ITS pair #487096732

details

property	value
status	complete
benchmark	neg-e-st88b-succeed.t2.smt2
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	From_T2

run statistics

property	value
solver	Ctrl
configuration	Transition
runtime (wallclock)	1.77745 seconds
cpu usage	1.79921
user time	0.950305
system time	0.848906
max virtual memory	723100.0
max residence set size	8548.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE

DP problem for innermost termination.
P =
  f18#(x1, x2, x3) -> f17#(x1, x2, x3) 
  f17#(I0, I1, I2) -> f16#(rnd1, rnd2, I2) [y1 = y1 /\ rnd2 = rnd2 /\ rnd1 = rnd2]
  f16#(I3, I4, I5) -> f1#(I3, I4, I5) [1 + I3 <= 0]
  f16#(I6, I7, I8) -> f3#(I6, I7, I8) [0 <= I6]
  f15#(I9, I10, I11) -> f3#(I9, I10, I11) 
  f5#(I12, I13, I14) -> f14#(I12, I13, I14) [I12 <= 5]
  f5#(I15, I16, I17) -> f13#(I15, I16, I17) [6 <= I15]
  f14#(I18, I19, I20) -> f10#(I18, I19, I20) [1 <= I20]
  f14#(I21, I22, I23) -> f13#(I21, I22, I23) [I23 <= 0]
  f13#(I24, I25, I26) -> f4#(1 + I24, I25, I26) [I24 <= 5]
  f13#(I27, I28, I29) -> f4#(1 + I27, I28, I29) [6 <= I27]
  f12#(I30, I31, I32) -> f9#(I30, I31, I32) 
  f9#(I33, I34, I35) -> f12#(I33, I34, I35) 
  f11#(I36, I37, I38) -> f3#(I36, I37, I38) [I36 <= 2]
  f11#(I39, I40, I41) -> f10#(-1 + I39, I40, I41) [3 <= I39]
  f10#(I42, I43, I44) -> f11#(I42, I43, I44) 
  f8#(I45, I46, I47) -> f9#(I45, I46, I47) 
  f4#(I51, I52, I53) -> f5#(I51, I52, rnd3) [rnd3 = rnd3]
  f3#(I54, I55, I56) -> f4#(I54, I55, I56) 
  f2#(I57, I58, I59) -> f1#(I57, I58, I59) 
  f1#(I60, I61, I62) -> f2#(I60, I61, I62) 
R = 
  f18(x1, x2, x3) -> f17(x1, x2, x3) 
  f17(I0, I1, I2) -> f16(rnd1, rnd2, I2) [y1 = y1 /\ rnd2 = rnd2 /\ rnd1 = rnd2]
  f16(I3, I4, I5) -> f1(I3, I4, I5) [1 + I3 <= 0]
  f16(I6, I7, I8) -> f3(I6, I7, I8) [0 <= I6]
  f15(I9, I10, I11) -> f3(I9, I10, I11) 
  f5(I12, I13, I14) -> f14(I12, I13, I14) [I12 <= 5]
  f5(I15, I16, I17) -> f13(I15, I16, I17) [6 <= I15]
  f14(I18, I19, I20) -> f10(I18, I19, I20) [1 <= I20]
  f14(I21, I22, I23) -> f13(I21, I22, I23) [I23 <= 0]
  f13(I24, I25, I26) -> f4(1 + I24, I25, I26) [I24 <= 5]
  f13(I27, I28, I29) -> f4(1 + I27, I28, I29) [6 <= I27]
  f12(I30, I31, I32) -> f9(I30, I31, I32) 
  f9(I33, I34, I35) -> f12(I33, I34, I35) 
  f11(I36, I37, I38) -> f3(I36, I37, I38) [I36 <= 2]
  f11(I39, I40, I41) -> f10(-1 + I39, I40, I41) [3 <= I39]
  f10(I42, I43, I44) -> f11(I42, I43, I44) 
  f8(I45, I46, I47) -> f9(I45, I46, I47) 
  f6(I48, I49, I50) -> f7(I48, I49, I50) 
  f4(I51, I52, I53) -> f5(I51, I52, rnd3) [rnd3 = rnd3]
  f3(I54, I55, I56) -> f4(I54, I55, I56) 
  f2(I57, I58, I59) -> f1(I57, I58, I59) 
  f1(I60, I61, I62) -> f2(I60, I61, I62) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 2, 3
  2 -> 20
  3 -> 18
  4 -> 18
  5 -> 7, 8
  6 -> 10
  7 -> 15
  8 -> 9, 10
  9 -> 17
  10 -> 17
  11 -> 12
  12 -> 11
  13 -> 18
  14 -> 15
  15 -> 13, 14
  16 -> 12
  17 -> 5, 6
  18 -> 17
  19 -> 20
  20 -> 19
Where:
  0) f18#(x1, x2, x3) -> f17#(x1, x2, x3) 
  1) f17#(I0, I1, I2) -> f16#(rnd1, rnd2, I2) [y1 = y1 /\ rnd2 = rnd2 /\ rnd1 = rnd2]
  2) f16#(I3, I4, I5) -> f1#(I3, I4, I5) [1 + I3 <= 0]
  3) f16#(I6, I7, I8) -> f3#(I6, I7, I8) [0 <= I6]
  4) f15#(I9, I10, I11) -> f3#(I9, I10, I11) 
  5) f5#(I12, I13, I14) -> f14#(I12, I13, I14) [I12 <= 5]
  6) f5#(I15, I16, I17) -> f13#(I15, I16, I17) [6 <= I15]
  7) f14#(I18, I19, I20) -> f10#(I18, I19, I20) [1 <= I20]
  8) f14#(I21, I22, I23) -> f13#(I21, I22, I23) [I23 <= 0]
  9) f13#(I24, I25, I26) -> f4#(1 + I24, I25, I26) [I24 <= 5]
  10) f13#(I27, I28, I29) -> f4#(1 + I27, I28, I29) [6 <= I27]
  11) f12#(I30, I31, I32) -> f9#(I30, I31, I32) 
  12) f9#(I33, I34, I35) -> f12#(I33, I34, I35) 
  13) f11#(I36, I37, I38) -> f3#(I36, I37, I38) [I36 <= 2]
  14) f11#(I39, I40, I41) -> f10#(-1 + I39, I40, I41) [3 <= I39]
  15) f10#(I42, I43, I44) -> f11#(I42, I43, I44) 
  16) f8#(I45, I46, I47) -> f9#(I45, I46, I47) 
  17) f4#(I51, I52, I53) -> f5#(I51, I52, rnd3) [rnd3 = rnd3]
  18) f3#(I54, I55, I56) -> f4#(I54, I55, I56) 
  19) f2#(I57, I58, I59) -> f1#(I57, I58, I59) 
  20) f1#(I60, I61, I62) -> f2#(I60, I61, I62) 

We have the following SCCs.
  { 11, 12 }
  { 5, 6, 7, 8, 9, 10, 13, 14, 15, 17, 18 }
  { 19, 20 }

DP problem for innermost termination.

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