I'm supposed to use recursion to output the total number of unique north-east paths ne(x, y) to get from point A to point B, where B is x rows north and y columns east of A. In addition, I am required to print the possible unique NE paths.
I know how to use recursion to get the total number of unique paths. However, I am stuck with using recursion to print all the NE paths correctly.
This is the given output of some test cases:
image of output
Anyway, here's a screenshot of my faulty recursive code.
Please do give me advice where I went wrong. I have been burning a lot of time on this, but still I can't reach a solution.
I think you should print if( rows == 0 && cols == 0 ), because it's the case when you've reached point B.
Why are you using path+="N" in the first ne call in return? this will add "N" to original path and then you'll get path+"N"+"E" in the second call.
Try following:
public static int ne( int rows, int cols, String path )
{
if( rows == 0 && cols == 0 )
{
System.out.println(path);
return 1;
}
int npats = 0, wpaths = 0;
if( rows != 0 )
npaths = ne( rows-1, cols, path+"N" );
if( cols != 0 )
wpaths = ne( rows, cols-1, path+"E" );
return npaths + wpaths;
}
Related
I am suppose to code the snake game in java with processing for IT classes and since I had no idea how to do it I searched for a YouTube tutorial. Now I did find one but he used the keys 'w','s','d','a' to move the snake around - I on the other hand want to use the arrow keys. Could someone explain to me how I transform this code:
if (keyPressed == true) {
int newdir = key=='s' ? 0 : (key=='w' ? 1 : (key=='d' ? 2 : (key=='a' ? 3 : -1)));
}
if(newdir != -1 && (x.size() <= 1 || !(x.get(1) ==x.get(0) + dx[newdir] && y.get (1) == y.get(0) + dy[newdir]))) dir = newdir;
}
into something like this:
void keyPressed () {
if (key == CODED) {
if (keyCode == UP) {}
else if (keyCode == RIGHT) {}
else if (keyCode == DOWN) {}
else if (keyCode == LEFT) {}
}
This is my entire coding so far:
ArrayList<Integer> x = new ArrayList<Integer> (), y = new ArrayList<Integer> ();
int w = 900, h = 900, bs = 20, dir = 1; // w = width ; h = height ; bs = blocksize ; dir = 2 --> so that the snake goes up when it starts
int[] dx = {0,0,1,-1} , dy = {1,-1,0,0};// down, up, right, left
void setup () {
size (900,900); // the 'playing field' is going to be 900x900px big
// the snake starts off on x = 5 and y = 30
x.add(5);
y.add(30);
}
void draw() {
//white background
background (255);
//
// grid
// vertical lines ; the lines are only drawn if they are smaller than 'w'
// the operator ++ increases the value 'l = 0' by 1
//
for(int l = 0 ; l < w; l++) line (l*bs, 0, l*bs, height);
//
// horizontal lines ; the lines are only drawn if they are smaller than 'h'
// the operator ++ increases the value 'l = 0' by 1
//
for(int l = 0 ; l < h; l++) line (0, l*bs, width, l*bs);
//
// snake
for (int l = 0 ; l < x.size() ; l++) {
fill (0,255,0); // the snake is going to be green
rect (x.get(l)*bs, y.get(l)*bs, bs, bs);
}
if(frameCount%5==0) { // will check it every 1/12 of a second -- will check it every 5 frames at a frameRate = 60
// adding points
x.add (0,x.get(0) + dx[dir]); // will add a new point x in the chosen direction
y.add (0,y.get(0) + dy[dir]); // will add a new point y in the chosen direction
// removing points
x.remove(x.size()-1); // will remove the previous point x
y.remove(y.size()-1); // will remove the previous point y
}
}
It's hard to answer general "how do I do this" type questions. Stack Overflow is designed for more specific "I tried X, expected Y, but got Z instead" type questions. That being said, I'll try to answer in a general sense:
You're going to have a very difficult time trying to take random code you find on the internet and trying to make it work in your sketch. That's not a very good way to proceed.
Instead, you need to take a step back and really think about what you want to happen. Instead of taking on your entire end goal at one time, try breaking your problem down into smaller steps and taking on those steps one at a time.
Step 1: Can you store the state of your game in variables? You might store things like the direction the snake is traveling the location of the snake, etc.
Step 2: Can you write code that just prints something to the console when you press the arrow keys? You might do this in a separate example sketch instead of trying to add it directly to your full sketch.
Step 3: Can you combine those two steps and change the state of your sketch when an arrow key is pressed? Maybe you change the direction the snake is traveling.
The point is that you need to try something instead of trying to copy-paste random code without really understanding it. Break your problem down into small steps, and then post an MCVE of that specific step if you get stuck. Good luck.
You should take a look into Java API KeyEvent VK_LEFT.
And as pczeus already told you, you need to implement a capturing of the keystrokes! This can be checked here (Link from this SO answer).
