details

property	value
status	complete
benchmark	p-60.t2_fixed.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n036.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	5.57353 seconds
cpu usage	5.20467
user time	2.60461
system time	2.60006
max virtual memory	730272.0
max residence set size	8748.0

stage attributes

key	value
starexec-result	YES

output

5.07/5.57 YES 5.07/5.57 5.07/5.57 DP problem for innermost termination. 5.07/5.57 P = 5.07/5.57 f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 5.07/5.57 f5#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3] 5.07/5.57 f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, 0, I14) 5.07/5.57 f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5.07/5.57 f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23] 5.07/5.57 R = 5.07/5.57 f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 5.07/5.57 f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3] 5.07/5.57 f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8] 5.07/5.57 f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 5.07/5.57 f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 5.07/5.57 f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23] 5.07/5.57 f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0] 5.07/5.57 5.07/5.57 The dependency graph for this problem is: 5.07/5.57 0 -> 2 5.07/5.57 1 -> 4 5.07/5.57 2 -> 1 5.07/5.57 3 -> 4 5.07/5.57 4 -> 3 5.07/5.57 Where: 5.07/5.57 0) f6#(x1, x2, x3, x4, x5) -> f4#(x1, x2, x3, x4, x5) 5.07/5.57 1) f5#(I0, I1, I2, I3, I4) -> f1#(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3] 5.07/5.57 2) f4#(I10, I11, I12, I13, I14) -> f5#(I10, I11, I12, 0, I14) 5.07/5.57 3) f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5.07/5.57 4) f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23] 5.07/5.57 5.07/5.57 We have the following SCCs. 5.07/5.57 { 3, 4 } 5.07/5.57 5.07/5.57 DP problem for innermost termination. 5.07/5.57 P = 5.07/5.57 f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5.07/5.57 f1#(I20, I21, I22, I23, I24) -> f3#(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23] 5.07/5.57 R = 5.07/5.57 f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 5.07/5.57 f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3] 5.07/5.57 f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8] 5.07/5.57 f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 5.07/5.57 f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 5.07/5.57 f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23] 5.07/5.57 f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0] 5.07/5.57 5.07/5.57 We use the reverse value criterion with the projection function NU: 5.07/5.57 NU[f1#(z1,z2,z3,z4,z5)] = 100 - z4 + -1 * 0 5.07/5.57 NU[f3#(z1,z2,z3,z4,z5)] = 100 - z4 + -1 * 0 5.07/5.57 5.07/5.57 This gives the following inequalities: 5.07/5.57 ==> 100 - I18 + -1 * 0 >= 100 - I18 + -1 * 0 5.07/5.57 0 <= -1 + I24 /\ 0 <= 100 - I23 ==> 100 - I23 + -1 * 0 > 100 - (1 + I23) + -1 * 0 with 100 - I23 + -1 * 0 >= 0 5.07/5.57 5.07/5.57 We remove all the strictly oriented dependency pairs. 5.07/5.57 5.07/5.57 DP problem for innermost termination. 5.07/5.57 P = 5.07/5.57 f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5.07/5.57 R = 5.07/5.57 f6(x1, x2, x3, x4, x5) -> f4(x1, x2, x3, x4, x5) 5.07/5.57 f5(I0, I1, I2, I3, I4) -> f1(I0, I1, I2, 1 + I3, I4) [0 <= -1 + I4 /\ 0 <= 100 - I3] 5.07/5.57 f5(I5, I6, I7, I8, I9) -> f2(rnd1, rnd2, 0, I8, I9) [rnd1 = rnd2 /\ rnd2 = 0 /\ I9 <= 0 /\ 0 <= 100 - I8] 5.07/5.57 f4(I10, I11, I12, I13, I14) -> f5(I10, I11, I12, 0, I14) 5.07/5.57 f3(I15, I16, I17, I18, I19) -> f1(I15, I16, I17, I18, I19) 5.07/5.57 f1(I20, I21, I22, I23, I24) -> f3(I20, I21, I22, 1 + I23, I24) [0 <= -1 + I24 /\ 0 <= 100 - I23] 5.07/5.57 f1(I25, I26, I27, I28, I29) -> f2(I30, I31, 0, I28, I29) [I30 = I31 /\ I31 = 0 /\ 101 - I28 <= 0] 5.07/5.57 5.07/5.57 The dependency graph for this problem is: 5.07/5.57 3 -> 5.07/5.57 Where: 5.07/5.57 3) f3#(I15, I16, I17, I18, I19) -> f1#(I15, I16, I17, I18, I19) 5.07/5.57 5.07/5.57 We have the following SCCs. 5.07/5.57 5.07/8.54 EOF popout

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actions

all output return to Integer_Transition_Systems 2019-03-29 01.54