You are given an array/list/vector of pairs of integers representing cartesian coordinates (x, y) of points on a 2D Euclidean plane; all coordinates are between −10⁴ and 10⁴, duplicates are allowed. Find the area of the convex hull of those points, rounded to the nearest integer; an exact midpoint should be rounded to the closest even integer. You may use floating-point numbers in intermediate computations, but only if you can guarantee that the final result will be always correct. This is a code-golf, the shortest correct program (ignoring non-significant whitespace, newlines and comments) wins.

Some test cases:

Input: [[50, -13]]
Result: 0

Input: [[-25, -26], [34, -27]]
Result: 0

Input: [[-6, -14], [-48, -45], [21, 25]]
Result: 400

Input: [[4, 30], [5, 37], [-18, 49], [-9, -2]]
Result: 562

Input: [[0, 16], [24, 18], [-43, 36], [39, -29], [3, -38]]
Result: 2978

Input: [[19, -19], [15, 5], [-16, -41], [6, -25], [-42, 1], [12, 19]]
Result: 2118

Input: [[-23, 13], [-13, 13], [-6, -7], [22, 41], [-26, 50], [12, -12], [-23, -7]]
Result: 2307

Input: [[31, -19], [-41, -41], [25, 34], [29, -1], [42, -42], [-34, 32], [19, 33], [40, 39]]
Result: 6037

Input: [[47, 1], [-22, 24], [36, 38], [-17, 4], [41, -3], [-13, 15], [-36, -40], [-13, 35], [-25, 22]]
Result: 3908

Input: [[29, -19], [18, 9], [30, -46], [15, 20], [24, -4], [5, 19], [-44, 4], [-20, -8], [-16, 34], [17, -36]]
Result: 2905

Java 10, 405 ...didn't fit anymore; see edit history.. 317 316 bytes

P->int n=P.length,l=0,i=0,p,q,t[],h[][]=P.clone(),s=0;for(;++i<n;)l=P[i][0]<P[l][0]?i:l;p=l;do for(h[s++]=P[p],q=-~p%n,i=-1;++i<n;q=(t[1]-P[p][1])*(P[q][0]-t[0])<(t[0]-P[p][0])*(P[q][1]-t[1])?i:q)t=P[i];while((p=q)!=l);for(p=i=0;i<s;p-=(t[0]+h[++i%s][0])*(t[1]-h[i%s][1]))t=h[i];return Math.round(.5*p/~(p%=2))*~p;

-52 bytes thanks to @OlivierGrégoire

-3 bytes thanks to @PeterTaylor

-7 bytes thanks to @ceilingcat

Try it online.

Or 299 bytes without rounding...

Explanation:

There are three steps to do:

1. Calculate the points for the Convex Hull based on the input-coordinates (using Jarvis' Algorithm/Wrapping)


2. Calculate the area of this Convex Hull


3. Banker's rounding..

To calculate the coordinates that are part of the Convex Hull, we use the following approach:

Set point $l$ and $p$ to the left-most coordinate. Then calculate the next point $p$ in a counterclockwise rotation; and continue doing so until we've reached back at the initial point $l$. Here a visual for this:

As for the code:

P-> // Method with 2D integer array as parameter & long return-type
 int n=P.length, // Integer `n`, the amount of points in the input
 l=0, // Integer `l`, to calculate the left-most point
 i=0, // Index-integer `i`
 p, // Integer `p`, which will be every next counterclockwise point
 q, // Temp integer `q`
 t[], // Temp integer-array/point
 h[][]=P.clone(), // Initialize an array of points `h` for the Convex Hull
 s=0; // And a size-integer for this Convex Hull array, starting at 0
 for(;++i<n;) // Loop `i` in the range [1, `n`):
 l= // Change `l` to:
 P[i][0]<P[l][0]? // If i.x is smaller than l.x:
 i // Replace `l` with the current `i`
 :l; // Else: leave `l` unchanged
 p=l; // Now set `p` to this left-most coordinate `l`
 do // Do:
 for(h[s++]=P[p], // Add the `p`'th point to the 2D-array `h`
 q=-~p%n, // Set `q` to `(p+1)` modulo-`n`
 i=-1;++i<n; // Loop `i` in the range [0, `n`):
 ;q= // After every iteration: change `q` to:
 // We calculate: (i.y-p.y)*(q.x-i.x)-(i.x-p.x)*(q.y-i.y), 
 // which results in 0 if the three points are collinear;
 // a positive value if they are clockwise;
 // or a negative value if they are counterclockwise
 (t[1]-P[p][1])*(P[q][0]-t[0])<(t[0]-P[p][0])*(P[q][1]-t[1])?
 // So if the three points are counterclockwise:
 i // Replace `q` with `i`
 :q) // Else: leave `q` unchanged
 t=P[i]; // Set `t` to the `i`'th Point (to save bytes)
 while((p=q) // And after every while-iteration: replace `p` with `q`
 !=l); // Continue the do-while as long as `p` is not back at the
 // left-most point `l` yet
 // Now step 1 is complete, and we have our Convex Hull points in the List `h`

 for(p=i=0; // Set `p` (the area) to 0
 i<s // Loop `i` in the range [0, `s`):
 ;p-= // After every iteration: Decrease the area `p` by:
 (t[0]+h[++i%s][0])// i.x+(i+1).x
 *(t[1]-h[i%s][1]))// Multiplied by i.y-(i+1).y
 t=h[i]; // Set `t` to the `i`'th point (to save bytes)
 return Math.round(.5*p/~(p%=2))*~p;
 // And return `p/2` rounded to integer with half-even

SQL Server 2012+, 84 bytes

SELECT Round(Geometry::ConvexHullAggregate(Geometry::Point(x,y,0)).STArea(),0)FROM A

Makes use of the geometry functions and aggregates in SQL Server.
Coordindates are from table A with columns x and y.

Wolfram Language (Mathematica), 27 bytes

Round@*Area@*ConvexHullMesh

Try it online!

JavaScript (ES6),  191  189 bytes

Implements the Jarvis march (aka gift wrapping algorithm).

P=>(r=(g=p=>([X,Y]=P[p],Y*h-X*v)+(P.map(([x,y],i)=>q=(y-Y)*(P[q][0]-x)<(x-X)*(P[q][1]-y)?i:q,q=P[++p]?p:0,h=X,v=Y)|q?g(q):V*h-H*v))(v=h=0,([[H,V]]=P.sort(([x],[X])=>x-X)))/2)+(r%1&&r&1)/2|0

Try it online!

Or 170 bytes without the cumbersome rounding scheme.

R, 85 81 78 bytes

function(i,h=chull(i),j=c(h,h[1]))round((i[h,1]+i[j[-1],1])%*%diff(-i[j,2])/2)

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Takes input as a 2-column matrix - first for x, second for y. R's round actually uses banker's rounding method, so we are quite lucky here.

The code uses a built-in function to determine, which points form the convex hull, and then applies the standard formula $sum_i(x_i-1+x)cdot(y_i-1-y_i)/2$ to get the polygon surface area.

Thanks to Giuseppe for -3 bytes.

[R + sp package], 55 bytes

function(x)round(sp::Polygon(x[chull(x),,drop=F])@area)

Try it at RDRR

A function which takes a n x 2 matrix and returns the rounded area. This uses the sp package. The drop=F is needed to handle the one co-ordinate case. RDRR used for demo since TIO lacks the sp package.