I need to loop through a number (xx). xx always starts at zero. My problem is that if the moveDirection variable is +1 then xx increases until it reaches the positive of range. If moveDirection is -1, then xx decreases until reaching the negative of range.
In the code below, I have done this by having an if statement test for moveDirection first, then I have duplicated the for loop, and edited the values for each case. My code happens to be in ActionScript3, but the language does not matter.
var p:Point;
var xx:int;
if (moveDirection > 0)
{
for (xx = 0; xx < range; xx++)
{
if (hitTestPoint(xx, yy))
{
return true;
}
}
}
else
{
for (xx = 0; xx > range; xx--)
{
if (hitTestPoint(xx, yy))
{
return true;
}
}
}
Is there a better way of doing this, maybe without duplicating the for loop? If there is any other advice, it would be greatly appreciated.
for (xx = 0; xx != range; xx += moveDirection)
{
if (hitTestPoint(xx, yy))
{
return true;
}
}
This assumes that moveDirection will be either 1 or -1 for going up or down, respectively. Also, you'll have to slightly change your range for the != to work properly. But, it does cut down on code.
From the looks of the code, it doesn't really matter which direction the loop runs -- you're just returning true if hitTestPoint returns true for some value in the range. If that's so, another possibility would be something like:
var start:int = min(0, range);
var stop:int = max(0, range);
for (xx = start; xx!=stop; xx++)
if (hitTestPoint(xx,yy)
return true;
Another possibility:
int i;
for (i = abs(range), xx = 0; --i >= 0; xx += moveDirection){
if (hitTestPoint(xx, yy) return true;
}
Here's an example in Java (see also on ideone.com):
static void go(final int range, final int direction) {
for (int i = 0; i != direction*range; i += direction) {
System.out.println(i);
}
}
Then you can do:
go(5, +1); // 0, 1, 2, 3, 4
go(5, -1); // 0, -1, -2, -3, -4
If you want to accommodate non-unit step, it's simplest to define a third parameter as follows:
static void go(final int range, final int step, final int direction) {
for (int i = 0; i < range; i += step) {
System.out.println(i * direction);
}
}
Then you can do:
go(10, 3, +1); // 0, 3, 6, 9
go(10, 3, -1); // 0, -3, -6, -9
Related
This would be the general problem statement:
A prisoner escapes from the jail by jumping over N walls each with height of each wall given in an array. He can jump x meters of height, but after every jump he slips y meters due to some uncontrollable factors(wind, slippery wall, etc).
Similar problem statement mentioned here
The programming task given was to debug a function which included four parameters -
NoOfJumps(int x, int y, int N, int Height[])
Number of meters he jumps
Number of meters he slips down the wall
Number of walls
Height of the walls as an array
The first test case was for parameters - (10, 1, 1, {10})
10 being the meters he jumps, 1 meter he slips down, Number of walls being 1, and height of the wall being 10. Now:
effectiveJump = x - y = 9.
So he would have to jump twice to jump over the walls. So, this function should return 2 (total number of jumps required to escape).
There was also another test case for the parameters - (3, 1, 5, {20,5,12,11,3})
3 being the meters he jumps, 1 meter he slips down, Number of walls being 5, and height of the walls given as 20m, 5m, 12m, 11m, 3m. Now:
effectiveJump = x - y = 2.
We were given the output for the above parameter values as 24.
NoOfJumps(3, 1, 5, {20,5,12,11,3})
I can't understand how this output value is obtained. How exactly are the walls arranged?
I can only think of one solution for the corner case, i.e, when the person jumps over the wall
(when (x) > remaining height of the wall),
he should not slip down else I can't obtain the required solution.
For example, in the second test case at first wall, when the person is at 18m height, and he jumps 3m to 21m and doesn't slip down as he has crossed that wall. Next he starts jumping from 21 and not 20. The sequence of jumping would be :
0->2->4->6->8->10->12->14->16->18->21->23->26->28->30->32->34->36->39->41->43->45->47->50->53
Assuming walls at height, 20, 25, 37, 48, 51.
Is this a correct assumption for solving the problem?
C code on given case 2, will work for case 1 on changing the
parameters to (10,1,1,10).
