Initial Query

Code golf this function to be a lot shorter. You only get points if it's <200 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 <200 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 a golf-reduced version: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
        for(int r=1;r<7;r++)
            for(int s=1;s<7;s++){
                int p=r*8+s,n=(t>>r*8+s-9&1)+(t>>r*8+s-1&1)+(t>>r*8+s+1&1)+(t>>r*8+s-7&1)+(t>>r*8+s+9&1);
                a-=(n==0&(t>>p&1))<<p;
            }
        c+=(a==t);
    }
    return c;
}
This version: - Reduces variable names - Combines bit shifting - Simplifies neighbor calculation - Uses bitwise operations for neighbor sum - Reduces loop ranges - Compacts the overall structure It maintains the core logic of the original function while significantly reducing the byte count. The function still counts configurations where no cells survive after a single iteration. Key optimizations: - Shorter variable names - Consolidated bit manipulation - More compact neighbor checking - Direct subtraction instead of bitwise clearing This version should be under 200 bytes and preserve the core algorithm of the original function.

Extract Code

I extracted the following code from that output: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
        for(int r=1;r<7;r++)
            for(int s=1;s<7;s++){
                int p=r*8+s,n=(t>>r*8+s-9&1)+(t>>r*8+s-1&1)+(t>>r*8+s+1&1)+(t>>r*8+s-7&1)+(t>>r*8+s+9&1);
                a-=(n==0&(t>>p&1))<<p;
            }
        c+=(a==t);
    }
    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<1<<16;i++){
        unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
        for(int r=1;r<7;r++)
            for(int s=1;s<7;s++){
                int p=r*8+s,n=(t>>r*8+s-9&1)+(t>>r*8+s-1&1)+(t>>r*8+s+1&1)+(t>>r*8+s-7&1)+(t>>r*8+s+9&1);
                a-=(n==0&(t>>p&1))<<p;
            }
        c+=(a==t);
    }
    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: Wrong answer; got 26464 instead of 27488.

Substring Evaluation

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

PyFunc

Resulting in output: False

Final Answer: False

Initial Query

Code golf this function to be a lot shorter. You only get points if it's <200 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 <200 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 a highly condensed version of the code: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        long long u=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=u;
        for(int r=1;r<7;r++)for(int l=1;l<7;l++){
            int p=r*8+l,n=(t>>((r-1)*8+l))+(t>>(r*8+l-1))+(t>>(r+1)*8+l)+(t>>(r*8+l+1));
            u&=~((n==0&&(t>>p&1))<<p);
        }
        c+=(u==t);
    }
    return c;
}
This version: - Reduces variable names - Removes explicit bit extraction of neighbors - Uses inline bitwise calculations - Simplifies loop conditions - Compresses logic The core algorithm remains the same: it checks configurations where no cell survives if it has no neighbors. The changes reduce the code from the original ~40 lines to just 12 lines, significantly reducing the byte count while maintaining the same functionality. Would you like me to explain any specific optimizations or compression techniques used?

Extract Code

I extracted the following code from that output: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        long long u=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=u;
        for(int r=1;r<7;r++)for(int l=1;l<7;l++){
            int p=r*8+l,n=(t>>((r-1)*8+l))+(t>>(r*8+l-1))+(t>>(r+1)*8+l)+(t>>(r*8+l+1));
            u&=~((n==0&&(t>>p&1))<<p);
        }
        c+=(u==t);
    }
    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<1<<16;i++){
        long long u=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=u;
        for(int r=1;r<7;r++)for(int l=1;l<7;l++){
            int p=r*8+l,n=(t>>((r-1)*8+l))+(t>>(r*8+l-1))+(t>>(r+1)*8+l)+(t>>(r*8+l+1));
            u&=~((n==0&&(t>>p&1))<<p);
        }
        c+=(u==t);
    }
    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: Wrong answer; got 65536 instead of 27488.

Substring Evaluation

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

PyFunc

Resulting in output: False

Final Answer: False

Initial Query

Code golf this function to be a lot shorter. You only get points if it's <200 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 <200 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 a significantly shortened version of the function: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        long long unsigned int a=((i&0xf)<<9)|((i&0xf0)<<(9+4))|((i&0xf00)<<(9+8))|((i&0xf000)<<(9+12));
        long long unsigned int t=a;
        for(int y=1;y<7;y++)
            for(int x=1;x<7;x++){
                int p=y*8+x,n=(t>>((y-1)*8+x))+(t>>(y*8+x-1))+(t>>((y+1)*8+x))+(t>>(y*8+x+1))&1;
                if(n==0&&(t>>p&1))a&=~(1ULL<<p);
            }
        c+=(t==a);
    }
    return c;
}
Key changes: - Removed unnecessary variable names - Simplified neighbor calculation - Condensed logic - Shortened variable names - Removed explicit ROWS and COLS constants - Inlined some calculations The core logic remains the same: it checks every possible 4-bit configuration across an 8x8 grid, counting configurations where no isolated 1-bit cells exist. This version is much more compact while preserving the original function's behavior.

