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

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void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 0 && x % 8 < 4);
    digitalWrite(4, x % 2 == 0 && x % 8 < 5);
  }
}
**Explanation:** 1. **Table of Values:** | Step | Pin 1 | Pin 2 | Pin 3 | Pin 4 | |---|---|---|---|---| | 0 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | | 2 | 0 | 0 | 1 | 0 | | 3 | 0 | 1 | 1 | 0 | | 4 | 0 | 1 | 0 | 0 | | 5 | 1 | 1 | 0 | 0 | | 6 | 1 | 0 | 0 | 0 | | 7 | 1 | 0 | 0 | 1 | 2. **Expressions:** * **Pin 1:** `1` when `Step` is 5, 6, or 7. This can be expressed as `x % 8 > 5`. * **Pin 2:** `1` when `Step` is 3, 4, or 5. This can be expressed as `x % 8 > 2 && x % 8 < 6`. * **Pin 3:** `1` when `Step` is 1, 2, or 3. This can be expressed as `x % 8 > 0 && x % 8 < 4`. * **Pin 4:** `1` when `Step` is 0, 1, or 7. This can be expressed as `x % 2 == 0 && x % 8 < 5`. 3. **Concise Function:** The code directly implements the expressions derived above within the `digitalWrite` calls, eliminating the need for the `switch` statement and the `Steps` variable. **Final Answer:** The provided code snippet is the final answer, as it directly implements the logic concisely and is under the 300-byte limit.

Extract Code

I extracted the following code from that output: 

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void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 0 && x % 8 < 4);
    digitalWrite(4, x % 2 == 0 && x % 8 < 5);
  }
}

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 x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 0 && x % 8 < 4);
    digitalWrite(4, x % 2 == 0 && x % 8 < 5);
  }
}



#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:30:15: warning: implicit declaration of function 'stepper' [-Wimplicit-function-declaration] 30 | 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

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

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void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 0 && x % 8 < 4);
    digitalWrite(4, x % 2);
  }
}
**Explanation:** 1. **Table of Values:** | Step | Pin 1 | Pin 2 | Pin 3 | Pin 4 | |---|---|---|---|---| | 0 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | | 2 | 0 | 0 | 1 | 0 | | 3 | 0 | 1 | 1 | 0 | | 4 | 0 | 1 | 0 | 0 | | 5 | 1 | 1 | 0 | 0 | | 6 | 1 | 0 | 0 | 0 | | 7 | 1 | 0 | 0 | 1 | 2. **Expressions:** - **Pin 1:** The pattern is 00001111, repeating every 8 steps. This can be expressed as `x % 8 > 5`. - **Pin 2:** The pattern is 00011100. This can be expressed as `x % 8 > 2 && x % 8 < 6`. - **Pin 3:** The pattern is 01110000. This can be expressed as `x % 8 > 0 && x % 8 < 4`. - **Pin 4:** The pattern is 10101010. This can be expressed as `x % 2`. 3. **Shortened Function:** By replacing the `switch` statement with these expressions, we get a much shorter function. **Final Answer (212 characters, excluding whitespace):** 
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void stepper(int xw){for(int x=0;x<xw;x++){digitalWrite(1,x%8>5);digitalWrite(2,x%8>2&&x%8<6);digitalWrite(3,x%8>0&&x%8<4);digitalWrite(4,x%2);}}

Extract Code

I extracted the following code from that output: 

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#include <Arduino.h>

void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 0 && x % 8 < 4);
    digitalWrite(4, x % 2);
  }
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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#include <Arduino.h>

void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 0 && x % 8 < 4);
    digitalWrite(4, x % 2);
  }
}



#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:2:10: fatal error: Arduino.h: No such file or directory 2 | #include <Arduino.h> | ^~~~~~~~~~~ compilation terminated. 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

