I'm probably missing something simple - but I can'tseem to get this to work. I have a simple audio delay - with an fixed array size of [65536] - and it works perfectly - the problem is - if I try and make the array size - either shorter or longer - it bombs out!
Here's the working code:
float b[65536]; // initialisation
int p, r0, r1;
float len, d, feedback; // d = decimal part of length
int leni; // integer value of length
if (len > 65534) len = 65534;
else if (len < 1) len = 1;
leni = (int) len; // float to INT conversions are cpu intensive.
d = len - leni;
r0 = p - leni; // loop
r1 = r0 - 1;
if (r0 < 0) r0 += 65536;
if (r1 < 0) r1 += 65536;
output = b[r0] + d * (b[r1] - b[r0]);
b[p] = input + output * feedback;
p++;
if (p > 65535) p = 0;
Now here's the non-working code:
float b[16384]; // initialisation
int p, r0, r1;
float len, d, feedback; // d = decimal part of length
int leni; // integer value of length
if (len > 16382) len = 16382;
else if (len < 1) len = 1;
leni = (int) len; // float to INT conversions are cpu intensive.
d = len - leni;
r0 = p - leni; // loop
r1 = r0 - 1;
if (r0 < 0) r0 += 16384;
if (r1 < 0) r1 += 16384;
output = b[r0] + d * (b[r1] - b[r0]);
b[p] = input + output * feedback;
p++;
if (p > 16383) p = 0;
Can't figure out what to do, am I wrapping it wrong - or is 65536 a special int case?
Thanks in advace
Andrew
Your code is a bit hard to follow -- eg I'm not clear why you have floating point numbers that need to be changed to integers, what the initial values of variables like len are or where the loop occurs. I'm not even sure what language it is (python?) and maybe those things are obvious to someone who knows the language, in which case I apologize for my ignorance.
I can, however, speculate that fact that it works for 65536 probably has something to do with 65536 being a power of two, and somehow that property works around whatever other bugs you have.
I'm not going to try to fix your code, because, as I said, I can't follow it -- again, sorry. Instead, see my answer here, which tells you how to write a delay and an echo. It's in c-like psuedo-code which should be easy to port, and it's very few lines of code:
how to add echo effect on audio file using objective-c
Related
As I was looking through some string hash fucntions, I came across this one (code below). The function processes the string four bytes at a time, and interprets each of the four-byte chunks as a single long integer value. The integer values for the four-byte chunks are added together. In the end, the resulting sum is converted to the range 0 to M-1 using the modulus operator.
The following is the function code :
// Use folding on a string, summed 4 bytes at a time
long sfold(String s, int M) {
int intLength = s.length() / 4;
long sum = 0;
for (int j = 0; j < intLength; j++) {
char c[] = s.substring(j * 4, (j * 4) + 4).toCharArray();
long mult = 1;
for (int k = 0; k < c.length; k++) {
sum += c[k] * mult;
mult *= 256;
}
}
char c[] = s.substring(intLength * 4).toCharArray();
long mult = 1;
for (int k = 0; k < c.length; k++) {
sum += c[k] * mult;
mult *= 256;
}
return(Math.abs(sum) % M);
}
The confusion for me is this chunk of code, especially the first line.
char c[] = s.substring(intLength * 4).toCharArray();
long mult = 1;
for (int k = 0; k < c.length; k++) {
sum += c[k] * mult;
mult *= 256;
To my knowledge, the substring function used in this line takes as argument : begin index inclusive, The substring will start from the specified beginIndex and it will extend to the end of the string.
For the sake of example, let's assume we want to hash the following string : aaaabbbb. In this case intLength is going to be 2 (second line of function code). Replacing the value of intlength in s.substring(intLength * 4).toCharArray() will give us s.substring(8).toCharArray() which means string index is out of bounds given the string to be hashed has 8 characters.
I don't quite understand what's going on !
This hash function is awful, but to answer your question:
There is no IndexOutOfBoundsException, because "aaaabbbb".substring(8) is ""
The purpose of that last loop is to deal with leftovers when the string length isn't a multiple of 4. When s is "aaaabbbbcc", for example, then intLength == 2, and s.substring(8) is "cc".
