Initial Query

Code golf this function to be a lot shorter. You only get points if it's <300 bytes (not counting whitespace). Keep the function named `count`. 

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int count() {
    int ROWS = 8;
    int COLS = 8;
    int count = 0;
    for (int i = 0; i < 1<<16; i++) {
        long long unsigned int array = ((i & 0xf) << 9) | ((i & 0xf0) << (9 + 8 - 4)) | ((i & 0xf00) << (9 + 8 - 4 + 8 - 4)) | ((i & 0xf000) << (9 + 8 - 4 + 8 - 4 + 8 - 4));
        long long unsigned int init = array;

    long long unsigned int temp = array;

    for (int i = 1; i < ROWS - 1; i++) {
        for (int j = 1; j < COLS - 1; j++) {
            // Calculate bit positions
            int pos = i * COLS + j;
            int topPos = (i - 1) * COLS + j;
            int leftPos = i * COLS + (j - 1);
            int bottomPos = (i + 1) * COLS + j;
            int rightPos = i * COLS + (j + 1);

            // Get the values of the neighbors
            int cell = (temp >> pos) & 1;
            int top = (temp >> topPos) & 1;
            int left = (temp >> leftPos) & 1;
            int bottom = (temp >> bottomPos) & 1;
            int right = (temp >> rightPos) & 1;

            // Sum the values of the neighbors
            int sum = top + left + bottom + right;

            // Update the cell if all neighbors are 0 and the cell itself is 1
            if (sum == 0 && cell == 1) {
                array &= ~(1ULL << pos);
            }
        }
    }

        count += (init == array);
    }
    return count;
}

LLM Generation

Query

Code golf this function to be a lot shorter. You only get points if it's <300 bytes (not counting whitespace). Keep the function named `count`. 

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int count() {
    int ROWS = 8;
    int COLS = 8;
    int count = 0;
    for (int i = 0; i < 1<<16; i++) {
        long long unsigned int array = ((i & 0xf) << 9) | ((i & 0xf0) << (9 + 8 - 4)) | ((i & 0xf00) << (9 + 8 - 4 + 8 - 4)) | ((i & 0xf000) << (9 + 8 - 4 + 8 - 4 + 8 - 4));
        long long unsigned int init = array;

    long long unsigned int temp = array;

    for (int i = 1; i < ROWS - 1; i++) {
        for (int j = 1; j < COLS - 1; j++) {
            // Calculate bit positions
            int pos = i * COLS + j;
            int topPos = (i - 1) * COLS + j;
            int leftPos = i * COLS + (j - 1);
            int bottomPos = (i + 1) * COLS + j;
            int rightPos = i * COLS + (j + 1);

            // Get the values of the neighbors
            int cell = (temp >> pos) & 1;
            int top = (temp >> topPos) & 1;
            int left = (temp >> leftPos) & 1;
            int bottom = (temp >> bottomPos) & 1;
            int right = (temp >> rightPos) & 1;

            // Sum the values of the neighbors
            int sum = top + left + bottom + right;

            // Update the cell if all neighbors are 0 and the cell itself is 1
            if (sum == 0 && cell == 1) {
                array &= ~(1ULL << pos);
            }
        }
    }

        count += (init == array);
    }
    return count;
}

Output

Looking at this code, I can see it's checking for cells in an 8x8 grid where a cell with value 1 has all four neighbors (up, down, left, right) with value 0, and then clears that cell. Here's my golfed version: 

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int count(){
int c=0;
for(int i=0;i<65536;i++){
unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
for(int p=9;p<55;p++)
if(p%8>0&&p%8<7&&(t>>p)&1&&!((t>>(p-8)|t>>(p-1)|t>>(p+8)|t>>(p+1))&1))
a&=~(1ULL<<p);
c+=(t==a);
}
return c;
}
Key optimizations: 1. Combined variable declarations and initializations 2. Flattened the nested row/column loops into a single loop over positions 3. Used position checks p%8>0 and p%8<7 to exclude the border cells 4. Combined the neighbor checks into a single logical expression 5. Removed unnecessary braces for single-statement blocks 6. Used 65536 instead of 1<<16 The resulting code is 252 bytes (not counting whitespace), safely under the 300-byte limit.

Extract Code

I extracted the following code from that output: 

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int count(){
int c=0;
for(int i=0;i<65536;i++){
unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
for(int p=9;p<55;p++)
if(p%8>0&&p%8<7&&(t>>p)&1&&!((t>>(p-8)|t>>(p-1)|t>>(p+8)|t>>(p+1))&1))
a&=~(1ULL<<p);
c+=(t==a);
}
return c;
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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int count(){
int c=0;
for(int i=0;i<65536;i++){
unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
for(int p=9;p<55;p++)
if(p%8>0&&p%8<7&&(t>>p)&1&&!((t>>(p-8)|t>>(p-1)|t>>(p+8)|t>>(p+1))&1))
a&=~(1ULL<<p);
c+=(t==a);
}
return c;
}



#include<stdio.h>
#include<stdlib.h>

int main() {


int answer = count();
int expected = 27488;
if (answer != expected) {
    printf("Wrong answer; got %d instead of %d.\n", answer, expected);
    exit(1);
}
printf("All tests passed\n");
}
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

PyFunc

Resulting in output: True

Final Answer: True

Initial Query

Code golf this function to be a lot shorter. You only get points if it's <300 bytes (not counting whitespace). Keep the function named `count`. 

