Geolocating a random island using geometry and CUDA programming

A developer known as gralhix solved an OSINT geolocation challenge by writing CUDA-accelerated geometry code instead of using Google Lens, identifying a resort island from a drone photo. The solution, documented in a GitHub writeup, used OpenStreetMap land polygon data (882 MB) and a series of filters—tropical latitude band (-30° to 30°), local density (≤10 neighbors within 5 km), and clustering (≥3 points within 20 km)—to narrow 141,131 land polygons down to 23,500 clusters, then generated candidate triangles for matching. The challenge, created by Sofia Santos, asked for the resort name, island coordinates, and camera direction.

gralhix004 | Geolocating Random Islet Image Using Geometry & CUDA GPU Programming 16-08-2026 NOTE: this is a genuine human work, didnt use LLM generation. I'm writing this page as a writeup for this challenge gralhix 004 made by Sofia Santos | Gralhix https://gralhix.com/list-of-osint-exercises/osint-exercise-004/ . You can view, clone and locally try all code files and the final report with all instructions here at github. Task briefing: This is a photo of a resort located on an island. a What is the name of the resort? b What are the coordinates of the island? c In which cardinal direction was the camera facing when the photo was taken? In my opinion, solving this challenge with google lens is wasting a fun opportunity, so decided to solve it with math and programming. a Metadata Of course, first thing u look for is the metadata . Ran that on my linux void : exiftool main.png File Type : WEBP lossless MIME Type : image/webp Image Width : 736 Image Height : 515 As expected, nothing useful here. No EXIF, no GPS, no camera make or model. b Building the fingerprint U can see from the img, there are 3 landmasses: - P0: the islet itself, - P1: the right island, - P2: the left front island having mountain peak I couldnt make a correct perspective model of birdview of this image, as clearly the image is taken by a drone and cant estimate the elevation at all and not found in the metadata . So I had to estimate that by intuition, I just want the relative distances between the 3 islands and angles of that triangle. I built a small click GUI 01 triangle gui.py that records pixel coordinates for each point in order and computes the triangle's geometry. Since clicking exact centers by eye isn't perfectly precise, I added a ±20% tolerance band around both values when searching. c SEARCH With the fingerprint locked in, the next step is checking every real landmass on Earth against it I used OpenStreetMap's split land polygon set as the dataset land-polygons-split-4326 , full global coastline vectors in WGS84 which has size of 882 MB . I created heuristic filters all by just intuition and non tangible proofs , spent days yea full days tweaking values and tons of trial and error 😭 untill I got this working filters recipe. 01 Tropical latitude bounding box $$ -30° \le latitude \le 30° $$ the islet in the photo reads as tropical, so I decided that anything outside the tropics is thrown out immediately, before doing any expensive geometry work. Exactly 141,131 land polygons survive that band filter. 02 Local density filter $$ N {5\text{km}} p \le 10 $$ $ N {5\text{km}} p $ counts how many other centroids fall within 5km of point p . Cap is 10 : if an islet has more than 10 neighbors that close, it's sitting in a dense reef field, a crowded coastline or a archipelago clutter, not a small isolated 3-4 island group like the photo shows. This dropped candidates down to 51,576 . 03 Clustering For every surviving point, find every other point within 20km heuristic, by eye from the image . If it has at least 2 neighbors that close 3 points total , it's a cluster. Points with no cluster of 3+ nearby are dropped, they can't form a triangle at all. tree = cKDTree f coords neigh = tree.query ball point f coords, CLUSTER RADIUS KM / 111.0 clusters = set tuple sorted n for n in neigh if len n = 3 $$ \left|\{q : \text{dist} p,q \le 20\,\text{km}\}\right| \ge 3 $$ That collapses down to 23,500 clusters. 