c# 10/2024
In “Test” mode, but not in “ATTEMPT” mode, this error message

appears after reset:

tests/Fixture.cs(10,55): error CS1503: Argument 1: cannot convert from 'uint' to 'int'

Should I change the type of the argument n from int to uint?

CS:
I think the labeling in the error message is reversed for the following tests: RandomBig, RandomFull, RandomTestsAlmostEmpty, and RandomTestsManyShare

Example:
RandomTestsAlmostEmpty

Test Failed
Expected: null

But was: < 66, 16, 4, 55, 76, 55, 36, 5, 34, 1, 35, 57, 60, 86, 6, 12, 16, 66, 60, 26, 19, 84, 40, 66, 26, 43, 26, 60, 26, 66, 43, 23, 77, 26, 50, 23, 66, 60, 74, 88, 43, 23, 12, 38, 6, 60, 85, 43, 57, 60, 6, 38, 21, 32, 20, 50, 74, 50, 50, 16, 40, 6, 2, 79, 43, 6, 66, 83, 77, 79, 12, 66, 2, 66, 32, 43, 40, 50, 26, 79, 86, 57, 6, 21, 79, 12, 43, 36 >

㊗

I suggest specifying how to sort if the sort key is not unique. I got this random test:

Expected: [[[99, [58, 39], 31, 19], 40, 3], 85, 46, [27, [17], 2], 2]

But was: [[[99, [58, 39], 31, 19], 40, 3], 85, [27, [17], 2], 46, 2]

I think the sentence in the description should read:

Prepare a code that given the number of digits n, may output the amount of palindromes of length equals to n and the total amount of palindromes below or equal to 10ⁿ.

the deepl-translation doesn't make any sense:

"I haven't decided this yet, but these people who praise kats that almost got divorced or spent 4+ hours on it are total freaks)))))"

what do you mean?

I don't understand why the correct answer is “true” in the test case "normalString3"

IEnumerable<string> columns = new string[] { "1:Good Morning.", "Good Morning." };
Assert.AreEqual(true, instance.testCompare(columns, 7));

When counting users, all numbers with digits 4 or 9 should be skipped. To rephrase it slightly, we need to determine which number will stand on the n-th place in the series of natural numbers after excluding from this series all numbers containing 4 and 9.

I have tested these keys without success to decrypt your message: agedyropulik politykarenu regulaminowy kaceminutowy koniecmatury arewybuhoki bawoletykiju

Should your message contain the Pound-Sign?

Let's assume the language is English and the alphabet consists of 18 ( = 26 (a, b, ..., z) - 8 (=p r o b l e m t) ) unknown characters. This results in 1.5e12 ( =Binomial(18*18, 6) ) keys. Then around 1.0e13 (=7*1.5e12) words would have to be checked to see whether they are valid English words. Rather difficult, isn't it?