I am attempting to write a function in Scala that takes a valid Latin Square as input and then returns a Boolean value indicating whether the input Latin Square is a valid instance of an operation-table for the Damm Algorithm, or not. The Damm Algorithm appears to be of the "Check digit" class of error detection mechanisms.
The desired function is an implementation of the implication described in the Design section of the Wikipedia article for the Damm Algorithm. The implication is captured by the assertion of:
a weak totally anti-symmetric quasigroup with the property x ∗ x = 0
I'm insufficient in both the math skills to properly read and interpret Damm's paper, and then having enough German (the language within which the paper was written) reading skills to be able to confidently interpret how I would encode a correct validation function.
Given the following function definition:
def validate(latinSquare: List[List[Int]]): Boolean =
???
With what would I replace the ????
Within the paper, a single instance of a valid 10x10 Latin Square is provided. It is reproduced within the Wikipedia article and looks like this:
|0 1 2 3 4 5 6 7 8 9
-+-------------------
0|0 3 1 7 5 9 8 6 4 2
1|7 0 9 2 1 5 4 8 6 3
2|4 2 0 6 8 7 1 3 5 9
3|1 7 5 0 9 8 3 4 2 6
4|6 1 2 3 0 4 5 9 7 8
5|3 6 7 4 2 0 9 5 8 1
6|5 8 6 9 7 2 0 1 3 4
7|8 9 4 5 3 6 2 0 1 7
8|9 4 3 8 6 1 7 2 0 5
9|2 5 8 1 4 3 6 7 9 0
I can see there are others who have sought an answer to this. However, no one has yet provided a requested code-based valid Latin Square solution.
I have coded up a generic Latin Square generator, but need the above validation function which will serve as the filter to eliminate each candidate Latin Square that does not meet the necessary condition(s) of the above implication.
In functional programming terms, the checksum algorithm is foldLeft with a carefully chosen binary operation. The requirements for this binary operation, in English:
In every two-digit input, if we change one of the digits, then the checksum changes (Latin square…);
In every three-digit input, if the latter two digits are distinct and we swap them, then the checksum changes (…with weak total antisymmetry);
The Latin square has zeros on the diagonal.
In Python 3:
def validate(latinSquare):
Q = range(len(latinSquare))
return all(
x == y
for c in Q
for x in Q
for y in Q
if latinSquare[latinSquare[c][x]][y] == latinSquare[latinSquare[c][y]][x]
) and all(latinSquare[x][x] == 0 for x in Q)
print(
validate(
[
[0, 3, 1, 7, 5, 9, 8, 6, 4, 2],
[7, 0, 9, 2, 1, 5, 4, 8, 6, 3],
[4, 2, 0, 6, 8, 7, 1, 3, 5, 9],
[1, 7, 5, 0, 9, 8, 3, 4, 2, 6],
[6, 1, 2, 3, 0, 4, 5, 9, 7, 8],
[3, 6, 7, 4, 2, 0, 9, 5, 8, 1],
[5, 8, 6, 9, 7, 2, 0, 1, 3, 4],
[8, 9, 4, 5, 3, 6, 2, 0, 1, 7],
[9, 4, 3, 8, 6, 1, 7, 2, 0, 5],
[2, 5, 8, 1, 4, 3, 6, 7, 9, 0],
]
)
)
This is a conversion into Scala of the Python 3 Answer that was provided by David Eisenstat's answer.
