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SPACE COMPLEXITY in .NET Writer Code 3/9 in .NET SPACE COMPLEXITY

SPACE COMPLEXITY use visual studio .net barcode 3 of 9 creation toconnect code 3/9 in .net PLANET Given Ri 1 and a guess g, the claim g Ri is veri ed barcode 3 of 9 for .NET as follows. Set c 0.

(initializing the main counter) For u = 1, . . .

, n do begin (the main scan) Guess whether or not u Ri . For a negative guess (i.e.

, u Ri ), do begin (Verify that u Ri via Eq. (5.1).

) Set c 0. (initializing a secondary counter) For w = 1, . .

. , n do begin (the secondary scan) Guess whether or not w Ri 1 . For a positive guess (i.

e., w Ri 1 ), do begin Verify that w Ri 1 (as in Step 1). Verify that u = w and (w, u) E.

If some veri cation failed then halt with output otherwise increment c . End (of handling a positive guess for w Ri 1 ). End (of secondary scan).

(c vertices in Ri 1 were checked) If c < . Ri 1 then halt with output . Otherwise (c = Ri 1 ), increment barcode 39 for .NET c. (u veri ed to be outside of Ri ) End (of handling a negative guess for u Ri ).

End (of main scan). (c vertices were shown outside of Ri ) If c < n g then halt with output . Otherwise g .

Ri is veri ed (since n Ri c n g).. Figure 5.3: The main step in proving N L = coN L. that u Ri .net vs 2010 ANSI/AIM Code 39 amounts to proving that for every w Ri 1 , it holds that u = w and (w, u) E. As hinted, the knowledge of .

Ri 1 allows for barcode 3/9 for .NET the enumeration of Ri 1 , and thus we merely check the aforementioned condition on each vertex in Ri 1 . Thus, verifying that u Ri is done as follows.

(a) We scan V guessing . Ri 1 vertices th barcode 3/9 for .NET at are in Ri 1 , and verify each such guess in the straightforward manner (i.e.

, as in Step 1).24 (b) For each w Ri 1 that was guessed and veri ed in Step 2a, we verify that both u = w and (w, u) E. By Eq.

(5.1), if u passes the foregoing veri cation then indeed u Ri . We use log2 n bits to store the number of vertices that were already veri ed to be in V \ Ri , another log2 n bits to store the current vertex u, another log2 n bits to count the number of vertices that are currently veri ed to be in Ri 1 , another log2 n bits to store such a vertex w, and another O(log n) bits for verifying that w Ri 1 (as in Step 1).

If any of the foregoing veri cations fails, then the procedure halts outputting the don t know symbol . Otherwise, it outputs g. Clearly, the foregoing non-deterministic procedure uses a logarithmic amount of space.

It can be veri ed that, when given the correct value of . Ri 1 , this procedure non-deterministically computes the value of Ri . That is, if all veri cations are 24 Note that implicit in Step 2a is a non-deterministic procedure that computes the mapping (G, v, i, . Ri 1 ) Ri 1 , w here Ri 1 denotes the set of vertices that are reachable in G by a path of length at most i from v.. 5.3. NON-DETERMINISTIC SPACE COMPLEXITY satis ed then it must hold that g = Ri , and if g = Ri then there exist adequate non-deterministic choices that satisfy all veri cations. Recall that Rn is computed iteratively, starting with . R0 . = 1 and computing Ri based on Ri 1 . Each itera tion i = 1, . .

. , n is non-deterministic, and is either completed with the correct value of . Ri (at which point Ri 1 is discarde d) or halts in failure (in which case we halt the entire process and output ). This yields a non-deterministic log-space machine for computing . Rn , and the th eorem follows. Digest. Step 2 is the heart of the proof (of Theorem 5.

14). In this step a non-deterministic procedure is used to verify non-membership in an NL-type set. Indeed, verifying membership in NL-type sets is the archetypical task of non-deterministic procedures (i.

e., they are de ned so as to t these tasks), and thus Step 1 is straightforward. In contrast, non-deterministic veri cation of non-membership is not a common phenomenon, and thus Step 2 is not straightforward at all.

Nevertheless, in the current context (of Step 2), the veri cation of non-membership is performed by an iterative (non-deterministic) process that consumes an admissible amount of resources (i.e., a logarithmic amount of space).

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