Given the following schedule, show the locks that will occur and the subsequent schedule. Assume that strict 2PL is in effect, w
ith no deadlock prevention and no starvation prevention. Assume that upgrading locks are allowed, downgrades are not. Assume that requests will be satisfied in the order of arrival if possible.
T1:R(X), T2:R(Q), T3:R(Q), T(2):R(X), T3:R(X), T2:W(X), T2:Commit, T3:Abort, T1:W(X), T1:Commit
T1:R(X), T2:R(Q), T3:R(Q), T(2):R(X), T3:R(X), T2:W(X), T2:Commit, T3:Abort, T1:W(X), T1:Commit
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Using the knowledge in computational language in C++ it is possible to write a code that cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits
<h3>Writting the code:</h3><em>#include <bits/stdc++.h></em>
<em>using namespace std;</em>
<em>// chracter to digit mapping, and the inverse</em>
<em>// (if you want better performance: use array instead of unordered_map)</em>
<em>unordered_map<char, int> c2i;</em>
<em>unordered_map<int, char> i2c;</em>
<em>int ans = 0;</em>
<em>// limit: length of result</em>
<em>int limit = 0;</em>
<em>// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit] </em>
<em>bool helper(vector<string>& words, string& result, int digit, int widx, int sum) { </em>
<em> if (digit == limit) {</em>
<em> ans += (sum == 0);</em>
<em> return sum == 0;</em>
<em> }</em>
<em> // if summation at digit position complete, validate it with result[digit].</em>
<em> if (widx == words.size()) {</em>
<em> if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {</em>
<em> if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result</em>
<em> return false;</em>
<em> c2i[result[digit]] = sum % 10;</em>
<em> i2c[sum%10] = result[digit];</em>
<em> bool tmp = helper(words, result, digit+1, 0, sum/10);</em>
<em> c2i.erase(result[digit]);</em>
<em> i2c.erase(sum%10);</em>
<em> ans += tmp;</em>
<em> return tmp;</em>
<em> } else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){</em>
<em> if (digit + 1 == limit && 0 == c2i[result[digit]]) {</em>
<em> return false;</em>
<em> }</em>
<em> return helper(words, result, digit+1, 0, sum/10);</em>
<em> } else {</em>
<em> return false;</em>
<em> }</em>
<em> }</em>
<em> // if word[widx] length less than digit, ignore and go to next word</em>
<em> if (digit >= words[widx].length()) {</em>
<em> return helper(words, result, digit, widx+1, sum);</em>
<em> }</em>
<em> // if word[widx][digit] already mapped to a value</em>
<em> if (c2i.count(words[widx][digit])) {</em>
<em> if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0) </em>
<em> return false;</em>
<em> return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);</em>
<em> }</em>
<em> // if word[widx][digit] not mapped to a value yet</em>
<em> for (int i = 0; i < 10; i++) {</em>
<em> if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;</em>
<em> if (i2c.count(i)) continue;</em>
<em> c2i[words[widx][digit]] = i;</em>
<em> i2c[i] = words[widx][digit];</em>
<em> bool tmp = helper(words, result, digit, widx+1, sum+i);</em>
<em> c2i.erase(words[widx][digit]);</em>
<em> i2c.erase(i);</em>
<em> }</em>
<em> return false;</em>
<em>}</em>
<em>void isSolvable(vector<string>& words, string result) {</em>
<em> limit = result.length();</em>
<em> for (auto &w: words) </em>
<em> if (w.length() > limit) </em>
<em> return;</em>
<em> for (auto&w:words) </em>
<em> reverse(w.begin(), w.end());</em>
<em> reverse(result.begin(), result.end());</em>
<em> int aa = helper(words, result, 0, 0, 0);</em>
<em>}</em>
<em />
<em>int main()</em>
<em>{</em>
<em> ans = 0;</em>
<em> vector<string> words={"GREEN" , "BLUE"} ;</em>
<em> string result = "BLACK";</em>
<em> isSolvable(words, result);</em>
<em> cout << ans << "\n";</em>
<em> return 0;</em>
<em>}</em>
See more about C++ code at brainly.com/question/19705654
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