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

To solve this problem, we need to optimize the given function `stepper` to be as concise as possible while maintaining its functionality. The goal is to reduce the code to under 300 bytes by leveraging bitwise operations and array lookups instead of a lengthy switch-case structure. ### Approach The original function uses a switch-case structure to set the states of four pins (1-4) in a specific sequence. By analyzing the sequence, we can determine that each case corresponds to a 4-bit binary number. These binary numbers can be stored in an array, and for each iteration, we can use bitwise operations to determine the state of each pin. This approach significantly reduces the code length by eliminating repetitive switch-case statements and using bitwise operations for efficiency. ### Solution Code 

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void stepper(int xw){for(int s=0;s<xw;s++){int v=(int[]){1,3,2,6,4,12,8,9}[s&7];for(int p=1;p<5;p++)digitalWrite(p,v>>(4-p)&1);}}
### Explanation 1. **Array Initialization**: The array `{1, 3, 2, 6, 4, 12, 8, 9}` represents the binary states of the pins for each step. Each element in the array corresponds to the combined state of pins 1-4 as a 4-bit binary number. 2. **Loop through Steps**: The outer loop runs `xw` times, simulating each step of the stepper motor. 3. **Modulo Operation**: The step index `s` is taken modulo 8 (using bitwise operation `s & 7` for efficiency) to cycle through the array indices 0 to 7 repeatedly. 4. **Bitwise Operations**: For each pin (1-4), the code uses bitwise right shift and bitwise AND to extract the corresponding bit from the array element, determining whether the pin should be set to HIGH (1) or LOW (0). This approach efficiently replaces the lengthy switch-case structure with a compact loop and bitwise operations, achieving the desired functionality in fewer bytes.

Extract Code

I extracted the following code from that output: 

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

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;s<xw;s++){int v=(int[]){1,3,2,6,4,12,8,9}[s&7];for(int p=1;p<5;p++)digitalWrite(p,v>>(4-p)&1);}}



#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:2:101: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 2 | void stepper(int xw){for(int s=0;s<xw;s++){int v=(int[]){1,3,2,6,4,12,8,9}[s&7];for(int p=1;p<5;p++)digitalWrite(p,v>>(4-p)&1);}} | ^~~~~~~~~~~~

Substring Evaluation

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

PyFunc

Resulting in output: True

Final Answer: True