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;
  }
}
To do this follow these steps: 1. start out by creating a table of the values that are being written to the pins. 2. Then find an expression for this. 3. Write the function as short as possible. 4. See if you can make it any shorter. Return this as the final answer and call it your final answer. Make sure to name it stepper still.

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;
  }
}
To do this follow these steps: 1. start out by creating a table of the values that are being written to the pins. 2. Then find an expression for this. 3. Write the function as short as possible. 4. See if you can make it any shorter. Return this as the final answer and call it your final answer. Make sure to name it stepper still.

Output

Okay, let's golf this function. **1. Create a table of pin values:** Here's the state of pins 1, 2, 3, and 4 for each step: | Steps | Pin 1 | Pin 2 | Pin 3 | Pin 4 | Pattern (Binary 4321) | Pattern (Decimal) | | :---- | :---- | :---- | :---- | :---- | :-------------------- | :---------------- | | 0 | 0 | 0 | 0 | 1 | 1000 | 8 | | 1 | 0 | 0 | 1 | 1 | 1100 | 12 | | 2 | 0 | 0 | 1 | 0 | 0100 | 4 | | 3 | 0 | 1 | 1 | 0 | 0110 | 6 | | 4 | 0 | 1 | 0 | 0 | 0010 | 2 | | 5 | 1 | 1 | 0 | 0 | 0011 | 3 | | 6 | 1 | 0 | 0 | 0 | 0001 | 1 | | 7 | 1 | 0 | 0 | 1 | 1001 | 9 | **2. Find an expression:** The sequence of decimal patterns is `8, 12, 4, 6, 2, 3, 1, 9`. This doesn't have an obvious simple mathematical formula. The best approach is to use a lookup table (array) for these patterns. We can represent the state of the 4 pins as a single integer, where bit 0 corresponds to Pin 1, bit 1 to Pin 2, bit 2 to Pin 3, and bit 3 to Pin 4. The lookup table will be: `int p[] = {8, 12, 4, 6, 2, 3, 1, 9};` Inside the loop, for a given step `s`, the pattern is `p[s]`. We can get the state for pin `i` (where `i` is 1, 2, 3, or 4) by checking the `(i-1)`-th bit of the pattern: `(p[s] >> (i - 1)) & 1`. **3. Write the function as short as possible:** We can replace the `switch` statement with the lookup table and a loop for `digitalWrite`. We can also use shorter variable names, replace the outer `for` loop with a `while` loop decrementing `xw`, and use bitwise AND (`& 7`) instead of modulo (`% 8`). 

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// Intermediate shorter version
void stepper(int xw) {
  int s = 0; // Step counter
  int p[] = {8, 12, 4, 6, 2, 3, 1, 9}; // Patterns
  while (xw--) { // Loop xw times
    int pattern = p[s]; // Get current pattern
    for (int i = 0; i < 4; i++) { // Loop through pins 0..3
      // Write to pin i+1 the value of bit i
      digitalWrite(i + 1, (pattern >> i) & 1);
    }
    s = ++s & 7; // Increment step, wrap around 0-7
  }
}
**4. Make it shorter:** We can further shorten this by: * Declaring loop variable `i` outside the `for` loop to shorten the `for` statement. * Removing the temporary `pattern` variable. * Removing unnecessary braces where possible (though the prompt implies counting bytes excluding whitespace, braces are still characters). 
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// Golfed version
void stepper(int xw){
  int s=0,i; // Declare s and i
  int p[]={8,12,4,6,2,3,1,9}; // Lookup table
  while(xw--){ // Loop xw times
    for(i=0;i<4;i++) // Loop pins 0..3
      digitalWrite(i+1,(p[s]>>i)&1); // Set pin i+1 based on bit i of pattern p[s]
    s=++s&7; // Next step (0-7)
  }
}
Let's count the bytes of the core logic (excluding `void stepper(int xw){` and `}`): `int s=0,i;int p[]={8,12,4,6,2,3,1,9};while(xw--){for(i=0;i<4;i++)digitalWrite(i+1,(p[s]>>i)&1);s=++s&7;}` * `int s=0,i;` = 10 bytes * `int p[]={8,12,4,6,2,3,1,9};` = 27 bytes * `while(xw--){` = 11 bytes * `for(i=0;i<4;i++)` = 14 bytes * `digitalWrite(i+1,(p[s]>>i)&1);` = 28 bytes * `s=++s&7;` = 8 bytes Total = 10 + 27 + 11 + 14 + 28 + 8 = **98 bytes**. This is well under the 300-byte limit. **Final Answer:** 
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void stepper(int xw){
int s=0,i;int p[]={8,12,4,6,2,3,1,9};while(xw--){for(i=0;i<4;i++)digitalWrite(i+1,(p[s]>>i)&1);s=++s&7;}
}

Extract Code

I extracted the following code from that output: 

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

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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void stepper(int xw){
int s=0,i;int p[]={8,12,4,6,2,3,1,9};while(xw--){for(i=0;i<4;i++)digitalWrite(i+1,(p[s]>>i)&1);s=++s&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: All tests passed main.c: In function 'stepper': main.c:3:66: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 3 | int s=0,i;int p[]={8,12,4,6,2,3,1,9};while(xw--){for(i=0;i<4;i++)digitalWrite(i+1,(p[s]>>i)&1);s=++s&7;} | ^~~~~~~~~~~~

Substring Evaluation

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

PyFunc

Resulting in output: True

Final Answer: True