The code runs correctly and it does what it is supposed to do, but I was told I could make it faster by using Boolean expressions instead but would not really know where to insert them here. The problem is:
Given a sequence of n points with their coordinates, write a program remote, which calculates the value of the smallest remoteness of a point, which is outside the square. A point is outside the square, if it is neither inner to the square, nor belongs to square contour. If there are no points outside the square, your program has to output 0.
Constraints:
1 ≤ n ≤ 10000 and 1 ≤ a ≤ 1000 ;
Example:
Input:
5 4
1 2
4 6
-3 2
-2 2
4 -1
Output: 5
Could someone suggest me any technique to make the code more efficient?
int remote(int x, int y) {
int z = abs(x) + abs(y);
return z;
}
int main() {
int n, a;
int x;
int y;
cin >> n >> a;
int z=20001;
for (int i = 1; i <= n; i++) {
cin >> x >> y;
if (x > a / 2 || y > a / 2) {
if (z > remote(x, y)) {
z = remote(x, y);
}
}
}
cout << z <<endl;
return 0;
}
For one, you are calling remote twice (in some cases) needlessly.
Consider using this:
#include <algorithm>
z = std::max(z, remote(x, y));
This will also shorten and clarify the code.
Also, it's possible the divisions are slow. Try (after profiling!) replacing
x > a / 2 || y > a / 2
by
(x << 1) > a || (y << 1) > a
Note #Donnie & others claims in the comments that compilers will do the latter optimization, and they are probably correct.
I would like to show you the timings on my machine:
Version 1:
for (int i = 1; i <= n; i++) {
cin >> x >> y;
if (x > a / 2 || y > a / 2) {
if (z > remote(x, y)) {
z = remote(x, y);
}
}
}
Version 2:
for (int i = 1; i <= n; i++) {
cin >> x >> y;
/* if (x > a / 2 || y > a / 2) {
if (z > remote(x, y)) {
z = remote(x, y);
}
}
*/
}
For n=10^5, compiled with -O3 both yield 60ms. Compiled without optimization: both 60ms.
First step for optimizing is to know where your program spends time. Reading/parsing the data is the bottle neck.
You could speed up it a little bit by adding as first line to your main:
ios_base::sync_with_stdio(false);
On my machine I'm down to 20ms.
1) Assign a temporary value to the remote function:
if (x > a / 2 || y > a / 2)
{
const int r = remote(x,y);
if (z > r)
{
z = r;
}
}
2) Replace the call to remote with the contents of remote, removing the overhead of a function call:
if (x > a / 2 || y > a / 2)
{
const int r = abs(x) + abs(y);
if (z > r)
{
z = r;
}
}
3) Replace a / 2 with a constant temporary variable:
const int midpoint = a >> 1;
if (x > midpoint || y > midpoint)
4) Change compiler optimization level to high - for speed.
5) The bottleneck is now in the input statement. Any gain by optimizing the remainder of the loop is wasted by the Input time. There is no more Return On Investment for further changes.
Related
I am trying to understand what is happening in my recursion step return x + times(x, y-1).
Specifically times(x, y-1) Since there is no equation in the function what is happening in the recursion? I don't see how the values are added.
#include <iostream>
using namespace std;
int times(int x, int y)
{
if (x == 0 || y == 0)
return 0;
else if (y == 1)
return x;
else
return x + times(x, y - 1);
}
int main()
{
int x, y;
cout << "Enter two numbers to be multiplied seperated by a space: ";
cin >> x >> y;
cout << "The product is " << times(x, y) << endl;
return 0;
}
The best way to understand recursion is to write all the stack trace on paper.
Let's take an example of times(3, 2):
Call 1: times(3, 2) -> returns 6
Call 2: 3 + times(3, 1) -> returns (3 + 3), that is 6
Call 3: times(3, 1) -> returns 3
So, the final answer is 6.
x + times(x, y-1): It's a way of representing multiplication using addition. For example:
1) 3 * 2 = (3 + 3)
2) 4 * 3 = (4 + 4 + 4 + 4)
#dxiv mentioned it perfectly, a * n = a + a * (n - 1).
