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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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

To significantly reduce the length of the `stepper` function, we can leverage bit manipulation based on the pattern observed in the `digitalWrite` sequences. Let's start by analyzing the pattern and then implement a more concise version. 1. **Creating a Table of Values:** 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
Steps | 1  2  3  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. **Finding an Expression:** The pattern closely resembles a binary count with a twist. The twist being the last step (7) which repeats the pattern from the first step but with the first pin also set to 1. A straightforward way to express each step is by using bit manipulation to set each pin based on the current step. 3. **Writing the Function as Short as Possible:** We can encode the pattern into an array and loop through each step, using the array values to set the pins directly. 
1
2
3
4
5
6
7
8
void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    int pattern[] = {1, 3, 2, 6, 4, 12, 8, 9};
    for (int pin = 1; pin <= 4; pin++) {
      digitalWrite(pin, (pattern[x % 8] >> (pin - 1)) & 1);
    }
  }
}
4. **Making It Shorter - Final Answer:** We can further shorten this by embedding the pattern directly into the loop, eliminating the need for the array declaration. 
1
2
3
4
5
void stepper(int xw) {
  for (int x = 0; x < xw; x++)
    for (int pin = 1; pin <= 4; pin++)
      digitalWrite(pin, ((int[]){1,3,2,6,4,12,8,9}[x%8]>>(pin-1))&1);
}
This final version of `stepper` is significantly shorter than the original, and it efficiently replicates the functionality using bit manipulation and a compact loop structure.

Extract Code

I extracted the following code from that output: 

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

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
void stepper(int xw) {
  for (int x = 0; x < xw; x++)
    for (int pin = 1; pin <= 4; pin++)
      digitalWrite(pin, ((int[]){1,3,2,6,4,12,8,9}[x%8]>>(pin-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: Wrong answer; got 370724802 instead of -1475775710. main.c: In function 'stepper': main.c:5:7: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 5 | digitalWrite(pin, ((int[]){1,3,2,6,4,12,8,9}[x%8]>>(pin-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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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

To code golf the given function, let's follow the steps outlined: 1. **Create a table of values:** 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
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. **Find an expression:** - Observing the pattern, the pins can be controlled by calculating their state based on the value of `Steps` modulo 8. - For pin 1: `digitalWrite(1, Steps >= 5 || Steps == 0 || Steps == 7);` - For pin 2: `digitalWrite(2, Steps >= 3 && Steps <= 5 || Steps == 7);` - For pin 3: `digitalWrite(3, Steps >= 1 && Steps <= 3 || Steps == 5);` - For pin 4: `digitalWrite(4, Steps % 2 == 0 && Steps <= 2 || Steps == 7);` 3. & 4. **Write the function as short as possible and then make it any shorter:** Given the above expressions, we can further simplify and compress our code by looping through the steps and directly setting the digital pins within the loop. Here's a much shorter version: 
1
2
3
4
5
6
7
8
void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 >= 5 || x % 8 == 0 || x % 8 == 7);
    digitalWrite(2, x % 8 >= 3 && x % 8 <= 5 || x % 8 == 7);
    digitalWrite(3, x % 8 >= 1 && x % 8 <= 3 || x % 8 == 5);
    digitalWrite(4, x % 8 % 2 == 0 && x % 8 <= 2 || x % 8 == 7);
  }
}
This version significantly shortens the original function by directly computing the state of each pin within the loop without a switch or if statements, thereby achieving the goal of being less than 300 bytes.

Extract Code

I extracted the following code from that output: 

1
2
3
4
5
6
7
8
void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 >= 5 || x % 8 == 0 || x % 8 == 7);
    digitalWrite(2, x % 8 >= 3 && x % 8 <= 5 || x % 8 == 7);
    digitalWrite(3, x % 8 >= 1 && x % 8 <= 3 || x % 8 == 5);
    digitalWrite(4, x % 8 % 2 == 0 && x % 8 <= 2 || x % 8 == 7);
  }
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    digitalWrite(1, x % 8 >= 5 || x % 8 == 0 || x % 8 == 7);
    digitalWrite(2, x % 8 >= 3 && x % 8 <= 5 || x % 8 == 7);
    digitalWrite(3, x % 8 >= 1 && x % 8 <= 3 || x % 8 == 5);
    digitalWrite(4, x % 8 % 2 == 0 && x % 8 <= 2 || 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: Wrong answer; got 1390203459 instead of -1475775710. main.c: In function 'stepper': main.c:4:5: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 4 | digitalWrite(1, x % 8 >= 5 || x % 8 == 0 || x % 8 == 7); | ^~~~~~~~~~~~

