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If a map labeling problem can be reduced to a situation in which each remaining label has only two potential positions in which it can be placed, then it may be solved efficiently by using an instance of 2-satisfiability to find a placement avoiding any conflicting pairs of placements ; several exact and approximate label placement algorithms for more complex types of problems are based on this principle.
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If and map
# If adiabats and isotherms are graphed severally at regular changes of entropy and temperature, respectively ( like altitude on a contour map ), then as the eye moves towards the axes ( towards the south-west ), it sees the density of isotherms stay constant, but it sees the density of adiabats grow.
If A is a fixed element of a ring ℜ, the first additional relation can also be interpreted as a Leibniz rule for the map given by B ↦.
If G is a group, and g is a fixed element of G, then the conjugation map
If the user is unable to identify what is being demonstrated in a reasonable fashion, the map may be regarded as useless.
If the four-color conjecture were false, there would be at least one map with the smallest possible number of regions that requires five colors.
If a map contains a reducible configuration, then the map can be reduced to a smaller map.
If f: X → Y is a continuous map, x < sub > 0 </ sub > ∈ X and y < sub > 0 </ sub > ∈ Y with f ( x < sub > 0 </ sub >) = y < sub > 0 </ sub >, then every loop in X with base point x < sub > 0 </ sub > can be composed with f to yield a loop in Y with base point y < sub > 0 </ sub >.
* If V is a normed vector space with linear subspace U ( not necessarily closed ) and if is continuous and linear, then there exists an extension of φ which is also continuous and linear and which has the same norm as φ ( see Banach space for a discussion of the norm of a linear map ).
* If V is a normed vector space with linear subspace U ( not necessarily closed ) and if z is an element of V not in the closure of U, then there exists a continuous linear map with ψ ( x ) = 0 for all x in U, ψ ( z ) = 1, and || ψ || = 1 / dist ( z, U ).
If an isomorphism can be found from a relatively unknown part of mathematics into some well studied division of mathematics, where many theorems are already proved, and many methods are already available to find answers, then the function can be used to map whole problems out of unfamiliar territory over to " solid ground " where the problem is easier to understand and work with.
If G is any subgroup of GL < sub > n </ sub >( R ), then the exponential map takes the Lie algebra of G into G, so we have an exponential map for all matrix groups.
Thus the set of all polynomials with coefficients in the ring R forms itself a ring, the ring of polynomials over R, which is denoted by R. The map from R to R sending r to rX < sup > 0 </ sup > is an injective homomorphism of rings, by which R is viewed as a subring of R. If R is commutative, then R is an algebra over R.
If a player dies all their weapons are lost and they receive the spawn weapons for the current map, usually the gauntlet and machine gun.
If a pushwall exits the boundaries of the level, the game quits with the error message " PushWall Attempting to escape off the edge of the map ".
If I is a right ideal of R, then R / I is simple if and only if I is a maximal right ideal: If M is a non-zero proper submodule of R / I, then the preimage of M under the quotient map is a right ideal which is not equal to R and which properly contains I.
If the tangent space is defined via curves, the map is defined as
More precisely, if A is a finite set of generators for G then the word problem is the membership problem for the formal language of all words in A and a formal set of inverses that map to the identity under the natural map from the free monoid with involution on A to the group G. If B is another finite generating set for G, then the word problem over the generating set B is equivalent to the word problem over the generating set A.
If Φ is a unital positive map, then for every normal element a in its domain, we have Φ ( a * a ) ≥ Φ ( a *) Φ ( a ) and Φ ( a * a ) ≥ Φ ( a ) Φ ( a *).
If a Borel function happens to be a section of some map, it is called a Borel section.
If K is a field, then for every vector space V over K we have a " natural " injective linear map from the vector space into its double dual.
If and labeling
His government has also been condemned for allegedly arming and financing the insurgency in Somalia ; the United States is considering labeling Eritrea a " State Sponsor of Terrorism ," however, many experts on the topic have shied from this assertion, stating that " If there is one country where the fighting of extremists and terrorists was a priority when it mattered, it was Eritrea.
If the labels are all congruent rectangles, the corresponding 2-SAT instance can be shown to have only linearly many constraints, leading to near-linear time algorithms for finding a labeling.
If a thiol group ,-SH, were swapped in for it, the labeling would, by its definition, not be affected by the substitution.
If and problem
If there were no West Berlin problem, imperialist quarters would have invented an excuse for stepping up the armaments race to try to solve the internal and external problems besetting the United States and its NATO partners.
If they are to be commended for foresight in their planning, what then is the judgment of a town council that compounds this problem during the planning stage??
If action is indicated, what kind of action is relevant to the problem??
If, as I suspect, the problem is largely of the second sort, then development of a theory better able to handle tone will result automatically in better theory for all phonologic subsystems.
If the problem is enlarged to require a complete coverage of feed states, Af operations are needed by the dynamic program and Af by the direct search.
If high accuracy is required in preflight leveling, it is usually necessary to integrate or doubly integrate the accelerometer outputs ( this also minimizes the noise problem ).
If he foresaw any problem because of the quality of the hymen, it was recommended that simple procedures be undertaken at once to incise the hymen or, preferably, to dilate it.
`` If you substitute ' atom ' for ' angel ', the problem is not insoluble, given the metallic content of the pin in question ''.
If we allow only quantifiers, it becomes the Co-NP-complete tautology problem.
If we allow both, the problem is called the quantified Boolean formula problem ( QBF ), which can be shown to be PSPACE-complete.
If so, then no NP-complete problem can be in co-NP and no co-NP-complete problem can be in NP.
If a problem can be shown to be in both NP and co-NP, that is generally accepted as strong evidence that the problem is probably not NP-complete ( since otherwise NP = co-NP ).
If a problem is suspected then a more comprehensive test using a leak-down tester can locate the leak.
If I told you my son's age, then there would no longer be two unknowns ( variables ), and the problem becomes a linear equation with just one variable, that can be solved as described above.
If at least 50 cars had been built, sportscars like the GT40 and the Lola T70 were allowed, with a maximum of 5. 0 L. John Wyer's revised 4. 7 litre ( Bored to 4. 9 litres, and o-rings cut and installed between the deck and head to prevent head gasket failure, a common problem found with the 4. 7 engine.
If the open string is in tune, but sharp or flat when frets are pressed, the bridge saddle position can be adjusted with a screwdriver or hex key to remedy the problem.
If the status code indicated a problem, the user agent might display the reason phrase to the user to provide further information about the nature of the problem.
If never, the problem is likely to be physiological ; if sometimes ( however rarely ), it could be physiological or psychological.
If the example with the colored bricks above is viewed as an unbounded knapsack problem, then the solution is to take three yellow boxes and three grey boxes.
(...) If someone asks me how I can sleep at night knowing that my arms have killed millions of people, I respond that I have no problem sleeping, my conscience is clean.
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