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Integer_Transition_Systems 2019-03-29 01.54 pair #432275940
details
property
value
status
complete
benchmark
java_Break.c.t2_fixed.smt2
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n053.star.cs.uiowa.edu
space
From_T2
run statistics
property
value
solver
Ctrl
configuration
Transition
runtime (wallclock)
1.74102 seconds
cpu usage
1.71471
user time
0.916157
system time
0.798557
max virtual memory
130192.0
max residence set size
8680.0
stage attributes
key
value
starexec-result
YES
output
1.66/1.74 YES 1.66/1.74 1.66/1.74 DP problem for innermost termination. 1.66/1.74 P = 1.66/1.74 f7#(x1, x2, x3) -> f6#(x1, x2, x3) 1.66/1.74 f6#(I0, I1, I2) -> f5#(I0, I1, I2) 1.66/1.74 f6#(I3, I4, I5) -> f2#(I3, I4, I5) 1.66/1.74 f6#(I6, I7, I8) -> f3#(I6, I7, I8) 1.66/1.74 f6#(I9, I10, I11) -> f1#(I9, I10, I11) 1.66/1.74 f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 1.66/1.74 f5#(I18, I19, I20) -> f2#(I20, I19, 0) 1.66/1.74 f2#(I21, I22, I23) -> f3#(I23, I22, I23) [11 <= I23] 1.66/1.74 f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10] 1.66/1.74 f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 1.66/1.74 R = 1.66/1.74 f7(x1, x2, x3) -> f6(x1, x2, x3) 1.66/1.74 f6(I0, I1, I2) -> f5(I0, I1, I2) 1.66/1.74 f6(I3, I4, I5) -> f2(I3, I4, I5) 1.66/1.74 f6(I6, I7, I8) -> f3(I6, I7, I8) 1.66/1.74 f6(I9, I10, I11) -> f1(I9, I10, I11) 1.66/1.74 f6(I12, I13, I14) -> f4(I12, I13, I14) 1.66/1.74 f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 1.66/1.74 f5(I18, I19, I20) -> f2(I20, I19, 0) 1.66/1.74 f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23] 1.66/1.74 f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10] 1.66/1.74 f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30] 1.66/1.74 f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 1.66/1.74 1.66/1.74 The dependency graph for this problem is: 1.66/1.74 0 -> 1, 2, 3, 4, 5 1.66/1.74 1 -> 6 1.66/1.74 2 -> 7, 8 1.66/1.74 3 -> 1.66/1.74 4 -> 9 1.66/1.74 5 -> 6 1.66/1.74 6 -> 8 1.66/1.74 7 -> 1.66/1.74 8 -> 9 1.66/1.74 9 -> 7, 8 1.66/1.74 Where: 1.66/1.74 0) f7#(x1, x2, x3) -> f6#(x1, x2, x3) 1.66/1.74 1) f6#(I0, I1, I2) -> f5#(I0, I1, I2) 1.66/1.74 2) f6#(I3, I4, I5) -> f2#(I3, I4, I5) 1.66/1.74 3) f6#(I6, I7, I8) -> f3#(I6, I7, I8) 1.66/1.74 4) f6#(I9, I10, I11) -> f1#(I9, I10, I11) 1.66/1.74 5) f6#(I15, I16, I17) -> f5#(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 1.66/1.74 6) f5#(I18, I19, I20) -> f2#(I20, I19, 0) 1.66/1.74 7) f2#(I21, I22, I23) -> f3#(I23, I22, I23) [11 <= I23] 1.66/1.74 8) f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10] 1.66/1.74 9) f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 1.66/1.74 1.66/1.74 We have the following SCCs. 1.66/1.74 { 8, 9 } 1.66/1.74 1.66/1.74 DP problem for innermost termination. 1.66/1.74 P = 1.66/1.74 f2#(I24, I25, I26) -> f1#(I26, I25, I26) [I26 <= 10] 1.66/1.74 f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 1.66/1.74 R = 1.66/1.74 f7(x1, x2, x3) -> f6(x1, x2, x3) 1.66/1.74 f6(I0, I1, I2) -> f5(I0, I1, I2) 1.66/1.74 f6(I3, I4, I5) -> f2(I3, I4, I5) 1.66/1.74 f6(I6, I7, I8) -> f3(I6, I7, I8) 1.66/1.74 f6(I9, I10, I11) -> f1(I9, I10, I11) 1.66/1.74 f6(I12, I13, I14) -> f4(I12, I13, I14) 1.66/1.74 f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 1.66/1.74 f5(I18, I19, I20) -> f2(I20, I19, 0) 1.66/1.74 f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23] 1.66/1.74 f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10] 1.66/1.74 f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30] 1.66/1.74 f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 1.66/1.74 1.66/1.74 We use the reverse value criterion with the projection function NU: 1.66/1.74 NU[f1#(z1,z2,z3)] = 10 + -1 * (1 + z3) 1.66/1.74 NU[f2#(z1,z2,z3)] = 10 + -1 * z3 1.66/1.74 1.66/1.74 This gives the following inequalities: 1.66/1.74 I26 <= 10 ==> 10 + -1 * I26 > 10 + -1 * (1 + I26) with 10 + -1 * I26 >= 0 1.66/1.74 ==> 10 + -1 * (1 + I34) >= 10 + -1 * (1 + I34) 1.66/1.74 1.66/1.74 We remove all the strictly oriented dependency pairs. 1.66/1.74 1.66/1.74 DP problem for innermost termination. 1.66/1.74 P = 1.66/1.74 f1#(I32, I33, I34) -> f2#(I34, I33, 1 + I34) 1.66/1.74 R = 1.66/1.74 f7(x1, x2, x3) -> f6(x1, x2, x3) 1.66/1.74 f6(I0, I1, I2) -> f5(I0, I1, I2) 1.66/1.74 f6(I3, I4, I5) -> f2(I3, I4, I5) 1.66/1.74 f6(I6, I7, I8) -> f3(I6, I7, I8) 1.66/1.74 f6(I9, I10, I11) -> f1(I9, I10, I11) 1.66/1.74 f6(I12, I13, I14) -> f4(I12, I13, I14) 1.66/1.74 f6(I15, I16, I17) -> f5(I17, rnd2, rnd3) [rnd3 = rnd2 /\ rnd2 = rnd2] 1.66/1.74 f5(I18, I19, I20) -> f2(I20, I19, 0) 1.66/1.74 f2(I21, I22, I23) -> f3(I23, I22, I23) [11 <= I23] 1.66/1.74 f2(I24, I25, I26) -> f1(I26, I25, I26) [I26 <= 10] 1.66/1.74 f3(I27, I28, I29) -> f4(I29, I30, I31) [I31 = I30 /\ I30 = I30] 1.66/1.74 f1(I32, I33, I34) -> f2(I34, I33, 1 + I34) 1.66/1.74 1.66/1.74 The dependency graph for this problem is:
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