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pruefungen:bachelor:aud:loesungss14 [22.03.2018 10:08] – Evren | pruefungen:bachelor:aud:loesungss14 [31.07.2019 09:30] – Anpassungen Aufgabe 4 dom | ||
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Zeile 5: | Zeile 5: | ||
**b)** 1te, 2te | **b)** 1te, 2te | ||
- | **c)**2te (O(n log n)) | + | **c)** 2te (O(n log n)) |
**d)** 2te, 3te Antwort | **d)** 2te, 3te Antwort | ||
Zeile 42: | Zeile 42: | ||
**b)** \\ | **b)** \\ | ||
- | **(i)** A schon vorhanden, Balancfaktoren stimmen auch => gar nichts machen ?? | + | (i) ^ |
- | \\ | + | {{: |
- | **(ii)** Nichts rotieren, da Balancfaktoren stimmen: | + | |
- | < | + | (ii) ^ |
- | | + | {{: |
- | / \ | + | |
- | A(-1) Y(-1) | + | (iii) ^ |
- | | + | {{: |
- | B(0) Z(0) | + | |
- | </ | + | |
- | \\ | + | |
- | **(iii)** | + | |
- | C einfügen und anschließend Knoten (V) nach rechts rotieren | + | |
- | < | + | |
- | | + | |
- | / | + | |
- | W(2!) Y(-1) V(0) Y(-1) | + | |
- | / | + | |
- | | + | |
- | / | + | |
- | | + | |
- | </ | + | |
\\ | \\ | ||
**c)** \\ | **c)** \\ | ||
**(i)** pre-order: E -> N -> J -> H -> O -> O -> D \\ | **(i)** pre-order: E -> N -> J -> H -> O -> O -> D \\ | ||
- | **(ii)** in-order: J -> N -> H -> O -> E -> D -> O \\ | + | **(ii)** in-order: J -> N -> O -> H -> E -> D -> O \\ |
**(iii)** post-order: J -> O -> H -> N -> D -> O -> E \\ | **(iii)** post-order: J -> O -> H -> N -> D -> O -> E \\ | ||
\\ | \\ | ||
Zeile 93: | Zeile 79: | ||
</ | </ | ||
+ | ====3) Graphalgorithmen==== | ||
- | ** | + | **a)** \\ |
- | Falsche Aufgabe. Bei SS14 / 2 muss man ausschließlich Bäume zeichnen. | + | {A, B, C, D, E} \\ |
- | ** | + | {A, B, D, E} \\ |
+ | {A, B, C, E} \\ | ||
+ | {A, B, C, D} \\ | ||
+ | {F, G, H} \\ | ||
+ | {A, B, C} \\ | ||
+ | {A, B, E} \\ | ||
- | Loesung uebernommen aus dem Forum, eventuell gegenchecken | + | **b)** |
- | https://fsi.informatik.uni-erlangen.de/forum/thread/12419-Backtracking-SS14 | + | ^ A ^ B ^ C ^ D ^ E ^ |
+ | | ∞ | [0] | ∞ | ∞ | ∞ | | ||
+ | | ∞ | | 5< | ||
+ | | 16< | ||
+ | | 14< | ||
+ | | [13< | ||
+ | Endergebnis: | ||
- | static boolean exists(boolean[][] brd, int r, int c) { | + | **c)** \\ |
- | if(brd[r] == null || brd[r].length < 1) { | + | F --(7)--> B --(5)--> C --(6)--> D --(2)--> A \\ |
- | return false; | + | Weglänge: 7 + 5 + 6 + 2 = 20 |
- | } | + | |
- | if(r >= brd.length || c >= brd[r].length || r < 0 || c < 0) { | + | **d)** \\ |
- | return false; | + | {{: |
- | } | + | |
- | return true; | + | |
- | } | + | |
- | static int[][] allocateSolutionArray(boolean[][] brd, int mrl) { | ||
- | int[][] sol = new int[brd.length][mrl]; | ||
- | for(int i = 0; i < brd.length; i++) { | ||
- | for(int j = 0; j < mrl; j++) { | ||
- | if(exists(brd, | ||
- | sol[i][j] = 0; | ||
- | } else { | ||
- | sol[i][j] = -1; | ||
- | } | ||
- | } | ||
- | } | ||
- | return sol; | ||
- | } | ||
- | static boolean solve(boolean[][] brd, int[][] sol, int tf, int r, int c, int n) { | ||
- | int[][] jumps = { { r + 2, c - 1 }, { r + 1, c - 2 }, { r - 1, c - 2 }, { r - 2, c - 1 }, { r - 2, c + 1 }, | ||
- | { r - 1, c + 2 }, { r + 1, c + 2 }, { r + 2, c + 1 } }; | ||
- | if(sol[r][c] >= 1) { | ||
- | return false; | ||
- | } | ||
- | if(n == tf) { | ||
- | sol[r][c] = n; | ||
- | return true; | ||
- | } | ||
- | sol[r][c] = n; | ||
- | for(int k = 0; k < jumps.length; | ||
- | if(exists(brd, | ||
- | if(solve(brd, | ||
- | return true; | ||
- | } | ||
- | } | ||
- | } | ||
- | sol[r][c] = 0; | ||