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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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

To significantly reduce the size of the `stepper` function, we can observe the patterns in the `digitalWrite` calls within each case. Let's start by creating a table of the values being written to each pin for each step: | 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 | Notice that the pattern is essentially a rotation of bits. We can use this observation to simplify the implementation drastically. Here is a more concise version of the `stepper` function: 

1
2
3
4
5
6
7
8
9
void stepper(int xw) {
  for (int x = 0; x < xw; x++) {
    int s = (0b0001 << (x % 4)) | (0b1000 >> (x % 4));
    digitalWrite(1, s & 0b0001);
    digitalWrite(2, s & 0b0010);
    digitalWrite(3, s & 0b0100);
    digitalWrite(4, s & 0b1000);
  }
}
However, we can make it shorter by further exploiting the pattern and the fact that the operations are cyclic with a period of 8 steps, similar to how bits are set in the original cases: Final answer: 
1
2
3
4
5
6
7
8
9
void stepper(int xw) {
  for (int x=0; x<xw; x++) {
    int s=0b10011001>>x%4;
    digitalWrite(1,s&1);
    digitalWrite(2,s>>1&1);
    digitalWrite(3,s>>2&1);
    digitalWrite(4,s>>3&1);
  }
}
This version compresses the original logic into a loop that cycles through the steps using bitwise operations to decide the state of each pin based on the step number modulo 4. This significantly reduces the size of the code while maintaining the functionality.

Extract Code

I extracted the following code from that output: 

1
2
3
4
5
6
7
8
9
void stepper(int xw) {
  for (int x=0; x<xw; x++) {
    int s=0b10011001>>x%4;
    digitalWrite(1,s&1);
    digitalWrite(2,s>>1&1);
    digitalWrite(3,s>>2&1);
    digitalWrite(4,s>>3&1);
  }
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
void stepper(int xw) {
  for (int x=0; x<xw; x++) {
    int s=0b10011001>>x%4;
    digitalWrite(1,s&1);
    digitalWrite(2,s>>1&1);
    digitalWrite(3,s>>2&1);
    digitalWrite(4,s>>3&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 408397632 instead of -1475775710. main.c: In function 'stepper': main.c:5:5: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 5 | digitalWrite(1,s&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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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

To code golf the `stepper` function, let's start by creating a table for the values being written to the pins for each step, from 0 to 7: 

1
2
3
4
5
6
7
8
9
Step | digitalWrite(1) | digitalWrite(2) | digitalWrite(3) | digitalWrite(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
The patterns for each pin can be represented as follows: - Pin 1 is HIGH when `Steps` is 5, 6, or 7. - Pin 2 is HIGH when `Steps` is 3, 4, or 5. - Pin 3 is HIGH when `Steps` is 1, 2, or 3. - Pin 4 is HIGH for `Steps` 0, 1, and 7. Finding expressions that match these patterns: 1. Pin 1: `Steps >= 5` 2. Pin 2: `Steps == 3 || Steps == 4 || Steps == 5` 3. Pin 3: `Steps == 1 || Steps == 2 || Steps == 3` 4. Pin 4: `Steps == 0 || Steps == 1 || Steps == 7` Since we need to loop `Steps` from 0 to 7, we can use modulo `%` to cycle `Steps` through these values. Here's a shorter version of the stepper function: 
1
2
3
4
5
6
7
8
void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) % 8) {
    digitalWrite(1, Steps >= 5);
    digitalWrite(2, Steps >= 3 && Steps <= 5);
    digitalWrite(3, Steps >= 1 && Steps <= 3);
    digitalWrite(4, Steps == 0 || Steps == 1 || Steps == 7);
  }
}
To further reduce the code size, we can take advantage of the sequential nature of the steps and the repeating pattern: 
1
2
3
4
5
6
7
8
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 < 6);
    digitalWrite(3, s > 0 && s < 4);
    digitalWrite(4, s < 2 || s == 7);
  }
}
This version combines the conditions for each `digitalWrite` to minimize the length further. Note that `& 7` is used instead of `% 8` for efficiency, as it achieves the same result for values of `s` in the range `[0, 7]`. This is a common optimization in code golf for powers of 2. Final answer: 
1
2
3
4
5
6
7
8
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 < 6);
    digitalWrite(3, s > 0 && s < 4);
    digitalWrite(4, s < 2 || s == 7);
  }
}
This version retains the original functionality in a much shorter form, adhering to the requirement of being under 300 bytes, even without counting whitespace.

