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comp330_hw1_sols - Solutions to Assignment 1 Sophia Knight...

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Solutions to Assignment 1 Sophia Knight September 29, 2006 Question 1 Consider the following relation on finite sequences of natural numbers: n 1 n 2 n 3 ...n k < m 1 m 2 m 3 ...m l if and only if i such that n i < m i and j < i n j = m j . The case where k < l and i k n i = m i is regarded as a special case of this definition. We define n 1 n 2 n 3 ...n k m 1 m 2 m 3 ...m l as n 1 n 2 n 3 ...n k < m 1 m 2 m 3 ...m l or n 1 n 2 n 3 ...n k = m 1 m 2 m 3 ...m l . Show that is a partial order. Solution: We must show that is antisymmetric, transitive and reflexive. In order to include the special case of the definition, we consider a blank space to be an element of a sequence which is smaller than any number. Antisymmetric : If a = a 1 a 2 ...a m < b = b 1 b 2 ...b n , then at the first position i where a and b differ, a i < b i . If we also have b < a , then at the first position i where b and a differ, b i < a i . Clearly, the i = i , so both cannot be true. Thus, if a < b , then it is not the case that b < a , so if a b and b a , then a = b . So is antisymmetric. Transitive : Assume that a = a 1 a 2 ...a l < b = b 1 b 2 ...b m < c = c 1 c 2 ...c n . So, i 1 such that a i 1 < b i 1 and for all j < i 1 , a j = b j . And i 2 such that b i 2 < c i 2 and for all j < i 2 , b j = c j . So, let i = min( i 1 , i 2 ). Then for all j < i , a j = b j = c j . Furthermore, if i = i 1 = i 2 , then a i < b i < c i . Otherwise, i = i 1 or i = i 2 . If i = i 1 , then i < i 2 and so a i < b i = c i . If i = i 2 , then a i = b i < c i . In any case, a i < c i , so a < c . Finally, if a = b and b < c or if a < b and b = c , or if a = b = c then it is still the case that a c . So, if a b c , then a c , and the relation is transitive. Reflexive This is obvious from the definition. Show that is a well-founded order. Solution: It turns out that it’s not a well-founded order! Here’s an infinite descending sequence: ((1);(0,1);(0,0,1);(0,0,0,1);...). We thank Artour Tomberg for this counter-example. It is well-founded if we bound the lengths of the sequences. Here is the proof: We must show that for every nonempty subset S of finite sequences of natural numbers, there is a minimal element with respect to our relation, that is an element x with no other element in the set less than x . Let S = { s 1 , s 2 , ... } = { ( a 1 , 1 a 1 , 2 a 1 , 3 ...a 1 ,n 1 ); ( a 2 , 1 a 2 , 2 ...a 2 ,n 2 ); ( a 3 , 1 a 3 , 2 ...a 3 ,n 3 ); ... } Now, let a 1 = min { a 1 , 1 , a 2 , 1 , a 3 , 1 , ... } . We know that a 1 exists because N is a well-ordered set. Let S 1 = { s m S | a m, 1 = a 1 } . If S 1 has an element of length one, then it is clear from the definition of 1
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that this element is the least element of S . If S 1 has no element of length 1, repeat this process starting with n = 2: let a n = min { a m,n | a m S n - 1 } , and let S n = { a m | a m S n - 1 and a m,n = a n } .
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