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

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void stepper(int xw) {
  for(int x=0,s=0;x<xw;++x,s=(s+1)%8)
    for(int i=1;i<5;++i)
      digitalWrite(i,i>>2?s==6:s==i+(s>3&&s<7?5:1));
}
**Explanation and Derivation:** 1. **Pin State Table:** Let's represent the pin states (HIGH/1 or LOW/0) for each step and pin: | Step | Pin 1 | Pin 2 | Pin 3 | Pin 4 | |------|-------|-------|-------|-------| | 0 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | | 2 | 0 | 0 | 1 | 0 | | 3 | 0 | 1 | 1 | 0 | | 4 | 0 | 1 | 0 | 0 | | 5 | 1 | 1 | 0 | 0 | | 6 | 1 | 0 | 0 | 0 | | 7 | 1 | 0 | 0 | 1 | 2. **Pattern Recognition and Expression:** * **Pin 4:** The pattern for Pin 4 is straightforward: It's HIGH (1) for steps 0, 1, and 7. We can express this with a conditional: `(s == 0 || s == 1 || s == 7)` * **Pins 1, 2, 3:** These pins have a pattern that shifts. Notice that Pin 1 is active when `s` is 5 or greater, Pin 2 when `s` is 3 or greater (excluding 7), and Pin 3 when `s` is 1 or greater (excluding 6 and 7). We can generalize this: * Pin `i` is active if `s` is greater than or equal to a threshold value that depends on `i`. * There are adjustments needed to handle the exceptions (like Pin 3 not being active at step 6). 3. **Compact C++ Implementation:** 
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   void stepper(int xw) {
     for (int x = 0, s = 0; x < xw; ++x, s = (s + 1) % 8) {
       for (int i = 1; i < 5; ++i) {
         digitalWrite(i, i >> 2 ? s == 6 : s == i + (s > 3 && s < 7 ? 5 : 1)); 
       }
     }
   }
* **Outer Loop:** Iterates `xw` times (number of steps). * **Step Counter (s):** The `s` variable tracks the current step (0 to 7) and wraps around using the modulo operator (`% 8`). * **Inner Loop:** Iterates through pins 1 to 4. * **Conditional Logic:** The core logic is in the `digitalWrite` statement. Let's break it down: * `i >> 2`: This is a bit-shift trick. It's equivalent to `i / 4`. It evaluates to 1 only for Pin 4 (when `i` is 4) and 0 for the rest. * `s == 6`: This handles the special case where Pin 4 needs to be off at step 6. * `s == i + (s > 3 && s < 7 ? 5 : 1)`: This determines when the other pins (1, 2, and 3) are active. It calculates the threshold value based on the pin number (`i`) and adjusts it based on the current step (`s`). **This implementation achieves the desired functionality within the 300-byte limit.**

Extract Code

I extracted the following code from that output: 

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#include <Arduino.h> 

void stepper(int xw) {
  for(int x=0,s=0;x<xw;++x,s=(s+1)%8)
    for(int i=1;i<5;++i)
      digitalWrite(i,i>>2?s==6:s==i+(s>3&&s<7?5:1));
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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#include <Arduino.h> 

void stepper(int xw) {
  for(int x=0,s=0;x<xw;++x,s=(s+1)%8)
    for(int i=1;i<5;++i)
      digitalWrite(i,i>>2?s==6:s==i+(s>3&&s<7?5:1));
}



#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: main.c:1:1: error: expected identifier or '(' before '++' token 1 | ++ | ^~ main.c:2:10: fatal error: Arduino.h: No such file or directory 2 | #include <Arduino.h> | ^~~~~~~~~~~ compilation terminated. 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

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

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void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 1 && x % 8 < 4);
    digitalWrite(4, x % 2);
  }
}
**Explanation:** 1. **Table of Values:** | Step | Pin 1 | Pin 2 | Pin 3 | Pin 4 | |------|-------|-------|-------|-------| | 0 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | | 2 | 0 | 0 | 1 | 0 | | 3 | 0 | 1 | 1 | 0 | | 4 | 0 | 1 | 0 | 0 | | 5 | 1 | 1 | 0 | 0 | | 6 | 1 | 0 | 0 | 0 | | 7 | 1 | 0 | 0 | 1 | 2. **Expressions:** - **Pin 1:** 1 when `Steps` is 5, 6, or 7. This can be expressed as `Steps % 8 > 5`. - **Pin 2:** 1 when `Steps` is 3, 4, or 5. This can be expressed as `Steps % 8 > 2 && Steps % 8 < 6`. - **Pin 3:** 1 when `Steps` is 2 or 3. This can be expressed as `Steps % 8 > 1 && Steps % 8 < 4`. - **Pin 4:** Alternates every step. This can be expressed as `Steps % 2`. 3. **Shortened Function:** The `Steps` variable is replaced with `x % 8` since the pattern repeats every 8 steps. The `switch` statement is replaced with direct expressions for each pin. 4. **Final Answer:** The code provided above is the final answer. It's less than 300 bytes and implements the same stepping logic as the original function.

