Let x in {10, 37, 96, 104} set.
Let f(x) a "select case" function:
int f1(int x) {
switch(x) {
case 10: return 3;
case 37: return 1;
case 96: return 0;
case 104: return 1;
}
assert(...);
}
Then, we can avoid conditional jumps writing f(x) as a "integer polynomial" like
int f2(int x) {
// P(x) = (x - 70)^2 / 1000
int q = x - 70;
return (q * q) >> 10;
}
In some cases (still including mul operations) would f2 better than f1 (eg. large conditional evaluations).
Are there methods to find P(x) from a switch injection?
Thank you very much!
I suggest you start reading the Wikipedia page about Polynomial Interpolation, if you do not know how to calculate the interpolation polynomial.
Note, that not all calculation methods are suitable for practical application, because of numerical issues (e.g. divisions in the Lagrange version). I am confident that you shold be able to find a libary providing this functionality. Note that the construction will take some time too, hence this makes only sence if your function will be called quite frequently.
Be aware that integer function values and integer points of support do not imply integer coefficients for your polynomial! Thus, in the general case, you will require O(n) floating point operations, and finally a round toward the nearest integer. It may depend on your input wether the interpolation method is reliable and faster than the approach using switch.
Further, I want to propose a differnt solution, assuming that n is rather large. Why dont you put your entries (the pairs (10,3), (37,1), (96,0), (104,1) for your example) inside a serchtree (e.g. std::map in C++ or SortedDictionary in C#)? Thus, your query cost would reduce from linear to O(log n)!
Related
The problem is derive from OJ.
The description is :
We are playing the Guess Game. The game is as follows:
I pick a number from 1 to n. You have to guess which number I picked.
Every time you guess wrong, I'll tell you whether the number I picked is higher or lower.
However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.
Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.
I write small snippet about MinMax problem in recursion. But it is slow and I want to rewrite it in a iterative way. Could anyone help with that and give me the idea about how you convert the recursive solution to iterative one? Any idea is appreciated. The code is showed below:
public int getMoneyAmount(int n) {
int[][] dp = new int[n + 1][n + 1];
for(int i = 0; i < dp.length; i++)
Arrays.fill(dp[i], -1);
return solve(dp, 1, n);
}
private int solve(int[][] dp, int left, int right){
if(left >= right){
return 0;
}
if(dp[left][right] != -1){
return dp[left][right];
}
dp[left][right] = Integer.MAX_VALUE;
for(int i = left; i <= right; i++){
dp[left][right] = Math.min(dp[left][right], i + Math.max(solve(dp, left, i - 1),solve(dp, i + 1, right)));
}
return dp[left][right];
}
In general, you convert using some focused concepts:
Replace the recursion with a while loop -- or a for loop, if you can pre-determine how many iterations you need (which you can do in this case).
Within the loop, check for the recursion's termination conditions; when you hit one of those, skip the rest of the loop.
Maintain local variables to replace the parameters and return value.
The loop termination is completion of the entire problem. In your case, this would be filling out the entire dp array.
The loop body consists of the computations that are currently in your recursion step: preparing the arguments for the recursive call.
Your general approach is to step through a nested (2-D) loop to fill out your array, starting from the simplest cases (left = right) and working your way to the far corner (left = 1, right = n). Note that your main diagonal is 0 (initialize that before you get into the loop), and your lower triangle is unused (don't even bother to initialize it).
For the loop body, you should be able to derive how to fill in each succeeding diagonal (one element shorter in each iteration) from the one you just did. That assignment statement is the body. In this case, you don't need the recursion termination conditions: the one that returns 0 is what you cover in initialization; the other you never hit, controlling left and right with your loop indices.
Are these enough hints to get you moving?
I am new to functional programming. Loops in imperative programming replaces recursion in FP. Another statement is FP gives high concurrency. The instructions being executed parallelly on multi-core/cpu systems as the data is immutable.
Whereas in recursion, steps cannot be executed parallelly due to a step execution is dependent on the previous steps result.
So, I am assuming that recursion in FP will not give high concurrency. Am I correct?
