simplification results in 4 + 3 = 7.
Hence the result is 7.
73. | |
A. | + |
B. | // |
C. | % |
D. | ** |
Answer» D. ** | |
Explanation: all of the operators shown above have associativity from left to right, except exponentiation operator (**) which has its associativity from right to left. |
74. | |
A. | 43 |
B. | 44 |
C. | 22 |
D. | 23 |
Answer» B. 44 | |
Explanation: The expression shown above is an example of explicit conversion. It is evaluated as int(43.55+1) = int(44.55) = 44. Hence the result of this expression is 44. |
75. | |
A. | (6.0, 16.0) |
B. | (6.00, 16.00) |
C. | (6, 16) |
D. | (6.00, 16.0) |
Answer» A. (6.0, 16.0) | |
Explanation: the result of the expression shown above is (6.0, 16.0). this is because the result is automatically rounded off to one decimal place. |
76. | |
A. | / |
B. | % |
C. | // |
Answer» C. // | |
Explanation: // is the operator for truncation division. it is called so because it returns only the integer part of the quotient, truncating the decimal part. for example: 20//3 = 6. |
77. | |
A. | 64, 512, 64 |
B. | 64, 64, 64 |
C. | 512, 512, 512 |
D. | 512, 64, 512 |
Answer» D. 512, 64, 512 | |
Explanation: Expression 1 is evaluated as: 2**9, which is equal to 512. Expression 2 is evaluated as 8**2, which is equal to 64. The last expression is evaluated as 2**(3**2). This is because the associativity of ** operator is from right to left. Hence the result of the third expression is 512. |
78. | |
A. | (1.0, 4.0) |
B. | (1.0, 1.0) |
C. | (4.0. 1.0) |
D. | (4.0, 4.0) |
Answer» A. (1.0, 4.0) | |
Explanation: the above expressions are evaluated as: 2/2, 8/2, which is equal to (1.0, 4.0). |
79. | |
A. | 30.0 |
B. | 30.8 |
C. | 28.4 |
D. | 27.2 |
Answer» D. 27.2 | |
Explanation: The expression shown above is evaluated as: 2+9*(36-8)/10, which simplifies to give 2+9*(2.8), which is equal to 2+25.2 =27.2. Hence the result of this expression is 27.2. |
80. | |
A. | (1,3) |
B. | (0,3) |
C. | (1,0) |
D. | (3,1) |
Answer» A. (1,3) | |
Explanation: the expressions are evaluated as: 4%3 and 6//2 respectively. this results in the answer (1,3). this is because the associativity of both of the expressions shown above is left to right. |
81. | |
A. | 5.0 |
B. | 5 |
C. | 4.0 |
D. | 4 |
Answer» C. 4.0 | |
Explanation: the above expression is an example of explicit conversion. it is evaluated as: float(4+int(2.39)%2) = float(4+2%2) = float(4+0) = 4.0. hence the result of this expression is 4.0. |
82. | |
A. | 3 |
B. | 7 |
C. | 77 |
D. | 0 |
Answer» B. 7 | |
Explanation: The order of precedence is: **, //, +. The expression 4+2**5//10 is evaluated as 4+32//10, which is equal to 4+3 = 7. Hence the result of the expression shown above is 7. |
83. | |
A. | true |
B. | false |
Answer» B. false | |
Explanation: the value of the expression (2**2)**3 = 4**3 = 64. when the expression 2**2**3 is evaluated in python, we get the result as 256, because this expression is evaluated as 2**(2**3). this is because the associativity of exponentiation operator (**) is from right to left and not from left to right. |
84. | |
A. | 1011 |
B. | 11 |
C. | 13 |
D. | 1101 |
Answer» A. 1011 | |
Explanation: the result of the expression shown will be 1011. this is because we have not specified the base in this expression. |
85. | |
A. | int(1011) |
B. | int(‘1011’,23) |
C. | int(1011,2) |
D. | int(‘1011’) |
Answer» C. int(1011,2) | |
Explanation: the expression int(1011,2) results in an error. had we written this expression as int(‘1011’,2), then there would not be an error. |
86. | |
A. | & |
B. | ^ |
D. | ! |
Answer» B. ^ | |
Explanation: the ^ operator represent bitwise xor operation. &: bitwise and, |
87. | |
A. | ‘0bx1000’ |
B. | 8 |
C. | 1000 |
D. | ‘0b1000’ |
Answer» D. ‘0b1000’ | |
Explanation: the prefix 0x specifies that the value is hexadecimal in nature. when we convert this hexadecimal value to binary form, we get the result as: ‘0b1000’. |
88. | |
A. | 115 |
B. | 116 |
C. | 117 |
D. | 118 |
Answer» C. 117 | |
Explanation: the binary value of 0x35 is 110101 and that of 0x75 is 1110101. on or- ing these two values we get the output as: 1110101, which is equal to 117. hence the result of the above expression is 117. |
89. | |
A. | true |
B. | false |
Answer» B. false | |
