this string matching algorithm is O(N/M)
as compared to O(N+M) for kmp
on any alphabet size.
daniel 2391++

num minusbit(num bits, num g) { if (g < one + fifthtau) { if (bits & ones[g]) bits &= diagovreason[g]; else { bits += ones[g]; bits = minusbit(bits, g + one); } } return bits;}
num minus(num more, num less) { for(num g = zero; g < one+fifthtau; g++) if (less & ones[g]) more = minusbit(more, g); return more; }
typedef void joy;
template < class tt > class set { // fibolacci 0 1 2 4 7 11 17 26 /// space growth of ordered sets
public: joy allocay(num get) {if(have<get){tt* tmp = new tt[(num)get]; if (tmp == (tt*)zero) {} else { for (num h = zero; h < have; h++)
{ tmp[h] = v[h]; v[h] = (tt)zero; } if(v == (tt*)zero) {} else { delete[]v; } { have = zero; v = tmp; have = get; tmp = (tt*)zero;}}}}
joy add(tt elt){if(y==have){allocay((one+(y>>one)+y));}v[y++]=elt;}
inline num size(){return y;} inline tt& operator[](num r){return v[r];} num setsize(num z){y=z;return z;}
joy clear(){for(num g=zero;g<have;g++)v[g]=(tt)zero;y=zero;}
~set(){clear();if(v==(tt*)zero){}else{delete[]v;}y=zero;have=zero;v=(tt*)zero;}
set(num beg):y(zero),have(zero),v((tt*)zero){allocay(beg);} set():y(zero),have(zero),v((tt*)zero){}
num y;num have;tt* v;
};/// ten lines
typedef set<num> nums; typedef set<nums*> numnums;
/// (time n over space m search)
num age = zero;// count all finds one by one ()

num map(nums&p,num&z,num&dep,num&lga){
/// map a few letters of p to a small number d
num d=zero;for(num h=zero;h<dep;h++)d=(d<<lga)+p.v[z+h];return d;}

num skip(nums&string,nums&book,num&lga,numnums*myn){
/// allocate proper memory space of empty sets O(constant)
num m=string.size();num logm=zero;num d=one;while(d<m){d=d<<lga;logm++;}
for(num g=(*myn).size();g<d;g++)(*myn).add(new nums);
/// the m part of partten are mapped precisely O(mlgm)
for(num g=zero;g+logm<m;g++)(*(*myn)[map(string,g,logm,lga)]).add(g);
/// map part of the book into pattern O(nlgm/(m+1-lgm))
num mm=minus(m+one,logm); for(num j=mm;j+logm<book.size();j=j+mm){
nums* d=(*myn)[map(book,j,logm,lga)]; for(num o=zero;o<(*d).size();o++)
{ num be=minus(j,(*d)[o]);num h=one;for(num g=zero;g<m;g++){
if(book[be+g]==string[g])continue;h=zero;break;}if(h)age++;}}
/// release resources from step two O(mlgm)
for(num g=zero;g+logm<m;g++)(*(*myn)[map(string,g,logm,lga)]).clear();return age;}

0 three planes make a point
1 one half dozen lines
2 define an idea
3 two halfspaces make a space
4 two halfplanes make a plane
5 two halflines make a line
6 three planes make a point
7 four lines make a square
8 numbers are two colorable
9 theory is two colorable
10 all is black information on white background