The following pseudocode constructs a graph from the “Words” dataset. A dictionary wordI is created. Each word in the dictionary, wordI, is considered as a node. The matrix W represents the graph. \mathbf{W[i][j]} = 1 if there is an edge from node i to node j. i = 0 wordI = { }, wordF = { } while (Table 1 has more rows) { Read the first row X in Table 1 if (isKey(wordF, X.Word)) { wordF[X.Word] = wordF[X.Word] + 1 } else { wordF[X.Word] = 1 wordI[X.Word] = i i = i + 1 } Move row X to Table 2 } n = length(keys(wordI)) W = createMatrix(n, n) foreach r in keys(wordI) { foreach c in keys(wordI) { if (r ≠ c and wordF[r] == wordF[c] and isDifferent(r, c)) { W[wordI[r]] [wordI[c]] = 1 } } } Procedure isDifferent(P, Q) if (P.LetterCount == Q.LetterCount) { return (False) } else { return (True) } End isDifferent Study the pseudocode given above and answer the subquestions: There will be an edge between word i and word j if and only if:

(B) The letter counts of word i and j are different but both have the same frequency.

(A) The letter counts of word i and j are the same and both have the same frequency.

(C) The letter counts of word i and j are different and both have different frequency.

(D) The letter counts of word i and j are the same but both have different frequency.

Computational Thinking · End Term · 8 Aug 2021 · May 2021 term

Question 18: The following pseudocode constructs a graph from the “Wo…

Question 18

+2 marksOne correct option

The following pseudocode constructs a graph from the “Words” dataset. A dictionary wordI is created. Each word in the dictionary, wordI, is considered as a node. The matrix W represents the graph. $\mathbf{W[i][j]} = 1$ if there is an edge from node i to node j.

text

1i = 0
2wordI = { }, wordF = { }
3while (Table 1 has more rows) {
4    Read the first row X in Table 1
5    if (isKey(wordF, X.Word)) {
6        wordF[X.Word] = wordF[X.Word] + 1
7    }
8    else {
9        wordF[X.Word] = 1
10        wordI[X.Word] = i
11        i = i + 1
12    }
13    Move row X to Table 2
14}
15
16n = length(keys(wordI))
17W = createMatrix(n, n)
18foreach r in keys(wordI) {
19    foreach c in keys(wordI) {
20        if (r ≠ c and wordF[r] == wordF[c] and isDifferent(r, c)) {
21            W[wordI[r]] [wordI[c]] = 1
22        }
23    }
24}
25
26Procedure isDifferent(P, Q)
27    if (P.LetterCount == Q.LetterCount) {
28        return (False)
29    }
30    else {
31        return (True)
32    }
33End isDifferent

Study the pseudocode given above and answer the subquestions:

There will be an edge between word i and word j if and only if:

A
The letter counts of word i and j are the same and both have the same frequency.
B
The letter counts of word i and j are different but both have the same frequency.
C
The letter counts of word i and j are different and both have different frequency.
D
The letter counts of word i and j are the same but both have different frequency.
E
None of these

Show answer

Correct answer

B
The letter counts of word i and j are different but both have the same frequency.

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Question 18 of 25 in the IIT Madras BS Computational Thinking (Computational Thinking (CT)) End Term paper sat on 8 Aug 2021, in the May 2021 term (IIT M TERM FINAL EXAM POD21TEFNQP1 8 Aug 2021). It carries 2 marks.

Question 18: The following pseudocode constructs a graph from the “Wo…

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