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

Let me help golf this down following those steps. 1. First, let's make a table of the pin values (pins 1-4): 

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Step  P1 P2 P3 P4
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. Looking at the pattern, we can treat these as bits in a 4-bit number and observe it's a circular pattern. 3. Here's the golfed solution: 
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void stepper(int xw){for(int i=0;i<xw;i++){int s=i&7;for(int p=1;p<5;p++)digitalWrite(p,s==(p<3?6-p:p-2)||s==(p<3?7-p:p-3));}}
This comes in at 129 bytes and maintains the same functionality by: - Using a single conditional expression to determine pin states - Eliminating the switch statement and lookup table - Using the bit pattern to determine when pins should be high Final answer: This is the shortest I could make it while maintaining readability of the function name and correct functionality. The main trick was recognizing that each pin follows a pattern that can be expressed with simple comparisons rather than needing a full switch statement.

Extract Code

I extracted the following code from that output: 

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#include <Arduino.h>

void stepper(int xw){for(int i=0;i<xw;i++){int s=i&7;for(int p=1;p<5;p++)digitalWrite(p,s==(p<3?6-p:7-p)||s==(p<3?7-p:p-3));}}

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 i=0;i<xw;i++){int s=i&7;for(int p=1;p<5;p++)digitalWrite(p,s==(p<3?6-p:7-p)||s==(p<3?7-p:p-3));}}



#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: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

Let me solve this step by step: 1. First, let's create a table of values (pins 1-4): 

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Step: 0 → 0001
Step: 1 → 0011
Step: 2 → 0010
Step: 3 → 0110
Step: 4 → 0100
Step: 5 → 1100
Step: 6 → 1000
Step: 7 → 1001
2. Looking at the pattern, these are actually binary numbers (0x1,0x3,0x2,0x6,0x4,0xC,0x8,0x9) 3. Here's my final golfed answer: 
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void stepper(int w){int s=0,p[]={1,3,2,6,4,12,8,9};while(w--){for(int i=1;i<5;i++)digitalWrite(i,p[s]>>(i-1)&1);s=++s%8;}}
This comes in at 115 bytes. It works by: - Storing the pattern in an array - For each step, bitshifting the appropriate amount and masking with 1 to get each pin's value - Using a while loop to count down instead of for loop counting up - Using minimal variable names and no unnecessary spaces This maintains the exact same functionality as the original but is much more compact.

Extract Code

I extracted the following code from that output: 

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

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

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void stepper(int w) {
    int s=0,p[]={1,3,2,6,4,12,8,9};
    while(w--) {
        for(int i=1;i<5;i++)
            digitalWrite(i,p[s]>>(i-1)&1);
        s=++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: Wrong answer; got 370724802 instead of -1475775710. main.c: In function 'stepper': main.c:6:13: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 6 | digitalWrite(i,p[s]>>(i-1)&1); | ^~~~~~~~~~~~

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

Let me help golf this code by following those steps. 1. First, let's make a table of the pin values (pins 1-4): 

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Step: 0  → 0001
Step: 1  → 0011
Step: 2  → 0010
Step: 3  → 0110
Step: 4  → 0100
Step: 5  → 1100
Step: 6  → 1000
Step: 7  → 1001
2. Looking at this pattern, it's a sequence of 8 values that can be represented as a binary number. 3. Here's my golfed solution: 
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void stepper(int w){int s=0,p[]={1,3,2,6,4,12,8,9};for(;w--;)for(int i=1;i<5;i++)digitalWrite(i,p[s%8]>>(i-1)&1),s+=i>3;}
4. Final version - I managed to golf it even more by removing the array and using bit shifts: 
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void stepper(int w){int s=0;for(;w--;)for(int i=1;i<5;i++)digitalWrite(i,1&(0x19962>>(((s%8)*4)+(i-1))),s+=i>3;}
This is my final answer. It's 119 bytes (not counting whitespace) and maintains the same functionality as the original code. The magic number 0x19962 contains all the pin states encoded in a compact form, and we use bit shifting to extract the right values for each pin at each step. The key tricks used: - Encoded all 8 states into a single hex number - Used modulo and bit shifting to extract values - Combined the step increment into the pin writing loop - Removed the need for an array or switch statement

Extract Code

I extracted the following code from that output: 

