This calculator helps you to score your official practice tests results. PersonoutlineTimurschedule 2018-05-31 21:50:39. If you are preparing for PSAT/NMSQT test, you most likely will simulate test day by printing and taking official PSAT/NMSQT practice tests from here. However, this page suggests you score by hand, and provides 9-page pdf. Jan 3, 2009 - One approach is to extend an online SAT solver with the possibility of generating models of satisfiable formulas and proofs of unsatisfiability for.
IntroductionSAT is short for 'satisfiability'. Chances are you have heard of it or one ofits variants like 3-SAT in passing, especially in discussions of complexity andNP-completeness. In this post, we will go into details of what it is all about,why it is of such importance from both a theoretical and practical perspective,and how to approach solving it by developing a simple Python SAT solver. Bythe end of this post, we will have a working SAT solver with a command-lineinterface. The code for it is on GitHub:. Feel free to fork and contributeimprovements. Of course, our implementation will not be anywhere close to morecomplicated SAT solvers implemented in C or C, such as.
The focus here is on simplicity sincethe code is to be an introduction to SAT and SAT solvers.Sections marked with. are more theoretical and not required forunderstanding the algorithm we will use. On the other hand, the rest of theintroduction section below can be skipped if you already know the problemdefinition and relevant technical terms. Technical TerminologyNow that the problem makes sense, let's define the technicalvocabulary. First, what we called 'candidates' are called variables. Thevariables in the above example are A A A, B B Band C C C. Avariable can be assigned true or false.
A literal is a variable or itsnegation. For example A A Aand ∼ A sim A ∼ Aare literals. Literalswithout the ∼ sim ∼are called positive, pure, or unnegated literals.Literals with ∼ sim ∼are called negated literals. A set of literals iscalled a clause. An assignment is a mapping of variables to true or false.For example, the assignment that satisfied the clauses in the previous examplewas given by A = t r u e A = true A = t r u e, B = f a l s e B = false B = f a l s eand C = t r u e C = true C = t r u e. A clauseis satisfied by an assignment if at least one of its unnegated literals isassigned true by the assignment, or one of its negated literals is assignedfalse in the assignment. It is customary, in logic notation, to separate theliterals in a clause using the ∨ vee ∨symbol, read 'or'.
For example, thefirst clause above is written as A ∨ B ∨ ∼ C A vee B vee sim C A ∨ B ∨ ∼ Cin mathematicalnotation.So SAT can be summarized as follows: given a list of clauses, determine ifthere exists an assignment that satisfies all of them simultaneously.It is also worthy of mention that there is a variation of SAT called 3-SAT withthe restriction that each clause consists of at most 3 (distinct) literals.It can be shown with relative ease that SAT is in fact reducible to 3-SAT. A Simple SAT Solver In PythonEven though SAT is NP-complete and therefore no known polynomial-time algorithmfor it is (yet) known, many improvements over the basic backtracking algorithmshave been made over the last few decades. However, here we will look at one ofthe most basic yet relatively efficient algorithms for solving SAT. The encodingand the algorithm are based on Knuth's SAT0W program which you can downloadfrom his.The algorithm is a watch-list based backtracking algorithm. What makes thewatch-list based algorithms particularly simple, as we will see, is that verylittle (practically nothing) needs to be done to 'undo' steps taken when weneed to backtrack. Parsing & Encoding The InputBefore we can approach solving a SAT instance, we need to be able to representthe instance in memory.
Let's remember that a SAT instance is a set of clauses,and each clause is a set of literals. Finally, a literal is a variable that iseither negated or not. Of course, we can just store the instance as a list ofclauses, with each clause being a list of strings that are the literals. Theproblem with this approach is that we will not be able to quickly look upvariables, and checking to see if a literal is negated or not, and negating itif not, would be rather slow string operations.Instead, we will first assign a unique number, starting from 0 0 0andcounting up, to each variable as we encounter them, using a dictionary to keeptrack of the mapping.
So variables will be encoded as numbers 0 0 0ton − 1 n-1 n − 1where n n nis the number of variables. Then for an unnegatedliteral with variable encoded as number x x xwe will encode the literal as2 x 2x 2 x, and the negated one will be 2 x + 1 2x + 1 2 x + 1. Then a clause willsimply be a list of numbers that are the encoded literals, andLet's look at an example first. For this, let's see how the code that we willlook at in a minute behaves. Class SATInstance ( object ): def parseandaddclause ( self, line ): clause = for literal in line.
Split : negated = 1 if literal. Startswith ( ' ) else 0 variable = literal negated: if variable not in self. Variabletable: self.
Variabletable variable = len ( self. Variables ) self.
Append ( variable ) encodedliteral = self. Variabletable variable 0 and not line. Startswith ( '#' ): instance. Parseandaddclause ( line ) return instance def literaltostring ( self, literal ): s = ' if literal & 1 else ' return s + self.
