Lecture 3 - Measuring Runtime and Pseudocode

void f(int *out, const int *in, int size) {
    int product = 1;
    for (int i = 0; i < size; i++) { 
        product *= in[i]; // multiplication
    }
    for (int i = 0; i < size; i++) {
        out[i] = product / in[i];
    }
} // f()

Measuring Runtime on Linux - produce a runtime report with this command, if you're in the current folder

(the % just means you're in the terminal, though usually it's $?)

% /usr/bin/time ./progName -options args

Kill it with control-C

Measuring Time in C++

http://stackoverflow.com/questions/27586001/how-getrusage-works-and-what-is-really-inside-the-rusage-struct

ru_utime This is the total amount of time spent executing in user mode, expressed in a timeval structure (seconds plus microseconds).

• user mode cannot see all of memory, it cannot communicate with I/O devices, and can only really do number-crunching

• the ru_utimestruct returns the amount of time your program has spent doing actual computation

ru_stime This is the total amount of time spent executing in kernel mode, expressed in a timeval structure (seconds plus microseconds).

• kernel can do: memory page allocation, filesystem read/write, printing things to the screen, etc.
• the ru_stime struct returns the amount of time your program has been waiting for an answer from the kernel when doing disk access, printing to the screen etc

#include <iostream>
#include <sys/resource.h>
#include <sys/time.h>

class Timer {
private:
    struct rusage startu;
    struct rusage endu;
    double duration;
public:
    Timer() {getrusage(RUSAGE_SELF, &startu); }

    void recordTime() {
        getrusage(RUSAGE_SELF, &endu);
        double start_sec = startu.ru_utime.tv_sec + startu.ru_utime.tv_usec/1e7;
        double end_sec = end.ru_utime.tv_sec + endu.ru_utime.tv_usec/1e7;
        duration = end_sec - start_sec; // get duration for elapsed time 
    } // recordTime()

    double getTime() { return duration; }
}; // Timer

Pseudocode

Algorithm arrayMax(A, n)
    Input array A of n integers
    Output max element of A

    current Max = A[0]
    for i = 1 to n - 1
        if A[i] > currentMax
            currentMax = A[i]
    return currentMax  # can also return multiple items in pseudocode

C++ Code

int arrayMax(int A[], int n) {
    int currentMax = A[0];

    for (int i = 1; i < n; ++i) {
        if (currentMax < A[i]) currentMax = A[i];
    }
    return currentMax;
} // arrayMax()

Exam Prep - Binomial Coefficient

(n, k) = (n-1, k-1) + (n-1, k) with all (n, 0) and (n, n) having a value of 1.

Tail-Recursive Solution (might be buggy)

double bin_coef_rekt(double n, double k) {
    if (n < 0 || k < 0) return 0;
    if (n == k ||k == 0) return 1;
    else if (n == 0 && k > 0) return 0;
    return bin_coef_iter(n - 1, k - 1) + bin_coef_iter(n - 1, k);
}