By Gregory J. Chaitin

During this mathematical autobiography, Gregory Chaitin offers a technical survey of his paintings and a non-technical dialogue of its importance. the quantity is a significant other to the sooner number of Chaitin's papers "Information, Randomness and Incompleteness" additionally released by means of international medical. The technical survey comprises many new effects, together with a close dialogue of LISP software dimension and new types of Chaitin's so much primary information-theoretic incompleteness theorems. The nontechnical half contains the lecture given by way of Chaitin in Goedel's lecture room on the college of Vienna, a transcript of a BBC television interview, and articles from the "New Scientist", "La Recherche", and the "Mathematical Intelligencer".

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**Extra info for Information Theoretic Incompleteness (World Scientific Series in Computer Science 35)**

**Example text**

C examines each T until it finds one of the form “Hb (s ) > j” that asserts that a specific bit string s has blank-endmarker complexity greater than a specific natural number j that is greater than or equal to |A| + 2k + 1. C then outputs s and halts. This shows that k 0’s |A| + 2k + 1 < Hb (s ) ≤ |A1 000 · · · 000 | + simC . , |A| + 2k + 1 < Hb (s ) ≤ |A| + 1 + k + simC . This implies k < simC . Taking k = simC , we have a contradiction. It follows that s cannot exist for this value of k. The theorem is therefore proved with c = 2k + 1 = 2simC + 1.

From the preceding discussion, we know that if k is fixed and we let n go to infinity, in the limit each of the 2k possible successive substrings will occur with equal relative frequency 2−k . 1), we see that n βn ≤ H(S) ≤ H(s)/2k + o(n). k |s|=k Multiplying through by k, dividing through by n, and letting n go to infinity, we see that βk ≤ H(s)/2k . |s|=k This piece of reasoning has a pleasant probabilistic flavor. We can go a bit farther. The maximum max|s|=n H(s) cannot be less than the average |s|=n H(s)/2n , which in turn cannot be less than βn.

Here we order all S-expressions and bit strings, first by size, and then within those of the same size, lexicographically. Using this notation, by the time we get to the 2k+1 th S-expression, we will have covered all ≤ k-bit strings, because there are not that many of them. Thus max|s|=k H(s) ≤ n + c if the number of ≤ n-character S-expressions is ≥ 2k+1 . Here c is the number of characters that must be added to the jth S-expression to obtain an expression whose value is the jth bit string. 52 Part I—Survey So the key to further progress is to study how smoothly the number of S-expressions of size n varies as a function of n; see Appendix B [1] for an example of such an analysis.