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 name `stepper`. 

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void stepper(int xw) {
  int Steps = 0;
  for (int x = 0; x < xw; x++) {
    switch (Steps) {
      case 0:
        digitalWrite(1, 0);
        digitalWrite(2, 0);
        digitalWrite(3, 0);
        digitalWrite(4, 1);
        break;
      case 1:
        digitalWrite(1, 0);
        digitalWrite(2, 0);
        digitalWrite(3, 1);
        digitalWrite(4, 1);
        break;
      case 2:
        digitalWrite(1, 0);
        digitalWrite(2, 0);
        digitalWrite(3, 1);
        digitalWrite(4, 0);
        break;
      case 3:
        digitalWrite(1, 0);
        digitalWrite(2, 1);
        digitalWrite(3, 1);
        digitalWrite(4, 0);
        break;
      case 4:
        digitalWrite(1, 0);
        digitalWrite(2, 1);
        digitalWrite(3, 0);
        digitalWrite(4, 0);
        break;
      case 5:
        digitalWrite(1, 1);
        digitalWrite(2, 1);
        digitalWrite(3, 0);
        digitalWrite(4, 0);
        break;
      case 6:
        digitalWrite(1, 1);
        digitalWrite(2, 0);
        digitalWrite(3, 0);
        digitalWrite(4, 0);
        break;
      case 7:
        digitalWrite(1, 1);
        digitalWrite(2, 0);
        digitalWrite(3, 0);
        digitalWrite(4, 1);
        break;
      default:
        digitalWrite(1, 0);
        digitalWrite(2, 0);
        digitalWrite(3, 0);
        digitalWrite(4, 0);
        break;
    }
    Steps = (Steps + 1) % 8;
  }
}

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 name `stepper`. 

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void stepper(int xw) {
  int Steps = 0;
  for (int x = 0; x < xw; x++) {
    switch (Steps) {
      case 0:
        digitalWrite(1, 0);
        digitalWrite(2, 0);
        digitalWrite(3, 0);
        digitalWrite(4, 1);
        break;
      case 1:
        digitalWrite(1, 0);
        digitalWrite(2, 0);
        digitalWrite(3, 1);
        digitalWrite(4, 1);
        break;
      case 2:
        digitalWrite(1, 0);
        digitalWrite(2, 0);
        digitalWrite(3, 1);
        digitalWrite(4, 0);
        break;
      case 3:
        digitalWrite(1, 0);
        digitalWrite(2, 1);
        digitalWrite(3, 1);
        digitalWrite(4, 0);
        break;
      case 4:
        digitalWrite(1, 0);
        digitalWrite(2, 1);
        digitalWrite(3, 0);
        digitalWrite(4, 0);
        break;
      case 5:
        digitalWrite(1, 1);
        digitalWrite(2, 1);
        digitalWrite(3, 0);
        digitalWrite(4, 0);
        break;
      case 6:
        digitalWrite(1, 1);
        digitalWrite(2, 0);
        digitalWrite(3, 0);
        digitalWrite(4, 0);
        break;
      case 7:
        digitalWrite(1, 1);
        digitalWrite(2, 0);
        digitalWrite(3, 0);
        digitalWrite(4, 1);
        break;
      default:
        digitalWrite(1, 0);
        digitalWrite(2, 0);
        digitalWrite(3, 0);
        digitalWrite(4, 0);
        break;
    }
    Steps = (Steps + 1) % 8;
  }
}

Output

Okay, here's a golfed version using a lookup table and bit manipulation. 

