show the output with screenshots of the prolog code below, state(5,1, [state(5,1)|_1).12start():- state(0,0, [state(0,0)]).state(X, Y,W): - X < 5,upd_state(5,Y, w, 'Rule 1: Fil1 Jug 1 full --->').7state(X, Y,W): - Y<2,upd_state(X, 2, W, 'Rule 2: Fill Jug 2 full --->').89.10state(X,Y,W): -X>0,upd_state(0, Y, W, 'Rule 3: Empty Jug 1 --->').111213state(X, Y,W): -Y>0,upd_state(X,0,W, 'Rule 4: Empty Jug 2 --->').141516state(X,Y,W) : -X+Y=<5,Y>0,X1 is X+Y,171819upd_state(X1,0,W, 'Rule 5: Pour all of Jug 2 to Jug 1 --->').20state(X, Y,W): -X+Y>=5,Y>0,Y1 is Y-(5-x),upd_state(5, Y1,W, 'Rule 6: Fill Jug 1 from Jug 2 --->').21222324state(X,Y,W) : -X+Y>=2, X>0,X1 is X-(2-Y),upd_state(X1,2, W, 'Rule 7: Fill Jug 2 from Jug 1 --->').25262728upd_state(X, Y,W, P): -not (member(state(X,Y),W)),print ([P,' Current State: (' ,x,',',Y,').\n']),state(X,Y, [state(X, Y) |W]).2930313233print([]).print([H|T]): -write(H), print(T).3435

This is Prolog code. The code is showing the water jug problem. We have to provide the output of the…

state(5,1, [state(5,1)|_1). start(): - state(0,0,[state(0,0)]). state(X,Y,W): - x < 5, upd_state(5, Y, w, 'Rule 1: Fil1 Jug 1 full --->'). state(X, Y,W): - Y<2, upd_state(X,2, W, 'Rule 2: Fill Jug 2 full --- >'). state(X,Y,W): -X>e, upd_state(0,Y, W, 'Rule 3: Empty Jug 1 --->'). state(X,Y,W):-Y>0, upd_state(X,0,w, 'Rule 4: Empty Jug 2 --->'). state(X,Y,W): -X+Y=<5,Y>0, X1 is X+Y, upd_state(X1,0,w, 'Rule 5: Pour all of Jug 2 to Jug 1 --->'). state(X,Y,W): -X+Y>=5,Y>0, Y1 is Y-(5-X), upd_state(5, Y1,W, 'Rule 6: Fill Jug 1 from Jug 2 --->'). state(X,Y,W) : -X+Y>=2, X>0, X1 is X-(2-Y), upd_state(X1,2, W, 'Rule 7: Fill Jug 2 from Jug 1 --->'). upd_state(X, Y, W, P):- not (member(state(X,Y),W)), print ([P,' Current State: (',x,',',Y,').\n'1), state(X,Y, [state(X,Y) |W]). print([]). print([H|T]): -write(H), print (T).

state(5,1, [state(5,1)|_1). start(): - state(0,0,[state(0,0)]). state(X,Y,W): - x < 5, upd_state(5, Y, w, 'Rule 1: Fil1 Jug 1 full --->'). state(X, Y,W): - Y<2, upd_state(X,2, W, 'Rule 2: Fill Jug 2 full --- >'). state(X,Y,W): -X>e, upd_state(0,Y, W, 'Rule 3: Empty Jug 1 --->'). state(X,Y,W):-Y>0, upd_state(X,0,w, 'Rule 4: Empty Jug 2 --->'). state(X,Y,W): -X+Y=<5,Y>0, X1 is X+Y, upd_state(X1,0,w, 'Rule 5: Pour all of Jug 2 to Jug 1 --->'). state(X,Y,W): -X+Y>=5,Y>0, Y1 is Y-(5-X), upd_state(5, Y1,W, 'Rule 6: Fill Jug 1 from Jug 2 --->'). state(X,Y,W) : -X+Y>=2, X>0, X1 is X-(2-Y), upd_state(X1,2, W, 'Rule 7: Fill Jug 2 from Jug 1 --->'). upd_state(X, Y, W, P):- not (member(state(X,Y),W)), print ([P,' Current State: (',x,',',Y,').\n'1), state(X,Y, [state(X,Y) |W]). print([]). print([H|T]): -write(H), print (T).

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Question

show the output with screenshots of the prolog code below,

state(5,1, [state(5,1)|_1).
1
2
start():- state(0,0, [state(0,0)]).
state(X, Y,W): - X < 5,
upd_state(5,Y, w, 'Rule 1: Fil1 Jug 1 full --->').
7
state(X, Y,W): - Y<2,
upd_state(X, 2, W, 'Rule 2: Fill Jug 2 full --->').
8
9.
10
state(X,Y,W): -X>0,
upd_state(0, Y, W, 'Rule 3: Empty Jug 1 --->').
11
12
13
state(X, Y,W): -Y>0,
upd_state(X,0,W, 'Rule 4: Empty Jug 2 --->').
14
15
16
state(X,Y,W) : -X+Y=<5,Y>0,
X1 is X+Y,
17
18
19
upd_state(X1,0,W, 'Rule 5: Pour all of Jug 2 to Jug 1 --->').
20
state(X, Y,W): -X+Y>=5,Y>0,
Y1 is Y-(5-x),
upd_state(5, Y1,W, 'Rule 6: Fill Jug 1 from Jug 2 --->').
21
22
23
24
state(X,Y,W) : -X+Y>=2, X>0,
X1 is X-(2-Y),
upd_state(X1,2, W, 'Rule 7: Fill Jug 2 from Jug 1 --->').
25
26
27
28
upd_state(X, Y,W, P): -
not (member(state(X,Y),W)),
print ([P,' Current State: (' ,x,',',Y,').\n']),
state(X,Y, [state(X, Y) |W]).
29
30
31
32
33
print([]).
print([H|T]): -write(H), print(T).
34
35

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