Explanation: in most cases the value of two’s |
90. | |
A. | or |
B. | and |
C. | xor |
D. | not |
Answer» C. xor | |
Explanation: bitwise xor gives 1 if either of the bits is 1 and 0 when both of the bits are 1. |
91. | |
A. | 2 |
B. | 4 |
C. | 8 |
D. | 12 |
Answer» C. 8 | |
Explanation: ^ is the xor operator. the binary form of 4 is 0100 and that of 12 is 1100. therefore, 0100^1100 is 1000, which is equal to 8. |
92. | |
A. | a<<2 |
B. | a<<4 |
C. | a>>2 |
D. | a>>4 |
Answer» A. a<<2 | |
Explanation: let us consider an example wherein a=2. the binary form of 2 is 0010. when we left shift this value by 2, we get 1000, the value of which is 8. hence if we want to multiply a given number ‘a’ by 4, we can use the expression: a<<2. |
93. | |
A. | 1011011 |
B. | 11010100 |
C. | 11101011 |
D. | 10110011 |
Answer» B. 11010100 | |
Explanation: the binary form of -44 is 00101100. the one’s complement of this value is 11010011. on adding one to this we get: 11010100 (two’s complement). |
94. | |
A. | true |
B. | false |
C. | error |
D. | no output |
Answer» B. false | |
Explanation: the expression not(10<20) returns false. the expression not(10>30) returns true. the and operation between false and true returns false. hence the output is false. |
95. | |
A. | 0 |
B. | no output |
C. | error |
D. | none of the mentioned |
Answer» B. no output | |
Explanation: range(0) is empty. |
96. | |
A. | 0.0 1.0 |
B. | 0 1 |
C. | error |
D. | none of the mentioned |
Answer» C. error | |
Explanation: object of type float cannot be interpreted as an integer. |
97. | |
A. | 0.0 1.0 |
B. | 0 1 |
C. | error |
D. | none of the mentioned |
Answer» B. 0 1 | |
Explanation: range(int(2.0)) is the same as range(2). |
98. | |
A. | 0 1 2 3 |
B. | 0 1 2 2 |
C. | 3 3 3 3 |
D. | error |
Answer» A. 0 1 2 3 | |
Explanation: the value of a[0] changes in each iteration. since the first value that it takes is itself, there is no visible error in the current example. |
99. | |
A. | a |
B. | bc |
C. | bca |
D. | abc |
Answer» D. abc | |
Explanation: + operator is concatenation operator. |
100. | |
A. | a |
B. | ab |
C. | cd |
D. | dc |
Answer» C. cd | |
Explanation: slice operation is performed on string. |
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Title: solving an industrially relevant quantum chemistry problem on quantum hardware.
Abstract: Quantum chemical calculations are among the most promising applications for quantum computing. Implementations of dedicated quantum algorithms on available quantum hardware were so far, however, mostly limited to comparatively simple systems without strong correlations. As such, they can also be addressed by classically efficient single-reference methods. In this work, we calculate the lowest energy eigenvalue of active space Hamiltonians of industrially relevant and strongly correlated metal chelates on trapped ion quantum hardware, and integrate the results into a typical industrial quantum chemical workflow to arrive at chemically meaningful properties. We are able to achieve chemical accuracy by training a variational quantum algorithm on quantum hardware, followed by a classical diagonalization in the subspace of states measured as outputs of the quantum circuit. This approach is particularly measurement-efficient, requiring 600 single-shot measurements per cost function evaluation on a ten qubit system, and allows for efficient post-processing to handle erroneous runs.
Comments: | 12 pages, 5 figures |
Subjects: | Quantum Physics (quant-ph) |
Cite as: | [quant-ph] |
(or [quant-ph] for this version) | |
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Algorithm and flowcharts mcqs set-3.
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This Algorithm and Flowcharts MCQs contains a carefully curated selection of objective questions, as well as multiple choice questions with answers, sourced from reputable reference books, university exams, and question papers. These resources are invaluable for individuals preparing for university exams,competitive exams and interviews
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2. the chart that contains only function flow and no code is called as.
8. the sequence logic will not be used while.
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