Imagine you have 3 buckets, but each of them has a hole in it. I'm trying to fill a bath tub. The bath tub has a minimum level of water it needs and a maximum level of water it can contain. By the time you reach the tub with the bucket it is not clear how much water will be in the bucket, but you have a range of possible values.
Is it possible to adequately fill the tub with water?
Pretty much you have 3 ranges (min,max), is there some sum of them that will fall within a 4th range?
For example:
Bucket 1 : 5-10L
Bucket 2 : 15-25L
Bucket 3 : 10-50L
Bathtub 100-150L
Is there some guaranteed combination of 1 2 and 3 that will fill the bathtub within the requisite range? Multiples of each bucket can be used.
EDIT: Now imagine there are 50 different buckets?
If the capacity of the tub is not very large ( not greater than 10^6 for an example), we can solve it using dynamic programming.
Approach:
Initialization: memo[X][Y] is an array to memorize the result. X = number of buckets, Y = maximum capacity of the tub. Initialize memo[][] with -1.
Code:
bool dp(int bucketNum, int curVolume){
if(curVolume > maxCap)return false; // pruning extra branches
if(curVolume>=minCap && curVolume<=maxCap){ // base case on success
return true;
}
int &ret = memo[bucketNum][curVolume];
if(ret != -1){ // this state has been visited earlier
return false;
}
ret = false;
for(int i = minC[bucketNum]; i < = maxC[bucketNum]; i++){
int newVolume = curVolume + i;
for(int j = bucketNum; j <= 3; j++){
ret|=dp(j,newVolume);
if(ret == true)return ret;
}
}
return ret;
}
Warning: Code not tested
Here's a naïve recursive solution in python that works just fine (although it doesn't find an optimal solution):
def match_helper(lower, upper, units, least_difference, fail = dict()):
if upper < lower + least_difference:
return None
if fail.get((lower,upper)):
return None
exact_match = [ u for u in units if u['lower'] >= lower and u['upper'] <= upper ]
if exact_match:
return [ exact_match[0] ]
for unit in units:
if unit['upper'] > upper:
continue
recursive_match = match_helper(lower - unit['lower'], upper - unit['upper'], units, least_difference)
if recursive_match:
return [unit] + recursive_match
else:
fail[(lower,upper)] = 1
return None
def match(lower, upper):
units = [
{ 'name': 'Bucket 1', 'lower': 5, 'upper': 10 },
{ 'name': 'Bucket 2', 'lower': 15, 'upper': 25 },
{ 'name': 'Bucket 3', 'lower': 10, 'upper': 50 }
]
least_difference = min([ u['upper'] - u['lower'] for u in units ])
return match_helper(
lower = lower,
upper = upper,
units = sorted(units, key = lambda u: u['upper']),
least_difference = min([ u['upper'] - u['lower'] for u in units ]),
)
result = match(100, 175)
if result:
lower = sum([ u['lower'] for u in result ])
upper = sum([ u['upper'] for u in result ])
names = [ u['name'] for u in result ]
print lower, "-", upper
print names
else:
print "No solution"
It prints "No solution" for 100-150, but for 100-175 it comes up with a solution of 5x bucket 1, 5x bucket 2.
Assuming you are saying that the "range" for each bucket is the amount of water that it may have when it reaches the tub, and all you care about is if they could possibly fill the tub...
Just take the "max" of each bucket and sum them. If that is in the range of what you consider the tub to be "filled" then it can.
Updated:
Given that buckets can be used multiple times, this seems to me like we're looking for solutions to a pair of equations.
Given buckets x, y and z we want to find a, b and c:
a*x.min + b*y.min + c*z.min >= bathtub.min
and
a*x.max + b*y.max + c*z.max <= bathtub.max
Re: http://en.wikipedia.org/wiki/Diophantine_equation
If bathtub.min and bathtub.max are both multiples of the greatest common divisor of a,b and c, then there are infinitely many solutions (i.e. we can fill the tub), otherwise there are no solutions (i.e. we can never fill the tub).
This can be solved with multiple applications of the change making problem.
Each Bucket.Min value is a currency denomination, and Bathtub.Min is the target value.
When you find a solution via a change-making algorithm, then apply one more constraint:
sum(each Bucket.Max in your solution) <= Bathtub.max
If this constraint is not met, throw out this solution and look for another. This will probably require a change to a standard change-making algorithm that allows you to try other solutions when one is found to not be suitable.
Initially, your target range is Bathtub.Range.
Each time you add an instance of a bucket to the solution, you reduce the target range for the remaining buckets.
For example, using your example buckets and tub:
Target Range = 100..150
Let's say we want to add a Bucket1 to the candidate solution. That then gives us
Target Range = 95..140
because if the rest of the buckets in the solution total < 95, then this Bucket1 might not be sufficient to fill the tub to 100, and if the rest of the buckets in the solution total > 140, then this Bucket1 might fill the tub over 150.