- | return false; | ||
- | } | ||
==== 4) Backtracking ==== | ==== 4) Backtracking ==== | ||
- | [[pruefungen: | + | **a)** |
+ | <code java> | ||
+ | static boolean exists(boolean[][] brd, int r, int c) { | ||
+ | if (brd[r] == null || brd[r].length < 1) { // Zeilen null oder leer | ||
+ | return false; | ||
+ | } | ||
+ | if (r >= brd.length || c >= brd[r].length || r < 0 || c < 0) { // Indizes ausserhalb der Grenzen | ||
+ | return false; | ||
+ | } | ||
+ | return true; | ||
+ | } | ||
+ | </code> | ||
+ | |||
+ | **b)** | ||
+ | |||
+ | <code=java> | ||
+ | static int[][] allocateSolutionArray(boolean[][] brd, int mrl) { | ||
+ | int[][] = null; | ||
+ | int[][] sol = new int[brd.length][mrl]; | ||
+ | |||
+ | for (int r = 0; r < brd.length; r++) { | ||
+ | for (int c = 0; c < mrl; c++) { | ||
+ | if (exists(brd, | ||
+ | sol[r][c] = 0; | ||
+ | } else { | ||
+ | sol[c][c] = -1; // Felder mit -1 vorbelegen | ||
+ | } | ||
+ | } | ||
+ | } | ||
+ | return sol; | ||
+ | } | ||
+ | </code> | ||
+ | **c)** | ||
+ | < | ||
+ | static boolean solve(boolean[][] brd, int[][] sol, int tf, int r, int c, int n) { | ||
+ | // possible targets from (r,c) | ||
+ | int[][] jumps = { { r + 2, c - 1 }, { r + 1, c - 2 }, { r - 1, c - 2 }, { r - 2, c - 1 }, | ||
+ | { r - 2, c + 1 }, { r - 1, c + 2 }, { r + 1, c + 2 }, { r + 2, c + 1 } }; | ||
+ | |||
+ | // | ||
+ | if (n == tf) { | ||
+ | sol[r][c] = n; | ||
+ | return true; | ||
+ | } | ||
+ | for (int j = 0; j < jumps.length; | ||
+ | |||
+ | // Wert setzen | ||
+ | sol[r][c] = n | ||
+ | |||
+ | // Brett brd und Loesungsfeld sol pruefen | ||
+ | if (exists(brd, | ||
+ | |||
+ | // Rekursion | ||
+ | if (solve(brd, sol, tf, jumps[j][0], | ||
+ | return true; | ||
+ | } | ||
+ | // Backtrack | ||
+ | sol[r][c] = 0; | ||
+ | } | ||
+ | } | ||
+ | return false; | ||
+ | } | ||
+ | </ | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Alternative: | ||
** a) *** | ** a) *** | ||
Zeile 198: | Zeile 226: | ||
if (exists(brd, | if (exists(brd, | ||
sol[r][c] = n; | sol[r][c] = n; | ||
- | + | if (solve(brd, sol, tf, jump[0], jump[1], n + 1)) { | |
- | boolean res = solve(brd, sol, tf, jump[0], jump[1], n + 1); | + | |
- | if (res) { | + | |
return true; | return true; | ||
} | } | ||
- | |||
sol[r][c] = 0; | sol[r][c] = 0; | ||
} | } | ||
- | |||
} | } | ||
- | |||
return false; | return false; | ||
} | } | ||
</ | </ | ||
+ | [[pruefungen: | ||
==== 5) Haldensortierung ==== | ==== 5) Haldensortierung ==== | ||
Zeile 223: | Zeile 247: | ||
\\ | \\ | ||
c) | c) | ||
+ | <code java> | ||
+ | // | ||
void reheap(W[] w, Comparator< | void reheap(W[] w, Comparator< | ||
int leftId = 2 * i + 1; | int leftId = 2 * i + 1; | ||
Zeile 240: | Zeile 266: | ||
} | } | ||
} | } | ||
+ | </ | ||
\\ | \\ | ||
d) | d) | ||
Zeile 260: | Zeile 287: | ||
| | ||
==== 6) Bäume und Rekursion==== | ==== 6) Bäume und Rekursion==== | ||
- | LinkedList< | + | <code java> |
+ | LinkedList< | ||
assert s != null : new IllegalArgumentException(); | assert s != null : new IllegalArgumentException(); | ||
HashMap< | HashMap< | ||
for(char c : s.toCharArray()) { | for(char c : s.toCharArray()) { | ||
- | if(map.get(c) | + | Node n = map.get(c); |
- | map.put(c, new Node(c, 1)); | + | if(n != null) { |
+ | //variable frequenz updaten | ||
+ | n.f++; | ||
} else { | } else { | ||
- | map.get(c).f += 1; | + | //neuer node in die map |
+ | map.put(c, new Node(c,1)); | ||
} | } | ||
} | } | ||
Zeile 311: | Zeile 342: | ||
} | } | ||
} | } | ||
+ | </ |