197 lines
6.2 KiB
C
197 lines
6.2 KiB
C
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/*
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THE COMPUTER CODE CONTAINED HEREIN IS THE SOLE PROPERTY OF PARALLAX
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SOFTWARE CORPORATION ("PARALLAX"). PARALLAX, IN DISTRIBUTING THE CODE TO
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END-USERS, AND SUBJECT TO ALL OF THE TERMS AND CONDITIONS HEREIN, GRANTS A
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ROYALTY-FREE, PERPETUAL LICENSE TO SUCH END-USERS FOR USE BY SUCH END-USERS
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IN USING, DISPLAYING, AND CREATING DERIVATIVE WORKS THEREOF, SO LONG AS
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SUCH USE, DISPLAY OR CREATION IS FOR NON-COMMERCIAL, ROYALTY OR REVENUE
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FREE PURPOSES. IN NO EVENT SHALL THE END-USER USE THE COMPUTER CODE
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CONTAINED HEREIN FOR REVENUE-BEARING PURPOSES. THE END-USER UNDERSTANDS
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AND AGREES TO THE TERMS HEREIN AND ACCEPTS THE SAME BY USE OF THIS FILE.
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COPYRIGHT 1993-1998 PARALLAX SOFTWARE CORPORATION. ALL RIGHTS RESERVED.
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*/
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/*
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*
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* Functions to make faces planar and probably other things.
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*
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <math.h>
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#include <string.h>
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#include "key.h"
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#include "gr.h"
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#include "inferno.h"
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#include "segment.h"
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#include "editor.h"
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#include "error.h"
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#include "gameseg.h"
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#define SWAP(a,b) {temp = (a); (a) = (b); (b) = temp;}
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// -----------------------------------------------------------------------------------------------------------------
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// Gauss-Jordan elimination solution of a system of linear equations.
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// a[1..n][1..n] is the input matrix. b[1..n][1..m] is input containing the m right-hand side vectors.
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// On output, a is replaced by its matrix inverse and b is replaced by the corresponding set of solution vectors.
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void gaussj(fix **a, int n, fix **b, int m)
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{
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int indxc[4], indxr[4], ipiv[4];
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int i, icol=0, irow=0, j, k, l, ll;
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fix big, dum, pivinv, temp;
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if (n > 4) {
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Int3();
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}
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for (j=1; j<=n; j++)
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ipiv[j] = 0;
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for (i=1; i<=n; i++) {
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big = 0;
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for (j=1; j<=n; j++)
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if (ipiv[j] != 1)
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for (k=1; k<=n; k++) {
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if (ipiv[k] == 0) {
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if (abs(a[j][k]) >= big) {
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big = abs(a[j][k]);
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irow = j;
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icol = k;
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}
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} else if (ipiv[k] > 1) {
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Int3();
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}
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}
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++(ipiv[icol]);
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// We now have the pivot element, so we interchange rows, if needed, to put the pivot
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// element on the diagonal. The columns are not physically interchanged, only relabeled:
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// indxc[i], the column of the ith pivot element, is the ith column that is reduced, while
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// indxr[i] is the row in which that pivot element was originally located. If indxr[i] !=
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// indxc[i] there is an implied column interchange. With this form of bookkeeping, the
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// solution b's will end up in the correct order, and the inverse matrix will be scrambled
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// by columns.
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if (irow != icol) {
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for (l=1; l<=n; l++)
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SWAP(a[irow][l], a[icol][l]);
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for (l=1; l<=m; l++)
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SWAP(b[irow][l], b[icol][l]);
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}
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indxr[i] = irow;
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indxc[i] = icol;
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if (a[icol][icol] == 0) {
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Int3();
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}
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pivinv = fixdiv(F1_0, a[icol][icol]);
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a[icol][icol] = F1_0;
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for (l=1; l<=n; l++)
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a[icol][l] = fixmul(a[icol][l], pivinv);
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for (l=1; l<=m; l++)
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b[icol][l] = fixmul(b[icol][l], pivinv);
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for (ll=1; ll<=n; ll++)
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if (ll != icol) {
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dum = a[ll][icol];
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a[ll][icol] = 0;
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for (l=1; l<=n; l++)
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a[ll][l] -= a[icol][l]*dum;
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for (l=1; l<=m; l++)
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b[ll][l] -= b[icol][l]*dum;
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}
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}
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// This is the end of the main loop over columns of the reduction. It only remains to unscramble
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// the solution in view of the column interchanges. We do this by interchanging pairs of
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// columns in the reverse order that the permutation was built up.
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for (l=n; l>=1; l--) {
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if (indxr[l] != indxc[l])
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for (k=1; k<=n; k++)
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SWAP(a[k][indxr[l]], a[k][indxc[l]]);
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}
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}
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// -----------------------------------------------------------------------------------------------------------------
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// Return true if side is planar, else return false.
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int side_is_planar_p(segment *sp, int side)
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{
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sbyte *vp;
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vms_vector *v0,*v1,*v2,*v3;
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vms_vector va,vb;
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vp = Side_to_verts[side];
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v0 = &Vertices[sp->verts[vp[0]]];
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v1 = &Vertices[sp->verts[vp[1]]];
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v2 = &Vertices[sp->verts[vp[2]]];
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v3 = &Vertices[sp->verts[vp[3]]];
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vm_vec_normalize(vm_vec_normal(&va,v0,v1,v2));
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vm_vec_normalize(vm_vec_normal(&vb,v0,v2,v3));
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// If the two vectors are very close to being the same, then generate one quad, else generate two triangles.
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return (vm_vec_dist(&va,&vb) < F1_0/1000);
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}
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// -------------------------------------------------------------------------------------------------
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// Return coordinates of a vertex which is vertex v moved so that all sides of which it is a part become planar.
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void compute_planar_vert(segment *sp, int side, int v, vms_vector *vp)
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{
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if ((sp) && (side > -3))
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*vp = Vertices[v];
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}
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// -------------------------------------------------------------------------------------------------
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// Making Cursegp:Curside planar.
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// If already planar, return.
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// for each vertex v on side, not part of another segment
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// choose the vertex v which can be moved to make all sides of which it is a part planar, minimizing distance moved
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// if there is no vertex v on side, not part of another segment, give up in disgust
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// Return value:
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// 0 curside made planar (or already was)
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// 1 did not make curside planar
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int make_curside_planar(void)
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{
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int v;
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sbyte *vp;
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vms_vector planar_verts[4]; // store coordinates of up to 4 vertices which will make Curside planar, corresponding to each of 4 vertices on side
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int present_verts[4]; // set to 1 if vertex is present
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if (side_is_planar_p(Cursegp, Curside))
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return 0;
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// Look at all vertices in side to find a free one.
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for (v=0; v<4; v++)
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present_verts[v] = 0;
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vp = Side_to_verts[Curside];
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for (v=0; v<4; v++) {
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int v1 = vp[v]; // absolute vertex id
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if (is_free_vertex(Cursegp->verts[v1])) {
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compute_planar_vert(Cursegp, Curside, Cursegp->verts[v1], &planar_verts[v]);
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present_verts[v] = 1;
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}
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}
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// Now, for each v for which present_verts[v] == 1, there is a vector (point) in planar_verts[v].
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// See which one is closest to the plane defined by the other three points.
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// Nah...just use the first one we find.
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for (v=0; v<4; v++)
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if (present_verts[v]) {
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med_set_vertex(vp[v],&planar_verts[v]);
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validate_segment(Cursegp);
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// -- should propagate tmaps to segments or something here...
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return 0;
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}
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// We tried, but we failed, to make Curside planer.
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return 1;
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}
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