So, this gives you a quick way to check if a candidate solution is valid:
TargetRange = Bathtub.Range
foreach Bucket in CandidateSolution
TargetRange.Min -= Bucket.Min
TargetRange.Max -= Bucket.Max
if TargetRange.Min == 0 AND TargetRange.Max >= 0 then solution found
if TargetRange.Min < 0 or TargetRange.Max < 0 then solution is invalid
This still leaves the question - How do you come up with the set of candidate solutions?
Brute force would try all possible combinations of buckets.
Here is my solution for finding the optimal solution (least number of buckets). It compares the ratio of the maximums to the ratio of the minimums, to figure out the optimal number of buckets to fill the tub.
private static void BucketProblem()
{
Range bathTub = new Range(100, 175);
List<Range> buckets = new List<Range> {new Range(5, 10), new Range(15, 25), new Range(10, 50)};
Dictionary<Range, int> result;
bool canBeFilled = SolveBuckets(bathTub, buckets, out result);
}
private static bool BucketHelper(Range tub, List<Range> buckets, Dictionary<Range, int> results)
{
Range bucket;
int startBucket = -1;
int fills = -1;
for (int i = buckets.Count - 1; i >=0 ; i--)
{
bucket = buckets[i];
double maxRatio = (double)tub.Maximum / bucket.Maximum;
double minRatio = (double)tub.Minimum / bucket.Minimum;
if (maxRatio >= minRatio)
{
startBucket = i;
if (maxRatio - minRatio > 1)
fills = (int) minRatio + 1;
else
fills = (int) maxRatio;
break;
}
}
if (startBucket < 0)
return false;
bucket = buckets[startBucket];
tub.Maximum -= bucket.Maximum * fills;
tub.Minimum -= bucket.Minimum * fills;
results.Add(bucket, fills);
return tub.Maximum == 0 || tub.Minimum <= 0 || startBucket == 0 || BucketHelper(tub, buckets.GetRange(0, startBucket), results);
}
public static bool SolveBuckets(Range tub, List<Range> buckets, out Dictionary<Range, int> results)
{
results = new Dictionary<Range, int>();
buckets = buckets.OrderBy(b => b.Minimum).ToList();
return BucketHelper(new Range(tub.Minimum, tub.Maximum), buckets, results);
}
I have these two find algorithm which look the same to me. Can anyone help me out why they are actually different?
Find ( x ) :
if x.parent = x then
return x
else
return Find ( x.parent )
vs
Find ( x ) :
if x.parent = x then
return x
else
x.parent <- Find(x.parent)
return x.parent
I interpret the first one as
int i = 0;
return i++;
while the second one as
int i = 0;
int tmp = i++;
return tmp
which are exactly the same to me.
This looks like Disjoint-set data structure.
Now to the question:
For the sake of clarity first version is FindA, second is FindB.
Suppose you have structure:
0
|
1
|
2
|
...
n
First call to FindA(n) will return 0 in O(n), second call will return 0 in O(n) and so on.
If you call FindB(n) it will return 0 in O(n), but will also modify structure:
0
/ /|\
1 2...n
Now second call to FindB(n) will return 0 in O(1). More over FindB(k) will return 0 in O(1).
The second one will change the value of x.parent as a side effect to the result of find
This is an interview question: "How to build a distributed algorithm to compute the balance of the parentheses ?"
Usually he balance algorithm scans a string form left to right and uses a stack to make sure that the number of open parentheses always >= the number of close parentheses and finally the number of open parentheses == the number of close parentheses.
How would you make it distributed ?
You can break the string into chunks and process each separately, assuming you can read and send to the other machines in parallel. You need two numbers for each string.
The minimum nesting depth achieved relative to the start of the string.
The total gain or loss in nesting depth across the whole string.
With these values, you can compute the values for the concatenation of many chunks as follows:
minNest = 0
totGain = 0
for p in chunkResults
minNest = min(minNest, totGain + p.minNest)
totGain += p.totGain
return new ChunkResult(minNest, totGain)
The parentheses are matched if the final values of totGain and minNest are zero.
I would apply the map-reduce algorithm in which the map function would compute a part of the string return either an empty string if parentheses are balanced or a string with the last parenthesis remaining.
Then the reduce function would concatenate the result of two returned strings by map function and compute it again returning the same result than map. At the end of all computations, you'd either obtain an empty string or a string containing the un-balanced parenthesis.
