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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def tiledict(D, sz=None):
 """Construct an image allowing visualization of dictionary content. Parameters D : array_like Dictionary matrix/array. sz : tuple Size of each block in dictionary. Returns ------- im : ndarray Image tiled with dictionary entries. """


    # Handle standard 2D (non-convolutional) dictionary
    if D.ndim == 2:
        D = D.reshape((sz + (D.shape[1],)))
        sz = None
    dsz = D.shape

    if D.ndim == 4:
        axisM = 3
        szni = 3
    else:
        axisM = 2
        szni = 2

    # Construct dictionary atom size vector if not provided
    if sz is None:
        sz = np.tile(np.array(dsz[0:2]).reshape([2, 1]), (1, D.shape[axisM]))
    else:
        sz = np.array(sum(tuple((x[0:2],) * x[szni] for x in sz), ())).T

    # Compute the maximum atom dimensions
    mxsz = np.amax(sz, 1)

    # Shift and scale values to [0, 1]
    D = D - D.min()
    D = D / D.max()

    # Construct tiled image
    N = dsz[axisM]
    Vr = int(np.floor(np.sqrt(N)))
    Vc = int(np.ceil(N / float(Vr)))
    if D.ndim == 4:
        im = np.ones((Vr*mxsz[0] + Vr - 1, Vc*mxsz[1] + Vc - 1, dsz[2]))
    else:
        im = np.ones((Vr*mxsz[0] + Vr - 1, Vc*mxsz[1] + Vc - 1))
    k = 0
    for l in range(0, Vr):
        for m in range(0, Vc):
            r = mxsz[0]*l + l
            c = mxsz[1]*m + m
            if D.ndim == 4:
                im[r:(r+sz[0, k]), c:(c+sz[1, k]), :] = D[0:sz[0, k],
                                                          0:sz[1, k], :, k]
            else:
                im[r:(r+sz[0, k]), c:(c+sz[1, k])] = D[0:sz[0, k],
                                                       0:sz[1, k], k]
            k = k + 1
            if k >= N:
                break
        if k >= N:
            break

    return im