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

1. **Table of Values:** | Steps | 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:** Observing the table, we can derive the following expressions using bitwise operations: * Pin 1: `(Steps >> 2) & 1` (equivalent to `Steps >= 4`) * Pin 2: `(Steps >> 1) & 1` (equivalent to `Steps == 3 || Steps == 4 || Steps == 5`) * Pin 3: `(Steps ^ (Steps >> 1)) & 1` (equivalent to `Steps == 1 || Steps == 2 || Steps == 7` ) * Pin 4: `Steps & 1` (equivalent to `Steps % 2 == 0`) 3. **Short Function:** 

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void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) % 8) {
    digitalWrite(1, (Steps >> 2) & 1);
    digitalWrite(2, (Steps >> 1) & 1);
    digitalWrite(3, (Steps ^ (Steps >> 1)) & 1);
    digitalWrite(4, Steps & 1);
  }
}
4. **Final Answer (Shorter):** 
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void stepper(int xw) {
  for (int x = 0, s = 0; x < xw; x++, s = ++s % 8) 
    for(int i=1; i<5; i++) digitalWrite(i, (s >> (4-i)) & (i!=3 ? 1 : (1 ^ (s>>1)&1) ));
}
This version combines the loop increment with the modulo operation, and uses a nested loop to iterate through the pins, calculating the `digitalWrite` argument within the loop. It handles the slightly different logic for pin 3 using the ternary operator. This version comes in at under 300 characters (not including whitespace). It can be even shorter if pin order doesn't matter (the pin numbers could be reordered to simplify the logic).

Extract Code

I extracted the following code from that output: 

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void stepper(int xw) {
  for (int x = 0, s = 0; x < xw; x++, s = ++s % 8) 
    for(int i=1; i<5; i++) digitalWrite(i, (s >> (4-i)) & (i!=3 ? 1 : (1 ^ (s>>1)&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 x = 0, s = 0; x < xw; x++, s = ++s % 8) 
    for(int i=1; i<5; i++) digitalWrite(i, (s >> (4-i)) & (i!=3 ? 1 : (1 ^ (s>>1)&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: In function 'main': main.c:26:15: warning: implicit declaration of function 'stepper' [-Wimplicit-function-declaration] 26 | 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

1. **Table of Values:** | Steps | 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. **Expression:** Observing the table, we can see patterns for each pin based on the `Steps` value: * Pin 1: `(Steps >> 2) & 1` (equivalent to `Steps >= 4`) * Pin 2: `(Steps >> 1) & 1` (equivalent to `Steps == 2 || Steps == 3 || Steps == 4 || Steps == 5`) * Pin 3: `Steps & 1 ^ (Steps >> 1) & 1` (equivalent to Steps == 1, 2, 4, 7) * Pin 4: `Steps & 1 ^ (Steps >> 2) & 1 ^ (Steps >> 3) & 1` (equivalent to 0, 1, 3, 7) 3. **Short Function:** 

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void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) % 8) {
    digitalWrite(1, (Steps >> 2) & 1);
    digitalWrite(2, (Steps >> 1) & 1);
    digitalWrite(3, Steps & 1 ^ (Steps >> 1) & 1);
    digitalWrite(4, Steps & 1 ^ (Steps >> 2) & 1 ^ (Steps >> 3) & 1);
  }
}
4. **Final Answer (Shorter):** 
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void stepper(int xw){
  for(int x=0,s=0;x<xw;x++,s=(s+1)&7) {
      digitalWrite(1,s/4);
      digitalWrite(2,s/2&1);
      digitalWrite(3,s&1^s/2&1);
      digitalWrite(4,s&1^s/4^s/8); 
  }
}
This version uses integer division (`/`) instead of bit shifts (`>>`) where applicable. It also uses the modulo operator as a bitwise AND with 7 (`&7`), which is equivalent for modulo 8 when the value is positive. This shaves off a few more characters. This final version is well under the 300-byte limit.