View in Scastie:
def isValidDammOperationTable(validLatinSquare: List[List[Int]]): Boolean = {
val indices = validLatinSquare.indices.toList
(
indices.forall(index => validLatinSquare(index)(index) == 0)
&& indices.forall(
c =>
indices.forall(
x =>
indices.forall(
y =>
(validLatinSquare(validLatinSquare(c)(x))(y) != validLatinSquare(validLatinSquare(c)(y))(x))
|| (x == y)
)
)
)
)
}
val exampleLatinSquareX10: List[List[Int]] =
List(
List(0, 3, 1, 7, 5, 9, 8, 6, 4, 2)
, List(7, 0, 9, 2, 1, 5, 4, 8, 6, 3)
, List(4, 2, 0, 6, 8, 7, 1, 3, 5, 9)
, List(1, 7, 5, 0, 9, 8, 3, 4, 2, 6)
, List(6, 1, 2, 3, 0, 4, 5, 9, 7, 8)
, List(3, 6, 7, 4, 2, 0, 9, 5, 8, 1)
, List(5, 8, 6, 9, 7, 2, 0, 1, 3, 4)
, List(8, 9, 4, 5, 3, 6, 2, 0, 1, 7)
, List(9, 4, 3, 8, 6, 1, 7, 2, 0, 5)
, List(2, 5, 8, 1, 4, 3, 6, 7, 9, 0)
)
println(isValidDammOperationTable(exampleLatinSquareX10)) //prints "true"
This question already has answers here:
Algorithm: optimal way to rearrange a list from one order to another?
(4 answers)
Closed 4 years ago.
Given two lists, for example:
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
I wish to find a series of moves which will transform list a into list b, where each move is an operation:
move(from_index, to_index)
which moves the element at location from_index and places it at location to_index. So if:
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
then the operation move(3,1) on the list a will transform a into:
a = [0, 3, 1, 2, 4, 5, 6, 7, 8, 9]
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
move(0, 8)
a = [1, 2, 3, 4, 5, 6, 7, 0, 8, 9]
move(0, 8)
a = [2, 3, 4, 5, 6, 7, 0, 1, 8, 9]
move(1, 8)
a = [2, 4, 5, 6, 7, 0, 1, 3, 8, 9]
move(2, 8)
a = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
a==b
Hopefully that's what you're looking for.
Basically, start with the left- most element and move it to where it should be. For example, I took 0 and placed it right after the value that it is supposed to eventually end up behind, which is 7. I continued moving from left to right until all of the elements were in the desired order.
I'd iterate over the second sequence (the sorted list) and swap items in the first. I wrote this pseudo-code in python:
>>> a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
>>> def swap(seq, i, j):
... a = seq[i]
... seq[i] = seq[j]
... seq[j] = a
...
>>> for index_in_b, value in enumerate(b):
... index_in_a = a.index(value)
... if index_in_b != index_in_a:
... swap(a, index_in_a, index_in_b)
... print('move {} to {}'.format(index_in_a, index_in_b))
move 0 to 2
move 1 to 4
move 2 to 6
move 3 to 7
move 4 to 6
move 5 to 6
move 6 to 7
In this case I'm moving the items in the first sequence by swapping them.
Update
We can slightly improve the performance in python by removing the move inside swap function and also removing the function call. Here is a performance comparison:
import timeit
s1 = """
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
def swap(seq, i, j):
a = seq[i]
seq[i] = seq[j]
seq[j] = a
for index_in_b, value in enumerate(b):
index_in_a = a.index(value)
if index_in_b != index_in_a:
swap(a, index_in_a, index_in_b)"""
s2 = """
a = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
b = [2, 4, 6, 7, 0, 1, 3, 5, 8, 9]
for index_in_b, value in enumerate(b):
index_in_a = a.index(value)
if index_in_b != index_in_a:
a[index_in_a], a[index_in_b] = a[index_in_b], a[index_in_a]"""
# on an i7 macbook pro
timeit.timeit(s1)
4.087386846542358
timeit.timeit(s2)
3.5381240844726562
Slightly better, but for sure there are better ways to achieve this.
Given an array of n elements, remove any adjacent pair of elements which are equal. Repeat this operation until there are no more adjacent pairs to remove; that will be the final array.
For e.g 1 2 2 3 4 should return the array 1 3 4.
please note array need not to be sorted.
check this test case also: 1,2,2,3,4,4,3,5 o/p should be 1,5.