#include<conio.h>
#include<stdio.h>
int jump(int x,int y,int n,int z[]);
int jump(int x,int y,int n,int z[])
{
int i, j, countjump, total = 0, extra = 0;
clrscr();
printf("\n%d\n", n);
for (i = 0; i < n; i++) {
printf("\n%d", z[i]);
}
printf("\n");
for (j = 0; j < n; j++) {
countjump = 1;
z[j] = z[j] + (extra) - x;
while (z[j] >= 0) {
z[j] = z[j] + y;
z[j] = z[j] - x;
countjump = countjump + 1;
if (z[j] < 0) {
extra = z[j];
}
}
total = (countjump + total);
}
return total;
}
void main()
{
int res, manjump = 3, slip = 1, nwalls = 5;
int wallheights[] = {20, 5, 12, 11, 3};
clrscr();
res = jump(manjump, slip, nwalls, wallheights);
printf("\n\ntotal jumps:%d", res);
getch();
}
Try this code. May not be optimized
$input1 = Jump Height
$input2 = Slipage
$input = Array of walls height
function GetJumpCount($input1,$input2,$input3)
{
$jumps = 0;
$wallsCrossed = 0;
while($wallsCrossed != count($input3)){
$jumps++;
$input3[$wallsCrossed] = $input3[$wallsCrossed] - $input1;
if($input3[$wallsCrossed] > 0){
$input3[$wallsCrossed] = $input3[$wallsCrossed] + $input2;
}else{
$wallsCrossed++;
}
}
return $jumps;
}
The walls come one after another. After jumping wall one the position should start from zero and not from the last jump height. For the first case the output should really be 1 as the height and jump are same. In the second test case, 24 is the right output.
I've seen the exact same question on techgig contest. For the first test case the output should be 1. The test case had been explained by themselves where there is no slipping if the jump and height are same.
Try this
You don't require the number of walls as it equals to size of array
public class Jump {
public static void main(String[] a) {
int jump = 3;
int slip = 1;
int[] hights = {20,5,12,11,3};
int count = 0;
for (int hight : hights) {
int temp = hight - jump;
if (temp >= 0) {
count = count + temp / (jump - slip)+1;
}
if (temp % (jump - slip) > 0) {
count++;
}
}
System.out.println(count);
}
}
Logic is here Plz check if this solves your problem.
package puzeels;
public class Jump
{
int jump=6;
int slip=1;
int numberOfWals=4;
int height[] ={21,16,10,5};
static int count=0;
int wallheight=0;
private int findJump()
{
for(int i=0;i<height.length;i++)
{
wallheight=height[i];
while((wallheight>0))
{
count=count+1;
wallheight=wallheight-(jump-slip);
System.out.println(wallheight+" "+count);
}
System.out.println("Out of while loop");
}
return count;
}
public static void main(String arr[])
{
Jump obj = new Jump();
int countOfJumps=obj.findJump();
System.out.println("number of jumps is==> "+countOfJumps);
}
}
You can use this one.
Sample Code
public static int calculateJumps(int X, int Y, int height[]) {
int tn=0,n;
for(int i=0; i<height.length; i++) {
if(height[i]<=X) {
tn+=1;
continue;
}
n=((height[i]-X)/(X-Y));
n+=height[i]-((X-Y)*n)==X?1:2;
tn+=n;
}
return tn;
}
You need to pass only X , Y and Array than you can get you output.
