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Hi A,

Below is an example to achieve what you want. Depending on your polygons etc. you may want to add a more fancy polygon analysis, but that is not Nazca specific.

import nazca as nd
import nazca.geometries as geom
from nazca.interconnects import Interconnect

def rect1(layer_n=1):
    rect_shape = nd.geom.rectangle(length=20, height=5)
    with nd.Cell('rect1') as cell:
        nd.Polygon(layer=1, points=rect_shape).put(0)
        nd.Polygon(layer=10, points=rect_shape).put(2, 10)
        # guide=ic.sbend_p2p(pin1=r1, pin2=r2.pin['a0']).put()
    return cell
rect1().put(array=[4, [30, 0], 2, [0, 20]])  # This could be a gds_load() based cell.

# 1. Extract all the polygon from cell into a dict with key (layer, x, y),
#    where x and y will be, just a choice, the geometrical average of the polygon points.
polygons = {}
for params in nd.cell_iter(nd.cfg.topcell, flat=True):
    for pgon, points, bbox in params.iters['polygon']:
        xp, yp = zip(*points)
        xavg, yavg = round(sum(xp) / len(xp), 2), round(sum(yp) / len(yp), 2)
        polygons[(pgon.layer, xavg, yavg)] = points
# 2. Analyse polygons: order them and extract positions for pins.
for i, ((layer, x, y), points) in enumerate(sorted(polygons.items())):
    nd.text(str(i), height=2, align='cc', layer=0).put(x, y)
    sort = sorted(points)
    x0 = sort[0][0] + 0.5 * (sort[1][0] - sort[0][0])
    y0 = sort[0][1] + 0.5 * (sort[1][1] - sort[0][1])
    w0 = abs(sort[1][1] - sort[0][1])
    nd.Pin(f'a{i}', width=w0).put(x0, y0, 180)
    x1 = sort[-1][0] + 0.5 * (sort[-2][0] - sort[-1][0])
    y1 = sort[-1][1] + 0.5 * (sort[-2][1] - sort[-1][1])
    nd.Pin(f'b{i}', width=w0).put(x1, y1, 0)

# 3. Add interconnects.
ic = Interconnect(radius=5.0, width=w0)
for i, (name, pin) in enumerate(nd.cfg.topcell.pin.items()):
    if i in [0, 1, 2, 3, 4, 5]: 

Note that it would make sense to put item 2 and 3 in the script above inside a new cell with nd.Cell("newcellname") as C:, and work with cell C, rather than working on the default topcell under nd.cfg.topcell.