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Integer_Transition_Systems 2019-03-29 01.54 pair #432274668
details
property
value
status
complete
benchmark
florian_pldi.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n032.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
25.6512 seconds
cpu usage
25.8287
user time
13.2444
system time
12.5843
max virtual memory
745940.0
max residence set size
11960.0
stage attributes
key
value
starexec-result
YES
output
25.80/25.65 YES 25.80/25.65 25.80/25.65 DP problem for innermost termination. 25.80/25.65 P = 25.80/25.65 f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 25.80/25.65 f9#(I0, I1, I2, I3) -> f2#(I0, 0, I2, I3) 25.80/25.65 f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] 25.80/25.65 f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] 25.80/25.65 f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 25.80/25.65 f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 25.80/25.65 f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 25.80/25.65 f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 25.80/25.65 f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 25.80/25.65 f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] 25.80/25.65 f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] 25.80/25.65 f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] 25.80/25.65 R = 25.80/25.65 f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 25.80/25.65 f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) 25.80/25.65 f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] 25.80/25.65 f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] 25.80/25.65 f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 25.80/25.65 f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) 25.80/25.65 f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) 25.80/25.65 f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 25.80/25.65 f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) 25.80/25.65 f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] 25.80/25.65 f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] 25.80/25.65 f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] 25.80/25.65 f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] 25.80/25.65 25.80/25.65 The dependency graph for this problem is: 25.80/25.65 0 -> 1 25.80/25.65 1 -> 8 25.80/25.65 2 -> 5 25.80/25.65 3 -> 5 25.80/25.65 4 -> 6 25.80/25.65 5 -> 6 25.80/25.65 6 -> 7 25.80/25.65 7 -> 10, 11 25.80/25.65 8 -> 9 25.80/25.65 9 -> 7 25.80/25.65 10 -> 2, 3, 4 25.80/25.65 11 -> 8 25.80/25.65 Where: 25.80/25.65 0) f10#(x1, x2, x3, x4) -> f9#(x1, x2, x3, x4) 25.80/25.65 1) f9#(I0, I1, I2, I3) -> f2#(I0, 0, I2, I3) 25.80/25.65 2) f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] 25.80/25.65 3) f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] 25.80/25.65 4) f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 25.80/25.65 5) f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 25.80/25.65 6) f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 25.80/25.65 7) f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 25.80/25.65 8) f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 25.80/25.65 9) f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] 25.80/25.65 10) f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] 25.80/25.65 11) f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] 25.80/25.65 25.80/25.65 We have the following SCCs. 25.80/25.65 { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 } 25.80/25.65 25.80/25.65 DP problem for innermost termination. 25.80/25.65 P = 25.80/25.65 f3#(I4, I5, I6, I7) -> f8#(I4, I5, I6, I7) [1 + I4 <= 0] 25.80/25.65 f3#(I8, I9, I10, I11) -> f8#(I8, I9, I10, I11) [1 <= I8] 25.80/25.65 f3#(I12, I13, I14, I15) -> f7#(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 25.80/25.65 f8#(I16, I17, I18, I19) -> f7#(I16, I17, -1 + I18, -1 + I19) 25.80/25.65 f7#(I20, I21, I22, I23) -> f6#(I20, I21, 1 + I22, I23) 25.80/25.65 f6#(I24, I25, I26, I27) -> f1#(I24, I25, I26, I27) 25.80/25.65 f2#(I28, I29, I30, I31) -> f4#(I28, I29, I30, I31) 25.80/25.65 f4#(I32, I33, I34, I35) -> f6#(I32, I33, 1 + I33, I35) [1 + I33 <= I35] 25.80/25.65 f1#(I40, I41, I42, I43) -> f3#(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] 25.80/25.65 f1#(I44, I45, I46, I47) -> f2#(I44, 1 + I45, I46, I47) [I47 <= I46] 25.80/25.65 R = 25.80/25.65 f10(x1, x2, x3, x4) -> f9(x1, x2, x3, x4) 25.80/25.65 f9(I0, I1, I2, I3) -> f2(I0, 0, I2, I3) 25.80/25.65 f3(I4, I5, I6, I7) -> f8(I4, I5, I6, I7) [1 + I4 <= 0] 25.80/25.65 f3(I8, I9, I10, I11) -> f8(I8, I9, I10, I11) [1 <= I8] 25.80/25.65 f3(I12, I13, I14, I15) -> f7(I12, I13, I14, I15) [0 <= I12 /\ I12 <= 0] 25.80/25.65 f8(I16, I17, I18, I19) -> f7(I16, I17, -1 + I18, -1 + I19) 25.80/25.65 f7(I20, I21, I22, I23) -> f6(I20, I21, 1 + I22, I23) 25.80/25.65 f6(I24, I25, I26, I27) -> f1(I24, I25, I26, I27) 25.80/25.65 f2(I28, I29, I30, I31) -> f4(I28, I29, I30, I31) 25.80/25.65 f4(I32, I33, I34, I35) -> f6(I32, I33, 1 + I33, I35) [1 + I33 <= I35] 25.80/25.65 f4(I36, I37, I38, I39) -> f5(I36, I37, I38, I39) [I39 <= I37] 25.80/25.65 f1(I40, I41, I42, I43) -> f3(rnd1, I41, I42, I43) [rnd1 = rnd1 /\ 1 + I42 <= I43] 25.80/25.65 f1(I44, I45, I46, I47) -> f2(I44, 1 + I45, I46, I47) [I47 <= I46] 25.80/25.65 25.80/25.65 We use the extended value criterion with the projection function NU: 25.80/25.65 NU[f4#(x0,x1,x2,x3)] = -x1 + x3 + 1 25.80/25.65 NU[f2#(x0,x1,x2,x3)] = -x1 + x3 + 1 25.80/25.65 NU[f1#(x0,x1,x2,x3)] = -x1 + x3 25.80/25.65 NU[f6#(x0,x1,x2,x3)] = -x1 + x3 25.80/25.65 NU[f7#(x0,x1,x2,x3)] = -x1 + x3 25.80/25.65 NU[f8#(x0,x1,x2,x3)] = -x1 + x3 25.80/25.65 NU[f3#(x0,x1,x2,x3)] = -x1 + x3 25.80/25.65 25.80/25.65 This gives the following inequalities: 25.80/25.65 1 + I4 <= 0 ==> -I5 + I7 >= -I5 + I7 25.80/25.65 1 <= I8 ==> -I9 + I11 >= -I9 + I11
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