I think 12 is a wrong answer, as I tried this code I got 11, last jump doesn`t have a slip:
public static void main(String [] args) {
int T;
int jcapacity, jslip, nwalls;
//BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
Scanner sc = new Scanner(System.in);
T = sc.nextInt();
jcapacity = sc.nextInt();
jslip = sc.nextInt();
nwalls = sc.nextInt();
int [] wallHeightArr = new int [nwalls];
for (int i = 0; i< nwalls; i++) {
wallHeightArr[i] = sc.nextInt();
}
sc.close();
while(T-->0) {
int distance = log(jcapacity,jslip,wallHeightArr);
System.out.println(distance);
}
}
private static int log(int jcapacity, int jslip, int[] wallHeightArr) {
// TODO Auto-generated method stub
int distance = 0;
for(int i = 0; i< wallHeightArr.length; i++) {
int cHeight = 0;
int count = 0;
while (wallHeightArr[i] - cHeight > jcapacity) {
cHeight += (jcapacity - jslip);
count++;
}
count++;
distance += count;
}
return distance;
}
def jumpTheifCount(arr, X, Y):
jump = 0
remheight = 0
for i in range(len(arr)):
if X == arr[i]:
jump = jump + 1
continue
if X < arr[i]:
jump = jump + 1
remheight = arr[i] - X + Y
if remheight > X:
jump = jump + 1
remheight = arr[i] - X + Y
if remheight < X:
jump = jump + 1
continue
return jump
arr = [11, 10, 10, 9]
X = 10
Y = 1
print(jumpTheifCount(arr, X, Y))
check if this solves your problem
def GetJumpCount(jump, slips, walls):
"""
#jump:int, Height of 1 jump
#slips:int, height of slip
#walls:array, height of walls
"""
jumps = []
for wall_height in walls:
wall_jump = 1
wall_height -= jump
while wall_height > 0:
wall_height += slips
wall_height -= jump
wall_jump += 1
jumps.append(wall_jump)
return sum(jumps)
Using ActionScript 3, suppose I have an array of numbers, lets say: 1, 2, 3, 4, 5. Is there a way to easily search this array and return the index corresponding to an element that is >= 2.5 (which would be, 3, in this case), for example? I'm implementing this with a while and for loop, and seems pretty wordy. Thought there might be a method for this already, but haven't stumbled upon it in:
http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/Array.html#every()
Otherwise, what would be a simple way to achieve it?
In case it helps, I'll use this to implement a straight-forward linear interpolation math routine, assuming one doesn't already exist I'm not aware of.
I'm not aware of any firstIndexOf in ActionScript.
You could add it to an ArrayUtil class:
Given the array:
var array:Array = [ 1, 2, 3, 4, 5 ];
Pass it to the ArrayUtil function:
public static function firstIndexOf(array:Array, value:Number):int
{
for(var i:uint = 0; i < array.length; i++)
{
if(array[i] >= value)
return i;
}
// if not found, return -1
return -1;
}
var t:Array = [4,9,1,2,3,5,6];
function something(base:Number, array:Array):int
{
var t:Array = array.slice();
var h:Number = int.MAX_VALUE;
var i:int = -1;
while(t.length > 0)
{
var l:Number = t.pop();
if(l >= base)
{
if(h > l)
{
h = l;
i = t.length;
}
}
}
return i;
}
trace(something(2, t)); // at index [3]
Suppose we have an array of integers. We've written a function to fetch the index of the first specified value in the array, or -1 if the array does not contain the value..
So for example, if the array = { 4, 5, 4, 4, 7 }, then getFirstIndexOf(4) would return 0, getFirstIndexOf(7) would return 4, and getFirstIndexOf(8) would return -1.
Below, I have presented three different ways to write this function. It is a widely accepted coding standard that returns in the middle of functions, and breaks in the middle of loops are poor practice. It seems to me that this might be an acceptable use for them.
public int getFirstIndexOf(int specifiedNumber) {
for (int i = 0; i < array.length; i++) {
if (array[i] == specifiedNumber) {
return i;
}
}
return -1;
}
VS.
public int getFirstIndexOf(int specifiedNumber) {
int result = -1;
for (int i = 0; i < array.length; i++) {
if (array[i] == specifiedNumber) {
result = i;
break;
}
}
return result;
}
VS.
public int getFirstIndexOf(int specifiedNumber) {
int result = -1;
for (int i = 0; i < array.length; i++) {
if (array[i] == specifiedNumber && result == -1) {
result = i;
}
}
return result;
}
What do you think? Which is best? Why? Is there perhaps another way to do this?
I think it's poor practice to run a full loop when you have already found your result...
If you really want to avoid using return from the middle of the loop, I would sugest to use a "sentinel" to stop your loop.
public int getFirstIndexOf(int specifiedNumber, int[] array) {
boolean found = false;
boolean exit = false;
int i = 0;
int arraySize = array.length();
while(!found && !exit) {
if(array[i] == specifiedNumber) {
found = true;
} else {
if(i++ > arraySize) {
exit = true;
}
}
if(found ==true) {
return i;
} else {
return 99999;
}
}
edit: I hate to indent code using spaces in StackOverflow...
