Cryptanalysis Cryptarithmetic problem

Q-1 (a) Solve The following Crypt arithmetic problem : C R O S S + R O A D S D E N G E R

Ans : D has to be 1 because when you do your addition by hand: CROSS +ROADS ----------DANGER You see that the D in DANGER must have be en carried over from the addition of C and R in CROSS and ROADS, respectively. Thus, we h ave: CROSS +ROA1S -----------1ANGER This, by some fairly weird algebra and modular arithmetic, leads to the following relationships: R = 2S mod 10 E = (3S + 1 - (2S mod 10)) mod 10 A = (C + 2R - (R + O mod 10)) mod 10 A = G - O mod 10 N = (R + 2O - (O + A mod 10)) mod 10 C = A - R mod 10 O = N - R mod 10 O = G - A mod 10 G = (O + A + S + 1 - (S + 1 mod 10)) mod 10 Obviously, this is of no help whatsoever to anyone, so we do trial and error. So now we try things. 12344 +25614 -----3NO! D has to be 1 92344 +25614 -----11NO! A cannot be 1 because D is 1

96233 +62513 -----158746

(b) Give the problem space for the following : (i)

Monkey – Banana problem

(ii)

Water – Jug problem

Ans: (i)

Monkey – Banana problem A monkey is in a cage and bananas are suspended f rom the ceiling, the monkey

wants to eat a banana but cannot reach them

–

in the room are a chair and a stick

–

if the monkey stands on the chair and waves the stick, he can knock a banana down to eat it

what are the actions the monkey should take? Initial state: monkey on ground with empty hand bananas suspended Goal state: monkey eating bananas. Actions: climb chair/get off  grab X wave X eat X

(ii)

Water – Jug problem The state space for this problem can be described as the set of ordered pairs of 

integers (x, y) such that x = 0, 1,2 , 3 or 4 and y = 0 ,1,2 or 3; x represents the number of  gallons of water in the 4-gallon jug and y represents the quantity of water in 3-gallon jug

•

The start state is (0,0)

•

The goal state is (2,n)

Production rules for Water Jug Problem

The operators to be used to solve the problem can be described as follows:

Sr. No

Current state

Next State

Description

1

(x, y) if x < 4

(4,y)

Fill the 4 gallon jug

2

(x, y) if y 0

(x-d, y)

Pour some water out of the 4 gallon jug

4

(x, y) if y > 0

(x, y-d)

Pour some water out of the 3-gallon jug

5

(x, y) if x>0

(0, y)

Empty the 4 gallon jug

6

(x, y) if y >0

(x,0)

Empty the 3 gallon jug on the ground

7

(x, y) if x + y

(4, y-(4-x))

Pour water from the 3 –gallon jug into the 4 –

>= 4 and y >0 8

(x, y) if x + y

gallon jug until the 4-gallon jug is full (x-(3-y), 3)

>= 3 and x>0 9

(x, y) if x + y

gallon jug until the 3-gallon jug is full (x + y, 0)

0 10

(x, y) if x + y

(0,2)

Pour all the water from the 3-gallon jug into the 4-gallon jug

(0, x + y)

0 11

Pour water from the 4-gallon jug into the 3-

Pour all the water from the 4-gallon jug into the 3-gallon jug

(2,0)

Pour the 2 gallons from 3-gallon jug into the 4gallon jug

12

(2,y)

(0,y)

Empty the 2 gallons in the 4-gallon jug on the ground

To solve the water jug problem

•

Required a control structure that loops through a simple cycle in which some rule whose left side matches the current state is chosen, the appropriate change to the state is made as described in the corresponding right side, and the resulting state is checked to see if it corresponds to goal state.

•

One solution to the water jug problem

•

Shortest such sequence will have a impact on the choice of appropriate mechanism to guide the search for solution. Gallons in the

Gallons in the 3-

4-gallon jug

gallon jug

0

0

2

0

3

9

3

0

2

3

3

7

4

2

5 or 12

0

2

9 0r 11

2

0

Rule applied

(c) Solve the following 8-puzzle using breadth f irst search 1

2

8

3 4

7

6

5

2

8

3

1

6

4

7

5

Ans :