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#include <arduino.h>

void stepper(int w){int s=0;for(;w--;)for(int i=1;i<5;i++)digitalWrite(i,1&(0x19962>>(((s%8)*4)+(i-1))),s+=i>3;}

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 w){int s=0;for(;w--;)for(int i=1;i<5;i++)digitalWrite(i,1&(0x19962>>(((s%8)*4)+(i-1))),s+=i>3;}



#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: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

Let me help code golf this function. 1. First, let's create the value table (pin 1-4): 

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Step  P1 P2 P3 P4
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. Looking at the pattern, we can use bit operations since each row represents a 4-bit number: 
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0001 (1)
0011 (3)
0010 (2)
0110 (6)
0100 (4)
1100 (12)
1000 (8)
1001 (9)
3. Final golfed answer: 
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void stepper(int xw){for(int i=0;i<xw;i++){int p=1<<((i&7)/2)<<(((i&7)>5)?0:((i&7)>1));digitalWrite(1,p&8);digitalWrite(2,p&4);digitalWrite(3,p&2);digitalWrite(4,p&1);}}
This is 179 bytes. Here's how it works: - Uses a single bit-shifting expression to generate the pattern - Avoids storing state in a separate variable - Uses the loop counter modulo 8 directly - Masks each bit and writes directly to pins I tried several other approaches but this was the shortest I could make it while maintaining the same functionality.

Extract Code

I extracted the following code from that output: 

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#include "Arduino.h"

void stepper(int xw){for(int i=0;i<xw;i++){int p=1<<((i&7)/2)<<(((i&7)>5)?0:((i&7)>1));digitalWrite(1,p&8);digitalWrite(2,p&4);digitalWrite(3,p&2);digitalWrite(4,p&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 i=0;i<xw;i++){int p=1<<((i&7)/2)<<(((i&7)>5)?0:((i&7)>1));digitalWrite(1,p&8);digitalWrite(2,p&4);digitalWrite(3,p&2);digitalWrite(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: 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

Let me help golf this down following those steps. 1. First, let's make a table of the pin values (pins 1-4): 

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Step: 0 → [0,0,0,1]
Step: 1 → [0,0,1,1]
Step: 2 → [0,0,1,0]
Step: 3 → [0,1,1,0]
Step: 4 → [0,1,0,0]
Step: 5 → [1,1,0,0]
Step: 6 → [1,0,0,0]
Step: 7 → [1,0,0,1]
2. Looking at the pattern, we can express each pin's state using bit operations on the step count. 3/4. Here's the final golfed version: 
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void stepper(int w){for(int i=0,s=0;i<w;i++,s=++s&7)for(int p=1;p<5;p++)digitalWrite(p,(s&4)>>2&!(p-4)||(s&2)>>1&!(p-3)||(s&1)&!(p-2)||s>>3&!(p-1));}
This is 158 bytes (not counting whitespace). The function uses bitwise operations to determine pin states. For each step: - It cycles through steps 0-7 using `s=++s&7` - For each pin (1-4), it calculates its state using bitwise operations - The complex expression determines if a pin should be HIGH based on the current step This is significantly shorter than the original switch-case version while maintaining the same functionality.

Extract Code

I extracted the following code from that output: 

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void stepper(int w){for(int i=0,s=0;i<w;i++,s=++s&7)for(int p=1;p<5;p++)digitalWrite(p,(s&4)>>2&!(p-4)||(s&2)>>1&!(p-3)||(s&1)&!(p-2)||s>>3&!(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 w){for(int i=0,s=0;i<w;i++,s=++s&7)for(int p=1;p<5;p++)digitalWrite(p,(s&4)>>2&!(p-4)||(s&2)>>1&!(p-3)||(s&1)&!(p-2)||s>>3&!(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: Wrong answer; got 2021291684 instead of -1475775710. main.c: In function 'stepper': main.c:2:73: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 2 | void stepper(int w){for(int i=0,s=0;i<w;i++,s=++s&7)for(int p=1;p<5;p++)digitalWrite(p,(s&4)>>2&!(p-4)||(s&2)>>1&!(p-3)||(s&1)&!(p-2)||s>>3&!(p-1));} | ^~~~~~~~~~~~

Substring Evaluation

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

PyFunc

Resulting in output: True

Final Answer: False