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Tuesday, September 29, 2009
About C / C++ / C#: A Big Fat D'oh...
Monday, September 28, 2009
About Today: Best Cheap Cell Phones by Carrier
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Friday, September 25, 2009
About Today: How to Make Gummy Worms
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Thursday, September 24, 2009
Welcome to ddankwa Programming Tutorials
Little Knowledge in.............
The Below codes are for illustrating what goes into programming
#include
#include "fibsrch.h"
/*
* If val is found in arr, return the index of its location in arr.
* Otherwise, return the index of the smallest element greater than val
*/
static int binsrch_geq(const int *arr, int n, int val)
{
register int low, high, mid;
int geq;
low=0; high=n-1; geq=-1;
/* binary search for finding the element with value val */
while(low<=high){
mid=(low+high)>>1; //(low+high)/2;
if(val<arr[mid]){
high=mid-1;
geq=mid;
}
else if(val>arr[mid])
low=mid+1;
else
return mid; /* found */
}
return geq; /* not found */
}
/*
Fibonaccian search for locating the index of "val" in an array "arr" of size "n"
that is sorted in ascending order. See http://doi.acm.org/10.1145/367487.367496
Algorithm description
-----------------------------------------------------------------------------
Let Fk represent the k-th Fibonacci number where Fk+2=Fk+1 + Fk for k>=0 and
F0 = 0, F1 = 1. To test whether an item is in a list of n = Fm ordered numbers,
proceed as follows:
a) Set k = m.
b) If k = 0, finish - no match.
c) Test item against entry in position Fk-1.
d) If match, finish.
e) If item is less than entry Fk-1, discard entries from positions Fk-1 + 1 to n.
Set k = k - 1 and go to b).
f) If item is greater than entry Fk-1, discard entries from positions 1 to Fk-1.
Renumber remaining entries from 1 to Fk-2, set k = k - 2 and go to b)
If Fm>n then the original array is augmented with Fm-n numbers larger
than val and the above algorithm is applied.
*/
int fibsrch(const int *arr, int n, int val)
{
register int k, idx, offs;
static int prevn=-1, prevk=-1;
/* Precomputed Fibonacci numbers F0 up to F46. This implementation assumes that the size n
* of the input array fits in 4 bytes. Note that F46=1836311903 is the largest Fibonacci
* number that is less or equal to the 4-byte INT_MAX (=2147483647). The next Fibonacci
* number, i.e. F47, is 2971215073 and is larger than INT_MAX, implying that it does not
* fit in a 4 byte integer. Note also that the last array element is INT_MAX rather than
* F47. This ensures correct operation for n>F46.
*/
const static int Fib[47+1]={0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765,
10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578,
5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296,
433494437, 701408733, 1134903170, 1836311903, INT_MAX};
/* find the smallest fibonacci number that is greater or equal to n. Store this
* number to avoid recomputing it in the case of repetitive searches with identical n.
*/
if(n!=prevn){
#if 1
k=(n>1)? binsrch_geq(Fib, sizeof(Fib)/sizeof(int), n) : 1;
#else /* the above binary search call can be substituted by the following code fragment: */
{
register int f0, f1, t;
for(f0=0, f1=1, k=1; f1<n; t=f1, f1+=f0, f0=t, ++k)
;
}
#endif
prevk=k;
prevn=n;
}
else k=prevk;
/* If the sought value is larger than the largest Fibonacci number less than n,
* care must be taken top ensure that we do not attempt to read beyond the end
* of the array. If we do need to do this, we pretend that the array is padded
* with elements larger than the sought value.
*/
for(offs=0; k>0; ){
idx=offs+Fib[--k];
/* note that at this point k has already been decremented once */
if(idx>=n || val<arr[idx]) // index out of bounds or val in 1st part
continue;
else if (val>arr[idx]){ // val in 2nd part
offs=idx;
--k;
}
else // val==arr[idx], found
return idx;
}
return -1; // not found
}
#if 0
/* Sample driver program for fibsrch() */
main()
{
int data[]={1, 4, 5, 7, 9, 11, 13, 16, 18, 20, 25, 27, 30, 32, 33, 36, 39, 41, 44, 47, 51, 53, 55};
int i, x, n;
x=30; n=sizeof(data)/sizeof(int);
i=fibsrch(data, n, x);
if(i>=0)
printf("%d found at index %d\n", x, i);
else
printf("%d was not found\n", x);
}
#endif
#include
#include "fibsrch.h"
/*
* If val is found in arr, return the index of its location in arr.