Sort of. You cannot get more execution parallelism than the data parallelism; this is Amdahl's law. However, you frequently have more data parallelism than is expressed in typical sequential algorithms, whether functional or imperative. Consider for example taking the scalar multiple of a vector: (note: this is some made-up algol-style language):1
function scalar_multiple(scalar c, vector v) {
vector v1;
for (int i = 0; i < length(v); i++) {
v1[i] = c * v[i];
}
return v1;
}
Obviously, this isn't going to run in parallel. The situation isn't improved if we re-write in a functional language, using recursion (you can think of this as Haskell):
scalar_multiple c [] = []
scalar_multiple c (x:xn) = c * x : scalar_multiple c xn
This is still a sequential algorithm!
However, you can notice that there is no data dependency --- you don't actually need the result of earlier / later multiplications to calculate later ones. So we have the potential for parallelization here. This can be accomplished in an imperative language:
function scalar_multiple(scalar c, vector v) {
vector v1;
parallel_for (int i in 0..length(v)-1) {
v1[i] = c * v[i];
}
return v1;
}
But this parallel_for is a dangerous construct. Consider a search function:
function first(predicate p, vector v) {
for (int i = 0; i < length(v); i++) {
if (p(v[i])) return i;
}
return -1;
}
If we try speeding this up by replacing for with parallel_for:
function first(predicate p, vector v) {
parallel_for (int i in 0..length(v)-1) {
if (p(v[i])) return i;
}
return -1;
}
Now we won't necessarily return the index of the first element to satisfy the condition, just an element that satisfies it. We broke the contract of the function by parallelizing it.
The obvious solution is 'don't allow return inside parallel_for. But there are lots of other dangerous constructs; in fact, you'll notice I had to abandon the C-style for loop because the increment-and-test pattern itself is dangerous in parallel languages. Consider:
function sequence(int n) {
vector v;
int c = 0;
parallel_for (int i = 0..n-1) {
v[i] = c++;
}
return v;
}
This is again a 'toy' example ("just use v[i] = i;!"), but it illustrates the point: this function initializes v in a random order, due to parallelism. So it turns out that the constructs that are 'safe' to use inside a construct like parallel_for are precisely the constructs that are allowed in purely-functional languages, which makes adding parallel constructs to those languages 'safer' than adding them to imperative languages.
1 This is just a very simple example; of course, real parallelism involves finding bigger chunks of work to parallize than this!
Not sure, if I understand you right, but it generally depends on what you want to accomplish.
One recursion alone cannot execute its subcalls parallel. But you CAN have 2 recursions working on the same dataset. i.e. processing an array from left AND from right simultaneosly trough two concurrent running recursive functions. Those (two) functions can then (theretically) run parallel.
In detail it does not matter if you have a recursive function or a function with a loop inside as long as there is a function who can run on its own. So in respect to your question:
No, a recursive function per definition does not give you any concurrency.
Loops are replaced by higher-order functions more frequently than by direct recursion. Recursion is sort of a catch-all measure in functional programming for when higher-order functions don't already exist for what you need to do.
For example, if you want to run the same calculation on all elements of a list, you use a map, which is highly parallelizable. Finding which elements meet certain criteria is a filter, also highly parallelizable.
Some algorithms just plain require the result of the previous iteration in order to proceed. Those are the ones that tend to require a recursive function, and you're right, they are not generally easy to make highly concurrent.
Is there a better mathematical way to multiply a number by 10 n times in Dart than the following (below). I don't want to use the math library, because it would be overkill. It's no big deal; however if there's a better (more elegant) way than the "for loop", preferably one line, I'd like to know.
int iDecimals = 3;
int iValue = 1;
print ("${iValue} to power of ${iDecimals} = ");
for (int iLp1 = 1; iLp1 <= iDecimals; iLp1++) {
iValue *= 10;
}
print ("${iValue}");
You are not raising to a power of ten, you are multiplying by a power of ten. That is in your code the answer will be iValue * 10^(iDecimals) while raising to a power means iValue^10.
