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Integer_Transition_Systems 2019-03-29 01.54 pair #432275334
details
property
value
status
complete
benchmark
simpleWhile.t2.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n001.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
6.56284 seconds
cpu usage
6.30356
user time
3.20596
system time
3.0976
max virtual memory
366708.0
max residence set size
8564.0
stage attributes
key
value
starexec-result
YES
output
6.24/6.56 YES 6.24/6.56 6.24/6.56 DP problem for innermost termination. 6.24/6.56 P = 6.24/6.56 f11#(x1, x2, x3, x4) -> f10#(x1, x2, x3, x4) 6.24/6.56 f10#(I0, I1, I2, I3) -> f5#(I0, 0, I2, 0) 6.24/6.56 f2#(I4, I5, I6, I7) -> f5#(I4, 1 + I5, I6, I7) 6.24/6.56 f4#(I8, I9, I10, I11) -> f7#(I8, I9, 1, I11) [I8 <= I11 /\ I11 <= I8] 6.24/6.56 f4#(I12, I13, I14, I15) -> f8#(I12, I13, I14, I15) [1 + I15 <= I12] 6.24/6.56 f4#(I16, I17, I18, I19) -> f8#(I16, I17, I18, I19) [1 + I16 <= I19] 6.24/6.56 f8#(I24, I25, I26, I27) -> f7#(I24, I25, 1, I27) [1 + I24 <= I27 /\ I27 <= 1 + I24] 6.24/6.56 f8#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) [1 + I31 <= 1 + I28] 6.24/6.56 f8#(I32, I33, I34, I35) -> f6#(I32, I33, I34, I35) [2 + I32 <= I35] 6.24/6.56 f6#(I36, I37, I38, I39) -> f7#(I36, I37, 0, I39) 6.24/6.56 f5#(I40, I41, I42, I43) -> f3#(I40, I41, I42, I43) 6.24/6.56 f3#(I44, I45, I46, I47) -> f1#(I44, I45, I46, I47) [1 + I45 <= I44] 6.24/6.56 f3#(I48, I49, I50, I51) -> f4#(I48, I49, I50, I51) [I48 <= I49] 6.24/6.56 f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) 6.24/6.56 f1#(I56, I57, I58, I59) -> f2#(I56, I57, I58, 2 + I59) 6.24/6.56 R = 6.24/6.56 f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 6.24/6.56 f10(I0, I1, I2, I3) -> f5(I0, 0, I2, 0) 6.24/6.56 f2(I4, I5, I6, I7) -> f5(I4, 1 + I5, I6, I7) 6.24/6.56 f4(I8, I9, I10, I11) -> f7(I8, I9, 1, I11) [I8 <= I11 /\ I11 <= I8] 6.24/6.56 f4(I12, I13, I14, I15) -> f8(I12, I13, I14, I15) [1 + I15 <= I12] 6.24/6.56 f4(I16, I17, I18, I19) -> f8(I16, I17, I18, I19) [1 + I16 <= I19] 6.24/6.56 f7(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 6.24/6.56 f8(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [1 + I24 <= I27 /\ I27 <= 1 + I24] 6.24/6.56 f8(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) [1 + I31 <= 1 + I28] 6.24/6.56 f8(I32, I33, I34, I35) -> f6(I32, I33, I34, I35) [2 + I32 <= I35] 6.24/6.56 f6(I36, I37, I38, I39) -> f7(I36, I37, 0, I39) 6.24/6.56 f5(I40, I41, I42, I43) -> f3(I40, I41, I42, I43) 6.24/6.56 f3(I44, I45, I46, I47) -> f1(I44, I45, I46, I47) [1 + I45 <= I44] 6.24/6.56 f3(I48, I49, I50, I51) -> f4(I48, I49, I50, I51) [I48 <= I49] 6.24/6.56 f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) 6.24/6.56 f1(I56, I57, I58, I59) -> f2(I56, I57, I58, 2 + I59) 6.24/6.56 6.24/6.56 The dependency graph for this problem is: 6.24/6.56 0 -> 1 6.24/6.56 1 -> 10 6.24/6.56 2 -> 10 6.24/6.56 3 -> 6.24/6.56 4 -> 7 6.24/6.56 5 -> 6, 8 6.24/6.56 6 -> 6.24/6.56 7 -> 9 6.24/6.56 8 -> 9 6.24/6.56 9 -> 6.24/6.56 10 -> 11, 12 6.24/6.56 11 -> 13, 14 6.24/6.56 12 -> 3, 4, 5 6.24/6.56 13 -> 2 6.24/6.56 14 -> 2 6.24/6.56 Where: 6.24/6.56 0) f11#(x1, x2, x3, x4) -> f10#(x1, x2, x3, x4) 6.24/6.56 1) f10#(I0, I1, I2, I3) -> f5#(I0, 0, I2, 0) 6.24/6.56 2) f2#(I4, I5, I6, I7) -> f5#(I4, 1 + I5, I6, I7) 6.24/6.56 3) f4#(I8, I9, I10, I11) -> f7#(I8, I9, 1, I11) [I8 <= I11 /\ I11 <= I8] 6.24/6.56 4) f4#(I12, I13, I14, I15) -> f8#(I12, I13, I14, I15) [1 + I15 <= I12] 6.24/6.56 5) f4#(I16, I17, I18, I19) -> f8#(I16, I17, I18, I19) [1 + I16 <= I19] 6.24/6.56 6) f8#(I24, I25, I26, I27) -> f7#(I24, I25, 1, I27) [1 + I24 <= I27 /\ I27 <= 1 + I24] 6.24/6.56 7) f8#(I28, I29, I30, I31) -> f6#(I28, I29, I30, I31) [1 + I31 <= 1 + I28] 6.24/6.56 8) f8#(I32, I33, I34, I35) -> f6#(I32, I33, I34, I35) [2 + I32 <= I35] 6.24/6.56 9) f6#(I36, I37, I38, I39) -> f7#(I36, I37, 0, I39) 6.24/6.56 10) f5#(I40, I41, I42, I43) -> f3#(I40, I41, I42, I43) 6.24/6.56 11) f3#(I44, I45, I46, I47) -> f1#(I44, I45, I46, I47) [1 + I45 <= I44] 6.24/6.56 12) f3#(I48, I49, I50, I51) -> f4#(I48, I49, I50, I51) [I48 <= I49] 6.24/6.56 13) f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) 6.24/6.56 14) f1#(I56, I57, I58, I59) -> f2#(I56, I57, I58, 2 + I59) 6.24/6.56 6.24/6.56 We have the following SCCs. 6.24/6.56 { 2, 10, 11, 13, 14 } 6.24/6.56 6.24/6.56 DP problem for innermost termination. 6.24/6.56 P = 6.24/6.56 f2#(I4, I5, I6, I7) -> f5#(I4, 1 + I5, I6, I7) 6.24/6.56 f5#(I40, I41, I42, I43) -> f3#(I40, I41, I42, I43) 6.24/6.56 f3#(I44, I45, I46, I47) -> f1#(I44, I45, I46, I47) [1 + I45 <= I44] 6.24/6.56 f1#(I52, I53, I54, I55) -> f2#(I52, I53, I54, I55) 6.24/6.56 f1#(I56, I57, I58, I59) -> f2#(I56, I57, I58, 2 + I59) 6.24/6.56 R = 6.24/6.56 f11(x1, x2, x3, x4) -> f10(x1, x2, x3, x4) 6.24/6.56 f10(I0, I1, I2, I3) -> f5(I0, 0, I2, 0) 6.24/6.56 f2(I4, I5, I6, I7) -> f5(I4, 1 + I5, I6, I7) 6.24/6.56 f4(I8, I9, I10, I11) -> f7(I8, I9, 1, I11) [I8 <= I11 /\ I11 <= I8] 6.24/6.56 f4(I12, I13, I14, I15) -> f8(I12, I13, I14, I15) [1 + I15 <= I12] 6.24/6.56 f4(I16, I17, I18, I19) -> f8(I16, I17, I18, I19) [1 + I16 <= I19] 6.24/6.56 f7(I20, I21, I22, I23) -> f9(I20, I21, I22, I23) 6.24/6.56 f8(I24, I25, I26, I27) -> f7(I24, I25, 1, I27) [1 + I24 <= I27 /\ I27 <= 1 + I24] 6.24/6.56 f8(I28, I29, I30, I31) -> f6(I28, I29, I30, I31) [1 + I31 <= 1 + I28] 6.24/6.56 f8(I32, I33, I34, I35) -> f6(I32, I33, I34, I35) [2 + I32 <= I35] 6.24/6.56 f6(I36, I37, I38, I39) -> f7(I36, I37, 0, I39) 6.24/6.56 f5(I40, I41, I42, I43) -> f3(I40, I41, I42, I43) 6.24/6.56 f3(I44, I45, I46, I47) -> f1(I44, I45, I46, I47) [1 + I45 <= I44] 6.24/6.56 f3(I48, I49, I50, I51) -> f4(I48, I49, I50, I51) [I48 <= I49] 6.24/6.56 f1(I52, I53, I54, I55) -> f2(I52, I53, I54, I55) 6.24/6.56 f1(I56, I57, I58, I59) -> f2(I56, I57, I58, 2 + I59) 6.24/6.56 6.24/6.56 We use the extended value criterion with the projection function NU: 6.24/6.56 NU[f1#(x0,x1,x2,x3)] = x0 - x1 - 2
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