I’ve been doing the Weekly
Challenges. The
latest
involved list processing and matrix testing. (Note that this ends today.)
Task 1: Smaller than Current
You are given a array of integers, @ints
.
Write a script to find out how many integers are smaller than
current i.e. foreach ints[i]
, count ints[j] < ints[i]
where i != j
.
There's an obvious O(n²) way of doing this, but I thought it was
cleaner to sort.
Scala:
def smallerthancurrent(a: List[Int]): List[Int] = {
Set up a sorted list of inputs, and a map that'll contain the solution.
val s: List[Int] = a.sortWith(_ < _)
var l = mutable.Map.empty[Int, Int]
For each item in the sorted list, store the index, which is the number
of items smaller than that. (Use the first occurrence of the item; I
added the test case with a duplicate value.)
for ((n, i) < s.zipWithIndex) {
if (!l.contains(n)) {
l += (n > i)
}
}
Then just read off the values in the order of the original list.
a.map { n => l(n) }.toList
}
Task 2: Reduced Row Echelon
Given a matrix M, check whether the matrix is in reduced row echelon form.
A matrix must have the following properties to be in reduced row
echelon form:

If a row does not consist entirely of zeros, then the first
nonzero number in the row is a 1. We call this the leading 1.

If there are any rows that consist entirely of zeros, then they
are grouped together at the bottom of the matrix.

In any two successive rows that do not consist entirely of zeros,
the leading 1 in the lower row occurs farther to the right than
the leading 1 in the higher row.

Each column that contains a leading 1 has zeros everywhere else
in that column.
I propsed a bunch of the test cases here because I wanted to test each
rule individually. Raku:
sub reducedrowechelon(@a) {
We'll be using the position of the first 1 in each row quite a bit.
That goes in @leadingone
.
my @leadingone;
Inspect each row:
for @a > @row {
Set the leadingone position to 1, indiciating no 1 found.
my $lp = 1;
For each item in the row:
for 0 .. @row.end > $cn {
my $cell = @row[$cn];
If it's a 1, store that position and exit.
if ($cell == 1) {
$lp = $cn;
last;
If it's not a zero, bail out (rule 1).
} elsif ($cell != 0) {
return False;
}
}
Add this row's leadingone position to the list.
@leadingone.push($lp);
}
If there are trailing rows of allzero, ignore them.
while (@leadingone[*1] == 1) {
@leadingone.pop;
}
Now sort the remaining leadingone positions.
my @c = @leadingone.sort({$^a <=> $^b});
If there are any allzero rows, they weren't trailing. Bail out by
rule 2.
if (@c[0] == 1) {
return False;
}
If the sorted list doesn't match the original, the leading ones aren't
progressively to the right. Bail out by rule 3.
if (!(@c eqv @leadingone)) {
return False;
}
Finally take a columnwise slice of each column that has a leading 1 in
it. Test for 1s and 0s in the right places (rule 4).
for @c > $i {
my @col = @a.map({$_[$i]}).sort({$^a <=> $^b});
if (@col[*1] != 1 
@col[*2] != 0 
@col[0] != 0) {
return False;
}
}
return True;
}
I spot two possible problems here:
(1) the rule3 test will pass two leadingones of equal value. But the
rule4 test will remove those because they won't be unique in the
column. I'd rather rule 3 be fully tested in its own segment of the
codex.
(2) the rule4 test should reject a negative value found in the same
column as a leading 1 (@col[0] != 0
above). The existing test cases
don't fail in that case (this is my fault since I proposed most of
them).
Full code on
github.
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