Extract Code

I extracted the following code from that output: 

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#include <Arduino.h> // Assuming Arduino environment

void stepper(int xw){
  for(int x=0,s=0;x<xw;x++,s=(s+1)&7) {
      digitalWrite(1,s/4);
      digitalWrite(2,s/2&1);
      digitalWrite(3,s&1^s/2&1);
      digitalWrite(4,s&1^s/4^s/8); 
  }
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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#include <Arduino.h> // Assuming Arduino environment

void stepper(int xw){
  for(int x=0,s=0;x<xw;x++,s=(s+1)&7) {
      digitalWrite(1,s/4);
      digitalWrite(2,s/2&1);
      digitalWrite(3,s&1^s/2&1);
      digitalWrite(4,s&1^s/4^s/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: 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> // Assuming Arduino environment | ^~~~~~~~~~~ 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

1. **Table of Values:** | Steps | 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:** Observing the table, we can derive the following expressions using bitwise operations: * Pin 1: `(Steps >> 2) & 1` * Pin 2: `(Steps >> 1) & 1` * Pin 3: `Steps & 1` * Pin 4: `(Steps == 0) || (Steps == 1) || (Steps == 7)` which simplifies to `(Steps & 1) ^ ((Steps >> 1) & 1)` 3. **Shorter Function:** 

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void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) % 8) {
    digitalWrite(1, (Steps >> 2) & 1);
    digitalWrite(2, (Steps >> 1) & 1);
    digitalWrite(3, Steps & 1);
    digitalWrite(4, (Steps & 1) ^ ((Steps >> 1) & 1));
  }
}
4. **Final Answer (Even Shorter):** 
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void stepper(int xw){
  for(int x=0,s=0;x<xw;x++,s=(s+1)&7) {
    digitalWrite(1,s>>2);
    digitalWrite(2,s>>1&1);
    digitalWrite(3,s&1);
    digitalWrite(4,(s&1)^(s>>1&1));
  }
}
This version uses the `&7` for the modulo 8 operation, eliminating the more verbose `% 8`. It also omits redundant `& 1` operations since `digitalWrite` implicitly handles truthiness. This final version is well 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,s=0;x<xw;x++,s=(s+1)&7) {
    digitalWrite(1,s>>2);
    digitalWrite(2,s>>1&1);
    digitalWrite(3,s&1);
    digitalWrite(4,(s&1)^(s>>1&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 x=0,s=0;x<xw;x++,s=(s+1)&7) {
    digitalWrite(1,s>>2);
    digitalWrite(2,s>>1&1);
    digitalWrite(3,s&1);
    digitalWrite(4,(s&1)^(s>>1&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: 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

1. **Table of Values:** | Steps | 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. **Expression:** Observing the table, we can see patterns for each pin based on the `Steps` value: * Pin 1: `(Steps >> 2) & 1` (equivalent to `Steps >= 4`) * Pin 2: `(Steps >> 1) & 1` (equivalent to `Steps == 3 || Steps == 4 || Steps == 5 || Steps ==6`) * Pin 3: `Steps & 1 ^ ((Steps >> 1) & 1)` (XOR - equivalent to `Steps == 1 || Steps == 2`) * Pin 4: `Steps & 1 ^ ((Steps >> 2) & 1)` (XOR - equivalent to `Steps == 0 || Steps == 1 || Steps == 7`) 3. **Shortened Function:** 

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void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) % 8) {
    digitalWrite(1, (Steps >> 2) & 1);
    digitalWrite(2, (Steps >> 1) & 1);
    digitalWrite(3, Steps & 1 ^ ((Steps >> 1) & 1));
    digitalWrite(4, Steps & 1 ^ (Steps >> 2)); //Simplified slightly
  }
}
4. **Final Answer (Further Shortened):** 
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void stepper(int xw){
  for(int x=0,s=0;x<xw;x++,s=(s+1)&7) { // Use &7 instead of %8
    digitalWrite(1,s/4);  // Integer division is equivalent to >> 2 and & 1
    digitalWrite(2,s/2&1); // Combined >> 1 and & 1
    digitalWrite(3,s&1^s/2&1); // Combined bitwise operations
    digitalWrite(4,s&1^s/4);  // Combined bitwise operations
  }
}
This version uses integer division combined with bitwise AND to replace the right shift and AND operations, making it slightly shorter. Using `&7` instead of `%8` also saves a character. This final version comes in at well under 300 bytes.