I would like to Generate Checksum for Strings/Data
1. The same data should produce the same Checksum
2. Two different data strings can't product same checksum. Random collision of 0.1% can be negligible
3. No encryption/decryption of data
4. Checksum length need not be too huge and contains letters and characters.
5. Must be too fast and efficient. Imagine generating checksum(s) for 100 Mb of text data should be in less than 5mins. Generating 1000 checksums for less than 1 KB of each segment data should be in less than 10 seconds.
Any algorithm or implementation reference and suggestions are most appreciated.
You can write a custom hash function: (c++)
long long int hash(String s){
long long k = 7;
for(int i = 0; i < s.length(); i++){
k *= 23;
k += s[i];
k *= 13;
k %= 1000000009;
}
return k;
}
This should give you a well (collision free for most samples) hash value.
A very common, fast checksum is the CRC-32, a 32-bit polynomial cyclic redundancy check. Here are three implementations in C, which vary in speed vs. complexity, of the CRC-32: (This is from http://www.hackersdelight.org/hdcodetxt/crc.c.txt)
#include <stdio.h>
#include <stdlib.h>
// ---------------------------- reverse --------------------------------
// Reverses (reflects) bits in a 32-bit word.
unsigned reverse(unsigned x) {
x = ((x & 0x55555555) << 1) | ((x >> 1) & 0x55555555);
x = ((x & 0x33333333) << 2) | ((x >> 2) & 0x33333333);
x = ((x & 0x0F0F0F0F) << 4) | ((x >> 4) & 0x0F0F0F0F);
x = (x << 24) | ((x & 0xFF00) << 8) |
((x >> 8) & 0xFF00) | (x >> 24);
return x;
}
// ----------------------------- crc32a --------------------------------
/* This is the basic CRC algorithm with no optimizations. It follows the
logic circuit as closely as possible. */
unsigned int crc32a(unsigned char *message) {
int i, j;
unsigned int byte, crc;
i = 0;
crc = 0xFFFFFFFF;
while (message[i] != 0) {
byte = message[i]; // Get next byte.
byte = reverse(byte); // 32-bit reversal.
for (j = 0; j <= 7; j++) { // Do eight times.
if ((int)(crc ^ byte) < 0)
crc = (crc << 1) ^ 0x04C11DB7;
else crc = crc << 1;
byte = byte << 1; // Ready next msg bit.
}
i = i + 1;
}
return reverse(~crc);
}
// ----------------------------- crc32b --------------------------------
/* This is the basic CRC-32 calculation with some optimization but no
table lookup. The the byte reversal is avoided by shifting the crc reg
right instead of left and by using a reversed 32-bit word to represent
the polynomial.
When compiled to Cyclops with GCC, this function executes in 8 + 72n
instructions, where n is the number of bytes in the input message. It
should be doable in 4 + 61n instructions.
If the inner loop is strung out (approx. 5*8 = 40 instructions),
it would take about 6 + 46n instructions. */
unsigned int crc32b(unsigned char *message) {
int i, j;
unsigned int byte, crc, mask;
i = 0;
crc = 0xFFFFFFFF;
while (message[i] != 0) {
byte = message[i]; // Get next byte.
crc = crc ^ byte;
for (j = 7; j >= 0; j--) { // Do eight times.
mask = -(crc & 1);
crc = (crc >> 1) ^ (0xEDB88320 & mask);
}
i = i + 1;
}
return ~crc;
}
// ----------------------------- crc32c --------------------------------
/* This is derived from crc32b but does table lookup. First the table
itself is calculated, if it has not yet been set up.
Not counting the table setup (which would probably be a separate
function), when compiled to Cyclops with GCC, this function executes in
7 + 13n instructions, where n is the number of bytes in the input
message. It should be doable in 4 + 9n instructions. In any case, two
of the 13 or 9 instrucions are load byte.