04 Generating Triplets For every cluster, every combination of 3 points inside it becomes a candidate triangle. That's $ C n, 3 $, which explodes fast for big clusters, for example: a cluster of 60 points already gives 34,220 triples on its own. So each cluster gets capped at 60 points first, sampled by size, not randomly. $$ \binom{n}{3} = \frac{n n-1 n-2 }{6} $$ python def stratified sample idx arr, area arr, cap : order = np.argsort area arr idx arr n small = cap // 3 n large = cap // 3 n mid = cap - n small - n large mid start = max 0, len idx arr - n large - n mid // 2 keep = np.unique np.concatenate order :n small , order -n large: , order mid start:mid start + n mid , return idx arr keep def gen cluster triples idx arr : local = np.array list itertools.combinations range len idx arr , 3 , dtype=np.int64 return idx arr local The sampling takes a third small islands, a third large, a third from the middle of the size distribution, instead of the full cluster or a random cut. 23,500 clusters produce 80,690,777 triples total 05 Matching, on the GPU I gave every triple one CUDA thread. Each thread sorts its 3 points by land area to pick out P0 smallest, the resort islet , then uses the winding direction of the other two to assign P1 and P2: long long i = blockIdx.x long long blockDim.x + threadIdx.x; if i = n triples return; int pos 3 = {0, 1, 2}; for int a1 = 1; a1 < 3; a1++ { int key = pos a1 ; double keyval = a key ; int j = a1 - 1; while j = 0 && a pos j keyval { pos j + 1 = pos j ; j--; } pos j + 1 = key; } P1 vs P2 comes from a 2D cross product, no branching on which cluster the triple came from, just the sign: $$ \text{cross} = x a y b - x b y a $$ $$ P1 = \begin{cases} a & \text{cross} 0 \\ b & \text{cross} \le 0 \end{cases} $$ Walk from P0 to a, then to b. If cross 0, that's a left turn counterclockwise . If cross < 0, it's a right turn clockwise . It's the same sign trick used to tell if 3 points curve one way or the other. then angle at P0 and the distance ratio, same formulas as the fingerprint step, computed independently by every thread: $$ \theta 0 = \arccos\left \frac{\vec{d 1} \cdot \vec{d 2}}{|\vec{d 1}||\vec{d 2}|}\right , \qquad r = \frac{|\vec{d 1}|}{|\vec{d 2}|} $$ A triple survives if angle, ratio, P0's size, the separation between P0 and P1, and both side lengths all land inside the fingerprint's tolerance windows. Threads that pass write their result into a shared output array using an atomic counter, so two threads finishing at the same time never overwrite each other: if hit { unsigned long long slot = atomicAdd out count, 1ULL ; out p0 slot = p0idx; out p1 slot = p1idx; out p2 slot = p2idx; } Now printed in the CLI directly from the kernel: gpu: NVIDIA GeForce RTX 3050 sm 86 vram used: 5169 MB kernel time: 204.1 ms 80.7 million triples go in, one thread each, in parallel. 158,784 pass the mask. 06 Dedup Since same physical triple can get hit by multiple GPU threads if it belonged to more than one overlapping cluster, so raw matches get collapsed by identity first: seen = set uniq = for i in range len p0 all : key = p0 all i , p1 all i , p2 all i if key not in seen: seen.add key uniq.append i 8,915 unique triples after dedup. 07 The Open Rectangle Every surviving triple gets one more test: is the space next to it actually open water, like the photo shows ? A rectangle gets built along the P0→P1 edge, on whichever side P2 is not on, then checked against the land dataset for anything else sitting inside it. width = np.hypot x1, y1 u = np.array x1, y1 / width v = np.array -u 1 , u 0 p2 sits on the +v side by construction, so the check goes on -v length = 2 width corners local = 0, 0 , x1, y1 , x1 - v 0 length, y1 - v 1 length , -v 0 length, -v 1 length , If anything other than the 3 candidate islands themselves intersects that rectangle, the candidate is dropped. Land sitting there means it's not the open, unobstructed water the photo actually shows. 