I have written a function which currently displays all of the prime numbers under 1000.
I can keep making 1000 bigger to generate more numbers, but I do not know how to make it keep going on forever once run.
func generatePrimes() {
let numbers = 2...1000
for n in numbers {
var prime = true
for i in 2..<n {
if n % i == 0 {
prime = false
break
}
}
if prime == true {
print(n)
}
}
}
You could use a while loop with the argument being true. That way, the loop would never end.
func generatePrimes() {
var n = 2
while true {
var prime = true
for i in 2..<n {
if n % i == 0 {
prime = false
break
}
}
if prime == true {
print(n)
}
n += 1
}
}
Note that we use a variable (n) and increment (add 1 to it) on each iteration of the while loop.
If you want it to end when it hits some n, you can have a conditional that checks against n and break's out of the loop:
var n = 2
while true {
if n == 1000 {
break
}
n += 1
}
The above code will stop execution when n hits 1000 and will not print anything for that case.
Use a while loop instead of for loop. It can also be done with a for loop, but for loop is preferred when you know the number or range of the iterations you are performing. If you're using a while loop you just have to set the condition for the loop, in this case you need to specify the condition as true since you want the loop to run forever.
Just replace your for loop in your code with while. It should look something like this
while(true)
{
if n % i == 0 {
prime = false
break
}
}
You also can build an unconditional infinite loop using do {} block and continue statement.
AGAIN:
do {
print("step...")
continue AGAIN
}
func generatePrimes() {
var prime = true
let numbers = 2...1000
for n in numbers {
prime = true
if n == 2 || n == 3 {
print(n)
} else {
for i in 2..<n-1 {
if n % i == 0 {
//not prime numbers
prime = false
}
}
if prime == true {
print(n)
}
}
}
}
You can do it like this. Just exclude 1 and n.
Related
I m making a card game where 3 random numbers are generated..I need to check are these numbers Row numbers...
like 4 6 5 and 23,24,22. are row numbers
I have made method but I think there should be easy arithmetic formulas
I have tried this and working well, but I need simple arithmatic formula to avoid use of array and for
bool isAllInRow(int num1, int num2,int num3)
{
//subject : tinpati
List<int> numbers=[num1,num2,num3];
bool is_in_row=true;
numbers.sort();
if(numbers[0]==1 && numbers[1]==12 && numbers[2]==13)
return true;
for(int x=0;x<numbers.length-1;x++)
{
if(numbers[x]-numbers[x+1]!=-1)
{
is_in_row=false;
break;
}
}
return is_in_row;
}
So you want to know if the cards form a straight, with aces both low and high.
Is the "three cards" fixed, or would you want to generalize to more cards?
Sorting should be cheap for such a short list, so that's definitely a good start. Then you just need to check the resulting sequence is increasing adjacent values.
I'd do it as:
bool isStraight(List<int> cards) {
var n = cards.length;
if (n < 2) return true;
cards.sort();
var first = cards.first;
if (first == 1 && cards[1] != 2) {
// Pretend Ace is Jack if n == 3.
// Accepts if remaining cards form a straight up to the King.
first = 14 - n;
}
for (var i = 1; i < n; i++) {
if (cards[i] != first + i) return false;
}
return true;
}
This code rejects card sets that have duplicates, or do not form a straight.
I think you are looking for Arithmetic Progression.
bool checkForAP(List<int> numberArr) {
numberArr.sort();
int diff = numberArr[1] - numberArr[0];
if (numberArr[2] - numberArr[1] != diff) {
return false;
}
return true;
}
And modify your function like
bool isAllInRow(int num1, int num2,int num3) {
//subject : tinpati
List<int> numbers=[num1,num2,num3];
bool is_in_row=true;
numbers.sort();
if(numbers[0]==1 && numbers[1]==12 && numbers[2]==13)
return true;
return checkForAP(numbers);
}
Note: remove sort in AP method as it is of no use. Since your numbers
list length is 3 I directly compared numbers for AP, the same can also
be written for n numbers with for.
bool checkForAp(numberArr) {
numberArr.sort();
int diff = numberArr[1] - numberArr[0];
for(int i = 2; i< numberArr.length ;i++) {
if (numberArr[i] - numberArr[i - 1] != diff) {
return false;
}
}
return true;
}
You could do it like this:
bool isAllInRow(int num1, int num2,int num3) {
if (num1 == num2 || num2 == num3) return false;
var maxNum = max(num1, max(num2, num3));
var minNum = min(num1, min(num2, num3));
return (maxNum - minNum == 2) || (minNum == 1 && maxNum == 13 && num1 + num2 + num3 == 26);
}
In Dart, is there a simple way to check whether the sum of a list will produce a 'real' value (a value that doesn't overflow or underflow)?