I'll try to have a more detailed explain on #jonderry's answer. Code first, in Scala
def parBalance(chars: Array[Char], chunkSize: Int): Boolean = {
require(chunkSize > 0, "chunkSize must be greater than 0")
def traverse(from: Int, until: Int): (Int, Int) = {
var count = 0
var stack = 0
var nest = 0
for (n <- from until until) {
val cur = chars(c)
if (cur == '(') {
count += 1
stack += 1
}
else if (cur == ')') {
count -= 1
if (stack > 0) stack -= 1
else nest -= 1
}
}
(nest, count)
}
def reduce(from: Int, until: Int): (Int, Int) = {
val m = (until + from) / 2
if (until - from <= chunkSize) {
traverse(from, until)
} else {
parallel(reduce(from, m), reduce(m, until)) match {
case ((minNestL, totGainL), (minNestR, totGainR)) => {
((minNestL min (minNestR + totGainL)), (totGainL + totGainR))
}
}
}
}
reduce(0, chars.length) == (0,0)
}
Given a string, if we remove balanced parentheses, what's left will be in a form )))(((, give n for number of ) and m for number of (, then m >= 0, n <= 0(for easier calculation). Here n is minNest and m+n is totGain. To make a true balanced string, we need m+n == 0 && n == 0.
In a parallel operation, how to we derive those for node from it's left and right? For totGain we just needs to add them up. When calculating n for node, it can just be n(left) if n(right) not contribute or n(right) + left.totGain whichever is smaller.
I have an array (of 9 elements, say) which I must treat as a (3 by 3) square.
For the sake of simplifying the question, this is a one-based array (ie, indexing starts at 1 instead of 0).
My goal is to determine valid adjacent squares relative to a starting point.
In other words, how it's stored in memory: 1 2 3 4 5 6 7 8 9
How I'm treating it:
7 8 9
4 5 6
1 2 3
I already know how to move up and down and test for going out of bounds (1 >= current_index <= 9)
edit: I know the above test is overly general but it's simple and works.
//row_size = 3, row_step is -1, 0 or 1 depending on if we're going left,
//staying put or going right respectively.
current_index += (row_size * row_step);
How do I test for an out of bounds condition when going left or right? Conceptually I know it involves determining if 3 (for example) is on the same row as 4 (or if 10 is even within the same square as 9, as an alternate example, given that multiple squares are in the same array back to back), but I can't figure out how to determine that. I imagine there's a modulo in there somewhere, but where?
Thanks very much,
Geoff
Addendum:
Here's the resulting code, altered for use with a zero-based array (I cleaned up the offset code present in the project) which walks adjacent squares.
bool IsSameSquare(int index0, int index1, int square_size) {
//Assert for square_size != 0 here
return (!((index0 < 0) || (index1 < 0))
&& ((index0 < square_size) && (index1 < square_size)))
&& (index0 / square_size == index1 / square_size);
}
bool IsSameRow(int index0, int index1, int row_size) {
//Assert for row_size != 0 here
return IsSameSquare(index0, index1, row_size * row_size)
&& (index0 / row_size == index1 / row_size);
}
bool IsSameColumn(int index0, int index1, int row_size) {
//Assert for row_size != 0 here
return IsSameSquare(index0, index1, row_size * row_size)
&& (index0 % row_size == index1 % row_size);
}
//for all possible adjacent positions
for (int row_step = -1; row_step < 2; ++row_step) {
//move up, down or stay put.
int row_adjusted_position = original_position + (row_size * row_step);
if (!IsSameSquare(original_position, row_adjusted_position, square_size)) {
continue;
}
for (int column_step = -1; column_step < 2; ++column_step) {
if ((row_step == 0) & (column_step == 0)) { continue; }
//hold on to the position that has had its' row position adjusted.
int new_position = row_adjusted_position;
if (column_step != 0) {
//move left or right
int column_adjusted_position = new_position + column_step;
//if we've gone out of bounds again for the column.
if (IsSameRow(column_adjusted_position, new_position, row_size)) {
new_position = column_adjusted_position;
} else {
continue;
}
} //if (column_step != 0)
//if we get here we know it's safe, do something with new_position
//...
} //for each column_step
} //for each row_step
This is easier if you used 0-based indexing. These rules work if you subtract 1 from all your indexes:
Two indexes are in the same square if (a/9) == (b/9) and a >= 0 and b >= 0.
Two indexes are in the same row if they are in the same square and (a/3) == (b/3).
Two indexes are in the same column if they are in the same square and (a%3) == (b%3).
There are several way to do this, I'm choosing a weird one just for fun. Use modulus.
Ase your rows are size 3 just use modulus of 3 and two simple rules.
If currPos mod 3 = 0 and (currPos+move) mod 3 = 1 then invalid
If currPos mod 3 = 1 and (currPos+move) mod 3 = 0 then invalid
this check for you jumping two a new row, you could also do one rule like this
if (currPos mod 3)-((currPos+move) mod 3)> 1 then invalid
Cheers
You should be using a multidimensional array for this.
If your array class doesn't support multidimensional stuff, you should write up a quick wrapper that does.