Left 1

2

8

3

1

6

4

7

5

Right 3

2

8

3

1

6

4

7

5

Up 2 2 1

8 7 6

3 5 4

2

8

3

1 7

4 6

5

Left 1

2

8

3

1

6

4

7

5

8

3

6

4

1

7

5

2

8

3

U 1.1 2

UP 2 1 7 2 Left 2.1 7

8

3

1

4

6

5

4 6

2 U 2 .2

5 3

1

8

4

7

6

5

Right 2.3

2

8

3

1

4

7

6

5

2

8

3

Right 3

Up 3.1

2

8

3

1

6

4

7

5

2

8

1

6

7

5

4

2

8

3

6

4

7

5

3

Up 1.1

1

2

8

3

6

4

Right 1.1.2

6 1

4 7

5

2 Left 2.1

Up 2.1.1

7

8

3

2

1

4

7

6

5

3

1

4

6

5

2

Up 2.2

Left 2.2.1

8

2

3

1

8

4

7

6

5

Right 2.3

2

8

3

Down 2.1.2

7

1

4

6

5

Right 2.2.2

2

3

1

8

4

7

6

5

3

1

8

4

7

6

5

2

8

1

4

7

6

3

5 Down 2.3.2

Up 2.3.1 2

8

1

4

3

7

6

5

Up 3.1 2 Left 3.1.1

8

1 7

3 6

5

4

2

8

1

6

7

5

2

8

3

1

4

5

7

6

2

8

1

6

3

7

5

4

3

4 Up 3.1.2

Up 1.1.1

8

3

2

6

4

1

7

5

Right 1.1.1.1 8

Right 1.1.2

3

2

6

4

1

7

5

2

8

3

6 1

2 6

3 8

4 7

Right 1.1.2.2

4

Left 2.2.1

5

2

8

6

4

1

7

5

2

3

8

4

6

5

2

3

1

8

4

7

6

5

Down 2.2.1.1

1

7

Right 2.2.1.1.1

1

2

8 7

3

3 4

6

5

Final goal state find and ans is Block Right 2.2.1.1.1

Q-2 Q2

(a)

Explain Problem Decomposability with example.

Down 1.1.2.3

2

8

3

6

7

4

1

Down 2.2.1.1.2

5

1

2

3

7

8

4

6

5

Suppose we want to solve the problem of computing the expression (x2 + 3x + 3x  + sin2;c

/

• cos2*) dx 

We can solve this problem by breaking it down into three smaller problems; each of which we can then solve by using a small collection of specific rules. Figure-2;9 shows the problem tree that will be generated by the process of  problem decomposition as it can be exploited by a simple recursive integration program that worksras follows: At each step, it checks tojsee whether the problem if is working on is immediately solvable.' If-so, then the answer is returned directly. If the problem is not easily solvable, the integrator checks to see whether it can decompose the problem into smaller problems. If it can, it creates those problems and calls itself recursively on them. Using: thistechnique of  problem  problem decomposition, decomposition, we can often solvevery large problemseasily;

v

.

Now consider the problem illustrated in Figure 2.10. This problem is drawn firom.the domain often referred to in AMiterature as the blocks world. Assume that the following operators are available:

Figure 2.10: A Simple Blocks World Problem

2 8 3 1 2 3 1 6 48 7

4

57 6 5

Figure 2.11 : A Proposed Solution for the Blocks Problem 1. CLEARQc) [block x  [block  x has has nothing on it] -+ ON(JC, Table) [pick up x  up  x and and put it

on the table]

CLEAR(x)andCLEAR(y)->ON(jt,;y) ->ON(jt,;y) [put* on y] 2. CLEAR(x)andCLEAR(y) Applying the technique of problem decomposition to this simple blocks world example would lead to a solution tree such as that shown in Figure 2.11. In the figure, goals are underlined. States that have been achieved are not underlined. The idea of this solution is to reduce the problem of getting B on C and A on B to two separate problems. The first of these new problems, getting B on C, is simple, given the start state. Simply put B on C. The second subgoal is not quite so simple. Since the only operators we have allow us to pick up single blocks at a time, we have to clear off A by removing C before we can pick up A and put it on B. This can easily be done. However, if we now try to combine the two subsolutions into one solution, we will fail. Regardless of which one we do first, we will not be able to do the second as we had planned. In this problem, the two subproblems are not independent. They interact and those interactions must be considered in order to arrive at a solution for the entire problem. These two examples, symbolic integration and the blocks world, illustrate the difference between decomposable and nondecomposable problems. In Chapter. 3^ we present a specific algorithm for problem decomjposition, and in Chapter 13, we look at what happens when decomposition' d ecomposition'is'impossible. is'impossible.

(b)

Write and explain Means-End Analysis algorithm.

So far, we have presented a collection of search strategies that can reason either forward or backward, but for a given problem, one direction or the other must be chosen. Often, however, a mixture of the two directions is appropriate. Such a mixed strategy would make it possible to solve the major parts of a problem first and then go back and solve the small problems that arise in "gluing" the big pieces together. A technique known as means-ends analysis allows us to do that. The means-ends analysis process centers around the'detection of differences between the current state and the goal state. Once such a difference is isolated, an operator that can reduce the difference must be found. But perhaps that operator cannot be applied to the current state. So we set up a subproblem of 

getting to a state-in which it can be applied. The kind of backward chaining in which operators are selected and then subgoals are set up to establish the preconditions of the'operators is called operator subgoaling. But maybe the operator does not produce exactly the goal state we want. Then we have a second subproblem of getting from the state it does produce to the goal. But if  the difference was chosen correctly and if the operator is really effective at reducing the difference,, then the two subproblems should be easier to solve than the

Algorithm: Means-Ends Analysis (CURRENT, GOAL) 1. Compare CURRENT to CURRENT  to GOAL. If there are no differences between them then

return. 2. Otherwise, select the most important difference and reduce it by doing the

following until success or failure is signaled:

(a) Select an as yet untried operator O that is applicable to the current

difference. difference. If there are no such s uch operators, then signal failure. (b) Attempt to apply O to CURRENT. Generate descriptions of two states: O-

START, a state in which O's preconditions are satisfied and O-RESULT, the state that would result if  if O O were applied in O-START. (c) If 

(FIRST-PART