Examples:
overflowSafeSum([0,1,2]) //3
overflowSafeSum([1,9223372036854775807]) //Over
overflowSafeSum([-1,-9223372036854775808]) //Under
I'm new to dart, this is the best I got right now:
import 'dart:math' show pow;
enum Overflow {
over,
under,
}
void main() {
//idea: Iterate through the elements of a list and add them,
//each time the sum overflows: increase overflowCounter by 1
//each time the sum underflows: decrease overflowCounter by 1
//if all the elements have been added and the overflowCounter == 0, the sum must be real
overflowSafeSum(List<int> userList) {
var sum = 0, overflowCounter = 0;
for (int index = 0, nextTerm;
index < userList.length;
index++, sum += nextTerm) {
nextTerm = userList[index];
if (sum.sign != nextTerm.sign) {
continue; //adding a postive and negative can't overflow or underflow
} else if (sum >= 0 && nextTerm >= 0) {
if ((sum + nextTerm) < 0) overflowCounter++;
} else {
if ((sum + nextTerm) >= 0) overflowCounter--;
}
}
if (overflowCounter == 0) {
return sum;
} else if (overflowCounter > 0) {
return Overflow.over;
} else {
return Overflow.under;
}
}
var myList = [1,0,(pow(2,63)-1).toInt()];
print(overflowSafeSum(myList)); //Overflow.over
}
(To be pedantic: "underflow" is not negative overflow. Overflow occurs when the magnitude of a number is too large to be represented, regardless of sign. Underflow is an issue with floating-point operations where the magnitude of a number is too small (too close to 0) to be represented.)
You can't generally detect overflow with Dart ints since Dart for the web is transpiled to JavaScript, where ints are backed by JavaScript numbers (IEEE-754 double-precision floating-point values). If you instead use Int32 or Int64 from package:fixnum (or if you restrict yourself to the Dart VM), then you could make a helper function like:
class OverflowException implements Exception {
OverflowException({this.positive = true});
bool positive;
}
Int64 checkedAdd(Int64 a, Int64 b) {
var sum = a + b;
if (a > 0 && b > 0 && sum < 0) {
throw OverflowException(positive: true);
}
if (a < 0 && b < 0 && sum > 0) {
throw OverflowException(positive: false);
}
return sum;
}
From there, you could trivially add a function that calls it in a loop:
Int64 overflowSafeSum(Iterable<int> numbers) {
var sum = Int64(0);
for (var number in numbers) {
sum = checkedAdd(sum, Int32(number));
}
return sum;
}
or if you prefer using Iterable.fold:
Int64 overflowSafeSum(Iterable<int> numbers) =>
numbers.fold<Int64>(Int64(0), (sum, i) => checkedAdd(sum, Int64(i)));
I want to do something like this with an optional in swift.
var opt:MyOptional?
func myfunction() {
if (opt == nil) || (opt?.x != 10 && opt?.y != 20)) {
opt = MyOptional()
opt.x = 10
opt.y = 20
}
}
My question is if this is a valid pattern, even though it compiles and runs. Does Swift compiler ensures condition 2 runs after condition 1 (opt!= nil)?
Well && and || operators in swift are Left Associative which means your evaluation of conditions goes from left hand side.
(opt != nil). // result 1
this condition will get evaluate first and as you are using the || operator.
Secondaly your (opt?.x != 10 && opt2?.y != 20) // result 2
will get now evaluate if your result 1 is false otherwise it would have gone in the loop because of || operator
final condition
if (result 1 || result 2) {
if only result 1 is true it not evaluate for result 2 due to || operator otherwise it would calculate result 2 and if result 2 is true its a success
Assuming you have got a typo this code should looks like this:
struct MyOptional {
var x: Int = 0
var y: Int = 0
}
class SomeClass {
var opt: MyOptional?
func myFunction() {
if let unwrappedOpt = opt,
unwrappedOpt.x != 10 && unwrappedOpt.y != 20 {
opt = MyOptional(x: 10, y: 20)
}
}
}
What about your question? You are right.
