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Integer_Transition_Systems 2019-03-29 01.54 pair #432275790
details
property
value
status
complete
benchmark
java_Factorial.c.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n007.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
2.15496 seconds
cpu usage
2.19505
user time
1.18033
system time
1.01472
max virtual memory
237856.0
max residence set size
8580.0
stage attributes
key
value
starexec-result
YES
output
2.11/2.15 YES 2.11/2.15 2.11/2.15 DP problem for innermost termination. 2.11/2.15 P = 2.11/2.15 f7#(x1, x2, x3) -> f6#(x1, x2, x3) 2.11/2.15 f6#(I0, I1, I2) -> f3#(I0, I1, I2) 2.11/2.15 f6#(I3, I4, I5) -> f5#(I3, I4, I5) 2.11/2.15 f6#(I6, I7, I8) -> f4#(I6, I7, I8) 2.11/2.15 f6#(I9, I10, I11) -> f1#(I9, I10, I11) 2.11/2.15 f3#(I15, I16, I17) -> f5#(I17, I16, I17) 2.11/2.15 f5#(I18, I19, I20) -> f4#(I20, I19, I20) [1 + I20 <= 0] 2.11/2.15 f5#(I21, I22, I23) -> f1#(I23, I22, I23) [0 <= I23] 2.11/2.15 f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 2.11/2.15 R = 2.11/2.15 f7(x1, x2, x3) -> f6(x1, x2, x3) 2.11/2.15 f6(I0, I1, I2) -> f3(I0, I1, I2) 2.11/2.15 f6(I3, I4, I5) -> f5(I3, I4, I5) 2.11/2.15 f6(I6, I7, I8) -> f4(I6, I7, I8) 2.11/2.15 f6(I9, I10, I11) -> f1(I9, I10, I11) 2.11/2.15 f6(I12, I13, I14) -> f2(I12, I13, I14) 2.11/2.15 f3(I15, I16, I17) -> f5(I17, I16, I17) 2.11/2.15 f5(I18, I19, I20) -> f4(I20, I19, I20) [1 + I20 <= 0] 2.11/2.15 f5(I21, I22, I23) -> f1(I23, I22, I23) [0 <= I23] 2.11/2.15 f4(I24, I25, I26) -> f2(I26, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 2.11/2.15 f1(I27, I28, I29) -> f3(I29, I28, -1 + I29) 2.11/2.15 f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33] 2.11/2.15 2.11/2.15 The dependency graph for this problem is: 2.11/2.15 0 -> 1, 2, 3, 4 2.11/2.15 1 -> 5 2.11/2.15 2 -> 6, 7 2.11/2.15 3 -> 2.11/2.15 4 -> 8 2.11/2.15 5 -> 6, 7 2.11/2.15 6 -> 2.11/2.15 7 -> 8 2.11/2.15 8 -> 5 2.11/2.15 Where: 2.11/2.15 0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 2.11/2.15 1) f6#(I0, I1, I2) -> f3#(I0, I1, I2) 2.11/2.15 2) f6#(I3, I4, I5) -> f5#(I3, I4, I5) 2.11/2.15 3) f6#(I6, I7, I8) -> f4#(I6, I7, I8) 2.11/2.15 4) f6#(I9, I10, I11) -> f1#(I9, I10, I11) 2.11/2.15 5) f3#(I15, I16, I17) -> f5#(I17, I16, I17) 2.11/2.15 6) f5#(I18, I19, I20) -> f4#(I20, I19, I20) [1 + I20 <= 0] 2.11/2.15 7) f5#(I21, I22, I23) -> f1#(I23, I22, I23) [0 <= I23] 2.11/2.15 8) f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 2.11/2.15 2.11/2.15 We have the following SCCs. 2.11/2.15 { 5, 7, 8 } 2.11/2.15 2.11/2.15 DP problem for innermost termination. 2.11/2.15 P = 2.11/2.15 f3#(I15, I16, I17) -> f5#(I17, I16, I17) 2.11/2.15 f5#(I21, I22, I23) -> f1#(I23, I22, I23) [0 <= I23] 2.11/2.15 f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 2.11/2.15 R = 2.11/2.15 f7(x1, x2, x3) -> f6(x1, x2, x3) 2.11/2.15 f6(I0, I1, I2) -> f3(I0, I1, I2) 2.11/2.15 f6(I3, I4, I5) -> f5(I3, I4, I5) 2.11/2.15 f6(I6, I7, I8) -> f4(I6, I7, I8) 2.11/2.15 f6(I9, I10, I11) -> f1(I9, I10, I11) 2.11/2.15 f6(I12, I13, I14) -> f2(I12, I13, I14) 2.11/2.15 f3(I15, I16, I17) -> f5(I17, I16, I17) 2.11/2.15 f5(I18, I19, I20) -> f4(I20, I19, I20) [1 + I20 <= 0] 2.11/2.15 f5(I21, I22, I23) -> f1(I23, I22, I23) [0 <= I23] 2.11/2.15 f4(I24, I25, I26) -> f2(I26, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 2.11/2.15 f1(I27, I28, I29) -> f3(I29, I28, -1 + I29) 2.11/2.15 f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33] 2.11/2.15 2.11/2.15 We use the extended value criterion with the projection function NU: 2.11/2.15 NU[f1#(x0,x1,x2)] = x2 2.11/2.15 NU[f5#(x0,x1,x2)] = x2 + 1 2.11/2.15 NU[f3#(x0,x1,x2)] = x2 + 1 2.11/2.15 2.11/2.15 This gives the following inequalities: 2.11/2.15 ==> I17 + 1 >= I17 + 1 2.11/2.15 0 <= I23 ==> I23 + 1 > I23 with I23 + 1 >= 0 2.11/2.15 ==> I29 >= (-1 + I29) + 1 2.11/2.15 2.11/2.15 We remove all the strictly oriented dependency pairs. 2.11/2.15 2.11/2.15 DP problem for innermost termination. 2.11/2.15 P = 2.11/2.15 f3#(I15, I16, I17) -> f5#(I17, I16, I17) 2.11/2.15 f1#(I27, I28, I29) -> f3#(I29, I28, -1 + I29) 2.11/2.15 R = 2.11/2.15 f7(x1, x2, x3) -> f6(x1, x2, x3) 2.11/2.15 f6(I0, I1, I2) -> f3(I0, I1, I2) 2.11/2.15 f6(I3, I4, I5) -> f5(I3, I4, I5) 2.11/2.15 f6(I6, I7, I8) -> f4(I6, I7, I8) 2.11/2.15 f6(I9, I10, I11) -> f1(I9, I10, I11) 2.11/2.15 f6(I12, I13, I14) -> f2(I12, I13, I14) 2.11/2.15 f3(I15, I16, I17) -> f5(I17, I16, I17) 2.11/2.15 f5(I18, I19, I20) -> f4(I20, I19, I20) [1 + I20 <= 0] 2.11/2.15 f5(I21, I22, I23) -> f1(I23, I22, I23) [0 <= I23] 2.11/2.15 f4(I24, I25, I26) -> f2(I26, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 2.11/2.15 f1(I27, I28, I29) -> f3(I29, I28, -1 + I29) 2.11/2.15 f1(I30, I31, I32) -> f2(I32, I33, I34) [I34 = I33 /\ I33 = I33] 2.11/2.15
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