Find gaps in sequence of numbers

Question

Hi,
I'm trying to scan through multiple records containing millions of numbers. What I need to do is to print out the numbers that are sequential as such:
If the record contains a sequence like this:
46859995600
46859995699
And there is no breakage in that sequence, then print:
46859995600 100
If that sequence is broken somewhere, then print series of 10 that are sequential, f.e:
46859995600 10
46859995620 10
46859995630 10
I already have a bash script for this, but it's way too slow. If anyone can help me out I'd be very grateful, perhaps a "awk" solution.
Thanks!

Answer 1

Indeed.  Also, note that the code above makes several assumptions: 1) monotonicity ... but "sort" can take care of that; 2) there is no zero.  This would be easy to fix, but in the first draft I didn't bother with it: 000-009 will not currently be counted; 3) the application is not mission critical; 4) there are no blank line; 5) each line contains *only* the number.
On #3, my code may be faster, but it hasn't been checked for accuracy.  Be sure to test for accuracy before trusting its output.  In fact, I just tested it ... it failed.
I tested it after I recoded it in C.  The speed went from 2min 3sec to 0.39sec ... that's nice.  But the results file was *much* smaller.
The results from the bash version includes several duplicates, and some erroneous inclusions.  Given that so far the C version seems more accurate (but still needs testing) and *much* faster, I'd go with patching this up further:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
unsigned long lt = 0, lh = 0;
unsigned long n, prev = 0, tens[10];
int nten=0;
void check_tens() {
int i;
if (lh) printf("%lu 100\n",lh);
else for (i = 0; i < 10 && tens[i]; i++)
printf("%lu 10\n",tens[i]);
memset(tens,0,10*sizeof(unsigned long));
nten = 0;
int main() {
int i;
memset(tens,0,10*sizeof(unsigned long));
while (fscanf(stdin,"%lu\n",&n) == 1) {
if ((n % 10) == 0) lt = n;
else if ((n % 10) == 9) {
if (!lt || n - 9 == lt) tens[nten++] = lt;
lt = 0;
if ((n % 100) == 0) lh = n;
else if ((n % 100) == 99) check_tens();
if (n - prev != 1) {
lt = lh = 0;
for (i = prev; i < n; i++)
if ((i % 100) == 99) { check_tens(); break; }
if ((n % 10) == 0) lt = n;
if ((n % 100) == 0) lh = n;
prev=n;
lh = 0;
check_tens();
return 0;
EDIT: I can actually confirm the C version is not yet accurate.  ARGH!
Last edited by Trilby (2014-05-02 00:11:09)