(2,2) and (4,4) gets removed, then (3,3) which became adjacent after the removal of (4,4)
Any time you remove a pair of elements, you also need to see if you generated another pair that you want to remove.
The algorithm should follow naturally from that observation.
In Python:
>>> l=[1,2,2,3,4,4,3,5]
>>> [x for x in l if not l.count(x) > 1]
[1, 5]
This removes all integers that occur more than once in the list. This is a correct result for your example but I think that you are really trying to state something different. I think you are saying:
list:=(an unsorted list of integers)
while adjacent_pairs(list) is True:
remove_adjacent_pairs(list)
Once again, in Python:
#!/usr/bin/env python
def dedupe_adjacent(l):
for i in xrange(len(l) - 1, 0, -1):
if l[i] == l[i-1]:
del l[i-1:i+1]
return True
return False
def process_list(l):
print "input list: ",l
i=1
while(dedupe_adjacent(l)):
print " loop ",i,":",l
i+=1
print "processed list=",l
print
process_list([1,2,2,3,4,4,3,5])
process_list([1,2,2,3,4,4,6,3,5])
Output:
input list: [1, 2, 2, 3, 4, 4, 3, 5]
loop 1 : [1, 2, 2, 3, 3, 5]
loop 2 : [1, 2, 2, 5]
loop 3 : [1, 5]
processed list= [1, 5]
input list: [1, 2, 2, 3, 4, 4, 6, 3, 5]
loop 1 : [1, 2, 2, 3, 6, 3, 5]
loop 2 : [1, 3, 6, 3, 5]
processed list= [1, 3, 6, 3, 5]
The following:
function compress(arr) {
var prev, res = [];
for (var i in arr) {
if (i == 0 || (arr[i] != arr[i - 1]))
res.push(arr[i]);
}
return res;
}
compress([1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]);
Returns:
[1, 2, 3, 4, 3, 5, 6, 7, 8]
Also (JavaScript 1.6 solution):
[1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8].filter(function(el, i, arr) {
return i == 0 || (el != arr[i - 1]);
})
Edit: Removing any item that appears in the array more than once requires a different solution:
function dedup(arr) {
var res = [], seen = {};
for (var i in arr)
seen[arr[i]] = seen[arr[i]] ? ++seen[arr[i]] : 1;
for (var j in arr) {
if (seen[arr[j]] == 1)
res.push(arr[j]);
}
return res;
}
The following:
dedup([1, 2, 2, 3, 4, 4, 3, 5]);
Produces:
[1, 5]
I have a solution to this in Java. You need to use replaceAll method in String class in Java. You can use regular expession to remove such adjacent redundant characters:
public class MyString {
public static void main(String[] args) {
String str = "12234435";
while(!str.replaceAll("(\\w)\\1+", "").equalsIgnoreCase(str))
str = str.replaceAll("(\\w)\\1+", "");
System.out.println(str);
}
}
You can find how to give a regular expression here
I would:
Sort the array.
From the start of the array, until you are at the last element of the array do:
`count` = count the number of array[i] elements.
remove the first `count` elements of the array if `count` > 1.
The following Python 3 code will remove duplicates from a list (array). It does this by scanning the array from start towards end and compares the target element with the element one larger. If they are the same they are removed. If the element pointer is not pointing at 0, then it is reduced by 1 in order to catch nested pairs. If the two compared elements are different then the pointer is incremented.
I'm sure there's a more pythonic way to remove two adjacent elements from a list, but I'm new to Python and haven't figured that out yet. Also, you'll want to get rid of the print(indx, SampleArray) statement--I left it in there to let you follow the progress in the output listing below.
