Last time we showed how you can take any two coprime positive integers x and m and compute a third positive integer y with the property that (x * y) % m == 1, and therefore that (x * z * y) % m == z % m for any positive integer z. That is, there always exists a “multiplicative inverse”, that “undoes” the results of multiplying by x modulo m. That’s pretty neat, but of what use is it?
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Here’s an interesting question: what is the behavior of this infinite sequence? WOLOG let’s make the simplifying assumption that x and m are greater than zero and that x is smaller than m. Continue reading
UPDATE: Coverity has removed the dev testing blog; I’ve put in a link to the internet archive.
In the latest episode of the Coverity Development Testing Blog‘s continuing series “Ask the Bug Guys”, I dig into an interesting question about inconsistencies between de facto and de jure unreachability — that is, the difference between code that is actually unreachable and the smaller set of code that the C# compiler detects as unreachable. The difference can cause some interesting inconsistencies in the compiler’s behavior. And my colleague Jon ponders the wisdom of fixing fragile, hard-to-understand code even if it is at present correct.
As always, if you have questions about a bug you’ve found in a C, C++, C# or Java program that you think would make a good episode of ATBG, please send your question along with a small reproducer of the problem to TheBugGuys@Coverity.com. We cannot promise to answer every question or solve every problem, but we’ll take a selection of the best questions that we can answer and address them on the dev testing blog every couple of weeks.
Now that we have an integer library, let’s see if we can do some 2300-year-old mathematics. The Euclidean Algorithm as you probably recall dates from at least 300 BC and determines the greatest common divisor (GCD) of two natural numbers. The Extended Euclidean Algorithm solves Bézout’s Identity: given non-negative integers a and b, it finds integers x and y such that:
a * x + b * y == d
where d is the greatest common divisor of a and b. Obviously if we can write an algorithm that finds x and y then we can get d, and x and y have many useful properties themselves that we’ll explore in future episodes.
Now we come to the controversial one. As I pointed out back in 2011, the % operator is the remainder operator. When extending the division and remainder operators from naturals to integers, I require that the following conditions be true for dividend x, non-zero divisor y, quotient q and remainder r:
x == y * q + rx and y are non-negative then 0 <= r < y(-x) / y == x / (-y) == -(x / y)Let’s take a look at the consequences of these three facts. Continue reading
I’m going to use the same technique I used for Natural comparisons to do Integer comparisons: make a helper function that returns -1 if x < y, 0 if x == y and 1 if x > y, and then have every entry point simply call that helper function. This way we know that all the operators are consistent. The big difference here of course is that I do not need to have a special case that says that null sorts before any value. Continue reading
In the latest episode of the Coverity Development Testing Blog‘s continuing series “Ask the Bug Guys”, I dig into two questions about the null coalescing operator. This handy little operator is probably the least well understood of all the C# operators, but it is quite useful. Unfortunately, it is easy to accidentally use it incorrectly due to precedence issues.
As always, if you have questions about a bug you’ve found in a C, C++, C# or Java program that you think would make a good episode of ATBG, please send your question along with a small reproducer of the problem to TheBugGuys@Coverity.com. We cannot promise to answer every question or solve every problem, but we’ll take a selection of the best questions that we can answer and address them on the dev testing blog every couple of weeks.
Almost all the Integer operations are simply wrappers around the Natural operations, plus some gear to deal with the signs. Continue reading
The integers are an extension to the naturals that closes them over subtraction. Rather than do anything fancy, I’m simply going to say that an integer is a natural with an associated sign, and that zero is always “positive”. We don’t want to end up in a situation where there are two forms of zero, since that is confusing. (We could of course have three signs, positive, negative and zero, but, meh. We don’t gain any great benefits from having a third sign.)
This time however I’m going to make a struct. C# requires that default(SomeStruct), where all the fields are their default values, be a valid value of the type. The default should logically be zero, as it is with all the built-in number types. So rather than checking for null, as we did with the reference type Natural, I’m going to check the sign field for null, which will only be possible if we’ve got a default struct. We’ll simply replace that with zero. (I am in general opposed to modifying the formal parameters of a method; I prefer to treat them as read-only variables. I’ll make an exception in this case because it keeps the code short and clear.) Continue reading
We have only two operators left, integer division and remainder. We’ve left the most complicated two operations for last, so let’s tread carefully here.
(Only two? I’m not going to do bit shifting operators; I don’t think of them as operations on natural numbers. If you want them, obviously they are trivial: shifting right is just taking the tail, shifting left is appending a zero bit as the new head. Similarly I am not going to implement the bitwise logical operators. Nor am I going to do the unary plus operator, which wins my award for World’s Most Useless Operator. It is a no-op on both naturals and integers; if you really want it, it’s not hard to write. The unary minus operator makes no sense on naturals; the only legal operand value would be zero.) Continue reading
Fabulous adventures in coding 