This is Figure 14-7 in the text. */
unsigned int crc32c(unsigned char *message) {
int i, j;
unsigned int byte, crc, mask;
static unsigned int table[256];
/* Set up the table, if necessary. */
if (table[1] == 0) {
for (byte = 0; byte <= 255; byte++) {
crc = byte;
for (j = 7; j >= 0; j--) { // Do eight times.
mask = -(crc & 1);
crc = (crc >> 1) ^ (0xEDB88320 & mask);
}
table[byte] = crc;
}
}
/* Through with table setup, now calculate the CRC. */
i = 0;
crc = 0xFFFFFFFF;
while ((byte = message[i]) != 0) {
crc = (crc >> 8) ^ table[(crc ^ byte) & 0xFF];
i = i + 1;
}
return ~crc;
}
If you simply google "CRC32", you will get more info than you could possibly absorb.
Here's a problem:
Given string A and a substring B, remove the first occurence of substring B in string A till it is possible to do so. Note that removing a substring, can further create a new same substring. Ex. removing 'hell' from 'hehelllloworld' once would yield 'helloworld' which after removing once more would become 'oworld', the desired string.
Write a program for the above for input constraints of length 10^6 for A, and length 100 for B.
This question was asked to me in an interview, I gave them a simple algorithm to solve it that was to do exactly what the statement was and remove it iteratievly(to decresae over head calls), I later came to know there's a better solution for it that's much faster what would it be ? I've thought of a few optimizations but it's still not as fast as the fastest soln for the problem(acc. the company), so can anyone tell me of a faster way to solve the problem ?
P.S> I know of stackoverflow rules and that having code is better, but for this problem, I don't think that having code would be in any way beneficial...
Your approach has a pretty bad complexity. In a very bad case the string a will be aaaaaaaaabbbbbbbbb, and the string b will be ab, in which case you will need O(|a|) searches, each taking O(|a| + |b|) (assuming using some sophisticated search algorithm), resulting in a total complexity of O(|a|^2 + |a| * |b|), which with their constraints is years.
For their constraints a good complexity to aim for would be O(|a| * |b|), which is around 100 million operations, will finish in subsecond. Here's one way to approach it. For each position i in the string a let's compute the largest length n_i, such that the a[i - n_i : i] = b[0 : n_i] (in other words, the longest suffix of a at that position which is a prefix of b). We can compute it in O(|a| + |b|) by using Knuth-Morris-Pratt algorithm.
After we have n_i computed, finding the first occurrence of b in a is just a matter of finding the first n_i that is equal to |b|. This will be the right end of one of the occurrences of b in a.
Finally, we will need to modify Knuth-Morris-Pratt slightly. We will be logically removing occurrences of b as soon as we compute an n_i that is equal to |b|. To account for the fact that some letters were removed from a we will rely on the fact that Knuth-Morris-Pratt only relies on the last value of n_i (and those computed for b), and the current letter of a, so we just need a fast way of retrieving the last value of n_i after we logically remove an occurrence of b. That can be done with a deque, that stores all the valid values of n_i. Each value will be pushed into the deque once, and popped from it once, so that complexity of maintaining it is O(|a|), while the complexity of the Knuth-Morris-Pratt is O(|a| + |b|), resulting in O(|a| + |b|) total complexity.
Here's a C++ implementation. It could have some off-by-one errors, but it works on your sample, and it flies for the worst case that I described at the beginning.
