OPERATIONS RESEARCH >INTERNATIONAL EDITI
4th Edition
ISBN: 9780534423629
Author: WINSTON
Publisher: CENGAGE L
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Expert Solution & Answer
Chapter 2.4, Problem 1P
Explanation of Solution
Determining the dependency of the given sets of
Consider the given sets of vectors,
A matrix A is formed as given below; whose rows are the above given vectors:
The Gauss-Jordan method is applied to find the dependency of the above given sets of vectors.
Exchange row 3 and row 1, then the following matrix is obtained,
Multiply row 1 with 0.5, then the following matrix is obtained,
Now, replace row 2 by (row 2 – row 1), then the following matrix is obtained,
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Exercise 1 Function and Structure [30 pts]
Please debug the following program and answer the following questions. There is a cycle in a linked
list if some node in the list can be reached again by continuously following the next pointer.
#include
typedef struct node {
int value;
struct node *next;
} node;
int 11_has_cycle (node *first)
if (first
==
node *head =
{
NULL) return 0;
first;
while (head->next != NULL) {
}
if (head
first) {
return 1; }
head = head->next;
return 0;
void test ll_has_cycle () {
int i;
node nodes [6];
for (i = 0; i < 6; i++) {
nodes [i] .next = NULL;
nodes [i].value = i;
}
nodes [0] .next
=
&nodes [1];
nodes [1] .next
=
&nodes [2];
nodes [2] .next
=
&nodes [3];
nodes [3] .next
nodes [4] .next
&nodes [4];
NULL;
nodes [5] .next = &nodes [0];
printf("1. Checking first list for cycles. \n Function 11_has_cycle says it
has s cycle\n\n", 11_has_cycle (&nodes [0])?"a":"no");
printf("2. Checking length-zero list for cycles. \n Function 11_has_cycle
says it has %s…
how to read log logs
Discrete Mathematics for Computer Engineering
Chapter 2 Solutions
OPERATIONS RESEARCH >INTERNATIONAL EDITI
Ch. 2.1 - Prob. 1PCh. 2.1 - Prob. 2PCh. 2.1 - Prob. 3PCh. 2.1 - Prob. 4PCh. 2.1 - Prob. 5PCh. 2.1 - Prob. 6PCh. 2.1 - Prob. 7PCh. 2.2 - Prob. 1PCh. 2.3 - Prob. 1PCh. 2.3 - Prob. 2P
Ch. 2.3 - Prob. 3PCh. 2.3 - Prob. 4PCh. 2.3 - Prob. 5PCh. 2.3 - Prob. 6PCh. 2.3 - Prob. 7PCh. 2.3 - Prob. 8PCh. 2.3 - Prob. 9PCh. 2.4 - Prob. 1PCh. 2.4 - Prob. 2PCh. 2.4 - Prob. 3PCh. 2.4 - Prob. 4PCh. 2.4 - Prob. 5PCh. 2.4 - Prob. 6PCh. 2.4 - Prob. 7PCh. 2.4 - Prob. 8PCh. 2.4 - Prob. 9PCh. 2.5 - Prob. 1PCh. 2.5 - Prob. 2PCh. 2.5 - Prob. 3PCh. 2.5 - Prob. 4PCh. 2.5 - Prob. 5PCh. 2.5 - Prob. 6PCh. 2.5 - Prob. 7PCh. 2.5 - Prob. 8PCh. 2.5 - Prob. 9PCh. 2.5 - Prob. 10PCh. 2.5 - Prob. 11PCh. 2.6 - Prob. 1PCh. 2.6 - Prob. 2PCh. 2.6 - Prob. 3PCh. 2.6 - Prob. 4PCh. 2 - Prob. 1RPCh. 2 - Prob. 2RPCh. 2 - Prob. 3RPCh. 2 - Prob. 4RPCh. 2 - Prob. 5RPCh. 2 - Prob. 6RPCh. 2 - Prob. 7RPCh. 2 - Prob. 8RPCh. 2 - Prob. 9RPCh. 2 - Prob. 10RPCh. 2 - Prob. 11RPCh. 2 - Prob. 12RPCh. 2 - Prob. 13RPCh. 2 - Prob. 14RPCh. 2 - Prob. 15RPCh. 2 - Prob. 16RPCh. 2 - Prob. 17RPCh. 2 - Prob. 18RPCh. 2 - Prob. 19RPCh. 2 - Prob. 20RPCh. 2 - Prob. 21RPCh. 2 - Prob. 22RP
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