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ITS pair #487097248
details
property
value
status
complete
benchmark
n-3.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n147.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
7.52903 seconds
cpu usage
7.65296
user time
4.03732
system time
3.61564
max virtual memory
741256.0
max residence set size
10256.0
stage attributes
key
value
starexec-result
MAYBE
output
MAYBE DP problem for innermost termination. P = f6#(x1, x2, x3) -> f5#(x1, x2, x3) f5#(I0, I1, I2) -> f1#(I0, I1, I2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0] The dependency graph for this problem is: 0 -> 1 1 -> 3, 5 2 -> 3, 5 3 -> 2 4 -> 3, 5 5 -> 4 Where: 0) f6#(x1, x2, x3) -> f5#(x1, x2, x3) 1) f5#(I0, I1, I2) -> f1#(I0, I1, I2) 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 3) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] 4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] We have the following SCCs. { 2, 3, 4, 5 } DP problem for innermost termination. P = f4#(I3, I4, I5) -> f1#(I3, I4, I5) f1#(I6, I7, I8) -> f4#(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0] We use the reverse value criterion with the projection function NU: NU[f3#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0 NU[f1#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0 NU[f4#(z1,z2,z3)] = -1 - z2 + z3 + -1 * 0 This gives the following inequalities: ==> -1 - I4 + I5 + -1 * 0 >= -1 - I4 + I5 + -1 * 0 0 <= -1 - I7 + I8 ==> -1 - I7 + I8 + -1 * 0 > -1 - (1 + I7) + I8 + -1 * 0 with -1 - I7 + I8 + -1 * 0 >= 0 ==> -1 - I10 + I11 + -1 * 0 >= -1 - I10 + I11 + -1 * 0 0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0 ==> -1 - I13 + I14 + -1 * 0 >= -1 - (1 + I13) + I14 + -1 * 0 We remove all the strictly oriented dependency pairs. DP problem for innermost termination. P = f4#(I3, I4, I5) -> f1#(I3, I4, I5) f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8] f3(I9, I10, I11) -> f1(I9, I10, I11) f1(I12, I13, I14) -> f3(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] f1(I15, I16, I17) -> f2(rnd1, I16, I17) [rnd1 = rnd1 /\ I16 - I17 <= 0 /\ -1 * I16 + I17 <= 0] The dependency graph for this problem is: 2 -> 5 4 -> 5 5 -> 4 Where: 2) f4#(I3, I4, I5) -> f1#(I3, I4, I5) 4) f3#(I9, I10, I11) -> f1#(I9, I10, I11) 5) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] We have the following SCCs. { 4, 5 } DP problem for innermost termination. P = f3#(I9, I10, I11) -> f1#(I9, I10, I11) f1#(I12, I13, I14) -> f3#(I12, 1 + I13, I14) [0 <= -1 + I13 - I14 /\ -1 * I13 + I14 <= 0] R = f6(x1, x2, x3) -> f5(x1, x2, x3) f5(I0, I1, I2) -> f1(I0, I1, I2) f4(I3, I4, I5) -> f1(I3, I4, I5) f1(I6, I7, I8) -> f4(I6, 1 + I7, I8) [0 <= -1 - I7 + I8]
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