8,915 unique triples down to 948 . and below is the map of places of the 948 candidates. d Coral Cay Shape Check In this stage, we look only at P0, the resort islet, and check whether its shape actually looks like a coral cay. 1 Compactness , how close to a circle the shape is: Polsby Popper Score: $$ PP = \frac{4\pi \cdot \text{area}}{\text{perimeter}^2} $$ python def compactness row : return 4 np.pi row.area km2 / row.perim km 2 + 1e-12 1.0 is a perfect circle, lower means a more jagged or elongated outline. Coral cays tend to be round from wave deposition, so anything < 0.5 gets dropped. 2 Micro Cay Halo Check: python def micro cay count gdf, sindex, lon, lat : dists km = nearby.geometry.distance pt 111.0 mask = dists km 0 & dists km <= HALO KM & nearby "area km2" .values < MICRO KM2 return int mask.sum We Count land fragments under 0.05 km² within 1.5km of P0 just heuristic . Real reef systems scatter tiny sandbars around the main island, not just one isolated landmass I knew that with the hardway 😭 . So we need at least 1. 213/948 candidates survive both checks. e Oval Shape Check Another geometric filter on P0's own polygon. Fits the minimum rotated rectangle around it and measures two ratios from that box. python def aspect and fill geom : mrr = geom.minimum rotated rectangle coords = list mrr.exterior.coords s1 = math.hypot coords 1 0 - coords 0 0 , coords 1 1 - coords 0 1 s2 = math.hypot coords 2 0 - coords 1 0 , coords 2 1 - coords 1 1 long side, short side = max s1, s2 , min s1, s2 return long side / short side, geom.area / mrr.area Aspect ratio is long side over short side of that box: $$ \text{aspect} = \frac{\text{long side}}{\text{short side}} \in 1.05,\ 2.2 $$ Too close to 1.0 and it's basically a perfect circle, not the slightly elongated shape in the photo. Too high are shapes too much elongated more than 2:1. Fill ratio is how much of that bounding box the shape actually fills, and this one has an identity behind it: any ellipse fills precisely $ \pi / 4 $ of its own minimum area bounding rectangle, regardless of how stretched it is. $$ \frac{\text{area} {\text{ellipse}}}{\text{area} {\text{box}}} = \frac{\pi}{4} \approx 0.785 $$ that's the theoretical ceiling for a perfectly smooth oval. Real coral cays aren't perfect ellipses, so the cutoff is set as a heuristic safe fraction of that ceiling: $$ \text{FILL\ RATIO\ MIN} = 0.75 \times \frac{\pi}{4} \approx 0.589 $$ A shape needs to retain at least 75% of a perfect ellipse's fill to survive. Crescents, rings, and notched coastlines fall well below that, solid rounded cays don't. 137/213 candidates survive. f NDVI Vegetation Check We reached the final API phase, I put it at the end, because it is network bound not compute bound. We gonna connect to Earth Search, run by Element84 , a public STAC API that indexes Sentinel-2 imagery hosted on AWS's Open Data program, free, no API key. You can look at it https://earth-search.aws.element84.com/v1 https://earth-search.aws.element84.com/v1 We now check whether P0 is actually vegetated, palm cover, not bare sand or rock. It pulls the most recent low cloud Sentinel-2 scene over the point from a public STAC catalog, samples the red and near infrared bands at that exact pixel. $$ \text{NDVI} = \frac{\text{NIR} - \text{Red}}{\text{NIR} + \text{Red}} $$ Live vegetation reflects strongly in near infrared and absorbs red light, so healthy palm cover pushes NDVI well above 0, bare sand or open water sits near 0 or negative. You can view this image I got from this nice Geoawesome Blog. https://geoawesome.com/eo-hub/understanding-aerial-data-normalized-difference-vegetation-index-ndvi/ Threshold is set at 0.6 , high enough to require real tree cover, not just scattered units. 