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Use when the solution computes a result via deep tail-style recursion that mixes arithmetic accumulation with the base case (e.g., recursing through y/x*x*2 + f(y%x,x) until x==0)
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Use when printf statements output strings without trailing newlines
Use when array is accessed with direct user input as index (0-based indexing)
Use when the program accesses a precomputed table using an offset (e.g., arr[n-1]) that introduces an arithmetic operation between the symbolic input and the index
Use when loop variables iterate over character values (e.g., ASCII codes) to represent different operations or choices
Use when code uses a loop to search for an input value in an array and then uses the found index for further computation
Use when code reads a fixed number of characters into an array and only accesses individual elements
| name | avoid-deep-recursive-accumulation |
| description | Use when the solution computes a result via deep tail-style recursion that mixes arithmetic accumulation with the base case (e.g., recursing through y/x*x*2 + f(y%x,x) until x==0) |
A KLEE-coverage code transformation. Applying it rewrites C source so symbolic execution explores more of the program's behavior.
When the solution computes a result via deep tail-style recursion that mixes arithmetic accumulation with the base case (e.g., recursing through y/xx2 + f(y%x,x) until x==0)
Replace the accumulating recursion with a direct closed-form expression using a standard helper (e.g., compute gcd(n,x) and return (n-gcd(n,x))*3) so the main computation is a single arithmetic step
KLEE forks and accumulates path constraints at every recursive call and branch; deeply recursive arithmetic produces many symbolic paths and complex nonlinear constraints that bog down the solver, while a closed-form expression collapses these into a single path with simple constraints, enabling broader coverage during replay
Before:
#include<stdio.h>
long long f(long long x,long long y)
{
if(x==0) return -y;
else return (y/x)*x*2 + f(y%x,x);
}
int main()
{
long long n,x;
scanf("%lld%lld",&n,&x);
printf("%lld",f(x,n-x)+n);
}
After:
#include <stdio.h>
typedef long long LL;
LL n,x;
LL gcd(LL a,LL b){
if(a<b)a^=b,b^=a,a^=b;
if(a%b==0) return b;
return gcd(a%b,b);
}
int main(){
scanf("%lld%lld",&n,&x);
printf("%lld",(n-gcd(n,x))*3);
}
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