Answer: provided in the explanation section
Explanation:
Given that:
Assume D(k) =║ true it is [1 : : : k] is valid sequence words or false otherwise
now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.
So, from here we have that D(k) is given by the following recorance relation:
D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1 : : : k] ∦ array of boolean variable.
for a ← 1 to m
do ;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1 : : : a])
d (a) ← True
(b). Algorithm Output
if D[k] = = True
stack = temp stack ∦stack is used to print the strings in order
c = k
while C > 0
stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c
P = W (p) - 1
output stack
= 0 =
cheers i hope this helps !!!
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Answer:
Total Memory= 4 KB = 4096 bytes = 32768 bits
Explanation:
<em><u>1. Data lines are 8 From D0 to D7</u></em>
so
Total memory at single address locations is 8 bits.
<em><u>2. Address lines are 12 (A0 to A11)</u></em>
There are 12 address lines but 3 out 12 are for selction of chip or memory bank.
so only 9 pins are there to address the locations on one chip.
Total No. of address locations on single chip = 2^9 = 512 locations
as 1 location is 1 byte so total memory of single chip is 512 bytes.
<u><em>3. Total Memory Bank </em></u>
There are total 3 selection pins for memory bank.
so
Total chips = 2^3 = 8.
<em><u>4. Total Memory </u></em>
Total size of 1 chip = 512 bytes
Total size of 8 chip = 8x512 bytes = 4096 bytes = 4096/1024 kb = 4 kb
<em>So total memory of system is 4 Kb = 4096 bytes = 32768 bits</em>
Answer:
Contiguous
Explanation:
A Contiguous memory allocation is known to be a classical memory allocation model. In this situation, we have a system which assigns consecutive memory blocks to a process. It is one of the oldest methods of memory allocation. If the process is in need of execution, the memory would be requested by the process. The processes size would then be compared to the amount of Contiguous memory that is available for the execution of the process.
