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Enhancing Tree Compression in Declarative Debugging

Enhancing Tree Compression in Declarative Debugging. David Insa Josep Silva C ésar Tomás. I did X I wanted Y Instead I got Z. Algorithmic Debugging. TWO PHASES: Generate the execution tree Traverse the execution tree asking questions until the bug is found.

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Enhancing Tree Compression in Declarative Debugging

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  1. EnhancingTreeCompression in DeclarativeDebugging David Insa Josep Silva César Tomás

  2. I didX I wantedY Instead I gotZ

  3. AlgorithmicDebugging • TWO PHASES: • Generatetheexecutiontree • Traversetheexecutiontreeaskingquestionsuntilthe bug isfound If a symptom of an error isdetected then AD willfindthe bug main = 4 WhatisanExecutionTree? listSum [1,2] = 4 Example: main = listSum [1,2] listSum [] = 1 listSum (x:xs) = x + (listSumxs) 1+3 = 4 listSum [2] = 3 2+1 = 3 listSum [] = 1

  4. AlgorithmicDebugging • Traversingtheexecutiontree • GOLDEN RULE:When a wrongnode has notanywrongchildrenthen • thisnodeis a buggynode. main = 4 listSum [1,2] = 4 Example: main = listSum [1,2] listSum [] = 1 listSum (x:xs) = x + (listSumxs) 1+3 = 4 listSum [2] = 3 2+1 = 3 listSum [] = 1

  5. AlgorithmicDebugging • Traversingtheexecutiontree • GOLDEN RULE:When a wrongnode has notanywrongchildrenthen • thisnodeis a buggynode. main = 5 listSum [1,2] = 5 Example: main = listSum [1,2] listSum [] = 0 listSum (x:xs) = x + (listSumxs) + 1 1+3+1 = 5 listSum [2] = 3 2+0+1 = 3 listSum [] = 0

  6. Debuggingsession Debuggingsession main = sqrTest[1,2] sqrTestx = test (squares (listSumx)) test (x,y,z) = (x==y) && (y==z) listSum[] = 0 listSum(x:xs) = x + (listSumxs) squares x = ((square1 x),(square2 x),(square3 x)) square1 x = square x square x = x*x square2 x = listSum(list x x) list x y | y==0 = [] | otherwise = x:list x (y-1) square3 x = listSum(partialSumsx) partialSumsx = [(sum1 x),(sum2 x)] sum1 x = div (x * (incr x)) 2 sum2 x = div (x + (decr x)) 2 incr x = x + 1 decr x = x - 1

  7. Debuggingsession DebuggingsessionusingDivide & Query(byHirunkitti). main = False Startingthedebuggingsession… square2 3 = 9? YES square3 3 = 8? NO partialSums 3 = [6,2]? NO sum1 3 = 6? YES sum2 3 = 2? NO decr 3 = 2? YES Bug found in rule: sum2 x = div (x + (decr x)) 2 sqrTest [1,2] = False test (9,9,8) = False squares 3 = (9,9,8) listSum [1,2] = 3 listSum [2] = 2 squares1 3 = 9 squares2 3 = 9 squares3 3 = 8 listSum [] = 0 square 3 = 9 listSum [3,3,3] = 9 list 3 3 = [3,3,3] partialSums 3 = [6,2] listSum [6,2] = 8 listSum [3,3] = 6 list 3 2 = [3,3] listSum [2] = 2 sum1 3 = 6 sum2 3 = 2 decr 3 = 2 listSum [3] = 3 list 3 1 = [3] listSum [] = 0 incr 3 = 4 listSum [] = 0 list 3 0 = []

  8. Debuggingsession Debuggingsession main = sqrTest[1,2] sqrTestx = test (squares (listSumx)) test (x,y,z) = (x==y) && (y==z) listSum[] = 0 listSum(x:xs) = x + (listSumxs) squares x = ((square1 x),(square2 x),(square3 x)) square1 x = square x square x = x*x square2 x = listSum(list x x) list x y | y==0 = [] | otherwise = x:list x (y-1) square3 x = listSum(partialSumsx) partialSumsx = [(sum1 x),(sum2 x)] sum1 x = div (x * (incr x)) 2 sum2 x = div (x + (decr x)) 2 incr x = x + 1 decr x = x - 1

  9. DeclarativeDebuggingStrategies Single Stepping Single Stepping Divide & Query Top Down Top Down - Left to Right Top Down - Heaviest First Top Down - More Rules First Top Down - Less Yes First Divide & Query (by Shapiro) Divide & Query (by Hirunkitti) Divide by Rules & Query Divide by Yes & Query Hat Delta Hat Delta - More Wrongs Hat Delta - Less Rights Hat Delta - Best Division

  10. AlgorithmicDebugging Isitpossibletooptimizean ET beforestartingthequestiongeneration? Yes, withTreeCompression

  11. 1 1 1 1 1 1 1 1 2 1 2 2 1 1 1 1 2 AlgorithmicDebugging • TREE COMPRESSION • Howmanyquestions do youneedtofind a bug in theseETs? • Whatstrategy are youusing? 1

  12. AlgorithmicDebugging • TREE COMPRESSION • ET transformation that can be done before exploring the ET. • It compresses all chains of nodes that represent recursive calls. • Given two nodes A,B where A recursivelly calls B, • Tree compression deletes node B and makes all children of B to be new children of A.