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)&7) { // Use &7 instead of %8
    digitalWrite(1,s/4);  // Integer division is equivalent to >> 2 and & 1
    digitalWrite(2,s/2&1); // Combined >> 1 and & 1
    digitalWrite(3,s&1^s/2&1); // Combined bitwise operations
    digitalWrite(4,s&1^s/4);  // Combined bitwise operations
  }
}

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)&7) { // Use &7 instead of %8
    digitalWrite(1,s/4);  // Integer division is equivalent to >> 2 and & 1
    digitalWrite(2,s/2&1); // Combined >> 1 and & 1
    digitalWrite(3,s&1^s/2&1); // Combined bitwise operations
    digitalWrite(4,s&1^s/4);  // Combined bitwise operations
  }
}



#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

1. **Truth Table:** | Steps | 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. **Bitwise Expressions:** Observing the table, we can derive the following: * Pin 1: `(Steps >> 2) & 1` (equivalent to `Steps >= 4`) * Pin 2: `(Steps >> 1) & 1` XOR `(Steps >> 2) & 1` * Pin 3: `Steps & 1` XOR `(Steps >> 1) & 1` * Pin 4: `(Steps & 1)` XOR `(Steps >> 2) & 1` 3. **Initial Shortened Function:** 

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void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) % 8) {
    digitalWrite(1, (Steps >> 2) & 1);
    digitalWrite(2, ((Steps >> 1) & 1) ^ ((Steps >> 2) & 1));
    digitalWrite(3, (Steps & 1) ^ ((Steps >> 1) & 1));
    digitalWrite(4, (Steps & 1) ^ ((Steps >> 2) & 1));
  }
}
4. **Final Answer (Further Optimized):** 
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void stepper(int xw){
  for(int x=0,s=0;x<xw;x++,s=(s+1)&7) {
      digitalWrite(1,s/4);
      digitalWrite(2,(s/2)^s/4);
      digitalWrite(3,(s&1)^(s/2));
      digitalWrite(4,(s&1)^s/4);
  }
}
**Explanation of Optimizations:** * **Modulo 8:** Replacing `% 8` with `& 7` is faster as it uses a bitwise AND operation, which is generally more efficient than modulo. This works because we're dealing with powers of 2. * **Bit Shifts:** Integer division by powers of 2 (like `/ 2`, `/ 4`) can be directly replaced by right bit shifts (`>> 1`, `>> 2`) which are also more efficient. This final version is significantly shorter and more efficient than the original while achieving the same functionality. It's well under the 300-byte limit.

Extract Code

I extracted the following code from that output: 

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#include <Arduino.h> // Assuming Arduino environment

void stepper(int xw){
  for(int x=0,s=0;x<xw;x++,s=(s+1)&7) {
      digitalWrite(1,s/4);
      digitalWrite(2,(s/2)^s/4);
      digitalWrite(3,(s&1)^(s/2));
      digitalWrite(4,(s&1)^s/4);
  }
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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#include <Arduino.h> // Assuming Arduino environment

void stepper(int xw){
  for(int x=0,s=0;x<xw;x++,s=(s+1)&7) {
      digitalWrite(1,s/4);
      digitalWrite(2,(s/2)^s/4);
      digitalWrite(3,(s&1)^(s/2));
      digitalWrite(4,(s&1)^s/4);
  }
}



#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> // Assuming Arduino environment | ^~~~~~~~~~~ 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