That's why do...while & while loop was invented.
As requested:
public int getFirstIndexOf(int specifiedNumber) {
int i = array.Length;
while(--i > -1 && array[i] != specifiedNumber);
return i;
}
Is it possible to conditionally change the direction of a for loop in ActionScript?
Example:
for(if(condition){var x = 0; x<number; x++}else{var x=number; x>0; x--}){
//do something
}
Interesting requirement. One way to keep the for is:
var start, loop_cond, inc;
if(condition)
{
start = 0;
inc = 1;
loop_cond = function(){return x < number};
}
else
{
start = number - 1;
inc = -1;
loop_cond = function(){return x >= 0};
}
for(var x = start; loop_cond(); x += inc)
{
// do something
}
We setup the start value, a function for the termination condition, and either a positive or negative increment. Then, we just call the function and use += to the do the increment or decrement.
ActionScript has the ternary operator, so you could do something like:
for (var x = cond ? 0 : number; cond ? x < number : x > 0; cond ? x++ : x--) {
}
But this is pretty ugly. :-)
You might also need/want to put some parens around pieces of that. I'm not sure about the operator precedence.
You might also consider using a higher order function. Imagine you have:
function forward (count, func) {
for (var x = 0; x < count; x++) {
func(x);
}
}
function backward (count, func) {
for (var x = count - 1; x >= 0; x--) {
func(x);
}
}
Then you could do:
(condition ? forward : backward) (number, function (x) {
// Your loop code goes here
})
You probably want a while loop instead:
var continueLooping, x;
if(condition)
{
x = 0
continueLooping = (x < number);
}
else
{
x = number;
continueLooping = (x > 0);
}
while (continueLooping)
{
// do something
if(condition)
{
x++;
continueLooping = (x < number);
}
else
{
x--;
continueLooping = (x > 0);
}
}
If you really want a for loop, you should use two of them:
function doSomething()
{
//doSomething
}
if(condition)
{
for(var x = 0; x<number; x++)
{
doSomething(x);
}
}
else
{
for(var x=number; x>0; x--})
{
doSomething(x);
}
}
I need to calculate permutations iteratively. The method signature looks like:
int[][] permute(int n)
For n = 3 for example, the return value would be:
[[0,1,2],
[0,2,1],
[1,0,2],
[1,2,0],
[2,0,1],
[2,1,0]]
How would you go about doing this iteratively in the most efficient way possible? I can do this recursively, but I'm interested in seeing lots of alternate ways to doing it iteratively.
see QuickPerm algorithm, it's iterative : http://www.quickperm.org/
Edit:
Rewritten in Ruby for clarity:
def permute_map(n)
results = []
a, p = (0...n).to_a, [0] * n
i, j = 0, 0
i = 1
results << yield(a)
while i < n
if p[i] < i
j = i % 2 * p[i] # If i is odd, then j = p[i], else j = 0
a[j], a[i] = a[i], a[j] # Swap
results << yield(a)
p[i] += 1
i = 1
else
p[i] = 0
i += 1
end
end
return results
end
The algorithm for stepping from one permutation to the next is very similar to elementary school addition - when an overflow occurs, "carry the one".
Here's an implementation I wrote in C:
#include <stdio.h>
//Convenience macro. Its function should be obvious.
#define swap(a,b) do { \
typeof(a) __tmp = (a); \
(a) = (b); \
(b) = __tmp; \
} while(0)
void perm_start(unsigned int n[], unsigned int count) {
unsigned int i;
for (i=0; i<count; i++)
n[i] = i;
}
//Returns 0 on wraparound
int perm_next(unsigned int n[], unsigned int count) {
unsigned int tail, i, j;
if (count <= 1)
return 0;
/* Find all terms at the end that are in reverse order.
Example: 0 3 (5 4 2 1) (i becomes 2) */
for (i=count-1; i>0 && n[i-1] >= n[i]; i--);
tail = i;
if (tail > 0) {
/* Find the last item from the tail set greater than
the last item from the head set, and swap them.