* Otherwise, return the index of the smallest element greater than val
*/
static int binsrch_geq(const int *arr, int n, int val)
{
register int low, high, mid;
int geq;
low=0; high=n-1; geq=-1;
/* binary search for finding the element with value val */
while(low<=high){
mid=(low+high)>>1; //(low+high)/2;
if(val<arr[mid]){
high=mid-1;
geq=mid;
}
else if(val>arr[mid])
low=mid+1;
else
return mid; /* found */
}
return geq; /* not found */
}
/*
Fibonaccian search for locating the index of "val" in an array "arr" of size "n"
that is sorted in ascending order. See http://doi.acm.org/10.1145/367487.367496
Algorithm description
-----------------------------------------------------------------------------
Let Fk represent the k-th Fibonacci number where Fk+2=Fk+1 + Fk for k>=0 and
F0 = 0, F1 = 1. To test whether an item is in a list of n = Fm ordered numbers,
proceed as follows:
a) Set k = m.
b) If k = 0, finish - no match.
c) Test item against entry in position Fk-1.
d) If match, finish.
e) If item is less than entry Fk-1, discard entries from positions Fk-1 + 1 to n.
Set k = k - 1 and go to b).
f) If item is greater than entry Fk-1, discard entries from positions 1 to Fk-1.
Renumber remaining entries from 1 to Fk-2, set k = k - 2 and go to b)
If Fm>n then the original array is augmented with Fm-n numbers larger
than val and the above algorithm is applied.
*/
int fibsrch(const int *arr, int n, int val)
{
register int k, idx, offs;
static int prevn=-1, prevk=-1;
/* Precomputed Fibonacci numbers F0 up to F46. This implementation assumes that the size n
* of the input array fits in 4 bytes. Note that F46=1836311903 is the largest Fibonacci
* number that is less or equal to the 4-byte INT_MAX (=2147483647). The next Fibonacci
* number, i.e. F47, is 2971215073 and is larger than INT_MAX, implying that it does not
* fit in a 4 byte integer. Note also that the last array element is INT_MAX rather than
* F47. This ensures correct operation for n>F46.
*/
const static int Fib[47+1]={0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765,
10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578,
5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296,
433494437, 701408733, 1134903170, 1836311903, INT_MAX};
/* find the smallest fibonacci number that is greater or equal to n. Store this
* number to avoid recomputing it in the case of repetitive searches with identical n.
*/
if(n!=prevn){
#if 1
k=(n>1)? binsrch_geq(Fib, sizeof(Fib)/sizeof(int), n) : 1;
#else /* the above binary search call can be substituted by the following code fragment: */
{
register int f0, f1, t;
for(f0=0, f1=1, k=1; f1<n; t=f1, f1+=f0, f0=t, ++k)
;
}
#endif
prevk=k;
prevn=n;
}
else k=prevk;
/* If the sought value is larger than the largest Fibonacci number less than n,
* care must be taken top ensure that we do not attempt to read beyond the end
* of the array. If we do need to do this, we pretend that the array is padded
* with elements larger than the sought value.
*/
for(offs=0; k>0; ){
idx=offs+Fib[--k];
/* note that at this point k has already been decremented once */
if(idx>=n || val<arr[idx]) // index out of bounds or val in 1st part
continue;
else if (val>arr[idx]){ // val in 2nd part
offs=idx;
--k;
}
else // val==arr[idx], found
return idx;
}
return -1; // not found
}
#if 0
/* Sample driver program for fibsrch() */
main()
{
int data[]={1, 4, 5, 7, 9, 11, 13, 16, 18, 20, 25, 27, 30, 32, 33, 36, 39, 41, 44, 47, 51, 53, 55};
int i, x, n;
x=30; n=sizeof(data)/sizeof(int);
i=fibsrch(data, n, x);
if(i>=0)
printf("%d found at index %d\n", x, i);
else
printf("%d was not found\n", x);
}
#endif
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