Explanation Given information: For m = 3, n = 3, and sets A and B represented as the arrays a [ 1 ] = u , a [ 2 ] = v , a [ 3 ] = w , b [ 1 ] = w , b [ 2 ] = u , and b [ 3 ] = v . Calculation: Input: m = 3 n = 3 a [ 1 ] = u a [ 2 ] = v a [ 3 ] = w b [ 1 ] = w b [ 2 ] = u b [ 3 ] = v We also initialize i as 1 and we initialize answer as A ⊆ B . i 1 j found answer A ⊆ B Next, the algorithm will enter the outer while-loop. We first assign the value 1 to j and the value “no” to found. i 1 1 j 1 found no answer A ⊆ B A ⊆ B Since a [ 1 ] = u ≠ w = b [ 1 ] , the value of found is not changed in the first iteration of the inner while-loop. We increase the value of j by 1. Since a [ 1 ] = u = b [ 2 ] , we change the value of found to “yes” i 1 1 1 j 1 2 found no yes answer A ⊆ B A ⊆ B A ⊆ B Since found is set to “yes”, the inner while-loop stops executing and we then increase i by 1. Next, we start the second iteration of the outer while-loop, which re-initializes j as 1 and found as no i 1 1 1 2 j 1 2 1 found no yes no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B Since a [ 2 ] = v ≠ w = b [ 1 ] , the value of found is not changed in the first iteration of the inner while-loop. We increase the value of j by 1. Since a [ 2 ] = v ≠ u = b [ 2 ] , the value of found remains “no” i 1 1 1 2 2 j 1 2 1 2 found no yes no no answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B We increase the value of j by 1. Since a [ 1 ] = u = b [ 3 ] , we change the value of found to “yes” i 1 1 1 2 2 2 j 1 2 1 2 3 found no yes no no yes answer A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B A ⊆ B Since found is set to “yes”, the inner while-loop stops executing and we then increase i by 1

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Chapter 6.1, Problem 36ES

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Trace the action of Algorithm 6.1.1 on the variables i, j, found, and answerfor m = 3, n = 3, and sets Aand Brepresented as the arrays. a[1]=u, a[2]=v, a[3]=w, b[1]=w, b[2]=u,and b[3] =v

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Chapter 6.1, Problem 35ES

Chapter 6.1, Problem 37ES

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٣:٥٣ النموذج الاول . . . O O O بشما ند الحمر الحمر الجمهورية الجنية وزارة التربية والتعليم اليوم التاريخ اللجنة العليا للاختبارات الزمن اختبار مادة الجبر والهندسة لجنة المطابع السرية المركزية للشهادة الثانوية العامة (القسم العلمي) الفترة %97 (1) ظلل في ورقة الإجابة الدائرة التي تحتوي على الحرف ( ص ) للإجابة الصحيحة والحرف ( خ ) للإجابة الخطأ بحسب رقم الفقرة لكل مما يأتي ( درجة لكل فقرة ) )1 ) 2 ) 3 ) 4 ) بؤرة القطع س" = ١٢ ص هي ( ۲ ) طول المحور الأصغر للقطع ٩ س + ص = ٩ يساوي 6 وحدات طول . ) إذا كان & عدد مركب ، 181 + 11 = ٦ ، فإن ١١ = ٣ . ) إذا كان م + ۳ ت = ۲ + ت ب م ، ب دع ، فإن م + ب = 5 ( ) إذا كان & = ۱ + ٣ ت ، فإن ٠ = ١٠ . 6 ( - ) إذا كان ٥٠ - ٣ - ١٢٠ ٤ - ٣ ، فإن قيمة ٧ = ٥ . 1 ) = N ) إذا كان ح هو الحد الخالي من س في المفكوك ( س + v. N 8 ( ( قيمة المقدار , = + ۱ ، * . . + ، فإن قيمة ٧ = ١٦ . ۱ + 9 ( ) المستقيمان المقاربان للقطع الذي معادلته س" = ١ هما ص = : ۹ 10 ( ) إذا كان ٥ + س = ٢٤ ، فإن قيمة س = - 1 س 11 ( ) إذا كانت النسبة بين الحدين الأوسطين تساوي 9 في المفكوك ( س + - ) ،…

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Discrete Mathematics With Applications

Ch. 6.1 - The notation is read”______” and means that___Ch. 6.1 - To use an element argument for proving that a set...Ch. 6.1 - Prob. 3TY Ch. 6.1 - An element x is in AB if , and only if,_______Ch. 6.1 - An element x in AB if, and only if,______Ch. 6.1 - An element x is in B-A if, and only if,______Ch. 6.1 - An elements x is in Acif, and only if.______Ch. 6.1 - The empty set is a set with ______Ch. 6.1 - The power set of a set A is _____Ch. 6.1 - Prob. 10TY

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Trace the action of Algorithm 6.1.1 on the variables i, j, found, and answerfor m = 3, n = 3, and sets Aand Brepresented as the arrays. a [ 1 ] = u , a [ 2 ] = v , a [ 3 ] = w , b [ 1 ] = w , b [ 2 ] = u , and b [ 3 ] = v

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Trace the action of Algorithm 6.1.1 on the variables i, j, found, and answerfor m = 3, n = 3, and sets Aand Brepresented as the arrays. a[1]=u, a[2]=v, a[3]=w, b[1]=w, b[2]=u,and b[3] =v

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