I followed a tut to help me build a tic tac toe game to keep learning Swift. In doing so I also wanted to add winning slashes for whatever combination that wins.
In my code I have
let winningCombinations = [0, 1, 2], [3, 4, 5], [6, 7, 8], [0, 3, 6], [1, 4, 7], [2, 5, 8], [0, 4, 8], [2, 4, 6]]
these are the buttons tags.
I have created labels that are just color bars with no text to be the slashes.
I connected them one as
#IBOutlet weak var lineFirstRow: UILabel!
then hid the label on the storyboard.
I have been trying things like
let lineRowOne = [0, 1, 2]
if lineRowOne = true {
lineFirstRow.isHidden = true
} else {
lineFirstRow.isHidden = false
}
I know this is not correct.
Not entirely sure what you're trying to achieve, but I suspect your solution will be on something like this:
let winningCombinations = [[0, 1, 2], [3, 4, 5], [6, 7, 8], [0, 3, 6], [1, 4, 7], [2, 5, 8], [0, 4, 8], [2, 4, 6]]
let lineOne = [0, 1, 2]
if winningCombinations.contains(where: { $0 == lineOne })
{
// Do Stuff
}
Related
I want to compare each element against all others like following. The number of variables like a, b, c is dynamic. However, each variable's array size is uniform.
let a = [1, 2, 3]
let b = [3, 4, 5]
let c = [4, 5, 6]
for i in a {
for j in b {
for k in c {
/// comparison
}
}
}
Instead looping from start to finish at once, what would be a way to make each comparison on call? For example:
compare(iteration: 0)
/// compares a[0], b[0], c[0]
compare(iteration: 1)
/// compares a[0], b[0], c[1]
/// all the way to
/// compares a[2], b[2], c[2]
Or it could even be like following:
next()
/// compares a[0], b[0], c[0]
next()
/// compares a[0], b[0], c[1]
almost like an iterator stepping through each cycle dictated by my invocation.
Let the number of arrays be n. And let the number of elements in each array, which is guaranteed the same for all of them, be k.
Then create an array consisting of the integers 0 through k-1, repeated n times. For example, in your case, n is 3, and k is 3, so generate the array
[0, 1, 2, 0, 1, 2, 0, 1, 2]
Now obtain all combinations of n elements of that array. You can do this using the algorithm at https://github.com/apple/swift-algorithms/blob/main/Guides/Combinations.md. Unique the result (by, for example, coercing to a Set and then back to an Array). This will give you a result equivalent, in some order or other, to
[[0, 1, 2], [0, 1, 0], [0, 1, 1], [0, 2, 0], [0, 2, 1], [0, 2, 2], [0, 0, 1], [0, 0, 2], [0, 0, 0], [1, 2, 0], [1, 2, 1], [1, 2, 2], [1, 0, 1], [1, 0, 2], [1, 0, 0], [1, 1, 2], [1, 1, 0], [1, 1, 1], [2, 0, 1], [2, 0, 2], [2, 0, 0], [2, 1, 2], [2, 1, 0], [2, 1, 1], [2, 2, 0], [2, 2, 1], [2, 2, 2]]
You can readily see that those are all 27 possible combinations of the numbers 0, 1, and 2. But that is exactly what you were doing with your for loops! So now, use those subarrays as indexes into each of your original arrays respectively.
So for instance, using my result and your original example, the first subarray [0, 1, 2] yields [1, 4, 6] — the first value from the first array, the second value from the second array, and the third value from the third array. And so on.
In this way you will have generated all possible n-tuples by choosing one value from each of your original arrays, which is the desired result; and we are in no way bound to fixed values of n and k, which was what you wanted to achieve. You will then be able to "compare" the elements of each n-tuple, whatever that may mean to you (you did not say in your question what it means).
In the case of your original values, we will get these n-tuples (expressed as arrays):
[1, 4, 6]
[1, 4, 4]
[1, 4, 5]
[1, 5, 4]
[1, 5, 5]
[1, 5, 6]
[1, 3, 5]
[1, 3, 6]
[1, 3, 4]
[2, 5, 4]
[2, 5, 5]
[2, 5, 6]
[2, 3, 5]
[2, 3, 6]
[2, 3, 4]
[2, 4, 6]
[2, 4, 4]
[2, 4, 5]
[3, 3, 5]
[3, 3, 6]
[3, 3, 4]
[3, 4, 6]
[3, 4, 4]
[3, 4, 5]
[3, 5, 4]
[3, 5, 5]
[3, 5, 6]
Those are precisely the triples of values you are after.