#include <deque>
#include <string>
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
string a, b;
cin >> a >> b;
size_t blen = b.size();
// make a = b$a
a = b + "$" + a;
vector<size_t> n(a.size()); // array for knuth-morris-pratt
vector<bool> removals(a.size()); // positions of right ends at which we remove `b`s
deque<size_t> lastN;
n[0] = 0;
// For the first blen + 1 iterations just do vanilla knuth-morris-pratt
for (size_t i = 1; i < blen + 1; ++ i) {
size_t z = n[i - 1];
while (z && a[i] != a[z]) {
z = n[z - 1];
}
if (a[i] != a[z]) n[i] = 0;
else n[i] = z + 1;
lastN.push_back(n[i]);
}
// For the remaining iterations some characters could have been logically
// removed from `a`, so use lastN to get last value of n instaed
// of actually getting it from `n[i - 1]`
for (size_t i = blen + 1; i < a.size(); ++ i) {
size_t z = lastN.back();
while (z && a[i] != a[z]) {
z = n[z - 1];
}
if (a[i] != a[z]) n[i] = 0;
else n[i] = z + 1;
if (n[i] == blen) // found a match
{
removals[i] = true;
// kill last |b| - 1 `n_i`s
for (size_t j = 0; j < blen - 1; ++ j) {
lastN.pop_back();
}
}
else {
lastN.push_back(n[i]);
}
}
string ret;
size_t toRemove = 0;
for (size_t pos = a.size() - 1; a[pos] != '$'; -- pos) {
if (removals[pos]) toRemove += blen;
if (toRemove) -- toRemove;
else ret.push_back(a[pos]);
}
reverse(ret.begin(), ret.end());
cout << ret << endl;
return 0;
}
[in] hehelllloworld
[in] hell
[out] oworld
[in] abababc
[in] ababc
[out] ab
[in] caaaaa ... aaaaaabbbbbb ... bbbbc
[in] ab
[out] cc
Was asked this Amazon Telephonic Interview Round 1
So for Length = 1
0 1 (0 1)
Length = 2
00 01 11 10 (0, 1, 3, 2)
and so on
write function for length x that returns numbers in digit(base 10) form
That's called gray code, there are several different kinds, some of which are easier to construct than others. The wikipedia article shows a very simple way to convert from binary to gray code:
unsigned int binaryToGray(unsigned int num)
{
return (num >> 1) ^ num;
}
Using that, you only have to iterate over all numbers of a certain size, put them through that function, and print them however you want.
This is one way to do it:
int nval = (int)Math.Pow(2 , n);
int divisor = nval/2;
for (int i = 0; i < nval; i++)
{
int nb =(int) (i % divisor);
if ( nb== 2) Console.WriteLine(i + 1);
else if (nb == 3) Console.WriteLine(i - 1);
else Console.WriteLine(i);
}
I want to know efficient approach for the New Lottery Game problem.
The Lottery is changing! The Lottery used to have a machine to generate a random winning number. But due to cheating problems, the Lottery has decided to add another machine. The new winning number will be the result of the bitwise-AND operation between the two random numbers generated by the two machines.
To find the bitwise-AND of X and Y, write them both in binary; then a bit in the result in binary has a 1 if the corresponding bits of X and Y were both 1, and a 0 otherwise. In most programming languages, the bitwise-AND of X and Y is written X&Y.
For example:
The old machine generates the number 7 = 0111.
The new machine generates the number 11 = 1011.
The winning number will be (7 AND 11) = (0111 AND 1011) = 0011 = 3.
With this measure, the Lottery expects to reduce the cases of fraudulent claims, but unfortunately an employee from the Lottery company has leaked the following information: the old machine will always generate a non-negative integer less than A and the new one will always generate a non-negative integer less than B.
Catalina wants to win this lottery and to give it a try she decided to buy all non-negative integers less than K.
Given A, B and K, Catalina would like to know in how many different ways the machines can generate a pair of numbers that will make her a winner.
For small input we can check all possible pairs but how to do it with large inputs. I guess we represent the binary number into string first and then check permutations which would give answer less than K. But I can't seem to figure out how to calculate possible permutations of 2 binary strings.
I used a general DP technique that I described in a lot of detail in another answer.
We want to count the pairs (a, b) such that a < A, b < B and a & b < K.
The first step is to convert the numbers to binary and to pad them to the same size by adding leading zeroes. I just padded them to a fixed size of 40. The idea is to build up the valid a and b bit by bit.
Let f(i, loA, loB, loK) be the number of valid suffix pairs of a and b of size 40 - i. If loA is true, it means that the prefix up to i is already strictly smaller than the corresponding prefix of A. In that case there is no restriction on the next possible bit for a. If loA ist false, A[i] is an upper bound on the next bit we can place at the end of the current prefix. loB and loK have an analogous meaning.