Variables literal 1 def clausetostring ( self, clause ): return ' '. Literaltostring ( l ) for l in clause ) def assignmenttostring ( self, assignment, brief = False, startingwith = ' ): literals = for a, v in (( a, v ) for a, v in zip ( assignment, self. Variables ) if v. Startswith ( startingwith )): if a 0 and not brief: literals. Append ( ' + v ) elif a: literals. Append ( v ) return ' '. Join ( literals )As you can see, we also include methods here to decode variables, literals,clauses, and assignments.
These are used for outputting logging messages aswell as the final solutions. Keeping Track Of The AssignmentOur algorithm will be a backtracking algorithm, in which we will assign true orfalse to all the variables, starting from variable 0 0 0and going in order tovariable n − 1 n-1 n − 1. Of course, the basic search space is of size 2 n 2^n 2 nbut bypruning, we will not explore the whole space (usually anyway). The assignmentwill be kept as a list of length n n n, with item at index i i ibeingNone if neither true or false has been assigned variable i i i, and0 0 0(false) or 1 1 1(true) otherwise, depending on the assignment.When we backtrack, we set the corresponding item in the assignment list back toNone to indicate it is no longer assigned. Watch-listsNow that we have the encoding in place, and know how to keep track of theassignment, let's look at the key idea of our algorithm.
For each clause to besatisfied, it needs to have at least one of its literals satisfied. As such, wecan make each clause watch one of its literals, and ensure that the followinginvariant is maintained throughout our algorithm: Invariant All watched literals are either not assigned yet, or they have beenassigned true.We then proceed to assign true or false to variables, starting from 0 0 0ton − 1 n-1 n − 1. If we successfully assign true or false to every variable whilemaintaining the above variant, then we have an assignment that satisfies everyclause.To maintain this invariant, any time we assign true or false to a variable, weensure to update the watch-list accordingly.
To do this efficiently, we need tokeep a list of clauses that are currently watching a given literal. This isdone in the code below using a list of length 2 n 2n 2 nof double-ended queue( collections.deque), with each clause initially watching the first literal init. The function below takes care of this setting up of the watch-list. Def setupwatchlist ( instance ): watchlist = deque for in range ( 2. len ( instance. Variables )) for clause in instance. Clauses: # Make the clause watch its first literal watchlist clause 0.
Append ( clause ) return watchlistWhy double-ended queues instead of just a list? Short answer is that afterexperimenting, I found out that double-ended queues provided the bestperformance.Back to the algorithm, whenever we assign true to a variable x x xwe mustmake clauses watching ∼ x sim x ∼ xwatch something else. Andsimilarly, whenever we assign false to a variable x x xwe make clauses watchingx x xwatch something else.
If we can not make a clause watch something, whichhappens when all the other literals in a clause have already been assignedfalse, then we know that the current assignment contradicts the clause, and westop and backtrack. We only need one clause to be contradicted to know not togo any further. As such, the heart of our algorithm will be where we update thewatch-list after an assignment has been made.
The Python function below, whichis in ( watchlist.py), implements this part of the algorithm. Making It Iterative.For fun, let's see if we can implement the above algorithm without recursion.This is in fact how Knuth implements the algorithm. (He seems to dislikerecursion, see for example.)The basic idea here is to manually keep track of the current state of thebacktrack tree.
![Online Online](/uploads/1/2/5/5/125503715/287631850.jpg)
When we use recursion, the state is kept implicitly using thestack and which instruction is executing in each of the function calls. In theiterative case, we will store the state using d which is the current depthof the backtrack tree we are currently in, and also the variable we are toassign to currently, and the state list which keeps track of whichassignments for each variable have been tried so far. Here is the code. NP-Completeness Of SAT.In previous section we saw how a problem regarding colouring of regions in amap can be reduced to SAT.
This can be further generalized to much larger classof problems: any decision problem that can be decided in polynomial time usinga non-deterministic Turing machine can be reduced in polynomial time toSAT. This was first proved in Stephen Cook's paper, which is the paperthat introduced the famous P = NP question as well. Let's go over the basicidea in the paper very briefly here. If you are interested in more details,make sure you have a look at the paper, as it is rather short and a pleasure toread.But before we go into detail, let us take a moment to discuss why it is of suchimportance. First, nobody has yet come up with an efficient (polynomial time)algorithm to solve SAT in its generality. (SAT with some restrictions, e.g.2-SAT, can be solved efficiently though.) Showing that a problem can be reducedto SAT means that if we find an efficient algorithm for SAT then we have foundan efficient algorithm for that problem as well.
