Moving the Origin

If p and q are two points in a Euclidean space, then q − p is a vector in a vector space of as many dimensions as the Euclidean space.
E is the set of points in the Euclidean space; V is the set of vectors.
+ : E×V → E
− : E×E → V
1 e − e = 0
2 e + (v₁ + v₂) = (e + v₁) + v₂
3 If a, b, c, d ∊ E and they are points around a parallelogram, then a − b = d − c.
4 (e₁ − e₂) + (e₂ − e₃) = (e₁ − e₃)
5 {2} ⊢ (e + v) − v = e

We take up this question.

In n space we have n points {p_i}. We have as well points x and y. Neither {x}∪{p_i} nor {y}∪{p_i} lie in any n−1 dimensional subspace. The n vectors p_i − x form a basis for V; likewise the set {p_i − y}. We may take x as an origin for a coordinate system in E such that the jth coordinate of p_i is δ_ij. Taking y as origin in place of x, we have another coordinate system. Given the coordinates of y in x’s system, what are the coordinates of x in y’s system?

Since for any given n+1 independent points and any other given n+1 points, there exists an affine transformation the maps the first onto the second, the answer does not depend on the locations of the p’s.

Let x_i be the coordinates of x in the y system and y_i be the coordinates of y in the x system. We have the vector equations
x = y + Σx_i(p_i − y)
y = x + Σy_i(p_i − x)
We want to express each x_i in terms of the y_i’s. Substituting x we have:
y = Σx_i(p_i − y) + y + Σy_i(p_i − x)
Σx_i(p_i − y) + Σy_i(p_i − x) = 0
Σx_ip_i + Σy_ip_i = xΣy_i + yΣx_i (type error!)

Another start to this problem. In the zbcc of this zone we compute the location of the far vertex V of the zone across the facet opposite the origin of this zone. Let v_i name the n zbc coordinates of V. We have as direct data d_k, the sd (squared distance) of V from the ends of each of the basis vectors. This gives us n equations, one for each value of k:
Σ_ijg_ij(v_i − δ_ik)(v_j − δ_jk) = d_k
Σ_ijg_ijv_iv_j − Σ_ij(g_ijv_iδ_jk) − Σ_ij(g_ijv_jδ_ik) + Σ_ij(g_ijδ_ikδ_jk) = d_k
Σ_ijg_ijv_iv_j − Σ_i(g_ikv_i) − Σ_j(g_kjv_j) + g_kk = d_k
Σ_ijg_ijv_iv_j − 2 Σ_i(g_ikv_i) = d_k − g_kk

let ts g v d =
let dl i j = if i=j then 1. else 0.
and aini = Array.init and len = Array.length in
let aa a b = aini (len a) (fun j -> a.(j) +. b.(j)) in
let n = len g in
let fd = (aini n (fun _ -> let s = ref 0. in for i=0 to n-1 do for j=0 to n-1 do
    s := !s +. g.(i).(j) *. v.(i) *. v.(j) done done; !s)) in
(aini n (fun k -> let s = ref 0. in for i=0 to n-1 do for j=0 to n-1 do
  s := !s +. g.(i).(j) *. (v.(i) -. (dl i k)) *. (v.(j) -. (dl j k)) done done; !s) = d,

(aa (aa fd (aini n (fun k -> let s = ref 0. in for i=0 to n-1 do for j=0 to n-1 do
    s := !s -. g.(i).(j) *. v.(i) *. (dl j k) done done; !s)))
  (aa (aini n (fun k -> let s = ref 0. in for i=0 to n-1 do for j=0 to n-1 do
    s := !s -. g.(i).(j) *. v.(j) *. (dl i k) done done; !s))
    (aini n (fun k -> let s = ref 0. in for i=0 to n-1 do for j=0 to n-1 do
    s := !s +. g.(i).(j) *. (dl i k) *. (dl j k) done done; !s)))) = d,

(aa (aa fd (aini n (fun k -> let s = ref 0. in for i=0 to n-1 do
   s := !s -. g.(i).(k) *. v.(i) done; !s)))
(aa (aini n (fun k -> let s = ref 0. in for j=0 to n-1 do
   s := !s -. g.(k).(j) *. v.(j) done; !s))
   (aini n (fun k -> g.(k).(k))))) = d,

(aa fd
  (aini n (fun k -> let s = ref 0. in for i=0 to n-1 do
    s := !s +. -2. *. g.(i).(k) *. v.(i) done; !s)))
= aini n (fun k -> d.(k) -. g.(k).(k)))
in [|ts [|[|1.; 0.|]; [|0.; 1.|]|] [|1.; 1.|] [|1.; 1.|];
    ts [|[|1.; 0.|]; [|0.; 1.|]|] [|2.; 2.|] [|5.; 5.|];
    ts [|[|1.; 0.|]; [|0.; 1.|]|] [|2.; 1.|] [|2.; 4.|];
    ts [|[|4.; 2.|]; [|2.; 2.|]|] [|0.5; 1.|] [|1.; 1.|];
    ts [|[|45.; 36.|]; [|36.; 45.|]|] [|2.; 2.|] [|9.; 9.|]|];;
--------
vertices at  (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, -1)
zones at (0, 1, 2, 3), (0, 1, 2, 4)
complementary coordinates: (2, 0, 0), (2, 0, 0)

See Lagrange multipliers!