\datethis
@*Intro. This program is a revision of {\mc ACHAIN1}, which you should
read first. I'm thinking that a few changes will speed that program up,
but as usual the proof of the pudding is in the eating.
The main changes here are:
(i)~If $a[j]=b[j]=2^j+2^k,ドル where $k<j-3,ドル one can prove that $a[j-k-2]$ and $b[j-k-2]$ must both be 2ドル^{j-k-2}$. This additional constraint makes more chain values ``exist'' at early stages. (ii)~Every addition chain corresponds to a directed graph, and many different addition chains can correspond to the same ``reduced'' digraph, as explained at the end of Section 4.6.3. Therefore I've worked out a scheme by which only one chain of each equivalence class is explored. These changes, and the various changes that distinguish {\mc ACHAIN1} from {\mc ACHAIN0}, are fairly independent. If I had time, I could therefore experiment with various subsets, in order to see which of them are really worth the effort. Who knows, maybe some of them actually cause a slowdown, taken individually. @d nmax 10000000 /* should be less than 2ドル^{24}$ on a 32-bit machine */ @c #include <stdio.h>
#include <stdlib.h>
#include <time.h>
char l[nmax];
int a[128],b[128];
unsigned int undo[128*128];
int ptr; /* this many items of the &#124;undo&#124; stack are in use */
struct {
 int lbp,ubp,lbq,ubq,r,ptrp,ptrq;
} stack[128];
int tail[128],outdeg[128],outsum[128],limit[128];
FILE *infile, *outfile;
int prime[1000]; /* 1000 primes will take us past 60 million */
int pr; /* the number of primes known so far */
char x[64]; /* exponents of the binary representation of $n,ドル less 1 */
int main(int argc, char* argv[])
{
 register int i,j,n,p,q,r,s,ubp,ubq,lbp,lbq,ptrp,ptrq;
 int lb,ub,timer=0;
 @<process the command line@>;
 prime[0]=2, pr=1;
 a[0]=b[0]=1, a[1]=b[1]=2; /* an addition chain always begins like this */
 for (n=1;n<nmax;n++) { @<input the next lower bound, &#124;lb&#124;@>;
 @<find an upper bound; or in simple cases, set $l(n)$ and &#124;goto done&#124;@>;
 @<backtrack until $l(n)$ is known@>;
done: @<output the value of $l(n)$@>;
 if (n%1000==0) {
 j=clock(); 
 printf("%d..%d done in %.5g minutes\n",
 n-999,n,(double)(j-timer)/(60*CLOCKS_PER_SEC));
 timer=j; 
 }
 }
}
@ @<process the command line@>=
if (argc!=3) {
 fprintf(stderr,"Usage: %s infile outfile\n",argv[0]);
 exit(-1);
}
infile=fopen(argv[1],"r");
if (!infile) {
 fprintf(stderr,"I couldn't open `%s' for reading!\n",argv[1]);
 exit(-2);
}
outfile=fopen(argv[2],"w");
if (!outfile) {
 fprintf(stderr,"I couldn't open `%s' for writing!\n",argv[2]);
 exit(-3);
}
@ @<output the value of $l(n)$@>=
fprintf(outfile,"%c",l[n]+' ');
fflush(outfile); /* make sure the result is viewable immediately */
@ At this point I compute the ``lower bound'' $\lfloor\lg n\rfloor+3,ドル
which is valid if $\nu n>4$. Simple cases where $\nu n\le 4$ will be
handled separately below.