66/137 survive the NDVI check. g Elevation & Mountain Check Last check before the final reveal. There are two conditions: - P0 itself must be low and flat, consistent with a small reef islet, - P2 must have real elevated terrain in the direction the camera was actually facing. The "front" of the shot is the bisector between the bearing to P1 and the bearing to P2: $$ \theta P 0, P i = $$ $$ \text{atan2}\Big \sin \Delta\lambda \cos\phi i,\ \cos\phi 0\sin\phi i - \sin\phi 0\cos\phi i\cos \Delta\lambda \Big $$ $$ \theta {\text{front}} = $$ $$ \theta P 0, P 2 + \frac{\big \theta P 0,P 1 - \theta P 0,P 2 + 180 \bmod 360\big - 180}{2} $$ That gives one heading, the direction the lens was pointed. From there, a fan of sample points gets swept ±50° around that heading, at radii from 2km out to 20km: $$ \text{lat}, \text{lon} = \Big \text{lat} 0 + \frac{r\cos\theta}{111},\ \ \text{lon} 0 + \frac{r\sin\theta}{111\cos \text{lat} 0 }\Big $$ Every one of those points gets sampled against real 30m Copernicus DEM tiles . Copernicus DEM GLO-30 , published by the EU's Copernicus program, hosted as free public Cloud-Optimized GeoTIFFs on AWS Open Data, no account or key needed. For more info, you can view https://registry.opendata.aws/copernicus-dem/ https://registry.opendata.aws/copernicus-dem/ Finally, those two simple heuristic conditions decide survival yea I know, everything became heuristic haha : $$ \text{elev} P 0 \le 50\text{m} $$ $$ 100\text{m} \le \max {\text{arc}} \text{elev} \le 500\text{m} $$ You can see from this abstract graph image, the dashed line is the camera's front bearing, the wedge is the ±50° search arc swept out to 20km for the elevation check. 26/66 survive the elevation check. You can see the 26 survivors, all are located in southern Asia, Australia and Oceania, except one near Brazil h Final Report Finally, last stage, it just makes the final candidates checkable by eye. Each survivor gets its country name via a point in polygon lookup against a country boundary file, then a direct Google Maps satellite link for P0, P1, and P2. Output is a plain HTML table, index, country, three clickable coordinate pairs per row. I got this final list, lets check each one by eye. Won't go one by one here, but those first 7 are totally off for me. Till I opened that 8th one in the table of country of Micronesia 😍 first time to know that a country named Micronesia : and ensured through P1 and P2: and that is the solution 🥳 ... you can view it here on google maps https://www.google.com/maps/@7.3633,151.755983,50m/data= 3m1 1e3 i FINALLY, ANSWERS ... a What is the name of the resort? $$ \text{Oan} $$ b What are the coordinates of the island? $$7^\circ\,21^\prime\,48.4^{\prime\prime}\,\text{N} \qquad 151^\circ\,45^\prime\,20.7^{\prime\prime}\,\text{E}$$ $$ \text{or} $$ $$7.363444^\circ,\ 151.755750^\circ$$ c In which cardinal direction was the camera facing when the photo was taken? $$ \because\quad \theta = \text{atan2}\Big \sin \Delta\lambda \cos\phi 1,\ \cos\phi 0\sin\phi 1 - \sin\phi 0\cos\phi 1\cos \Delta\lambda \Big $$ $$ P 0 = 7.3633,\ 151.755983 , \quad P 1 = 7.386573,\ 151.739534 $$ $$ \therefore\quad \theta = 324.97^\circ \implies \textbf{NW} $$ j Data & Licenses Coastline polygons : land-polygons-split-4326 https://osmdata.openstreetmap.de/data/land-polygons.html © OpenStreetMap contributors, available under the Open Database License ODbL 1.0 https://opendatacommons.org/licenses/odbl/ . The candidate sets and final report in the repo are a Derived Database and are published under the same license. Elevation : Copernicus DEM GLO-30. © DLR e.V. 2010-2014 and © Airbus Defence and Space GmbH 2014-2018 provided under COPERNICUS by the European Union and ESA; all rights reserved. Satellite imagery : Contains modified Copernicus Sentinel data 2025-2026, accessed through Earth Search https://earth-search.aws.element84.com/v1 by Element 84 on AWS Open Data. Country boundaries : Natural Earth https://www.naturalearthdata.com/ 10m admin-0, public domain. Challenge & source photo : OSINT Exercise 004 https://gralhix.com/list-of-osint-exercises/osint-exercise-004/ by Sofia Santos gralhix https://gralhix.com/ . Satellite screenshots in section h are from Google Maps / Google Earth