Question

Suppose we define a square matrix. Let a[0,0] = 1, a[0, i] = [infinity] for i > 0, and a[i,0] = [infinity] for i> 0. Recursively define a[i, j] min(a[i-1, j - 1], 2. a[i-1, j]). What is the value of a[5, 1]?

Answer 1

The recursive definition of the matrix elements allows us to determine that a[5, 1] is 2, using the initial conditions provided and applying the minimum function to select the non-infinity value.

Step-by-step explanation:

The student question pertains to the recursive definition of the elements of a square matrix. We are given that a[0,0] = 1, a[0, i] = ∞ (infinity) for i > 0, and a[i,0] = ∞ (infinity) for i > 0. The recursive definition for other elements is a[i, j] = min(a[i-1, j - 1], 2 × a[i-1, j]). To determine the value of a[5, 1], we will apply the given recursive definition step by step.

• Since a[0, i] is infinity for i > 0, a[1,0] is also infinity.
• To find a[1,1], we use the minimum of a[0,0] and 2 × a[0,1]. a[0,0] is 1, and a[0,1] is infinity, so a[1,1] = min(1, 2 × infinity) = 1.
• For i > 1, since a[i,0] is infinity, a[i,1] will use a[i-1,0] which is infinity, thus a[i,1] will always be 2 × 1 = 2 due to the minimum function picking the non-infinity value from a[i-1, 0].

Following these steps up to a[5, 1], we see that the value would be 2 because it takes the minimum of infinity and 2 (since a[4,1] would have been 2).