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Step 1 of 1

For This proof we tack contradiction, and assume binary locks for simplicity.

Let n transactions T1, T2, ..., Tn such that they all obey the basic two-phase locking rule which is no transaction has an unlock operation followed by a lock operation. And Suppose that a non-(conflict)-serializable schedule S for T1, T2, ..., Tn does occur; then, according to the precedence (serialization) graph for S must have a cycle. Hence, there must be some sequence within the schedule of the form:

S: ...; [o1(X); ...; o2(X);] ...; [ o2(Y); ...; o3(Y);] ... ; [on(Z); ...; o1(Z);]...

where each pair of operations between square brackets [o,o] are conflicting (either [w,w], or [w, r], or [r,w]) in order to create an arc in the serialization graph. This implies that in transaction T1, Than a sequence of the following form occurs:

T1: ...; o1(X); ... ; o1(Z); ...

Furthermore, T1 has to unlock item X (so T2 can lock it before applying o2(X) to follow

the rules of locking) and has to lock item Z (before applying o1(Z), but this must occur

after Tn has unlocked it). Hence, a sequence in T1 of the following form occurs:

T1: ...; o1(X); ...; unlock(X); ... ; lock(Z); ...; o1(Z); ...

This implies that T1 does not obey the two-phase locking protocol (since lock(Z) follows

unlock(X)), contradicting our assumption that all transactions in S follow the two-phase

locking protocol.

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A palindrome is a number or a text phrase that reads the same backwards as forwards. For example, each of the following five-digit integers is a palindrome: 12321, 55555, 45554 and 11611. Write a C++ program that reads in a five-digit integer and determines whether it is a palindrome .

Question: A palindrome is a number or a text phrase that reads the same backwards as forwards. For example, each of the following five-digit integers is a palindrome: 12321, 55555, 45554 and 11611. Write a C++ program that reads in a five-digit integer and determines whether it is a palindrome . Answer: Step 1 PROGRAMING CODE:   #include <stdio.h> int main() {   int n, a, b, c, d, e;   int digit = 0;   printf( "Enter one 5 digit number: \n" );   scanf( "%d", &n );   a = n / 10000;   b = n / 1000 % 10;   c = n / 100 % 10;   d = n / 10 % 10;   e = n % 10;   if ( a == e && b == d ) {     printf( "Palindrome!\n" );   }   if ( a != e || b != d ) {     printf( "Not a palindrome.. :(\n" );   }   return 0; } PROGRAM SCREENSHOT: Step 2

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