Extract Code

I extracted the following code from that output: 

1
2
3
4
5
6
7
8
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 < 6);
    digitalWrite(3, s > 0 && s < 4);
    digitalWrite(4, s < 2 || s == 7);
  }
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
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 < 6);
    digitalWrite(3, s > 0 && s < 4);
    digitalWrite(4, s < 2 || s == 7);
  }
}



#include<stdio.h>
#include<stdlib.h>

#include <stdio.h>

int out = 0;
int digitalWrite(int pin, int value) {
  out *= 101;
  out += pin;
  out *= 97;
  out += value;
  return out;
}

int main() {


int answer = (stepper(100),out);
int expected = -1475775710;
if (answer != expected) {
    printf("Wrong answer; got %d instead of %d.\n", answer, expected);
    exit(1);
}
printf("All tests passed\n");
}
And got the output: All tests passed main.c: In function 'stepper': main.c:4:5: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 4 | digitalWrite(1, s > 4); | ^~~~~~~~~~~~

Substring Evaluation

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

PyFunc

Resulting in output: True

Final Answer: True

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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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`. 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
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

To golf the `stepper` function, we first analyze the pattern in the calls to `digitalWrite` within the switch-case structure. The goal is to find a more compact way to represent this repetitive structure. ### Step 1: Create a 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 | ### Step 2: Find an Expression Observing the table, we can see that the pattern repeats every 8 steps. Each pin can be controlled by a specific bitwise operation on the `Steps` variable. ### Step 3 & 4: Write the Function as Short as Possible We can use bitwise operations to directly calculate the values for each pin based on the current step. The final golfed version of the stepper function is: 

1
2
3
4
5
6
7
8
void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) & 7) {
    digitalWrite(1, Steps < 2 || Steps == 7);
    digitalWrite(2, Steps > 2 && Steps < 6);
    digitalWrite(3, Steps > 0 && Steps < 4 || Steps == 5);
    digitalWrite(4, Steps % 4 < 2);
  }
}
This version drastically reduces the size of the original function by eliminating the switch-case structure and directly computing the state for each pin based on the current step `Steps`. It uses logical operations to determine the state of each pin for every step in a compact manner. ### Final Answer The final answer, adhering to the code golf challenge to be under 300 bytes and not counting whitespace, is the code snippet provided above.

Extract Code

I extracted the following code from that output: 

1
2
3
4
5
6
7
8
void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) & 7) {
    digitalWrite(1, Steps < 2 || Steps == 7);
    digitalWrite(2, Steps > 2 && Steps < 6);
    digitalWrite(3, Steps > 0 && Steps < 4 || Steps == 5);
    digitalWrite(4, Steps % 4 < 2);
  }
}

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
void stepper(int xw) {
  for (int x = 0, Steps = 0; x < xw; x++, Steps = (Steps + 1) & 7) {
    digitalWrite(1, Steps < 2 || Steps == 7);
    digitalWrite(2, Steps > 2 && Steps < 6);
    digitalWrite(3, Steps > 0 && Steps < 4 || Steps == 5);
    digitalWrite(4, Steps % 4 < 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: Wrong answer; got 221689492 instead of -1475775710. main.c: In function 'stepper': main.c:4:5: warning: implicit declaration of function 'digitalWrite' [-Wimplicit-function-declaration] 4 | digitalWrite(1, Steps < 2 || Steps == 7); | ^~~~~~~~~~~~

Substring Evaluation

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

PyFunc

Resulting in output: True

Final Answer: False