Now, your code still contains exponentiation and what it does is raises ten to the power iDecimals and then multiplies by iValue. Raising may be made way more efficient. (Disclaimer: I've never written a line of dart code before and I don't have an interpreter to test, so this might not work right away.)
int iValue = 1;
int p = 3;
int a = 10;
// The following code raises `a` to the power of `p`
int tmp = 1;
while (p > 1) {
if (p % 2 == 0) {
p /= 2;
} else {
c *= a;
p = (p - 1) / 2;
}
a *= a;
}
a *= t;
// in our example now `a` is 10^3
iValue *= a;
print ("${iValue}");
This exponentiation algorithm is very straightforward and it is known as Exponentiation by squaring.
Use the math library. Your idea of doing so being "overkill" is misguided. The following is easier to write, easier to read, fewer lines of code, and most likely faster than anything you might replace it with:
import 'dart:math';
void main() {
int iDecimals = 3;
int iValue = 1;
print("${iValue} times ten to the power of ${iDecimals} = ");
iValue *= pow(10, iDecimals);
print(iValue);
}
Perhaps you're deploying to JavaScript, concerned about deployment size, and unaware that dart2js does tree shaking?
Finally, if you do want to raise a number to the power of ten, as you asked for but didn't do, simply use pow(iValue, 10).
Considering that you don't want to use any math library, i think this is the best way to compute the power of a number. The time complexity of this code snippet also seems minimal. If you need a one line solution you will have to use some math library function.
Btw, you are not raising to the power but simply multiplying a number with 10 n times.
Are you trying to multiply something by a power of 10? If so, I believe Dart supports scientific notation. So the above value would be written as: iValue = 1e3;
Which is equal to 1000. If you want to raise the number itself to the power of ten, I think your only other option is to use the Math library.
Because the criteria was that the answer needed to not require the math library and needed to be fast and ideally a mathematical-solution (not String), and because using the exponential solution requires too much overhead - String, double, integer, I think that the only answer that meets the criteria is as follows :
for (int iLp1=0; iLp1<iDecimal; iLp1++, iScale*=10);
It is quite fast, doesn't require the "math" library, and is a one-liner
I'm trying to work on a demonstration about multithreading. I need an example of a computationally-intensive function/method. But at the same time, the code that does the computing should be simple.
For example, I'm looking for a function that maybe does something like calculate the nth digit of pi or e:
function calculatePiToNthDecimalDigit(digits) {
var pi = "3.";
for (var i = 1; i < digits; i++) {
pi += digitOfPiAtDecimalPlace(i);
}
return pi;
}
function digitOfPiAtDecimalPlace(decimalPlace) {
...
}
Can anyone give me an example of a function that is relatively simple but can be used in succession (e.g. tight loop) to generate a very hard-to-compute (takes a long time) value?
The simplest I can think of is summing a huge list of numbers. Addition is obviously easy, but if the list is huge, that will make it computationally-intensive, and the problem lends itself well to multi-threading.
Real tests come from real problems. How about the numerical integration of a function using a simple formula such as the trapezoidal rule:
Lets try to prove that using C#
void Main(string[] args)
{
int N = 2097153;
double two = Integral(0, Math.PI, N);
double err = (2.0 - two) / 2.0;
Console.WriteLine("N={0} err={1}", N, err);
}
double f(double x) { return Math.Sin(x); }
double Integral(double a, double b, int N)
{
double h = (b - a) / N;
double res = (f(a) + f(b)) / 2;
for (int j = 1; j < N; j++)
{
double x = a + j*h;
res += f(x);
}
return h * res;
}
at which point I get N=2097153 and err=2.1183055309848E-13 after several milliseconds. If you go much higher in accuracy then the error starts to up as round-off errors start to creep in. I think something similar might happen with a calculation for Pi whereas you will reach you machine accuracy within a few milliseconds and beyond that you are really calculating garbage. You could just repeat the integral several times for a longer overall effect.
So you might be ok to show a drop in time from lets say 140 ms down to 90 ms and count it as a victory.