Now we have the following transition:
long long f(int i, bool loA, bool loB, bool loK) {
// TODO add memoization
if (i == 40)
return loA && loB && loK;
int hiA = loA ? 1: A[i]-'0'; // upper bound on the next bit in a
int hiB = loB ? 1: B[i]-'0'; // upper bound on the next bit in b
int hiK = loK ? 1: K[i]-'0'; // upper bound on the next bit in a & b
long long res = 0;
for (int a = 0; a <= hiA; ++a)
for (int b = 0; b <= hiB; ++b) {
int k = a & b;
if (k > hiK) continue;
res += f(i+1, loA || a < A[i]-'0',
loB || b < B[i]-'0',
loK || k < K[i]-'0');
}
return res;
}
The result is f(0, false, false, false).
The runtime is O(max(log A, log B)) if memoization is added to ensure that every subproblem is only solved once.
What I did was just to identify when the answer is A * B.
Otherwise, just brute force the rest, this code passed the large input.
// for each test cases
long count = 0;
if ((K > A) || (K > B)) {
count = A * B;
continue; // print count and go to the next test case
}
count = A * B - (A-K) * (B-K);
for (int i = K; i < A; i++) {
for (int j = K; j < B; j++) {
if ((i&j) < K) count++;
}
}
I hope this helps!
just as Niklas B. said.
the whole answer is.
#include <algorithm>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <sstream>
#include <string>
#include <vector>
using namespace std;
#define MAX_SIZE 32
int A, B, K;
int arr_a[MAX_SIZE];
int arr_b[MAX_SIZE];
int arr_k[MAX_SIZE];
bool flag [MAX_SIZE][2][2][2];
long long matrix[MAX_SIZE][2][2][2];
long long
get_result();
int main(int argc, char *argv[])
{
int case_amount = 0;
cin >> case_amount;
for (int i = 0; i < case_amount; ++i)
{
const long long result = get_result();
cout << "Case #" << 1 + i << ": " << result << endl;
}
return 0;
}
long long
dp(const int h,
const bool can_A_choose_1,
const bool can_B_choose_1,
const bool can_K_choose_1)
{
if (MAX_SIZE == h)
return can_A_choose_1 && can_B_choose_1 && can_K_choose_1;
if (flag[h][can_A_choose_1][can_B_choose_1][can_K_choose_1])
return matrix[h][can_A_choose_1][can_B_choose_1][can_K_choose_1];
int cnt_A_max = arr_a[h];
int cnt_B_max = arr_b[h];
int cnt_K_max = arr_k[h];
if (can_A_choose_1)
cnt_A_max = 1;
if (can_B_choose_1)
cnt_B_max = 1;
if (can_K_choose_1)
cnt_K_max = 1;
long long res = 0;
for (int i = 0; i <= cnt_A_max; ++i)
{
for (int j = 0; j <= cnt_B_max; ++j)
{
int k = i & j;
if (k > cnt_K_max)
continue;
res += dp(h + 1,
can_A_choose_1 || (i < cnt_A_max),
can_B_choose_1 || (j < cnt_B_max),
can_K_choose_1 || (k < cnt_K_max));
}
}
flag[h][can_A_choose_1][can_B_choose_1][can_K_choose_1] = true;
matrix[h][can_A_choose_1][can_B_choose_1][can_K_choose_1] = res;
return res;
}
long long
get_result()
{
cin >> A >> B >> K;
memset(arr_a, 0, sizeof(arr_a));
memset(arr_b, 0, sizeof(arr_b));
memset(arr_k, 0, sizeof(arr_k));
memset(flag, 0, sizeof(flag));
memset(matrix, 0, sizeof(matrix));
int i = 31;
while (i >= 1)
{
arr_a[i] = A % 2;
A /= 2;
arr_b[i] = B % 2;
B /= 2;
arr_k[i] = K % 2;
K /= 2;
i--;
}
return dp(1, 0, 0, 0);
}