I was trying to solve this problem and from the comments section in the editorial, I was directed to the following solution :
#include <bits/stdc++.h>
using namespace std;
#define MAX(a,b,c) max(a,max(b,c))
int n,a,b,c,dp[4001];
int f(int x)
{
if (x == 0) return 0;
if (x < 0 || (x > 0 && x < a && x < b && x < c))
return 0xACCE97ED; // <- **I have doubt here**
if (!dp[x]) dp[x] = MAX(f(x-a),f(x-b),f(x-c)) + 1;
return dp[x];
}
int main()
{
cin >> n >> a >> b >> c;
memset(dp,0,sizeof(dp));
cout << f(n) << endl;
}
I wanted to know:
What is the need of the if statement that returns 0xACCE97ED for the test case:
4000 1 2 3. This test case dosen't work when that specific if statement is missing.
Why specifically 0xACCE97ED is being returned? Because when I tried to return any other number (say 9999), then the output is expected output + 9999.
if (x < 0 || (x > 0 && x < a && x < b && x < c))
return 0xACCE97ED; // -1395746835
Well looking at the dp function, it is basically maximizing values and this specific if statement is saying:
if x < 0
the length of the ribbon you cut is negative (which should be impossible)
or if x > 0 and x < a, b, c which means you can still cut X but all available sizes would result into having a ribbon of negative length
return 0xACCE97ED; return a random negative value which happens to spell out ACCEPTED because this state is invalid
And since the third if statement will try to get the max value, 0xACCE97ED will never be selected as the max value.
0xACCE97ED means "ACCEPTED" in the 1ee7 speech. nothing else specific about this value.
What is the need of the if statement that returns 0xACCE97ED for the test case: 4000 1 2 3
if (x < 0 || (x > 0 && x < a && x < b && x < c))
return 0xACCE97ED; // <- **I have doubt here**
because the function f is recursive, in the next line it calls itself:
if (!dp[x]) dp[x] = MAX(f(x-a),f(x-b),f(x-c)) + 1;
return dp[x];
with a smaller values for x so presumable it will eventually make that if statement true and will return "accepted" (0xACCE97ED).
I'm new to the Z3-Solver with API in C++ and want to solve a group of inequalities and find the results.
I've read the answer which written in Python and try to write it in C++, but it repeating prints one model.
5 <= x + y + z <= 16
AND -4 <= x - y <= 6
AND 1 <= y - z <= 3
AND -1 <= x - z <= 7
AND x >= 0 AND y >= 0 AND z >= 0
The inequalities were added into the solver, and have a lot of evaluations.
c is context and s is the solver.
vector<const char*> variables {"x", "y", "z"};
// ...
// till here, s was added into several constraints
while(s.check() == sat){
model m = s.get_model();
cout << m << "\n######\n";
expr tmp = c.bool_val(false);
for(int i = 0; i < variables.size(); ++ i){
tmp = tmp || (m[c.int_const(variables[i])] != c.int_const(variables[i]));
}
s.add(tmp);
}
And the result:
(define-fun z () Int
0)
(define-fun y () Int
2)
(define-fun x () Int
3)
######
(define-fun z () Int
0)
(define-fun y () Int
2)
(define-fun x () Int
3)
######
(define-fun z () Int
0)
...
And it just print one model. I'm not sure where is wrong.
How can I get all models or get one or more convex sets (such as {l1 <= x <= u1 and l2 <= x - y <= u2 and ...}), but not to traverse all evaluations.
BTW, there's a lot tutorials on python(such as this), where I can learn z3 in c++ as the example and api doc. is not easy to get started.