The multiplication of two NxN matrices has complexity proportional to N^3, so it is relatively easy to create a "computationally intensive" task, just by squaring a sufficiently large matrix. For example, as size goes from N=10 to N=100 to N=1000, the number of (scalar) multiplications required by the classic algorithm for matrix multiplication goes from one thousand to one million to one billion.
Also such a task has plenty of opportunities for parallel processing, if your multi-threading demonstration is meant to take advantage of such opportunities. E.g. the same row can be multiplied by more than one column in parallel.
I know this is probably a very simple question but how would I do something like
n2 in a programming language?
Is it n * n? Or is there another way?
n * n is the easiest way.
For languages that support the exponentiation operator (** in this example), you can also do n ** 2
Otherwise you could use a Math library to call a function such as pow(n, 2) but that is probably overkill for simply squaring a number.
n * n will almost always work -- the couple cases where it won't work are in prefix languages (Lisp, Scheme, and co.) or postfix languages (Forth, Factor, bc, dc); but obviously then you can just write (* n n) or n n* respectively.
It will also fail when there is an overflow case:
#include <limits.h>
#include <stdio.h>
int main()
{
volatile int x = INT_MAX;
printf("INT_MAX squared: %d\n", x * x);
return 0;
}
I threw the volatile quantifier on there just to point out that this can be compiled with -Wall and not raise any warnings, but on my 32-bit computer this says that INT_MAX squared is 1.
Depending on the language, you might have a power function such as pow(n, 2) in C, or math.pow(n, 2) in Python... Since those power functions cast to floating-point numbers, they are more useful in cases where overflow is possible.
There are many programming languages, each with their own way of expressing math operations.
Some common ones will be:
x*x
pow(x,2)
x^2
x ** 2
square(x)
(* x x)
If you specify a specific language, we can give you more guidance.
If n is an integer :p :
int res=0;
for(int i=0; i<n; i++)
res+=n; //res=n+n+...+n=n*n
For positive integers you may use recursion:
int square(int n){
if (n>1)
return square(n-1)+(n-1)+n;
else
return 1;
}
Calculate using array allocation (extremely sub-optimal):
#include <iostream>
using namespace std;
int heapSquare(int n){
return sizeof(char[n][n]);
}
int main(){
for(int i=1; i<=10; i++)
cout << heapSquare(i) << endl;
return 0;
}
Using bit shift (ancient Egyptian multiplication):
int sqr(int x){
int i=0;
int result = 0;
for (;i<32;i++)
if (x>>i & 0x1)
result+=x << i;
return result;
}
Assembly:
int x = 10;
_asm_ __volatile__("imul %%eax,%%eax"
:"=a"(x)
:"a"(x)
);
printf("x*x=%d\n", x);
Always use the language's multiplication, unless the language has an explicit square function. Specifically avoid using the pow function provided by most math libraries. Multiplication will (except in the most outrageous of circumstances) always be faster, and -- if your platform conforms to the IEEE-754 specification, which most platforms do -- will deliver a correctly-rounded result. In many languages, there is no standard governing the accuracy of the pow function. It will generally give a high-quality result for such a simple case (many library implementations will special-case squaring to save programmers from themselves), but you don't want to depend on this[1].
I see a tremendous amount of C/C++ code where developers have written:
double result = pow(someComplicatedExpression, 2);
presumably to avoid typing that complicated expression twice or because they think it will somehow slow down their code to use a temporary variable. It won't. Compilers are very, very good at optimizing this sort of thing. Instead, write:
const double myTemporaryVariable = someComplicatedExpression;
double result = myTemporaryVariable * myTemporaryVariable;
To sum up: Use multiplication. It will always be at least as fast and at least as accurate as anything else you can do[2].
1) Recent compilers on mainstream platforms can optimize pow(x,2) into x*x when the language semantics allow it. However, not all compilers do this at all optimization settings, which is a recipe for hard to debug rounding errors. Better not to depend on it.
2) For basic types. If you really want to get into it, if multiplication needs to be implemented in software for the type that you are working with, there are ways to make a squaring operation that is faster than multiplication. You will almost never find yourself in a situation where this matters, however.