@<input the next lower bound, &#124;lb&#124;@>=
for (q=n,i=-1,j=0;q;q>>=1,i++)
 if (q&1) x[j++]=i;
/* now $i=\lfloor\lg n\rfloor$ and $j=\nu n$ */
lb=fgetc(infile)-' '; /* &#124;fgetc&#124; will return a negative value after EOF */
if (lb<i+3) lb=i+3; @ Three elementary and well-known upper bounds are considered: (i)~$l(n)\le\lfloor\lg n\rfloor+\nu n-1$; (ii)~$l(n)\le l(n-1)+1$; (iii)~$l(n)\le l(p)+l(q)$ if $n=pq$. Furthermore, there are four special cases when Theorem 4.6.3C tells us we can save a step. In this regard, I had to learn (the hard way) to avoid a curious bug: Three of the four cases in Theorem 4.6.3C arise when we factor $n,ドル so I thought I needed to test only the other case here. But I got a surprise when $n=165$: Then $n=3\cdot55,ドル so the factor method gave the upper bound $l(3)+l(55)=10$; but another factorization, $n=5\cdot33,ドル gives the better bound $l(5)+l(33)=9$. @<find an upper bound...@>=
ub=i+j-1;
if (ub>l[n-1]+1) ub=l[n-1]+1;
@<div class="naked_tr">;
l[n]=ub;
if (j<=3) goto done; if (j==4) { p=x[3]-x[2], q=x[1]-x[0]; if (p==q &#124;&#124; p==q+1 &#124;&#124; (q==1 && (p==3 &#124;&#124; (p==5 && x[2]==x[1]+1)))) l[n]=i+2; goto done; } @ It's important to try the factor method even when &#124;j<=4&#124;, because of the way prime numbers are recognized here: We would miss the prime~3, for example. On the other hand, we don't need to remember large primes that will never arise as factors of any future~$n$. @<div class="naked_tr">=
if (n>2) for (s=0;;s++) {
 p=prime[s];
 q=n/p;
 if (n==p*q) {
 if (l[p]+l[q]<ub) ub=l[p]+l[q]; break; } if (q<=p) { /* $n$ is prime */ if (pr<1000) prime[pr++]=n; break; } } @*The interesting part. Nontrivial changes begin to occur when we get into the guts of the backtracking structure carried over from the previous versions of this program, but the controlling loop at the outer level remains intact. @<backtrack until $l(n)$ is known@>=
l[n]=lb;
while (lb<ub) { for (i=0;i<=lb;i++) outdeg[i]=outsum[i]=0; a[lb]=b[lb]=n; for (i=2;i<lb;i++) a[i]=a[i-1]+1, b[i]=b[i-1]<<1; for (i=lb-1;i>=2;i--) {
 if ((a[i]<<1)<a[i+1]) a[i]=(a[i+1]+1)>>1;
 if (b[i]>=b[i+1]) b[i]=b[i+1]-1;
 }
 @<div class="naked_tr">;
 l[n]=++lb;
}
@ The main new idea implemented below is related to the reduced digraph
representation of an addition chain, in which each element $a_s$
of the chain is effectively expressed as a sum of two or
more previous elements. For example, we might have $a_s=a_i+a_j+a_k+a_l$
where $i\le j\le k\le l,ドル represented by four arrows coming in to
node $a_s$ from nodes $a_i,ドル $a_j,ドル $a_k,ドル and $a_l$. The
colexicographically largest chain with this reduced digraph has
elements $a_k+a_l,ドル $a_j+a_k+a_l,ドル and $a_i+a_j+a_k+a_l,ドル and
we want to avoid redundant work by restricting consideration to such chains.
Fortunately there's an easy way to do this: For each chain element~$x$
we remember its ``tail,'' which is the smallest node that points to~$x$;
and we also keep track of the outdegree of each fixed element. If
$a_s=p+q$ with $p\ge q,ドル the tail of~$a_s$ is~$q$. And if $a_s$ has
outdegree~1 because it is used only as an input to $a_t,ドル we
insist that $q$ be at least as large as the tail of~$a_t$.
Array element &#124;outdeg[j]&#124; is the current number of elements
$a_{s+1},ドル $a_{s+2},ドル \dots, that make use of $a_j,ドル and
array element &#124;outsum[j]&#124; is the sum of all their subscripts.
These two arrays are easily updated as we move forward and backward
while backtracking; and &#124;outsum[j]&#124; turns out to be exactly the
value~&#124;t&#124; that we need to know when &#124;outdeg[j]=1&#124;.
(Has anybody seen this idea before? At my age I thought I had
seen all such simple tricks long ago, but maybe there still
are many more waiting to be discovered.)
(Note added 08 Oct 2005: I just found this idea attributed to
David Wood, by Eric Bach and Marcos Kiwi in {\sl Theoretical
Computer Science\/ \bf235} (2000), 5. Bach and Kiwi go on
to generalize the idea so that more than one element can be
identified, using power sums.)
One consequence: If &#124;outdeg[s-1]==0&#124;, we must either have
$a[s-1]={1\over2}a[s]$ or $a[s-1]\ge{2\over3}a[s]$. Because
after setting &#124;a[s]&#124; to a value other than ${1\over2}a[s]$
we will have &#124;outdeg[s-1]=1&#124;, and 
${1\over2}a[s-1]\ge&#124;tail&#124;[s-1]\ge&#124;tail&#124;[s]=a[s]-a[s-1]$.
@ We maintain a stack of subproblems, as usual when backtracking.
Suppose &#124;a[t]&#124; is the sum of two items already present, for all
$t>s$; we want to make sure that &#124;a[s]&#124; is legitimate too.
For this purpose we try all combinations &#124;a[s]=p+q&#124; where
$p\ge a[s]/2,ドル trying to make both $p$ and $q$ present.