Your "model refutation" loop isn't quite correct. Since you didn't post your whole code, it's hard to tell if there might be other issues, but this is how I would go about it:
#include<vector>
#include"z3++.h"
using namespace std;
using namespace z3;
int main(void) {
context c;
expr_vector variables(c);
variables.push_back(c.int_const("x"));
variables.push_back(c.int_const("y"));
variables.push_back(c.int_const("z"));
expr x = variables[0];
expr y = variables[1];
expr z = variables[2];
solver s(c);
s.add(5 <= x+y+z);
s.add(x+y+z <= 16);
s.add(-4 <= x-y);
s.add(x-y <= 6);
s.add(-1 <= x-z);
s.add(x-z <= 7);
s.add(x >= 0);
s.add(y >= 0);
s.add(z >= 0);
while (s.check() == sat) {
model m = s.get_model();
cout << m << endl;
cout << "#######" << endl;
expr tmp = c.bool_val(false);
for(int i = 0; i < variables.size(); ++i) {
tmp = tmp || (variables[i] != m.eval(variables[i]));
}
s.add(tmp);
}
return 0;
}
This code runs and enumerates all the "concrete" models. From your question, I surmise you're also wondering if you can get "symbolic" models: That's not possible with an SMT solver. SMT solvers only produce concrete (i.e., all ground term) models, so if you need to generalize from them, you'll have to do that outside of the z3 boundary.
I need to set thresholds on an x,y coords so that i can analyse numbers surrounding those coordinates. basically i need set a low threshold 1 below x and 1 below y and set a high threshold of 1 above x and 1 above y.
I also have limits so that if either x or y is 0 the low threshold is 0 and if x or y is 9 the high threshold is 9
The code i have so far is as follows: - its the same for y so i only posted finding thresholds for x...
bool CheckNextTo(int X, int Y){
int LowThreshX,LowthreshY;
int HighThreshX,HighThreshY;
if (X == 0){
LowThreshX =0;
HighThreshX =X+1;
}
else if (X== 9){
LowThreshX =X-1;
HighThreshX=9;
}
else{
LowThreshX=X-1;
HighThreshX=X+1;
}
}
This works i'm just wandering if there is an easier way to do this as i will have to write all this again to work out Y thresholds. I know i could write function that returns thresholds too but besides that is there anything else?
Thanks
You can use the ternary if operator, which is built-in in many languages (C and C++ included)
(condition ? then : else)
LowThresholdX= (X == 0 ? 0 : X-1);
HighThresholdX= (X == 9 ? 9 : X+1);
LowThresholdY= (Y == 0 ? 0 : Y-1);
HighThresholdY= (Y == 9 ? 9 : Y+1);
Min and Max functions come in handy:
LowThresholdX= Max(0, X-1);
HighThresholdX= Min(9, X+1);
LowThresholdY= Max(0, Y-1);
HighThresholdY= Min(9, Y+1);
I haven't really understood what you want but writing things without repetition might make things clearer to you.
Your code becomes :
bool CheckNextTo(int X, int Y){
int LowThreshX = X-1;
int HighThreshX = X+1;
if (X == 0){
LowThreshX = 0;
}
else if (X == 9){
HighThreshX = 9;
}
}
which might as well be written :
bool CheckNextTo(int X, int Y){
int LowThreshX = X-1;
int HighThreshX = X+1;
if (X == MIN){
LowThreshX = X;
}
else if (X == MAX){
HighThreshX = X;
}
}
Try this:
bool CheckNextTo(int X, int Y){
int LowThreshX = MAX(0,X-1);
int LowThreshY = MAX(0,Y-1);
int HighThreshX = MIN(9,X+1);
int HighThreshX = MIN(9,Y+1);
}
MAX, MIN are either the macro functions or std::max/min.
You could write this as a function, but that would depend on how your other code works, and you haven't shown us that.
This question already has answers here:
Mathematically Find Max Value without Conditional Comparison
(18 answers)
Closed 9 years ago.
So i have too get two numbers from user input, and find the max of the two numbers without using if statements.
The class is a beginner class, and we have too use what we already know. I kinda worked something out, but it only works if the numbers are inputted with the max number first.
