-/* nj.c 2009-07-09 */
+/* nj.c 2011-10-20 */
-/* Copyright 2006-2009 Emmanuel Paradis
+/* Copyright 2006-2011 Emmanuel Paradis
/* This file is part of the R-package `ape'. */
/* See the file ../COPYING for licensing issues. */
-#include <R.h>
-
-#define DINDEX(i, j) n*(i - 1) - i*(i - 1)/2 + j - i - 1
-
-int give_index(int i, int j, int n)
-{
- if (i > j) return(DINDEX(j, i));
- else return(DINDEX(i, j));
-}
+#include "ape.h"
double sum_dist_to_i(int n, double *D, int i)
/* returns the sum of all distances D_ij between i and j
5 3 7 10 12
6 4 8 11 13 14
- so that we sum the values of the ith column-1st loop-and those of
- (i - 1)th row (labelled 'i')-2nd loop */
+ so that we sum the values of the ith column--1st loop--and those of
+ (i - 1)th row (labelled 'i')--2nd loop */
double sum = 0;
int j, start, end;
return(sum);
}
-#define GET_I_AND_J \
-/* Find the 'R' indices of the two corresponding OTUs */ \
-/* The indices of the first element of the pair in the \
- distance matrix are n-1 times 1, n-2 times 2, n-3 times 3, \
- ..., once n-1. Given this, the algorithm below is quite \
- straightforward.*/ \
- i = 0; \
- for (OTU1 = 1; OTU1 < n; OTU1++) { \
- i += n - OTU1; \
- if (i >= smallest + 1) break; \
- } \
- /* Finding the second OTU is easier! */ \
- OTU2 = smallest + 1 + OTU1 - n*(OTU1 - 1) + OTU1*(OTU1 - 1)/2
-
-#define SET_CLADE \
-/* give the node and tip numbers to edge */ \
- edge2[k] = otu_label[OTU1 - 1]; \
- edge2[k + 1] = otu_label[OTU2 - 1]; \
- edge1[k] = edge1[k + 1] = cur_nod
-
void nj(double *D, int *N, int *edge1, int *edge2, double *edge_length)
{
- double SUMD, *S, Sdist, Ndist, *new_dist, A, B, *DI, d_i, x, y;
+ double *S, Sdist, Ndist, *new_dist, A, B, smallest_S, x, y;
int n, i, j, k, ij, smallest, OTU1, OTU2, cur_nod, o_l, *otu_label;
S = &Sdist;
new_dist = &Ndist;
otu_label = &o_l;
- DI = &d_i;
n = *N;
cur_nod = 2*n - 2;
- S = (double*)R_alloc(n*(n - 1)/2, sizeof(double));
+ S = (double*)R_alloc(n + 1, sizeof(double));
new_dist = (double*)R_alloc(n*(n - 1)/2, sizeof(double));
- otu_label = (int*)R_alloc(n, sizeof(int));
- DI = (double*)R_alloc(n - 2, sizeof(double));
+ otu_label = (int*)R_alloc(n + 1, sizeof(int));
+
+ for (i = 1; i <= n; i++) otu_label[i] = i; /* otu_label[0] is not used */
- for (i = 0; i < n; i++) otu_label[i] = i + 1;
k = 0;
while (n > 3) {
- SUMD = 0;
- for (i = 0; i < n*(n - 1)/2; i++) SUMD += D[i];
+ for (i = 1; i <= n; i++)
+ S[i] = sum_dist_to_i(n, D, i); /* S[0] is not used */
ij = 0;
+ smallest_S = 1e50;
+ B = n - 2;
for (i = 1; i < n; i++) {
for (j = i + 1; j <= n; j++) {
- A = sum_dist_to_i(n, D, i) - D[ij];
- B = sum_dist_to_i(n, D, j) - D[ij];
- S[ij] = (A + B)/(2*n - 4) + 0.5*D[ij]
- + (SUMD - A - B - D[ij])/(n - 2);
+ A = B*D[ij] - S[i] - S[j];
+ if (A < smallest_S) {
+ OTU1 = i;
+ OTU2 = j;
+ smallest_S = A;
+ smallest = ij;
+ }
ij++;
}
}
- /* find the 'C' index of the smallest value of S */
- smallest = 0;
- for (i = 1; i < n*(n - 1)/2; i++)
- if (S[smallest] > S[i]) smallest = i;
-
- GET_I_AND_J;
- SET_CLADE;
+ edge2[k] = otu_label[OTU1];
+ edge2[k + 1] = otu_label[OTU2];
+ edge1[k] = edge1[k + 1] = cur_nod;
/* get the distances between all OTUs but the 2 selected ones
and the latter:
a) get the sum for both
b) compute the distances for the new OTU */
- A = B = ij = 0;
+
+ A = D[smallest];
+ ij = 0;
for (i = 1; i <= n; i++) {
if (i == OTU1 || i == OTU2) continue;
x = D[give_index(i, OTU1, n)]; /* dist between OTU1 and i */
y = D[give_index(i, OTU2, n)]; /* dist between OTU2 and i */
- new_dist[ij] = (x + y)/2;
- A += x;
- B += y;
+ new_dist[ij] = (x + y - A)/2;
ij++;
}
/* compute the branch lengths */
- A /= n - 2;
- B /= n - 2;
- edge_length[k] = (D[smallest] + A - B)/2;
- edge_length[k + 1] = (D[smallest] + B - A)/2;
- DI[cur_nod - *N - 1] = D[smallest];
-
- /* update before the next loop */
- if (OTU1 > OTU2) { /* make sure that OTU1 < OTU2 */
- i = OTU1;
- OTU1 = OTU2;
- OTU2 = i;
- }
+ B = (S[OTU1] - S[OTU2])/B; /* don't need B anymore */
+ edge_length[k] = (A + B)/2;
+ edge_length[k + 1] = (A - B)/2;
+
+ /* update before the next loop
+ (we are sure that OTU1 < OTU2) */
if (OTU1 != 1)
- for (i = OTU1 - 1; i > 0; i--) otu_label[i] = otu_label[i - 1];
+ for (i = OTU1; i > 1; i--)
+ otu_label[i] = otu_label[i - 1];
if (OTU2 != n)
- for (i = OTU2; i <= n; i++) otu_label[i - 1] = otu_label[i];
- otu_label[0] = cur_nod;
+ for (i = OTU2; i < n; i++)
+ otu_label[i] = otu_label[i + 1];
+ otu_label[1] = cur_nod;
for (i = 1; i < n; i++) {
if (i == OTU1 || i == OTU2) continue;
for (i = 0; i < 3; i++) {
edge1[*N*2 - 4 - i] = cur_nod;
- edge2[*N*2 - 4 - i] = otu_label[i];
+ edge2[*N*2 - 4 - i] = otu_label[i + 1];
}
edge_length[*N*2 - 4] = (D[0] + D[1] - D[2])/2;
edge_length[*N*2 - 5] = (D[0] + D[2] - D[1])/2;
edge_length[*N*2 - 6] = (D[2] + D[1] - D[0])/2;
-
- for (i = 0; i < *N*2 - 3; i++) {
- if (edge2[i] <= *N) continue;
- /* In case there are zero branch lengths: */
- if (DI[edge2[i] - *N - 1] == 0) continue;
- edge_length[i] -= DI[edge2[i] - *N - 1]/2;
- }
}