Created
September 1, 2010 10:04
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| New file: inline_levenshtein | |
| inline do |builder| | |
| builder.add_link_flags '-lruby' | |
| builder.c_raw <<-EOT | |
| long minimum(long a,long b,long c) | |
| /*Gets the minimum of three values*/ | |
| { | |
| long min=a; | |
| if(b<min) | |
| min=b; | |
| if(c<min) | |
| min=c; | |
| return min; | |
| } | |
| EOT | |
| builder.c <<-EOT | |
| /****************************************/ | |
| /*Implementation of Levenshtein distance*/ | |
| /****************************************/ | |
| long levenshtein_distance(char *s,char*t) | |
| /*Compute levenshtein distance between s and t*/ | |
| { | |
| //Step 1 | |
| long k,i,j,n,m,cost,*d,distance; | |
| n=strlen(s); | |
| m=strlen(t); | |
| if(n!=0&&m!=0) | |
| { | |
| d=malloc((sizeof(long))*(m+1)*(n+1)); | |
| m++; | |
| n++; | |
| //Step 2 | |
| for(k=0;k<n;k++) | |
| d[k]=k; | |
| for(k=0;k<m;k++) | |
| d[k*n]=k; | |
| //Step 3 and 4 | |
| for(i=1;i<n;i++) | |
| for(j=1;j<m;j++) | |
| { | |
| //Step 5 | |
| if(s[i-1]==t[j-1]) | |
| cost=0; | |
| else | |
| cost=1; | |
| //Step 6 | |
| d[j*n+i]=minimum(d[(j-1)*n+i]+1,d[j*n+i-1]+1,d[(j-1)*n+i-1]+cost); | |
| } | |
| distance=d[n*m-1]; | |
| free(d); | |
| return distance; | |
| } | |
| else | |
| return -1; //a negative return value means that one or both strings are empty. | |
| } | |
| EOT | |
| end |
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