Minimum number of steps using only multiply A by 2, or divide A by 2 or increment A by one to go from number A to B - algorithm

Question

Given two numbers A and B, what is the minimum number of steps to transform number A to become number B.
A step can either be A *= 2, A++ or A /= 2 if and only if A is an even number.
What is the most efficient algorithm to achieve this?
Suppose A and B can be really large numbers.

Answer 1

Here's my take, done in C#.
var a = 2;
var b = 15;
var found = new HashSet<int>() { a };
var operations = new (string operation, Func<int, bool> condition, Func<int, int> projection)[]
{
("/2", x => x % 2 == 0, x => x / 2),
("*2", x => x <= int.MaxValue / 2, x => x *2),
("+1", x => true, x => x + 1),
};
IEnumerable<(int count, string operations, int value)> Project((int count, string operations, int value) current)
{
foreach (var operation in operations)
{
if (operation.condition(current.value))
{
var value = operation.projection(current.value);
if (!found.Contains(value))
{
found.Add(value);
yield return (current.count + 1, $"{current.operations}, {operation.operation}", value);
}
}
}
}
var candidates = new[] { (count: 0, operations: $"{a}", value: a) };
while (!found.Contains(b))
{
candidates =
candidates
.SelectMany(c => Project(c))
.ToArray();
}
var result = candidates.Where(x => x.value == b).First();
Console.WriteLine($"{result.count} operations: {result.operations} = {result.value}");
That outputs:
5 operations: 2, +1, *2, +1, *2, +1 = 15
Basically, this is starting with a at the zeroth step. It then takes this generation and produces all possible values from the operations to create the next generation. If it produces a value that it has already seen it discards the value as there is an equal or faster operation to produce the value. It keeps repeating until b is found.