# Algorithm to remove duplicates in a semi-sorted list
def CompressArray(SampleArray):
indx=0
while(indx < len(SampleArray)-1):
print(indx, SampleArray)
if(SampleArray[indx]==SampleArray[indx+1]):
del(SampleArray[indx])
del(SampleArray[indx])
if(indx>0):
indx-=1
else:
indx+=1
return SampleArray
Here are sample runs for:
[1, 2, 2, 3, 4]
[1, 2, 2, 3, 4, 4, 3, 5]
[1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
[1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
[1, 1, 2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
================================
0 [1, 2, 2, 3, 4]
1 [1, 2, 2, 3, 4]
0 [1, 3, 4]
1 [1, 3, 4]
[1, 3, 4]
================================
0 [1, 2, 2, 3, 4, 4, 3, 5]
1 [1, 2, 2, 3, 4, 4, 3, 5]
0 [1, 3, 4, 4, 3, 5]
1 [1, 3, 4, 4, 3, 5]
2 [1, 3, 4, 4, 3, 5]
1 [1, 3, 3, 5]
0 [1, 5]
[1, 5]
================================
0 [1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 3, 3, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 3, 3, 4, 3, 3, 5, 6, 7, 8, 8]
0 [1, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 4, 3, 3, 5, 6, 7, 8, 8]
2 [1, 4, 3, 3, 5, 6, 7, 8, 8]
1 [1, 4, 5, 6, 7, 8, 8]
2 [1, 4, 5, 6, 7, 8, 8]
3 [1, 4, 5, 6, 7, 8, 8]
4 [1, 4, 5, 6, 7, 8, 8]
5 [1, 4, 5, 6, 7, 8, 8]
[1, 4, 5, 6, 7]
================================
0 [1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 2, 2, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
2 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
3 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
4 [1, 3, 4, 6, 7, 7, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
3 [1, 3, 4, 6, 6, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
2 [1, 3, 4, 4, 3, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 3, 3, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 8, 8, 5, 9, 10, 10, 9, 11]
1 [1, 8, 8, 5, 9, 10, 10, 9, 11]
0 [1, 5, 9, 10, 10, 9, 11]
1 [1, 5, 9, 10, 10, 9, 11]
2 [1, 5, 9, 10, 10, 9, 11]
3 [1, 5, 9, 10, 10, 9, 11]
2 [1, 5, 9, 9, 11]
1 [1, 5, 11]
[1, 5, 11]
================================
0 [1, 1, 2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [2, 3, 3, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [2, 2, 4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
0 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
2 [4, 5, 6, 6, 5, 7, 8, 8, 7, 4, 9]
1 [4, 5, 5, 7, 8, 8, 7, 4, 9]
0 [4, 7, 8, 8, 7, 4, 9]
1 [4, 7, 8, 8, 7, 4, 9]
2 [4, 7, 8, 8, 7, 4, 9]
1 [4, 7, 7, 4, 9]
0 [4, 4, 9]
[9]
================================
I love Java, but functional solutions should get more time on this site.
In Haskell, doing things the way the question asks:
compress lst = if (length lst == length b) then lst else (compress b) where
b = helper lst
helper [] = []
helper [x] = [x]
helper (x:y:xs) = if (x == y) then (helper xs) else (x:helper (y:xs))
You can solve this problem in O(n) time, although it is a bit more complicated
compress' lst = reverse (helper [] lst) where
helper xs [] = xs
helper [] (x:xs) = helper [x] xs
helper (a:as) (x:xs)
| a == x = helper as xs
| otherwise = helper (x:a:as) xs
I think we could use a stack to check adjacent duplicated elements.
Scan the array. For each new element, if it is equal to the top element in the stack, drop it and pop the top element from the stack. Otherwise, push it into the stack.
Here is the stack based algorithm based upon the edited question.
// pseudo code, not tested
void RemoveDupp(vector<int> & vin, vector<int> & vout)
{
int i = 0, int j = -1;
vout.resize(vin.size());
while (i < vin.size())
{
if (j == -1 || vout[j] != vin[i])
vout[++j] = vin[i++]; //push
else
j--, i++; //pop
}
vout.resize(j + 1);
}