#include <iostream>
using namespace std;
int main()
{
int x = 0, y = 0, max = 0;
int smallest, largest;
cout << "Please enter 2 integer numbers, and i will show you which one is larger: ";
cin >> x >> y;
smallest = (x < y == 1) + (x - 1);
smallest = (y < x == 1) + (y - 1);
largest = (x < y == 1) + (y - 1);
largest = (y > x == 1) + (x + 1 - 1);
cout << "Smallest: " << smallest << endl;
cout << "Largest: " << largest << endl;
return 0;
}
Thats what i have so far, but after putting different test data in, i found out it only works for numbers such as 4,5 or 6,7. But numbers with more then 2 spaces between eachother they dont such as, 4,8 or 5, 7. Any help would be appreciated.
I saw this question in Cracking the Coding interview book.
Let’s try to solve this by “re-wording” the problem We will re-word the problem until we get something that has removed all if statements
Rewording 1: If a > b, return a; else, return b
Rewording 2: If (a - b) is negative, return b; else, return a
Rewording 3: If (a - b) is negative, let k = 1; else, let k = 0 Return a - k * (a - b)
Rewording 4: Let c = a - b Let k = the most significant bit of c Return a - k * c
int getMax(int a, int b) {
int c = a - b;
int k = (c >> ((sizeof(int) * CHAR_BIT) - 1)) & 0x1;
int max = a - k * c;
return max;
}
Source: http://www.amazon.com/Cracking-Coding-Interview-Programming-Questions/dp/098478280X
Edit: This code works even when a-b overflows.
Let k equal the sign of a-b such that if a-b >=0, then k is 1, else k=0.Let q be the inverse of k. Above code overflows when a is positive or b is negative, or the other way around. If a and b have different signs, then we want the k to equal sign(a).
/* Flips 1 to 0 and vice-versa */
public static int flip(int bit){
return 1^bit;
}
/* returns 1 if a is positive, and 0 if a is negative */
public static int sign(int a){
return flip((a >> ((sizeof(int) * CHAR_BIT) - 1)) & 0x1);
}
public static int getMax(int a, int b){
int c = a - b;
int sa = sign(a-b); // if a>=0, then 1 else 0
int sb = sign(a-b); // if b>=1, then 1 else 0
int sc = sign(c); // depends on whether or not a-b overflows
/* If a and b have different signs, then k = sign(a) */
int use_sign_of_a = sa ^ sb;
/* If a and b have the same sign, then k = sign(a - b) */
int use_sign_of_c = flip(sa ^ sb);
int k = use_sign_of_a * sa + use_sign_of_c * sc;
int q = flip(k); //opposite of k
return a * k + b * q;
}
Here is a funny solution:
int max_num = (x>y)*x + (y>=x)*y;
Assuming that you have covered bitwise operators already you can do this:
max = a-((a-b)&((a-b)>>(sizeof(int)*8-1)));
This is based off of the solution from Mathematically Find Max Value without Conditional Comparison that #user93353 pointed out in the comments above.
This may be overkill if you really are just trying to avoid if statements, not comparisons in general.
You can try this code to find max and min for two input variables.
((a > b) && (max = a)) || (max=b);
((a < b) && (min = a)) || (min=b);
For three input variables you can use similar method like this:
int main()
{
int a = 10, b = 9 , c = 8;
cin >> a >> b >> c;
int max = a, min = a;
// For Max
((a > b) && (a > c) && (max=a)) ||
((b > c) && (b > a) && (max=b)) ||
(max=c) ;
// For min
((a < b) && (a < c) && (min=a)) ||
((b < c) && (b < a) && (min=b)) ||
(min=c) ;
cout << "max = " << max;
cout << "and min = " << min;
return 1;
}
One run is:
:~$ ./a.out
1
2
3
max = 3 and min = 1
Edit
Thanks to #Tony D: This code will fail for negative numbers.
One may try this for negative numbers for two inputs to find max(not sure for this):
((a > b) && ( a > 0 && (max = a))) || ((b > a) && (max = b)) || (max = a);