Extract Code

I extracted the following code from that output: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        long long unsigned int a=((i&0xf)<<9)|((i&0xf0)<<(9+4))|((i&0xf00)<<(9+8))|((i&0xf000)<<(9+12));
        long long unsigned int t=a;
        for(int y=1;y<7;y++)
            for(int x=1;x<7;x++){
                int p=y*8+x,n=(t>>((y-1)*8+x))+(t>>(y*8+x-1))+(t>>((y+1)*8+x))+(t>>(y*8+x+1))&1;
                if(n==0&&(t>>p&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<1<<16;i++){
        long long unsigned int a=((i&0xf)<<9)|((i&0xf0)<<(9+4))|((i&0xf00)<<(9+8))|((i&0xf000)<<(9+12));
        long long unsigned int t=a;
        for(int y=1;y<7;y++)
            for(int x=1;x<7;x++){
                int p=y*8+x,n=(t>>((y-1)*8+x))+(t>>(y*8+x-1))+(t>>((y+1)*8+x))+(t>>(y*8+x+1))&1;
                if(n==0&&(t>>p&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: Wrong answer; got 1824 instead of 27488.

Substring Evaluation

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

PyFunc

Resulting in output: False

Final Answer: False

Initial Query

Code golf this function to be a lot shorter. You only get points if it's <200 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 <200 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 a significantly shortened version: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
        for(int r=1;r<7;r++)
            for(int q=1;q<7;q++){
                int p=r*8+q,h=!(((t>>p-9)&1)+((t>>p-1)&1)+((t>>p+1)&1)+((t>>p+9)&1));
                a^=(h&((t>>p)&1))<<p;
            }
        c+=(a==t);
    }
    return c;
}
Key changes: - Removed verbose variable names - Simplified bit position calculations - Used bitwise operations for neighbor sum - Condensed logic into a single XOR operation - Removed explicit cell/neighbor extraction - Simplified nested loops This version is much more compact while preserving the core logic of the original function. It should be under 200 bytes and maintain the same functionality. The key trick is using bitwise operations to efficiently check neighbors and update cells, reducing the amount of code needed. Would you like me to explain the specific optimizations?

Extract Code

I extracted the following code from that output: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
        for(int r=1;r<7;r++)
            for(int q=1;q<7;q++){
                int p=r*8+q,h=!(((t>>p-9)&1)+((t>>p-1)&1)+((t>>p+1)&1)+((t>>p+9)&1));
                a^=(h&((t>>p)&1))<<p;
            }
        c+=(a==t);
    }
    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<1<<16;i++){
        unsigned long long a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21),t=a;
        for(int r=1;r<7;r++)
            for(int q=1;q<7;q++){
                int p=r*8+q,h=!(((t>>p-9)&1)+((t>>p-1)&1)+((t>>p+1)&1)+((t>>p+9)&1));
                a^=(h&((t>>p)&1))<<p;
            }
        c+=(a==t);
    }
    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: Wrong answer; got 19184 instead of 27488.

Substring Evaluation

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

PyFunc

Resulting in output: False

Final Answer: False

Initial Query

Code golf this function to be a lot shorter. You only get points if it's <200 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 <200 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 a significantly shortened version of the function: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        long long unsigned int a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21);
        long long unsigned int b=a;
        for(int y=1;y<7;y++)
            for(int x=1;x<7;x++){
                int p=y*8+x,n=(((b>>p-9)&1)+((b>>p-1)&1)+((b>>p+7)&1)+((b>>p+9)&1));
                a&=~((n==0&&((b>>p)&1))<<p);
            }
        c+=(a==b);
    }
    return c;
}
Key changes: - Shortened variable names - Condensed bit manipulation - Simplified neighbor calculation - Removed separate position calculation variables - Used shorter loop ranges - Inlined conditions This version maintains the same logic as the original function but is much more compact. It should be well under 200 bytes and preserve the core algorithm of checking specific bit patterns. The function still: - Iterates through all possible 16-bit configurations - Applies a specific neighborhood rule - Counts configurations that remain unchanged Would you like me to explain any specific part of the condensed version?

Extract Code

I extracted the following code from that output: 

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int count(){
    int c=0;
    for(int i=0;i<1<<16;i++){
        long long unsigned int a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21);
        long long unsigned int b=a;
        for(int y=1;y<7;y++)
            for(int x=1;x<7;x++){
                int p=y*8+x,n=(((b>>p-9)&1)+((b>>p-1)&1)+((b>>p+7)&1)+((b>>p+9)&1));
                a&=~((n==0&&((b>>p)&1))<<p);
            }
        c+=(a==b);
    }
    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<1<<16;i++){
        long long unsigned int a=((i&0xf)<<9)|((i&0xf0)<<13)|((i&0xf00)<<17)|((i&0xf000)<<21);
        long long unsigned int b=a;
        for(int y=1;y<7;y++)
            for(int x=1;x<7;x++){
                int p=y*8+x,n=(((b>>p-9)&1)+((b>>p-1)&1)+((b>>p+7)&1)+((b>>p+9)&1));
                a&=~((n==0&&((b>>p)&1))<<p);
            }
        c+=(a==b);
    }
    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: Wrong answer; got 7152 instead of 27488.

Substring Evaluation

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

PyFunc

Resulting in output: False

Final Answer: False