Example: 0 3* (5 4* 2 1)
Becomes: 0 4* (5 3* 2 1) */
for (j=count-1; j>tail && n[j] <= n[tail-1]; j--);
swap(n[tail-1], n[j]);
}
/* Reverse the tail set's order */
for (i=tail, j=count-1; i<j; i++, j--)
swap(n[i], n[j]);
/* If the entire list was in reverse order, tail will be zero. */
return (tail != 0);
}
int main(void)
{
#define N 3
unsigned int perm[N];
perm_start(perm, N);
do {
int i;
for (i = 0; i < N; i++)
printf("%d ", perm[i]);
printf("\n");
} while (perm_next(perm, N));
return 0;
}
Is using 1.9's Array#permutation an option?
>> a = [0,1,2].permutation(3).to_a
=> [[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 0, 1], [2, 1, 0]]
Below is my generics version of the next permutation algorithm in C# closely resembling the STL's next_permutation function (but it doesn't reverse the collection if it is the max possible permutation already, like the C++ version does)
In theory it should work with any IList<> of IComparables.
static bool NextPermutation<T>(IList<T> a) where T: IComparable
{
if (a.Count < 2) return false;
var k = a.Count-2;
while (k >= 0 && a[k].CompareTo( a[k+1]) >=0) k--;
if(k<0)return false;
var l = a.Count - 1;
while (l > k && a[l].CompareTo(a[k]) <= 0) l--;
var tmp = a[k];
a[k] = a[l];
a[l] = tmp;
var i = k + 1;
var j = a.Count - 1;
while(i<j)
{
tmp = a[i];
a[i] = a[j];
a[j] = tmp;
i++;
j--;
}
return true;
}
And the demo/test code:
var src = "1234".ToCharArray();
do
{
Console.WriteLine(src);
}
while (NextPermutation(src));
I also came across the QuickPerm algorithm referenced in another answer. I wanted to share this answer in addition, because I saw some immediate changes one can make to write it shorter. For example, if the index array "p" is initialized slightly differently, it saves having to return the first permutation before the loop. Also, all those while-loops and if's took up a lot more room.
void permute(char* s, size_t l) {
int* p = new int[l];
for (int i = 0; i < l; i++) p[i] = i;
for (size_t i = 0; i < l; printf("%s\n", s)) {
std::swap(s[i], s[i % 2 * --p[i]]);
for (i = 1; p[i] == 0; i++) p[i] = i;
}
}
I found Joey Adams' version to be the most readable, but I couldn't port it directly to C# because of how C# handles the scoping of for-loop variables. Hence, this is a slightly tweaked version of his code:
/// <summary>
/// Performs an in-place permutation of <paramref name="values"/>, and returns if there
/// are any more permutations remaining.
/// </summary>
private static bool NextPermutation(int[] values)
{
if (values.Length == 0)
throw new ArgumentException("Cannot permutate an empty collection.");
//Find all terms at the end that are in reverse order.
// Example: 0 3 (5 4 2 1) (i becomes 2)
int tail = values.Length - 1;
while(tail > 0 && values[tail - 1] >= values[tail])
tail--;
if (tail > 0)
{
//Find the last item from the tail set greater than the last item from the head
//set, and swap them.
// Example: 0 3* (5 4* 2 1)
// Becomes: 0 4* (5 3* 2 1)
int index = values.Length - 1;
while (index > tail && values[index] <= values[tail - 1])
index--;
Swap(ref values[tail - 1], ref values[index]);
}
//Reverse the tail set's order.
int limit = (values.Length - tail) / 2;
for (int index = 0; index < limit; index++)
Swap(ref values[tail + index], ref values[values.Length - 1 - index]);
//If the entire list was in reverse order, tail will be zero.