For example, if we find apolynomial-time algorithm for SAT then we immediately have a polynomial-timealgorithm for the graph colouring problem given above.Now, the class of decision problems that can be solved in polynomial-time usinga non-deterministic Turing machine is known as NP (which stands forNon-deterministic Polynomial). This is a very large class of problems, sinceTuring machines are one of the most general computational models we have, andeven though we are limited to polynomial-time Turing machines, the fact thatthe Turing machine does not have to be deterministic allows us much morefreedom. Some examples of problems that are in NP are:. all problems in P, e.g. Determining if a number is prime or not (PRIMES), anddecision versions of shortest path, network flow, etc.,. integer factorization,.
graph colouring,. SAT,.
and all NP-complete problems (see for a ratherlarge list of examples).A problem is said to be NP-complete if it, in addition to being in NP, also hasthe property that any other problem in NP can be reduced to it inpolynomial-time. Cook's paper proved SAT to be NP-complete. In fact, since thatpaper introduced the concept of NP-completeness, SAT was the first problemto be proved NP-complete.
Since then, many other problems have been shown to beNP-complete, often by showing that SAT (or 3-SAT) can be reduced inpolynomial-time to those problems (converse of what we proved earlier forgraph colouring).Now, as promised, let's briefly look at why SAT is NP-complete. For this, weneed to know more precisely what a Turing machine is.
Unfortunately, this wouldinvolve a bit more detail than I want to include in this section. So instead, Iam going to show that if a problem can be solved using a finite-state machine(FSM) then it be reduced in polynomial-time to SAT. The case for Turingmachines, which are a generalizations of finite-state machines (Turing machinesare basically FSM's with the addition of a tape that they can read from andwrite to), is quite similar, just more complicated. I encourage you to readCook's original paper for details of the proof with Turing machines.First, let's define what an FSM is. In simplest terms, an FSM is a program thathas a finite number of states, and that when fed an input character, moves toanother state (or possibly stays in the same state) based on a fixed set ofrules. Also, some states are taken as 'accepting' states.
Given an inputstring, we feed the string character by character into the FSM, and if at theend the FSM is in an accepting state, the answer to our decision problem isyes. If not, the answer is no.The below code shows how an FSM could can be implemented in Python. Note thatin this implementation, we are forced to have a deterministic FSM. Let'signore this detail for now though. This particular example implements an FSMthat accepts input strings that contain an even number of ones. # No more than one state at each stepQ0-0 Q1-0Q0-1 Q1-1Q0-2 Q1-2Q0-3 Q1-3# At least one state in each stepQ0-0 Q1-0Q0-1 Q1-1Q0-2 Q1-2Q0-3 Q1-3# Add the rules # (EVEN, 1) - ODDQ0-0 P1 Q1-1Q0-1 P2 Q1-2Q0-2 P3 Q1-3# (ODD, 1) - EVENQ1-0 P1 Q0-1Q1-1 P2 Q0-2Q1-2 P3 Q0-3# (EVEN, 0) - EVENQ0-0 P1 Q0-1Q0-1 P2 Q0-2Q0-2 P3 Q0-3# (ODD, 0) - ODDQ1-0 P1 Q1-1Q1-1 P2 Q1-2Q1-2 P3 Q1-3# Start in state 0Q0-0# End in an accepting stateQ0-3Let's see the output of running a SAT solver on this, and another file fort = 3 t=3 t = 3and t = 4 t=4 t = 4.
Calculators are not provided at the test center, and you can’t share a calculator. Quick Facts. Only the calculators listed on this page are acceptable for the Math Test—Calculator portion of the test.
![Solver Solver](/uploads/1/2/5/5/125503715/955359867.jpg)
You may not use a calculator while working on the Reading, Writing and Language, or Math Test—No Calculator portions, and must put the calculator away during these sections of the test. You will be dismissed and your scores canceled if you use your calculator to share information during the test, or to remove test questions or answers from the test room.Calculator Smarts. Bring your own calculator.
You can’t share one. Don’t bring a brand-new calculator. Bring one you know. Practice for the test using the same calculator you’ll use on test day. The Math Test includes some questions where it’s better not to use a calculator, even though you’re allowed to. It may help to do scratch work in the test book. Get your thoughts down before using your calculator.
Make sure your calculator is in good working order and that batteries are fresh. The test center will not have batteries or extra calculators. If your calculator fails during testing and you have no backup, you'll have to complete the test without it. You can use calculators only for the Subject Tests in Mathematics.Acceptable CalculatorsIf you have a calculator with characters that are one inch or higher, or if your calculator has a raised display that might be visible to other test-takers, you will be seated at the discretion of the test coordinator.Only battery-operated, handheld equipment can be used for testing. No power cords are allowed.Calculators permitted during testing include:. Most graphing calculators (see chart).
All scientific calculators. All four-function calculators (not recommended)Brands and ModelsThese calculators all pass the test.