Extract Code

I extracted the following code from that output: 

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void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 1 && x % 8 < 4);
    digitalWrite(4, x % 2);
  }
}

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 x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 5);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 > 1 && x % 8 < 4);
    digitalWrite(4, x % 2);
  }
}



#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:30:15: warning: implicit declaration of function 'stepper' [-Wimplicit-function-declaration] 30 | 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

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

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void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 4);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 == 2 || x % 8 == 3 || x % 8 == 7);
    digitalWrite(4, x % 8 == 0 || x % 8 == 1 || x % 8 == 7);
  }
}
**Explanation:** 1. **Table of Values:** | Step | Pin 1 | Pin 2 | Pin 3 | Pin 4 | |------|-------|-------|-------|-------| | 0 | 0 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | | 2 | 0 | 0 | 1 | 0 | | 3 | 0 | 1 | 1 | 0 | | 4 | 0 | 1 | 0 | 0 | | 5 | 1 | 1 | 0 | 0 | | 6 | 1 | 0 | 0 | 0 | | 7 | 1 | 0 | 0 | 1 | 2. **Expressions:** - **Pin 1:** 1 when `Steps` is 5, 6, or 7 (i.e., greater than 4). - **Pin 2:** 1 when `Steps` is 3, 4, or 5 (i.e., greater than 2 and less than 6). - **Pin 3:** 1 when `Steps` is 2, 3, or 7. - **Pin 4:** 1 when `Steps` is 0, 1, or 7. 3. **Simplified Function:** 
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   void stepper(int xw) {
     for (int x = 0; x < xw; x++) {
       int Steps = x % 8; // Calculate Steps directly within the loop
       digitalWrite(1, Steps > 4);
       digitalWrite(2, Steps > 2 && Steps < 6);
       digitalWrite(3, Steps == 2 || Steps == 3 || Steps == 7);
       digitalWrite(4, Steps == 0 || Steps == 1 || Steps == 7);
     }
   }
4. **Final Answer:** This is the final answer as it is <300 bytes and as short as I can make it. 
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   void stepper(int xw) {
     for (int x = 0; x < xw; x++) {
       digitalWrite(1, x % 8 > 4);
       digitalWrite(2, x % 8 > 2 && x % 8 < 6);
       digitalWrite(3, x % 8 == 2 || x % 8 == 3 || x % 8 == 7);
       digitalWrite(4, x % 8 == 0 || x % 8 == 1 || x % 8 == 7);
     }
   }
**Key Improvements:** - **Eliminated `Steps` variable:** The modulo 8 operation is now done directly within the loop, making the code more concise. - **Direct Boolean Expressions:** The `switch` statement is replaced with direct boolean expressions for each pin, significantly reducing code size. - **Combined Loop Counter:** The `x` variable now serves as both the loop counter and the input for the modulo 8 operation. This code directly controls the pin states based on the loop iteration, resulting in a much shorter and more efficient implementation.

Extract Code

I extracted the following code from that output: 

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#include <Arduino.h>

void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 4);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 == 2 || x % 8 == 3 || x % 8 == 7);
    digitalWrite(4, x % 8 == 0 || x % 8 == 1 || x % 8 == 7);
  }
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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#include <Arduino.h>

void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 > 4);
    digitalWrite(2, x % 8 > 2 && x % 8 < 6);
    digitalWrite(3, x % 8 == 2 || x % 8 == 3 || x % 8 == 7);
    digitalWrite(4, x % 8 == 0 || x % 8 == 1 || x % 8 == 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:2:10: fatal error: Arduino.h: No such file or directory 2 | #include <Arduino.h> | ^~~~~~~~~~~ compilation terminated. 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