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 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 <200 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

Here's a significantly shorter version of the `stepper` function, under 200 bytes (excluding whitespace): 

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void stepper(int xw){for(int s=0,x=0;x<xw;x++){int p[]={0,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,0,0,0,1,0,0,1};for(int i=0;i<4;i++)digitalWrite(i+1,p[s*4+i]);s=(s+1)%8;}}
### Explanation: 1. **Array of Patterns**: The `p` array stores the patterns for each step. Each step is represented by 4 values (one for each pin). 2. **Loop**: The outer loop iterates `xw` times. 3. **Inner Loop**: The inner loop iterates over the 4 pins and sets their state based on the current step `s`. 4. **Step Update**: After setting the pins, the step `s` is updated to the next step, wrapping around after 8 steps. This version is concise and achieves the same functionality as the original.

Extract Code

I extracted the following code from that output: 

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void stepper(int xw){for(int s=0,x=0;x<xw;x++){int p[]={0,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,0,0,0,1,0,0,1};for(int i=0;i<4;i++)digitalWrite(i+1,p[s*4+i]);s=(s+1)%8;}}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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void stepper(int xw){for(int s=0,x=0;x<xw;x++){int p[]={0,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,0,0,0,1,0,0,1};for(int i=0;i<4;i++)digitalWrite(i+1,p[s*4+i]);s=(s+1)%8;}}



#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: All tests passed main.c: In function 'stepper': main.c:2:142: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 2 | void stepper(int xw){for(int s=0,x=0;x<xw;x++){int p[]={0,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,0,1,0,0,0,1,0,0,1};for(int i=0;i<4;i++)digitalWrite(i+1,p[s*4+i]);s=(s+1)%8;}} | ^~~~~~~~~~~~

Substring Evaluation

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

PyFunc

Resulting in output: True

Final Answer: True