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ITS pair #487097743
details
property
value
status
complete
benchmark
p-40.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n137.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
8.26171 seconds
cpu usage
7.96972
user time
4.20966
system time
3.76006
max virtual memory
685496.0
max residence set size
8788.0
stage attributes
key
value
starexec-result
YES
output
YES DP problem for innermost termination. P = f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) f2#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] R = f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8] f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29] f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] The dependency graph for this problem is: 0 -> 4 1 -> 3 2 -> 3 3 -> 2 4 -> 1 Where: 0) f6#(x1, x2, x3, x4, x5, x6, x7) -> f1#(x1, x2, x3, x4, x5, x6, x7) 1) f2#(I0, I1, I2, I3, I4, I5, I6) -> f3#(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] 2) f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) 3) f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] 4) f1#(I35, I36, I37, I38, I39, I40, I41) -> f2#(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] We have the following SCCs. { 2, 3 } DP problem for innermost termination. P = f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) f3#(I21, I22, I23, I24, I25, I26, I27) -> f5#(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] R = f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8] f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29] f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3,z4,z5,z6,z7)] = z1 + -1 * (1 + z2) NU[f5#(z1,z2,z3,z4,z5,z6,z7)] = z1 + -1 * (1 + z2) This gives the following inequalities: ==> I14 + -1 * (1 + I15) >= I14 + -1 * (1 + I15) 1 + I22 <= I21 ==> I21 + -1 * (1 + I22) > I21 + -1 * (1 + (1 + I22)) with I21 + -1 * (1 + I22) >= 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) R = f6(x1, x2, x3, x4, x5, x6, x7) -> f1(x1, x2, x3, x4, x5, x6, x7) f2(I0, I1, I2, I3, I4, I5, I6) -> f3(I0, 1 + I1, I2, I3, I4, I5, I6) [1 + I1 <= I0] f2(I7, I8, I9, I10, I11, I12, I13) -> f4(I7, I8, I9, I13, I11, I12, I13) [I7 <= I8] f5(I14, I15, I16, I17, I18, I19, I20) -> f3(I14, I15, I16, I17, I18, I19, I20) f3(I21, I22, I23, I24, I25, I26, I27) -> f5(I21, 1 + I22, I23, I24, I25, I26, I27) [1 + I22 <= I21] f3(I28, I29, I30, I31, I32, I33, I34) -> f4(I28, I29, I30, I34, I32, I33, I34) [I28 <= I29] f1(I35, I36, I37, I38, I39, I40, I41) -> f2(I35, rnd2, I39, I38, I39, I40, I41) [y1 = I40 /\ rnd2 = I39] The dependency graph for this problem is: 2 -> Where: 2) f5#(I14, I15, I16, I17, I18, I19, I20) -> f3#(I14, I15, I16, I17, I18, I19, I20) We have the following SCCs.
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