Here we show a sample answer from subject 5125, after obtaining consent.

Consent form, questions and survey

This file shows the consent form, questions and survey, that need to be signed and answered by subject 5125. The question order of this subject is E/D/G/B.

Hand-written answers

The subject's hand-wrtten answers to the above questions can be found in this file. In order to protect the subject's privacy, we do not include the consent form (with the subject's signature), and the demographic survey (with the subject's age and programming experience).

Transcribed answers

Program E:
#include <stdio.h>

double V4;
int V5;

int F1(int V1, int V2, int V3) {
  printf("a: %d %d %d %f\n", V1, V2, V3, V4);
  return printf(
      "b: %c\n",
      (V1-- + 1 && V1 + 4 && F1(V1, -1, V1), V1 && V2)
          ? (F1(-1, ++V2, V3),
             ((V5 = (int)(V4 = (int)(V2 + 1) /
                               (1 - (int)V3 * 2 - (int)V3 * (int)V3),
                          V4 * V4 >= 1 && ((2 % 3) / 4 - 2 + (V4 / 2)) < 1),
               printf("c: %d %d %d %f %d\n", V1, V2, V3, V4, V5), V5)["ab"]))
          : 'c');
}

int main() {
  F1(-1, -2, 0);
  printf("d\n");
}
Answer:

Start Time: 14:06

a: -1 -2 0 0.0

b: c

d

End Time: 14:17


Program D
#include <stdio.h>

void F1(int V1, int V2, int V3, int V4) {
  V1 = V1 + 1;
  V2 = V1;
  while (V2 < 4) {
    V3 = 0;

    printf("a: %d %d %d %d\n", V1, V2, V3, V4);
    int V9;
    if (V3 < V2) {
      V3 = (V3 * 8) + (V2 % 8);
      V2 /= 8;
      V4 = (V3 == V2) | ((V3 / 8) == V2);
      V9 = 1;
    } else {
      V4 = V4 + 1;
      if ((V1 - V4) != 0) {
        V9 = V1 % V4;
      } else {
        printf("b: %d\n", V1);
        V9 = 2 && (V1 % V4);
      }
    }

    for (; V9;) {
      printf("c: %d %d %d %d\n", V1, V2, V3, V4);
      if (V3 < V2) {
        V3 = (V3 * 8) + (V2 % 8);
        V2 /= 8;
        V4 = (V3 == V2) | ((V3 / 8) == V2);
        V9 = 1;
      } else {
        V4 = V4 + 1;
        if ((V1 - V4) != 0) {
          V9 = V1 % V4;
        } else {
          printf("d: %d\n", V1);
          V9 = 2 && (V1 % V4);
        }
      }
    }

    V1 = V1 + 1;
    V2 = V1;
  }
}

int main() {
  F1(1, 0, 0, 0);
  printf("e\n");
}
Answer:

Start Time: 14:17

a: 2 2 0 0

c: 2 0 0 1

a: 3 3 0 1

c: 3 0 0 1

d: 4

End Time: 14:36


Program G
#include <stdio.h>

int F1(int V1, int V2) {
  int V3, V4;

  printf("a: %d %d\n", V1, V2);

  for (V3 = V4 = 1; V3* V3 <= V1 ? V4 = V1 % V3 ? V4 : V3 : V2 + 1
                    ? V4 < 2 ? V1 && F1(V2, 0) : F1(V4, V2),
      printf("b: %d\n", V2 ? 10 : 32 << !V1),
      V1 -= V4 * !!V1 : (F1(V4, V1 / V4), 0);) {
    printf("c: %d %d\n", V1, V4);
    V3++;
  }

  return 0;
}

int main() {
  F1(1, 0);
  printf("d\n");
}
Answer:

Start Time: 14:37

a: 1 0

c: 1 0

a: 0 0

d

End Time: 14:49


Program B
#include <stdio.h>

void F1(int V1, char *V2, int V3) {
  printf("a: %d %s %d\n", V1, V2, V3);
  int V4 = (V1 / V3) + V3;
  char *V5 = V2 - V1;
  V2 = V2 - 1;
  int V6 = (int)V2 / (int)V2;
  printf("b: %d %s %d\n", V4, V5, V6);
}
int V7;
int main() {
  for (; "ab"[V7] != 0;) {
    F1(97 - 97, V7 + "zy", 122 / 122);
    V7 = V7 + 1;
  }
  printf("c\n");
}
Answer:

Start Time: 14:50

a: 0 zy 1

b: 1 zy 1

a: 0 y 1

b: 0 y 1

c

End Time: 14:54