(By the nature of the algorithm, we'll have &#124;a[s]=b[s]&#124; at
the time we choose &#124;p&#124; and~&#124;q&#124;, because shorter addition
chains have been ruled out.)
As elements of &#124;a&#124; and &#124;b&#124; are changed, we record their previous
values on the &#124;undo&#124; stack, so that we can easily restore them
later. Pointers &#124;ptrp&#124; and &#124;ptrq&#124; contain the limiting indexes
for undo information.
@<div class="naked_tr">=
ptr=0; /* clear the &#124;undo&#124; stack */
for (r=s=lb;s>2;s--) {
 if (outdeg[s]==1) limit[s]=tail[outsum[s]];@+ else limit[s]=1;
 for (;r>1&&a[r-1]==b[r-1];r--);
 if (outdeg[s-1]==0 && (a[s]&1)) q=a[s]/3;@+ else q=a[s]>>1;
 for (p=a[s]-q; p<=b[s-1];) { if (p>b[r-1]) {
 while (p>a[r]) r++; /* this step keeps &#124;r<s&#124;, since &#124;a[s-1]=b[s-1]&#124; */ p=a[r], q=a[s]-p, r++; } if (q<limit[s]) goto backup; @<find bounds $(&#124;lbp&#124;,&#124;ubp&#124;)$ and $(&#124;lbq&#124;,&#124;ubq&#124;)$ on where &#124;p&#124; and &#124;q&#124; can be inserted; but go to &#124;failpq&#124; if they can't both be accommodated@>;
 ptrp=ptr;
 for (; ubp>=lbp; ubp--) {
 @<put &#124;p&#124; into the chain at location &#124;ubp&#124;; &#124;goto failp&#124; if there's a problem@>;
 if (p==q) goto happiness;
 if (ubq>=ubp) ubq=ubp-1;
 ptrq=ptr;
 for (; ubq>=lbq; ubq--) {
 @<put &#124;q&#124; into the chain at location &#124;ubq&#124;; &#124;goto failq&#124; if there's a problem@>;
 happiness: @<put local variables on the stack and update outdegrees@>;
 goto onward; /* now &#124;a[s]&#124; is covered; try to fill in &#124;a[s-1]&#124; */
 backup: s++;
 if (s>lb) goto impossible;
 @<restore local variables from the stack and downdate outdegrees@>;
 if (p==q) goto failp;
 failq:@+ while (ptr>ptrq) @<undo a change@>;
 }
 failp:@+ while (ptr>ptrp) @<undo a change@>;
 }
 failpq:@+ if (p==q) {
 if (outdeg[s-1]==0) q=a[s]/3+1; /* will be decreased momentarily */
 if (q>b[s-2]) q=b[s-2];
 else q--;
 p=a[s]-q;
 }@+else p++,q--;
 }
 goto backup;
onward: continue;
}
goto done;
impossible:@;
 
@ @<put local variables on the stack and update outdegrees@>=
tail[s]=q, stack[s].r=r;
outdeg[ubp]++, outsum[ubp]+=s;
outdeg[ubq]++, outsum[ubq]+=s;
stack[s].lbp=lbp,stack[s].ubp=ubp;
stack[s].lbq=lbq,stack[s].ubq=ubq;
stack[s].ptrp=ptrp,stack[s].ptrq=ptrq;
@ @<restore local variables from the stack and downdate outdegrees@>=
ptrq=stack[s].ptrq,ptrp=stack[s].ptrp;
lbq=stack[s].lbq,ubq=stack[s].ubq;
lbp=stack[s].lbp,ubp=stack[s].ubp;
outdeg[ubq]--, outsum[ubq]-=s;
outdeg[ubp]--, outsum[ubp]-=s;
q=tail[s], p=a[s]-q, r=stack[s].r;
@ After the test in this step is passed, we'll have &#124;ubp>ubq&#124; and &#124;lbp>lbq&#124;.
@<find bounds...@>=
lbp=l[p];
if (lbp>=s) goto failpq;
while (b[lbp]<p) lbp++; if (a[lbp]>p) goto failpq;
for (ubp=lbp;a[ubp+1]<=p;ubp++); if (ubp==s-1) lbp=ubp; if (p==q) lbq=lbp,ubq=ubp; else { lbq=l[q]; if (lbq>=ubp) goto failpq;
 while (b[lbq]<q) lbq++; if (lbq>=ubp) goto failpq;
 if (a[lbq]>q) goto failpq;
 for (ubq=lbq;a[ubq+1]<=q && ubq+1<ubp;ubq++); if (lbp==lbq) lbp++; } @ The undoing mechanism is very simple: When changing &#124;a[j]&#124;, we put &#124;(j<<24)+x&#124; on the &#124;undo&#124; stack, where &#124;x&#124; was the former value. Similarly, when changing &#124;b[j]&#124;, we stack the value &#124;(1<<31)+(j<<24)+x&#124;. @d newa(j,y) undo[ptr++]=(j<<24)+a[j], a[j]=y @d newb(j,y) undo[ptr++]=(1<<31)+(j<<24)+b[j], b[j]=y @<undo a change@>=
{
 i=undo[--ptr];
 if (i>=0) a[i>>24]=i&0xffffff;
 else b[(i&0x3fffffff)>>24]=i&0xffffff;
}
@ At this point we know that $a[ubp]\le p\le b[ubp]$.