I've got a tree structure.
The task is to find the biggest sum/weight of path nodes, but i can only move n times. Thats ok, but going "up"/"back" cost nothing.
How can i accomplish that?
Below is my code, but the problem is that the each node can only be accessed once, so it doesnt work.
int mSum(Node* node, int mvLeft) {
if (node == nullptr) { return 0; }
if (mvLeft == 0) { return node->value; }
mvLeft--;
int sum = max(mSum(node->left, mvLeft), mSum(node->right, mvLeft));
return node->value + max(sum, mSum(node->parent, mvLeft + 1));
}
Here is the example graph. The numbers on the nodes represent the cost of getting to it. Each node can be visited only once except going "back".
The n step limit here is 3, we're counting entering the graph too, so the proper result is 21 because: 2->8->11.
If we would have limit of 4 steps the result would be 31: 2->10->8->11
My friend tried to do it with DFS, is he right? What's the best algorithm?
Good answer is taking multiple routes at the same time.
I mean we could go with 2-length limit:
2 left 0 right
1 left 1 right
0 left 2 right
Working, but somewhat slow, code :)
Its working time is 28s while other solutions can go with 2s (10 not known tests)
int mSum(Node* node, int mvLeft) {
mvLeft--;
if (mvLeft < 0) {
return 0;
}
else if (mvLeft == 0) {
return node->value;
}
if (node->left != nullptr && node->right != nullptr) {
int max = 0;
for (int i = 0; i <= mvLeft; i++) {
max = Max(max, mSum(node->left, i) + mSum(node->right, mvLeft - i));
}
return max + node->value;
}
else if (node->left != nullptr) {
return mSum(node->left, mvLeft) + node->value;
}
else if (node->right != nullptr) {
return mSum(node->right, mvLeft) + node->value;
}
return node->value;
}
The program works fine with var dig = 0 and it doesn't work with var dig:Int I get an error: Variable "dig" used before being initialized Could you explain me why?
func myFunc(a:Int, b:Int) {
var c = a / b
var o = a % b
var v = 0
var dig = 0
if o != 0 {println("\(a)/\(b) = \(c) и \(o)/\(b)")}
else {println("\(a)/\(b) = \(c)")}
if a > b {
v = b
}
else {
v = a
}
for var i = 1; i <= v; ++i {
if a % i == 0 && b % i == 0 {dig = i}
}
println("\(dig) - greatest common denominator of \(a) and \(b)")
}
myFunc(27,81)
The only place you set the value of dig is inside of an if statement that is inside of a for loop. The Swift compiler does not know if the body of the for loop will be executed, and it doesn't know if the if statement will ever be true, so it has to assume that there is a path in which dig is not initialized.
Consider this simpler example:
func myFunc(a:Int, b:Int) {
var dig: Int
if a >= b {
dig = 3
}
if a < b {
dig = 4
}
println("\(dig) - greatest common denominator of \(a) and \(b)")
}
This example also gives the same error, because Swift considers each if separately. It is obvious to us that a is either greater than or equal to b or it is less than b, but Swift doesn't go that far in evaluating the situation. It just considers that each if may not be true, and dig is only set inside of ifs, so it is possible (as far as Swift is concerned) that dig may not be set.
func myFunc(a:Int, b:Int) {
var dig: Int
if a >= b {
dig = 3
} else {
dig = 4
}
println("\(dig) - greatest common denominator of \(a) and \(b)")
}
If you change the second condition to an else, Swift is then happy because it can reason that the if must be true or false and dig is set in each path, so it will certainly have a value before the println statement.
The compiler does not know mathematics good enough to
recognize that the statement
if a % i == 0 && b % i == 0 {dig = i}
is actually executed at least once (for i == 1). Therefore
the compiler assumes that dig might be undefined at
println("\(dig) - greatest common denominator of \(a) and \(b)")
Assigning an initial value in
var dig = 0
is the correct solution.
Btw., the Euclidean Algorithm is a much more effective method to
compute the greatest common divisor, see for example
http://rosettacode.org/wiki/Greatest_common_divisor#Swift.