Stream hex number as decimal

Question

I would like to use the BBP formula to calculate Pi in a C-program's pthread process while another process prints the result for as far as it has got. However BBP gives a base 16 answer while I would like to stream a base 10 answer to the user.

How can I determine whether it's safe to print the n-th digit of a base 10 converted base 16 number?

Thanks in advance!

Answer 1

One solution is to test whether increasing the last hexadecimal digit currently available changes the decimal digit you are considering displaying.

Consider a number x with hexadecimal representation …h₃h₂h₁h₀.h_-1h_-2… and decimal representation …d₃d₂d₁d₀.d_-1d_-2…

Suppose we have a truncated numeral, so that we know only the digits from h_∞ to h_j. Let y be the number represented by these digits. Let z be y + 16^j, which is y plus one in the j digit position. Then the value of x might be any value from y (inclusive) to z (exclusive).

Now consider a candidate decimal numeral, with digits d_∞ to d_i. Let y' be the number represented by these digits. Let z' be y + 10ⁱ. Iff y' ≤ y and z ≤ z', then the decimal digits d_∞ to d_i must be a prefix of the complete decimal numeral for x (that is, these decimal digits are known to appear in the decimal numeral for x; they will not change as more hexadecimal digits are discovered).

This is because the value of x, being in [y, z), can be formed by adding some zero or positive value to y' and that value needed is less than 1 in the i digit position. Conversely, if the inequalities do not hold, then x could be outside the interval spanned by the candidate digits.

Answer 2

@Eric Postpischil posted a fine general purpose algorithm.

In implementing OP goal's, some short-cuts may be realized.
Handle the integer portion of Pi separate and only deal with the fraction.
Assume input is base 16 and then add 1 bit at a time.

Implementation notes:
I cheated by using fixed memory allocation and byte-array (string) handling. Certainly one would save the array length rather than strlen() and use byte 0 - 9 rather than char '0' to '9', but this was tossed together quickly and was easier to debug this way. Array size s/b dynamic, but that is easy to add.

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

typedef struct {
  char *Sum;
  char *Add;
} Pi_T;

void Pi_Print(const char *Title, Pi_T *State) {
  printf("%s\n", Title);
  printf("  Sum: '%s'\n", State->Sum);
  printf("  Add: '%s'\n", State->Add);
}

// Sum += Add
void Pi_Add(char *Sum, char *Add) {
  size_t LenS = strlen(Sum);
  size_t LenA = strlen(Add);
  while (LenS > LenA) {
    Add[LenA++] = '0';
    Add[LenA] = '\0';
  }
  while (LenA > LenS) {
    Sum[LenS++] = '0';
    Sum[LenS] = '\0';
  }
  unsigned Accumulator = 0;
  while (LenA > 0) {
    LenA--;
    Accumulator += Add[LenA] - '0';
    Accumulator += Sum[LenA] - '0';
    Sum[LenA] = Accumulator % 10 + '0';
    Accumulator /= 10;
    assert(Accumulator <= 9);
  }
  assert(Accumulator == 0);
}

// Divide the `Add` by 2
void Pi_Div2(char *Add) {
  size_t LenS = strlen(Add);
  size_t i;
  unsigned Accumulator = 0;
  for (i = 0; i < LenS; i++) {
    Accumulator += Add[i] - '0';
    Add[i] = Accumulator / 2 + '0';
    Accumulator %= 2;
    Accumulator *= 10;
    assert ((Accumulator == 0) ||  (Accumulator == 10));
  }
  if (Accumulator > 0) {
    Add[i++] = Accumulator / 2 + '0';
    Add[i] = '\0';
    Accumulator %= 2;
    Accumulator *= 10;
    assert(Accumulator == 0);
  }
}

void Pi_PutHex(Pi_T *State, unsigned HexDigit) {
  // Add HexDigit, 1 bit at a time.
  for (unsigned i = 4; i-- > 0;) {
    if (HexDigit & (1 << i)) {
      Pi_Add(State->Sum, State->Add);
    }
    // Should the Sum[0] be extracted?
    if (State->Add[0] == '0') {
      for (size_t i = 1; State->Sum[i] && State->Add[i]; i++) {
        unsigned Accumulator = State->Sum[i] - '0' + State->Add[i] - '0';
        if (Accumulator > 9)
          break;
        if (Accumulator < 9) {

          // !!!!!!!!!!!!!!!!!!!!!!!!!!!!
          // Print the decimal digit!
          printf("%c", State->Sum[0]);
          // !!!!!!!!!!!!!!!!!!!!!!!!!!!!

          memmove(&State->Sum[0], &State->Sum[1], strlen(State->Sum));
          memmove(&State->Add[0], &State->Add[1], strlen(State->Add));
          break;
        }
      }
    }
    Pi_Div2(State->Add);
  }
}

void Pi_Test(void) {
  Pi_T State;
  State.Sum = malloc(500);
  State.Add = malloc(500);
  State.Sum[0] = '\0';
  State.Add[0] = '5';
  State.Add[1] = '\0';
  // http://calccrypto.wikidot.com/math:pi-hex
  static const char *pi = "3.243F6A8885A308D313198A2E03707344A4093822299F31D0082EFA98EC4E6C89452821E638D01378";
  // http://www.miniwebtool.com/first-n-digits-of-pi/?number=100
  // 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679
  // Output
  // 3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117
  const char *p = &pi[2];
  // Pi_Print("Init", &State);
  printf("3.");
  // add each hex digit, one at a time.
  while (*p) {
    unsigned HexDigit = (*p <= '9')  ? (*p - '0') : (*p - 'A' + 10);

    // !!!!!!!!!!!!!!!!!!!!!!!!!!!!
    // Put in the hexadecimal digit
    Pi_PutHex(&State, HexDigit);
    // !!!!!!!!!!!!!!!!!!!!!!!!!!!!

    p++;
  }
  printf("\n");
  // Pi_Print("End", &State);
}