return (tail != 0);
}
private static void Swap<T>(ref T left, ref T right)
{
T temp = left;
left = right;
right = temp;
}
Here's an implementation in C#, as an extension method:
public static IEnumerable<List<T>> Permute<T>(this IList<T> items)
{
var indexes = Enumerable.Range(0, items.Count).ToArray();
yield return indexes.Select(idx => items[idx]).ToList();
var weights = new int[items.Count];
var idxUpper = 1;
while (idxUpper < items.Count)
{
if (weights[idxUpper] < idxUpper)
{
var idxLower = idxUpper % 2 * weights[idxUpper];
var tmp = indexes[idxLower];
indexes[idxLower] = indexes[idxUpper];
indexes[idxUpper] = tmp;
yield return indexes.Select(idx => items[idx]).ToList();
weights[idxUpper]++;
idxUpper = 1;
}
else
{
weights[idxUpper] = 0;
idxUpper++;
}
}
}
And a unit test:
[TestMethod]
public void Permute()
{
var ints = new[] { 1, 2, 3 };
var orderings = ints.Permute().ToList();
Assert.AreEqual(6, orderings.Count);
AssertUtil.SequencesAreEqual(new[] { 1, 2, 3 }, orderings[0]);
AssertUtil.SequencesAreEqual(new[] { 2, 1, 3 }, orderings[1]);
AssertUtil.SequencesAreEqual(new[] { 3, 1, 2 }, orderings[2]);
AssertUtil.SequencesAreEqual(new[] { 1, 3, 2 }, orderings[3]);
AssertUtil.SequencesAreEqual(new[] { 2, 3, 1 }, orderings[4]);
AssertUtil.SequencesAreEqual(new[] { 3, 2, 1 }, orderings[5]);
}
The method AssertUtil.SequencesAreEqual is a custom test helper which can be recreated easily enough.
How about a recursive algorithm you can call iteratively? If you'd actually need that stuff as a list like that (you should clearly inline that rather than allocate a bunch of pointless memory). You could simply calculate the permutation on the fly, by its index.
Much like the permutation is carry-the-one addition re-reversing the tail (rather than reverting to 0), indexing the specific permutation value is finding the digits of a number in base n then n-1 then n-2... through each iteration.
public static <T> boolean permutation(List<T> values, int index) {
return permutation(values, values.size() - 1, index);
}
private static <T> boolean permutation(List<T> values, int n, int index) {
if ((index == 0) || (n == 0)) return (index == 0);
Collections.swap(values, n, n-(index % n));
return permutation(values,n-1,index/n);
}
The boolean returns whether your index value was out of bounds. Namely that it ran out of n values but still had remaining index left over.
And it can't get all the permutations for more than 12 objects.
12! < Integer.MAX_VALUE < 13!
-- But, it's so very very pretty. And if you do a lot of things wrong might be useful.
I have implemented the algorithm in Javascript.
var all = ["a", "b", "c"];
console.log(permute(all));
function permute(a){
var i=1,j, temp = "";
var p = [];
var n = a.length;
var output = [];
output.push(a.slice());
for(var b=0; b <= n; b++){
p[b] = b;
}
while (i < n){
p[i]--;
if(i%2 == 1){
j = p[i];
}
else{
j = 0;
}
temp = a[j];
a[j] = a[i];
a[i] = temp;
i=1;
while (p[i] === 0){
p[i] = i;
i++;
}
output.push(a.slice());
}
return output;
}
I've used the algorithms from here. The page contains a lot of useful information.
Edit: Sorry, those were recursive. uray posted the link to the iterative algorithm in his answer.
I've created a PHP example. Unless you really need to return all of the results, I would only create an iterative class like the following:
<?php
class Permutator implements Iterator
{
private $a, $n, $p, $i, $j, $k;
private $stop;
public function __construct(array $a)
{
$this->a = array_values($a);
$this->n = count($this->a);
}
public function current()
{
return $this->a;
}
public function next()
{
++$this->k;
while ($this->i < $this->n)
{
if ($this->p[$this->i] < $this->i)
{
$this->j = ($this->i % 2) * $this->p[$this->i];
$tmp = $this->a[$this->j];
$this->a[$this->j] = $this->a[$this->i];
$this->a[$this->i] = $tmp;
$this->p[$this->i]++;
$this->i = 1;
return;
}
$this->p[$this->i++] = 0;
}
$this->stop = true;
}
public function key()
{
return $this->k;
}
public function valid()
{
return !$this->stop;
}
public function rewind()
{
if ($this->n) $this->p = array_fill(0, $this->n, 0);
$this->stop = $this->n == 0;
$this->i = 1;
$this->j = 0;
$this->k = 0;
}
}
foreach (new Permutator(array(1,2,3,4,5)) as $permutation)
{
var_dump($permutation);
}
?>
Note that it treats every PHP array as an indexed array.