Actual code:
// your original conditions
let a = [1, 2, 3]
let b = [3, 4, 5]
let c = [4, 5, 6]
let originals = [a, b, c]
// The actual solution starts here. Note that I never use any hard
// coded numbers.
let n = originals.count
let k = originals[0].count
var indices = [Int]()
for _ in 0..<n {
for i in 0..<k {
indices.append(i)
}
}
let combos = Array(indices.combinations(ofCount: n))
var combosUniq = [[Int]]()
var combosSet = Set<[Int]>()
for combo in combos {
let success = combosSet.insert(combo)
if success.inserted {
combosUniq.append(combo)
}
}
// And here's how to generate your actual desired values.
for combo in combosUniq {
var tuple = [Int]()
for (outerIndex, innerIndex) in combo.enumerated() {
tuple.append(originals[outerIndex][innerIndex])
}
print(tuple) // in real life, do something useful here
}
}
I have a 1d list of integers like:
List<int> x = [1, 4, 2, 8, 9, 3, 6, 5, 7];
I want to convert this list to a 2d list like this:
List<List<int>> y = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
List<int> x = [1, 4, 2, 8, 9, 3, 6, 5, 7];
// make ascending list
List<int> x = (b, a) => a.compareTo(b);
// install package matrix 2d dart package link : //https://pub.dev/packages/matrix2d
list = X.reshape(3,3);
i have written the code below,it works for shortest distance but not for shortest path,
import math
def floyd(dist_mat):
n=len(dist_mat)
p=[[0]*n]*n
for k in range(n):
for i in range(n):
for j in range(n):
if dist_mat[i][j]>dist_mat[i][k]+dist_mat[k][j]:
dist_mat[i][j] = dist_mat[i][k] + dist_mat[k][j]
p[i][j] = k+1
return p
if __name__ == '__main__':
print(floyd([[0,5,9999,9999],
[50,0,15,5],
[30,9999,0,15],
[15,9999,5,0]]))
result of this code is: [[4, 1, 4, 2], [4, 1, 4, 2], [4, 1, 4, 2], [4, 1, 4, 2]]
true result is: [[0, 0, 4, 2], [4, 0, 4, 0], [0, 1, 0, 0], [0, 1, 0, 0]],
I will be happy to receive your ideas about why it works wrong soon
I want to access a specific row and column restriction of a 2d numpy array.
> x
array([[1, 2, 0],
[3, 4, 0],
[0, 0, 1]])
If I do what seems natural, I just get the diagonal elements of the restricted array.
> x[[0,1], [0,1]]
array([1, 4])
Instead I can do this to read what I want -
> x[[0,1],:][:,[0,1]]
array([[1, 2],
[3, 4]])
..but it doesn't let me write/assign the values.
> x[[0,1],:][:,[0,1]] = np.array([[1,0],[0,1]])
> x
array([[1, 2, 0],
[3, 4, 0],
[0, 0, 1]])
How can I write to a matrix here?
Use np.ix_ to map that grid of elements and then assign -
x[np.ix_([0,1], [0,1])] = np.array([[1,0],[0,1]])
This works too:
x[:2, :2] = np.array([[1, 0], [0, 1]])
I'm taking Princeton's algorithms-divide-conquer course - 3rd week, and trying to implement the quicksort.
Here's my current implementation with some tests ready to run:
import unittest
def quicksort(x):
if len(x) <= 1:
return x
pivot = x[0]
xLeft, xRight = partition(x)
print(xLeft, xRight)
quicksort(xLeft)
quicksort(xRight)
return x
def partition(x):
j = 0
print('partition', x)
for i in range(0, len(x)):
if x[i] < x[0]:
n = x[j + 1]
x[j + 1] = x[i]
x[i] = n
j += 1
p = x[0]
x[0] = x[j]
x[j] = p
return x[:j + 1], x[j + 1:]
class Test(unittest.TestCase):
def test_partition_pivot_first(self):
arrays = [
[3, 1, 2, 5],
[3, 8, 2, 5, 1, 4, 7, 6],
[10, 100, 3, 4, 2, 101]
]
expected = [
[[2, 1, 3], [5]],
[[1, 2, 3], [5, 8, 4, 7, 6]],
[[2, 3, 4, 10], [100, 101]]
]
for i in range(0, len(arrays)):
xLeft, xRight = partition(arrays[i])
self.assertEqual(xLeft, expected[i][0])
self.assertEqual(xRight, expected[i][1])
def test_quicksort(self):
arrays = [
[1, 2, 3, 4, 5, 6],
[3, 5, 6, 10, 2, 4]
]
expected = [
[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 6, 10]
]
for i in range(0, len(arrays)):
arr = arrays[i]
quicksort(arr)
self.assertEqual(arr, expected[i])
if __name__ == "__main__":
unittest.main()
so for array = [3, 5, 6, 10, 2, 4] I get [2, 3, 6, 10, 5, 4] as a result... I can't figure what's wrong with my code. It partitions just fine, but the results are off...
Can anyone chip in? :) Thank you!
it's actually so minor problem that you'd be laughing
the problem resides with quicksort function
the correct one is:
def quicksort(x):
if len(x) <= 1:
return x
pivot = x[0]
xLeft, xRight = partition(x)
print(xLeft, xRight)
quicksort(xLeft)
quicksort(xRight)
x=xLeft+xRight #this one!
return x
what happens is python created a new object out of these xleft and xright they were never an in place-sort
so this is one solution(which is not in place)
the other one is to pass the list,the start_index,end_index
and do it in place
well done fella!
edit:
and actually if you'd print xleft and xright you'd see it performed perfectly:)