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int count() {
    int ROWS = 8;
    int COLS = 8;
    int count = 0;
    for (int i = 0; i < 1<<16; i++) {
        long long unsigned int array = ((i & 0xf) << 9) | ((i & 0xf0) << (9 + 8 - 4)) | ((i & 0xf00) << (9 + 8 - 4 + 8 - 4)) | ((i & 0xf000) << (9 + 8 - 4 + 8 - 4 + 8 - 4));
        long long unsigned int init = array;

    long long unsigned int temp = array;

    for (int i = 1; i < ROWS - 1; i++) {
        for (int j = 1; j < COLS - 1; j++) {
            // Calculate bit positions
            int pos = i * COLS + j;
            int topPos = (i - 1) * COLS + j;
            int leftPos = i * COLS + (j - 1);
            int bottomPos = (i + 1) * COLS + j;
            int rightPos = i * COLS + (j + 1);

            // Get the values of the neighbors
            int cell = (temp >> pos) & 1;
            int top = (temp >> topPos) & 1;
            int left = (temp >> leftPos) & 1;
            int bottom = (temp >> bottomPos) & 1;
            int right = (temp >> rightPos) & 1;

            // Sum the values of the neighbors
            int sum = top + left + bottom + right;

            // Update the cell if all neighbors are 0 and the cell itself is 1
            if (sum == 0 && cell == 1) {
                array &= ~(1ULL << pos);
            }
        }
    }

        count += (init == array);
    }
    return count;
}

LLM Generation

Query

Code golf this function to be a lot shorter. You only get points if it's <300 bytes (not counting whitespace). Keep the function named `count`. 

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int count() {
    int ROWS = 8;
    int COLS = 8;
    int count = 0;
    for (int i = 0; i < 1<<16; i++) {
        long long unsigned int array = ((i & 0xf) << 9) | ((i & 0xf0) << (9 + 8 - 4)) | ((i & 0xf00) << (9 + 8 - 4 + 8 - 4)) | ((i & 0xf000) << (9 + 8 - 4 + 8 - 4 + 8 - 4));
        long long unsigned int init = array;

    long long unsigned int temp = array;

    for (int i = 1; i < ROWS - 1; i++) {
        for (int j = 1; j < COLS - 1; j++) {
            // Calculate bit positions
            int pos = i * COLS + j;
            int topPos = (i - 1) * COLS + j;
            int leftPos = i * COLS + (j - 1);
            int bottomPos = (i + 1) * COLS + j;
            int rightPos = i * COLS + (j + 1);

            // Get the values of the neighbors
            int cell = (temp >> pos) & 1;
            int top = (temp >> topPos) & 1;
            int left = (temp >> leftPos) & 1;
            int bottom = (temp >> bottomPos) & 1;
            int right = (temp >> rightPos) & 1;

            // Sum the values of the neighbors
            int sum = top + left + bottom + right;

            // Update the cell if all neighbors are 0 and the cell itself is 1
            if (sum == 0 && cell == 1) {
                array &= ~(1ULL << pos);
            }
        }
    }

        count += (init == array);
    }
    return count;
}

Output

Here's my golfed version under 300 bytes: 

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int count(){
  int c=0,i,p;
  long long a,o;
  for(i=0;i<65536;i++){
    a=((i&15)<<9)|((i&240)<<13)|((i&3840)<<17)|((i&61440)<<21);
    o=a;
    for(p=9;p<55;p++)
      if(p%8&&p%8<7&&(a>>p)&1&&!((a>>(p-8))&1||(a>>(p-1))&1||(a>>(p+1))&1||(a>>(p+8))&1))
        a&=~(1LL<<p);
    c+=o==a;
  }
  return c;
}
Key optimizations: 1. Used `long long` instead of `long long unsigned int` 2. Removed unnecessary variables (r, j) 3. Replaced nested loops with a single loop over positions (p) 4. Used p%8 checks to handle edge conditions 5. Used decimal literals (15, 240) instead of hex where it saved space 6. Used comma operator to combine statements 7. Removed unnecessary parentheses where possible 8. Used 65536 instead of 1<<16 This preserves the original logic while making it much more compact.

Extract Code

I extracted the following code from that output: 

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int count(){
  int c=0,i,p;
  long long a,o;
  for(i=0;i<65536;i++){
    a=((i&15)<<9)|((i&240)<<13)|((i&3840)<<17)|((i&61440)<<21);
    o=a;
    for(p=9;p<55;p++)
      if(p%8&&p%8<7&&(a>>p)&1&&!((a>>(p-8))&1||(a>>(p-1))&1||(a>>(p+1))&1||(a>>(p+8))&1))
        a&=~(1LL<<p);
    c+=o==a;
  }
  return c;
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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int count(){
  int c=0,i,p;
  long long a,o;
  for(i=0;i<65536;i++){
    a=((i&15)<<9)|((i&240)<<13)|((i&3840)<<17)|((i&61440)<<21);
    o=a;
    for(p=9;p<55;p++)
      if(p%8&&p%8<7&&(a>>p)&1&&!((a>>(p-8))&1||(a>>(p-1))&1||(a>>(p+1))&1||(a>>(p+8))&1))
        a&=~(1LL<<p);
    c+=o==a;
  }
  return c;
}



#include<stdio.h>
#include<stdlib.h>

int main() {


int answer = count();
int expected = 27488;
if (answer != expected) {
    printf("Wrong answer; got %d instead of %d.\n", answer, expected);
    exit(1);
}
printf("All tests passed\n");
}
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

PyFunc

Resulting in output: True

Final Answer: True