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- /* segvhunt.cFind and eliminate all code that generates Segmentation Fault*/#include <stdio.h>int main() {char **s;char foo[] = "Hello World";*s = foo;printf("s is %s\n",s);s[0] = foo;printf("s[0] is %s\n",s[0]);return(0);}def find_root4(x, epsilon): ''' IN PYTHON Assume: x, epsilon are floating point numbers and epsilon > 0 Use bisection search to find the following root of x such that If x >=0, return y such that x - epsilon <= y ** 2 <= x + epsilon Else, return y such that x - epsilon <= y ** 7 <= x + epsilon Note: You must use bisection search to implement the function. ''' passA(n) __________ contains 8 __________.
- ce from the user t using recursion.Consider the given pmgqgh finclude(iostren} using namespace std; int main(){ //int n=2 take of y for(int i=0;ipython Q3: Repeatedly Cube Implement the following function, which cubes the given value x some number n times, based on the given skeleton. (define (repeatedly-cube n x) (if (zero? n) x (let (_________________________) (* y y y))))
- Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for the rugged hero to tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The number of ways that a positive integer n can be represented as a sum of consecutive integers is called its politeness, and can also be computed by tallying up the number of odd divisors of that number. However, note that the linked Wikipedia de0inition…Write a function that takes in an integer n and computes n!. Do this without recursion. In [ ]: deffactorial_iter(n):"""Takes in an integer n>0 and returns the product of all integers from 1 to n."""# YOUR CODE HEREraiseNotImplementedError() In [ ]: In [ ]: assert factorial_iter(6) == 720 assert factorial_iter(7) == 5040 assert factorial_iter(10) == 36288001. Let L = {w {a, b}* : w contains bba as a substring that starts in an odd numbered position in the string (where numbering starts at 1)}. For example, bbbbaaaab L, while bbbaaaab L. Show an NDFSM that accepts L
- Count consecutive summers def count_consecutive_summers(n): Like a majestic wild horse waiting for someone to come and tame it, positive integers can be broken down as sums of consecutive positive integers in various ways. For example, the integer 42 often used as placeholder in this kind of discussions can be broken down into such a sum in four different ways: (a) 3 + 4 + 5 + 6 + 7 + 8 + 9, (b) 9 + 10 + 11 + 12, (c) 13 + 14 + 15 and (d) 42. As the last solution (d) shows, any positive integer can always be trivially expressed as a singleton sum that consists of that integer alone. Given a positive integer n, determine how many different ways it can be expressed as a sum of consecutive positive integers, and return that count. The count of how many different ways a positive integer n can be represented as a sum of consecutive integers is also called its politeness, and can be alternatively computed by counting how many odd divisors that number has. However, note that the linked…Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0. This time, your problem has additional details: Constraints: The solution set must not contain duplicate triplets. The order of the output and the order of the triplets does not matter. 3 <= nums.length <= 3000 -105 <= nums[i] <= 105 Function definition for Java: public List<List<Integer>> threeSum(int[] nums) { // Your code here } Function definition for Python: def threeSum(self, nums: List[int]) -> List[List[int]]: #Your code here Announced Test Cases: Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0.Input: nums = [-5,0,5,10,-10,0] Output: [[-10,0,10],[-5,0,5]] Explanation: There are two possible combinations of triplets that satisfy: (-5,0,5) and (-10,0,10). Hint: There are 3 well-known ways to solve this problem!Edit this program so that it returns ecg data instead of heart beat in BPM. #include <ArduinoBLE.h> #define BLE_UUID_HEART_RATE_SERVICE "180D"#define BLE_UUID_HEART_RATE_MEASURMENT "2A37"#define ECG_SERV_UUID 0x2D0D#define ECG_MEAS_UUID 0x2D37#define ECG_NUM_CHANS_UUID 0x2D38#define ECG_SAMPLE_SETS_UUID 0x2D39#define ECG_COMMAND_UUID 0x2D3A #define HRM_VALUE_FORMAT_8BIT 0#define HRM_VALUE_FORMAT_16BIT 1#define HRM_SENSOR_CONTACT_NOT_DETECTED ( 2 << 1 )#define HRM_SENSOR_CONTACT_DETECTED ( 3 << 1 ) typedef struct {uint8_t flags;uint8_t heartRate;} heart_rate_measurment_t; union heart_rate_measurment_u{struct {heart_rate_measurment_t values;};uint8_t bytes[ sizeof( heart_rate_measurment_t ) ];}; union heart_rate_measurment_u heartRate = { .values = { .flags = 0, .heartRate = 0 } }; #define BLE_DEVICE_NAME "Arduino HRM"#define BLE_LOCAL_NAME "Arduino HRM" BLEService heartRateService( BLE_UUID_HEART_RATE_SERVICE );BLECharacteristic heartRateCharacteristic(…