  13. TreeCompression 1 1 2 6 1 2 4 6 6 2 2 3 3 4 5 3 4

  14. TreeCompression 32 / 9 = 3,55 1 2 1 1 4 5 2 2 2 2 2 2 3 4 5 2 3 4 26 / 8 = 3,25 27 / 7 = 3,86 1 4 1 6 1 2 2 2 4 2 2 2 2 3 4 2 2 2 4 5 6 1 2 3 2 2 2 2 3 4

  15. TreeCompression A metrictomeasuretheimpact of ET transformations: 13/4 17/5 1 1 < 2 2 3 4 3 4 4 AVG = Total Questions (Q) / Number of Nodes (W) Q = 4 + 2 + 3 + 4 = 13 W = 4 AVG = Q/W = 13/4

  16. TreeCompression Generalizationforany ET: +1 1 +3 +2 2 3 4 … … … + 4 3 2 Q = (AVG2 + ) * W2+ (AVG3 + ) * W3 + (AVG4 + ) * W4 1 (AVG2 * W2 + W2) + (AVG3 * W3 + 2W3) + (AVG4 * W4 + 3W4) + 4 AVG1 = W1 AVG1 * W1 = (AVG2 * W2 + W2) + (AVG3 * W3 + 2W3) + (AVG4 * W4 + 3W4) + 4 Q1 = (Q2 + W2) + (Q3 + 2W3) + (Q4 + 3W4) + 4

  17. TreeCompression Generalizationforany ET: 0 6 1 5 7 2 3 4 Q1 = (Q2 + W2) + (Q3 + 2W3) + (Q4 + 3W4) + 4 Q0 = (Q1 + W1) + (Q5 + 2W5) + (Q6 + 3W6) + (Q7+ 4W7) + 5 Q0 = (Q2 + Q3 + Q4) + Q5+ Q6 + Q7 + W1 + (W2 + 2W3 + 3W4 + 4WI1) + 2W5 + 3W6 + 4W7 + 5

  18. TreeCompression Generalizationforany ET: 0 2 6 5 3 4 7 Q0 = (Q2 + Q3 + Q4) + Q5+ Q6 + Q7 + W1 + (W2 + 2W3 + 3W4 + 4WI1) + 2W5 + 3W6 + 4W7 + 5 Q’0 = (Q2 + W2) + (Q3 + 2W3) + (Q4 + 3W4) + (Q5+ 4W5) + (Q6 + 5W6) + (Q7 + 6W7) + 5 Q’0 = Q2 + Q3 + Q4 + Q5+ Q6 + Q7 + W2 + 2W3 + 3W4 + 4W5 + 5W6 + 6W7 + 7

  19. TreeCompression Generalizationforany ET: 0 0 6 1 5 7 2 6 5 3 4 7 2 3 4 (Q2 + Q3 + Q4) + Q5 + Q6 + Q7 + Q0 = W1 + (W2 + 2W3 + 3W4 + 4WI1) + 2W5 + 3W6 + 4W7 + 5 Q2 + Q3 + Q4 + Q5 + Q6 + Q7 + Q’0 = Cost(nodes) = Σ{Pos(node, nodes) * Wnodo | node∈nodes} + |nodes| + 1 W2 + 2W3 + 3W4 + 4W5 + 5W6 + 6W7 + 7WI0 Cost(children0) + Cost(children1) Q0 = Cost(children0) + Cost(children1) AVG0 = - W0 > Cost(children’0) Q’0 = Cost(children’0) AVG’0 = W’0

  20. TreeCompression recs = {n | n,n’ ∈ N ∧ (n → n’) ∈ E } ∧ l(n) = l(n’)} while (recs≠ ∅) taken ∈ recssuchthat ∃/ n’ ∈ recswith (n → n’) ∈ E+ recs = recs \ {n} parent = n do maxImpr = 0 children = {c | (n → c) ∈ E ∧ l(n) = l(c)} foreachchild∈ children pchildren = Sort(Children(parent)) cchildren = Sort(Children(child)) comb = Sort((pchildren ∪ cchildren) \ {child}) impr = if (impr > maxImpr) maxImpr = impr bestNode = child endforeach if (maxImpr ≠ 0) T’ = Compress(T’, parent, bestNode) while (maxImpr ≠ 0) endwhile 1 1 2 6 1 2 4 6 6 2 2 3 3 4 5 Cost(comb) Cost(pchildren) + Cost(cchildren) - Wparent - 1 Wparent 3 4 Coste(nodos) = Σ{Pos(nodo, nodos) * Wnodo | nodo ∈ nodos} + |nodos| + 1

  21. Experiments Questionsreductionwiththe new algorithm9.72%

  22. Future Work Reduce the number of questions EXPANDING the structure of the ET Applicable to iterative loops: loop expansion Reduce the granularity level of bugs (from functions to loops) Combine treecompression and loopexpansion L 1 2 3 2 3 2 3 L L

  23. Thankyou

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