queue was found in this paper, authored by the creator of Dafny.
The error in question is:
unless an initializer is provided for the array elements, a new array of 'Data' must have empty size
which relates to both lines allocating an array via new Data[whatever] in the constructor and the enqueue method.
Dafny version: Dafny 2.0.0.00922 technical preview 0
Full code for reference.
class {:autocontracts} SimpleQueue<Data>
{
ghost var Contents: seq<Data>;
var a: array<Data>;
var m: int, n: int;
predicate Valid() {
a != null && a.Length != 0 && 0 <= m <= n <= a.Length && Contents == a[m..n]
}
constructor ()
ensures Contents == [];
{
a := new Data[10];
m := 0;
n := 0;
Contents := [];
}
method Enqueue(d: Data)
ensures Contents == old(Contents) + [d];
{
if n == a.Length {
var b := a;
if m == 0 {
b := new Data[2 * a.Length];
}
forall (i | 0 <= i < n - m) {
b[i] := a[m + i];
}
a, m, n := b, 0, n - m;
}
a[n], n, Contents := d, n + 1, Contents + [d];
}
method Dequeue() returns (d: Data)
requires Contents != [];
ensures d == old(Contents)[0] && Contents == old(Contents)[1..];
{
assert a[m] == a[m..n][0];
d, m, Contents := a[m], m + 1, Contents[1..];
}
}
method Main()
{
var q := new SimpleQueue();
q.Enqueue(5); q.Enqueue(12);
var x := q.Dequeue();
assert x == 5;
}
Since the time of writing that paper, Dafny's type system has been generalized to support types that are not "default initializable". This has led to some backwards incompatibilities.
The easiest fix is to change
class SimpleQueue<Data>
to
class SimpleQueue<Data(0)>
which means that the type variable Data can only be instantiated with default-initializable types.
Another fix is to change the constructor to accept a default value for type Data as an argument. Then you can allocate an array using an initializer function, as in
new Data[10] (_ => d)
This scripting language doesn't have a % or Mod(). I do have a Fix() that chops off the decimal part of a number. I only need positive results, so don't get too robust.
Will
// mod = a % b
c = Fix(a / b)
mod = a - b * c
do? I'm assuming you can at least divide here. All bets are off on negative numbers.
a mod n = a - (n * Fix(a/n))
For posterity, BrightScript now has a modulo operator, it looks like this:
c = a mod b
If someone arrives later, here are some more actual algorithms (with errors...read carefully)
https://eprint.iacr.org/2014/755.pdf
There are actually two main kind of reduction formulae: Barett and Montgomery. The paper from eprint repeat both in different versions (algorithms 1-3) and give an "improved" version in algorithm 4.
Overview
I give now an overview of the 4. algorithm:
1.) Compute "A*B" and Store the whole product in "C" that C and the modulus $p$ is the input for that algorithm.
2.) Compute the bit-length of $p$, say: the function "Width(p)" returns exactly that value.
3.) Split the input $C$ into N "blocks" of size "Width(p)" and store each in G. Start in G[0] = lsb(p) and end in G[N-1] = msb(p). (The description is really faulty of the paper)
4.) Start the while loop:
Set N=N-1 (to reach the last element)
precompute $b:=2^{Width(p)} \bmod p$
while N>0 do:
T = G[N]
for(i=0; i<Width(p); i++) do: //Note: that counter doesn't matter, it limits the loop)
T = T << 1 //leftshift by 1 bit
while is_set( bit( T, Width(p) ) ) do // (N+1)-th bit of T is 1
unset( bit( T, Width(p) ) ) // unset the (N+1)-th bit of T (==0)
T += b
endwhile
endfor
G[N-1] += T
while is_set( bit( G[N-1], Width(p) ) ) do
unset( bit( G[N-1], Width(p) ) )
G[N-1] += b
endwhile
N -= 1
endwhile
That does alot. Not we only need to recursivly reduce G[0]:
while G[0] > p do
G[0] -= p
endwhile
return G[0]// = C mod p
The other three algorithms are well defined, but this lacks some information or present it really wrong. But it works for any size ;)
What language is it?
A basic algorithm might be:
hold the modulo in a variable (modulo);
hold the target number in a variable (target);
initialize modulus variable;
while (target > 0) {
if (target > modulo) {
target -= modulo;
}
else if(target < modulo) {
modulus = target;
break;
}
}
This may not work for you performance-wise, but:
while (num >= mod_limit)
num = num - mod_limit
In javascript:
function modulo(num1, num2) {
if (num2 === 0 || isNaN(num1) || isNaN(num2)) {
return NaN;
}
if (num1 === 0) {
return 0;
}
var remainderIsPositive = num1 >= 0;
num1 = Math.abs(num1);
num2 = Math.abs(num2);
while (num1 >= num2) {
num1 -= num2
}
return remainderIsPositive ? num1 : 0 - num1;
}