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void stepper(int n){int s=0;char p[]={8,12,4,6,2,3,1,9};while(n--){for(int j=1;j<5;j++)digitalWrite(j,(p[s]>>(4-j))&1);s=(s+1)&7;}}
**Explanation:** 1. **`void stepper(int n)`**: Renamed `xw` to `n` for brevity. 2. **`int s=0;`**: Renamed `Steps` to `s` and initialized it. 3. **`char p[]={8,12,4,6,2,3,1,9};`**: This is the core optimization. * It stores the 8 distinct output patterns as numbers in a `char` array (`p`). `char` is used as the values are small (0-12). * Each number represents the state of pins 1-4. We map Pin 1 to the most significant bit (MSB) and Pin 4 to the least significant bit (LSB). * Case 0: `0001` -> `digitalWrite(1,0); digitalWrite(2,0); digitalWrite(3,0); digitalWrite(4,1);` -> Binary `1000` -> Decimal `8` * Case 1: `0011` -> `digitalWrite(1,0); digitalWrite(2,0); digitalWrite(3,1); digitalWrite(4,1);` -> Binary `1100` -> Decimal `12` * Case 2: `0010` -> `digitalWrite(1,0); digitalWrite(2,0); digitalWrite(3,1); digitalWrite(4,0);` -> Binary `0100` -> Decimal `4` * Case 3: `0110` -> `digitalWrite(1,0); digitalWrite(2,1); digitalWrite(3,1); digitalWrite(4,0);` -> Binary `0110` -> Decimal `6` * Case 4: `0100` -> `digitalWrite(1,0); digitalWrite(2,1); digitalWrite(3,0); digitalWrite(4,0);` -> Binary `0010` -> Decimal `2` * Case 5: `1100` -> `digitalWrite(1,1); digitalWrite(2,1); digitalWrite(3,0); digitalWrite(4,0);` -> Binary `0011` -> Decimal `3` * Case 6: `1000` -> `digitalWrite(1,1); digitalWrite(2,0); digitalWrite(3,0); digitalWrite(4,0);` -> Binary `0001` -> Decimal `1` * Case 7: `1001` -> `digitalWrite(1,1); digitalWrite(2,0); digitalWrite(3,0); digitalWrite(4,1);` -> Binary `1001` -> Decimal `9` * The array `p` now holds the sequence `8, 12, 4, 6, 2, 3, 1, 9`. 4. **`while(n--)`**: A shorter way to loop `n` times compared to `for(int x=0; x<n; x++)`. 5. **`for(int j=1;j<5;j++)`**: Loop through the pin numbers (1, 2, 3, 4). `j` is shorter than `pin`. 6. **`digitalWrite(j,(p[s]>>(4-j))&1);`**: This sets the pin state. * `p[s]` fetches the pattern number for the current step `s`. * `(p[s]>>(4-j))` shifts the pattern bits right so the bit corresponding to pin `j` is at the LSB position (position 0). * If `j=1`, shift by `4-1=3` bits. * If `j=2`, shift by `4-2=2` bits. * If `j=3`, shift by `4-3=1` bit. * If `j=4`, shift by `4-4=0` bits. * `&1` masks the result to get just the least significant bit (0 or 1). * `digitalWrite(j, ...)` writes this bit value to the pin `j`. 7. **`s=(s+1)&7;`**: Increments the step counter `s` and wraps it around using bitwise AND with 7 (binary `111`), which is equivalent to `% 8` for positive numbers and often shorter/faster. 8. **Removed Braces**: Braces for single-statement loops (`while`, `for`) are removed to save bytes. **Byte Count (excluding whitespace):** `voidstepper(intn){ints=0;charp[]={8,12,4,6,2,3,1,9};while(n--){for(intj=1;j<5;j++)digitalWrite(j,(p[s]>>(4-j))&1);s=(s+1)&7;}}` Counting the characters: 18 + 7 + 27 + 10 + 19 + 31 + 10 + 2 = **124 bytes**. This is well under the 300-byte limit.

Extract Code

I extracted the following code from that output: 

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++
void stepper(int n){int s=0;char p[]={8,12,4,6,2,3,1,9};while(n--){for(int j=1;j<5;j++)digitalWrite(j,(p[s]>>(4-j))&1);s=(s+1)&7;}}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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++
void stepper(int n){int s=0;char p[]={8,12,4,6,2,3,1,9};while(n--){for(int j=1;j<5;j++)digitalWrite(j,(p[s]>>(4-j))&1);s=(s+1)&7;}}



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

#include <stdio.h>

int out = 0;
int digitalWrite(int pin, int value) {
  out *= 101;
  out += pin;
  out *= 97;
  out += value;
  return out;
}

int main() {


int answer = (stepper(100),out);
int expected = -1475775710;
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: main.c:1:1: error: expected identifier or '(' before '++' token 1 | ++ | ^~ main.c: In function 'main': main.c:23:15: warning: implicit declaration of function 'stepper' [-Wimplicit-function-declaration] 23 | int answer = (stepper(100),out); | ^~~~~~~ main.sh: line 2: ./a.out: No such file or directory

Substring Evaluation

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

PyFunc

Resulting in output: True

Final Answer: False