Suppose An $8 \times 8$ Image Block Happens to Have the Following Entries

Suppose an $8 \times 8$ image block happens to have the following entries:
$\begin{array}{rrrrrrrr}183&160&94&0&0&0&0&0\\183&153&0&0&0&0&0&0\\179&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0\\\end{array}$
(Note that this is a greylevel, 8-bit image, not DCT output).
Now suppose we decide to encode this image into the frequency domain as follows:
• First we go down each column, and carry out a 1-dimensional $D C T$ transform, replacing each column by its set of DCT coefficients.
• However, for the first column we use only a length-3 DCT (i.e., $N=3$ ); for the second column we use a length-2 DCT, and for the third column we use a length-1 DCT, always leaving zeros in the transform domain just where they appeared in the original, image domain.
• We leave the DC coefficient always at the top of each column processed.
• Then we use the output from the above stage and go on to do the same procedure for rows 1 to 3.
Question:
(a) Which takes more calculations, the above procedure, or the ordinary 2-D DCT transform? Explain.
(b) Broadly, what is the difference, if any, in the output DCT Image between the new transform and the standard one, for this particular image?
Note: One need not do any calculations for this question but, for reference, recall that the 2-D DCT for an $M \times N$ block size is defined as
$$F(u, v)=\frac{2 C(u) C(v)}{\sqrt{M N}} \sum_{i=0}^{M-1} \sum_{j=0}^{N-1} \cos \frac{(2 i+1) \cdot u \pi}{2 M} \cos \frac{(2 j+1) \cdot v \pi}{2 N} f(i, j),$$
where $i, u \in[0, M-1], j, v \in[0, N-1]$ , and the constants $C(u)$ and $C(v)$ are determined by
$$C(\xi)=\left\{\begin{array}{cl}\frac{\sqrt{2}}{2} & \text { if } \xi=0 \\1 & \text { otherwise. }\end{array}\right.$$
The 1-D DCT is given by
$$F(u)=\frac{2 C(u)}{\sqrt{N}} \sum_{i=0}^{N-1} \cos \frac{(2 i+1) u \pi}{2 N} f(i)$$