@<put &#124;p&#124; into the chain at location &#124;ubp&#124;...@>=
if (a[ubp]!=p) {
 newa(ubp,p);
 for (j=ubp-1;(a[j]<<1)<a[j+1];j--) { i=(a[j+1]+1)>>1;
 if (i>b[j]) goto failp;
 newa(j,i);
 }
 for (j=ubp+1;a[j]<=a[j-1];j++) { i=a[j-1]+1; if (i>b[j]) goto failp;
 newa(j,i);
 }
}
if (b[ubp]!=p) {
 newb(ubp,p);
 for (j=ubp-1;b[j]>=b[j+1];j--) {
 i=b[j+1]-1;
 if (i<a[j]) goto failp; newb(j,i); } for (j=ubp+1;b[j]>b[j-1]<<1;j++) { i=b[j-1]<<1; if (i<a[j]) goto failp; newb(j,i); } } @<make forced moves if &#124;p&#124; has a special form@>;
@ If, say, we've just set &#124;a[8]=b[8]=132&#124;, special considerations apply,
because the only addition chains of length~8 for 132 are
$$\eqalign{
&1,2,4,8,16,32,64,128,132;\cr
&1,2,4,8,16,32,64,68,132;\cr
&1,2,4,8,16,32,64,66,132;\cr
&1,2,4,8,16,32,34,66,132;\cr
&1,2,4,8,16,32,33,66,132;\cr
&1,2,4,8,16,17,33,66,132.\cr}$$
The values of &#124;a[4]&#124; and &#124;b[4]&#124; must therefore be 16; and then, of course,
we also must have &#124;a[3]=b[3]=8&#124;, etc. Similar reasoning applies
whenever we set $a[j]=b[j]=2^j+2^k$ for $k\le j-4$.
Such cases may seem extremely special. But my hunch is that they are
important, because efficient chains need such values. When we try
to prove that no efficient chain exists, we want to show that
such values can't be present. Numbers with small &#124;l[p]&#124; are harder
to rule out, so it should be helpful to penalize them.
@<make forced moves if &#124;p&#124; has a special form@>=
i=p-(1<<(ubp-1)); if (i && ((i&(i-1))==0) && (i<<4)<p) { for (j=ubp-2;(i&1)==0;i>>=1,j--);
 if (b[j]<(1<<j)) goto failp; for (;a[j]<(1<<j);j--) newa(j,1<<j); } @ At this point we had better not assume that $a[ubq]\le q\le b[ubq],ドル because &#124;p&#124; has just been inserted. That insertion can mess up the bounds that we looked at when &#124;lbq&#124; and &#124;ubq&#124; were computed. @<put &#124;q&#124; into the chain at location &#124;ubq&#124;...@>=
if (a[ubq]!=q) {
 if (a[ubq]>q) goto failq;
 newa(ubq,q);
 for (j=ubq-1;(a[j]<<1)<a[j+1];j--) { i=(a[j+1]+1)>>1;
 if (i>b[j]) goto failq;
 newa(j,i);
 }
 for (j=ubq+1;a[j]<=a[j-1];j++) { i=a[j-1]+1; if (i>b[j]) goto failq;
 newa(j,i);
 }
}
if (b[ubq]!=q) {
 if (b[ubq]<q) goto failq; newb(ubq,q); for (j=ubq-1;b[j]>=b[j+1];j--) {
 i=b[j+1]-1;
 if (i<a[j]) goto failq; newb(j,i); } for (j=ubq+1;b[j]>b[j-1]<<1;j++) { i=b[j-1]<<1; if (i<a[j]) goto failq; newb(j,i); } } @<make forced moves if &#124;q&#124; has a special form@>;
@ @<make forced moves if &#124;q&#124; has a special form@>=
i=q-(1<<(ubq-1)); if (i && ((i&(i-1))==0) && (i<<4)<q) { for (j=ubq-2;(i&1)==0;i>>=1,j--);
 if (b[j]<(1<<j)) goto failq;
 for (;a[j]<(1<<j);j--) newa(j,1<<j);
}
@*Index.
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